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inequality $(a+c)(a+b+c)<0$, prove $(b-c)^2>4a(a+b+c)$ If $(a+c)(a+b+c)<0,$ prove $$(b-c)^2>4a(a+b+c)$$ I will use the constructor method that want to know can not directly prove it?

Consider the quadratic $$ f(x) = ax^2 - (b-c)x + (a+b+c) $$ $$f(1)f(0) = 2(a+c)(a+b+c) \lt 0$$ Thus if $a \neq 0$, then this has a real root in $(0,1)$ and so $$(b-c)^2 \ge 4a(a+b+c)$$ If $(b-c)^2 = 4a(a+b+c)$, then we have a double root in $(0,1)$ in which case, $f(0)$ and $f(1)$ will have the same sign. Thus $$(b-c)^2 \gt 4a(a+b+c)$$ If $a = 0$, then $c(b+c) \lt 0$, and so we cannot have $b=c$ and thus $(b-c)^2 \gt 0 = 4a(a+b+c)$ And if you want a more "direct" approach, we show that $(p+q+r)r \lt 0 \implies q^2 \gt 4pr$ using the following identity: $$(p+q+r)r = \left(p\left(1 + \frac{q}{2p}\right)\frac{q}{2p}\right)^2 + \left(r - \frac{q^2}{4p}\right)^2 + p\left(r - \frac{q^2}{4p}\right)\left(\left(1 + \frac{q}{2p}\right)^2 + \left(\frac{q}{2p}\right)^2\right)$$ If $(p+q+r)r \lt 0$, then we must have have $p\left(r - \frac{q^2}{4p}\right) \lt 0$, as all the other terms on the right side are non-negative. Of course, this was gotten by completing the square in $px^2 + qx + r$ and setting $x=0$ and $x=1$ and multiplying.

Critical points of $f(x,y)=x^2+xy+y^2+\frac{1}{x}+\frac{1}{y}$ I would like some help finding the critical points of $f(x,y)=x^2+xy+y^2+\frac{1}{x}+\frac{1}{y}$. I tried solving $f_x=0, f_y=0$ (where $f_x, f_y$ are the partial derivatives) but the resulting equation is very complex. The exercise has a hint: think of $f_x-f_y$ and $f_x+f_y$. However, I can't see where to use it. Thanks!

After doing some computations I found the following (lets hope I didn't make any mistakes). You need to solve the equations $$f_x = 2x + y - \frac{1}{x^2} = 0 \quad f_y = 2y + x -\frac{1}{y^2} = 0$$ therefore after subtracting and adding them as in the hint we get $$\begin{align} f_x - f_y &= x - y - \frac{1}{x^2} + \frac{1}{y^2} = 0 \\ f_x + f_y &= 3x + 3y -\frac{1}{x^2} - \frac{1}{y^2} = 0 \end{align} $$ but you can factor them a little bit to get $$ \begin{align} f_x - f_y &= x - y + \frac{x^2 - y^2}{x^2 y^2} = (x - y) \left ( 1 + \frac{x+ y}{x^2 y^2}\right ) = 0\\ f_x + f_y &= 3(x + y) -\frac{x^2 + y^2}{x^2 y^2} = 0 \end{align} $$ Now from the first equation you get two conditions, either $x = y$ or $x+y = -x^2 y^2$. If $x = y$ you can go back to your first equation for $f_x$ and substitute to get $$2x + x - \frac{1}{x^2} = 0 \implies 3x = \frac{1}{x^2} \implies x = \frac{1}{\sqrt[3]{3}}$$ and then you get the critical point $\left ( \dfrac{1}{\sqrt[3]{3}}, \dfrac{1}{\sqrt[3]{3}} \right )$ Now if instead $x + y = -x^2 y^2$ then if you substitute into the equation $f_x + f_y = 0$ we get the following $$ 3(-x^2 y^2) - \frac{x^2 + y^2}{x^2 y^2} = 0 \implies 3x^4 y^4 + x^2 + y^2 = 0 \implies x = y = 0 $$ But this is actually one of the points where the partial derivatives or even your original function are not defined.

Taking the derivative of $y = \dfrac{x}{2} + \dfrac {1}{4} \sin(2x)$ Again a simple problem that I can't seem to get the derivative of I have $\frac{x}{2} + \frac{1}{4}\sin(2x)$ I am getting $\frac{x^2}{4} + \frac{4\sin(2x)}{16}$ This is all very wrong, and I do not know why.

You deal with the sum of functions, $f(x) = \frac{x}{2}$ and $g(x)= \frac{1}{4} \sin(2 x)$. So you would use linearity of the derivative: $$ \frac{d}{d x} \left( f(x) + g(x) \right) = \frac{d f(x)}{d x} + \frac{d g(x)}{d x} $$ To evaluate these derivatives, you would use $\frac{d}{d x}\left( c f(x) \right) = c \frac{d f(x)}{d x}$, for a constant $c$. Thus $$ \frac{d}{d x} \left( \frac{x}{2} + \frac{1}{4} \sin(2 x) \right) = \frac{1}{2} \frac{d x}{d x} + \frac{1}{4} \frac{d \sin(2 x)}{d x} $$ To evaluate derivative of the sine function, you would need a chain rule: $$ \frac{d}{d x} y(h(x)) = y^\prime(h(x)) h^\prime(x) $$ where $y(x) = \sin(x)$ and $h(x) = 2x$. Now finish it off using table of derivatives.

Is my Riemann Sum correct (Example # 2)? Possible Duplicate: Is my Riemann Sum correct? This is my second attempt, the answer seems rather odd so I thought I would have it checked as well. For the integral: $$\int_{-5}^{2} \left( x^{2} -4 \right) dx$$ My calculations: $$\begin{align*}\Delta x &= \frac7n\\\\ x_i &= -5 + \frac{7i}n\\\\ f(x_i) &= 21 - \frac{70i}{n} + \frac{49i^2}{n^2} \\\\ A&=-738 \end{align*}$$

The preliminary computations are fine. That means that the $n$th right hand Riemann sum will be: $$\begin{align*} \text{RHS} &= \sum_{i=1}^n f(x_i)\Delta x\\ &= \sum_{i=1}^n\left(21 - \frac{70i}{n} +\frac{49i^2}{n^2}\right)\frac{7}{n}\\ &= \frac{7(21)}{n}\sum_{i=1}^n1 - \frac{7(70)}{n^2}\sum_{i=1}^n i + \frac{7(49)}{n^3}\sum_{i=1}^ni^2\\ &= \frac{147}{n}(n) - \frac{490}{n^2}\left(\frac{n(n+1)}{2}\right) + \frac{343}{n^3}\left(\frac{n(n+1)(2n+1)}{6}\right)\\ &= 147 - 245\frac{n^2}{n^2+n} + \frac{343}{6}\frac{n(n+1)(2n+1)}{n^3}, \end{align*}$$ using the formulas that say that $$\begin{align*} 1+2+3+\cdots + n &= \frac{n(n+1)}{2}\\ 1^2+2^2+3^2+\cdots+n^2 &= \frac{n(n+1)(2n+1)}{6}. \end{align*}$$ Now, if we take the limit as $n\to\infty$, we have $$\begin{align*} \lim\limits_{n\to\infty}\frac{n^2}{n^2+n} &= 1\\ \lim\limits_{n\to\infty}\frac{n(n+1)(2n+1)}{n^3} &= 2, \end{align*}$$ which means the area should be $$147 -245 +\frac{343}{3} = -98 + 114+\frac{1}{3} = 16+\frac{1}{3} = \frac{49}{3}.$$

How can one solve the equation $\sqrt{x\sqrt{x} - x} = 1-x$? $$\sqrt{x\sqrt{x} - x} = 1-x$$ I know the solution but have no idea how to solve it analytically.

Just writing out Robert's manipulation: $$\eqalign{ & \sqrt {x\sqrt x - x} = 1 - x \cr & x\sqrt x - x = {\left( {1 - x} \right)^2} \cr & x\left( {\sqrt x - 1} \right) = {\left( {1 - x} \right)^2} \cr & \sqrt x - 1 = \frac{{{{\left( {1 - x} \right)}^2}}}{x} \cr & \sqrt x = \frac{{{{\left( {1 - x} \right)}^2}}}{x} + 1 \cr & x = {\left( {\frac{{{{\left( {1 - x} \right)}^2}}}{x} + 1} \right)^2} \cr & x = {\left( {\frac{{1 - 2x + {x^2}}}{x} + 1} \right)^2} \cr & x = {\left( {\frac{1}{x} + x - 1} \right)^2} \cr & x = {x^2} - 2x + 3 - \frac{2}{x} + \frac{1}{{{x^2}}} \cr & {x^3} = {x^4} - 2{x^3} + 3{x^2} - 2x + 1 \cr & 0 = {x^4} - 3{x^3} + 3{x^2} - 2x + 1 \cr & 0 = \left( {x - 1} \right)\left( {{x^3} - 2{x^2} + x + 1} \right) \cr} $$ Note you will most probably have two complex solutions apart from $x=1$.

Alternative proof of the limitof the quotient of two sums. I found the following problem by Apostol: Let $a \in \Bbb R$ and $s_n(a)=\sum\limits_{k=1}^n k^a$. Find $$\lim_{n\to +\infty} \frac{s_n(a+1)}{ns_n(a)}$$ After some struggling and helpless ideas I considered the following solution. If $a > -1$, then $$\int_0^1 x^a dx=\frac{1}{a+1}$$ is well defined. Thus, let $$\lambda_n(a)=\frac{s_n(a)}{n^{a+1}}$$ It is clear that $$\lim\limits_{n\to +\infty} \lambda_n(a)=\int_0^1 x^a dx=\frac{1}{a+1}$$ and thus $$\lim_{n\to +\infty} \frac{s_n(a+1)}{ns_n(a)}=\lim_{n \to +\infty} \frac{\lambda_n(a+1)}{\lambda_n(a)}=\frac{a+1}{a+2}$$ Can you provide any other proof for this? I used mostly integration theory but maybe there are other simpler ideas (or more complex ones) that can be used. (If $a=-1$ then the limit is zero, since it is simply $H_n^{-1}$ which goes to zero since the harmonic series is divergent. For the case $a <-1$, the simple inequalities $s_n(a+1) \le n\cdot n^{a+1} = n^{a+2}$ and $s_n(a) \ge 1$ show that the limit is also zero.)

The argument below works for any real $a > -1$. We are given that $$s_n(a) = \sum_{k=1}^{n} k^a$$ Let $a_n = 1$ and $A(t) = \displaystyle \sum_{k \leq t} a_n = \left \lfloor t \right \rfloor$. Hence, $$s_n(a) = \int_{1^-}^{n^+} t^a dA(t)$$ The integral is to be interpreted as the Riemann Stieltjes integral. Now integrating by parts, we get that $$s_n(a) = \left. t^a A(t) \right \rvert_{1^-}^{n^+} - \int_{1^-}^{n^+} A(t) a t^{a-1} dt = n^a \times n - a \int_{1^-}^{n^+} \left \lfloor t \right \rfloor t^{a-1} dt\\ = n^{a+1} - a \int_{1^-}^{n^+} (t -\left \{ t \right \}) t^{a-1} dt = n^{a+1} - a \int_{1^-}^{n^+} t^a dt + a \int_{1^-}^{n^+}\left \{ t \right \} t^{a-1} dt\\ = n^{a+1} - a \left. \dfrac{t^{a+1}}{a+1} \right \rvert_{1^-}^{n^+} + a \int_{1^-}^{n^+}\left \{ t \right \} t^{a-1} dt\\ =n^{a+1} - a \dfrac{n^{a+1}-1}{a+1} + a \int_{1^-}^{n^+}\left \{ t \right \} t^{a-1} dt\\ = \dfrac{n^{a+1}}{a+1} + \dfrac{a}{a+1} + \mathcal{O} \left( a \times 1 \times \dfrac{n^a}{a}\right)\\ = \dfrac{n^{a+1}}{a+1} + \mathcal{O} \left( n^a \right)$$ Hence, we get that $$\lim_{n \rightarrow \infty} \dfrac{s_n(a)}{n^{a+1}/(a+1)} = 1$$ Hence, now $$\dfrac{s_{n}(a+1)}{n s_n(a)} = \dfrac{\dfrac{s_n(a+1)}{n^{a+2}/(a+2)}}{\dfrac{s_n(a)}{n^{a+1}/(a+1)}} \times \dfrac{a+1}{a+2}$$ Hence, we get that $$\lim_{n \rightarrow \infty} \dfrac{s_{n}(a+1)}{n s_n(a)} = \dfrac{\displaystyle \lim_{n \rightarrow \infty} \dfrac{s_n(a+1)}{n^{a+2}/(a+2)}}{\displaystyle \lim_{n \rightarrow \infty} \dfrac{s_n(a)}{n^{a+1}/(a+1)}} \times \dfrac{a+1}{a+2} = \dfrac11 \times \dfrac{a+1}{a+2} = \dfrac{a+1}{a+2}$$ Note that the argument needs to be slightly modified for $a = -1$ or $a = -2$. However, the two cases can be argued directly itself. If $a=-1$, then we want $$\lim_{n \rightarrow \infty} \dfrac{s_n(0)}{n s_n(-1)} = \lim_{n \rightarrow \infty} \dfrac{n}{n H_n} = 0$$ If $a=-2$, then we want $$\lim_{n \rightarrow \infty} \dfrac{s_n(-1)}{n s_n(-2)} = \dfrac{6}{\pi^2} \lim_{n \rightarrow \infty} \dfrac{H_n}{n} = 0$$ In general, for $a <-2$, note that both $s_n(a+1)$ and $s_n(a)$ converge. Hence, the limit is $0$. For $a \in (-2,-1)$, $s_n(a)$ converges but $s_n(a+1)$ diverges slower than $n$. Hence, the limit is again $0$. Hence to summarize $$\lim_{n \rightarrow \infty} \dfrac{s_n(a+1)}{n s_n(a)} = \begin{cases} \dfrac{a+1}{a+2} & \text{ if }a>-1\\ 0 & \text{ if } a \leq -1 \end{cases}$$

An alternating series ... Find the limit of the following series: $$ 1 - \frac{1}{4} + \frac{1}{6} - \frac{1}{9} + \frac{1}{11} - \frac{1}{14} + \cdot \cdot \cdot $$ If i go the integration way all is fine for a while but then things become pretty ugly. I'm trying to find out if there is some easier way to follow.

Let $S = 1 - x^{3} + x^{5} -x^{8} + x^{10} - x^{13} + \cdots$. Then what you want is $\int_{0}^{1} S \ dx$. But we have \begin{align*} S &= 1 - x^{3} + x^{5} -x^{8} + x^{10} - x^{13} + \cdots \\\ &= -(x^{3}+x^{8} + x^{13} + \cdots) + (1+x^{5} + x^{10} + \cdots) \\\ &= -\frac{x^{3}}{1-x^{5}} + \frac{1}{1-x^{5}} \end{align*} Now you have to evaluate: $\displaystyle \int_{0}^{1}\frac{1-x^{3}}{1-x^{5}} \ dx$ And wolfram gives the answer as:

What is the Taylor series for $g(x) =\frac{ \sinh(-x^{1/2})}{(-x^{1/2})}$, for $x < 0$? What is the Taylor series for $$g(x) = \frac{\sinh((-x)^{1/2})}{(-x)^{1/2}}$$, for $x < 0$? Using the standard Taylor Series: $$\sinh x = x + \frac{x^3}{3!} + \frac{x^5}{5!} + \frac{x^7}{7!}$$ I substituted in $x = x^{1/2}$, since $x < 0$, it would simply be $x^{1/2}$ getting, $$\sinh(x^{1/2}) = x^{1/2} + \frac{x^{3/2}}{3!} + \frac{x^{5/2}}{5!} + \frac{x^{7/2}}{7!}$$ Then to get the Taylor series for $\sinh((-x)^{1/2})/((-x)^{1/2})$, would I just divide each term by $x^{1/2}$? This gives me, $1+\frac{x}{3!}+\frac{x^2}{5!}+\frac{x^3}{7!}$ Is this correct? Thanks for any help!

As Arturo pointed out in a comment, It has to be $(-x)^{\frac{1}{2}}$ to be defined for $x<0$, then you have: $$\sinh x = x + \frac{x^3}{3!} + \frac{x^5}{5!} + \frac{x^7}{7!}+\dots$$ Substituting $x$ with $(-x)^{\frac{1}{2}}$ we get: $$\sinh (-x)^{\frac{1}{2}} = (-x)^{\frac{1}{2}} + \frac{({(-x)^{\frac{1}{2}}})^3}{3!} + \frac{({(-x)^{\frac{1}{2}}})^5}{5!} + \frac{({(-x)^{\frac{1}{2}}})^7}{7!}+\dots$$ Dividing by $(-x)^{\frac{1}{2}}$: $$\frac{\sinh (-x)^{\frac{1}{2}}}{(-x)^{\frac{1}{2}}} = 1 + \frac{({(-x)^{\frac{1}{2}}})^2}{3!} + \frac{({(-x)^{\frac{1}{2}}})^4}{5!} + \frac{({(-x)^{\frac{1}{2}}})^6}{7!}+\dots$$ And after simplification: $$\frac{\sinh (-x)^{\frac{1}{2}}}{(-x)^{\frac{1}{2}}} = 1 - \frac{x}{3!} + \frac{x^2}{5!} - \frac{x^3}{7!}+\dots$$

Given an alphabet with 6 non-distinct integers, how many distinct 4-digit integers are there? How many distinct four-digit integers can one make from the digits $1$, $3$, $3$, $7$, $7$ and $8$? I can't really think how to get started with this, the only way I think might work would be to go through all the cases. For instance, two $3$'s and two $7$'s as one case, one $1$, two $3$'s and one $8$ as another. This seems a bit tedious though (especially for a larger alphabet) and so I'm here to ask if there's a better way. Thanks.

Distinct numbers with two $3$s and two $7$s: $\binom{4}{2}=6$. Distinct numbers with two $3$s and one or fewer $7$s: $\binom{4}{2}3\cdot2=36$. Distinct numbers with two $7$s and one or fewer $3$s: $\binom{4}{2}3\cdot2=36$. Distinct numbers with one or fewer $7$s and one or fewer $3$s: $4\cdot3\cdot2\cdot1=24$. Total: $6+36+36+24=102$ With larger alphabets, Suppose there are $a$ numbers with 4 or more in the list, $b$ numbers with exactly 3 in the list, $c$ numbers with exactly 2 in the list, and $d$ numbers with exactly 1 in the list. Distinct numbers with all 4 digits the same: $a$ Distinct numbers with 3 digits the same: $\binom{4}{3}(a+b)(a+b+c+d-1)$ Distinct numbers with 2 pairs of digits: $\binom{4}{2}\binom{a+b+c}{2}$ Distinct numbers with exactly 1 pair of digits: $\binom{4}{2}(a+b+c)\binom{a+b+c+d-1}{2}2!$ Distinct numbers with no pair of digits: $\binom{a+b+c+d}{4}4!$ Total: $a+4(a+b)(a+b+c+d-1)+6\binom{a+b+c}{2}+12(a+b+c)\binom{a+b+c+d-1}{2}+24\binom{a+b+c+d}{4}$ Apply to the previous case: $a=b=0$, $c=2$, and $d=2$: $0+0+6\binom{2}{2}+12(2)\binom{3}{2}+24\binom{4}{4}=102$

Sum with binomial coefficients: $\sum_{k=1}^m \frac{1}{k}{m \choose k} $ I got this sum, in some work related to another question: $$S_m=\sum_{k=1}^m \frac{1}{k}{m \choose k} $$ Are there any known results about this (bounds, asymptotics)?

Consider a random task as follows. First, one chooses a nonempty subset $X$ of $\{1,2,\ldots,m\}$, each with equal probability. Then, one uniformly randomly selects an element $n$ of $X$. The event of interest is when $n=\max(X)$. Fix $k\in\{1,2,\ldots,m\}$. The probability that $|X|=k$ is $\frac{1}{2^m-1}\,\binom{m}{k}$. The probability that the maximum element of $X$, given $X$ with $|X|=k$, is chosen is $\frac{1}{k}$. Consequently, the probability that the desired event happens is given by $$\sum_{k=1}^m\,\left(\frac{1}{2^m-1}\,\binom{m}{k}\right)\,\left(\frac{1}{k}\right)=\frac{S_m}{2^m-1}\,.$$ Now, consider a fixed element $n\in\{1,2,\ldots,m\}$. Then, there are $\binom{n-1}{j-1}$ possible subsets $X$ of $\{1,2,\ldots,m\}$ such that $n=\max(X)$ and $|X|=j$. The probability of getting such an $X$ is $\frac{\binom{n-1}{j-1}}{2^m-1}$. The probability that $n=\max(X)$, given $X$, is $\frac{1}{j}$. That is, the probability that $n=\max(X)$ is $$\begin{align} \sum_{j=1}^{n}\,\left(\frac{\binom{n-1}{j-1}}{2^m-1}\right)\,\left(\frac{1}{j}\right) &=\frac{1}{2^m-1}\,\sum_{j=1}^n\,\frac1j\,\binom{n-1}{j-1}=\frac{1}{2^m-1}\,\left(\frac{1}{n}\,\sum_{j=1}^n\,\frac{n}{j}\,\binom{n-1}{j-1}\right) \\ &=\frac{1}{2^m-1}\,\left(\frac{1}{n}\,\sum_{j=1}^n\,\binom{n}{j}\right)=\frac{1}{2^m-1}\left(\frac{2^n-1}{n}\right)\,. \end{align}$$ Finally, it follows that $\frac{S_m}{2^m-1}=\sum_{n=1}^m\,\frac{1}{2^m-1}\left(\frac{2^n-1}{n}\right)$. Hence, $$\sum_{k=1}^m\,\frac{1}{k}\,\binom{m}{k}=S_m=\sum_{n=1}^m\,\left(\frac{2^n-1}{n}\right)\,.$$ It can then be shown by induction on $m>3$ that $\frac{2^{m+1}}{m}<S_m<\frac{2^{m+1}}{m}\left(1+\frac{2}{m}\right)$.

How to solve this quartic equation? For the quartic equation: $$x^4 - x^3 + 4x^2 + 3x + 5 = 0$$ I tried Ferrari so far and a few others but I just can't get its complex solutions. I know it has no real solutions.

$$x^4 - x^3 + 4x^2 + \underbrace{3x}_{4x-x} + \overbrace{5}^{4+1} = \\\color{red}{x^4-x^3}+4x^2+4x+4\color{red}{-x+1}\\={x^4-x^3}-x+1+4(x^2+x+1)\\={x^3(x-1)}-(x-1)+4(x^2+x+1)\\=(x-1)(x^3-1)+4(x^2+x+1)\\=(x-1)(x-1)(x^2+x+1)+4(x^2+x+1)\\=(x^2+x+1)(x-1)^2+4)=(x^2+x+1)(x^2-2x+5)$$ As for the roots, I assume you could solve those two quadratic equations and you could find the results on Wolfram|Alpha.

Simplify these expressions with radical sign 2 My question is 1) Rationalize the denominator: $$\frac{1}{\sqrt{2}+\sqrt{3}+\sqrt{5}}$$ My answer is: $$\frac{\sqrt{12}+\sqrt{18}-\sqrt{30}}{18}$$ My question is 2) $$\frac{1}{\sqrt{2}+\sqrt{3}-\sqrt{5}}+\frac{1}{\sqrt{2}-\sqrt{3}-\sqrt{5}}$$ My answer is: $$\frac{1}{\sqrt{2}}$$ I would also like to know whether my solutions are right. Thank you,

$\begin{eqnarray*} (\sqrt{2}+\sqrt{3}+\sqrt{5})(\sqrt{12}+\sqrt{18}-\sqrt{30}) & = & (\sqrt{2}+\sqrt{3}+\sqrt{5})(2\sqrt{3}+3\sqrt{2}-\sqrt{2}\sqrt{3}\sqrt{5})\\& = & 12, \end{eqnarray*}$ if you expand out the terms, so your first answer is incorrect. The denominator should be $12$. $\begin{eqnarray*} (\sqrt{2}+\sqrt{3}-\sqrt{5})(\sqrt{2}-\sqrt{3}-\sqrt{5}) & = & (\sqrt{2}-\sqrt{5})^2-\sqrt{3}^2\\& = & 7-2\sqrt{10}-3\\& = & 2\sqrt{2}(\sqrt{2}-\sqrt{5}), \end{eqnarray*}$ and so when your fractions in the second part are given common denominators, you'll have exactly $\cfrac{1}{\sqrt{2}}$ after cancellation, so your second answer is correct. Note: In general, if you want to see if two fractions are the same (as in the first problem), cross-multiplication is often a useful way to see it.

Evaluation of $\lim\limits_{x\rightarrow0} \frac{\tan(x)-x}{x^3}$ One of the previous posts made me think of the following question: Is it possible to evaluate this limit without L'Hopital and Taylor? $$\lim_{x\rightarrow0} \frac{\tan(x)-x}{x^3}$$

Here is a different approach. Let $$L = \lim_{x \to 0} \dfrac{\tan(x) - x}{x^3}$$ Replacing $x$ by $2y$, we get that \begin{align} L & = \lim_{y \to 0} \dfrac{\tan(2y) - 2y}{(2y)^3} = \lim_{y \to 0} \dfrac{\dfrac{2 \tan(y)}{1 - \tan^2(y)} - 2y}{(2y)^3}\\ & = \lim_{y \to 0} \dfrac{\dfrac{2 \tan(y)}{1 - \tan^2(y)} - 2 \tan(y) + 2 \tan(y) - 2y}{(2y)^3}\\ & = \lim_{y \to 0} \dfrac{\dfrac{2 \tan^3(y)}{1 - \tan^2(y)} + 2 \tan(y) - 2y}{(2y)^3}\\ & = \lim_{y \to 0} \left(\dfrac{2 \tan^3(y)}{8y^3(1 - \tan^2(y))} + \dfrac{2 \tan(y) - 2y}{8y^3} \right)\\ & = \lim_{y \to 0} \left(\dfrac{2 \tan^3(y)}{8y^3(1 - \tan^2(y))} \right) + \lim_{y \to 0} \left(\dfrac{2 \tan(y) - 2y}{8y^3} \right)\\ & = \dfrac14 \lim_{y \to 0} \left(\dfrac{\tan^3(y)}{y^3} \dfrac1{1 - \tan^2(y)} \right) + \dfrac14 \lim_{y \to 0} \left(\dfrac{\tan(y) - y}{y^3} \right)\\ & = \dfrac14 + \dfrac{L}4 \end{align} Hence, $$\dfrac{3L}{4} = \dfrac14 \implies L = \dfrac13$$ EDIT In Hans Lundmark answer, evaluating the desired limit boils down to evaluating $$S=\lim_{x \to 0} \dfrac{\sin(x)-x}{x^3}$$ The same idea as above can be used to evaluate $S$ as well. Replacing $x$ by $2y$, we get that \begin{align} S & = \lim_{y \to 0} \dfrac{\sin(2y) - 2y}{(2y)^3} = \lim_{y \to 0} \dfrac{2 \sin(y) \cos(y) - 2y}{8y^3}\\ & = \lim_{y \to 0} \dfrac{2 \sin(y) \cos(y) - 2 \sin(y) + 2 \sin(y) - 2y}{8y^3}\\ & = \lim_{y \to 0} \dfrac{2 \sin(y) - 2y}{8y^3} + \lim_{y \to 0} \dfrac{2 \sin(y) \cos(y)-2 \sin(y)}{8y^3}\\ & = \dfrac14 \lim_{y \to 0} \dfrac{\sin(y) - y}{y^3} - \dfrac14 \lim_{y \to 0} \dfrac{\sin(y) (1 - \cos(y))}{y^3}\\ & = \dfrac{S}4 - \dfrac14 \lim_{y \to 0} \dfrac{\sin(y) 2 \sin^2(y/2)}{y^3}\\ & = \dfrac{S}4 - \dfrac18 \lim_{y \to 0} \dfrac{\sin(y)}{y} \dfrac{\sin^2(y/2)}{(y/2)^2}\\ & = \dfrac{S}4 - \dfrac18 \lim_{y \to 0} \dfrac{\sin(y)}{y} \lim_{y \to 0} \dfrac{\sin^2(y/2)}{(y/2)^2}\\ & = \dfrac{S}4 - \dfrac18\\ \dfrac{3S}4 & = - \dfrac18\\ S & = - \dfrac16 \end{align}

What is the result of sum $\sum\limits_{i=0}^n 2^i$ Possible Duplicate: the sum of powers of $2$ between $2^0$ and $2^n$ What is the result of $$2^0 + 2^1 + 2^2 + \cdots + 2^{n-1} + 2^n\ ?$$ Is there a formula on this? and how to prove the formula? (It is actually to compute the time complexity of a Fibonacci recursive method.)

Let us take a particular example that is large enough to illustrate the general situation. Concrete experience should precede the abstract. Let $n=8$. We want to show that $2^0+2^1+2^2+\cdots +2^8=2^9-1$. We could add up on a calculator, and verify that the result holds for $n=8$. However, we would not learn much during the process. We will instead look at the sum written backwards, so at $$2^8+2^7+2^6+2^5+2^4+2^3+2^2+2^1+2^0.$$ A kangaroo is $2^9$ feet from her beloved $B$. She takes a giant leap of $2^8$ feet. Now she is $2^8$ feet from $B$. She takes a leap of $2^7$ feet. Now she is $2^7$ feet from $B$. She takes a leap of $2^6$ feet. And so on. After a while she is $2^1$ feet from $B$, and takes a leap of $2^0$ feet, leaving her $2^0$ feet from $B$. The total distance she has covered is $2^8+2^7+2^6+\cdots+2^0$. It leaves her $2^0$ feet from $B$, and therefore $$2^8+2^7+2^6+\cdots+2^0+2^0=2^9.$$ Since $2^0=1$, we obtain by subtraction that $2^8+2^7+\cdots +2^0=2^9-1$. We can write out the same reasoning without the kangaroo. Note that $2^0+2^0=2^1$, $2^1+2^1=2^2$, $2^2+2^2=2^3$, and so on until $2^8+2^8=2^9$. Therefore $$(2^0+2^0)+2^1+2^2+2^3+2^4+\cdots +2^8=2^9.$$ Subtract the front $2^0$ from the left side, and $2^0$, which is $1$, from the right side, and we get our result.

Integration of $\int\frac{1}{x^{4}+1}\mathrm dx$ I don't know how to integrate $\displaystyle \int\frac{1}{x^{4}+1}\mathrm dx$. Do I have to use trigonometric substitution?

I think you can do it this way. \begin{align*} \int \frac{1}{x^4 +1} \ dx & = \frac{1}{2} \cdot \int\frac{2}{1+x^{4}} \ dx \\\ &= \frac{1}{2} \cdot \int\frac{(1-x^{2}) + (1+x^{2})}{1+x^{4}} \ dx \\\ &=\frac{1}{2} \cdot \int \frac{1-x^2}{1+x^{4}} \ dx + \frac{1}{2} \int \frac{1+x^{2}}{1+x^{4}} \ dx \\\ &= \frac{1}{2} \cdot -\int \frac{1-\frac{1}{x^2}}{\Bigl(x+\frac{1}{x})^{2} - 2} \ dx + \text{same trick} \end{align*}

Probably simple factoring problem I came across this in a friend's 12th grade math homework and couldn't solve it. I want to factor the following trinomial: $$3x^2 -8x + 1.$$ How to solve this is far from immediately clear to me, but it is surely very easy. How is it done?

A standard way of factorizing, when it is hard to guess the factors, is by completing the square. \begin{align} 3x^2 - 8x + 1 & = 3 \left(x^2 - \dfrac83x + \dfrac13 \right)\\ & (\text{Pull out the coefficient of $x^2$})\\ & = 3 \left(x^2 - 2 \cdot \dfrac43 \cdot x + \dfrac13 \right)\\ & (\text{Multiply and divide by $2$ the coefficient of $x$})\\ & = 3 \left(x^2 - 2 \cdot \dfrac43 \cdot x + \left(\dfrac43 \right)^2 - \left(\dfrac43 \right)^2 + \dfrac13 \right)\\ & (\text{Add and subtract the square of half the coefficient of $x$})\\ & = 3 \left(\left(x - \dfrac43 \right)^2 - \left(\dfrac43 \right)^2 + \dfrac13 \right)\\ & (\text{Complete the square})\\ & = 3 \left(\left(x - \dfrac43 \right)^2 - \dfrac{16}9 + \dfrac13 \right)\\ & = 3 \left(\left(x - \dfrac43 \right)^2 - \dfrac{16}9 + \dfrac39 \right)\\ & = 3 \left(\left(x - \dfrac43 \right)^2 - \dfrac{13}9\right)\\ & = 3 \left(\left(x - \dfrac43 \right)^2 - \left(\dfrac{\sqrt{13}}3 \right)^2\right)\\ & = 3 \left(x - \dfrac43 + \dfrac{\sqrt{13}}3\right) \left(x - \dfrac43 - \dfrac{\sqrt{13}}3\right)\\ & (\text{Use $a^2 - b^2 = (a+b)(a-b)$ to factorize}) \end{align} The same idea works in general. \begin{align} ax^2 + bx + c & = a \left( x^2 + \dfrac{b}ax + \dfrac{c}a\right)\\ & = a \left( x^2 + 2 \cdot \dfrac{b}{2a} \cdot x + \dfrac{c}a\right)\\ & = a \left( x^2 + 2 \cdot \dfrac{b}{2a} \cdot x + \left( \dfrac{b}{2a}\right)^2 - \left( \dfrac{b}{2a}\right)^2 + \dfrac{c}a\right)\\ & = a \left( \left( x + \dfrac{b}{2a}\right)^2 - \left( \dfrac{b}{2a}\right)^2 + \dfrac{c}a\right)\\ & = a \left( \left( x + \dfrac{b}{2a}\right)^2 - \dfrac{b^2}{4a^2} + \dfrac{c}a\right)\\ & = a \left( \left( x + \dfrac{b}{2a}\right)^2 - \left(\dfrac{b^2-4ac}{4a^2} \right)\right)\\ & = a \left( \left( x + \dfrac{b}{2a}\right)^2 - \left(\dfrac{\sqrt{b^2-4ac}}{2a} \right)^2\right)\\ & = a \left( x + \dfrac{b}{2a} + \dfrac{\sqrt{b^2-4ac}}{2a}\right) \left( x + \dfrac{b}{2a} - \dfrac{\sqrt{b^2-4ac}}{2a}\right)\\ \end{align}

Compute: $\sum_{k=1}^{\infty}\sum_{n=1}^{\infty} \frac{1}{k^2n+2nk+n^2k}$ I try to solve the following sum: $$\sum_{k=1}^{\infty}\sum_{n=1}^{\infty} \frac{1}{k^2n+2nk+n^2k}$$ I'm very curious about the possible approaching ways that lead us to solve it. I'm not experienced with these sums, and any hint, suggestion is very welcome. Thanks.

Here's another approach. It depends primarily on the properties of telescoping series, partial fraction expansion, and the following identity for the $m$th harmonic number $$\begin{eqnarray*} \sum_{k=1}^\infty \frac{1}{k(k+m)} &=& \frac{1}{m}\sum_{k=1}^\infty \left(\frac{1}{k} - \frac{1}{k+m}\right) \\ &=& \frac{1}{m}\sum_{k=1}^m \frac{1}{k} \\ &=& \frac{H_m}{m}, \end{eqnarray*}$$ where $m=1,2,\ldots$. Then, $$\begin{eqnarray*} \sum_{k=1}^{\infty}\sum_{n=1}^{\infty} \frac{1}{k^2n+2nk+n^2k} &=& \sum_{k=1}^{\infty} \frac{1}{k} \sum_{n=1}^{\infty} \frac{1}{n(n+k+2)} \\ &=& \sum_{k=1}^{\infty} \frac{1}{k} \frac{H_{k+2}}{k+2} \\ &=& \frac{1}{2} \sum_{k=1}^{\infty} \left( \frac{H_{k+2}}{k} - \frac{H_{k+2}}{k+2} \right) \\ &=& \frac{1}{2} \sum_{k=1}^{\infty} \left( \frac{H_k +\frac{1}{k+1}+\frac{1}{k+2}}{k} - \frac{H_{k+2}}{k+2} \right) \\ &=& \frac{1}{2} \sum_{k=1}^{\infty} \left( \frac{H_k}{k} - \frac{H_{k+2}}{k+2} \right) + \frac{1}{2} \sum_{k=1}^{\infty} \left(\frac{1}{k(k+1)} + \frac{1}{k(k+2)}\right) \\ &=& \frac{1}{2}\left(H_1 + \frac{H_2}{2}\right) + \frac{1}{2}\left(H_1 + \frac{H_2}{2}\right) \\ &=& \frac{7}{4}. \end{eqnarray*}$$

Limit of exponentials Why is $n^n (n+m)^{-{\left(n+m\over 2\right)}}(n-m)^{-{\left(n-m\over 2\right)}}$ asymptotically equal to $\exp\left(-{m^2\over 2n}\right)$ as $n,m\to \infty$?

Let $m = n x$. Take the logarithm: $$ n \log(n) - n \frac{1+x}{2} \left(\log n + \log\left(1+x\right) \right) - n \frac{1-x}{2} \left( \log n + \log\left(1-x\right) \right) $$ Notice that all the terms with $\log(n)$ cancel out, so we are left with $$ -\frac{n}{2} \left( (1+x) \log(1+x) - (1-x) \log(1-x) \right) $$ It seems like the you need to assume that $x$ is small here, meaning that $ m \ll n$. Then, using Taylor series expansion of the logarithm: $$ (1+x) \log(1+x) + (1-x) \log(1-x) = (1+x) \left( x - \frac{x^2}{2} + \mathcal{o}(x^2) \right) + (1-x) \left(-x - \frac{x^2}{2} + \mathcal{o}(x^2)\right) = x^2 + \mathcal{o}(x^3) $$ Hence the original expression, asymptotically, equals $$ \exp\left( -\frac{n}{2} x^2 + \mathcal{o}(n x^3)\right) = \exp\left(- \frac{m^2}{2n} + \mathcal{o}\left(\frac{m^3}{n^2}\right) \right) $$

$p=4n+3$ never has a Decomposition into $2$ Squares, right? Primes of the form $p=4k+1\;$ have a unique decomposition as sum of squares $p=a^2+b^2$ with $0<a<b\;$, due to Thue's Lemma. Is it correct to say that, primes of the form $p=4n+3$, never have a decomposition into $2$ squares, because sum of the quadratic residues $a^2+b^2$ with $a,b\in \Bbb{N}$ $$ a^2 \bmod 4 +b^2 \bmod 4 \le 2? $$ If so, are there alternate ways to prove it?

Yes, as it has been pointed out, if $a^2+b^2$ is odd, one of $a$ and $b$ must be even and the other odd. Then $$ (2m)^2+(2n+1)^2=(4m^2)+(4n^2+4n+1)=4(m^2+n^2+n)+1\equiv1\pmod{4} $$ Thus, it is impossible to have $a^2+b^2\equiv3\pmod{4}$. In fact, suppose that a prime $p\equiv3\pmod{4}$ divides $a^2+b^2$. Since $p$ cannot be written as the sum of two squares, $p$ is also a prime over the Gaussian integers. Therefore, since $p\,|\,(a+ib)(a-ib)$, we must also have that $p\,|\,a+ib$ or $p\,|\,a-ib$, either of which implies that $p\,|\,a$ and $p\,|\,b$. Thus, the exponent of $p$ in the factorization of $a^2+b^2$ must be even. Furthermore, each positive integer whose prime factorization contains each prime $\equiv3\pmod{4}$ to an even power is a sum of two squares. Using the result about quadratic residues in this answer, for any prime $p\equiv1\pmod{4}$, we get that $-1$ is a quadratic residue $\bmod{p}$. That is, there is an $x$ so that $$ x^2+1\equiv0\pmod{p}\tag{1} $$ This means that $$ p\,|\,(x+i)(x-i)\tag{2} $$ since $p$ can divide neither $x+i$ nor $x-i$, $p$ is not a prime in the Gaussian integers, so it must be the product of two Gaussian primes (any more, and we could find a non-trivial factorization of $p$ over the integers). That is, we can write $$ p=(u+iv)(u-iv)=u^2+v^2\tag{3} $$ Note also that $$ 2=(1+i)(1-i)=1^2+1^2\tag{4} $$ Suppose $n$ is a positive integer whose prime factorization contains each prime $\equiv3\pmod{4}$ to even power. Each factor of $2$ and each prime factor $\equiv1\pmod{4}$ can be split into a pair of conjugate Gaussian primes. Each pair of prime factors $\equiv3\pmod{4}$ can be split evenly. Thus, we can split the factors into conjugate pairs: $$ n=(a+ib)(a-ib)=a^2+b^2\tag{5} $$ For example, $$ \begin{align} 90 &=2\cdot3^2\cdot5\\ &=(1+i)\cdot(1-i)\cdot3\cdot3\cdot(2+i)\cdot(2-i)\\ &=[(1+i)3(2+i)]\cdot[(1-i)3(2-i)]\\ &=(3+9i)(3-9i)\\ &=3^2+9^2 \end{align} $$ Thus, we have shown that a positive integer is the sum of two squares if and only if each prime $\equiv3\pmod{4}$ in its prime factorization occurs with even exponent.

${10 \choose 4}+{11 \choose 4}+{12 \choose 4}+\cdots+{20 \choose 4}$ can be simplified as which of the following? ${10 \choose 4}+{11 \choose 4}+{12 \choose 4}+\cdots+{20 \choose 4}$ can be simplified as ? A. ${21 \choose 5}$ B. ${20 \choose 5}-{11 \choose 4}$ C. ${21 \choose 5}-{10 \choose 5}$ D. ${20 \choose 4}$ Please give me a hint. I'm unable to group the terms. By brute force, I'm getting ${21 \choose 5}-{10 \choose 5}$

What is the problem in this solution? If $S=\binom{10}{4} + \binom{11}{4} + \cdots + \binom{20}{4}$ we have \begin{eqnarray} S&=& \left \{ \binom{4}{4} + \binom{5}{4} \cdots + \binom{20}{4}\right \} - \left \{\binom{4}{4} + \binom{5}{4} \cdots + \binom{9}{4}\right \} &=& \binom{21}{5} - \binom{10}{5} \end{eqnarray}

proving :$\frac{ab}{a^2+3b^2}+\frac{cb}{b^2+3c^2}+\frac{ac}{c^2+3a^2}\le\frac{3}{4}$. Let $a,b,c>0$ how to prove that : $$\frac{ab}{a^2+3b^2}+\frac{cb}{b^2+3c^2}+\frac{ac}{c^2+3a^2}\le\frac{3}{4}$$ I find that $$\ \frac{ab}{a^{2}+3b^{2}}=\frac{1}{\frac{a^{2}+3b^{2}}{ab}}=\frac{1}{\frac{a}{b}+\frac{3b}{a}} $$ By AM-GM $$\ \frac{ab}{a^{2}+3b^{2}} \leq \frac{1}{2 \sqrt{3}}=\frac{\sqrt{3}}{6} $$ $$\ \sum_{cyc} \frac{ab}{a^{2}+3b^{2}} \leq \frac{\sqrt{3}}{2} $$ But this is obviously is not working .

I have a Cauchy-Schwarz proof of it,hope you enjoy.:D first,mutiply $2$ to each side,your inequality can be rewrite into $$ \sum_{cyc}{\frac{2ab}{a^2+3b^2}}\leq \frac{3}{2}$$ Or $$ \sum_{cyc}{\frac{(a-b)^2+2b^2}{a^2+3b^2}}\geq \frac{3}{2}$$ Now,Using Cauchy-Schwarz inequality,we have $$ \sum_{cyc}{\frac{(a-b)^2+2b^2}{a^2+3b^2}}\geq \frac{\left(\sum_{cyc}{\sqrt{(a-b)^2+2b^2}}\right)^2}{4(a^2+b^2+c^2)}$$ Therefore,it's suffice to prove $$\left(\sum_{cyc}{\sqrt{(a-b)^2+2b^2}}\right)^2\geq 6(a^2+b^2+c^2) $$ after simply expand,it's equal to $$ \sum_{cyc}{\sqrt{[(a-b)^2+2b^2][(b-c)^2+2c^2]}}\geq a^2+b^2+c^2+ab+bc+ca $$ Now,Using Cauchy-Schwarz again,we notice that $$ \sqrt{[(a-b)^2+2b^2][(b-c)^2+2c^2]}\geq (b-a)(b-c)+2bc=b^2+ac+bc-ab$$ sum them up,the result follows. Hence we are done! Equality occurs when $a=b=c$

A problem dealing with even perfect numbers. Question: Show that all even perfect numbers end in 6 or 8. This is what I have. All even perfect numbers are of the form $n=2^{p-1}(2^p -1)$ where $p$ is prime and so is $(2^p -1)$. What I did was set $2^{p-1}(2^p -1)\equiv x\pmod {10}$ and proceeded to show that $x=6$ or $8$ were the only solutions. Now, $2^{p-1}(2^p -1)\equiv x\pmod {10} \implies 2^{p-2}(2^p -1)\equiv \frac{x}{2}\pmod {5}$, furthermore there are only two solutions such that $0 \le \frac{x}{2} < 5$. So I plugged in the first two primes and only primes that satisfy. That is if $p=2$ then $\frac{x}{2}=3$ when $p=3$ then $\frac{x}{2}=4$. These yield $x=6$ and $x=8$ respectively. Furthermore All solutions are $x=6+10r$ or $x=8+10s$. I would appreciate any comments and or alternate approaches to arrive at a good proof.

$p$ is prime so it is 1 or 3$\mod 4$. So, the ending digit of $2^p$ is (respectively) 2 or 8 (The ending digits of powers of 2 are $2,4,8,6,2,4,8,6,2,4,8,6...$ So, the ending digit of $2^{p-1}$ is (respectively) 6 or 4; and the ending digit of $2^p-1$ is (respectively) 1 or 7. Hence the ending digit of $2^{p-1}(2^p-1)$ is (respectively) $6\times1$ or $4\times7$ modulo 10, i.e., $6$ or $8$.

When is $(6a + b)(a + 6b)$ a power of two? Find all positive integers $a$ and $b$ for which the product $(6a + b)(a + 6b)$ is a power of $2$. I havnt been able to get this one yet, found it online, not homework! any help is appreciated thanks!

Assume, (6a + b)(a + 6b) = 2^c where, c is an integer ge 0 Assume, (6a + b) = 2^r where, r is an integer ge 0 Assume, (a + 6b) = 2^s where, s is an integer ge 0 Now, (2^r)(2^s) = 2^c i.e. r + s = c Now, (6a + b) + (a + 6b) = 2^r + 2^s i.e. 7(a + b) = 2^r + 2^s i.e. a + b = (2^r)/7 + (2^s)/7 Now, (6a + b) - ( a + 6b) = 2^r - 2^s i.e. a - b = (2^r)/5 - (2^s)/5 Now we have, a + b = (2^r)/7 + (2^s)/7 a - b = (2^r)/5 - (2^s)/5 Here solving for a we get, a = ((2^r)(6) - 2^s)/35 Now solving for b we get, b = ((2^s)(6) - 2^r)/35 A careful observation reveals that there are no positive integers a and b for which the product (6a + b)(a + 6b) is a power of 2.

Factorize $f$ as product of irreducible factors in $\mathbb Z_5$ Let $f = 3x^3+2x^2+2x+3$, factorize $f$ as product of irreducible factors in $\mathbb Z_5$. First thing I've used the polynomial reminder theorem so to make the first factorization: $$\begin{aligned} f = 3x^3+2x^2+2x+3 = (3x^2-x+3)(x+1)\end{aligned}$$ Obviously then as second step I've taken care of that quadratic polynomial, so: $x_1,x_2=\frac{-b\pm\sqrt{\Delta}}{2a}=\frac{1\pm\sqrt{1-4(9)}}{6}=\frac{1\pm\sqrt{-35}}{6}$ my question is as I've done calculations in $\mathbb Z_5$, was I allowed to do that: as $-35 \equiv_5 100 \Rightarrow \sqrt{\Delta}=\sqrt{-35} = \sqrt{100}$ then $x_1= \frac{11}{6} = 1 \text { (mod 5)}$, $x_2= -\frac{3}{2} = 1 \text { (mod 5)}$, therefore my resulting product would be $f = (x+1)(x+1)(x+1)$. I think I have done something illegal, that is why multiplying back $(x+1)(x+1)$ I get $x^2+2x+1 \neq 3x^2-x+3$. Any ideas on how can I get to the right result?

If $f(X) = aX^2 + bX + c$ is a quadratic polynomial with roots $x_1$ and $x_2$ then $f(X) = a(X-x_1)(X-x_2)$ (the factor $a$ is necessary to get the right leading coefficient). You found that $3x^2-x+3$ has a double root at $x_1 = x_2 = 1$, so $3x^2-x+3 = 3(x-1)^2$. Your mistakes were * *You forgot to multiply by the leading coefficient $3$. *You concluded that a root in $1$ corresponds to the linear factor $(x+1)$, but this would mean a root in $-1$. The right linear factor is $(x-1)$.

Solve for $x$; $\tan x+\sec x=2\cos x;-\infty\lt x\lt\infty$ Solve for $x$; $\tan x+\sec x=2\cos x;-\infty\lt x\lt\infty$ $$\tan x+\sec x=2\cos x$$ $$\left(\dfrac{\sin x}{\cos x}\right)+\left(\dfrac{1}{\cos x}\right)=2\cos x$$ $$\left(\dfrac{\sin x+1}{\cos x}\right)=2\cos x$$ $$\sin x+1=2\cos^2x$$ $$2\cos^2x-\sin x+1=0$$ Edit: $$2\cos^2x=\sin x+1$$ $$2(1-\sin^2x)=\sin x+1$$ $$2\sin^2x+\sin x-1=0$$ $\sin x=a$ $$2a^2+a-1=0$$ $$(a+1)(2a-1)=0$$ $$a=-1,\dfrac{1}{2}$$ $$\arcsin(-1)=-90^\circ=-\dfrac{\pi}{2}$$ $$\arcsin\left(\dfrac{1}{2}\right)=30^\circ=\dfrac{\pi}{6}$$ $$180^\circ-30^\circ=150^\circ=\dfrac{5 \pi}{6}$$ $$x=\dfrac{\pi}{6},-\dfrac{\pi}{2},\dfrac{5 \pi}{6}$$ I actually do not know if those are the only answers considering my range is infinite:$-\infty\lt x\lt\infty$

\begin{eqnarray} \left(\dfrac{\sin x}{\cos x}\right)+\left(\dfrac{1}{\cos x}\right)&=&2\cos x\\ \sin x + 1 &=& 2 \cos^2 x \\ \sin x + 1 &=& 2(1-\sin^2 x)\\ \end{eqnarray} Then $\sin x = -1$ or $\sin x = 1/2$ and the solutions are $-\pi + 2k \pi, \pi/6 + 2 k\pi$ and $5\pi/6 + 2k \pi$.

Proving:$\tan(20^{\circ})\cdot \tan(30^{\circ}) \cdot \tan(40^{\circ})=\tan(10^{\circ})$ how to prove that : $\tan20^{\circ}.\tan30^{\circ}.\tan40^{\circ}=\tan10^{\circ}$? I know how to prove $ \frac{\tan 20^{0}\cdot\tan 30^{0}}{\tan 10^{0}}=\tan 50^{0}, $ in this way: $ \tan{20^0} = \sqrt{3}.\tan{50^0}.\tan{10^0}$ $\Longleftrightarrow \sin{20^0}.\cos{50^0}.\cos{10^0} = \sqrt{3}.\sin{50^0}.\sin{10^0}.\cos{20^0}$ $\Longleftrightarrow \frac{1}{2}\sin{20^0}(\cos{60^0}+\cos{40^0}) = \frac{\sqrt{3}}{2}(\cos{40^0}-\cos{60^0}).\cos{20^0}$ $\Longleftrightarrow \frac{1}{4}\sin{20^0}+\frac{1}{2}\sin{20^0}.\cos{40^0} = \frac{\sqrt{3}}{2}\cos{40^0}.\cos{20^0}-\frac{\sqrt{3}}{4}.\cos{20^0}$ $\Longleftrightarrow \frac{1}{4}\sin{20^0}-\frac{1}{4}\sin{20^0}+\frac{1}{4}\sin{60^0} = \frac{\sqrt{3}}{4}\cos{60^0}+\frac{\sqrt{3}}{4}\cos{20^0}-\frac{\sqrt{3}}{4}\cos{20^0}$ $\Longleftrightarrow \frac{\sqrt{3}}{8} = \frac{\sqrt{3}}{8}$ Could this help to prove the first one and how ?Do i just need to know that $ \frac{1}{\tan\theta}=\tan(90^{\circ}-\theta) $ ?

Another approach: Lets, start by arranging the expression: $$\tan(20°) \tan(30°) \tan(40°) = \tan(30°) \tan(40°) \tan(20°)$$$$=\tan(30°) \tan(30°+10°) \tan(30° - 10°)$$ Now, we will express $\tan(30° + 10°) $ and $\tan(30° - 10°)$ as the ratio of Prosthaphaeresis Formulas, giving us: $$\tan(30°) \left( \frac{\tan(30°) + \tan(10°)}{1 - \tan(30°) \tan(10°)}\right) \left( \frac{\tan(30°) - \tan(10°)}{1 + \tan(30°) \tan(10°)}\right) $$ $$= \tan(30°) \left( \frac{\tan^2(30°) - \tan^2(10°)}{1 - \tan^2(30°) \tan^2(10°)}\right) $$ Substituting the value of $\color{blue}{\tan(30°)}$, $$ = \tan(30°) \left(\frac{1 - 3\tan²(10°)}{3 - \tan²(10°)}\right) $$ Multiplying and dividing by $\color{blue}{\tan(10°)}$, $$=\tan(30°) \tan(10°) \left(\frac{1 - 3\tan²(10°)}{3 \tan(10°) - \tan^3(10°)}\right) $$ It can be easily shown that: $\color{blue}{\tan 3A =\large \frac{3 \tan A - \tan^3A}{1-3\tan^2A}}$, Thus, our problem reduces to: $$=\tan(30°) \tan(10°) \frac{1}{\tan(3\times 10°)}= \tan(10°)$$ QED!

Prove $ (r\sin A \cos A)^2+(r\sin A \sin A)^2+(r\cos A)^2=r^2$ How would I verify the following trig identity? $$(r\sin A \cos A)^2+(r\sin A \sin A)^2+(r\cos A)^2=r^2$$ My work thus far is $$(r^2\cos^2A\sin^2A)+(r^2\sin^2A\sin^2A)+(r^2\cos^2A)$$ But how would I continue? My math skills fail me.

Just use the distributive property and $\sin^2(x)+\cos^2(x)=1$: $$ \begin{align} &(r\sin(A)\cos(A))^2+(r\sin(A)\sin(A))^2+(r\cos(A))^2\\ &=r^2\sin^2(A)(\cos^2(A)+\sin^2(A))+r^2\cos^2(A)\\ &=r^2\sin^2(A)+r^2\cos^2(A)\\ &=r^2(\sin^2(A)+\cos^2(A))\\ &=r^2\tag{1} \end{align} $$ This can be generalized to $$ \begin{align} &(r\sin(A)\cos(B))^2+(r\sin(A)\sin(B))^2+(r\cos(A))^2\\ &=r^2\sin^2(A)(\cos^2(B)+\sin^2(B))+r^2\cos^2(A)\\ &=r^2\sin^2(A)+r^2\cos^2(A)\\ &=r^2(\sin^2(A)+\cos^2(A))\\ &=r^2\tag{2} \end{align} $$ $(2)$ verifies that spherical coordinates have the specified distance from the origin.

Prove $\frac{1}{1 \cdot 3} + \frac{1}{3 \cdot 5} + \frac{1}{5 \cdot 7} + \cdots$ converges to $\frac 1 2 $ Show that $$\frac{1}{1 \cdot 3} + \frac{1}{3 \cdot 5} + \frac{1}{5 \cdot 7} + \cdots = \frac{1}{2}.$$ I'm not exactly sure what to do here, it seems awfully similar to Zeno's paradox. If the series continues infinitely then each term is just going to get smaller and smaller. Is this an example where I should be making a Riemann sum and then taking the limit which would end up being $1/2$?

Solution as per David Mitra's hint in a comment. Write the given series as a telescoping series and evaluate its sum: $$\begin{eqnarray*} S &=&\frac{1}{1\cdot 3}+\frac{1}{3\cdot 5}+\frac{1}{5\cdot 7}+\cdots \\ &=&\sum_{n=1}^{\infty }\frac{1}{\left( 2n-1\right) \left( 2n+1\right) } \\ &=&\sum_{n=1}^{\infty }\left( \frac{1}{2\left( 2n-1\right) }-\frac{1}{ 2\left( 2n+1\right) }\right)\quad\text{Partial fractions decomposition} \\ &=&\frac{1}{2}\sum_{n=1}^{\infty }\left( \frac{1}{2n-1}-\frac{1}{2n+1} \right) \qquad \text{Telescoping series} \\ &=&\frac{1}{2}\sum_{n=1}^{\infty }\left( a_{n}-a_{n+1}\right), \qquad a_{n}= \frac{1}{2n-1},a_{n+1}=\frac{1}{2\left( n+1\right) -1}=\frac{1}{2n+1} \\ &=&\frac{1}{2}\left( a_{1}-\lim_{n\rightarrow \infty }a_{n}\right) \qquad\text{see below} \\ &=&\frac{1}{2}\left( \frac{1}{2\cdot 1-1}-\lim_{n\rightarrow \infty }\frac{1 }{2n-1}\right) \\ &=&\frac{1}{2}\left( 1-0\right) \\ &=&\frac{1}{2}. \end{eqnarray*}$$ Added: The sum of the telescoping series $\sum_{n=1}^{\infty }\left( a_{n}-a_{n+1}\right)$ is the limit of the telescoping sum $\sum_{n=1}^{N}\left( a_{n}-a_{n+1}\right) $ as $N$ tends to $\infty$. Since $$\begin{eqnarray*} \sum_{n=1}^{N}\left( a_{n}-a_{n+1}\right) &=&\left( a_{1}-a_{2}\right) +\left( a_{2}-a_{3}\right) +\ldots +\left( a_{N-1}-a_{N}\right) +\left( a_{N}-a_{N+1}\right) \\ &=&a_{1}-a_{2}+a_{2}-a_{3}+\ldots +a_{N-1}-a_{N}+a_{N}-a_{N+1} \\ &=&a_{1}-a_{N+1}, \end{eqnarray*}$$ we have $$\begin{eqnarray*} \sum_{n=1}^{\infty }\left( a_{n}-a_{n+1}\right) &=&\lim_{N\rightarrow \infty }\sum_{n=1}^{N}\left( a_{n}-a_{n+1}\right) \\ &=&\lim_{N\rightarrow \infty }\left( a_{1}-a_{N+1}\right) \\ &=&a_{1}-\lim_{N\rightarrow \infty }a_{N+1} \\ &=&a_{1}-\lim_{N\rightarrow \infty }a_{N} \\ &=&a_{1}-\lim_{n\rightarrow \infty }a_{n}.\end{eqnarray*}$$

Trigonometry proof involving sum difference and product formula How would I solve the following trig problem. $$\cos^5x = \frac{1}{16} \left( 10 \cos x + 5 \cos 3x + \cos 5x \right)$$ I am not sure what to really I know it involves the sum and difference identity but I know not what to do.

$$\require{cancel} \frac1{16} [ 5(\cos 3x+\cos x)+\cos 5x+5\cos x ]\\ =\frac1{16}[10\cos x \cos 2x+ \cos 5x +5 \cos x]\\ =\frac1{16} [5\cos x(2\cos 2x+1)+\cos 5x]\\ =\frac1{16} [5\cos x(2(2\cos^2 x-1)+1)+\cos 5x]\\ =\frac1{16} [5\cos x(4\cos^2 x-1)+\cos 5x]\\ =\frac1{16} [5\cos x(4\cos^2 x-1)+\cos 5x]\\ =\frac1{16} [20\cos^3 x-5\cos x+\cos 5x]\\ =\frac1{16} [20\cos^3 x-5\cos x+\cos 5x]\\ =\frac1{16} [\cancel{20\cos^3 x}\cancel{-5\cos x}+16\cos^5 x\cancel{-20\cos^3 x}+\cancel{5\cos x}]\\ =\frac1{16} (16\cos^5 x)\\=\cos^5 x$$

Find the domain of $f(x)=\frac{3x+1}{\sqrt{x^2+x-2}}$ Find the domain of $f(x)=\dfrac{3x+1}{\sqrt{x^2+x-2}}$ This is my work so far: $$\dfrac{3x+1}{\sqrt{x^2+x-2}}\cdot \sqrt{\dfrac{x^2+x-2}{x^2+x-2}}$$ $$\dfrac{(3x+1)(\sqrt{x^2+x-2})}{x^2+x-2}$$ $(3x+1)(\sqrt{x^2+x-2})$ = $\alpha$ (Just because it's too much to type) $$\dfrac{\alpha}{\left[\dfrac{-1\pm \sqrt{1-4(1)(-2)}}{2}\right]}$$ $$\dfrac{\alpha}{\left[\dfrac{-1\pm \sqrt{9}}{2}\right]}$$ $$\dfrac{\alpha}{\left[\left(\dfrac{-1+3}{2}\right)\left(\dfrac{-1-3}{2}\right)\right]}$$ $$\dfrac{\alpha}{(1)(-2)}$$ Now, I checked on WolframAlpha and the domain is $x\in \mathbb R: x\lt -2$ or $x\gt 1$ But my question is, what do I do with the top of the problem? Or does it just not matter at all.

Note that $x^2+x-2=(x-1)(x+2)$. There is a problem only if $(x-1)(x+2)$ is $0$ or negative. (If it is $0$, we have a division by $0$ issue, and if it is negative we have a square root of a negative issue.) Can you find where $(x-1)(x+2)$ is $0$? Can you find where it is negative? Together, these are the numbers which are not in the domain of $f(x)$. Or, to view things more positively, the function $f(x)$ is defined precisely for all $x$ such that $(x-1)(x+2)$ is positive. Remark: Let $g(x)=x^2+x-2$. We want to know where $g(x)$ is positive. By factoring, or by using the Quadratic Formula, we can see that $g(x)=0$ at $x=-2$ and at $x=1$. It is a useful fact that a nice continuous function can only change sign by going throgh $0$. This means that in the interval $(-\infty, -2)$, $g(x)$ has constant sign. It also has constant sign in $(-2,1)$, and also in $(1,\infty)$. We still don't know which signs. But this can be determined by finding $g(x)$ at a test point in each interval. For example, let $x=-100$. Clearly $g(-100)$ is positive, so $g(x)$ is positive for all $x$ in the interval $(-\infty,0)$. For the interval $(-2,1)$, $x=0$ is a convenient test point. Note that $g(0) \lt 0$, so $g(x)$ is negative in the whole interval $(-2,1)$. A similar calculation will settle things for the remaining interval $(1,\infty)$. There are many other ways to handle the problem. For example, you know that the parabola $y=(x+2)(x-1)$ is upward facing, and crosses the $x$-axis at $x=-2$ and $x=1$. So it is below the $x$-axis (negative) only when $-2 \lt x \lt 1$. Or we can work with pure inequalities. The product $(x+2)(x-1)$ is positive when $x+2$ and $x-1$ are both positive. This happens when $x \gt 1$. The product is also positive when $x+2$ and $x-1$ are both negative. This happens when $x \lt -2$.

Solving linear system of equations when one variable cancels I have the following linear system of equations with two unknown variables $x$ and $y$. There are two equations and two unknowns. However, when the second equation is solved for $y$ and substituted into the first equation, the $x$ cancels. Is there a way of re-writing this system or re-writing the problem so that I can solve for $x$ and $y$ using linear algebra or another type of numerical method? $2.6513 = \frac{3}{2}y + \frac{x}{2}$ $1.7675 = y + \frac{x}{3}$ In the two equations above, $x=3$ and $y=0.7675$, but I want to solve for $x$ and $y$, given the system above. If I subtract the second equation from the first, then: $2.6513 - 1.7675 = \frac{3}{2}y - y + \frac{x}{2} - \frac{x}{3}$ Can the equation in this alternate form be useful in solving for $x$ and $y$? Is there another procedure that I can use? In this alternate form, would it be possible to limit $x$ and $y$ in some way so that a solution for $x$ and $y$ can be found by numerical optimization?

$$\begin{equation*} \left\{ \begin{array}{c} 2.6513=\frac{3}{2}y+\frac{x}{2} \\ 1.7675=y+\frac{x}{3} \end{array} \right. \end{equation*}$$ If we multiply the first equation by $2$ and the second by $3$ we get $$\begin{equation*} \left\{ \begin{array}{c} 5.3026=3y+x \\ 5.3025=3y+x \end{array} \right. \end{equation*}$$ This system has no solution because $$\begin{equation*} 5.3026\neq 5.3025 \end{equation*}$$ However if the number $2.6513$ resulted from rounding $2.65125$, then the same computation yields $$\begin{equation*} \left\{ \begin{array}{c} 5.3025=3y+x \\ 5.3025=3y+x \end{array} \right. \end{equation*}$$ which is satisfied by all $x,y$. A system of the form $$\begin{equation*} \begin{pmatrix} a_{11} & a_{12} \\ a_{21} & a_{22} \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} b_{1} \\ b_{2} \end{pmatrix} \end{equation*}$$ has the solution (Cramer's rule) $$\begin{equation*} \begin{pmatrix} x \\ y \end{pmatrix} =\frac{1}{\det \begin{pmatrix} a_{11} & a_{12} \\ a_{21} & a_{22} \end{pmatrix} } \begin{pmatrix} a_{22}b_{1}-a_{12}b_{2} \\ a_{11}b_{2}-a_{21}b_{1} \end{pmatrix} = \begin{pmatrix} \frac{a_{22}b_{1}-a_{12}b_{2}}{a_{11}a_{22}-a_{21}a_{12}} \\ \frac{a_{11}b_{2}-a_{21}b_{1}}{a_{11}a_{22}-a_{21}a_{12}} \end{pmatrix} \end{equation*}$$ if $\det \begin{pmatrix} a_{11} & a_{12} \\ a_{21} & a_{22}% \end{pmatrix}% \neq 0$. In the present case, we have $$\begin{equation*} \begin{pmatrix} \frac{1}{2} & \frac{3}{2} \\ \frac{1}{3} & 1 \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} 2.6513 \\ 1.7675 \end{pmatrix} \end{equation*}$$ and $$\det \begin{pmatrix} \frac{1}{2} & \frac{3}{2} \\ \frac{1}{3} & 1 \end{pmatrix} =0$$

About the sequence satisfying $a_n=a_{n-1}a_{n+1}-1$ "Consider sequences of positive real numbers of the form x,2000,y,..., in which every term after the first is 1 less than the product of its two immediate neighbors. For how many different values of x does the term 2001 appear somewhere in the sequence? (A) 1 (B) 2 (C) 3 (D) 4 (E) More than 4" Can anyone suggest a systematic way to solve this problem? Thanks!

(This is basically EuYu's answer with the details of periodicity added; took a while to type up.) Suppose that $a_0 , a_1 , \ldots$ is a generalised sequence of the type described, so that $a_i = a_{i-1} a_{i+1} - 1$ for all $i > 0$. Note that this condition is equivalent to demanding that $$a_{i+1} = \frac{ a_i + 1 }{a_{i-1}}.$$ Using this we find the following recurrences: $$ a_2 = \frac{ a_1 + 1}{a_0}; \\ a_3 = \frac{ a_2 + 1}{a_1} = \frac{ \frac{ a_1 + 1}{a_0} }{a_1} = \frac{ a_0 + a_1 + 1 }{ a_0a_1 }; \\ a_4 = \frac{ a_3 + 1 }{a_2} = \frac{\frac{ a_0 + a_1 + 1 }{ a_0a_1 } + 1}{\frac{ a_1 + 1}{a_0}} = \frac{ ( a_0 + 1 )( a_1 + 1) }{ a_1 ( a_1 + 1 ) } = \frac{a_0 + 1}{a_1};\\ a_5 = \frac{ a_4 + 1 }{ a_3 } = \frac{ \frac{a_0 + 1}{a_1} + 1}{\frac{ a_0 + a_1 + 1 }{ a_0a_1 }} = \frac{ \left( \frac{a_0 + a_1 + 1}{a_1} \right) }{ \left( \frac{a_0+a_1+1}{a_0a_1} \right) } = a_0 \\ a_6 = \frac{ a_5 + 1 }{a_4} = \frac{ a_0 + 1}{ \left( \frac{ a_0 + 1 }{a_1} \right) } = a_1. $$ Thus every such sequence is periodic with period 5, so if 2001 appears, it must appear as either $a_0, a_1, a_2, a_3, a_4$. * *Clearly if $a_0 = 2001$, we're done. *As we stipulate that $a_1 = 2000$, it is impossible for $a_1 = 2001$. *If $a_2 = 2001$, then it must be that $2001 = \frac{ 2000 + 1 }{a_0}$ and so $a_0 = 1$. *If $a_3 = 2001$, then it must be that $2001 = \frac{a_0 + 2000 + 1}{a_0 \cdot 2000}$, and it follows that $a_0 = \frac{2001}{2000 \cdot 2001 - 1}$. *If $a_4 = 2001$, then it must be that $2001 = \frac{ a_0 + 1 }{2000}$, and so $a_0 = 2001 \cdot 2000 - 1$. There are thus exactly four values of $a_0$ such that 2001 appears in the sequence.

Can you construct a field with 6 elements? Possible Duplicate: Is there anything like GF(6)? Could someone tell me if you can build a field with 6 elements.

If such a field $F$ exists, then the multiplicative group $F^\times$ is cyclic of order 5. So let $a$ be a generator for this group and write $F = \{ 0, 1, a, a^2, a^3, a^4\}$. From $a(1 + a + a^2 + a^3 + a^4) = 1 + a + a^2 + a^3 + a^4$, it immediately follows that $1 + a + a^2 + a^3 + a^4 = 0$. Let's call this (*). Since $0$ is the additive inverse of itself and $F^\times$ has odd number of elements, at least one element of $F^\times$ is its own additive inverse. Since $F$ is a field, this implies $1 = -1$. So, in fact, every element of $F^\times$ is its own additive inverse (**). Now, note that $1 + a$ is different from $0$, $1$ and $a$. So it is $a^i$, where i = 2, 3 or 4. Then, $1 + a - a^i = 1 + a + a^i = 0$. Hence, by $(*)$ one of $a^2 + a^3$, $a^2 + a^4$ and $a^3 + a^4$ must be $0$, a contradiction with (**).

Rigorous proof of the Taylor expansions of sin $x$ and cos $x$ We learn trigonometric functions in high school, but their treatment is not rigorous. Then we learn that they can be defined by power series in a college. I think there is a gap between the two. I'd like to fill in the gap in the following way. Consider the upper right quarter of the unit circle $C$ = {$(x, y) \in \mathbb{R}^2$; $x^2 + y^2 = 1$, $0 \leq x \leq 1$, $y \geq 0$}. Let $\theta$ be the arc length of $C$ from $x = 0$ to $x$, where $0 \leq x \leq 1$. By the arc length formula, $\theta$ can be expressed by a definite integral of a simple algebraic function from 0 to $x$. Clearly $\sin \theta = x$ and $\cos\theta = \sqrt{1 - \sin^2 \theta}$. Then how do we prove that the Taylor expansions of $\sin\theta$ and $\cos\theta$ are the usual ones?

Since $x^2 + y^2 = 1$, $y = \sqrt{1 - x^2}$, $y' = \frac{-x}{\sqrt{1 - x^2}}$ By the arc length formula, $\theta = \int_{0}^{x} \sqrt{1 + y'^2} dx = \int_{0}^{x} \frac{1}{\sqrt{1 - x^2}} dx$ We consider this integral on the interval [$-1, 1$] instead of [$0, 1$]. Then $\theta$ is a monotone strictly increasing function of $x$ on [$-1, 1$]. Hence $\theta$ has the inverse function defined on [$\frac{-\pi}{2}, \frac{\pi}{2}$]. We denote this function also by $\sin\theta$. We redefine $\cos\theta = \sqrt{1 - \sin^2 \theta}$ on [$\frac{-\pi}{2}, \frac{\pi}{2}$]. Since $\frac{d\theta}{dx} = \frac{1}{\sqrt{1 - x^2}}$, (sin $\theta)' = \frac{dx}{d\theta} = \sqrt{1 - x^2} =$ cos $\theta$. On the other hand, $(\cos\theta)' = \frac{d\sqrt{1 - x^2}}{d\theta} = \frac{d\sqrt{1 - x^2}}{dx} \frac{dx}{d\theta} = \frac{-x}{\sqrt{1 - x^2}} \sqrt{1 - x^2} = -x = -\sin\theta$ Hence $(\sin\theta)'' = (\cos\theta)' = -\sin\theta$ $(\cos\theta)'' = -(\sin\theta)' = -\cos\theta$ Hence by the induction on $n$, $(\sin\theta)^{(2n)} = (-1)^n\sin\theta$ $(\sin\theta)^{(2n+1)} = (-1)^n\cos\theta$ $(\cos\theta)^{(2n)} = (-1)^n\cos\theta$ $(\cos\theta)^{(2n+1)} = (-1)^{n+1}\sin\theta$ Since $\sin 0 = 0, \cos 0 = 1$, $(\sin\theta)^{(2n)}(0) = 0$ $(\sin\theta)^{(2n+1)}(0) = (-1)^n$ $(\cos\theta)^{(2n)}(0) = (-1)^n$ $(\cos\theta)^{(2n+1)}(0) = 0$ Note that $|\sin\theta| \le 1$, $|\cos\theta| \le 1$. Hence, by Taylor's theorem, $\sin\theta = \sum_{n=0}^{\infty} (-1)^n \frac{\theta^{2n+1}}{(2n+1)!}$ $\cos\theta = \sum_{n=0}^{\infty} (-1)^n \frac{\theta^{2n}}{(2n)!}$ QED Remark: When you consider the arc length of the lemniscate instead of the circle, you will encounter $\int_{0}^{x} \frac{1}{\sqrt{1 - x^4}} dx$. You may find interesting functions like we did with $\int_{0}^{x} \frac{1}{\sqrt{1 - x^2}} dx$. This was young Gauss's approach and he found elliptic functions.

What is the best way to solve a problem given remainders and divisors? $x$ is a positive integer less than $100$. When $x$ is divided by $5$, its remainder is $4$. When $x$ is divided by $23$, its remainder is $7$. What is $x$ ?

So, $x=5y+4$ and $x=23z+7$ for some integers $y,z$ So, $5y+4=23z+7=>5y+20=23z+23$ adding 16 to either sides, $5(y+4)=23(z+1)=>y+4=\frac{23(z+1)}{5}$, so,$5|(z+1)$ as $y+4$ is integer and $(5,23)=1$ $=>z+1=5w$ for some integer $w$. $x=23(5w-1)+7=115w-16$ As $0<x<100$ so,$x=99$ putting $w=1$ Alternatively, We have $5y=23z+3$ Using convergent property of continued fractions, (i)$\frac{23}{5}=4+\frac{3}{5}=4+\frac{1}{\frac{5}{3}}=4+\frac{1}{1+\frac{2}{3}}$ $=4+\frac{1}{1+\frac{1}{\frac{3}{2}}}=4+\frac{1}{1+\frac{1}{1+\frac{1}{2}}}$ So, the last but one convergent is $4+\frac{1}{1+\frac{1}{1}}=\frac{9}{2}$ $=>23\cdot2-5\cdot9=1$ $5y=23z+3(23\cdot2-5\cdot9)=>5(y+27)=23(z+6)=>5|(z+6)=>5|(z+1)$ (ii)$\frac{23}{5}=4+\frac{3}{5}=4+\frac{1}{\frac{5}{3}}=4+\frac{1}{1+\frac{2}{3}}$ $=4+\frac{1}{1+\frac{1}{\frac{3}{2}}}=4+\frac{1}{1+\frac{1}{1+\frac{1}{2}}}$ $=4+\frac{1}{1+\frac{1}{1+\frac{1}{1+1}}}$ So, the last convergent is $4+\frac{1}{1+\frac{1}{1+1}}=\frac{14}{3}$ $=>23\cdot3-5\cdot14=-1$ $5y=23z-3(23\cdot3-5\cdot14)=5(y-42)=23(z-9)=>5|(z-9)=>5|(z+1)$

how can one find the value of the expression, $(1^2+2^2+3^2+\cdots+n^2)$ Possible Duplicate: Proof that $\sum\limits_{k=1}^nk^2 = \frac{n(n+1)(2n+1)}{6}$? Summation of natural number set with power of $m$ How to get to the formula for the sum of squares of first n numbers? how can one find the value of the expression, $(1^2+2^2+3^2+\cdots+n^2)$ Let, $T_{2}(n)=1^2+2^2+3^2+\cdots+n^2$ $T_{2}(n)=(1^2+n^2)+(2^2+(n-1)^2)+\cdots$ $T_{2}(n)=((n+1)^2-2(1)(n))+((n+1)^2-2(2)(n-1))+\cdots$

The claim is that $\sum_{i = 1}^n i^2 = \frac{n(n + 1)(2n + 1)}{6}$ We will verify this by induction. Clearly $n = 1$ holds. Suppose the formula holds for $n$. Lets verify it holds for $n + 1$. $$\sum_{i = 1}^{n + 1} i^2 = \sum_{i = 1}^n i^2 + (n + 1)^2 = \frac{n(n + 1)(2n + 1)}{6} + (n + 1)^2 \\ = \frac{n(n + 1)(2n + 1)}{6} + \frac{6(n^2 + 2n + 1)}{6} \\ = \frac{2n^3 + 3n^2 + n}{6} + \frac{6n^2 + 12 n + 6}{6} \\ = \frac{2n^3 + 9n^2 + 13n + 6}{6}$$ If you factor you get $$= \frac{(n + 1)(n + 2)(2n + 3)}{6} = \frac{(n + 1)((n + 1) + 1)(2(n + 1) + 1)}{6}$$ The result follows for $n + 1$. So by induction the formula holds.

Why is $a^n - b^n$ divisible by $a-b$? I did some mathematical induction problems on divisibility * *$9^n$ $-$ $2^n$ is divisible by 7. *$4^n$ $-$ $1$ is divisible by 3. *$9^n$ $-$ $4^n$ is divisible by 5. Can these be generalized as $a^n$ $-$ $b^n$$ = (a-b)N$, where N is an integer? But why is $a^n$ $-$ $b^n$$ = (a-b)N$ ? I also see that $6^n$ $- 5n + 4$ is divisible by $5$ which is $6-5+4$ and $7^n$$+3n + 8$ is divisible by $9$ which is $7+3+8=18=9\cdot2$. Are they just a coincidence or is there a theory behind? Is it about modular arithmetic?

Another proof: Denote $r=b/a$. We know that the sum of a geometric progression of the type $1+r+r^2+\ldots+r^{n-1}$ is equal to $\frac{1-r^n}{1-r}$. Thus, we have \begin{align} 1-r^n&=(1-r)(1+r+r^2+\ldots+r^{n-1}),\quad\text{substituting $r=b/a$ gives:}\\ a^n-b^n &= (a-b)\color{red}{(a^{n-1}+a^{n-2}b+\ldots+b^{n-1})}\\ a^n-b^n &= (a-b)N \end{align} The last step follows since $a,b$ are integers and a polynomial expression of the type in $\color{red}{red}$ font is also an integer.

Proving inequality $\frac{a}{b}+\frac{b}{c}+\frac{c}{a}+\frac{3\sqrt[3]{abc}}{a+b+c} \geq 4$ I started to study inequalities - I try to solve a lot of inequlites and read interesting .solutions . I have a good pdf, you can view from here . The inequality which I tried to solve and I didn't manage to find a solution can be found in that pdf but I will write here to be more explicitly. Exercise 1.3.4(a) Let $a,b,c$ be positive real numbers. Prove that $$\displaystyle \frac{a}{b}+\frac{b}{c}+\frac{c}{a}+\frac{3\sqrt[3]{abc}}{a+b+c} \geq 4.$$ (b) For real numbers $a,b,c \gt0$ and $n \leq3$ prove that: $$\displaystyle \frac{a}{b}+\frac{b}{c}+\frac{c}{a}+n\left(\frac{3\sqrt[3]{abc}}{a+b+c} \right)\geq 3+n.$$

Write $$\frac ab+\frac ab+\frac bc\geq \frac{3a}{\sqrt[3]{abc}}$$ by AM-GM. You get $$\operatorname{LHS} \geq \frac{a+b+c}{\sqrt[3]{abc}}+n\left(\frac{\sqrt[3]{abc}}{a+b+c}\right).$$ Set $$z:=\frac{a+b+c}{\sqrt[3]{abc}}$$ and then notice that for $n\leq 3$, $$z+\frac{3n}{z}\geq 3+n.$$ Indeed the minimum is reached for $z=\sqrt{3n}\leq 3$; since $z\geq 3$, the minimum is reached in fact for $z=3$.

solving for a coefficent term of factored polynomial. Given: the coefficent of $x^2$ in the expansion of $(1+2x+ax^2)(2-x)^6$ is $48,$ find the value of the constant $a.$ I expanded it and got $64-64\,x-144\,{x}^{2}+320\,{x}^{3}-260\,{x}^{4}+108\,{x}^{5}-23\,{x}^{ 6}+2\,{x}^{7}+64\,a{x}^{2}-192\,a{x}^{3}+240\,a{x}^{4}-160\,a{x}^{5}+ 60\,a{x}^{6}-12\,a{x}^{7}+a{x}^{8} $ because of the given info $48x^2=64x^2-144x^2$ solve for $a,$ $a=3$. Correct? P.S. is there an easier method other than expanding the terms? I have tried using the bionomal expansion; however, one needs still to multiply the terms. Expand $(2-x)^6$ which is not very fast.

It would be much easier to just compute the coefficient at $x^2$ in the expansion of $(1+2x+ax^2)(2-x)^6$. You can begin by computing: $$ (2-x)^6 = 64 - 6 \cdot 2^5 x + 15 \cdot 2^4 x^2 + x^3 \cdot (...) = 64 - 192 x + 240 x^2 + x^3 \cdot (...) $$ Now, multiply this by $(1+2x+ax^2)$. Again, you're only interested in the term at $x^2$, so you can spare yourself much effort by just computing this coefficient: to get $x^2$ in the product, you need to take $64, \ -192 x, \ 240 x^2$ from the first polynomial, and $ax^2,\ 2x, 1$ from the second one (respectively). $$(1+2x+ax^2)(2-x)^6 = (...)\cdot 1 + (...) \cdot x + (64a - 2\cdot 192 + 240 )\cdot x^2 + x^3 \cdot(...) $$ Now, you get the equation: $$ 64a - 2\cdot 192 + 240 = 48 $$ whose solution is indeed $a = 3$. As an afterthought: there is another solution, although it might be an overkill. Use that the term at $x^2$ in polynomial $p$ is $p''(0)/2$. Your polynomial is: $$ p(x) = (1+2x+ax^2)(2-x)^6$$ so you can compute easily enough: $$ p'(x) = (2+2ax)(2-x)^6 + 6(1+2x+ax^2)(2-x)^5 $$ and then: $$ p''(x) = 2a(2-x)^6 + 2 \cdot 6(2+2ax)(2-x)^5 + 30(1+2x+ax^2)(2-x)^4 $$ You can now plug in $x=0$: $$ p''(0) = 2a \cdot 2^6 + 2 \cdot 6 \cdot 2 \cdot 2^5 + 30 \cdot 2^4 $$ On the other hand, you have $$p''(0) = 2 \cdot 48$$ These two formulas for $p''(0)$ let you write down an equation for $a$.

Evaluate $\int\frac{dx}{\sin(x+a)\sin(x+b)}$ Please help me evaluate: $$ \int\frac{dx}{\sin(x+a)\sin(x+b)} $$

The given integral is: $$\int\frac{dx}{\sin(x+a)\sin(x+b)}$$ The given integral can write: $$\int\frac{dx}{\sin(x+a)\sin(x+b)}=\int\frac{\sin(x+a)}{\sin(x+b)}\cdot\frac{dx}{\sin^2(x+a)}$$ We substition $$\frac{\sin(x+a)}{\sin(x+b)}=t$$ By the substition of the above have: $$\frac{dx}{\sin^2(x+a)}=\frac{dt}{\sin(a-b)}$$ Now have: $$\int\frac{dx}{\sin(x+a)\sin(x+b)}=\int\frac{\sin(x+a)}{\sin(x+b)}\cdot\frac{dx}{\sin^2(x+a)}=\frac{1}{\sin(a-b)}\int\frac{dt}{t}=\frac{1}{\sin(a-b)}\ln |t|=\frac{1}{\sin(a-b)}\ln|\frac{\sin(x+a)}{\sin(x+b)}|+C$$

Representations of integers by a binary quadratic form Let $\mathfrak{F}$ be the set of binary quadratic forms over $\mathbb{Z}$. Let $f(x, y) = ax^2 + bxy + cy^2 \in \mathfrak{F}$. Let $\alpha = \left( \begin{array}{ccc} p & q \\ r & s \end{array} \right)$ be an element of $SL_2(\mathbb{Z})$. We write $f^\alpha(x, y) = f(px + qy, rx + sy)$. Since $(f^\alpha)^\beta$ = $f^{\alpha\beta}$, $SL_2(\mathbb{Z})$ acts on $\mathfrak{F}$ from right. Let $f, g \in \mathfrak{F}$. If $f$ and $g$ belong to the same $SL_2(\mathbb{Z})$-orbit, we say $f$ and $g$ are equivalent. Let $f = ax^2 + bxy + cy^2 \in \mathfrak{F}$. We say $D = b^2 - 4ac$ is the discriminant of $f$. Let $m$ be an integer. If $m = ax^2 + bxy + cy^2$ has a solution in $\mathbb{Z}^2$, we say $m$ is represented by $ax^2 + bxy + cy^2$. If $m = ax^2 + bxy + cy^2$ has a solution $(s, t)$ such that gcd$(s, t) = 1$, we say $m$ is properly represented by $ax^2 + bxy + cy^2$. Is the following proposition true? If yes, how do we prove it? Proposition Let $ax^2 + bxy + cy^2 \in \mathfrak{F}$. Suppose its discriminant is not a square. Let $m$ be an integer. Then $m$ is properly represented by $ax^2 + bxy + cy^2$ if and only if there exist integers $l, k$ such that $ax^2 + bxy + cy^2$ and $mx^2 + lxy + ky^2$ are equivalent.

Lemma 1 Let $f = ax^2 + bxy + cy^2 \in \mathfrak{F}$. Let $\alpha = \left( \begin{array}{ccc} p & q \\ r & s \end{array} \right)$ be an element of $SL_2(\mathbb{Z})$. Then $f^\alpha(x, y) = f(px + qy, rx + sy) = kx^2 + lxy + my^2$, where $k = ap^2 + bpr + cr^2$ $l = 2apq + b(ps + qr) + 2crs$ $m = aq^2 + bqs + cs^2$. Proof: Clear. Proof of the proposition Let $f(x, y) = ax^2 + bxy + cy^2$. Suppose $m$ is properly represented by $f(x, y)$. There exist integers $p, r$ such that gcd$(p, r) = 1$ and $m = f(p, r)$. Since gcd$(p, r) = 1$, there exist integers $s, r$ such that $ps - rq = 1$. By Lemma 1, $f(px + qy, rx + sy) = mx^2 + lxy + ky^2$, where $m = ap^2 + bpr + cr^2$ $l = 2apq + b(ps + qr) + 2crs$ $k = aq^2 + bqs + cs^2$. Hence, $ax^2 + bxy + cy^2$ and $mx^2 + lxy + ky^2$ are equivalent. Conversely suppose $ax^2 + bxy + cy^2$ and $mx^2 + lxy + ky^2$ are equivalent. There exists integer $p, q, r, s$ such that $ps - rq = 1$ and $f(px + qy, rx + sy) = mx^2 + lxy + ky^2$. Letting $x = 1, y = 0$, we get $f(p, r) = m$. Since $ps - rq = 1$, gcd$(p, r) = 1$. Hence $m$ is properly represented by $ax^2 + bxy + cy^2$.

Solve $\sqrt{x-4} + 10 = \sqrt{x+4}$ Solve: $$\sqrt{x-4} + 10 = \sqrt{x+4}$$ Little help here? >.<

Square both sides, and you get $$x - 4 + 20\sqrt{x - 4} + 100 = x + 4$$ This simplifies to $$20\sqrt{x - 4} = -92$$ or just $$\sqrt{x - 4} = -\frac{92}{20}$$ Since square roots of numbers are always nonnegative, this cannot have a solution.

Sum of the sequence What is the sum of the following sequence $$\begin{align*} (2^1 - 1) &+ \Big((2^1 - 1) + (2^2 - 1)\Big)\\ &+ \Big((2^1 - 1) + (2^2 - 1) + (2^3 - 1) \Big)+\ldots\\ &+\Big( (2^1 - 1)+(2^2 - 1)+(2^3 - 1)+\ldots+(2^n - 1)\Big) \end{align*}$$ I tried to solve this. I reduced the equation into the following equation $$n(2^1) + (n-1)\cdot2^2 + (n-2)\cdot2^3 +\ldots$$ but im not able to solve it further. Can any one help me solve this equation out. and btw its not a Home work problem. This equation is derived from some puzzle. Thanks in advance

Let's note that $$(2^1 - 1) + (2^2 - 1) + \cdots + (2^k - 1) = 2^{k+1} - 2-k$$ where we have used the geometric series. Thus, the desired sum is actually $$\sum_{k=1}^n{2^{k+1}-2-k}$$. As this is a finite sum, we can evaluate each of the terms separately. We get the sum is $$2\left(\frac{2^{n+1}-1}{2-1}-1\right) - 2n- \frac{n(n+1)}{2} = 2^{n+2}-4 - 2n-\frac{n(n+1)}{2} $$

Prove that $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\geq \frac{9}{a+b+c} : (a, b, c) > 0$ Please help me for prove this inequality: $$\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\geq \frac{9}{a+b+c} : (a, b, c) > 0$$

Inequality can be written as: $$\left(a+b+c\right) \cdot \left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right) \geq 9 .$$ And now we apply the $AM-GM$ inequality for both of parenthesis. So: $\displaystyle \frac{a+b+c}{3} \geq \sqrt[3]{abc} \tag{1}$ and $\displaystyle \frac{\frac{1}{a}+\frac{1}{b}+\frac{1}{c}}{3} \geq \frac{1}{\sqrt[3]{abc}} \tag{2}.$ Now multiplying relation $(1)$ with relation $(2)$ we obtained that : $$\left(\frac{a+b+c}{3}\right) \cdot \left(\frac{\frac{1}{a}+\frac{1}{b}+\frac{1}{c}}{3}\right) \geq \frac{\sqrt[3]{abc}}{\sqrt[3]{abc}}=1. $$ So, we obtained our inequality.

The integral relation between Perimeter of ellipse and Quarter of Perimeter Ellipse Equation $$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$$ $x=a\cos t$ ,$y=b\sin t$ $$L(\alpha)=\int_0^{\alpha}\sqrt{\left(\frac{dx}{dt}\right)^2+\left(\frac{dy}{dt}\right)^2}\,dt$$ $$L(\alpha)=\int_0^\alpha\sqrt{a^2\sin^2 t+b^2 \cos^2 t}\,dt $$ $$L(2\pi)=\int_0^{2\pi}\sqrt{b^2+(a^2-b^2)\sin^2 t}\,dt \tag{Perimeter of ellipse}$$ $$L(\pi/2)=\int_0^{\pi/2}\sqrt{b^2+(a^2-b^2)\sin^2 t}\,dt \tag {Quarter of Perimeter }$$ Geometrically, we can write $L(2\pi)=4L(\pi/2)$ $$4\int_0^{\pi/2}\sqrt{b^2+(a^2-b^2)\sin^2 t}\,dt=\int_0^{2\pi}\sqrt{b^2+(a^2-b^2)\sin^2 t}\,dt \tag1$$ If I change variable in integral of $L(2\pi)$ $$L(2\pi)=\int_0^{2\pi}\sqrt{b^2+(a^2-b^2)\sin^2 t}\,dt \tag{Perimeter of ellipse}$$ $t=4u$ $$L(2\pi)=4\int_0^{\pi/2}\sqrt{b^2+(a^2-b^2)\sin^2 4u}\,du$$ According to result (1), $$L(2\pi)=4\int_0^{\pi/2}\sqrt{b^2+(a^2-b^2)\sin^2 4u},du=4\int_0^{\pi/2}\sqrt{b^2+(a^2-b^2)\sin^2 t}\,dt$$ $$\int_0^{\pi/2}\sqrt{b^2+(a^2-b^2)\sin^2 4u}\,du=\int_0^{\pi/2}\sqrt{b^2+(a^2-b^2)\sin^2 t}\,dt \tag2$$ How to prove the relation $(2)$ analytically? Thanks a lot for answers

I proved the relation via using analytic way. I would like to share the solution with you. $$\int_0^{\pi/2}\sqrt{b^2+(a^2-b^2)\sin^2 4u}\,du=K$$ $u=\pi/4-z$ $$K=\int_{-\pi/4}^{\pi/4}\sqrt{b^2+(a^2-b^2)\sin^2 (\pi-4z)}\,dz$$ $\sin (\pi-4z)=\sin \pi \cos 4z-\cos \pi \sin 4z= \sin 4z$ $$\int_{-\pi/4}^{\pi/4}\sqrt{b^2+(a^2-b^2)\sin^2 (\pi-4z)}\,dz=\int_{-\pi/4}^{\pi/4}\sqrt{b^2+(a^2-b^2)\sin^2 (4z)}\,dz$$ $$\int_{-\pi/4}^{\pi/4}\sqrt{b^2+(a^2-b^2)\sin^2 (4z)}\,dz=\int_{-\pi/4}^{0}\sqrt{b^2+(a^2-b^2)\sin^2 (4z)}\,dz+\int_{0}^{\pi/4}\sqrt{b^2+(a^2-b^2)\sin^2 (4z)}\,dz$$ $$\int_{-\pi/4}^{0}\sqrt{b^2+(a^2-b^2)\sin^2 (4z)}\,dz$$ $z=-p$ $$\int_{\pi/4}^{0}\sqrt{b^2+(a^2-b^2)\sin^2 (-4p)}\,(-dp)=\int_{0}^{\pi/4}\sqrt{b^2+(a^2-b^2)\sin^2 (4p)}\,dp$$ $$K=\int_{-\pi/4}^{\pi/4}\sqrt{b^2+(a^2-b^2)\sin^2 (4z)}\,dz=2\int_{0}^{\pi/4}\sqrt{b^2+(a^2-b^2)\sin^2 (4z)}\,dz$$ $$K=2\int_{0}^{\pi/4}\sqrt{b^2+(a^2-b^2)\sin^2 (4z)}\,dz$$ $z=\pi/8-v$ $$K=2\int_{-\pi/8}^{\pi/8}\sqrt{b^2+(a^2-b^2)\cos^2 (4v)}\,dv$$ $$K=2\int_{-\pi/8}^{0}\sqrt{b^2+(a^2-b^2)\cos^2 (4v)}\,dv+2\int_{0}^{\pi/8}\sqrt{b^2+(a^2-b^2)\cos^2 (4v)}\,dv$$ $$\int_{-\pi/8}^{0}\sqrt{b^2+(a^2-b^2)\cos^2 (4v)}\,dv$$ $v=-h$ $$\int_{\pi/8}^{0}\sqrt{b^2+(a^2-b^2)\cos^2 (-4h)}\,(-dh)=\int_{0}^{\pi/8}\sqrt{b^2+(a^2-b^2)\cos^2 (4h)}\,dh$$ $$K=2\int_{-\pi/8}^{0}\sqrt{b^2+(a^2-b^2)\cos^2 (4v)}\,dv+2\int_{0}^{\pi/8}\sqrt{b^2+(a^2-b^2)\cos^2 (4v)}\,dv$$ $$K=4\int_{0}^{\pi/8}\sqrt{b^2+(a^2-b^2)\cos^2 (4v)}\,dv$$ $v=\pi/8-t/4$ $$K=4\int_{\pi/2}^{0}\sqrt{b^2+(a^2-b^2)\cos^2 (4(\pi/8-t/4))}\,(-dt/4)$$ $$K=\int_{0}^{\pi/2}\sqrt{b^2+(a^2-b^2)\cos^2 (\pi/2-t)}\,dt$$ $$K=\int_{0}^{\pi/2}\sqrt{b^2+(a^2-b^2)\sin^2 (t)}\,dt$$

Inequality. $\sum{(a+b)(b+c)\sqrt{a-b+c}} \geq 4(a+b+c)\sqrt{(-a+b+c)(a-b+c)(a+b-c)}.$ Let $a,b,c$ be the side-lengths of a triangle. Prove that: I. $$\sum_{cyc}{(a+b)(b+c)\sqrt{a-b+c}} \geq 4(a+b+c)\sqrt{(-a+b+c)(a-b+c)(a+b-c)}.$$ What I have tried: \begin{eqnarray} a-b+c&=&x\\ b-c+a&=&y\\ c-a+b&=&z. \end{eqnarray} So $a+b+c=x+y+z$ and $2a=x+y$, $2b=y+z$, $2c=x+z$ and our inequality become: $$\sum_{cyc}{\frac{\sqrt{x}\cdot(x+2y+z)\cdot(x+y+2)}{4}} \geq 4\cdot(x+y+z)\cdot\sqrt{xyz}. $$ Or if we make one more notation : $S=x+y+z$ we obtain that: $$\sum_{cyc}{\sqrt{x}(S+y)(S+z)} \geq 16S\cdot \sqrt{xyz} \Leftrightarrow$$ $$S^2(\sqrt{x}+\sqrt{y}+\sqrt{z})+S(y\sqrt{x}+z\sqrt{x}+x\sqrt{y}+z\sqrt{y}+x\sqrt{z}+y\sqrt{z})+xy\sqrt{z}+yz\sqrt{x}+xz\sqrt{y} \geq 16S\sqrt{xyz}.$$ To complete the proof we have to prove that: $$y\sqrt{x}+z\sqrt{x}+x\sqrt{y}+z\sqrt{y}+x\sqrt{z}+y\sqrt{z} \geq 16\sqrt{xyz}. $$ Is this last inequality true ? II. Knowing that: $$p=\frac{a+b+c}{2}$$ we can rewrite the inequality: $$\sum_{cyc}{(2p-c)(2p-a)\sqrt{2(p-b)}} \geq 8p \sqrt{2^3 \cdot (p-a)(p-b)(p-c)} \Leftrightarrow$$ $$\sum_{cyc}{(2p-c)(2p-a)\sqrt{(p-b)}} \geq 16p \sqrt{(p-a)(p-b)(p-c)}$$ This help me ? Thanks :)

Notice that the inequality proposed is proved once we estabilish $$\frac{(a+b)(b+c)}{\sqrt{a+b-c}\sqrt{-a+b+c}}+\frac{(b+c)(c+a)}{\sqrt{a-b+c}\sqrt{-a+b+c}}+\frac{(c+a)(a+b)}{\sqrt{a-b+c}\sqrt{a+b-c}}\geq 4(a+b+c).$$ Using AM-GM on the denominators in the LHS, we estabilish that * *$\sqrt{a+b-c}\sqrt{-a+b+c}\leq b$, *$\sqrt{a-b+c}\sqrt{-a+b+c}\leq c$, *$\sqrt{a-b+c}\sqrt{a+b-c}\leq a$. Therefore $$\operatorname{LHS}\geq \frac{(a+b)(b+c)}{b}+\frac{(b+c)(c+a)}{c}+\frac{(c+a)(a+b)}{a}=3(a+b+c)+\frac{ca}{b}+\frac{ab}{c}+\frac{bc}{a}$$ To finish with the proof it suffices then to prove $$\frac{ca}{b}+\frac{ab}{c}+\frac{bc}{a}\geq a+b+c.$$ This follows from AM-GM, indeed * *$$\frac{\frac{ac}{b}+\frac{ab}{c}}{2}\geq a,$$ *$$\frac{\frac{bc}{a}+\frac{ab}{c}}{2}\geq b,$$ *$$\frac{\frac{ac}{b}+\frac{bc}{a}}{2}\geq c;$$ summing up these last inequalities, we get the desired result, hence finishing the proof. Notice that the condition of $a,b,c$ being the sides of a triangle is essential in the first usage of AM-GM.

Why does $\tan^{-1}(1)+\tan^{-1}(2)+\tan^{-1}(3)=\pi$? Playing around on wolframalpha shows $\tan^{-1}(1)+\tan^{-1}(2)+\tan^{-1}(3)=\pi$. I know $\tan^{-1}(1)=\pi/4$, but how could you compute that $\tan^{-1}(2)+\tan^{-1}(3)=\frac{3}{4}\pi$ to get this result?

Consider, $z_1= \frac{1+2i}{\sqrt{5}}$, $z_2= \frac{1+3i}{\sqrt{10} }$, and $z_3= \frac{1+i}{\sqrt{2} }$, then: $$ z_1 z_2 z_3 = \frac{1}{10} (1+2i)(1+3i)(1+i)=-1 $$ Take arg of both sides and use property that $\arg(z_1 z_2 z_3) = \arg(z_1) + \arg(z_2) + \arg(z_3)$: $$ \arg(z_1) + \arg(z_2) + \arg(z_3) = -1$$ The LHS we can write as: $$ \tan^{-1} ( \frac{2}{1}) +\tan^{-1} ( \frac{3}{1} ) + \tan^{-1} (1) = \pi$$ Tl;dr: Complex number multiplication corresponds to tangent angle addition

Question: Find all values of real number a such that $ \lim_{x\to1}\frac{ax^2+a^2x-2}{x^3-3x+2} $ exists. Thanks in advance for looking at my question. I was tackling this limits problem using this method, but I can't seem to find any error with my work. Question: Find all values of real number a such that $$ \lim_{x\to1}\frac{ax^2+a^2x-2}{x^3-3x+2} $$ exists. My Solution: Suppose $\lim_{x\to1}\frac{ax^2+a^2x-2}{x^3-3x+2}$ exists and is equals to $L$. We have $$\lim_{x\to1}{ax^2+a^2x-2}=\frac{\lim_{x\to1}ax^2+a^2x-2}{\lim_{x\to1}x^3-3x+2}*\lim_{x\to1}x^3-3x+2=L*0=0$$ Therefore, $\lim_{x\to1}{ax^2+a^2x-2}=0$ implying $a(1)^2+a^2(1)-2=0$. Solving for $a$, we get $a=-2$ or $a=1$. Apparently, the answer is only $a=-2$. I understand where they are coming from, but I can't see anything wrong with my solution either.

Since the denominator's limit is 0, the numerator cannot have a nonzero limit if the limit of the quotient is to be defined. The only hope is that the numerator's limit is also 0, and that after analyzing the indeterminate form, it does have a limit. So, it must be the case that $\lim_{x\to1} ax^2+a^2x-2=0$, and consequently $a^2+a-2=0$. The solutions to that are $a=-2$ and $a=1$, and if you substitute them into the expression, you will find that the numerator now factors into $-2(x-1)^2$ in the first case, and $(x+2)(x-1)$ in the second case. In either one, the $(x-1)$ can be cancelled with the $(x-1)$ factor in the denominator, so that the singularity (a pole of order 2) might disappear. $\lim_{x\to1}\frac{x^2+x-2}{x^3-3x+2}=\lim_{x\to1}\frac{x^2+x-2}{(x-1)(x^2+x-2)}=\lim_{x\to 1}\frac{1}{x-1}$ does not exist, so the $a=1 $ case is a false positive. In the other case: $\lim_{x\to1}\frac{-2(x-1)^2}{(x-1)^2(x+2)}=\lim_{x\to1}\frac{-2}{(x+2)}=\frac{-2}{3}$

Is it possible to take the absolute value of both sides of an equation? I have a problem that says: Suppose $3x^2+bx+7 > 0$ for every number $x$, Show that $|b|<2\sqrt21$. Since the quadratic is greater than 0, I assume that there are no real solutions since $y = 3x^2+bx+7$, and $3x^2+bx+7 > 0$, $y > 0$ since $y>0$ there are no x-intercepts. I would use the discriminant $b^2-4ac<0$. I now have $b^2-4(3)(7)<0$ $b^2-84<0$ $b^2<84$ $b<\pm\sqrt{84}$ Now how do I change $b$ to $|b|$? Can I take the absolute value of both sides of the equation or is there a proper way to do this?

What you've written is an inequality, not an equation. If you have an equation, say $a=b$, you can conclude that $|a|=|b|$. But notice that $3>-5$, although $|3|\not>|-5|$. If $3x^2+bx+7>0$ for every value of $x$, then the quadratic equation $3x^2+bx+7=0$ has no solutions that are real numbers. THat implies that the discriminant $b^2-4ac=b^2-4\cdot3\cdot7$ is negative. If $b^2-84<0$ then $b^2<84$, so $|b|<\sqrt{84}$. Now observe that $\sqrt{84}=\sqrt{4}\sqrt{21}=2\sqrt{21}$.

Stuck with solving a polynomial I am doing a problem for homework that says: Suppose $s(x)=3x^3-2$. Write the expression $\frac{s(2+x)-s(2)}{x}$ as a sum of terms, each of which is a constant times power of $x$. I was able to do the following work for this problem: $\frac{3(2+x)^3-3(2)^3-2}{x}$ $\frac{3(x^3+6x^2+12x+8)-24-2}{x}$ $\frac {3x^3+18x^2+36x+24-24-2}{x}$ $\frac {3x^3+18x^2+36x-2}{x}$ This is where I got stuck. I am not sure what I am supposed to do next. The multiple choice answers are: a) $2x^2-36x+18$ b) $3x^2+18x+36$ c) $18x^2+18x+36$ d) $x^3+18x^2+36x$ e) $-3x^2-18x-36$ The closest answer to the answer I got was d, does anyone know how I would solve this?

You went astray at the first step, when you got $$\frac{3(2+x)^3-3(2)^3-2}{x}\;;$$ in fact $$s(2+x)-s(2)=\Big(3(2+x)^3-2\Big)-\Big(3\cdot2^3-2\Big)=3(2+x)^3-3\cdot2^3\;.$$ Can you straighten the rest out from there? When you do it correctly, there will be no constant term in the numerator.

Why does this equation have different number of answers? I have a simple equation: $$\frac{x}{x-3} - \frac{2}{x-1} = \frac{4}{x^2-4x+3}$$ By looking at it, one can easily see that $x \not= 1$ because that would cause $\frac{2}{x-1} $ to become $\frac{2}{0}$, which is illegal. However, if you do some magic with it. First I factorized the last denominator to be able to simplify this: $$\frac{-b\pm\sqrt{b^2-4ac}}{2a}$$ $$\frac{-(-4)\pm\sqrt{(-4)^2-4\times1\times3}}{2 \times 1}$$ $$x=1 \vee x=3$$ Then we can multiply everything with the common factor, which is $(x-1)(x-3)$ and get: $$x(x-1) - 2(x-3) - 4 = 0$$ If we multiply out these brackets, we get: $$x^2-x-2x+6-4=0$$ $$x^2-3x+2=0$$ The quadratic formula gives $x = 1 \vee x=2$. We already know that $x$ CANNOT equal to 1, but we still get it as an answer. Have I done anything wrong here, because as I see it, this is the same as saying that: $$\frac{x}{x-3} - \frac{2}{x-1} = \frac{4}{x^2-4x+3}$$ $$=$$ $$x(x-1) - 2(x-3) - 4 = 0$$ which cannot be true, because the two doesn't have the same answers. What am I missing here?

If $\dfrac AB = 0$ then $A=0\cdot B$. But you can't say that if $A=0\cdot B$ then $\dfrac AB=0$ unless you know that $B\ne 0$. So if $A$ and $B$ are complicated expressions that can be solved for $x$, there may be values of $x$ that make $B$ equal to $0$, and if they also make $A$ equal to $0$, then they are solutions of the equation $A=0\cdot B$, but not of the equation $\dfrac AB=0$. "If P then Q" is not the same as "If Q then P". Another way of putting it is that this explains why "clearing fractions" is one of the operations that can introduce "extraneous roots". Perhaps more well known is that squaring both sides of an equation can do that.

For integers $a$ and $b \gt 0$, and $n^2$ a sum of two square integers, does this strategy find the largest integer $x | x^2 \lt n^2(a^2 + b^2)$? Here is some background information on the problem I am trying to solve. I start with the following equation: $n^2(a^2 + b^2) = x^2 + y^2$, where $n, a, b, x, y \in \mathbb Z$, and $a \ge b \gt 0$, $n \gt 0$, and $x \ge y \gt 0$. For given values of $a$ and $b$ and some $n$, I need to find $x$ and $y$ such that $x$ is as large as possible (and $y$ as small as possible). The value of $n$ is up to me as long as allowable values are linear ($n = kn_0$). A naive approach is to use the distributive property to set $x = na$ and $y = nb$, but this causes $x$ and $y$ to grow at the same rate and doesn't guarantee $x$ is as large as possible. I realized that if $n^2$ is the sum of two square integers (the length of the hypotenuse of a Pythagorean triangle), then I can use the Brahmagupta–Fibonacci identity to find values for $x$ and $y$ that are as good or better than using the distributive property: $(a^2 + b^2)(c^2 + d^2) = (ac + bd)^2 + (ad - bc)^2$ $n^2 = c^2 + d^2$, where $c \gt d$ So for example, if $a = 2$, $b = 1$, and $n = 5$, then instead of $5^2(2^2 + 1^2) = 10^2 + 5^2 = 125$, so that $x = 10$ and $y = 5$ we get $(4^2 + 3^2)(2^2 + 1^2) = (4\cdot2 +3\cdot1)^2 + (4\cdot1 - 3\cdot2)^2 = 11^2 + (-2)^2 = 125$, so $x = 11$ and $y=2$ My question is, does this strategy always find $x$ to be the largest possible integer whose square is less than $n^2(a^2 + b^2)$, for $n$ a sum of two square integers?

Let $N = n^2(a^2+b^2)$ and consider its prime factorization. Let $p$ be a prime divisor of $N$. If $p\equiv 3\pmod4$, then necessarily $p^2|N$, $p|x$ and $p|y$ (and the factor comes from $p|n$). If $p\equiv 1\pmod 4$, find $u,v$ such that $u^2+v^2=p$. Let $k$ be the exponet of $p$ in $N$, i.e. $p^k||N$. Then $x+iy$ must be a multiple of $(u+iv)^r(u-iv)^{k-r}$ with $0\le r\le k$. This gives you $k+1$ choices for each uch $p$, and in fact the optimal choice may be influenced by the choices for other primes. (The case $p=2$ is left as an exercise). The number of candidates to test is then the product of the $k+1$, which may become quite large, but of course much, much smaller than $N$.

Prove by induction $\sum_{i=1}^ni^3=\frac{n^2(n+1)^2}{4}$ for $n\ge1$ Prove the following statement $S(n)$ for $n\ge1$: $$\sum_{i=1}^ni^3=\frac{n^2(n+1)^2}{4}$$ To prove the basis, I substitute $1$ for $n$ in $S(n)$: $$\sum_{i=1}^11^3=1=\frac{1^2(2)^2}{4}$$ Great. For the inductive step, I assume $S(n)$ to be true and prove $S(n+1)$: $$\sum_{i=1}^{n+1}i^3=\frac{(n+1)^2(n+2)^2}{4}$$ Considering the sum on the left side: $$\sum_{i=1}^{n+1}i^3=\sum_{i=1}^ni^3+(n+1)^3$$ I make use of $S(n)$ by substituting its right side for $\sum_{i=1}^ni^3$: $$\sum_{i=1}^{n+1}i^3=\frac{n^2(n+1)^2}{4}+(n+1)^3$$ This is where I get a little lost. I think I expand the equation to be $$=\frac{(n^4+2n^3+n^2)}{4}+(n+1)^3$$ but I'm not totally confident about that. Can anyone provide some guidance?

Here's yet another way: You have $\frac{n^2(n+1)^2}{4}+(n+1)^3$ and you want $\frac{(n+1)^2(n+2)^2}{4}$ So manipulate it to get there; $\frac{n^2(n+1)^2}{4}+(n+1)^3 =$ $\frac{(n^2 + 4n + 4)(n+1)^2 - (4n + 4)(n+1)^2}{4}+(n+1)^3 =$ $\frac{(n+2)^2(n+1)^2}{4}- \frac{ (4n + 4)(n+1)^2}{4}+(n+1)^3 =$ $\frac{(n+2)^2(n+1)^2}{4}-(n + 1)(n+1)^2+(n+1)^3 =$ $\frac{(n+2)^2(n+1)^2}{4}-(n+1)^3+(n+1)^3 =$ $\frac{(n+2)^2(n+1)^2}{4}$

Solving trigonometric equations of the form $a\sin x + b\cos x = c$ Suppose that there is a trigonometric equation of the form $a\sin x + b\cos x = c$, where $a,b,c$ are real and $0 < x < 2\pi$. An example equation would go the following: $\sqrt{3}\sin x + \cos x = 2$ where $0<x<2\pi$. How do you solve this equation without using the method that moves $b\cos x$ to the right side and squaring left and right sides of the equation? And how does solving $\sqrt{3}\sin x + \cos x = 2$ equal to solving $\sin (x+ \frac{\pi}{6}) = 1$

Riffing on @Yves' "little known" solutions ... The above trigonograph shows a scenario with $a^2 + b^2 = c^2 + d^2$, for $d \geq 0$, and we see that $$\theta = \operatorname{atan}\frac{a}{b} + \operatorname{atan}\frac{d}{c} \tag{1}$$ (If the "$a$" triangle were taller than the "$b$" triangle, the "$+$" would become "$-$". Effectively, we can take $d$ to be negative to get the "other" solution.) Observe that both $c$ and $d$ are expressible in terms of $a$, $b$, $\theta$: $$\begin{align} a \sin\theta + b \cos\theta &= c \\ b \sin\theta - a\cos\theta &= d \quad\text{(could be negative)} \end{align}$$ Solving that system for $\sin\theta$ and $\cos\theta$ gives $$\left.\begin{align} \sin\theta &= \frac{ac+bd}{a^2+b^2} \\[6pt] \cos\theta &= \frac{bc-ad}{a^2+b^2} \end{align}\quad\right\rbrace\quad\to\quad \tan\theta = \frac{ac+bd}{bc-ad} \tag{2}$$ We can arrive at $(2)$ in a slightly-more-geometric manner by noting $$c d = (a\sin\theta + b \cos\theta)d = c( b\sin\theta - a \cos\theta ) \;\to\; ( b c - a d)\sin\theta = \left( a c + b d \right)\cos\theta \;\to\; (2) $$ where each term in the expanded form of the first equation can be viewed as the area of a rectangular region in the trigonograph. (For instance, $b c \sin\theta$ is the area of the entire figure.)

The intersection of a line with a circle Get the intersections of the line $y=x+2$ with the circle $x^2+y^2=10$ What I did: $y^2=10-x^2$ $y=\sqrt{10-x^2}$ or $y=-\sqrt{10-x^2}$ $ x+ 2 = y=\sqrt{10-x^2}$ If you continue, $x=-3$ or $x=1$ , so you get 2 points $(1,3)$, $(-3,-1)$ But then, and here is where the problems come: $x+2=-\sqrt{10-x^2}$ I then, after a while get the point $(-3\dfrac{1}{2}, -1\dfrac{1}{2}$) but this doesn't seem to be correct. What have I done wrong at the end?

Let the intersection be $(a,b)$, so it must satisfy both the given eqaution. So, $a=b+2$ also $a^2+b^2=10$ Putting $b=a+2$ in the given circle $a^2+(a+2)^2=10$ $2a^2+4a+4=10\implies a=1$ or $-3$ If $a=1,b=a+2=3$ If $a=-3,b=-3+2=-1$ So, the intersections are $(-3,-1)$ and $(1,3)$

Find the limit without l'Hôpital's rule Find the limit $$\lim_{x\to 1}\frac{(x^2-1)\sin(3x-3)}{\cos(x^3-1)\tan^2(x^2-x)}.$$ I'm a little rusty with limits, can somebody please give me some pointers on how to solve this one? Also, l'Hôpital's rule isn't allowed in case you were thinking of using it. Thanks in advance.

$$\dfrac{\sin(3x-3)}{\tan^2(x^2-x)} = \dfrac{\sin(3x-3)}{3x-3} \times \left(\dfrac{x^2-x}{\tan(x^2-x)} \right)^2 \times \dfrac{3(x-1)}{x^2(x-1)^2}$$ Hence, $$\dfrac{(x^2-1)\sin(3x-3)}{\cos(x^3-1)\tan^2(x^2-x)} = \dfrac{\sin(3x-3)}{3x-3} \times \left(\dfrac{x^2-x}{\tan(x^2-x)} \right)^2 \times \dfrac{3(x-1)(x^2-1)}{x^2(x-1)^2 \cos(x^3-1)}\\ = \dfrac{\sin(3x-3)}{3x-3} \times \left(\dfrac{x^2-x}{\tan(x^2-x)} \right)^2 \times \dfrac{3(x+1)}{x^2 \cos(x^3-1)}$$ Now the first and second term on the right has limit $1$ as $x \to 1$. The last term limit can be obtained by plugging $x=1$, to give the limit as $$1 \times 1 \times \dfrac{3 \times (1+1)}{1^2 \times \cos(0)} = 6$$

About the homework of differentiation This problem is not solved. $$ \begin{align} f(x) &=\log\ \sqrt{\frac{1+\sqrt{2}x +x^2}{1-\sqrt{2}x +x^2}}+\tan^{-1}\left(\frac{\sqrt{2}x}{1-x^2}\right) \cr \frac{df}{dx}&=\mathord? \end{align} $$

The answer is $$\frac{2\sqrt{2}}{1+x^4}$$ Hints: $$\frac{d\left(\log(1+x^2\pm\sqrt{2}x\right)}{dx}=\frac{2x\pm\sqrt{2}}{1+x^2\pm\sqrt{2}x}$$ and $$\left(1+x^2+\sqrt{2}x\right)\left(1+x^2-\sqrt{2}x\right)=\left(1+x^2\right)^2-\left(\sqrt{2}x\right)^2$$ Similarly, $$\frac{d\left(\tan^{-1}\left(\frac{\sqrt{2}x}{1-x^2}\right)\right)}{dx}=\frac{1}{1+\left(\frac{\sqrt{2}x}{1-x^2}\right)^2}\frac{d\left(\frac{\sqrt{2}x}{1-x^2}\right)}{dx}$$ $$=\frac{1}{1+\left(\frac{\sqrt{2}x}{1-x^2}\right)^2}\frac{\sqrt{2}}{2}\left(\frac{1}{\left(1-x\right)^2}+\frac{1}{\left(1+x\right)^2}\right)$$ The rest is by calculations.

Given $f(x+1/x) = x^2 +1/x^2$, find $f(x)$ Given $f(x+1/x) = x^2 +1/x^2$, find $f(x)$. Please show me the way you find it. The answer in my textbook is $f(x)=\frac{1+x^2+x^4}{x\cdot \sqrt{1-x^2}}$

$$f\left(x+\frac{1}{x}\right)=x^2+\frac{1}{x^2} = \left(x+\frac{1}{x}\right)^2-2$$ Let $x+\frac{1}{x}=z$. Then we get, $$f(z)=z^2-2.$$ Hence we put x on the place of z. And we get $f(x)=x^2-2$.

Do these $\delta-\epsilon$ proofs work? I'm new to $\delta, \epsilon$ proofs and not sure if I've got the hand of them quite yet. $$ \lim_{x\to -2} (2x^2+5x+3)= 1 $$ $|2x^2 + 5x + 3 - 1| < \epsilon$ $|(2x + 1)(x + 2)| < \epsilon$ $|(2x + 4 - 3)(x + 2)| < \epsilon$ $|(2(x+2)^2 -3(x + 2)| \leq 2|x+2|^2 +3|x + 2| < \epsilon$ (via the triangle inequality) Let $|x+2| < 1$ Then $\delta=\min\left(\dfrac{\epsilon}{5}, 1\right)$ and $$\lim_{x\to -2} (3x^2+4x-2)= 2$$ $|3x^2 + 4x - 2 - 2| < \epsilon$ $|3x^2 - 12 + 4x + 8| < \epsilon$ $|3(x+2)(x-2) + 4(x + 2)| < \epsilon$ $|3(x+2)(x + 2 -4) + 4(x + 2)| < \epsilon$ $|3[(x+2)^2 -4(x+2)] + 4(x + 2)| < \epsilon$ $|3(x+2)^2 - 8(x+2)| \leq 3|x+2|^2 + 8|x+2| < \epsilon$ Let $|x + 2| < 1$ Then $\delta =\min\left(\dfrac{\epsilon}{11}, 1\right)$ and to make sure I'm understanding this properly, when we assert that $|x+2| < 1$, this means that $\delta \leq 1$ as well, because if $\delta \geq 1$, this would allow for $|x+2| \geq 1$, which violates that condition we just imposed? Edit: Apologies for the bad tex

$$ \lim_{x\to -2} (2x^2+5x+3)= 1 $$ Finding $\delta$: $|2x^2 + 5x + 3 - 1| < \epsilon$ $|(2x + 1)(x + 2)| < \epsilon$ $|x - (-2)| < \delta $, pick $\delta = 3$ $|x+2| < 3 \Rightarrow -5 < x < 1 \Rightarrow -9 < 2x + 1 < 3$ This implies $|2x + 1||x + 2| < 3 \cdot |x + 2| < \epsilon \Rightarrow |x+2| < \frac{\epsilon}{3} $ $\\[22pt]$ Actual Proof: Let $\epsilon > 0 $. Choose $\delta = min\{3,\frac{\epsilon}{3}\}$ and assume that $0 < |x + 2| < \delta \Rightarrow |2x + 1||x+2| < 3 \cdot |x + 2| < 3 \cdot \frac{\epsilon}{3} = \epsilon$

Find the volume of the solid under the plane $x + 2y - z = 0$ and above the region bounded by $y=x$ and $y = x^4$. Find the volume of the solid under the plane $x + 2y - z = 0$ and above the region bounded by $y = x$ and $y = x^4$. $$ \int_0^1\int_{x^4}^x{x+2ydydx}\\ \int_0^1{x^2-x^8dx}\\ \frac{1}{3}-\frac{1}{9} = \frac{2}{9} $$ Did I make a misstep? The answer book says I am incorrect.

I think it should be calculated as \begin{eqnarray*} V&=&\int_0^1\int_{x^4}^x\int_0^{x+2y}dzdydx\\ &=&\int_0^1\int_{x^4}^x(x+2y)dydx\\ &=&\int_0^1\left.\left(xy+y^2\right)\right|_{x^4}^xdx\\ &=&\int_0^1(2x^2-x^5-x^8)dx\\ &=&\left.(\frac{2}{3}x^3-\frac{1}{6}x^6-\frac{1}{9}x^9)\right|_0^1\\ &=&\frac{7}{18} \end{eqnarray*}

Find all Laurent series of the form... Find all Laurent series of the form $\sum_{-\infty} ^{\infty} a_n $ for the function $f(z)= \frac{z^2}{(1-z)^2(1+z)}$ There are a lot of problems similar to this. What are all the forms? I need to see this example to understand the idea.

Here is related problem. First, convert the $f(z)$ to the form $$f(z) = \frac{1}{4}\, \frac{1}{\left( 1+z \right)}+\frac{3}{4}\, \frac{1}{\left( -1+z \right) }+\frac{1}{2}\,\frac{1}{\left( -1+z \right)}$$ using partial fraction. Factoring out $z$ gives $$ f(z)= \frac{1}{4z}\frac{1}{(1+\frac{1}{z})}- \frac{3}{4z}\frac{1}{(1-\frac{1}{z})}-\frac{1}{2z^2}\frac{1}{(1-\frac{1}{z})^2}\,. $$ Now, using the series expansion of each term yields the Laurent series for $|z|>1$ $$ = \frac{1}{4z}(1-\frac{1}{z}+ \frac{1}{z^2}+\dots) -\frac{3}{4z}(1+\frac{1}{z}+ \frac{1}{z^2}+\dots)-\frac{1}{2z}\sum_{k=0}^{\infty}{-2\choose k}\frac{(-1)^k}{z^k} \,, $$ $$ \sum_{k=0}^{\infty}\left( \frac{(-1)^k}{4}-\frac{3}{4}-\frac{(-1)^k{-2\choose k}}{2} \right)\frac{1}{z^{k+1}}\,. $$ Note that, $$(1+x)^{-1}=1-x+x^2-\dots \,,$$ $$ (1-x)^{-1}=1+x+x^2+\dots \,,$$ $$ (1-x)^{m}=\sum_{k=0}^{\infty}{m\choose k}(-1)^kx^k $$

sines and cosines law Can we apply the sines and cosines law on the external angles of triangle ?

This answer assumes a triangle with angles $A, B, C$ with sides $a,b,c$. Law of sines states that $$\frac{\sin A}{a} = \frac{\sin B}{b} = \frac{\sin C}{c}$$ Knowing the external angle is $\pi - \gamma$ if the angle is $\gamma$, $\sin(\pi-\gamma) = \sin \gamma$ because in the unit circle, you are merely reflecting the point on the circle over the y-axis and the sine value represents the $y$ value of the point, so the $y$ value will remain the same. (Also, $\sin(\pi-\gamma) = \sin\pi \cos \gamma - \cos \pi \sin \gamma = \sin \gamma$), so $$\frac{\sin A}{a} = \frac{\sin B}{b} = \frac{\sin C}{c} = \frac{\sin (\pi -A)}{a} = \frac{\sin (\pi -B)}{b} = \frac{\sin(\pi - C)}{c}$$ The law of cosines state that $$c^2=a^2+b^2-2ab\cos\gamma \Longrightarrow \cos \gamma = \frac{a^2 + b^2 - c^2}{2ab}$$ But $\cos(\pi-\gamma) = -\cos\gamma$ for pretty much the same reasons I used for the sine above. So if you are using external angles, the law of cosines will not work. However, if $\pi - \gamma$ is the external angle, then $$c^2=a^2+b^2+2ab\cos(\pi-\gamma) \Longrightarrow \cos(\pi- \gamma) = -\frac{a^2 + b^2 - c^2}{2ab}$$. So in short, the answer is yes for law of sines, and no for law of cosines (unless you make the slight modification I made).

Squeeze Theorem Problem I'm busy studying for my Calculus A exam tomorrow and I've come across quite a tough question. I know I shouldn't post such localized questions, so if you don't want to answer, you can just push me in the right direction. I had to use the squeeze theorem to determine: $$\lim_{x\to\infty} \dfrac{\sin(x^2)}{x^3}$$ This was easy enough and I got the limit to equal 0. Now the second part of that question was to use that to determine: $$\lim_{x\to\infty} \dfrac{2x^3 + \sin(x^2)}{1 + x^3}$$ Obvously I can see that I'm going to have to sub in the answer I got from the first limit into this equation, but I can't seem to figure how how to do it. Any help would really be appreciated! Thanks in advance!

I assume you meant $$\lim_{x \to \infty} \dfrac{2x^3 + \sin(x^2)}{1+x^3}$$ Note that $-1 \leq \sin(\theta) \leq 1$. Hence, we have that $$\dfrac{2x^3 - 1}{1+x^3} \leq \dfrac{2x^3 + \sin(x^2)}{1+x^3} \leq \dfrac{2x^3 + 1}{1+x^3}$$ Note that $$\dfrac{2x^3 - 1}{1+x^3} = \dfrac{2x^3 +2 -3}{1+x^3} = 2 - \dfrac3{1+x^3}$$ $$\dfrac{2x^3 + 1}{1+x^3} = \dfrac{2x^3 + 2 - 1}{1+x^3} = 2 - \dfrac1{1+x^3}$$ Hence, $$2 - \dfrac3{1+x^3} \leq \dfrac{2x^3 + \sin(x^2)}{1+x^3} \leq 2 - \dfrac1{1+x^3}$$ Can you now find the limit? EDIT If you want to make use of the fact that $\lim_{x \to \infty} \dfrac{\sin(x^2)}{x^3} = 0$, divide the numerator and denominator of $\dfrac{2x^3 + \sin(x^2)}{1+x^3}$ by $x^3$ to get $$\dfrac{2x^3 + \sin(x^2)}{1+x^3} = \dfrac{2 + \dfrac{\sin(x^2)}{x^3}}{1 + \dfrac1{x^3}}$$ Now make use of the fact that $\lim_{x \to \infty} \dfrac{\sin(x^2)}{x^3} = 0$ and $\lim_{x \to \infty} \dfrac1{x^3} = 0$ to get your answer.

Derivative of a split function We have the function: $$f(x) = \frac{x^2\sqrt[4]{x^3}}{x^3+2}.$$ I rewrote it as $$f(x) = \frac{x^2{x^{3/4}}}{x^3+2}.$$ After a while of differentiating I get the final answer: $$f(x)= \frac{- {\sqrt[4]{\left(\frac{1}{4}\right)^{19}} + \sqrt[4]{5.5^7}}}{(x^3+2)^2}$$(The minus isn't behind the four) But my answer sheet gives a different answer, but they also show a wrong calculation, so I don't know what is the right answer, can you help me with this?

Let $y=\frac{x^2\cdot x^{3/4}}{x^3+2}$ so $y=\frac{x^{11/4}}{x^3+2}$ and therefore $y=x^{11/4}\times(x^3+2)^{-1}$. Now use the product rule of two functions: $$(f\cdot g)'=f'\cdot g+f\cdot g'$$ Here $f(x)=x^{11/4}$ and $g(x)=(x^3+2)^{-1}$. So $f'(x)=\frac{11}{4}x^{7/4}$ and $g'(x)=(-1)(3x^2)(x^3+2)^{-2}$. But thinking of your answer in the body, I cannot see where did it come from.

Using induction to prove $3$ divides $\left \lfloor\left(\frac {7+\sqrt {37}}{2}\right)^n \right\rfloor$ How can I use induction to prove that $$\left \lfloor\left(\cfrac {7+\sqrt {37}}{2}\right)^n \right\rfloor$$ is divisible by $3$ for every natural number $n$?

Consider the recurrence given by $$x_{n+1} = 7x_n - 3 x_{n-1}$$ where $x_0 = 2$, $x_1 = 7$. Note that $$x_{n+1} \equiv (7x_n - 3 x_{n-1}) \pmod{3} \equiv (x_n + 3(2x_n - x_{n-1})) \pmod{3} \equiv x_n \pmod{3}$$ Since $x_1 \equiv 1 \pmod{3}$, we have that $x_n \equiv 1 \pmod{n}$. Ths solution to this recurrence is given by $$x_n = \left( \dfrac{7+\sqrt{37}}{2} \right)^n + \left( \dfrac{7-\sqrt{37}}{2}\right)^n \equiv 1 \pmod{3}$$ Further, $0 < \dfrac{7-\sqrt{37}}{2} < 1$. This means $$3M < \left( \dfrac{7+\sqrt{37}}{2} \right)^n < 3M+1$$ where $M \in \mathbb{Z}$. Hence, we have that $$3 \text{ divides }\left \lfloor \left (\dfrac{7+\sqrt{37}}{2}\right)^n \right \rfloor$$

The positive integer solutions for $2^a+3^b=5^c$ What are the positive integer solutions to the equation $$2^a + 3^b = 5^c$$ Of course $(1,\space 1, \space 1)$ is a solution.

There are three solutions which can all be found by elementary means. If $b$ is odd $$2^a+3\equiv 1 \bmod 4$$ Therefore $a=1$ and $b$ is odd. If $b>1$, then $2\equiv 5^c \bmod 9$ and $c\equiv 5 \bmod 6$ Therefore $2+3^b\equiv 5^c\equiv3 \bmod 7$ and $b\equiv 0 \bmod 6$, a contradiction. The only solution is $(a,b,c)=(1,1,1)$. If $b$ is even, $c$ is odd $$2^a+1\equiv 5 \bmod 8$$ Therefore $a=2$. If $b\ge 2$, then $4\equiv 5^c \bmod 9$ and $c\equiv 4 \bmod 6$, a contradiction. The only solution is $(a,b,c)=(2,0,1)$. If $b$ and $c$ are even Let $b=2B$ and $c=2C$. Then $$2^a=5^{2C}-3^{2B}=(5^C-3^B)(5^C+3^B)$$ Therefore $5^C-3^B$ is also a (smaller) power of 2. A check of $(B,C)=(0,1)$ and $(1,1)$ yields the third solution $(a,b,c)=(4,2,2)$. $(B,C)=(2,2)$ does not yield a further solution and we are finished.

For which values of $\alpha \in \mathbb R$ is the following system of linear equations solvable? The problem I was given: Calculate the value of the following determinant: $\left| \begin{array}{ccc} \alpha & 1 & \alpha^2 & -\alpha\\ 1 & \alpha & 1 & 1\\ 1 & \alpha^2 & 2\alpha & 2\alpha\\ 1 & 1 & \alpha^2 & -\alpha \end{array} \right|$ For which values of $\alpha \in \mathbb R$ is the following system of linear equations solvable? $\begin{array}{lcl} \alpha x_1 & + & x_2 & + & \alpha^2 x_3 & = & -\alpha\\ x_1 & + & \alpha x_2 & + & x_3 & = & 1\\ x_1 & + & \alpha^2 x_2 & + & 2\alpha x_3 & = & 2\alpha\\ x_1 & + & x_2 & + & \alpha^2 x_3 & = & -\alpha\\ \end{array}$ I got as far as finding the determinant, and then I got stuck. So I solve the determinant like this: $\left| \begin{array}{ccc} \alpha & 1 & \alpha^2 & -\alpha\\ 1 & \alpha & 1 & 1\\ 1 & \alpha^2 & 2\alpha & 2\alpha\\ 1 & 1 & \alpha^2 & -\alpha \end{array} \right|$ = $\left| \begin{array}{ccc} \alpha - 1 & 0 & 0 & 0\\ 1 & \alpha & 1 & 1\\ 1 & \alpha^2 & 2\alpha & 2\alpha\\ 1 & 1 & \alpha^2 & -\alpha \end{array} \right|$ = $(\alpha - 1)\left| \begin{array}{ccc} \alpha & 1 & 1\\ \alpha^2 & 2\alpha & 2\alpha \\ 1 & \alpha^2 & -\alpha \end{array} \right|$ = $(\alpha - 1)\left| \begin{array}{ccc} \alpha & 1 & 0\\ \alpha^2 & 2\alpha & 0 \\ 1 & \alpha^2 & -\alpha - \alpha^2 \end{array} \right|$ = $-\alpha^3(\alpha - 1) (1 + \alpha)$ However, now I haven't got a clue on solving the system of linear equations... It's got to do with the fact that the equations look like the determinant I calculated before, but I don't know how to connect those two. Thanks in advance for any help. (:

Let me first illustrate an alternate approach. You're looking at $$\left[\begin{array}{ccc} \alpha & 1 & \alpha^2\\ 1 & \alpha & 1\\ 1 & \alpha^2 & 2\alpha\\ 1 & 1 & \alpha^2 \end{array}\right]\left[\begin{array}{c} x_1\\ x_2\\ x_3\end{array}\right]=\left[\begin{array}{c} -\alpha\\ 1\\ 2\alpha\\ -\alpha\end{array}\right].$$ We can use row reduction on the augmented matrix $$\left[\begin{array}{ccc|c} \alpha & 1 & \alpha^2 & -\alpha\\ 1 & \alpha & 1 & 1\\ 1 & \alpha^2 & 2\alpha & 2\alpha\\ 1 & 1 & \alpha^2 & -\alpha \end{array}\right].$$ In particular, for the system to be solvable, it is necessary and sufficient that none of the rows in the reduced matrix is all $0$'s except for in the last column. Subtract the bottom row from the other rows, yielding $$\left[\begin{array}{ccc|c} \alpha-1 & 0 & 0 & 0\\ 0 & \alpha-1 & 1-\alpha^2 & 1+\alpha\\ 0 & \alpha^2-1 & 2\alpha-\alpha^2 & 3\alpha\\ 1 & 1 & \alpha^2 & -\alpha \end{array}\right].$$ It's clear then that if $\alpha=1$, the second row has all $0$s except in the last column, so $\alpha=1$ doesn't give us a solvable system. Suppose that $\alpha\neq 1$, multiply the top row by $\frac1{\alpha-1}$, and subtract the new top row from the bottom row, giving us $$\left[\begin{array}{ccc|c} 1 & 0 & 0 & 0\\ 0 & \alpha-1 & 1-\alpha^2 & 1+\alpha\\ 0 & \alpha^2-1 & 2\alpha-\alpha^2 & 3\alpha\\ 0 & 1 & \alpha^2 & -\alpha \end{array}\right].$$ Swap the second and fourth rows and add the new second row to the last two rows, giving us $$\left[\begin{array}{ccc|c} 1 & 0 & 0 & 0\\ 0 & 1 & \alpha^2 & -\alpha\\ 0 & \alpha^2 & 2\alpha & 2\alpha\\ 0 & \alpha & 1 & 1 \end{array}\right],$$ whence subtracting $\alpha$ times the fourth row from the third row gives us $$\left[\begin{array}{ccc|c} 1 & 0 & 0 & 0\\ 0 & 1 & \alpha^2 & -\alpha\\ 0 & 0 & \alpha & \alpha\\ 0 & \alpha & 1 & 1 \end{array}\right].$$ Note that $\alpha=0$ readily gives us the solution $x_1=x_2=0$, $x_3=1$. Assume that $\alpha\neq 0,$ multiply the third row by $\frac1\alpha$, subtract the new third row from the fourth row, and subtract $\alpha^2$ times the new third row from the second row, yielding $$\left[\begin{array}{ccc|c} 1 & 0 & 0 & 0\\ 0 & 1 & 0 & -\alpha^2-\alpha\\ 0 & 0 & 1 & 1\\ 0 & \alpha & 0 & 0 \end{array}\right],$$ whence subtracting $\alpha$ times the second row from the fourth row yields $$\left[\begin{array}{ccc|c} 1 & 0 & 0 & 0\\ 0 & 1 & 0 & -\alpha^2-\alpha\\ 0 & 0 & 1 & 1\\ 0 & 0 & 0 & \alpha^3+\alpha^2 \end{array}\right].$$ The bottom right entry has to be $0$, so since $\alpha\neq 0$ by assumption, we need $\alpha=-1$, giving us $$\left[\begin{array}{ccc|c} 1 & 0 & 0 & 0\\ 0 & 1 & 0 & 0\\ 0 & 0 & 1 & 1\\ 0 & 0 & 0 & 0 \end{array}\right].$$ Hence, the two values of $\alpha$ that give the system a solution are $\alpha=0$ and $\alpha=-1$, and in both cases, the system has solution $x_1=x_2=0$, $x_3=1$. (I think all my calculations are correct, but I'd recommend double-checking them.) The major upside of the determinant approach is that it saves you time and effort, since you've already calculated it. If we assume that $\alpha$ is a constant that gives us a solution, then since we're dealing with $4$ equations in only $3$ variables, we have to have at least one of the rows in the reduced echelon form of the augmented matrix be all $0$s--we simply don't have enough degrees of freedom otherwise. The determinant of the reduced matrix will then be $0$, and since we obtain it by invertible row operations on the original matrix, then the determinant of the original matrix must also be $0$. By your previous work, then, $-\alpha^3(\alpha-1)(1+\alpha)=0$, so the only possible values of $\alpha$ that can give us a solvable system are $\alpha=0$, $\alpha=-1$, and $\alpha=1$. We simply check the system in each case to see if it actually is solvable. If $\alpha=0$, we readily get $x_1=x_2=0$, $x_3=1$ as the unique solution; similarly for $\alpha=-1$. However, if we put $\alpha=1$, then the second equation becomes $$x_1+x_2+x_3=1,$$ but the fourth equation becomes $$x_1+x_2+x_3=-1,$$ so $\alpha=1$ does not give us a solvable system.

Prove $\frac{1}{a^3} + \frac{1}{b^3} +\frac{1}{c^3} ≥ 3$ Prove inequality $$\frac{1}{a^3} + \frac{1}{b^3} +\frac{1}{c^3} ≥ 3$$ where $a+b+c=3abc$ and $a,b,c>0$

If $a, b, c >0$ then $a+b+c=3abc \ \Rightarrow \ \cfrac 1{ab} + \cfrac 1{bc}+ \cfrac 1{ca} = 3$ See that $2\left(\cfrac 1{a^3} +\cfrac 1{b^3}+ \cfrac 1{c^3}\right) +3 =\left(\cfrac 1{a^3} +\cfrac 1{b^3}+ 1\right)+\left(\cfrac 1{b^3} +\cfrac 1{c^3}+ 1\right)+\left(\cfrac 1{c^3} +\cfrac 1{a^3}+ 1\right) $ Use $AM-GM$ inequality on each of them and you've got your proof.

Prove $3^{2n+1} + 2^{n+2}$ is divisible by $7$ for all $n\ge0$ Expanding the equation out gives $(3^{2n}\times3)+(2^n\times2^2) \equiv 0\pmod{7}$ Is this correct? I'm a little hazy on my index laws. Not sure if this is what I need to do? Am I on the right track?

Note that $$3^{2n+1} = 3^{2n} \cdot 3^1 = 3 \cdot 9^n$$ and $$2^{n+2} = 4 \cdot 2^n$$ Note that $9^{3k} \equiv 1 \pmod{7}$ and $2^{3k} \equiv 1 \pmod{7}$. If $n \equiv 0 \pmod{3}$, then $$3 \cdot 9^n + 4 \cdot 2^n \equiv (3+4) \pmod{7} \equiv 0 \pmod{7}$$ If $n \equiv 1 \pmod{3}$, then $$3 \cdot 9^n + 4 \cdot 2^n \equiv (3 \cdot 9 + 4 \cdot 2) \pmod{7} \equiv 35 \pmod{7} \equiv 0 \pmod{7}$$ If $n \equiv 2 \pmod{3}$, then $$3 \cdot 9^n + 4 \cdot 2^n \equiv (3 \cdot 9^2 + 4 \cdot 2^2) \pmod{7} \equiv 259 \pmod{7} \equiv 0 \pmod{7}$$ EDIT What you have written can be generalized a bit. In general, $$(x^2 + x + 1) \vert \left((x+1)^{2n+1} + x^{n+2} \right)$$ The case you are interested in is when $x=2$. The proof follows immediately from the factor theorem. Note that $\omega$ and $\omega^2$ are roots of $(x^2 + x + 1)$. If we let $f(x) = (x+1)^{2n+1} + x^{n+2}$, then $$f(\omega) = (\omega+1)^{2n+1} + \omega^{n+2} = (-\omega^2)^{2n+1} + \omega^{n+2} = \omega^{4n} (-\omega^2) + \omega^{n+2} = \omega^{n+2} \left( 1 - \omega^{3n}\right) = 0$$ Similarly, $$f(\omega^2) = (\omega^2+1)^{2n+1} + \omega^{2(n+2)} = (-\omega)^{2n+1} + \omega^{2n+4} = -\omega^{2n+1} + \omega^{2n+1} \omega^3 = -\omega^{2n+1} + \omega^{2n+1} = 0$$

A limit $\lim_{n\rightarrow \infty}\sum_{k=1}^{n}\frac{k\sin\frac{k\pi}{n}}{1+(\cos\frac{k\pi}{n})^2}$ maybe relate to riemann sum Find $$\lim_{n\rightarrow \infty}\sum_{k=1}^{n}\frac{k\sin\frac{k\pi}{n}}{1+(\cos\frac{k\pi}{n})^2}$$ I think this maybe relate to Riemann sum. but I can't deal with $k$ before $\sin$

If there is no typo, then the answer is $\infty$. Indeed, let $m$ be any fixed positive integer and consider the final $m$ consecutive terms: $$ \sum_{k=n-m}^{n-1} \frac{k \sin \frac{k \pi}{n}}{1 + \cos^2 \frac{k \pi}{n}} = \sum_{k=1}^{m} \frac{(n-k) \sin \frac{k \pi}{n}}{1 + \cos^2 \frac{k \pi}{n}}. $$ As $n \to \infty$, each term converges to $k \pi$, in view of the substitution $x = \frac{k\pi}{n}$ and the following limit $$ \lim_{x\to 0}\frac{\sin x}{x(1 + \cos^2 x)} = 1. $$ Thus $$ \liminf_{n\to\infty} \sum_{k=1}^{n} \frac{k \sin \frac{k \pi}{n}}{1 + \cos^2 \frac{k \pi}{n}} \geq \lim_{n\to\infty} \sum_{k=n-m}^{n-1} \frac{k \sin \frac{k \pi}{n}}{1 + \cos^2 \frac{k \pi}{n}} = \sum_{k=1}^{m} k \pi = \frac{m(m+1)}{2}\pi. $$ Now letting $m \to \infty$, we obtain the desired result. Indeed, we have $$ \lim_{n\to\infty} \sum_{k=1}^{n} \frac{\frac{k}{n} \sin \frac{k \pi}{n}}{1 + \cos^2 \frac{k \pi}{n}} \frac{1}{n} = \frac{1}{\pi^2} \int_{0}^{\pi} \frac{x \sin x}{1 + \cos^2 x} \, dx = \frac{1}{4}. $$

Are there five complex numbers satisfying the following equalities? Can anyone help on the following question? Are there five complex numbers $z_{1}$, $z_{2}$ , $z_{3}$ , $z_{4}$ and $z_{5}$ with $\left|z_{1}\right|+\left|z_{2}\right|+\left|z_{3}\right|+\left|z_{4}\right|+\left|z_{5}\right|=1$ such that the smallest among $\left|z_{1}\right|+\left|z_{2}\right|-\left|z_{1}+z_{2}\right|$, $\left|z_{1}\right|+\left|z_{3}\right|-\left|z_{1}+z_{3}\right|$, $\left|z_{1}\right|+\left|z_{4}\right|-\left|z_{1}+z_{4}\right|$, $\left|z_{1}\right|+\left|z_{5}\right|-\left|z_{1}+z_{5}\right|$, $\left|z_{2}\right|+\left|z_{3}\right|-\left|z_{2}+z_{3}\right|$, $\left|z_{2}\right|+\left|z_{4}\right|-\left|z_{2}+z_{4}\right|$, $\left|z_{2}\right|+\left|z_{5}\right|-\left|z_{2}+z_{5}\right|$, $\left|z_{3}\right|+\left|z_{4}\right|-\left|z_{3}+z_{4}\right|$, $\left|z_{3}\right|+\left|z_{5}\right|-\left|z_{3}+z_{5}\right|$ and $\left|z_{4}\right|+\left|z_{5}\right|-\left|z_{4}+z_{5}\right|$is greater than $8/25$? Thanks!

Suppose you have solutions and express $z_i$ as $r_i e^{\theta_i}$. (I use $s_i = \sin( \theta_i )$ and $c_i = \sin( \theta_i )$ to make notations shorter) Then$$\begin{align*} |z_i| + |z_j| - |z_i + z_j| & = |r_i e^{\theta_i}| + |r_j e^{\theta_j}| - |r_i e^{\theta_i} + r_j e^{\theta_j}| \\ & = r_i + r_j - |r_i(c_i +is_i) + r_j(c_j +is_j) | \\ & = r_i + r_j - |(r_ic_i+r_jc_j) + i(r_is_i+r_js_j)| \\ & = r_i + r_j - \sqrt{(r_ic_i+r_jc_j)^2 + (r_is_i+r_js_j)^2} \\ & = r_i + r_j - \sqrt{ ( r_i^2c_i^2 + 2r_ir_jc_ic_j + r_j^2c_j^2 ) + ( r_i^2s_i^2 + 2r_ir_js_is_j + r_j^2s_j^2 ) } \\ & = r_i + r_j - \sqrt{ r_i^2( c_i^2 + s_i ^2 ) + r_j^2(c_j^2 + s_j^2) + 2r_ir_j(c_ic_j+s_is_j) } \\ |z_i| + |z_j| - |z_i + z_j| & = r_i + r_j - \sqrt{r_i^2 + r_j^2+2r_ir_j\cos(\theta_i-\theta_j) } \end{align*} $$ Then $$\begin{align*} |z_i| + |z_j| - |z_i + z_j| > \frac{8}{25} & \Leftrightarrow r_i + r_j - \sqrt{r_i^2 + r_j^2+2r_ir_j\cos(\theta_i-\theta_j) } > \frac{8}{25} \\ & \Leftrightarrow r_i + r_j - \frac{8}{25} > \sqrt{r_i^2 + r_j^2+2r_ir_j\cos(\theta_i-\theta_j) } \\ & \Leftrightarrow ( r_i + r_j - \frac{8}{25} ) ^ 2 > r_i^2 + r_j^2+2r_ir_j\cos(\theta_i-\theta_j) \\ & \Leftrightarrow r_i^2 + r_j^2 + (\frac{8}{25})^2 + 2r_ir_j - 2\frac{8}{25} r_i - 2\frac{8}{25} r_j > r_i^2 + r_j^2+2r_ir_j\cos(\theta_i-\theta_j) \\ & \Leftrightarrow (\frac{8}{25})^2 + 2r_ir_j - 2\frac{8}{25} r_i - 2\frac{8}{25} r_j > 2r_ir_j\cos(\theta_i-\theta_j) \\ & \Leftrightarrow (\frac{8}{25})^2 + 2r_ir_j( 1 - \cos(\theta_i-\theta_j) ) - 2\frac{8}{25} r_i - 2\frac{8}{25} r_j > 0 \\ \end{align*} $$ I might update later but that's all I have for now :/ But I would try to express $r_i$ as a function of $r_{\sigma(i)}$ with $\sigma$ a permutation. And by doing that, you would probably get an ugly way of calculating any $r_j$ from all the $\theta_k$ where $j,k \in \{\sigma(i)^n, n\in \mathbb{N}\}$.

Generating function for binomial coefficients $\binom{2n+k}{n}$ with fixed $k$ Prove that $$ \frac{1}{\sqrt{1-4t}} \left(\frac{1-\sqrt{1-4t}}{2t}\right)^k = \sum\limits_{n=0}^{\infty}\binom{2n+k}{n}t^n, \quad \forall k\in\mathbb{N}. $$ I tried already by induction over $k$ but i have problems showing the statement holds for $k=0$ or $k=1$.

Due to a recent comment on my other answer, I took a second look at this question and tried to apply a double generating function. $$ \begin{align} &\sum_{n=0}^\infty\sum_{k=-n}^\infty\binom{2n+k}{n}x^ny^k\\ &=\sum_{n=0}^\infty\sum_{k=n}^\infty\binom{k}{n}\frac{x^n}{y^{2n}}y^k\\ &=\sum_{n=0}^\infty\frac{x^n}{y^{2n}}\frac{y^n}{(1-y)^{n+1}}\\ &=\frac1{1-y}\frac1{1-\frac{x}{y(1-y)}}\\ &=\frac{y}{y(1-y)-x}\\ &=\frac1{\sqrt{1-4x}}\left(\frac{1+\sqrt{1-4x}}{1+\sqrt{1-4x}-2y}-\frac{1-\sqrt{1-4x}}{1-\sqrt{1-4x}-2y}\right)\\ &=\frac1{\sqrt{1-4x}}\left(\frac{1+\sqrt{1-4x}}{1+\sqrt{1-4x}-2y}+\color{#C00000}{\frac{2x/y}{1+\sqrt{1-4x}-2x/y}}\right)\tag{1} \end{align} $$ The term in red contains those terms with negative powers of $y$. Eliminating those terms yields $$ \begin{align} \sum_{n=0}^\infty\sum_{k=0}^\infty\binom{2n+k}{n}x^ny^k &=\frac1{\sqrt{1-4x}}\frac{1+\sqrt{1-4x}}{1+\sqrt{1-4x}-2y}\\ &=\frac1{\sqrt{1-4x}}\sum_{k=0}^\infty\left(\frac{2y}{1+\sqrt{1-4x}}\right)^k\\ &=\frac1{\sqrt{1-4x}}\sum_{k=0}^\infty\left(\frac{1-\sqrt{1-4x}}{2x}\right)^ky^k\tag{2} \end{align} $$ Equating identical powers of $y$ in $(2)$ shows that $$ \sum_{n=0}^\infty\binom{2n+k}{n}x^n=\frac1{\sqrt{1-4x}}\left(\frac{1-\sqrt{1-4x}}{2x}\right)^k\tag{3} $$

numbers' pattern It is known that $$\begin{array}{ccc}1+2&=&3 \\ 4+5+6 &=& 7+8 \\ 9+10+11+12 &=& 13+14+15 \\\ 16+17+18+19+20 &=& 21+22+23+24 \\\ 25+26+27+28+29+30 &=& 31+32+33+34+35 \\\ldots&=&\ldots \end{array}$$ There is something similar for square numbers: $$\begin{array}{ccc}3^2+4^2&=&5^2 \\ 10^2+11^2+12^2 &=& 13^2+14^2 \\ 21^2+22^2+23^2+24^2 &=& 25^2+26^2+27^2 \\ \ldots&=&\ldots \end{array}$$ As such, I wonder if there are similar 'consecutive numbers' for cubic or higher powers. Of course, we know that there is impossible for the following holds (by Fermat's last theorem): $$k^3+(k+1)^3=(k+2)^3 $$

Let's start by proving the basic sequences and then where and why trying to step it up to cubes fails. I don't prove anything just reduce the problem to a two variable quartic Diophantine equation. Lemma $1 + 2 + 3 + 4 + \ldots + n = T_1(n) = \frac{n(n+1)}{2}$. Corollary $(k+1) + (k+2) + \ldots + (k+n) = -T_1(k) + T_1(k+n)$ The first sequence of identities is $$-T_1(s(n)) + T_1(s(n)+n+1) = -T_1(s(n)+n+1) + T_1(s(n)+2n+1)$$ so computing ? f(x) = (x*(x+1))/2 ? (-f(s)+f(s+n+1))-(-f(s+n+1)+f(s+2*n+1)) % = -n^2 + (s + 1) we find $s(n) = n^2-1$ and prove it. Lemma $1^2 + 2^2 + 3^2 + 4^2 + \ldots + n^2 = T_2(n) = \frac{n(n+1)(2n+1)}{6}$. The second sequence of identities is $$-T_2(s(n)) + T_2(s(n)+n+1) = -T_2(s(n)+n+1) + T_2(s(n)+2n+1)$$ so computing ? f(x) = (x*(x+1)*(2*x+1))/6 ? (-f(s-1)+f(s-1+n+1))-(-f(s-1+n+1)+f(s-1+2*n+1)) % = -2*n^3 + (-2*s - 1)*n^2 + s^2 this is a weird quadratic equation in two integers with some solutions (n,s) = (1,3), (2,10), (3,21), (4,36), (5,55), (6,76), ... the discriminant of the polynomial (as a polynomial in $s$) is $2^2 n^2 (n+1)^2$ so actually we can solve it and that explains where there's one solution for each $n$. Now lets try for cubes.. but at this point we know it's not going to work ? f(x) = ((x^2+x)/2)^2 ? (-f(s-1)+f(s-1+n+1))-(-f(s-1+n+1)+f(s-1+2*n+1)) % = -7/2*n^4 + (-6*s - 3)*n^3 + (-3*s^2 - 3*s - 1/2)*n^2 + s^3 so this is too complicated to actually solve but if anyone proves this doesn't have solutions for positive $n$ that will show there are no such cubic sequences. For reference $$7n^4 + (12s + 6)n^3 + (6s^2 + 6s + 1)n^2 - 2s^3 = 0$$ is the Diophantine equation that obstructs a cubic sequence from existing. Maybe you could conclude by the Mordell Conjecture that there's no infinite family of sequences of identities for cubic and higher power sums, if you can show these polynomials are always irreducible.

Solve the recurrence relation:$ T(n) = \sqrt{n} T \left(\sqrt n \right) + n$ $$T(n) = \sqrt{n} T \left(\sqrt n \right) + n$$ Master method does not apply here. Recursion tree goes a long way. Iteration method would be preferable. The answer is $Θ (n \log \log n)$. Can anyone arrive at the solution.

Let $n = m^{2^k}$. We then get that $$T(m^{2^k}) = m^{2^{k-1}} T (m^{2^{k-1}}) + m^{2^{k}}$$ \begin{align} f_m(k) & = m^{2^{k-1}} f_m(k-1) + m^{2^k} = m^{2^{k-1}}(m^{2^{k-2}} f_m(k-2) + m^{2^{k-1}}) + m^{2^k}\\ & = 2 m^{2^k} + m^{3 \cdot 2^{k-2}} f_m(k-2) \end{align} $$m^{3 \cdot 2^{k-2}} f_m(k-2) = m^{3 \cdot 2^{k-2}} (m^{2^{k-3}} f_m(k-3) + m^{2^{k-2}}) = m^{2^k} + m^{7 \cdot 2^{k-3}} f_m(k-3)$$ Hence, $$f_m(k) = 2 m^{2^k} + m^{3 \cdot 2^{k-2}} f_m(k-2) = 3m^{2^k} + m^{7 \cdot 2^{k-3}} f_m(k-3)$$ In general, it is not hard to see that $$f_m(k) = \ell m^{2^k} + m^{(2^{\ell}-1)2^{k-\ell}} f_m(k-\ell)$$ $\ell$ can go up to $k$, to give us $$f_m(k) = km^{2^k} + m^{(2^{k}-1)} f_m(0) = km^{2^k} + m^{(2^{k}-1)} m^{2^0} = (k+1) m^{2^k}$$ This gives us $$f_m(k) = (k+1) m^{2^k} = n \left(\log_2(\log_m n) + 1\right) = \mathcal{O}(n \log_2(\log_2 n))$$ since $$n=m^{2^k} \implies \log_m(n) = 2^k \implies \log_2(\log_m(n)) = k$$

$p$ an odd prime, $p \equiv 3 \pmod 8$. Show that $2^{(\frac{p-1}{2})}*(p-1)! \equiv 1 \pmod p$ $p$ an odd prime, $p \equiv 3 \pmod 8$. Show that $2^{\left(\frac{p-1}{2}\right)}\cdot(p-1)! \equiv 1 \pmod p$ From Wilson's thm: $(p-1)!= -1 \pmod p$. hence, need to show that $2^{\left(\frac{p-1}{2}\right)} \equiv -1 \pmod p. $ we know that $2^{p-1} \equiv 1 \pmod p.$ Hence: $2^{\left(\frac{p-1}{2}\right)} \equiv \pm 1 \pmod p. $ How do I show that this must be the negative option?

All the equalities below are in the ring $\mathbb{Z}/p\mathbb{Z}$. Note that $-1 = (p-1)! = \prod_{i=1}^{p-1}i = \prod_{i=1}^{\frac{p-1}{2}}(2i-1)\prod_{i=1}^{\frac{p-1}{2}} 2i = 2^{\frac{p-1}{2}} \prod_{i=1}^{\frac{p-1}{2}}(2i-1)\prod_{i=1}^{\frac{p-1}{2}} i$ Now let $S_1, S_2$ be the set of respectively all odd and even numbers in $\left \{ 1, \cdots, \frac{p-1}{2} \right \}$ and $S_3$ be the set of all even numbers in $\left \{ \frac{p+1}{2}, \ldots, p-1 \right \}$. Note that $\prod_{i=1}^{\frac{p-1}{2}} i = \prod _{j \in S_1}j \prod _{k \in S_2}k = (-1)^{|S_1|} \prod _{j \in S_1}(-j)\prod _{k \in S_2}k $ $= (-1)^{|S_1|} \prod _{t \in S_3}t \prod _{k \in S_2}k =(-1)^{|S_1|} \prod_{i=1}^{\frac{p-1}{2}}(2i)$ So $\prod_{i=1}^{\frac{p-1}{2}}(2i-1)\prod_{i=1}^{\frac{p-1}{2}} i = (-1)^{|S_1|}\prod_{i=1}^{\frac{p-1}{2}}(2i-1) \prod_{i=1}^{\frac{p-1}{2}} 2i = (-1)^{|S_1|} (p-1)! = (-1)^{|S_1| +1} $ Now we have $-1 = 2^{\frac{p-1}{2}} \prod_{i=1}^{\frac{p-1}{2}}(2i-1)\prod_{i=1}^{\frac{p-1}{2}} i = (-1)^{|S_1| + 1} \cdot 2^{\frac{p-1}{2}} $ i.e $\boxed{2^{\frac{p-1}{2}} = (-1)^{|S_1|}} $ Now $|S_1| = \frac{p+1}{4}$ if $\frac{p-1}{2}$ is odd and $|S_1| = \frac{p-1}{4}$ if $\frac{p-1}{2}$ is even. So if $p \equiv 3, 5 \mod 8 $ we have $2^{\frac{p-1}{2}} = -1$. if $p = 1,7 \mod 8$ we have $2^{\frac{p-1}{2}} = 1$.

Need someone to show me the solution Need someone to show me the solution. and tell me how ! $$(P÷N) × (N×(N+1)÷2) + N×(1-P) = N×(1-(P÷2)) + (P÷2)$$

\begin{align} \dfrac{P}{N} \times \dfrac{N(N+1)}2 + N\times (1-P) & = \underbrace{P \times \dfrac{N+1}{2} + N \times (1-P)}_{\text{Cancelling out the $N$ from the first term}}\\ & = \underbrace{\dfrac{PN + P}2 + N - NP}_{\text{$P \times (N+1) = PN + P$ and $N \times (1-P) = N - NP$}}\\ & = \underbrace{\dfrac{PN + P +2N - 2NP}2}_{\text{Take the lcm $2$.}}\\ & = \underbrace{\dfrac{P + 2N -NP}2}_{PN - 2NP = -NP}\\ & = \dfrac{P}2 + \dfrac{2N - NP}2\\ & = \underbrace{N\dfrac{2-P}2 + \dfrac{P}2}_{\text{Factor out $N$ from $2N-NP$}}\\ & = \underbrace{N \left(1 - \dfrac{P}2\right) + \dfrac{P}2}_{\text{Making use of the fact that $\dfrac{2-P}2 = \dfrac22 - \dfrac{P}2 = 1 - \dfrac{P}2$}} \end{align}

Example 2, Chpt 4 Advanced Mathematics (I) $$\int \frac{x+2}{2x^3+3x^2+3x+1}\, \mathrm{d}x$$ I can get it down to this: $$\int \frac{2}{2x+1} - \frac{x}{x^2+x+1}\, \mathrm{d}x $$ I can solve the first part but I don't exactly follow the method in the book. $$ = \ln \vert 2x+1 \vert - \frac{1}{2}\int \frac{\left(2x+1\right) -1}{x^2+x+1}\, \mathrm{d}x $$ $$= \ln \vert 2x+1 \vert - \frac{1}{2} \int \frac{\mathrm{d}\left(x^2+x+1\right)}{x^2+x+1} + \frac{1}{2}\int \dfrac{\mathrm{d}x}{\left(x+\dfrac{1}{2}\right)^2 + \frac{3}{4}} $$ For the 2nd part: I tried $ u = x^2+x+1 $ and $\mathrm{d}u = 2x+1\, \mathrm{d}x$ that leaves me with $\frac{\mathrm{d}u - 1}{2} = x\, \mathrm{d}x$ which seems wrong. because $x^2+x+1$ doesn't factor, I don't see how partial fractions again will help. $x = Ax+B$ isn't helpful.

The post indicates some difficulty with finding $\int \frac{dx}{x^2+x+1}$. We solve a more general problem. But I would suggest for your particular problem, you follow the steps used, instead of using the final result. Suppose that we want to integrate $\dfrac{1}{ax^2+bx+c}$, where $ax^2+bx+c$ is always positive, or always negative. We complete the square. In order to avoid fractions, note that equivalently we want to find $$\int \frac{4a\,dx}{4a^2 x+4abx+4ac}.$$ So we want to find $$\int \frac{4a\,dx}{(2ax+b)^2 + (4ac-b^2)}.$$ Let $$2ax+b=u\sqrt{4ac-b^2}.$$ Then $2a\,dx=\sqrt{4ac-b^2}\,du$. Our integral simplifies to $$\frac{2}{\sqrt{4ac-b^2}}\int\frac{du}{u^2+1},$$ and we are finished.

Two questions with mathematical induction First hello all, we have a lecture. It has 10 questions but I'm stuck with these two about 3 hours and I can't solve them. Any help would be appreciated. Question 1 Given that $T(1)=1$, and $T(n)=2T(\frac{n}{2})+1$, for $n$ a power of $2$, and greater than $1$. Using mathematical induction, prove that $T(n) = 2^k.T(\frac{n}{2^k}) + 2^k - 1$ for $k=0, 1, 2, \dots, \log_2 n$. Question 2 Definition: $H(j) = \frac{1}{1} + \frac{1}{2} + \frac{1}{3} + \dots + \frac{1}{j}$. Facts: $H(16) > 3.38$, $\frac{1}{3} + \frac{1}{4} = \frac{7}{12}, \frac{1}{4} + \frac{1}{5} + \frac{1}{7} = \frac{319}{420} $ a) Using induction on $n$, prove that $H(2^n) > 1 + 7n/12$, for $n\geq 4$.

1) Prove by induction on $k$: If $0\le k\le m$, then $T(2^m)=2^kT(2^{m-k})+2^k-1$. The case $k=0$ is trivial. If we already know that $T(2^m)=2^{k-1}T(2^{m-(k-1)})+2^{k-1}-1$ for all $m\ge k-1$, then for $m\ge 2^k$ we have $$\begin{align}T(2^m)&=2^{k-1}T(2^{m-(k-1)})+2^{k-1}-1\\ &=2^{k-1}\left(2\cdot T\left(\tfrac{2^{m-(k-1)}}2\right)+1\right)+2^{k-1}-1\\ &=2^k T(2^{m-k})+2^k-1\end{align}$$ 2) Note that $$H(2(m+1))-H(2m)=\frac 1{2m+1}+\frac1{2(m+2)}>\frac2{2m+2}=H(m+1)-H(m)$$ and therefore (by induction on $d$) for $d>0$ $$H(2(m+d))-H(2m)>H(m+d)-H(m)$$ hence with $m=d=2^{n-1}$ $$H(2^{n+1})-H(2^n)>H(2^n)-H(2^{n-1})$$ thus by induction on $n$ $$H(2^n)-H(2^{n-1})>H(4)-H(2)=\frac7{12}\text{ if }n\ge2$$ and finally by induction on $n$ $$H(2^n)>H(16)+\frac7{12}(n-4)\text{ for }n\ge 4.$$

Partial fractions integration Re-express $\dfrac{6x^5 + x^2 + x + 2}{(x^2 + 2x + 1)(2x^2 - x + 4)(x+1)}$ in terms of partial fractions and compute the indefinite integral $\dfrac{1}5{}\int f(x)dx $ using the result from the first part of the question.

Hint Use $$\dfrac{6x^5 + x^2 + x + 2}{(x^2 + 2x + 1)(2x^2 - x + 4)(x+1)}=\frac{A}{(x+1)^3}+\frac{B}{(x+1)^2}+\frac{C}{x+1}+\frac{Dx+E}{2x^2-x+4}+F$$ and solve for $A,B,C,D,E$ and $F$. Explanation Note that this partial fraction decomposition is a case where the degree of the numerator and denominator are the same. (Just notice that the sum of the highest degree terms in the denominator is $2+2+1=5.$) This means that the typical use of partial fraction decomposition does not apply. Furthermore, notice that $(x^2+2x+1)$ factorizes as $(x+1)^2$, which means the denominator is really $(x+1)^3(2x^2-x+4)$. This means the most basic decomposition would involve denominators of $(x+1)^3$ and $(2x^2-x+4)$. However, we can escape that by using a more complicated decomposition involving the denominators $(x+1)^3$, $(x+1)^2$, $(x+1)$, and $(2x^2-x+4)$. The $F$ term is necessary for the $6x^5$ term to arise in the multiplication, intuitively speaking. More thoroughly stated, the $F$ term is needed because of the following equivalency between $6x^5+x^2+x+2$ and the following expression: $$ \begin{align} &A(2x^2-x+4)\\ +&B(2x^2-x+4)(x+1)\\ +&C(2x^2-x+4)(x+1)^2\\ +&[Dx+E](x+1)^3\\ +&F(2x^2-x+4)(x+1)^3. \end{align} $$ Notice how the term $F(2x^2-x+4)(x+1)^3$ is the only possible term that can give rise to $6x^5$? That is precisely why it is there. Hint 2 Part 1 In the integration of $\frac{1}{5}\int f(x)dx$, first separate the integral, $\int f(x)dx$ into many small integrals with the constants removed from the integrals. That is, $\int f(x) dx$ is $$A\int \frac{1}{(x+1)^3}dx+B\int \frac{1}{(x+1)^2}dx+C\int \frac{1}{x+1}dx+D\int \frac{x}{2x^2-x+4}dx\quad+E\int \frac{1}{2x^2-x+4}dx+F\int 1dx.$$ Part 2 Next, use the substitution $u=x+1$ with $du=dx$ on the first three small integrals: $$A\int \frac{1}{u^3}du+B\int \frac{1}{u^2}du+C\int \frac{1}{u}du+D\int \frac{x}{2x^2-x+4}dx+E\int \frac{1}{2x^2-x+4}dx\quad +F\int 1dx.$$ Part 3 To deal with the integral $\int \frac{1}{2x^2-x+4}dx$, you must complete the square. This is done as follows: $$ \begin{align} \int \frac{1}{2x^2-x+4}dx&=\int \frac{\frac{1}{2}}{x^2-\frac{1}{2}x+2}dx\\ &=\frac{1}{2} \int \frac{1}{\left(x-\frac{1}{4}\right)^2+\frac{31}{16}}dx. \end{align} $$ To conclude, you make the trig substitution $x-\frac{1}{4}=\frac{\sqrt{31}}{4}\tan \theta$ with $dx=\left(\frac{\sqrt{31}}{4}\tan \theta+\frac{1}{4}\right)'d\theta=\frac{\sqrt{31}}{4}\sec^2\theta d\theta$. This gives you: $$\frac{1}{2} \int \frac{1}{\left(x-\frac{1}{4}\right)^2+\frac{31}{16}}dx=\frac{1}{2} \int \frac{\frac{\sqrt{31}}{4}\sec^2 \theta}{\frac{31}{16}\tan^2\theta+\frac{31}{16}}d\theta.$$

Showing $\sqrt{2}\sqrt{3} $ is greater or less than $ \sqrt{2} + \sqrt{3} $ algebraically How can we establish algebraically if $\sqrt{2}\sqrt{3}$ is greater than or less than $\sqrt{2} + \sqrt{3}$? I know I can plug the values into any calculator and compare the digits, but that is not very satisfying. I've tried to solve $$\sqrt{2}+\sqrt{3}+x=\sqrt{2}\sqrt{3} $$ to see if $x$ is positive or negative. But I'm just getting sums of square roots whose positive or negative values are not obvious. Can it be done without the decimal expansion?

Method 1: $\sqrt{2}+\sqrt{3}>\sqrt{2}+\sqrt{2}=2\sqrt{2}>\sqrt{3}\sqrt{2}$. Method 2: $(\sqrt{2}\sqrt{3})^2=6<5+2<5+2\sqrt{6}=2+3+2\sqrt{2}\sqrt{3}=(\sqrt{2}+\sqrt{3})^2$, so $\sqrt{2}\sqrt{3}<\sqrt{2}+\sqrt{3}$. Method 3: $\frac{196}{100}<2<\frac{225}{100}$ and $\frac{289}{100}<3<\frac{324}{100}$, so $\sqrt{2}\sqrt{3}<\frac{15}{10}\frac{18}{10}=\frac{270}{100}<\frac{14}{10}+\frac{17}{10}<\sqrt{2}+\sqrt{3}$.

Where is the mistake in the calculation of $y'$ if $ y = \Bigl( \dfrac{x^2+1}{x^2-1} \Bigr)^{1/4} $? Plase take a look here. If $ y = \Bigl( \dfrac{x^2+1}{x^2-1} \Bigr)^{1/4} $ \begin{eqnarray} y'&=& \dfrac{1}{4} \Bigl( \dfrac{x^2+1}{x^2-1} \Bigr)^{-3/4} \left \{ \dfrac{2x(x^2-1) - 2x(x^2+1) }{(x^2-1)^2} \right \}\\ &=& \Bigl( \dfrac{x^2+1}{x^2-1} \Bigr)^{-3/4} \dfrac{-x}{(x^2-1)^2}. \end{eqnarray} By the other hand, we have \begin{equation} \log y = \dfrac{1}{4} \left \{ \log (x^2+1) - \log (x^2-1) \right \} \end{equation} Then, \begin{eqnarray} \dfrac{dy}{dx} &=& y \dfrac{1}{4} \left \{ \dfrac{2x}{(x^2+1)} -\dfrac{ 2x}{(x^2-1)} \right \} \\ &=& \dfrac{1}{4} \dfrac{x^2+1}{x^2-1} \cdot 2x \dfrac{(x^2-1) - (x^2+1)}{(x^2+1)(x^2-1)} \\ &=& \dfrac{x^2+1}{x^2-1} \dfrac{-x}{(x^2+1)(x^2-1)} \\ &=& \dfrac{-x}{(x^2-1)^2}. \end{eqnarray} But this implies, \begin{equation} \dfrac{-x}{(x^2-1)^2} = \Bigl( \dfrac{x^2+1}{x^2-1} \Bigr)^{-3/4} \dfrac{-x}{(x^2-1)^2}. \end{equation} Where is the mistake?

I believe you forgot a power 1/4 when substituting for $y$ (in the calculation using logarithms). Edited to explain further: In your calculation, you write \begin{align} \frac{dy}{dx} &= y\frac14 \left\{ \frac{2x}{(x^2+1)} - \frac{2x}{(x^2-1)} \right\} \\ &= \frac14 \frac{x^2+1}{x^2-1} \cdot 2x\frac{(x^2-1)-(x^2+1)}{(x^2+1)(x^2-1)}. \end{align} However, this should be \begin{align} \frac{dy}{dx} &= y\frac14 \left\{ \frac{2x}{(x^2+1)} - \frac{2x}{(x^2-1)} \right\} \\ &= \frac14 \color{red}{\left(\color{black}{\frac{x^2+1}{x^2-1}}\right)^{\frac14}} \cdot 2x\frac{(x^2-1)-(x^2+1)}{(x^2+1)(x^2-1)}. \end{align}

How to simplify polynomials I can't figure out how to simplify this polynominal $$5x^2+3x^4-7x^3+5x+8+2x^2-4x+9-6x^2+7x$$ I tried combining like terms $$5x^2+3x^4-7x^3+5x+8+2x^2-4x+9-6x^2+7x$$ $$(5x^2+5x)+3x^4-(7x^3+7x)+2x^2-4x-6x^2+(8+9)$$ $$5x^3+3x^4-7x^4+2x^2-4x-6x^2+17$$ It says the answer is $$3x^4-7x^3+x^2+8x+17$$ but how did they get it?

You cannot combine terms like that, you have to split your terms by powers of $x$. So for example $$5x^2+5x+2x^2 = (5+2)x^2+5x = 7x^2+5x$$ and not $5x^3+2x^2$. Using this, you should end up with your answer.

Advanced integration, how to integrate 1/polynomial ? Thanks I have been trying to integrate a function with a polynomial as the denominator. i.e, how would I go about integrating $$\frac{1}{ax^2+bx+c}.$$ Any help at all with this would be much appreciated, thanks a lot :) ps The polynomial has NO real roots $${}{}$$

If $a=0$, then $$I = \int \dfrac{dx}{bx+c} = \dfrac1b \log (\vert bx+c \vert) + \text{constant}$$ $$I = \int \dfrac{dx}{ax^2 + bx + c} = \dfrac1a \int\dfrac{dx}{\left(x + \dfrac{b}{2a}\right)^2 + \left(\dfrac{c}a- \dfrac{b^2}{4a^2} \right)}$$ If $b^2 < 4ac$, then recall that $$\int \dfrac{dt}{t^2 + a^2} = \dfrac1a\arctan \left(\dfrac{t}a \right) + \text{constant}$$ Hence, if $b^2 < 4ac$, then $$I = \dfrac2{\sqrt{4ac-b^2}} \arctan \left( \dfrac{2ax + b}{\sqrt{4ac-b^2}} \right) + \text{constant}$$ If $b^2 = 4ac$, then $$I =\dfrac1a \dfrac{-1}{\left(x + \dfrac{b}{2a}\right)} + \text{constant} = - \dfrac2{2ax+b} + \text{constant}$$ If $b^2 > 4ac$, then $$I = \dfrac1a \int\dfrac{dx}{\left(x + \dfrac{b}{2a}\right)^2 - \sqrt{\left( \dfrac{b^2}{4a^2} -\dfrac{c}a\right)}^2}$$ Now $$\int \dfrac{dt}{t^2 - k^2} = \dfrac1{2k} \left(\int \dfrac{dt}{t-k} - \int \dfrac{dt}{t+k} \right) = \dfrac1{2k} \log \left(\left \vert \dfrac{t-k}{t+k} \right \vert \right) + \text{constant}$$ Hence, if $b^2 > 4ac$, then $$I = \dfrac1{\sqrt{b^2-4ac}} \log \left(\left \vert \dfrac{2ax + b - \sqrt{b^2-4ac}}{2ax + b + \sqrt{b^2-4ac}} \right \vert \right) + \text{constant}$$

Closed form for $\sum_{k=0}^{n} \binom{n}{k}\frac{(-1)^k}{(k+1)^2}$ How can I calculate the following sum involving binomial terms: $$\sum_{k=0}^{n} \binom{n}{k}\frac{(-1)^k}{(k+1)^2}$$ Where the value of n can get very big (thus calculating the binomial coefficients is not feasible). Is there a closed form for this summation?

Apparently I'm a little late to the party, but my answer has a punchline! We have $$ \frac{1}{z} \int_0^z \sum_{k=0}^{n} \binom{n}{k} s^k\,ds = \sum_{k=0}^{n} \binom{n}{k} \frac{z^k}{k+1}, $$ so that $$ - \int_0^z \frac{1}{t} \int_0^t \sum_{k=0}^{n} \binom{n}{k} s^k\,ds\,dt = - \sum_{k=0}^{n} \binom{n}{k} \frac{z^{k+1}}{(k+1)^2}. $$ Setting $z = -1$ gives an expression for your sum, $$ \sum_{k=0}^{n} \binom{n}{k} \frac{(-1)^k}{(k+1)^2} = \int_{-1}^{0} \frac{1}{t} \int_0^t \sum_{k=0}^{n} \binom{n}{k} s^k\,ds\,dt. $$ Now, $\sum_{k=0}^{n} \binom{n}{k} s^k = (1+s)^n$, so $$ \begin{align*} \sum_{k=0}^{n} \binom{n}{k} \frac{(-1)^k}{(k+1)^2} &= \int_{-1}^{0} \frac{1}{t} \int_0^t (1+s)^n \,ds\,dt \\ &= \frac{1}{n+1}\int_{-1}^{0} \frac{1}{t} \left[(1+t)^{n+1} - 1\right]\,dt \\ &= \frac{1}{n+1}\int_{0}^{1} \frac{u^{n+1}-1}{u-1}\,du \\ &= \frac{1}{n+1}\int_0^1 \sum_{k=0}^{n} u^k \,du \\ &= \frac{1}{n+1}\sum_{k=1}^{n+1} \frac{1}{k} \\ &= \frac{H_{n+1}}{n+1}, \end{align*} $$ where $H_n$ is the $n^{\text{th}}$ harmonic number.

greatest common divisor is 7 and the least common multiple is 16940 How many such number-pairs are there for which the greatest common divisor is 7 and the least common multiple is 16940?

Let the two numbers be $7a$ and $7b$. Note that $16940=7\cdot 2^2\cdot 5\cdot 11^2$. We make a pair $(a,b)$ with gcd $1$ and lcm $2^2\cdot 5\cdot 11^2$ as follows. We "give" $2^2$ to one of $a$ and $b$, and $2^0$ to the other. We give $5^1$ to one of $a$ and $b$, and $5^0$ to the other. Finally, we give $11^2$ to one of $a$ and $b$, and $11^0$ to the other. There are $2^3$ choices, and therefore $2^3$ ordered pairs such that $\gcd(a,b)=1$ and $\text{lcm}(a,b)=2^2\cdot 5\cdot 11^2$. If we want unordered pairs, divide by $2$. Here we used implicitly the Unique Factorization Theorem: Every positive integer can be expressed in an essentially unique way as a product of primes. There was nothing really special about $7$ and $16940$: any problem of this shape can be solved in basically the same way.

How to prove the given series is divergent? Given series $$ \sum_{n=1}^{+\infty}\left[e-\left(1+\frac{1}{n}\right)^{n}\right], $$ try to show that it is divergent! The criterion will show that it is the case of limits $1$, so maybe need some other methods? any suggestions?

Let $x > 1$. Then the inequality $$\frac{1}{t} \leq \frac{1}{x}(1-t) + 1$$ holds for all $t \in [1,x]$ (the right hand side is a straight line between $(1,1)$ and $(x, \tfrac{1}{x})$ in $t$) and in particular $$\log(x) = \int_1^x \frac{dt}{t} \leq \frac{1}{2} \left(x - \frac{1}{x} \right)$$ for all $x > 1$. Substitute $x \leftarrow 1 + \tfrac{1}{n}$ to get $$ \log \left(1 + \frac{1}{n} \right) \leq \frac{1}{2n} + \frac{1}{2(n+1)}$$ and after multiplying by $n$ $$\log\left(1 + \frac{1}{n} \right)^n \leq 1 - \frac{1}{2(n+1)}.$$ Use this together with the estimate $e^x \leq (1-x)^{-1}$ for all $x < 1$ to get $$\left(1 + \frac{1}{n} \right)^n \leq e \cdot e^{-\displaystyle\frac{1}{2(n+1)}} \leq e \cdot \left(1 - \frac{1}{2n+3} \right)$$ or $$e - \left(1 + \frac{1}{n} \right)^n \geq \frac{e}{2n+3}.$$ This shows that your series diverges.

Number of Divisor which perfect cubes and multiples of a number n = $2^{14}$$3^9$$5^8$$7^{10}$$11^3$$13^5$$37^{10}$ How many positive divisors that are perfect cubes and multiples of $2^{10}$$3^9$$5^2$$7^{5}$$11^2$$13^2$$37^{2}$. I'm able to solve number of perfect square and number of of perfect cubes. But the extra condition of multiples of $2^{10}$$3^9$$5^2$$7^{5}$$11^2$$13^2$$37^{2}$ is confusing, anyone can give me a hint?

The numbers you are looking for must be perfect cubes. If you split them into powers of primes, they can have a factor $2^0$, $2^3$, $2^6$, $2^9$ and so on but not $2^1, 2^2, 2^4$ etc. because these are not cubes. The same goes for powers of $3, 5, 7$ and any other primes. The numbers must also be multiples of $2^{10}$ so can have factors $2^{12}, 2^{15}, 2^{18}$ etc. because $2^9, 2^6$ and so on are not multiples of $2^{10}$. The numbers must divide $2^{14}$, which leaves only $2^{12}$ because $2^{15}, 2^{18}$ and so on don't divide $2^{14}$. You get another factor $3^9, 5^3$ or $5^6, 7^6$ or $7^9, 11^3, 13^3$, and $37^3, 37^6$ or $37^9$. For most powers you have one choice, for $5, 7$ and $11$ you have two choices, for $37$ you have three choices - total is $2 \times 2\times 2\times 3$ numbers.

$\sqrt{(a+b-c)(b+c-a)(c+a-b)} \le \frac{3\sqrt{3}abc}{(a+b+c)\sqrt{a+b+c}}$ Suppose $a, b, c$ are the lengths of three triangular edges. Prove that: $$\sqrt{(a+b-c)(b+c-a)(c+a-b)} \le \frac{3\sqrt{3}abc}{(a+b+c)\sqrt{a+b+c}}$$

As the hint give in the comment says (I denote by $S$ the area of $ABC$ and by $R$ the radius of its circumcircle), if you multiply your inequality by $\sqrt{a+b+c}$ you'll get $$4S \leq \frac{3\sqrt{3}abc}{a+b+c}$$ which is eqivalent to $$a+b+c \leq 3\sqrt{3}\frac{abc}{4S}=3\sqrt{3}R.$$ This inequality is quite known. If you want a proof, you can write $a=2R \sin A$ (and the other two equalities) and get the equivalent inequality $$ \sin A +\sin B +\sin C \leq \frac{3\sqrt{3}}{2}$$ which is an easy application of the Jensen inequality for the concave function $\sin : [0,\pi] \to [0,1]$.

Prove that $\frac{a^2}{a+b}+\frac{b^2}{b+c}+\frac{c^2}{c+a}\geq\frac{1}{2}(a+b+c)$ for positive $a,b,c$ Prove the following inequality: for $a,b,c>0$ $$\frac{a^2}{a+b}+\frac{b^2}{b+c}+\frac{c^2}{c+a}\geq\frac{1}{2}(a+b+c)$$ What I tried is using substitution: $p=a+b+c$ $q=ab+bc+ca$ $r=abc$ But I cannot reduce $a^2(b+c)(c+a)+b^2(a+b)(c+a)+c(a+b)(b+c) $ interms of $p,q,r$

Hint: $ \sum \frac{a^2 - b^2}{a+b} = \sum (a-b) = 0$. (How is this used?) Hint: $\sum \frac{a^2 + b^2}{a+b} \geq \sum \frac{a+b}{2} = a+b+c$ by AM-GM. Hence, $\sum \frac{ a^2}{ a+b} \geq \frac{1}{2}(a+b+c)$.

Proving a Geometric Progression Formula, Related to Geometric Distribution I am trying to prove a geometric progression formula that is related to the formula for the second moment of the geometric distribution. Specifically, I am wondering where I am going wrong, so I can perhaps learn a new technique. It is known, and I wish to show: $$ m^{(2)} = \sum_{k=0}^\infty k^2p(1-p)^{k-1} = p\left(\frac{2}{p^3} - \frac{1}{p^2}\right) = \frac{2-p}{p^2} $$ Now, dividing by $p$ both sides, and assigning $a = 1-p$ yields: $$ \sum_{k=1}^\infty k^2a^{k-1}=\frac{2}{(1-a)^3}-\frac{1}{(1-a)^2} \qquad \ldots \text{(Eq. 1)} $$ I want to derive the above formula. I know: $$ \sum_{k=0}^\infty ka^{k-1}=\frac{1}{(1-a)^2} $$ Multiplying the left side by $1=\frac aa$, and multiplying both sides by $a$, $$ \sum_{k=0}^\infty ka^k = \frac{a}{(1-a)^2} $$ Taking the derivative of both sides with respect to $a$, the result is: $$ \sum_{k=0}^{\infty}\left[a^k + k^2 a^{k-1}\right] = \frac{(1-a)^2 - 2(1-a)(-1)a}{(1-a^4)} = \frac{1}{(1-a)^2}+\frac{2a}{(1-a)^3} $$ Moving the known formula $\sum_{k=0}^\infty a^k = \frac{1}{1-a}$ to the right-hand side, the result is: $$ \sum_{k=0}^\infty k^2 a^{k-1} = \frac{1}{(1-a)^2} + \frac{2a}{(1-a^3)} - \frac{1}{1-a} $$ Then, this does not appear to be the same as the original formula (Eq. 1). Where did I go wrong? Thanks for assistance.

You have $$\sum_{k=0}^{\infty} ka^k = \dfrac{a}{(1-a)^2}$$ Differentiating with respect to $a$ gives us $$\sum_{k=0}^{\infty} k^2 a^{k-1} = \dfrac{(1-a)^2 - a \times 2 \times (a-1)}{(1-a)^4} = \dfrac{1-a + 2a}{(1-a)^3} = \dfrac{a-1+2}{(1-a)^3}\\ = \dfrac2{(1-a)^3} - \dfrac1{(1-a)^2}$$

A non-linear maximisation We know that $x+y=3$ where x and y are positive real numbers. How can one find the maximum value of $x^2y$? Is it $4,3\sqrt{2}, 9/4$ or $2$?

By AM-GM $$\sqrt[3]{2x^2y} \leq \frac{x+x+2y}{3}=2 $$ with equality if and only if $x=x=2y$. Second solution This one is more complicated, and artificial (since I needed to know the max)$. $$x^2y=3x^2-x^3=-4+3x^2-x^3+4=4- (x-2)^2(x+1)\leq 4$$ since $(x-2)^2(x+1) \geq 0$.

how many number like $119$ How many 3-digits number has this property like $119$: $119$ divided by $2$ the remainder is $1$ 119 divided by $3$ the remainderis $2 $ $119$ divided by $4$ the remainder is $3$ $119$ divided by $5$ the remainder is $4$ $119$ divided by $6$ the remainder is $5$

You seek numbers which, when divided by $k$ (for $k=2,3,4,5,6$) gives a remainder of $k-1$. Thus the numbers you seek are precisely those which are one less than a multiple of $k$ for each of these values of $k$. To find all such numbers, consider the lowest common multiple of $2$, $3$, $4$, $5$ and $6$, and count how many multiples of this have three digits.

$ \displaystyle\lim_{n\to\infty}\frac{1}{\sqrt{n^3+1}}+\frac{2}{\sqrt{n^3+2}}+\cdots+\frac{n}{\sqrt{n^3+n}}$ $$ \ X_n=\frac{1}{\sqrt{n^3+1}}+\frac{2}{\sqrt{n^3+2}}+\cdots+\frac{n}{\sqrt{n^3+n}}$$ Find $\displaystyle\lim_{n\to\infty} X_n$ using the squeeze theorem I tried this approach: $$ \frac{1}{\sqrt{n^3+1}}\le\frac{1}{\sqrt{n^3+1}}<\frac{n}{\sqrt{n^3+1}} $$ $$ \frac{1}{\sqrt{n^3+1}}<\frac{2}{\sqrt{n^3+2}}<\frac{n}{\sqrt{n^3+1}}$$ $$\vdots$$ $$\frac{1}{\sqrt{n^3+1}}<\frac{n}{\sqrt{n^3+n}}<\frac{n}{\sqrt{n^3+1}}$$ Adding this inequalities: $$\frac{n}{\sqrt{n^3+1}}\leq X_n<\frac{n^2}{\sqrt{n^3+1}}$$ And this doesn't help me much. How should i proced?

Hint: use $\frac{i}{\sqrt{n^3+n}} \le \frac{i}{\sqrt{n^3+i}} \le \frac{i}{\sqrt{n^3+1}}$. I even think the Squeeze theorem can be avoided.

If both roots of the Quadratic Equation are similar then prove that If both roots of the equation $(a-b)x^2+(b-c)x+(c-a)=0$ are equal, prove that $2a=b+c$. Things should be known: * *Roots of a Quadratic Equations can be identified by: The roots can be figured out by: $$\frac{-b \pm \sqrt{d}}{2a},$$ where $$d=b^2-4ac.$$ *When the equation has equal roots, then $d=b^2-4ac=0$. *That means $d=(b-c)^2-4(a-b)(c-a)=0$

As the two roots are equal the discriminant must be equal to $0$. $$(b-c)^2-4(a-b)(c-a)=(a-b+c-a)^2-4(a-b)(c-a)=\{a-b-(c-a)\}^2=(2a-b-c)^2=0 \iff 2a-b-c=0$$ Alternatively, solving for $x,$ we get $$x=\frac{-(b-c)\pm\sqrt{(b-c)^2-4(a-b)(c-a)}}{2(a-b)}=\frac{c-b\pm(2a-b-c)}{2(a-b)}=\frac{c-a}{a-b}, 1$$ as $a-b\ne 0$ as $a=b$ would make the equation linear. So, $$\frac{c-a}{a-b}=1\implies c-a=a-b\implies 2a=b+c$$

Find intersection of two 3D lines I have two lines $(5,5,4) (10,10,6)$ and $(5,5,5) (10,10,3)$ with same $x$, $y$ and difference in $z$ values. Please some body tell me how can I find the intersection of these lines. EDIT: By using the answer given by coffemath I would able to find the intersection point for the above given points. But I'm getting a problem for $(6,8,4) (12,15,4)$ and $(6,8,2) (12,15,6)$. I'm unable to calculate the common point for these points as it is resulting in Zero. Any ideas to resolve this? Thanks, Kumar.

The direction numbers $(a,b,c)$ for a line in space may be obtained from two points on the line by subtracting corresponding coordinates. Note that $(a,b,c)$ may be rescaled by multiplying through by any nonzero constant. The first line has direction numbers $(5,5,2)$ while the second line has direction numbers $(5,5,-2).$ Once one has direction numbers $(a,b,c)$, one can use either given point of the line to obtain the symmetric form of its equation as $$\frac{x-x_0}{a}=\frac{y-y_0}{b}=\frac{z-z_0}{c}.$$ Note that if one or two of $a,b,c$ are $0$ the equation for that variable is obtained by setting the top to zero. That doesn't happen in your case. Using the given point $(5,5,4)$ of the first line gives its symmetric equation as $$\frac{x-5}{5}=\frac{y-5}{5}=\frac{z-4}{2}.$$ And using the given point $(5,5,5)$ of the second line gives its symmetric form $$\frac{x-5}{5}=\frac{y-5}{5}=\frac{z-5}{-2}.$$ Now if the point $(x,y,z)$ is on both lines, the equation $$\frac{z-4}{2}=\frac{z-5}{-2}$$ gives $z=9/2$, so that the common value for the fractions is $(9/2-4)/2=1/4$. This value is then used to find $x$ and $y$. In this example the equations are both of the same form $(t-5)/5=1/4$ with solution $t=25/4$. So we may conclude the intersection point is $$(25/4,\ 25/4,\ 9/2).$$ ADDED CASE: The OP has asked about another case, which illustrates what happens when one of the direction numbers of one of the two lines is $0$. Line 1: points $(6,8,4),\ (12,15,4);$ directions $(6,7,0)$, "equation" $$\frac{x-6}{6}=\frac{y-8}{7}=\frac{z-4}{0},$$ where I put equation in quotes because of the division by zero, and as noted the zero denominator of the last fraction means $z=4$ (so $z$ is constant on line 1). Line 2: points $(6,8,2),\ (12,15,6);$ directions $(6,7,4)$, equation $$\frac{x-6}{6}=\frac{y-8}{8}=\frac{z-2}{4}.$$ Now since we know $z=4$ from line 1 equation, we can use $z=4$ in $(z-2)/4$ of line 2 equation, to get the common fraction value of $(4-2)/4=1/2$. Then from either line, $(x-6)/6=1/2$ so $x=9$, and $(y-8)/7=1/2$ so $y=23/2.$ So for these lines the intersection point is $(9,\ 23/2,\ 4).$ It should be pointed out that two lines in space generally do not intersect, they can be parallel or "skew". This would come out as some contradictory values in the above mechanical procedure.

Calculate value of expression $(\sin^6 x+\cos^6 x)/(\sin^4 x+\cos^4 x)$ Calculate the value of expresion: $$ E(x)=\frac{\sin^6 x+\cos^6 x}{\sin^4 x+\cos^4 x} $$ for $\tan(x) = 2$. Here is the solution but I don't know why $\sin^6 x + \cos^6 x = ( \cos^6 x(\tan^6 x + 1) )$, see this: Can you explain to me why they solved it like that?

You can factor the term $\cos^6(x)$ from $\sin^6(x)+\cos^6(x)$ in the numerator to find: $$\cos^6(x)\left(\frac{\sin^6(x)}{\cos^6(x)}+1\right)=\cos^6(x)\left(\tan^6(x)+1\right)$$ and factor $\cos^4(x)$ from the denominator to find: $$\cos^4(x)\left(\frac{\sin^4(x)}{\cos^4(x)}+1\right)=\cos^4(x)\left(\tan^4(x)+1\right)$$ and so $$E(x)=\frac{\cos^6(x)\left(\tan^6(x)+1\right)}{\cos^4(x)\left(\tan^4(x)+1\right)}$$ So if we assume $x\neq (2k+1)\pi/2$ then we can simplify the term $\cos^4(x)$ and find: $$E(x)=\frac{\cos^2(x)\left(\tan^6(x)+1\right)}{\left(\tan^4(x)+1\right)}$$ You know that $\cos^2(x)=\frac{1}{1+\tan^2(x)}$ as an trigonometric identity. So set it in $E(x)$ and find the value.

Check convergence $\sum\limits_{n=1}^\infty\left(\sqrt{1+\frac{7}{n^{2}}}-\sqrt[3]{1-\frac{8}{n^{2}}+\frac{1}{n^{3}}}\right)$ Please help me to check convergence of $$\sum_{n=1}^{\infty}\left(\sqrt{1+\frac{7}{n^{2}}}-\sqrt[3]{1-\frac{8}{n^{2}}+\frac{1}{n^{3}}}\right)$$

(Presumably with tools from Calc I...) Using the conjugate identities $$ \sqrt{a}-1=\frac{a-1}{1+\sqrt{a}},\qquad1-\sqrt[3]{b}=\frac{1-b}{1+\sqrt[3]{b}+\sqrt[3]{b^2}}, $$ one gets $$ \sqrt{1+\frac{7}{n^{2}}}-\sqrt[3]{1-\frac{8}{n^{2}}+\frac{1}{n^{3}}}=x_n+y_n, $$ with $$ x_n=\sqrt{1+a_n}-1=\frac{a_n}{1+\sqrt{1+a_n}},\qquad a_n=\frac{7}{n^2}, $$ and $$ y_n=1-\sqrt[3]{1-b_n}=\frac{b_n}{1+\sqrt[3]{1-b_n}+(\sqrt[3]{1-b_n})^2},\qquad b_n=\frac{8}{n^{2}}-\frac{1}{n^{3}}. $$ The rest is easy: when $n\to\infty$, the denominator of $x_n$ is at least $2$ hence $x_n\leqslant\frac12a_n$, likewise the denominator of $y_n$ is at least $1$ hence $y_n\leqslant b_n$. Thus, $$ 0\leqslant x_n+y_n\leqslant\tfrac12a_n+b_n\leqslant\frac{12}{n^2}, $$ since $a_n=\frac7{n^2}$ and $b_n\leqslant\frac8{n^2}$. The series $\sum\limits_n\frac1{n^2}$ converges, hence all this proves that the series $\sum\limits_nx_n$ converges (absolutely).

If x,y,z are positive reals, then the minimum value of $x^2+8y^2+27z^2$ where $\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=1$ is what If $x,y, z$ are positive reals, then the minimum value of $x^2+8y^2+27z^2$ where $\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=1$ is what? $108$ , $216$ , $405$ , $1048$

As $x,y,z$ are +ve real, we can set $\frac 1x=\sin^2A,\frac1y+\frac1z=\cos^2A$ again, $\frac1{y\cos^2A}+\frac1{z\cos^2A}=1$ we can put $\frac1{y\cos^2A}=\cos^2B, \frac1{z\cos^2A}=\sin^2B\implies y=\cos^{-2}A\cos^{-2}B,z=\cos^{-2}A\sin^{-2}B$ So, $$x^2+8y^2+27z^2=\sin^{-4}A+\cos^{-4}A(8\cos^{-4}B+27\sin^{-4}B)$$ We need $8\cos^{-4}B+27\sin^{-4}B$ to be minimum for the minimum value of $x^2+8y^2+27z^2$ Let $F(B)=p^3\cos^{-4}B+q^3\sin^{-4}B$ where $p,q$ are positive real numbers. then $F'(B)=p^3(-4)\cos^{-5}B(-\sin B)+q^3(-4)\sin^{-5}B(\cos B)$ for the extreme values of $F(B),F'(B)=0\implies (p\sin^2B)^3=(q\cos^2B)^3\implies \frac{\sin^2B}q=\frac{\cos^2B}p=\frac1{p+q}$ Observe that $F(B)$ can not have any finite maximum value. So, $F(B)_{min}=\frac{p^3}{\left(\frac p{p+q}\right)^2}+\frac{q^3}{\left(\frac q{p+q}\right)^2}=(p+q)^3$ So, the minimum value of $8\cos^{-4}B+27\sin^{-4}B$ is $(2+3)^3=125$ (Putting $p=2,q=3$) So, $$x^2+8y^2+27z^2=\sin^{-4}A+\cos^{-4}A(8\cos^{-4}B+27\sin^{-4}B)\ge \sin^{-4}A+125\cos^{-4}A\ge (1+5)^3=216$$ (Putting $p=1,q=5$)

How to solve $|x-5|=|2x+6|-1$? $|x-5|=|2x+6|-1$. The answer is $0$ or $-12$, but how would I solve it by algebraically solving it as opposed to sketching a graph? $|x-5|=|2x+6|-1\\ (|x-5|)^2=(|2x+6|-1)^2\\ ...\\ 9x^4+204x^3+1188x^2+720x=0?$

Consider different cases: Case 1: $x>5$ In this case, both $x-5$ and $2x+6$ are positive, and you can resolve the absolute values positively. hence $$ x-5=2x+6-1 \Rightarrow x = -10, $$ which is not compatible with the assumption that $x>5$, hence no solution so far. Case 2: $-3<x\leq5$ In this case, $x-5$ is negative, while $2x+6$ is still positive, so you get $$ -(x-5)=2x+6-1\Rightarrow x=0; $$ Since $0\in[-3,5]$, this is our first solution. Case 3: $x\leq-3$ In this final case, the arguments of both absolute values are negative and the equation simplifies to $$ -(x-5) = -(2x+6)-1 \Rightarrow x = -12, $$ in agreement with your solution by inspection of the graph.

$x^4 + y^4 = z^2$ $x, y, z \in \mathbb{N}$, $\gcd(x, y) = 1$ prove that $x^4 + y^4 = z^2$ has no solutions. It is true even without $\gcd(x, y) = 1$, but it is easy to see that $\gcd(x, y)$ must be $1$

This has been completely revised to match the intended question. The proof is by showing that there is no minimal positive solution, i.e., by infinite descent. It’s from some old notes; I’ve no idea where I cribbed it from in the first place. Suppose that $x^4+y^4=z^2$, where $z$ is the smallest positive integer for which there is a solution in positive integers. Then $(x^2,y^2,z)$ is a primitive Pythagorean triple, so there are relatively prime integers $m,n$ with $m>n$ such that $x^2=m^2-n^2,y^2=2mn$, and $z=m^2+n^2$. Since $2mn=y^2$, one of $m$ and $n$ is an odd square, and the other is twice a square. In particular, one is odd, and one is even. Now $x^2+n^2=m^2$, and $\gcd(x,n)=1$ (since $m$ and $n$ are relatively prime), so $(x,n,m)$ is a primitive Pythagorean triple, and it must be $n$ that is even: there must be integers $a$ and $b$ such that $a>b$, $a$ and $b$ are relatively prime, $x=a^2-b^2$, $n=2ab$, and $m=a^2+b^2$. It must be $m$ that is the odd square, so there are integers $r$ and $s$ such that $m=r^2$ and $n=2s^2$. Now $2s^2=n=2ab$, so $s^2=ab$, and we must have $a=c^2$ and $b=d^2$ for some integers $c$ and $d$, since $\gcd(a,b)=1$. The equation $m=a^2+b^2$ can then be written $r^2=c^4+d^4$.