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probability of sum of two integers less than an integer Two integers [not necessarily distinct] are chosen from the set {1,2,3,...,n}. What is the probability that their sum is <=k? My approach is as follows. Let a and b be two integers. First we calculate the probability of the sum of a+b being equal to x [1<=x<=n]. WLOG let a be chosen first. For b= x-a to be positive, we must have 1<=a < x. This gives (x-1) possible values for a out of total n possible values. Probability of valid selection of a= (x-1)/n. For each valid selection of a, we have one and only one possible value of b. Only 1 value of b is then valid out of total n possible values. Thus probability of valid selection of b= 1/n. Thus probability of (a+b= x) = (x-1)/n(n-1). Now probability of (a+b<=k) = Probability of (a+b= 2) + probability of (a+b= 3) + ... + probability of (a+b= k) = {1+2+3+4+5+...+(k-1)}n(n-1) = k(k-1)/n(n-1). Can anybody please check if my approach is correct here?

Notice if $k\le 1$ the probability is $0$, and if $k\ge 2n$ the probability is $1$, so let's assume $2\le k\le 2n-1$. For some $i$ satisfying $2\le i\le 2n-1$, how many ways can we choose $2$ numbers to add up to $i$? If $i\le n+1$, there are $i-1$ ways. If $i\ge n+2$, there are $2n-i+1$ ways. Now, suppose $k\le n+1$, so by summing we find: $$\sum_{i=2}^{k}i-1=\frac{k(k-1)}{2}$$ If $k\ge n+2$, if we sum from $i=2$ to $i=n+1$ we get $\frac{(n+1)n}{2}$, and then from $n+1$ to $k$ we get: $$\sum_{i=n+2}^k2n-i+1=\frac{1}{2}(3n-k)(k-n-1)$$ Adding the amount for $i\le n+1$ we get: $$2kn-\frac{k^2}{2}+\frac{k}{2}-n^2-n$$ Since there are $n^2$ choices altogether, we arrive at the following probabilities: $$\begin{cases}\frac{k(k-1)}{2n^2}&1\le k\le n+1\\\frac{4kn-k^2+k-2n^2-2n}{2n^2}&n+2\le k\le 2n\end{cases}$$

Prove $x^2+y^2+z^2 \ge 14$ with constraints Let $0<x\le y \le z,\ z\ge 3,\ y+z \ge 5,\ x+y+z = 6.$ Prove the inequalities: $I)\ x^2 + y^2 + z^2 \ge 14$ $II)\ \sqrt x + \sqrt y + \sqrt z \le 1 + \sqrt 2 + \sqrt 3$ My teacher said the method that can solve problem I can be use to solve problem II. But I don't know what method that my teacher talking about, so the hint is useless, please help me. Thanks

Hint: $$x^2+y^2+z^2 \ge 14 = 1^2+2^2+3^2\iff (x-1)(x+1)+(y-2)(y+2)+(z-3)(z+3) \ge 0$$ $$\iff (z-3)[(z+3)-(y+2)] + (y+z-5)[(y+2)-(x+1)] + (a+b+c-6)(a+1) \ge 0$$ (alway true)

How to prove $(a-b)^3 + (b-c)^3 + (c-a)^3 -3(a-b)(b-c)(c-a) = 0$ without calculations I read somewhere that I can prove this identity below with abstract algebra in a simpler and faster way without any calculations, is that true or am I wrong? $$(a-b)^3 + (b-c)^3 + (c-a)^3 -3(a-b)(b-c)(c-a) = 0$$ Thanks

To Prove: $$(a-b)^3 + (b-c)^3 + (c-a)^3 =3(a-b)(b-c)(c-a)$$ we know, $x^3 + y^3 = (x + y)(x^2 - xy + y^2)$ so, $$(a-b)^3 + (b-c)^3 = (a -c)((a-b)^2 - (a-b)(b-c) + (b-c)^2)$$ now, $$(a-b)^3 + (b-c)^3 + (c-a)^3 = (a -c)((a-b)^2 - (a-b)(b-c) + (b-c)^2) + (c-a)^3 = (c-a)(-(a-b)^2 + (a-b)(b-c)- (b-c)^2 +(c-a)^2)$$ now, $(c-a)^2 - (a-b)^2 = (c-a+a-b)(c-a-a+b) = (c-b)(c-2a+b)$ the expression becomes, $$(c-a)((c-b)(c-2a+b) + (b-c)(a-2b+c)) = (c-a)(b-c)(-c+2a-b+a-2b+c)=3(c-a)(b-c)(a-b)$$ Hence proved

Algebra simplification in mathematical induction . I was proving some mathematical induction problems and came through an algebra expression that shows as follows: $$\frac{k(k+1)(2k+1)}{6} + (k + 1)^2$$ The final answer is supposed to be: $$\frac{(k+1)(k+2)(2k+3)}{6}$$ I walked through every possible expansion; I combine like terms, simplify, factor, but never arrived at the answer. Could someone explain the steps?

First, let's write the expression as a sum of fractions with a common denominator. $$\dfrac{k(k+1)(2k+1)}{6} + (k + 1)^2 = \dfrac{k(k+1)(2k+1)}{6} + \dfrac{6(k+1)^2}{6}\tag{1}$$ Now expand $6(k+1)^2 = 6k^2 + 12k + 6\tag{2}$ and expand $k(k+1)(2k+1) = k(2k^2 + 3k + 1) = 2k^3 + 3k^2 + k\tag{3}$ So now, $(1)$ becomes $$\dfrac{k(k+1)(2k+1)}{6} + \dfrac{6(k+1)^2}{6} = \dfrac{(2k^3 + 3k^2 + k) + (6k^2 +12 k + 6)}{6} $$ $$= \dfrac{\color{blue}{\bf 2k^3 + 9k^2 +13k + 6}}{6}\tag{4}$$ We can factor the numerator in $(4)$, or we can expand the numerator of our "goal"... $$\frac{(k+1)(k+2)(2k+3)}{6} = \dfrac{\color{blue}{\bf 2k^3 + 9k^2 + 13k + 6}}{6}\tag{goal}$$

Evaluate $\int_0^\infty \frac{\log(1+x^3)}{(1+x^2)^2}dx$ and $\int_0^\infty \frac{\log(1+x^4)}{(1+x^2)^2}dx$ Background: Evaluation of $\int_0^\infty \frac{\log(1+x^2)}{(1+x^2)^2}dx$ We can prove using the Beta-Function identity that $$\int_0^\infty \frac{1}{(1+x^2)^\lambda}dx=\sqrt{\pi}\frac{\Gamma \left(\lambda-\frac{1}{2} \right)}{\Gamma(\lambda)} \quad \lambda>\frac{1}{2}$$ Differentiating the above equation with respect to $\lambda$, we obtain an expression involving the Digamma Function $\psi_0(z)$. $$\int_0^\infty \frac{\log(1+x^2)}{(1+x^2)^\lambda}dx = \sqrt{\pi}\frac{\Gamma \left(\lambda-\frac{1}{2} \right)}{\Gamma(\lambda)} \left(\psi_0(\lambda)-\psi_0 \left( \lambda-\frac{1}{2}\right) \right)$$ Putting $\lambda=2$, we get $$\int_0^\infty \frac{\log(1+x^2)}{(1+x^2)^2}dx = -\frac{\pi}{4}+\frac{\pi}{2}\log(2)$$ Question: But, does anybody know how to evaluate $\displaystyle \int_0^\infty \frac{\log(1+x^3)}{(1+x^2)^2}dx$ and $\displaystyle \int_0^\infty \frac{\log(1+x^4)}{(1+x^2)^2}dx$? Mathematica gives the values * *$\displaystyle \int_0^\infty \frac{\log(1+x^3)}{(1+x^2)^2}dx = -\frac{G}{6}+\pi \left(-\frac{3}{8}+\frac{1}{8}\log(2)+\frac{1}{3}\log \left(2+\sqrt{3} \right) \right)$ *$\displaystyle \int_0^\infty \frac{\log(1+x^4)}{(1+x^2)^2}dx = -\frac{\pi}{2}+\frac{\pi \log \left( 6+4\sqrt{2}\right)}{4}$ Here, $G$ denotes the Catalan's Constant. Initially, my approach was to find closed forms for $$\int_0^\infty \frac{1}{(1+x^2)^2(1+x^3)^\lambda}dx \ \ , \int_0^\infty \frac{1}{(1+x^2)^2(1+x^4)^\lambda}dx$$ and then differentiate them with respect to $\lambda$ but it didn't prove to be of any help. Please help me prove these two results.

I hope it is not too late. Define \begin{eqnarray} I(a)=\int_0^\infty\frac{\log(1+ax^4)}{(1+x^2)^2}dx. \end{eqnarray} Then \begin{eqnarray} I'(a)&=&\int_0^\infty \frac{x^4}{(1+ax^4)(1+x^2)^2}dx\\ &=&\frac{1}{(1+a)^2}\int_0^\infty\left(-\frac{2}{1+x^2}+\frac{1+a}{(1+x^2)^2}+\frac{1-a+2ax^2}{1+a x^4}\right)dx\\ &=&\frac{1}{(1+a)^2}\left(-\pi+\frac{1}{4}(1+a)\pi+\frac{(1-a)\pi}{2\sqrt2a^{1/4}}+\frac{\pi a^{1/4}}{\sqrt2}\right)\\ &=&\frac{1}{4(1+a)^2}\left(a-3+\frac{\sqrt2(1-a)}{a^{1/4}}+2\sqrt2 a^{1/4}\right). \end{eqnarray} and hence \begin{eqnarray} I(1)&=&\int_0^1\frac{1}{4(1+a)^2}\left(a-3+\frac{\sqrt2(1-a)}{a^{1/4}}+2\sqrt2 a^{1/4}\right)da\\ &=&-\frac{\pi}{2}+\frac{1}{4}\log(6+4\sqrt2). \end{eqnarray} For the other integral, we can do the same thing to define $$ J(a)=\int_0^\infty\frac{\log(1+ax^3)}{(1+x^2)^2}dx. $$ The calculation is similar and more complicated and here I omit the detail.

A recursive formula for $a_n$ = $\int_0^{\pi/2} \sin^{2n}(x)dx$, namely $a_n = \frac{2n-1}{2n} a_{n-1}$ Where does the $\frac{2n-1}{2n}$ come from? I've tried using integration by parts and got $\int \sin^{2n}(x)dx = \frac {\cos^3 x}{3} +\cos x +C$, which doesn't have any connection with $\frac{2n-1}{2n}$. Here's my derivation of $\int \sin^{2n}(x)dx = \frac {\cos^3 x}{3} +\cos x +C$: $\sin^{2n+1}xdx=\int(1-\cos^2x)\sin xdx=\int -(1-u^2)du=\int(u^2-1)du=\frac{u^3}{3}+u+C=\frac{\cos^3x}{3}+\cos x +C$ where $u=\cos x$;$du=-\sin x dx$ Credits to Xiang, Z. for the question.

Given the identity $$\int \sin^n x dx = - \frac{\sin^{n-1} x \cos x}{n}+\frac{n-1}{n}\int \sin^{n-2} xdx$$ plugging in $2n$ yields $$\int \sin^{2n} x dx = - \frac{\sin^{2n-1} x \cos x}{2n}+\frac{2n-1}{2n}\int \sin^{2n-2} xdx$$ Since $$\int_0^{\pi/2} \sin^{2n} x dx = - \frac{\sin^{2n-1} x \cos x}{2n}|_0^{\pi/2}+\frac{2n-1}{2n}\int_0^{\pi/2} \sin^{2n-2} xdx$$ and $\frac{\sin^{2n-1} x \cos x}{2n}|_0^{\pi/2}=0$ for $n \ge 1$, we get $$\int_0^{\pi/2} \sin^{2n} x dx = \frac{2n-1}{2n}\int_0^{\pi/2} \sin^{2n-2} xdx$$ (We only care about $n \ge 1$ because in the original question, $a_0=\frac{\pi}{2}$ is given and only integer values of n with $n \ge 1$ need to satisfy $a_n=\frac{2n-1}{2n}a_{n-1}$.)

How to find out X in a trinomial How can I find out what X equals in this? $$x^2 - 2x - 3 = 117$$ How would I get started? I'm truly stuck.

Hint:1.$$ax^2 +bx +c=0 \to D=b^2-4ac\ge0 $$$$\ \to\begin{cases} \color{green}{x_1=\frac{-b+\sqrt{D}}{2a}} \\ \color{red}{x_2=\frac{-b-\sqrt{D}}{2a}} \\ \end{cases} $$$$$$ 2. $$x^2 +bx +c=(x+\frac{b}{2})^2=\frac{b^2}{4}-c\ge0\quad \text{then} x=\pm\sqrt{\frac{b^2}{4}-c}-\frac{b}{2}$$ 3. find $x_1$and $x_2 $by solving following system \begin{cases} x_1+x_2=\frac{-b}{a} \\ x_1x_2=\frac{c}{a} \\ \end{cases}

Limit of two variables, proposed solution check: $\lim_{(x,y)\to(0,0)}\frac{xy}{\sqrt{x^2+y^2}}$ Does this solution make sense, The limit in question: $$ \lim_{(x,y)\to(0,0)}\frac{xy}{\sqrt{x^2+y^2}} $$ My solution is this: Suppose, $$ \sqrt{x^2+y^2} < \delta $$ therefore $$xy<\delta^2$$ So by the Squeeze Theorem the limit exists since $$\frac{xy}{\sqrt{x^2+y^2}}<\frac{\delta^2}{\delta}=\delta$$ Is this sufficient?

Here's a more direct solution. We know $|x|,|y|\le\sqrt{x^2+y^2}$, so if $\sqrt{x^2+y^2}<\delta$, then $$\left|\frac{xy}{\sqrt{x^2+y^2}}\right|\le\frac{\big(\sqrt{x^2+y^2}\big)^2}{\sqrt{x^2+y^2}}=\sqrt{x^2+y^2}<\delta.$$

Distinguishable telephone poles being painted Each of n (distinguishable) telephone poles is painted red, white, blue or yellow. An odd number are painted blue and an even number yellow. In how many ways can this be done? Can some give me a hint how to approach this problem?

Consider the generating function given by $( R + W + B + Y )^n$ Without restriction, the sum of all coefficients would give the number of ways to paint the distinguishable posts in any of the 4 colors. We substitute $R=W=B=Y = 1$ to find this sum, and it is (unsurprisingly) $4^n$. Since there are no restrictions on $R$ and $W$, we may replace them with $1$, and still consider the coefficients. If we have the restriction that we are only interested in cases where the degree of $B$ is odd (ignore $Y$ for now), then since $ \frac{1^k - (-1)^k}{2} = \begin{cases} 1 & k \equiv 1 \pmod{2} \\ 0 & k \equiv 0 \pmod{2} \\ \end{cases}$ the sum of the coefficients when the degree of $B$ is odd, is just the sum of the coefficients of $ \frac{ (R + W + 1 + Y) ^n - ( R + W + (-1) + Y) ^n} { 2} $. Substituting in $R=W=Y=1$, we get that the number of ways is $$ \frac{ (1 + 1 + 1 + 1)^n - (1 + 1 + (-1) +1)^n}{2} = \frac {4^n - 2^n} {2}$$ Now, how do we add in the restriction that the degree of $Y$ is even? Observe that since $ \frac{1^k + (-1)^k}{2} = \begin{cases} 1 & k \equiv 0 \pmod{2} \\ 0 & k \equiv 1 \pmod{2} \\ \end{cases}$ the sum of the coefficients when the degree of $B$ is odd and Y is even, is just the sum of the coefficients of $$ \frac{ \frac{ (R + W + 1 + 1) ^n - ( R + W + (-1) + 1) ^n} { 2} + \frac{ (R + W + 1 + (-1)) ^n - ( R + W + (-1) + (-1)) ^n} { 2} } { 2} $$ Now substituting in $R=W=1$, we get $\frac{ 4^n - 2 ^n + 2^n - 0^n } { 4} = 4^{n-1}$

If $x+{1\over x} = r $ then what is $x^3+{1\over x^3}$? If $$x+{1\over x} = r $$ then what is $$x^3+{1\over x^3}$$ Options: $(a) 3,$ $(b) 3r,$ $(c)r,$ $(d) 0$

$\displaystyle r^3=\left(x+\frac{1}{x}\right)^3=x^3+\frac{1}{x^3}+3(x)\frac{1}{x}\left(x+\frac{1}{x}\right)=x^3+\frac{1}{x^3}+3r$ $\displaystyle \Rightarrow r^3-3r=x^3+\frac{1}{x^3}$ Your options are incorrect.For a quick counter eg. you can take $x=1/2$ to get $r=\frac{5}{2}$ and $x^3+\frac{1}{x^3}=\frac{65}{8}$ but none of the options result in $65/8$

Show that $\frac{x}{1-x}\cdot\frac{y}{1-y}\cdot\frac{z}{1-z} \ge 8$. If $x,y,z$ are positive proper fractions satisfying $x+y+z=2$, prove that $$\dfrac{x}{1-x}\cdot\dfrac{y}{1-y}\cdot\dfrac{z}{1-z}\ge 8$$ Applying $GM \ge HM$, I get $$\left[\dfrac{x}{1-x}\cdot\dfrac{y}{1-y}\cdot\dfrac{z}{1-z}\right]^{1/3}\ge \dfrac{3}{\frac 1x-1+\frac 1y-1+\frac 1z-1}\\=\dfrac{3}{\frac 1x+\frac 1y+\frac 1z-3}$$ Then how to proceed. Please help.

Write $(1-x)=a, (1-y)=b \text { and} (1-z)=c$ $x=2-(y+z)=b+c$ $y=2-(z+x)=a+c$ $z=2-(x+y)=a+b$ Thus we have the same expression in simpler form: $\dfrac{b+c}{a} \cdot \dfrac{a+c}{b} \cdot \dfrac{a+b}{c}$ Now we have AM-GM: $b+c \ge 2 \sqrt{bc}$ $a+c \ge 2 \sqrt{ac}$ $b+a \ge 2 \sqrt{ba}$ $\dfrac{b+c}{a} \cdot \dfrac{a+c}{b} \cdot \dfrac{a+b}{c} \ge \dfrac{2^3 abc}{abc} =8$, Done.

Show that $(1+x)(1+y)(1+z)\ge 8(1-x)(1-y)(1-z)$. If $x>0,y>0,z>0$ and $x+y+z=1$, prove that $$(1+x)(1+y)(1+z)\ge 8(1-x)(1-y)(1-z).$$ Trial: Here $$(1+x)(1+y)(1+z)\ge 8(1-x)(1-y)(1-z) \\ \implies (1+x)(1+y)(1+z)\ge 8(y+z)(x+z)(x+y)$$ I am unable to solve the problem. Please help.

$$(1+x)(1+y)(1+z) \ge 8(1-x)(1-y)(1-z) \Leftrightarrow $$ $$(2x+y+z)(x+2y+z)(x+y+2z) \ge 8(y+z)(x+z)(x+y)$$ Let $a=x+y, b=x+z, c=y+z$. Then the inequality to prove is $$(a+b)(a+c)(b+c) \ge 8abc \,,$$ Which follows immediately from AM-GM: $$a+b \ge 2 \sqrt{ab}$$ $$a+c \ge 2 \sqrt{ac}$$ $$b+c \ge 2 \sqrt{bc}$$ Simplification The solution above can be simplified the following way: By AM-GM $$2\sqrt{(1-y)(1-z)}\le 1-y+1-z=1+x \,.$$ Similarly $$2\sqrt{(1-x)(1-z)}\le 1+y \,.$$ $$2\sqrt{(1-x)(1-y)}\le 1+z \,.$$

Function problem Show that function $f(x) =\frac{x^2+2x+c}{x^2+4x+3c}$ attains any real value if $0 < c \leq 1$ Problem : Show that function $f(x)=\dfrac{x^2+2x+c}{x^2+4x+3c}$ attains any real value if $0 < c \leq 1$ My approach : Let the given function $f(x) =\dfrac{x^2+2x+c}{x^2+4x+3c} = t $ where $t$ is any arbitrary constant. $\Rightarrow (t-1)x^2+2(2t-1)x+c(3t-1)=0$ The argument $x$ must be real, therefore $(2t-1)^2-(t-1)(3tc-c) \geq 0$. Now how to proceed further? Please guide. Thanks.

$(2t-1)^2-(t-1)(3tc-c) \geq 0\implies 4t^2+1-4t-(3t^2c-4tc+c)\geq 0\implies t^2(4-3c)+4(c-1)t+(1-c)\geq 0$ Now a quadratic polynomial $\geq 0$ $\forall t\in \Bbb R$ iff coefficient of second power of variable is positive and Discriminant $\leq 0$ which gives $4-3c>0\implies c<\frac{4}{3}$ and $D=16(c-1)^2+4(4-3c)(c-1)\leq 0\implies 4(c-1)(4c-4+4-3c)\leq 0\implies 4(c-1)c\leq 0\implies 0\leq c\leq 1$ So $c<\frac{4}{3}$ and $0\leq c\leq 1\implies 0\leq c\leq 1$

Evaluating $\int_0^\infty \frac{dx}{1+x^4}$. Can anyone give me a hint to evaluate this integral? $$\int_0^\infty \frac{dx}{1+x^4}$$ I know it will involve the gamma function, but how?

Following is a computation that uses Gamma function: For any real number $k > 1$, let $I_k$ be the integral: $$I_k = \int_0^\infty \frac{dx}{1+x^k}$$ Consider two steps in changing the variable. First by $y = x^k$ and then by $z = \frac{y}{1+y}$. Notice: $$\frac{1}{1+y} = 1 - z,\quad y = \frac{z}{1-z}\quad\text{ and }\quad dy = \frac{dz}{(1-z)^2}$$ We get: $$\begin{align} I_k = & \int_0^{\infty}\frac{1}{1 + y} d y^{\frac{1}{k}} = \frac{1}{k}\int_0^\infty \frac{1}{1+y}y^{\frac{1}{k}-1} dy\\ = & \frac{1}{k}\int_0^\infty (1-z) \left(\frac{z}{1-z}\right)^{\frac{1}{k}-1} \frac{dz}{(1-z)^2} = \frac{1}{k}\int_0^\infty z^{\frac{1}{k}-1} (1-z)^{-\frac{1}{k}} dz\\ = & \frac{1}{k} \frac{\Gamma(\frac{1}{k})\Gamma(1 - \frac{1}{k})}{\Gamma(1)} = \frac{\pi}{k \sin\frac{\pi}{k}} \end{align}$$ For $k = 4$, we get: $$I_4 = \int_0^\infty \frac{dx}{1+x^4} = \frac{\pi}{4\sin \frac{\pi}{4}} = \frac{\pi}{2\sqrt{2}}$$

Need help in proving that $\frac{\sin\theta - \cos\theta + 1}{\sin\theta + \cos\theta - 1} = \frac 1{\sec\theta - \tan\theta}$ We need to prove that $$\dfrac{\sin\theta - \cos\theta + 1}{\sin\theta + \cos\theta - 1} = \frac 1{\sec\theta - \tan\theta}$$ I have tried and it gets confusing.

$$\frac{\sin\theta-\cos\theta+1}{\sin\theta+\cos\theta-1}$$ $$=\frac{\tan\theta-1+\sec\theta}{\tan\theta+1-\sec\theta}(\text{ dividing the numerator & the denominator by} \cos\theta )$$ $$=\frac{\tan\theta-1+\sec\theta}{\tan\theta-\sec\theta+(\sec^2\theta-\tan^2\theta)} (\text{ putting } 1=\sec^2\theta-\tan^2\theta) $$ $$=\frac{\tan\theta+\sec\theta-1}{\tan\theta-\sec\theta-(\tan\theta-\sec\theta)(\tan\theta+\sec\theta)}$$ $$=\frac{\tan\theta+\sec\theta-1}{-(\tan\theta-\sec\theta)(\tan\theta+\sec\theta-1)}$$ $$=\frac1{\sec\theta-\tan\theta}$$ Alternatively using Double-angle formula by putting $\tan\frac\theta2=t,$ $$\text{ LHS= }\frac{\sin\theta-\cos\theta+1}{\sin\theta+\cos\theta-1}=\frac{\frac{2t}{1+t^2}-\frac{1-t^2}{1+t^2}+1}{\frac{2t}{1+t^2}+\frac{1-t^2}{1+t^2}-1}$$ $$=\frac{2t-(1-t^2)+1+t^2}{2t+(1-t^2)-(1+t^2)} =\frac{2t+2t^2}{2t-2t^2}=\frac{1+t}{1-t}\text{assuming }t\ne0$$ $$\text{ RHS= }\frac1{\sec\theta-\tan\theta}=\frac1{\frac{1+t^2}{1-t^2}-\frac{2t}{1-t^2}}=\frac{1-t^2}{(1-t)^2}=\frac{1+t}{1-t} \text{assuming }t-1\ne0$$

A proposed proof by induction of $1+2+\ldots+n=\frac{n(n+1)}{2}$ Prove: $\displaystyle 1+2+\ldots+n=\frac{n(n+1)}{2}$. Proof When $n=1,1=\displaystyle \frac{1(1+1)}{2}$,equality holds. Suppose when $n=k$, we have $1+2+\dots+k=\frac{k(k+1)}{2}$ When $n = k + 1$: \begin{align} 1+2+\ldots+k+(k+1) &=\frac{k(k+1)}{2}+k+1 =\frac{k(k+1)+2k+2}{2}\\ &=\frac{k^2+3k+2}{2}\\ \text{[step]}&=\displaystyle\frac{(k+1)(k+2)}{2}=\displaystyle\frac{(k+1)((k+1)+1)}{2} \end{align} equality holds. So by induction, the original equality holds $\forall n\in \mathbb{N}$. Question 1: any problems in writing? Question 2: Why [step] happen to equal? i.e., why does $k^2+3k+2=(k+1)(k+2)$ hold?

Q1: No problems, that's the way induction works. Q2: go back one step: $$k(k+1)+2k+2=k(k+1)+2(k+1)=(k+1)(k+2)$$

Calculating 7^7^7^7^7^7^7 mod 100 What is $$\large 7^{7^{7^{7^{7^{7^7}}}}} \pmod{100}$$ I'm not much of a number theorist and I saw this mentioned on the internet somewhere. Should be doable by hand.

Reading the other answers, I realize this is a longer way than necessary, but it gives a more general approach for when things are not as convenient as $7^4\equiv 1\bmod 100$. Note that, for any integer $a$ that is relatively prime to $100$, we have $$a^{40}\equiv 1\bmod 100$$ because $\varphi(100)=40$, and consequently $$a^m\equiv a^n\bmod 100$$ whenever $m\equiv n\bmod 40$. Thus, we need to find $7^{7^{7^{7^{7^{7}}}}}$ modulo $40$. By the Chinese remainder theorem, it is equivalent to know what it is modulo $8$ and modulo $5$. Modulo $8$, we have $7\equiv -1\bmod 8$, and $-1$ to an odd power is going to be $-1$, so we see that $$7^{7^{7^{7^{7^{7}}}}}\equiv (-1)^{7^{7^{7^{7^{7}}}}} \equiv -1\equiv 7\bmod 8.$$ Modulo $5$, we have $7^4\equiv 1\bmod 5$ (again by Euler's theorem), so we need to know $7^{7^{7^{7^{7}}}}\bmod 4$. But $7\equiv -1\bmod 4$, and $7^{7^{7^{7}}}$ is odd, so that $7^{7^{7^{7^{7}}}}\equiv -1\equiv 3\bmod 4$, so that $$7^{7^{7^{7^{7^{7}}}}}\equiv 7^3\equiv 343\equiv 3\bmod 5.$$ Applying the Chinese remainder theorem, we conclude that $$7^{7^{7^{7^{7^{7}}}}}\equiv 23\bmod 40,$$ and hence $$7^{7^{7^{7^{7^{7^{7}}}}}}\equiv 7^{23}\bmod 100.$$ This is tractable by again using the Chinese remainder theorem to find $7^{23}\bmod 4$ and $7^{23}\bmod 25$.

Induction and convergence of an inequality: $\frac{1\cdot3\cdot5\cdots(2n-1)}{2\cdot4\cdot6\cdots(2n)}\leq \frac{1}{\sqrt{2n+1}}$ Problem statement: Prove that $\frac{1*3*5*...*(2n-1)}{2*4*6*...(2n)}\leq \frac{1}{\sqrt{2n+1}}$ and that there exists a limit when $n \to \infty $. , $n\in \mathbb{N}$ My progress LHS is equivalent to $\frac{(2n-1)!}{(2n)!}=\frac{(2n-1)(2n-2)(2n-3)\cdot ....}{(2n)(2n-1)(2n-2)\cdot ....}=\frac{1}{2n}$ So we can rewrite our inequality as: $\frac{1}{2n}\leq \frac{1}{\sqrt{2n+1}}$ Let's use induction: For $n=1$ it is obviously true. Assume $n=k$ is correct and show that $n=k+1$ holds. $\frac{1}{2k+2}\leq \frac{1}{\sqrt{2k+3}}\Leftrightarrow 2k+2\geq\sqrt{2k+3}\Leftrightarrow 4(k+\frac{3}{4})^2-\frac{5}{4}$ after squaring and completing the square. And this does not hold for all $n$ About convergence: Is it not enough to check that $\lim_{n \to \infty}\frac{1}{2n}=\infty$ and conclude that it does not converge?

There's a direct proof to the inequality of $\frac{1}{\sqrt{2n+1}}$, though vadim has improved on the bound. Consider $A = \frac{1}{2} \times \frac{3}{4} \times \ldots \times \frac{2n-1} {2n}$ and $B = \frac{2}{3} \times \frac{4}{5} \times \ldots \times \frac{2n}{2n+1}$. Then $AB = \frac{1}{2n+1}$. Since each term of $A$ is smaller than the corresponding term in $B$, hence $A < B$. Thus $A^2 < AB = \frac{1}{2n+1}$, so $$A < \frac{1}{\sqrt{2n+1}}$$ Of course, the second part that a limit exists follows easily, and is clearly 0.

Integral $\int_0^\frac{\pi}{2} \sin^7x \cos^5x\, dx $ im asked to find the limited integral here but unfortunately im floundering can someone please point me in the right direction? $$\int_0^\frac{\pi}{2} \sin^7x \cos^5x\, dx $$ step 1 brake up sin and cos so that i can use substitution $$\int_0^\frac{\pi}{2} \sin^7(x) \cos^4(x) \cos(x) \, dx$$ step 2 apply trig identity $$\int_0^\frac{\pi}{2} \sin^7x\ (1-\sin^2 x)^2 \, dx$$ step 3 use $u$-substitution $$ \text{let}\,\,\, u= \sin(x)\ du=\cos(x) $$ step 4 apply use substitution $$\int_0^\frac{\pi}{2} u^7 (1-u^2)^2 du $$ step 5 expand and distribute and change limits of integration $$\int_0^1 u^7-2u^9+u^{11}\ du $$ step 6 integrate $$(1^7-2(1)^9+1^{11})-0$$ i would just end up with $1$ however the book answer is $$\frac {1}{120}$$ how can i be so far off?

$$ \begin{aligned} \int_0^{\frac{\pi}{2}} \sin ^7 x \cos ^5 x & \stackrel{x\mapsto\frac{\pi}{2}-x}{=} \int_0^{\frac{\pi}{2}} \cos ^5 x \sin ^7 x d x \\ &=\frac{1}{2} \int_0^{\frac{\pi}{2}} \sin ^5 x\cos ^5 x\left(\sin ^2 x+\cos ^2 x\right) d x \\ &=\frac{1}{64} \int_0^{\frac{\pi}{2}} \sin ^5(2 x)d x \\ &=\frac{1}{128} \int_0^{\frac{\pi}{2}}\left(1-\cos ^2 2 x\right)^2 d(\cos 2 x) \\ &=\frac{1}{128}\left[\cos 2 x-\frac{2 \cos ^3 2 x}{3}+\frac{\cos ^5 2 x}{5}\right]_0^{\frac{\pi}{2}} \\ &=\frac{1}{120} \end{aligned} $$

Triangle integral with vertices Evaluate $$I=\iint\limits_R \sin \left(\frac{x+y}{2}\right)\cos\left(\frac{x-y}{2}\right)\, dA,$$ where $R$ is the triangle with vertices $(0,0),(2,0)$ and $(1,1)$. Hint: use $u=\dfrac{x+y}{2},v=\dfrac{x-y}{2}$. Can anyone help me with this question I am very lost. Please help I know you can make the intergal $\sin(u)\cos(v)$, but then what to do?

$\sin\left(\frac{x+y}{2}\right)\cos\left(\frac{x-y}{2}\right)=\frac{1}{2}\left(\sin x+\sin y\right)$ The line joining $(0,0)$ and $(2,0)$ has an equation $y=0$ and $0\leq x\leq 2$ The second line: $y=-x+2$ The third line: $y=x$ The integral becomes: $$I=\frac{1}{2}\left(\int\limits_{0}^{1}\int\limits_{0}^{x}+\int\limits_{1}^{2}\int\limits_{0}^{2-x}\right)(\sin x+\sin y)dy~dx=\frac{1}{2}\left(\int\limits_{0}^{1}\int\limits_{0}^{x}(\sin x+\sin y)dy~dx+\int\limits_{1}^{2}\int\limits_{0}^{2-x}(\sin x+\sin y)dy~dx\right)=\frac{1}{2}\left(\int\limits_0^1\left[y\sin x-\cos y\right]_0^xdx+\int\limits_1^2\left[y\sin x-\cos y\right]_0^{2-x}dx\right)=\frac{1}{2}\left(\left[\sin x-x\cos x+x-\cos x\right]_0^1+\left[(x-2)\cos x-\sin x+x+\sin(2-x)\right]_1^2\right)$$

Solving for X in a simple matrix equation system. I am trying to solve for X in this simple matrix equation system: $$\begin{bmatrix}7 & 7\\2 & 4\\\end{bmatrix} - X\begin{bmatrix}5 & -1\\6 & -4\\\end{bmatrix} = E $$ where $E$ is the identity matrix. If I multiply $X$ with $\begin{bmatrix}5 & -1\\6 & -4\\\end{bmatrix}$ I get the following system: $$\begin{bmatrix}5x_1 & -1x_2\\6x_3 & -4x_4\\\end{bmatrix}$$ By subtracting this from $\begin{bmatrix}7 & 7\\2 & 4\\\end{bmatrix}$ I get $\begin{bmatrix}7 - 5x_1 & 7 + 1x_2\\2 - 6x_3 & 4 + 4x_4\\\end{bmatrix} = \begin{bmatrix}1 & 0\\0 & 1\\\end{bmatrix}$ Which gives me: $7-5x_1 = 1$ $7+1x_2 = 0$ $2-6x_3 = 0$ $4+4x_4 = 1$ These are not the correct answers, can anyone help me out here? Thank you!

Since $\begin{pmatrix}7&7\\2&4\end{pmatrix}-X\begin{pmatrix}5&-1\\6&-4\end{pmatrix}=\begin{pmatrix}1&0\\0&1\end{pmatrix}$, we obtain: $\begin{pmatrix}6&7\\2&3\end{pmatrix}=\begin{pmatrix}5x_1+6x_2&-x_1-4x_2\\5x_3+6x_4&-x_3-4x_4\end{pmatrix}$, where $X=\begin{pmatrix}x_1&x_2\\x_3&x_4\end{pmatrix}$. Now you can multiply both sides of the equation by $\frac{1}{-14}\begin{pmatrix}-4&1\\-6&5\end{pmatrix}$ =(inverse of $\begin{pmatrix}5&-1\\6&-4\end{pmatrix}$), to find: $X=\frac{1}{-14}\begin{pmatrix}6&7\\2&3\end{pmatrix}\begin{pmatrix}-4&1\\-6&5\end{pmatrix}=\frac{1}{-14}\begin{pmatrix}-66&41\\-26&17\end{pmatrix}$. Hope this helps.

simple limit but I forget how to prove it I have to calculate the following limit $$\lim_{x\rightarrow -\infty} \sqrt{x^2+2x+2} - x$$ it is in un undeterminated form. I tried to rewrite it as follows: $$\lim_{x\rightarrow -\infty} \sqrt{x^2+2x+2} - \sqrt{|x|^2}$$ but seems a dead road. Can anyone suggest a solution? thanks for your help

Clearly $$\lim_{x\rightarrow -\infty} \sqrt{x^2+2x+2} - x=+\infty+\infty=+\infty$$ But \begin{gather*}\lim_{x\rightarrow +\infty} \sqrt{x^2+2x+2} - x="\infty-\infty"=\\ =\lim_{x\rightarrow +\infty} \frac{(\sqrt{x^2+2x+2} - x)(\sqrt{x^2+2x+2} + x)}{\sqrt{x^2+2x+2} + x}=\lim_{x\rightarrow +\infty} \frac{2x+2}{\sqrt{x^2+2x+2} + x}=\lim_{x\rightarrow +\infty} \frac{2+2/x}{\sqrt{1+2/x+2/x^2} + 1}=1 \end{gather*}

Partial fraction integration $\int \frac{dx}{(x-1)^2 (x-2)^2}$ $$\int \frac{dx}{(x-1)^2 (x-2)^2} = \int \frac{A}{x-1}+\frac{B}{(x-1)^2}+\frac{C}{x-2}+\frac{D}{(x-2)^2}\,dx$$ I use the cover up method to find that B = 1 and so is C. From here I know that the cover up method won't really work and I have to plug in values for x but that won't really work either because I have two unknowns. How do I use the coverup method?

To keep in line with the processes you are learning, we have: $$\frac{1}{(x-1)^2 (x-2)^2} = \frac{A}{x-1}+\frac{B}{(x-1)^2}+\frac{C}{x-2}+\frac{D}{(x-2)^2}$$ So we want to find $A, B, C, D$ given $$A(x-1)(x-2)^2 + B(x-2)^2 + C(x-1)^2(x-2) + D(x-1)^2 = 1$$ As you found, when $x = 1$, we have $B = 1$, and when $x = 2$, we have $D = 1$. Now, we need to solve for the other two unknowns by creating a system of two equations and two unknowns: $A, C$, given our known values of $B, D = 1$. Let's pick an easy values for $x$: $x = 0$, $x = 3$ $$A(x-1)(x-2)^2 + B(x-2)^2 + C(x-1)^2(x-2) + D(x-1)^2 = 1\quad (x = 0) \implies$$ $$A(-1)((-2)^2) + B\cdot (-2)^2 + C\cdot (1)\cdot(-2) + D\cdot (-1)^2 = 1$$ $$\iff - 4A + 4B - 2C + D = 1 $$ $$B = D = 1 \implies -4A + 4 - 2C + 1 = 1 \iff 4A + 2C = 4\tag{x = 0}$$ Similarly, $x = 3 \implies $ $2A + 1 + 4C + 4 = 1 \iff 2A + 4C = -4 \iff A + 2C = -2\tag{x = 3}$ Now we have a system of two equations and two unknowns and can solve for A, C. And solving this way, gives of $A = 2, C= -2$ Now we have $$\int\frac{dx}{(x-1)^2 (x-2)^2} = \int \frac{2}{x-1}+\frac{1}{(x-1)^2}+\frac{-2}{x-2}+\frac{1}{(x-2)^2}\,dx$$

What went wrong? Calculate mass given the density function Calculate the mass: $$D = \{1 \leq x^2 + y^2 \leq 4 , y \leq 0\},\quad p(x,y) = y^2.$$ So I said: $M = \iint_{D} {y^2 dxdy} = [\text{polar coordinates}] = \int_{\pi}^{2\pi}d\theta {\int_{1}^{2} {r^3sin^2\theta dr}}$. But when I calculated that I got the answer $0$ which is wrong, it should be $\frac{15\pi}{8}$. Can someone please tell me what I did wrong?

You have the set-up correct, but you have incorrectly computed the integral Let's work it out together. $\int_{\pi}^{2\pi}d\theta {\int_{1}^{2} {r^3\sin^2\theta dr}}$ $\int_{\pi}^{2\pi} {\int_{1}^{2} {r^3\sin^2\theta drd\theta}}$ $\int_{\pi}^{2\pi} \sin^2\theta d\theta {\int_{1}^{2} {r^3dr}}$ $\int_{\pi}^{2\pi} \sin^2\theta d\theta (\frac{2^4}{4} - \frac{1^4}{4})$ $\int_{\pi}^{2\pi} \sin^2\theta d\theta (3\frac{3}{4})$ $\frac{1}{2}((2\pi - \sin(2\pi)\cos(2\pi) - \pi +\sin(\pi)\cos(\pi)) (3\frac{3}{4})$ note that the integral of $\sin^2(x)$ = $\frac{1}{2}(x - \sin(x)\cos(x))$ $\frac{1}{2}(\pi)(3\frac{3}{4}) = \frac{15\pi}{8}$

Proof that $ \lim_{x \to \infty} x \cdot \log(\frac{x+1}{x+10})$ is $-9$ Given this limit: $$ \lim_{x \to \infty} x \cdot \log\left(\frac{x+1}{x+10}\right) $$ I may use this trick: $$ \frac{x+1}{x+1} = \frac{x+1}{x} \cdot \frac{x}{x+10} $$ So I will have: $$ \lim_{x \to \infty} x \cdot \left(\log\left(\frac{x+1}{x}\right) + \log\left(\frac{x}{x+10}\right)\right) = $$ $$ = 1 + \lim_{x \to \infty} x \cdot \log\left(\frac{x}{x+10}\right) $$ But from here I am lost, I still can't make it look like a fondamental limit. How to solve it?

I'll use the famous limit $$\left(1+\frac{a}{x+1}\right)^x\approx\left(1+\frac{a}{x}\right)^x\to e^a$$ We have $$x \ln \frac{x+1}{x+10}=x \ln \frac{x+1}{x+1+9}=-x\ln\left( 1+\frac{9}{x+1} \right)=-\ln\left( 1+\frac{9}{x+1} \right)^x\to-9$$

Find the coefficient of $x^{20}$ in $(x^{1}+⋯+x^{6} )^{10}$ I'm trying to find the coefficient of $x^{20}$ in $$(x^{1}+⋯+x^{6} )^{10}$$ So I did this : $$\frac {1-x^{m+1}} {1-x} = 1+x+x^2+⋯+x^{m}$$ $$(x^1+⋯+x^6 )=x(1+x+⋯+x^5 ) = \frac {x(1-x^6 )} {1-x} = \frac {x-x^7} {1-x}$$ $$(x^1+⋯+x^6 )^{10} =\left(\dfrac {x-x^7} {1-x}\right)^{10}$$ But what do I do from here ? any hints ? Thanks

Since $(x+x^2+\cdots+x^6)^{10}=x^{10}(1+x+\cdots+x^5)^{10}$ and $1+x+\cdots+x^5=\frac{1-x^6}{1-x},$ we need to find the coefficient of $x^{10}$ in $(\frac{1-x^6}{1-x})^{10}=(1-x^6)^{10}(1-x)^{-10}.$ Since $(1-x^6)^{10}(1-x)^{-10} = (1-10x^6+45x^{12}+\cdots) \sum_{m=0}^{\infty}\binom{m+9}{9}x^{m},$ the coefficient of $x^{10}$ will be $\binom{19}{9}-10\binom{13}{9}. $

Show $\cos(x+y)\cos(x-y) - \sin(x+y)\sin(x-y) = \cos^2x - \sin^2x$ Show $\cos(x+y)\cos(x-y) - \sin(x+y)\sin(x-y) = \cos^2x - \sin^2x$ I have got as far as showing that: $\cos(x+y)\cos(x-y) = \cos^2x\cos^2y -\sin^2x\sin^2y$ and $\sin(x+y)\sin(x-y) = \sin^2x\cos^2y - \cos^2x\sin^2y$ I get stuck at showing: $\cos^2x\cos^2y -\sin^2x\sin^2y - \sin^2x\cos^2y - \cos^2x\sin^2y = \cos^2x - \sin^2x$ I know that $\sin^2x + \cos^2x = 1$ and I have tried rearranging this identity in various ways, but this has not helped me so far.

Hint: $$\cos(a+b)=\cos a \cos b-\sin a \sin b$$$$\cos(2a)=\cos^2 a -\sin ^2a $$ $$\begin{align}\cos(x+y)\cos(x-y) - \sin(x+y)\sin(x-y) &= \cos((x+y)+(x-y))\\&=\cos2x\\&=\cos^2x - \sin^2x\end{align}$$

Solving the trigonometric equation $A\cos x + B\sin x = C$ I have a simple equation which i cannot solve for $x$: $$A\cos x + B\sin x = C$$ Could anyone show me how to solve this. Is this a quadratic equation?

$A\cos x+B\sin x=C$ so if $A\neq 0, B\neq 0$ then $$\frac{A}{\sqrt{A^2+B^2}}\cos x+\frac{B}{\sqrt{A^2+B^2}}\sin x=\frac{C}{\sqrt{A^2+B^2}}$$ in which $$\frac{A}{\sqrt{A^2+B^2}}\le1,~~\frac{B}{\sqrt{A^2+B^2}}\le1,~~\frac{C}{\sqrt{A^2+B^2}}\le1$$ This means you can suppose there is a $\xi$ such that $\cos(\xi)=\frac{A}{\sqrt{A^2+B^2}},\sin(\xi)=\frac{B}{\sqrt{A^2+B^2}}$ and so...

If $f(t) = 1+ \frac{1}{2} +\frac{1}{3}+....+\frac{1}{t}$, find $\sum^n_{r=1} (2r+1)f(r)$ in terms of $f(n)$ If $f(t) = 1+ \frac{1}{2} +\frac{1}{3}+....+\frac{1}{t}$, Find $x$ and $y$ such that $\sum^n_{r=1} (2r+1)f(r) =xf(r) -y$

Since $(n+1)^2 - n^2 = 2n+1$, we'd expect $\sum_{r=1}^n (2r+1)f(r)$ to be something like $n^2 f(n)$. A little experimentation shows that it's actually $(n+1)^2 f(n) - n(n+1)/2$. $\textbf{Proof}$: At $n=1$, this is $4 - 1 = 3 = (2+1)f(1)$. The forward difference of $(n+1)^2 f(n) - n(n+1)/2$ is $(n+2)^2 f(n+1) - (n+1)(n+2)/2 - [(n+1)^2 f(n) - n(n+1)/2] = (n+2)^2 f(n+1) - (n+1)^2 (f(n+1)-1/(n+1)) - (n+1) = (2n+3)f(n+1) = (2(n+1)+1)f(n+1)$, as required.

solve congruence $x^{59} \equiv 604 \pmod{2013}$ This is an exercise from my previous exam; how should I approach this? Solve congruence $\;x^{59} \equiv 604 \pmod{2013}$ Thanks in advance :)

Hint We have that $3 \cdot 11\cdot 61=2013$. Break up your congruence into three. By $x^2\equiv 1\mod 3$, the first one turns into $x\equiv 1\mod 3$, for example, since we can deduce $3\not\mid x$. Glue back using CRT. ADD Just in case you want the solution. First we may write $x^{59}\equiv 604\mod 3$ as $ x^{2\cdot 27}x\equiv 1\mod 3$. The last equation reveals$3\not\mid x$, so $x^2\equiv 1\mod 3$ and $x\equiv 1\mod 3$. The second one can be reduced to $x^{59}\equiv 10\mod 11$ which again reveals $11\not\mid x$. Thus $x^{10}\equiv 1\mod 11$ and then $x^{-1}\equiv 10\mod 11$ which gives $x\equiv 10\mod 11$. Finally we have $x^{59}\equiv 55\mod 61$. Again $61\not\mid x$ so $x^{60}\equiv 1\mod 61$ and we get $x^{-1}\equiv 55\mod 61$. Using the Euclidean algorithm, we find $55\cdot 10-61\cdot 9=1$ so $x\equiv 10\mod 61$. Thus we have that $$\begin{cases}x\equiv 1\mod 3\\x\equiv 10\mod 11\\x\equiv 10\mod 61\end{cases}$$ One may apply now the Chinese Remainder Theorem, or note $x=10$ is a solution of the above.

Number Theory divisibilty How can I check if $$12^{2013} + 7^{2013}$$ is divisible by $19$? Also, how can I format my questions to allow for squares instead of doing the ^ symbol.

Proposition : $a+b$ divides $a^m+b^m$ if $m$ is odd Some proofs : $1:$ Let $a+b=c,$ $a^m+b^m=a^m+(c-a)^m\equiv a^m+(-a)^m\pmod c\equiv \begin{cases} 2a^m &\mbox{if } m \text{ is even } \\ 0 & \mbox{if } m \text{ is odd } \end{cases}\pmod c $ $2:$ If $m$ is odd, $a^m+b^m=a^m-(-b)^m$ is divisible by $a-(-b)=a+b$ as $\frac{A^r-B^r}{A-B}=A^{r-1}+A^{r-2}B+A^{r-3}B^2+\cdots+A^2B^{r-3}+AB^{r-2}+B^{r-1}$ which is an integer if $A,B$ are integers and integer $r\ge0$ $3:$ Inductive proof: $\underbrace{a^{2n+3}+b^{2n+3}}=a^2\underbrace{(a^{2n+1}+b^{2n+1})}-b^{2n+1}(a^2-b^2)\equiv a^2\underbrace{(a^{2n+1}+b^{2n+1})}\pmod {(a+b)}$ So, $a^{2(n+1)+1}+b^{2(n+1)+1}$ will be divisible by $a+b$ if $(a^{2n+1}+b^{2n+1})$ is divisible by $a+b$ Now clearly,$(a^{2n+1}+b^{2n+1})$ is divisible by $a+b$ for $n=0,1$ Hence the proposition will hold for all positive integer $n$ (By induction)

Simplifying compound fraction: $\frac{3}{\sqrt{5}/5}$ I'm trying to simplify the following: $$\frac{3}{\ \frac{\sqrt{5}}{5} \ }.$$ I know it is a very simple question but I am stuck. I followed through some instructions on Wolfram which suggests that I multiply the numerator by the reciprocal of the denominator. The problem is I interpreted that as: $$\frac{3}{\ \frac{\sqrt{5}}{5} \ } \times \frac{5}{\sqrt{5}},$$ Which I believe is: $$\frac{15}{\ \frac{5}{5} \ } = \frac{15}{1}.$$ What am I doing wrong?

This means $$ 3\cdot \frac{5}{\sqrt{5}}=3\cdot\frac{(\sqrt{5})^2}{\sqrt{5}} =3\sqrt{5} $$ You're multiplying twice for the reciprocal of the denominator. Another way to see it is multiplying numerator and denominator by the same number: $$ \frac{3}{\frac{\sqrt{5}}{5}}=\frac{3\sqrt{5}}{\frac{\sqrt{5}}{5}\cdot\sqrt{5}} =\frac{3\sqrt{5}}{1} $$

Power series for $(1+x^3)^{-4}$ I am trying to find the power series for the sum $(1+x^3)^{-4}$ but I am not sure how to find it. Here is some work: $$(1+x^3)^{-4} = \frac{1}{(1+x^3)^{4}} = \left(\frac{1}{1+x^3}\right)^4 = \left(\left(\frac{1}{1+x}\right)\left(\frac{1}{x^2-x+1}\right)\right)^4$$ I can now use $$\frac{1}{(1-ax)^{k+1}} = \left(\begin{array}{c} k \\ 0 \end{array}\right)+\left(\begin{array}{c} k+1 \\ 1 \end{array}\right)ax+\left(\begin{array}{c} k+2 \\ 2 \end{array}\right)a^2x^2+\dots$$ on the $\frac{1}{1+x}$ part but I am not sure how to cope with the rest of formula.

You could use the binomial expansion as noted in the previous answer but just for fun here's an alternative Note that: $$ \dfrac{1}{\left( 1+y \right) ^{4}}=-\dfrac{1}{6}\,{\frac {d^{3}}{d{y}^{3}}} \dfrac{1}{ 1+y } $$ and that the geometric series gives: $$ \dfrac{1}{ 1+y }=\sum _{n=0}^{\infty } \left( -y \right) ^{n}$$ so by: $${\frac {d ^{3}}{d{y}^{3}}} \left( -y \right) ^{ n} = \left( -1 \right) ^{n}{y}^{n-3} \left( n-2 \right) \left( n-1 \right) n$$ we have: $$ \dfrac{1}{\left( 1+y \right) ^{4}}=-\dfrac{1}{6}\,\sum _{n=0}^{\infty } \left( -1 \right) ^{n}{y}^{n-3} \left( n-2 \right) \left( n-1 \right) n$$ and putting $y=x^3$ gives: $$\dfrac{1}{\left( 1+x^3 \right) ^{4}}=-\dfrac{1}{6}\,\sum _{n=0}^{\infty }{x}^{3\,n-9 } \left( -1 \right) ^{n} \left( n-2 \right) \left( n-1 \right) n$$ then noting that the first 3 terms are zero because of the $n$ factors we can shift the index by letting $n\rightarrow m+3$ to get: $$\dfrac{1}{\left( 1+x^3 \right) ^{4}}=\dfrac{1}{6}\,\sum _{m=0}^{\infty }{x}^{3m} \left( -1 \right) ^{m} \left( 1+m \right) \left( m+2 \right) \left( m+3 \right) $$ For comparison the binomial expansion would tell you that: $$\dfrac{1}{\left( 1+x^3 \right) ^{4}}=\sum _{m=0}^{\infty }{-4\choose m}{x}^ {3\,m}$$ from which it follows, by the uniqueness of Taylor series, that: $${-4\choose m}=\dfrac{1}{6} \left( -1 \right) ^{m} \left( 1+m \right) \left( m+2 \right) \left( m+3 \right)=(-1)^m\dfrac{(3+m)!}{3!\,m!}$$

we need to show gcd is $1$ I need to show if $(a,b)=1,n$ is an odd positive integer then $\displaystyle \left(a+b,{a^n+b^n\over a+b}\right)\mid n.$ let $\displaystyle \left(a+b,{a^n+b^n\over a+b}\right)=d$ $\displaystyle d\mid {a^n+b^n\over a+b}=(a+b)^2(a^{n-3}-a^{n-4}b\dots+b^{n-3})-2ab(a^{n-3}\dots+b^{n-3})-ab^{n-2}$ $d\mid (a+b)$ so from the rest can I conclude $d=1$ as $(a,b)=1$?

Modulo $a+b$, you have $a^kb^{n-k}\equiv (-1)^kb^n$, so $$ \frac{a^n+b^n}{a+b}=a^{n-1}-a^{n-2}b+a^{n-3}b^2\mp\cdots -ab^{n.2}+b^{n-1}\equiv n$$ so $n=\frac{a^n+b^n}{a+b}+(\ldots)\cdot (a+b)$.

Testing for convergence of this function For the integral $$\int_2^\infty \dfrac{x+1}{(x-1)(x^2+x+1)}dx .$$ Can I know if it's convergent or not? If it does can I know how to evaluate it? I tried to use $u$ substitution but it didn't work.

Hint: Apply partial fractions. $$\frac{x+1}{(x-1)(x^2+x+1)}=\frac A{x-1}+\frac{Bx+C}{x^2+x+1}$$ $$x+1=(A+B)x^2+(A-B+C)x+(A-C)$$ $$\therefore A=\frac23,B=-\frac23,C=-\frac13$$ Now we know that: $$\int \frac{x+1}{(x-1)(x^2+x+1)}dx=\frac23\int\frac1{x-1}dx-\frac13\int\frac{2x-1}{x^2+x+1}dx\\ =\frac23\ln|x-1|-\frac13\ln(x^2+x+1)=\frac13\ln\frac{(x-1)^2}{x^2+x+1}$$ Therefore, \begin{align*} \int^\infty_2\frac{x+1}{(x-1)(x^2+x+1)}dx&=\lim_{n\to \infty}\int^n_2\frac{x+1}{(x-1)(x^2+x+1)}dx \\ \\ &=\lim_{n\to\infty}\left(\frac13\ln\frac{(n-1)^2}{n^2+n+1}-\frac13\ln\frac17\right)\\ \\ &=-\frac13\ln\frac17\\ \\ &=\boxed{\dfrac13\ln7} \end{align*}

show that $\int_{0}^{\infty} \frac {\sin^3(x)}{x^3}dx=\frac{3\pi}{8}$ show that $$\int_{0}^{\infty} \frac {\sin^3(x)}{x^3}dx=\frac{3\pi}{8}$$ using different ways thanks for all

Using the formula found in my answer, $$ \begin{aligned} \int_{0}^{\infty} \frac{\sin ^{3} x}{x^{3}} &=\frac{\pi}{2^{3} \cdot 2 !}\left[\left(\begin{array}{l} 3 \\ 0 \end{array}\right) 3^{2}-\left(\begin{array}{l} 3 \\ 1 \end{array}\right) 1^{2}\right] \\ &=\frac{3 \pi}{8} \end{aligned} $$

GRE test prep question [LCM and divisors] Let $S$ be the set of all positive integers $n$ such that $n^2$ is a multiple of both $24$ and $108$. Which of the following integers are divisors of every integer $n$ in $S$ ? Indicate all such integers: $A:12$ $B:24$ $C:36$ $D:72$ The answers are $A$ and $C$ First I took the lcm of $24$ and $108$ which is $2^3\times3^3$ but then it says that "the prime factorization of a square number must contain only even exponents. Thus, the least multiple of $(2^3)(3^3)$ that is a square is $(2^4)(3^4)$" Can somebody explain why that is true? What if the lcm was $2^3\times3^4$ ? Would I just make it $2^4\times3^4$ ? Help!

Let $n=p_1^{a_1}\cdots p_r^{a_r}$ where the $p_i$ are primes, so $n^2=p_1^{2a_1}\cdots p_r^{2a_r}$. As you observed, $n^2$ must be a multiple of $LCM(24, 108)=2^{3} 3^{3}$, so $2a_1\ge 3$ and $2a_2\ge 3$ with $p_1=2$ and $p_2=3$. Therefore $a_1\ge 2$ and $a_2\ge 2$, so n is a multiple of $2^{2} 3^{2}=36$. Thus S consists of all positive multiples of 36, so the integers which divide every integer in S are simply the divisors of 36.

show that $\int_{0}^{\pi/2}\tan^ax \, dx=\frac {\pi}{2\cos(\frac{\pi a}{2})}$ show that $$\int_{0}^{\pi/2}\tan^ax \, dx=\frac {\pi}{2\cos(\frac{\pi a}{2})}$$ I think we can solve it by contour integration but I dont know how. If someone can solve it by two way using complex and real analysis its better for me. thanks for all.

Sorry for being late $$ \begin{aligned} \int_0^{\frac{\pi}{2}} \tan ^a x d x&=\int_0^{\frac{\pi}{2}} \sin ^a x \cos ^{-a} x d x \\ & =\int_0^{\frac{\pi}{2}} \sin ^{2\left(\frac{a+1}{2}\right)-1} x \cos ^{2\left(\frac{-a+1}{2}\right)-1} x d x \\ & =\frac{1}{2} B\left(\frac{a+1}{2}, \frac{-a+1}{2}\right) \\ & =\frac{1}{2} \pi \csc \frac{(a+1) \pi}{2} \\ & \end{aligned} $$ Applying the Euler-reflection property $$ B(x, 1-x)=\pi \csc (\pi x) \quad x \notin \mathbb{Z}, $$ we have $$\boxed{\int_0^{\frac{\pi}{2}} \tan ^a x d x =\frac{\pi}{2 \cos \left(\frac{\pi a}{2}\right)}} $$

Show that $\sin3\alpha \sin^3\alpha + \cos3\alpha \cos^3\alpha = \cos^32\alpha$ Show that $\sin3\alpha \sin^3\alpha + \cos3\alpha \cos^3\alpha = \cos^32\alpha$ I have tried $\sin^3\alpha(3\sin\alpha - 4 \sin^3\alpha) = 3\sin^4\alpha - 4\sin^6\alpha$ and $\cos^3\alpha(4\cos^3\alpha - 3\cos\alpha) = 4\cos^6\alpha - 3\cos^4\alpha$ to give $$\sin3\alpha \sin^3\alpha + \cos3\alpha \cos^3\alpha = 3\sin^4\alpha - 4\sin^6\alpha + 4\cos^6\alpha - 3\cos^4\alpha$$ I can't work out how to simplify this to $\cos^32\alpha$. I also noticed that the LHS of the question resembles $\cos(A-B)$, but I can't figure a way of making that useful.

\begin{align} L.H.S=& \sin 3\alpha \sin \alpha\sin^2\alpha+\cos 3\alpha\cos \alpha \cos^2 \alpha\\ \ =& \frac{1}{2}\left(\sin 3\alpha \sin \alpha (1-\cos 2\alpha)+\cos 3\alpha\cos \alpha(1+\cos 2\alpha)\right)\\ \ =& \frac{1}{2}\left(\sin 3\alpha \sin \alpha+ \cos 3\alpha\cos \alpha\right)+\frac{1}{2}\left(\cos 3\alpha\cos \alpha-\sin 3\alpha \sin \alpha\right)\cos 2\alpha\\ \ =& \frac{1}{2}\cos(3\alpha-\alpha)+\frac{1}{2}\cos(3\alpha+\alpha)\cos 2\alpha\\ \ =& \frac{1}{2}\cos 2\alpha(1+\cos 4\alpha)\\ \ =& \frac{1}{2}\cos 2\alpha \cdot 2\cos^2 2\alpha\\ \ =& \cos^3 2\alpha\hspace{6cm}\Box \end{align}

find the inverse of $x^2 + x + 1$ In $\mathbb{F}_2[x]$ modulo $x^4 + x + 1$ find the inverse of $x^2 + x + 1$ not 100% sure but here what i have: user euclid algorithm: $x^4 + x + 1 = (x^3 + 1)(x + 1) + x$ $(x^3 + 1) = x * x * x + 1$ $1 = (x^3 + 1) - x * x * x $

Using that $\;x^4=x+1\; $ in $\,\Bbb F_2[x]/(x^4+x+1)\;$ , prove that $$x^2+x=(x^2+x+1)^{-1}\;\;\; (\text{ further hint:}\;(x^2+x+1)^3=1)$$

Show that if $a$ has order $3\bmod p$ then $a+1$ has order $6\bmod p$. Show that if $a$ has order $3\bmod p$ then $a+1$ has order $6\bmod p$. I know I am supposed to use primitive roots but I think this is where I am getting caught up. The definition of primitive root is "if $a$ is a least residue and the order of $a\bmod p$ is $\phi(m)$ then $a$ is a primitive root of $m$. But I really am not sure how to use this to my advantage in solving anything. Thanks!

Note that if $a=1$, $a$ has order $1$. Thus, we can assume $a\ne1$. Furthermore, $p\ne2$ since no element mod $2$ has order $3$. Therefore, $-1\ne1\pmod{p}$. $$ \begin{align} (a+1)^3 &=a^3+3a^2+3a+1\\ &=1+3a^2+3a+1\\ &=-1+3(1+a+a^2)\\ &=-1+3\frac{a^3-1}{a-1}\\ &=-1 \end{align} $$ Therefore, $(a+1)^3=-1$ and $(a+1)^6=1$. (Shortened a la ccorn). $$ \begin{align} (a+1)^2 &=a^2+2a+1\\ &=a+(a^2+a+1)\\ &=a+\frac{a^3-1}{a-1}\\ &=a\\ &\ne1 \end{align} $$ Therefore, $(a+1)^2\ne1$. Thus, $(a+1)$ has order $6$.

Proof read from "A problem seminar" May you help me judging the correctness of my proof?: Show that the if $a$ and $b$ are positive integers, then $$\left(a+\frac{1}{2}\right)^n+\left(b+\frac{1}{2}\right)^n$$ is integer for only finintely many positive integers $n$ We want $n$ so that $$\left(a+\frac{1}{2}\right)^n+\left(b+\frac{1}{2}\right)^n\equiv0\pmod{1}$$ So we know by the binomial theorem that $(an+b)^k\equiv b^n\pmod{n}$ for positive $k$ Then, $$\left(a+\frac{1}{2}\right)^n\equiv(1/2)^n\pmod{1}$$ and similarly with the $b$ So $$\left(a+\frac{1}{2}\right)^n+\left(b+\frac{1}{2}\right)^n\equiv 2*(1/2)^n\pmod{1}$$ Therefore, we want $2*(1/2)^n$ to be integer, so that $2^n|2$ clearly, the only positive option is $n=1$ (Editing, my question got prematurely posted. Done)

Our expression can be written as $$\frac{(2a+1)^n+(2b+1)^n}{2^n}.$$ If $n$ is even, then $(2a+1)^n$ and $(2b+1)^n$ are both the squares of odd numbers. Any odd perfect square is congruent to $1$ modulo $8$. So their sum is congruent to $2$ modulo $8$, and therefore cannot be divisible by any $2^n$ with $n\gt 1$. So we can assume that $n$ is odd. For odd $n$, we have the identity $$x^n+y^n=(x+y)(x^{n-1}-x^{n-2}y+\cdots +y^{n-1}).$$ Let $x=2a+1$ and $y=2b+1$. Note that $x^{n-1}-x^{n-2}y+\cdots +y^{n-1}$ is a sum of an odd number of terms, each odd, so it is odd. Thus the highest power of $2$ that divides $(2a+1)^n+(2b+1)^n$ is the highest power of $2$ that divides $(2a+1)+(2b+1)$. Since $(2a+1)+(2b+1)\ne 0$, there is a largest $n$ such that our expression is an integer. Remark: The largest $n$ such that our expression is an integer can be made quite large. You might want to see for example what happens if we let $2a+1=2049$ and $2b+1=2047$. Your proposed proof suggests, in particular, that $n$ cannot be greater than $1$. I suggest that when you are trying to write out a number-theoretic argument, you avoid fractions as much as possible and deal with integers only.

Evaluating a 2-variable limit Could you help me evaluating this limit? $$ \lim_{x\to 0}\frac{1}{x}\cdot\left[\arccos\left(\frac{1}{x\sqrt{x^{2}- 2x\cdot \cos(y)+1}}-\frac{1}{x}\right)-y\right] $$

Notice: I changed what I think a typo otherwise the limit is undefined. By the Taylor series we have (and we denote $a=\cos(y)$) $$\frac{1}{\sqrt{x^{2}-2xa+1}}=1+xa+x^2(\frac{3}{2}a^2-\frac{1}{2})+O(x^3)$$ so $$\frac{1}{x\sqrt{x^{2}-2xa+1}}-\frac{1}{x}=a+x(\frac{3}{2}a^2-\frac{1}{2})+O(x^2)$$ Now using $$\arccos(a+\alpha x)=\arccos(a)-\frac{\alpha}{\sqrt{1-a^2}}x+O(x^2)$$ we have $$\arccos(\frac{1}{x\sqrt{x^{2}-2xa+1}}-\frac{1}{x})=\arccos(a)-\frac{\frac{3}{2}a^2-\frac{1}{2}}{\sqrt{1-a^2}}x+O(x^2)$$ so if we suppose that $y\in[-\frac{\pi}{2},\frac{\pi}{2}]$ then $$\lim_{x\to 0}\frac{1}{x}\cdot\left[\arccos\left(\frac{1}{x\sqrt{x^{2}-2x\cdot \cos(y)+1}}-\frac{1}{x}\right)-y\right]=-\frac{\frac{3}{2}a^2-\frac{1}{2}}{\sqrt{1-a^2}}$$

Does $\det(A + B) = \det(A) + \det(B)$ hold? Well considering two $n \times n$ matrices does the following hold true: $$\det(A+B) = \det(A) + \det(B)$$ Can there be said anything about $\det(A+B)$? If $A/B$ are symmetric (or maybe even of the form $\lambda I$) - can then things be said?

Although the determinant function is not linear in general, I have a way to construct matrices $A$ and $B$ such that $\det(A + B) = \det(A) + \det(B)$, where neither $A$ nor $B$ contains a zero entry and all three determinants are nonzero: Suppose $A = [a_{ij}]$ and $B = [b_{ij}]$ are 2 x 2 real matrices. Then $\det(A + B) = (a_{11} + b_{11})(a_{22} + b_{22}) - (a_{12} + b_{12})(a_{21} + b_{21})$ and $\det(A) + \det(B) = (a_{11} a_{22} - a_{12} a_{21}) + (b_{11} b_{22} - b_{12} b_{21})$. These two determinant expressions are equal if and only if $a_{11} b_{22} + b_{11} a_{22} - a_{12} b_{21} - b_{12} a_{21} = $ $\det \left[ \begin{array}{cc} a_{11} & a_{12}\\ b_{21} & b_{22} \end{array} \right]$ + $\det \left[ \begin{array}{cc} b_{11} & b_{12}\\ a_{21} & a_{22} \end{array} \right]$ = 0. Therefore, if we choose any nonsingular 2 x 2 matrix $ A = [a_{ij}]$ with nonzero entries and then create $B = [b_{ij}]$ such that $b_{11} = - a_{21}, b_{12} = - a_{22}, b_{21} = a_{11},$ and $b_{22} = a_{12}$, we have solved our problem. For example, if we take $$A = \begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix} \quad \text{and}\quad B = \begin{bmatrix} -3 & -4 \\ 1 & 2\end{bmatrix} ,$$ then $\det(A) = -2, \det(B) = -2, $ and $\det(A + B) = -4$, as required.

How find this $3\sqrt{x^2+y^2}+5\sqrt{(x-1)^2+(y-1)^2}+\sqrt{5}(\sqrt{(x-1)^2+y^2}+\sqrt{x^2+(y-1)^2})$ find this follow minimum $$3\sqrt{x^2+y^2}+5\sqrt{(x-1)^2+(y-1)^2}+\sqrt{5}\left(\sqrt{(x-1)^2+y^2}+\sqrt{x^2+(y-1)^2}\right)$$ I guess This minimum is $6\sqrt{2}$ But I can't prove,Thank you

If $v_1 = (0,0), v_2 = (1,1), v_3 = (0,1)$, and $v_4 = (1,0)$ and $p = (x,y)$, then you are trying to minimize $$3|p - v_1| + 5|p - v_2| + \sqrt{5}|p - v_3| + \sqrt{5}|p - v_4|$$Note that if $p$ is on the line $y = x$, moving it perpendicularly away from the line will only increase $|p - v_1|$ and $|p - v_2|$, and it is not too hard to show it also increases $|p - v_3| + |p - v_4|$. So the minimum has to occur on the line $y = x$. So letting $p = (t,t)$ your problem becomes to minimize $$3\sqrt{2}t + 5\sqrt{2}(1 - t) + 2\sqrt{5}\sqrt{2t^2 - 2t + 1}$$ This can be minimized through calculus... maybe there's a slick geometric way too.

Find the 12th term and the sum of the first 12 terms of a geometric sequence. A geometric series has a first term $\sqrt{2}$ and a second term $\sqrt{6}$ . Find the 12th term and the sum of the first 12 terms. I can get to the answers as irrational numbers using a calculator but how can I can obtain the two answers in radical form $243 * \sqrt{6}$ and $364 \left(\sqrt{6}+\sqrt{2}\right)$ ? The closest I get with the 12th term is $\sqrt{2} \left(\sqrt{6} \over \sqrt{2}\right)^{(12-1)}$ or $\sqrt{2} * 3^\left({11\over 2}\right)$ And for the sum ${\sqrt{2}-\sqrt{2}*(\sqrt{3})^{12} \over 1 - \sqrt{3}}$

So, the common ratio $=\frac{\sqrt6}{\sqrt2}=\sqrt3$ So, the $n$ th term $=\sqrt2(\sqrt3)^{n-1}\implies 12$th term $=\sqrt2(\sqrt3)^{12-1}=\sqrt2(\sqrt3)^{11}$ Now, $\displaystyle(\sqrt3)^{11}=\sqrt3 \cdot 3^5=243\sqrt3$ The sum of $n$ term is $\displaystyle \sqrt2\cdot\frac{(\sqrt3)^n-1}{\sqrt3-1}$ $\implies 12$th term $=\displaystyle \sqrt2\cdot\frac{(\sqrt3)^{12}-1}{\sqrt3-1}=\sqrt2\cdot\frac{(3^6-1)(\sqrt3+1)}{(\sqrt3-1)(\sqrt3+1)}$ (rationalizing the denominator ) $\displaystyle=\frac{(3^3-1)(3^3+1)\sqrt2(\sqrt3+1)}2=364(\sqrt6+\sqrt2)$

Calculate $\int \frac{dx}{x\sqrt{x^2-1}}$ I am trying to solve the following integral $$\int \frac{dx}{x\sqrt{x^2-1}}$$ I did the following steps by letting $u = \sqrt{x^2-1}$ so $\text{d}u = \dfrac{x}{\sqrt{{x}^{2}-1}}$ then \begin{align} &\int \frac{\sqrt{x^2-1} \, \text{d}u}{x \sqrt{x^2-1}} \\ &\int \frac{1}{x} \text{d}u \\ &\int \frac{1}{\sqrt{u^2+1}} \text{d}u\\ \end{align} Now, this is where I am having trouble. How can I evaluate that? Please provide only hints Thanks! EDIT: The problem specifically states that one must use substitution with $u = \sqrt{x^2-1}$. This problem is from the coursera course for Single Variable Calculus.

$$ \begin{aligned}\int \frac{d x}{x \sqrt{x^{2}-1}} =\int \frac{1}{x^{2}} d\left(\sqrt{x^{2}-1}\right) =\int \frac{d\left(\sqrt{x^{2}-1}\right)}{\left(\sqrt{x^{2}-1}\right)^{2}+1} =\tan ^{-1}\left(\sqrt{x^{2}-1}\right)+C \end{aligned} $$

Proof: for a pretty nasty limit Let $$ f(x) = \lim_{n\to\infty} \dfrac {[(1x)^2]+[(2x)^2]+\ldots+[(nx)^2]} {n^3}$$. Prove that f(x) is continuous function. Edit: $[.] $ is the greatest integer function.

In general, if $[z]$ is the integer part of $z$, $D(n) =\sum_{k=1}^n f(kx) -\sum_{k=1}^n [f(kx)] =\sum_{k=1}^n (f(kx)-[f(kx)]) $. Since $0 \le z-[z] < 1$, $0 \le D(n) < n$. For your case, $f(x) = x^2$, so $0 \le \sum_{k=1}^n (kx)^2 -\sum_{k=1}^n [(kx)^2] < n$ or $0 \le \dfrac1{n^3}\sum_{k=1}^n (kx)^2 -\dfrac1{n^3}\sum_{k=1}^n [(kx)^2] < \frac{n}{n^3} =\frac1{n^2} $. Therefore, since $\lim_{n \to \infty} \frac1{n^3}\sum_{k=1}^n (kx)^2 =\lim_{n \to \infty} \frac{x^2}{n^3}\sum_{k=1}^n k^2 =\lim_{n \to \infty} \frac{x^2}{n^3}\frac{n(n+1)(2n+1)}{6} =\frac{x^2}{3} $, $\lim_{n \to \infty} \frac1{n^3}\sum_{k=1}^n [(kx)^2] =\frac{x^2}{3} $.

Prove that: $ \cot7\frac12 ^\circ = \sqrt2 + \sqrt3 + \sqrt4 + \sqrt6$ How to prove the following trignometric identity? $$ \cot7\frac12 ^\circ = \sqrt2 + \sqrt3 + \sqrt4 + \sqrt6$$ Using half angle formulas, I am getting a number for $\cot7\frac12 ^\circ $, but I don't know how to show it to equal the number $\sqrt2 + \sqrt3 + \sqrt4 + \sqrt6$. I would however like to learn the technique of dealing with surds such as these, especially in trignometric problems as I have a lot of similar problems and I don't have a clue as to how to deal with those. Hints please! EDIT: What I have done using half angles is this: (and please note, for convenience, I am dropping the degree symbols. The angles here are in degrees however). I know that $$ \cos 15 = \dfrac{\sqrt3+1}{2\sqrt2}$$ So, $$\sin7.5 = \sqrt{\dfrac{1-\cos 15} {2}}$$ $$\cos7.5 = \sqrt{\dfrac{1+\cos 15} {2}} $$ $$\implies \cot 7.5 = \sqrt{\dfrac{2\sqrt2 + \sqrt3 + 1} {2\sqrt2 - \sqrt3 + 1}} $$

$$\text{As } \cot x =\frac{\cos x}{\sin x}$$ $$ =\frac{2\cos^2x}{2\sin x\cos x}(\text{ multiplying the numerator & the denominator by }2\cos7\frac12 ^\circ)$$ $$=\frac{1+\cos2x}{\sin2x}(\text{using }\sin2A=2\sin A\cos A,\cos2A=2\cos^2A-1$$ $$ \cot7\frac12 ^\circ =\frac{1+\cos15^\circ}{\sin15^\circ}$$ $\cos15^\circ=\cos(45-30)^\circ=\cos45^\circ\cos30^\circ+\sin45^\circ\sin30^\circ=\frac{\sqrt3+1}{2\sqrt2}$ $\sin15^\circ=\sin(45-30)^\circ=\sin45^\circ\cos30^\circ-\cos45^\circ\sin30^\circ=\frac{\sqrt3-1}{2\sqrt2}$ Method $1:$ $$\frac{1+\cos15^\circ}{\sin15^\circ}=\csc15^\circ+\cot15^\circ$$ $$\cot15^\circ=\frac{\cos15^\circ}{\sin15^\circ}=\frac{\sqrt3+1}{\sqrt3-1}=\frac{(\sqrt3+1)^2}{(\sqrt3-1)(\sqrt3+1)}=2+\sqrt3$$ $$\csc15^\circ=\frac{2\sqrt2}{\sqrt3-1}=\frac{2\sqrt2(\sqrt3+1)}{(\sqrt3-1)(\sqrt3+1)}=\sqrt2(\sqrt3+1)=\sqrt6+\sqrt2$$ Method $2:$ $$\implies \frac{1+\cos15^\circ}{\sin15^\circ}=\frac{1+\frac{\sqrt3+1}{2\sqrt2}}{\frac{\sqrt3-1}{2\sqrt2}}=\frac{2\sqrt2+\sqrt3+1}{\sqrt3-1}=\frac{(2\sqrt2+\sqrt3+1)(\sqrt3+1)}{(\sqrt3-1)(\sqrt3+1)}(\text{ rationalizing the denominator })$$ $$=\frac{2\sqrt6+4+2\sqrt3+2\sqrt2}2$$

I need to calculate $x^{50}$ $x=\begin{pmatrix}1&0&0\\1&0&1\\0&1&0\end{pmatrix}$, I need to calculate $x^{50}$ Could anyone tell me how to proceed? Thank you.

The Jordan Decomposition yields $$ \left[ \begin{array}{r} 1 & 0 & 0 \\ 1 & 0 & 1 \\ 0 & 1 & 0 \end{array} \right] = \left[ \begin{array}{r} 0 & 0 & 2 \\ -1 & 1 & 1 \\ 1 & 1 & 0 \end{array} \right] \left[ \begin{array}{r} -1 & 0 & 0 \\ 0 & 1 & 1 \\ 0 & 0 & 1 \end{array} \right] \left[ \begin{array}{r} 0 & 0 & 2 \\ -1 & 1 & 1 \\ 1 & 1 & 0 \end{array} \right]^{-1} $$ Block matrices are easier to raise to a power: $$ \begin{align} \left[ \begin{array}{r} 1 & 0 & 0 \\ 1 & 0 & 1 \\ 0 & 1 & 0 \end{array} \right]^{50} &= \left[ \begin{array}{r} 0 & 0 & 2 \\ -1 & 1 & 1 \\ 1 & 1 & 0 \end{array} \right] \left[ \begin{array}{r} -1 & 0 & 0 \\ 0 & 1 & 1 \\ 0 & 0 & 1 \end{array} \right]^{50} \left[ \begin{array}{r} 0 & 0 & 2 \\ -1 & 1 & 1 \\ 1 & 1 & 0 \end{array} \right]^{-1}\\[6pt] &= \left[ \begin{array}{r} 0 & 0 & 2 \\ -1 & 1 & 1 \\ 1 & 1 & 0 \end{array} \right] \left[ \begin{array}{r} 1 & 0 & 0 \\ 0 & 1 & 50 \\ 0 & 0 & 1 \end{array} \right] \left[ \begin{array}{r} 0 & 0 & 2 \\ -1 & 1 & 1 \\ 1 & 1 & 0 \end{array} \right]^{-1}\\[6pt] &= \left[ \begin{array}{r} 1 & 0 & 0 \\ 25 & 1 & 0 \\ 25 & 0 & 1 \end{array} \right] \end{align} $$

Show (via differentiation) $1-2+3-4+\cdots+(-1)^{n-1}n$ is $-\frac{n}{2}$ for $n$ even, $\frac{(n+1)}{2}$ for $n$ odd. i) By considering $(1+x+x^2+\cdots+x^n)(1-x)$ show that, if $x\neq 1$, $$1+x+x^2+\cdots+x^n=\frac{(1-x^{n+1})}{1-x}$$ ii) By differentiating both sides and setting $x=-1$ show that $$1-2+3-4+\cdots+(-1)^{n-1}n$$ takes the value $-\frac{n}{2}$ if n is even and the value $\frac{(n+1)}{2}$ if n is odd. For part i) I just simplified the LHS, divided by $(1-x)$ and got the desired result. For the next part I found the derivative of both sides, and set $x=-1$ giving me: $$1-2+3-4+\cdots+(-1)^{n-1}n = \frac{(2)(-1(n+1)(-1)^n)-(1-x^{n+1})(-1)}{4} = \frac{-2(n+1)(-1)^n+1+(-1)^{n+2}}{4}$$ However I'm not understanding the part about n being even and odd. If n is even, does this mean that $n = 2n$ and if it is odd, $n = 2n+1/2n-1$? What would be the next step? Thanks

If $n$ is even, $n = 2k$ for some integer $k$. Then $(-1)^n = (-1)^{2k} = ((-1)^2)^k = 1^k = 1$ and $(-1)^{n + 2} = (-1)^n\times(-1)^2 = 1\times 1 = 1$. Therefore, we have \begin{align*} \frac{-2(n+1)(-1)^n + 1 +(-1)^{n+2}}{4} &= \frac{-2(n+1)\times 1 + 1 + 1}{4}\\ &= \frac{-2(n+1) + 2}{4}\\ &= \frac{-2n -2 + 2}{4}\\ &= \frac{-2n}{4}\\ &= -\frac{n}{2}. \end{align*} Can you follow the steps to do the odd case?

Calculate $ \lim_{x \to 4} \frac{3 -\sqrt{5 -x}}{1 -\sqrt{5 -x}} $ How evaluate the following limit? $$ \lim_{x \to 4} \frac{3 -\sqrt{5 -x}}{1 -\sqrt{5 -x}} $$ I cannot apply L'Hopital because $ f(x) = 3 -\sqrt{5 -x} \neq 0 $ at $x = 5$

Let $x=5\cos4y$ where $0\le4y\le\pi$ $x\to4\implies\cos4y\to\dfrac45$ But as $\cos4y=2\cos^22y-1, 2y\to\arccos\dfrac3{\sqrt{10}}=\arcsin\dfrac1{\sqrt{10}}$ $$F=\lim_{x\to4}\frac{3-\sqrt{5+x}}{1-\sqrt{5-x}}=\lim_{2y\to\arcsin\frac1{\sqrt{10}}}\dfrac{\dfrac3{\sqrt{10}}-\cos2y}{\dfrac1{\sqrt{10}}-\sin2y}$$ If we set $\arccos\dfrac3{\sqrt{10}}=\arcsin\dfrac1{\sqrt{10}}=2A\implies\cos2A=?,\sin2A=?,\tan2A=\dfrac13$ $$F=\lim_{2y\to2A}\dfrac{\cos2A-\cos2y}{\sin2A-\sin2y}=\lim_{y\to A}\dfrac{-2\sin(A-y)\sin(A+y)}{2\sin(A-y)\cos(A+y)}=-\tan(A+A)=?$$

How to solve a system of trigonometric equations This question today appeared in my maths olympiad paper: If $\cos x + \cos y + \cos z = \sin x + \sin y + \sin z = 0$, then, prove that $\cos 2x + \cos 2y + \cos 2z = \sin 2x + \sin 2y + \sin 2z = 0$. Can anyone please help me in finding out the solution of this equation ? I have not gone any far in this solution.

Putting $a=\cos x+i\sin x$ etc, we have $a+b+c=0$ and $a^{-1}=\frac1{\cos x+i\sin x}=\cos x-i\sin x$ $\implies a^{-1}+b^{-1}+c^{-1}=0\implies ab+bc+ca=0$ $a^2+b^2+c^2=(a+b+c)^2-2(ab+bc+ca)=0$ Now, $a^2=(\cos x+i\sin x)^2=\cos^2x-\sin^2x+i2\sin x\cos x=\cos2x+i\sin2x$ which is a special case of de Moivre's formula

Getting the $x$-intercept of $f(x) = -16x^2 + 80x + 5$ $$f(x) = -16x^2 + 80x + 5$$ I need to find the bigger value of $x$ that makes $f(x) = 0$. Naturally, I thought to do: $$0=-16x^2+80x+5$$ and I applied the quadratic formula $$0=\frac{-80\pm\sqrt{6080}}{-32}$$ but the answer doesn't seem like it would be correct. Did I do something wrong?

First of all, please remember that the $x$-intercept is where the graph $y = \operatorname{f}(x)$ meets the $x$-axis. If you're not plotting a graph then it doesn't make sense to talk about $x$- and $y$-intercepts. You're looking for the solutions to the equation $\operatorname{f}(x)=0$. If $\operatorname{f}(x)=-16x^2+80x+5$ then you need to solve $-16x^2+80x+5=0$. The quadratic formula can be used where $a=-16$, $b=80$ and $c=5$. We have: $$\begin{array} .x &=& \frac{-b\pm\sqrt{b^2-4ac}}{2a} \\ \\ &=& \frac{-80 \pm \sqrt{(80)^2-4(-16)(5)}}{2(-16)} \\ \\ &=& \frac{-80\pm\sqrt{6720}}{-32} \\ \\ &=& \tfrac{5}{2} \pm \tfrac{1}{4}\sqrt{105} \end{array}$$

Number of divisors of $9!$ which are of the form $3m+2$ Total number of divisors of $9!$ which are is in the form of $3m+2$, where $m\in \mathbb{N}$ My Try: Let $ N = 9! = 1\times 2 \times 3 \times 2^2 \times 5 \times 2 \times 3 \times 7 \times 2^3 \times 3^2 = 2^7 \times 3^4 \times 5 \times 7$ Now If Here $N$ must be a mutiple of $3m+2$, means when $N$ is divided by $3$ It will gave a remainder $2$ But I did not understand how can i proceed further, thanks in advance

Hint: If the factor is of the form $3m+2$, then the prime factorization must be of the form $$ 2^a \times 3^0 \times 5^b \times 7^c, $$ where $a+b \equiv 1 \pmod{2} $ and $ c= 0$ or $1$. Count the number of possibilities.

Matrix generating $\operatorname{SL}_n(\mathbb{R})$ How do I show that the following matrices generate $\operatorname{SL}_2(\mathbb{R})$ $\begin{pmatrix} 1 & a \\ 0 & 1 \\ \end{pmatrix}$ or $\begin{pmatrix} 1 & 0 \\ a & 1 \\ \end{pmatrix}$

Let $G$ be the span of the matrices $$ \begin{pmatrix} 1 & a \\ 0 & 1 \end{pmatrix} \quad\text{and}\quad \begin{pmatrix} 1 & 0 \\ a & 1 \end{pmatrix} $$ with $a\in \mathbb R$. We have $$ \begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix} \begin{pmatrix} 1 & 0 \\ -1 & 1 \end{pmatrix} \begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix} = \begin{pmatrix} 0 & 1 \\ -1 & 0 \end{pmatrix} \in G $$ and for all $a\in \mathbb{R}^\times$ $$ \begin{pmatrix} 1 & a \\ 0 & 1 \end{pmatrix} \begin{pmatrix} 0 & 1 \\ -1 & 0 \end{pmatrix} \begin{pmatrix} 1 & a^{-1} \\ 0 & 1 \end{pmatrix} \begin{pmatrix} 0 & 1 \\ -1 & 0 \end{pmatrix} \begin{pmatrix} 1 & a \\ 0 & 1 \end{pmatrix} \begin{pmatrix} 0 & 1 \\ -1 & 0 \end{pmatrix} = \begin{pmatrix} a & 0 \\ 0 & a^{-1} \end{pmatrix} \in G $$ Now by a slightly modified Gauss elimination, any matrix $A\in \operatorname{SL}_2(\mathbb R)$ can be transformed into the unit matrix using the above matrices. So $G = \operatorname{SL}_2(\mathbb R)$

Prove that $ \frac{a^2}{a+b}+ \frac{b^2}{b+c} \geq \frac{3a+2b-c}{4} $ Prove that: $$ \frac{a^2}{a+b} + \frac{b^2}{b+c} \geq \frac{3a+2b-c}{4} : (a, b, c)\in \mathbb{R}^+$$ This is just one of these questions where you just have no idea how to start. First impressions, I don't see how any known inequality can be used, and I also don't want to go just make everything as a sum then solve it. I always was bad at inequalities and I don't know why. I did the other exercise just fine, but inequalities are hard for me. This is from a high-school olympiad.

$$\begin{align} \frac{a^2}{a+b} + \frac{b^2}{b+c} &\geqslant \frac{3a + 2b - c}{4}\\ \iff \frac{a^2}{a+b} - a + \frac{b^2}{b+c} - b &\geqslant - \frac{a + 2b + c}{4}\\ \iff -\frac{ab}{a+b} - \frac{bc}{b+c} &\geqslant - \frac{a+b}{4} - \frac{b+c}{4}\\ \iff \frac{(a+b)^2 - 4ab}{4(a+b)} + \frac{(b+c)^2 - 4bc}{4(b+c)} &\geqslant 0. \end{align}$$ Since all numbers are positive, the denominators are positive, and $(x+y)^2 - 4xy = (x-y)^2 \geqslant 0$.

Prove $\sum_n^{\infty} \prod_{k=0}^n \dfrac{1}{x+k} = e \sum_ n^{\infty} \dfrac{(-1)^n}{(x+n)n!}$ Let $$f_n(x) = \prod_{k=0}^n \dfrac{1}{x+k}.$$ Show that $$\sum_{n=0}^{\infty} f_n(x) = e \sum_ {n=0}^{\infty} \dfrac{(-1)^n}{(x+n)n!}.$$

Using the partial fraction identity that was proved by a straightforward induction technique at this MSE post, we have that $$\prod_{k=0}^n \frac{1}{x+k} = \frac{1}{n!} \sum_{k=0}^n (-1)^k \binom{n}{k} \frac{1}{x+k}.$$ Now to compute $$\sum_{n=0}^\infty \prod_{k=0}^n \frac{1}{x+k}$$ we ask about the coefficient of $$\frac{1}{x+k}$$ taking into account all terms of the series. We see immediately that all products with $n\ge k$ include this term, so the coefficient is $$(-1)^k \sum_{n\ge k} \frac{1}{n!} \binom{n}{k} = \frac{(-1)^k}{k!} \sum_{n\ge k} \frac{1}{(n-k)!} = e\frac{(-1)^k}{k!}.$$ Now summing for $k\ge 0$ we get the desired result $$e\sum_{k=0}^\infty \frac{(-1)^k}{k!} \frac{1}{x+k}.$$ Interestingly enough the term $$g_n(x) = \prod_{k=1}^n \frac{1}{x+k}$$ can also be evaluated using Mellin transforms. (Drop the factor $1/x$ for the moment to keep the Mellin integral simple -- no pole at zero.) We get $$g_n^*(s) = \mathfrak{M}(g_n(x); s) = \int_0^\infty \prod_{k=1}^n \frac{1}{x+k} x^{s-1} dx$$ which gives (use a keyhole contour with the slot on the real axis, which is also the branch cut of the logarithm for $x^{s-1}$) $$g_n^*(s) (1-e^{2\pi i(s-1)}) = 2\pi i \sum_{q=1}^n \operatorname{Res}(g_n(x); x=-q).$$ The sum of the residues is $$ \sum_{q=1}^n \operatorname{Res}(g_n(x); x=-q) \\= \sum_{q=1}^n (-q)^{s-1} \prod_{k=1}^{q-1} \frac{1}{-q+k} \prod_{k=q+1}^n \frac{1}{-q+k} = \sum_{q=1}^n e^{i\pi(s-1)} q^{s-1} \frac{(-1)^{q-1}}{(q-1)!} \frac{1}{(n-q)!} \\= - e^{i\pi s} \sum_{q=1}^n q^s \frac{(-1)^{q-1}}{q!} \frac{1}{(n-q)!} = -\frac{e^{i\pi s}}{n!}\sum_{q=1}^n q^s (-1)^{q-1} \binom{n}{q}.$$ This gives $$g_n^*(s) = - \frac{1}{n!} \frac{2\pi i \times e^{i\pi s}}{1-e^{2\pi i(s-1)}} \sum_{q=1}^n q^s (-1)^{q-1} \binom{n}{q} \\ = - \frac{1}{n!} \frac{2\pi i }{e^{-\pi i s} - e^{\pi i s}} \sum_{q=1}^n q^s (-1)^{q-1} \binom{n}{q} = \frac{1}{n!} \frac{\pi}{\sin(\pi s)} \sum_{q=1}^n q^s (-1)^{q-1} \binom{n}{q}.$$ We apply Mellin inversion to recover $g_n(x)$ with the Mellin inversion integral being $$\frac{1}{2\pi i}\int_{1/2-i\infty}^{1/2+i\infty} g_n^*(s)/x^s ds,$$ shifting to the left to recover an expansion about zero. The residue at $s=0$ is special, it has the value $$\frac{1}{n!} \sum_{q=1}^n (-1)^{q-1} \binom{n}{q} = \frac{1}{n!} .$$ The remaining residues at the negative integers $-p$ contribute $$\frac{1}{n!} \sum_{p=1}^\infty (-1)^p x^p \sum_{q=1}^n \frac{1}{q^p} (-1)^{q-1} \binom{n}{q} = \frac{1}{n!} \sum_{q=1}^n (-1)^{q-1} \binom{n}{q} \sum_{p=1}^\infty \frac{(-1)^p\times x^p}{q^p} \\= \frac{1}{n!} \sum_{q=1}^n (-1)^{q-1} \binom{n}{q} \frac{-x/q}{1+x/q} = \frac{1}{n!} \sum_{q=1}^n (-1)^q \binom{n}{q} \frac{x}{x+q}.$$ Including the residue at zero we thus obtain $$ g_n(x) = \frac{1}{n!} + \frac{1}{n!} \sum_{q=1}^n (-1)^q \binom{n}{q} \frac{x}{x+q} = \frac{1}{n!} \sum_{q=0}^n (-1)^q \binom{n}{q} \frac{x}{x+q}.$$ Since $f_n(x) = 1/x \times g_n(x)$ we get that $$f_n(x) = \frac{1}{n!} \sum_{q=0}^n (-1)^q \binom{n}{q} \frac{1}{x+q}.$$ Observation Aug 25 2014. The Mellin transform calculation is little more than a reworked computation of the partial fraction decomposition of the product term by residues and is in fact not strictly necessary here. An example of this very simple technique (no transforms) is at this MSE link.

Integration of $x^2 \sin(x)$ by parts How would I integrate the following? $$\int_0^{\pi/2} x^2\sin(x)\,dx$$ I did $u=x^2$ and $dv=\sin(x)$ I got $x^2-\cos(x)+2\int x\cos(x)\,dx.\quad$ I then used $u=x$ and $dv=\cos(x).$ I got $$x^2-\cos(x)+2[x-\sin(x)-\int\sin(x)]$$ then $x^2-\cos(x)+-2 \sin(x)(x)-\cos(x)\Big|_0^{\pi/2} =\dfrac{\pi^2}{4}-0-2$

You need to multiply $u$ and $v$, then subtract the subsequent integral: So you should have $$\begin{align} \int_0^{\pi/2} x^2\sin(x)\,dx & = -x^2\cos(x)+2\int x\cos(x)\,dx \\ \\ & = -x^2 \cos x + 2\Big[x \sin x - \int \sin x\,dx\Big]\\ \\ & = -x^2\cos x + 2x \sin x - (-2\cos x)\\ \\ & = -x^2 \cos x + 2x \sin x + 2\cos x \Big|_0^{2\pi}\end{align}$$ And proceed from there.

Trigonometric problem using basic trigonometry If $x$ is a solution of the equation: $$\tan^3 x = \cos^2 x - \sin^2 x$$ Then what is the value of $\tan^2 x$? This is the problem you are supposed to do it just with highschool trigonometry , but i can't manage to do it please help Here are the possible answers: $$a) \sqrt{2}-1, b) \sqrt{2}+1, c) \sqrt{3}-1, d) \sqrt{3}+1, e)\sqrt{2}+3$$

$\tan^2 x=\cos^2 x-\sin^2 x$ $\sin^2 x=\cos^4 x -\sin^2 x \cos^2 x$ $0=\cos^4 x -\sin^2 x \cos^2 x -\sin^2 x$ $0=\cos^4 x - \sin^2 x(1+\cos^2 x)$ $0=\cos^4 x - (1-\cos^2 x)(1+\cos^2 x)$ $0=\cos^4 x -(1- \cos^4 x)$ $0=2 \cos^4 x -1$ $\cos^2 x=\frac {\sqrt 2}{2}$ $\large \frac {1}{\cos^2 x}=\sqrt 2$ $\tan^2 x=\sqrt 2 \sin^2 x$ Looks like we're stuck, but from above we have $\cos^2 x=\frac {\sqrt 2}{2}$ , so $\sin^2 x=1-\frac {\sqrt 2}{2}$ By substitution, $\tan^2 x=\sqrt 2 (1-\frac {\sqrt 2}{2})$ $\tan^2 x=\sqrt 2 -1$

How to compute the integral $ \int\frac{1}{x\sqrt{x^2 +3x}}dx$ Given a problem : $$ \int\frac{1}{x\sqrt{x^2 +3x}}dx, $$ what is the best solution for this? I am thinking about solving this problem by using : $$ u = x+3;\qquad x = u-3; $$ So that we get : $ \int\frac{1}{x\sqrt{x}\sqrt{x+3}} dx$, then $ \int\frac{1}{(u-3)^{3/2}(u)^{1/2}} du$, then $ \int (u)^{-1/2} (u-3)^{-3/2} du$. Am I right so far? or is there a better method? Thanks.

I have not done an integral this cumbersome in quite a while. The technique that immediately stands out to me is trigonometric substitution. While I have performed the integration correctly on paper, I would appreciate someone being on the hunt for typesetting errors. Buckle seatbelt... We have $$\int \frac{1}{x\sqrt{x^2+3x}} dx.$$ Now we complete the square on the radicand to get $$\int \frac{1}{x\sqrt{x^2+3x}} dx=\int \frac{1}{x\sqrt{\left( x+\frac{3}{2} \right)^2-\frac{9}{4}}} dx.$$ For our substitution, let \begin{align*} &x+\frac{3}{2}=\frac{3}{2}\sec(\theta) \\ \Rightarrow & x=\frac{3}{2}(\sec(\theta)-1) \\ \Rightarrow & dx=\frac{3}{2}\sec(\theta)\tan(\theta)d \theta. \\ &\left( x+\frac{3}{2} \right)^2=\frac{9}{4}\sec^2(\theta) \\ \Rightarrow &\left( x+\frac{3}{2} \right)^2-\frac{9}{4}=\frac{9}{4}\sec^2(\theta)-\frac{9}{4} \\ & \qquad \qquad \qquad \quad= \frac{9}{4}\tan^2(\theta). \end{align*} We now make our substitutions and then integrate with respect to $\theta$. \begin{align*} \int \frac{1}{x\sqrt{\left( x+\frac{3}{2} \right)^2-\frac{9}{4}}} dx &=\int\frac{\frac{3}{2}\sec(\theta)\tan(\theta) d \theta}{\frac{3}{2}(\sec(\theta)-1)\sqrt{\frac{9}{4}\tan(\theta)}} \\ &=\frac{2}{3}\int \frac{\sec(\theta)\tan(\theta) d \theta}{\tan(\theta)(\sec(\theta)-1)} \\ &=\frac{2}{3}\int\frac{\sec(\theta)d \theta}{\sec(\theta)-1} \\ &=... \\ &=\frac{2}{3}\int \left( \csc^2(\theta)+\csc(\theta)\cot(\theta) \right) d\theta \\ &=\frac{2}{3}\left( -\cot(\theta)-\csc(\theta) \right) +c. \end{align*} Finally we back substitute (a right triangle helps). We started with the substitution $$x+\frac{3}{2}=\frac{3}{2}\sec(\theta) \Rightarrow \sec(\theta)=\frac{2x+3}{3}.$$ If we form a right triangle with $3$ adjacent to $\theta$, and $2x+3$ on the hypoteneuse (this follows as $\sec(\theta)=\frac{2x+3}{3}$), then we find that the opposite side is $2\sqrt{x^2+3x}.$ Thus the back substitution becomes (reading straight from the right triangle), \begin{align*} &\frac{2}{3}\left( -\cot(\theta)-\csc(\theta) \right) +c \\ =&\frac{2}{3}\left( -\frac{3}{2\sqrt{x^2+3x}} -\frac{2x+3}{\sqrt{x^2+3x}}\right) +c \\ =& ... \\ =& -\frac{2\sqrt{x^2+3x}}{3x}+c. \end{align*} I will be happy to include the derivations of the trigonometric moves, and this final simplification upon request where I inserted "..." into the process. Transforming $$\frac{\sec(\theta)}{\sec(\theta)-1}=\csc^2(\theta)+\csc(\theta)\cot(\theta)$$ was not what I would call trivial (took me 6 steps or so). A picture of the right triangle for the back substitution would also enhance this, and I could also add that later unless someone else feels more motivated right now.

How to evaluate this limit: $\lim_{x\to 0}\frac{\sqrt{x+1}-1}{x} = \frac12$? How do I evaluate the limit of $$\lim_{x\to 0}\frac{\sqrt{x+1}-1}{x} = \frac{1}{2}$$? As $x$ approaches $0$, I know the answer is $\frac{1}{2}$, but I got this question wrong. I think you have to multiply by the conjugate of the numerator? Can someone explain the steps to this, I must be doing something wrong when I multiply.

Method 1 (basic) $$\frac{\sqrt{x+1} - 1}{x} \stackrel{\sqrt{x+1}^2 = |x+1|}{=} \frac{|x + 1| - 1}{x (\sqrt{x+1} + 1)} \stackrel{x+1 \geq 0, \text{ for well-def.}}{=} \frac{1}{\sqrt{x + 1} + 1} \to \frac{1}{1+1} = \frac{1}{2}$$ as $x\to 0$, because $x\mapsto \sqrt{x}$ is continuous (and so the limit can be "used as input") Method 2 (derivative) $$\frac{\sqrt{x+1} - 1}{x} = \frac{f(1 + x) - f(1)}{x}$$ where $f(y) = \sqrt{y}$, thus the limit is $$f'(1) = \frac{1}{2 \sqrt{1}} = \frac{1}{2}$$ Method 3 (l'Hospital) Since the form is $\frac{0}{0}$, l'Hospital can be applied and gives $$\lim_{x\to 0} \frac{\sqrt{x+1} - 1}{x} = \lim_{x\to 0} \frac{\frac{1}{2\sqrt{x+1}}}{1} = \lim_{x\to 0} \frac{1}{2\sqrt{x + 1}} = \frac{1}{2 \sqrt{1}} = \frac{1}{2}$$

What should be added to $x^4 + 2x^3 - 2x^2 + x - 1$ to make it exactly divisible by $x^2 + 2x - 3$? I'm a ninth grader so please try to explain the answer in simple terms . I cant fully understand the explanation in my book . It just assumes that the expression that should be added has a degree of 1. I apologize if this question is too simple or just stupid but this is a genuine doubt.

You can get to a quartic divisible by $x^2+2x-3$ by writing $$\begin{align} x^4+2x^3-2x^2+x-1+\text{something}&=(x^2+2x-3)(x^2+ax+b)\cr &=x^4+(a+2)x^3+(2a+b-3)x^2+(2b-3a)x-3b\cr \end{align}$$ which leads to $$\text{something} = ax^3+(2a+b-1)x^2+(2b-3a-1)x+(1-3b)$$ for any coefficients $a$ and $b$ (of the quotient) that your heart desires. What the book presumably has in mind is to make the "something" of degree as small as possible. You can obviously get rid of the $ax^3$ by setting $a=0$ and then the $(2a+b-1)x^2$ by setting $b=1$. This leaves $$\text{something}=(2b-3a-1)x+(1-3b)= x-2$$

Limit $\lim_{n\to\infty}\frac{n^3+2}{n^2+3}$ Find the limit and prove your answer is correct $$\lim_{n\to\infty}\frac{n^3+2}{n^2+3}$$ By divide everything by $n^3$ I got $$\lim_{n\to\infty}\frac{n^3+2}{n^2+3}=\frac10 $$ which is undefined. So I conclude that there is no limit for this sequence. However, I don't know how to prove it

Hint: $$\frac{n^3+2}{n^2+3}=\frac{n^3+3n-3n +2 }{n^2+3}= \frac{n(n^2+3)}{n^2+3} +\frac{2-3n}{n^2+3}=n +\frac{2-3n}{n^2+3} $$

Evaluating $\int_a^b \frac12 r^2\ \mathrm d\theta$ to find the area of an ellipse I'm finding the area of an ellipse given by $\frac{x^2}{a^2}+\frac{y^2}{b^2} = 1$. I know the answer should be $\pi ab$ (e.g. by Green's theorem). Since we can parameterize the ellipse as $\vec{r}(\theta) = (a\cos{\theta}, b\sin{\theta})$, we can write the polar equation of the ellipse as $r = \sqrt{a^2 \cos^2{\theta}+ b^2\sin^2{\theta}}$. And we can find the area enclosed by a curve $r(\theta)$ by integrating $$\int_{\theta_1}^{\theta_2} \frac12 r^2 \ \mathrm d\theta.$$ So we should be able to find the area of the ellipse by $$\frac12 \int_0^{2\pi} a^2 \cos^2{\theta} + b^2 \sin^2{\theta} \ \mathrm d\theta$$ $$= \frac{a^2}{2} \int_0^{2\pi} \cos^2{\theta}\ \mathrm d\theta + \frac{b^2}{2} \int_0^{2\pi} \sin^2{\theta} \ \mathrm d\theta$$ $$= \frac{a^2}{4} \int_0^{2\pi} 1 + \cos{2\theta}\ \mathrm d\theta + \frac{b^2}{4} \int_0^{2\pi} 1- \cos{2\theta}\ \mathrm d\theta$$ $$= \frac{a^2 + b^2}{4} (2\pi) + \frac{a^2-b^2}{4} \underbrace{\int_0^{2\pi} \cos{2\theta} \ \mathrm d\theta}_{\text{This is $0$}}$$ $$=\pi\frac{a^2+b^2}{2}.$$ First of all, this is not the area of an ellipse. Second of all, when I plug in $a=1$, $b=2$, this is not even the right value of the integral, as Wolfram Alpha tells me. What am I doing wrong?

HINT: Putting $x=r\cos\theta,y=r\sin\theta$ $$\frac {x^2}{a^2}+\frac{y^2}{b^2}=1,$$ $$r^2=\frac{a^2b^2}{b^2\cos^2\theta+a^2\sin^2\theta}=b^2\frac{\sec^2\theta}{\frac{b^2}{a^2}+\tan^2\theta}$$

Need help with $\int_0^\infty\left(\pi\,x+\frac{S(x)\cos\frac{\pi x^2}2-C(x)\sin\frac{\pi x^2}2}{S(x)^2+C(x)^2}\right)dx$ Let $$I=\int_0^\infty\left(\pi\,x+\frac{S(x)\cos\frac{\pi x^2}2-C(x)\sin\frac{\pi x^2}2}{S(x)^2+C(x)^2}\right)dx,\tag1$$ where $$S(x)=-\frac12+\int_0^x\sin\frac{\pi t^2}2dt,\tag2$$ $$C(s)=-\frac12+\int_0^x\cos\frac{\pi t^2}2dt\tag3$$ are shifted Fresnel integrals. Mathematica and Maple return the integral $I$ unevaluated. Numerical integration suggests that $$I\stackrel?=-\frac\pi4,\tag4$$ but I was not able to prove it. So, I ask for your help with this problem.

Notice the integrand can be rewritten as: $$\pi x + \frac{S C' - C S'}{S^2 + C^2} = \pi x + \frac{1}{2i}\left(\frac{C' - iS'}{C - iS} - \frac{C' + iS'}{C + iS}\right) = \pi x + \frac{1}{2i} \log\left(\frac{C-iS}{C+iS}\right)' $$ The integral is equal to $$\lim_{x\to\infty} \left[\frac{\pi t^2}{2} + \frac{1}{2i}\log\left(\frac{C(t)-iS(t)}{C(t)+iS(t)}\right)\right]_0^x =\lim_{x\to\infty} \left(\frac{\pi x^2}{2} + \Im ( \log M(x) )\right) $$ where $\displaystyle\quad M(x) = \frac{C(x)-iS(x)}{C(0)-iS(0)}$. Using $$\int_0^{\infty} \cos \frac{\pi t^2}{2} dt = \int_0^{\infty} \sin \frac{\pi t^2}{2} dt = \frac12,$$ we can simplify $M(x)$ to $$\begin{align} \frac{\color{red}{-}\int_{x}^{\infty} e^{-i\frac{\pi}{2} t^2}dt}{\color{red}{-\frac12+\frac12 i}} = & \sqrt{2}e^{i(\color{red}{\frac{\pi}{4}}-\color{blue}{\frac{\pi}{2}x^2})}\int_0^{\infty} e^{-i(\frac{\pi}{2} t^2 + \pi x t)} dt\\ \sim & \frac{\sqrt{2}}{\color{green}{i} \pi x} e^{i(\color{red}{\frac{\pi}{4}}-\color{blue}{\frac{\pi}{2}x^2})}( 1 + O(\frac{1}{x}) ) \quad\text{ for large } x. \end{align} $$ From this, we find the integral equals to $$\lim_{x\to\infty} \left( \frac{\pi x^2}{2} + \color{red}{\frac{\pi}{4}} - \color{green}{\frac{\pi}{2}} - \color{blue}{\frac{\pi x^2}{2}} + O(\frac{1}{x}) \right) = -\frac{\pi}{4}.$$

Finding the Determinant of a $3\times 3$ matrix. Show that: $$ \begin{vmatrix} x & a & b \\ x^2 & a^2 & b^2 \\ a+b & x+b & x+a \\ \end{vmatrix} = (b - a)(x-a)(x-b)(x+a+b) $$ I tried expanding the whole matrix out, but it looks like a total mess. Does anyone have an idea how this could be simplified?

you can expand the matrix determinant and look for "shortcuts" , my solution : $$ \begin{vmatrix} x & a & b \\ x^2 & a^2 & b^2 \\ a+b & x+b & x+a \\ \end{vmatrix} = x \begin{vmatrix} x & a\\ x^2 & a^2\\ \end{vmatrix}-a\begin{vmatrix} x^2 & b^2\\ a+b & x+a\\ \end{vmatrix}+b\begin{vmatrix} x^2 & a^2\\ a+b & x+b\\ \end{vmatrix} $$ now just expand the matrices determinant: $$=x(x^2a^2-ax^2)-a(x^2(x+a)-b^2(a+b))+b(x^2(x+b)-a^2(a+b))=$$ open the brackets: $$x^3a^2-ax^3-ax^3-a^2x^3-a^2b^2+ab^3+bx^3+x^2b^2-a^3b-a^2b^2= $$ eliminate opposite sign expressions : $$ a^3x-b^3x-ax^3+ab^3+bx^3-a^3b= $$ $$x(a^3-b^3)-x^3(a-b)-ab(a^2-b^3)=$$ extract common divisor : $$[b-a](x^3-x(a^2+b^2+ab)+ab(a+b))=$$ $$[b-a](x^3-x((a+b)^2-ab)+ab(a+b))=$$ $$[b-a](x(x^2-(a+b)^2)+ab(x+a+b))=[b-a][x+a+b](x(x-a-b)+ab)=[b-a][x+a+b](x^2-x(a+b)+ab)=[b-a][x+a+b][x-a][x-b]. $$ $$ \square $$

Show that $\frac{1}{(n+1)!}(1+\frac{1}{(n+1)}+\frac{1}{(n+1)^2}+\frac{1}{(n+1)^3}+\cdots)=\frac{1}{n!n}$ Show that $\frac{1}{(n+1)!}(1+\frac{1}{(n+1)}+\frac{1}{(n+1)^2}+\frac{1}{(n+1)^3}+\cdots)=\frac{1}{n!n}$, (here $n$ is a natural number) Maybe easy, but I cannot see it. Thanks in advance! Alexander

Since $$1+x+x^2+\cdots=\frac1{1-x}$$ Therefore, $$1+\frac1{n+1}+\frac1{(n+1)^2}+\cdots=\frac1{1-\frac1{n+1}}=\frac1{\frac{n+1-1}{n+1}}=\frac{n+1}{n}$$ Now, $$\frac1{(n+1)!}\cdot(1+\frac1{n+1}+\frac1{(n+1)^2}+\cdots)=\frac1{(n+1)!}\cdot\frac{n+1}{n}=\frac1{n!n}$$

Show that the equation $x^2+xy-y^2=3$ does not have integer solutions. Show that the equation $$x^2+xy-y^2=3$$ does not have integer solutions. I solved the equation for $x$: $x=\displaystyle \frac{-y\pm\sqrt{y^2+4(y^2+3)}}{2}$ $\displaystyle =\frac{-y\pm\sqrt{5y^2+12}}{2}$ I was then trying to show that $\sqrt{5y^2+12}$ can not be an integer using $r^2\equiv 12 \pmod{5y^2}$. I got stuck here.

Note that $$x^2+xy-y^2=(x-2y)^2+5(xy-y^2)=(x-2y)^2\qquad({\rm mod}\>5)\ .$$ But $$0^2=0,\quad(\pm1)^2=1,\quad(\pm2)^2=-1\qquad({\rm mod}\>5)\ ,$$ which implies that $3$ is not a quadratic residue modulo $5$.

How to prove $\cos 36^{\circ} = (1+ \sqrt 5)/4$? Given $4 \cos^2 x -2\cos x -1 = 0$. Use this to show that $\cos 36^{\circ} = (1+ \sqrt 5)/4$, $\cos 72^{\circ} = (-1+\sqrt 5)/4$ Your help is greatly appreciated! Thanks

Hint: Look at the Quadratic Formula: The solution to $ax^2+bx+c=0$ is $x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}$ The equation is based on the fact that $$ \cos(5x)=16\cos^5(x)-20\cos^3(x)+5\cos(x) $$ and that $\cos(5\cdot36^\circ)=-1$ to get $$ 16\cos^5(36^\circ)-20\cos^3(36^\circ)+5\cos(36^\circ)+1=0 $$ Factoring yields $$ (\cos(36^\circ)+1)(4\cos^2(36^\circ)-2\cos(36^\circ)-1)^2=0 $$ We know that $\cos(36^\circ)+1\ne0$; therefore, $$ 4\cos^2(36^\circ)-2\cos(36^\circ)-1=0 $$ Deciding between the two roots of this equation is a matter of looking at the signs of the roots. For $\cos(72^\circ)$, use the identity $\cos(2x)=2\cos^2(x)-1$.

How can I show that $\begin{pmatrix} 1 & 1 \\ 0 & 1\end{pmatrix}^n = \begin{pmatrix} 1 & n \\ 0 & 1\end{pmatrix}$? Well, the original task was to figure out what the following expression evaluates to for any $n$. $$\begin{pmatrix} 1 & 1 \\ 0 & 1\end{pmatrix}^{\large n}$$ By trying out different values of $n$, I found the pattern: $$\begin{pmatrix} 1 & 1 \\ 0 & 1\end{pmatrix}^{\large n} = \begin{pmatrix} 1 & n \\ 0 & 1\end{pmatrix}$$ But I have yet to figure out how to prove it algebraically. Suggestions?

Powers of matrices occur in solving recurrence relations. If you write $$ \begin{pmatrix} x_{n+1} \\ y_{n+1} \end{pmatrix} = \begin{pmatrix} 1 & 1 \\ 0 & 1\end{pmatrix} \begin{pmatrix} x_n \\ y_n \end{pmatrix} $$ then clearly $$ \begin{pmatrix} x_{n} \\ y_{n} \end{pmatrix} = \begin{pmatrix} 1 & 1 \\ 0 & 1\end{pmatrix}^n \begin{pmatrix} x_0 \\ y_0 \end{pmatrix} $$ and also $$ x_{n+1}=x_n+y_n, \qquad y_{n+1}=y_n $$ from which you get $$ x_n = x_0 + n\ y_0, \qquad y_n = y_0 $$ The first column of $A^n$ is given by taking $x_0=1$ and $y_0=0$, and so is $(1 \ 0)^T$. The second column is given by taking $x_0=0$ and $y_0=1$, and so is $(n \ 1)^T$.

Explicit Formula for a Recurrence Relation for {2, 5, 9, 14, ...} (Chartrand Ex 6.46[b]) Consider the sequence $a_1 = 2, a_2 = 5, a_3 = 9, a_4 = 14,$ etc... (a) The recurrence relation is: $a_1 = 2$ and $a_n = a_{n - 1} + (n + 1) \; \forall \;n \in [\mathbb{Z \geq 2}]$. (b) Conjecture an explicit formula for $a_n$. (Proof for conjecture pretermitted here) I wrote out some $a_n$ to compass to cotton on to an idea or pattern. It seems bootless. $\begin{array}{cc} a_2 = 5 & a_3 = 9 & a_4 = 14 & a_5 = 20 & a_6 = 27\\ \hline \\ 5 = 2 + (2 + 1) & 9 = 5 + (3 + 1) & 14 = 9 + (4 + 1) & 20 = 14 + (5 + 1) & 27 = 20 + (7 + 1) \\ \end{array}$ The snippy answer only says $a_n = (n^2 + 3n)/2$. Thus, could someone please explicate the (missing) motivation or steps towards this conjecture? How and why would one envision this?

Write out the series for $a_{n}$ to start with. We have that $a_{n} = a_{n-1} + (n+1)\\ \quad = a_{n-2} + n + (n+1) \\ \quad = \ldots \\ \quad = a_{1} + 3 + 4 + \ldots + (n+1) \\ \quad = 2 + 3 + 4 + \ldots + (n+1) \\ \quad = \displaystyle \left(\sum_{i=1}^{n+1} i \right) - 1 \\ \quad = (n+1)(n+2)/2 - 1, \quad (\text{using the value for the sum of the integers between 1 and}\ n + 1) \\ \quad = (n^{2} + 3n)/2 + 1 -1 \\ \quad = (n^{2} + 3n)/2$

Re-write $1 \cdot x$ to $x$. Given the following bi-directional re-write rules (where $1$ is a constant, $^{-1}$ is a unary operator, $\cdot$ is a binary operator, and $x,y,z$ are arbitrary terms): $$\begin{align*} x \cdot 1 &= x \\ x \cdot (y \cdot z) &= (x\cdot y) \cdot z \\ x \cdot x^{-1} &= 1 \end{align*}$$ we're asked to prove that $1 \cdot x = x$ (ie, there is a chain $t_0 \to t_1 \to t_3 \to \cdots \to t_n$ with $t_0 = 1\cdot x$, $t_n = x$, and $t_i \to t_{i+1}$ meaning one of the 3 equations above re-writes $t_i$ to $t_{i+1}$ (in either direction)). After staring at this for a while I'm beginning to doubt whether or not this is possible... can anyone a) confirm this is indeed possible and b) potentially nudge me in the right direction?

Another method of proof would be: Using the fact that $x = (x^{-1})^{-1}$, then $x^{-1} \cdot x = 1$ from axiom 3. \begin{align*} &1 \cdot 1 = 1 \\ &1 \cdot (x \cdot x^{-1}) = 1 \\ &(1 \cdot x) \cdot x^{-1} = x \cdot x^{-1} \\ &((1 \cdot x) \cdot x^{-1}) \cdot x = (x \cdot x^{-1}) \cdot x \\ &(1 \cdot x) \cdot (x^{-1} \cdot x) = x \cdot (x^{-1} \cdot x) \\ &(1 \cdot x) \cdot 1 = x \cdot 1 \\ &1 \cdot x = x \\ \end{align*}

How to prove: $a+b+c\le a^2+b^2+c^2$, if $abc=1$? Let $a,b,c \in \mathbb{R}$, and $abc=1$. What is the simple(st) way to prove inequality $$ a+b+c \le a^2+b^2+c^2. $$ (Of course, it can be generalized to $n$ variables).

By replacing $a, b, c$ by $|a|, |b|, |c|$ if needed, we may assume they are non-negative. Then just apply Jensen inequality (or AM-GM inequality) to deduce that $$ a+b+c = \sum_{\text{cyclic}} a^{4/3}b^{1/3}c^{1/3} \leq \sum_{\text{cyclic}} \frac{4}{6}a^{2} + \frac{1}{6}b^{2} + \frac{1}{6}c^{2} = a^2 + b^2 + c^2. $$

solving second order differential equation Bonsoir je cherche les solutions de l'équation differentielle de type $$x^3y''(x)+(ax^3+bx^2+cx+d)y(x) =0$$ Merci d'avance Good evening, I'm searching solutions of a differential equation of the type: $x^3y''(x) + (ax^3+bx^2+cx+d)y(x) = 0.$ Thanks is advance.

Hint: $x^3y''(x)+(ax^3+bx^2+cx+d)y(x)=0$ $\dfrac{d^2y}{dx^2}+\left(a+\dfrac{b}{x}+\dfrac{c}{x^2}+\dfrac{d}{x^3}\right)y=0$ Let $r=\dfrac{1}{x}$ , Then $\dfrac{dy}{dx}=\dfrac{dy}{dr}\dfrac{dr}{dx}=-\dfrac{1}{x^2}\dfrac{dy}{dr}=-r^2\dfrac{dy}{dr}$ $\dfrac{d^2y}{dx^2}=\dfrac{d}{dx}\left(-r^2\dfrac{dy}{dr}\right)=\dfrac{d}{dr}\left(-r^2\dfrac{dy}{dr}\right)\dfrac{dr}{dx}=\left(-r^2\dfrac{d^2y}{dr^2}-2r\dfrac{dy}{dr}\right)\left(-\dfrac{1}{x^2}\right)=\left(-r^2\dfrac{d^2y}{dr^2}-2r\dfrac{dy}{dr}\right)(-r^2)=r^4\dfrac{d^2y}{dr^2}+2r^3\dfrac{dy}{dr}$ $\therefore r^4\dfrac{d^2y}{dr^2}+2r^3\dfrac{dy}{dr}+(dr^3+cr^2+br+a)y=0$

Find zeros of this function: $$(3\tan(x)+4\cot(x))\cdot\sin(2x)$$ Do I have to multiply them and solve, or one by one, like: $$(3\tan(x)+4\cot(x))=0$$and$$\sin(2x)=0.$$

Observe that if $\displaystyle(3\tan x+4\cot x)=0\implies 3\tan x+\frac4{\tan x}=0\iff3\tan^2x+4=0$ which is impossible for real $x$ If $\sin2x=0, 2x=n\pi$ where $n$ is any integer If $n$ is even $=2m$(say) $2x=2m\pi, x=m\pi,\cot x=\cot m\pi=\frac{\cos m\pi}{\sin m\pi}=\frac{(-1)^m}0$ hence not finite If $n$ is odd $=2m+1$(say) $2x=(2m+1)\pi, x=\frac{(2m+1)\pi}2,\tan x=\tan\frac{(2m+1)\pi}2=\frac{(-1)^m}0$ hence not finite So, we don't have any real solution which will be more evident below Method $1:$ On multiplication, $\displaystyle(3\tan x+4\cot x)\sin2x=\left(3\frac{\sin x}{\cos x}+4\frac{\cos x}{\sin x}\right)2\sin x\cos x=6\sin^2x+8\cos^2x$ Using $\cos2A=2\cos^2A-1=1-2\sin^2A$ $6\sin^2x+8\cos^2x=3(1-\cos2x)+4(1+\cos2x)=7+\cos2x$ Do you know for real $y,-1\le \cos y\le 1$ Method $2:$ Using $\displaystyle\sin2A=\frac{2\tan x}{1+\tan^2x}$ and $\displaystyle\cot x=\frac1{\tan x}$ $\displaystyle(3\tan x+4\cot x)\sin2x=\left(3t+\frac4t\right)\frac{2t}{1+t^2}=\frac{2(3t^2+4)}{1+t^2}$ where $t=\tan x$

Calculations with permutations: show that $(1,2,3)^2(5,7)^2=(1,3,2)$ How can I show $(1,2,3)^2(5,7)^2=(1,3,2)$? And, specifically, what does $(1,2,3)^2$ and $(5,7)^2$ equal individually?

Note that with $(5, 7)^2 = (5, 7)(5, 7)$, $\quad (5 \to 7 \to 5)$ and $(7 \to 5 \to 7),\quad $ which gives us $(5, 7)(5, 7) = (1)$, the identity permutation. Any two-cycle, squared, gives us the identity permutation: it's an order two permutation. Since the first two squared cycles are disjoint from one another, you can simply compute $$(1, 2, 3)^2 \cdot (5, 7)^2 = (1, 2, 3)(1, 2, 3)\cdot(1) = (1, 2, 3)(1, 2, 3) = (1, 3, 2)$$

Show that $\gcd(a + b, a^2 + b^2) = 1$ or $2$ if $\gcd(a, b)=1$ Show that $\gcd(a + b, a^2 + b^2) = 1$ or $2$ if $\gcd(a, b)=1$. I have absolutely no clue where to start and what to do, please provide complete proof and answer.

We first show that there is no odd prime $p$ that divides both $a+b$ and $a^2+b^2$. For if $p$ divides both, then $p$ divides $(a+b)^2-(a^2+b^2)$, so $p$ divides $2ab$. Since $p$ is odd, it divides one of $a$ or $b$, say $a$. But then since $p$ divides $a+b$, it must divide $b$. This contradicts the fact that $a$ and $b$ are relatively prime. So the only possible common divisors of $a+b$ and $a^2+b^2$ are powers of $2$. If $a$ is even and $b$ is odd (or the other way) then the greatest common divisor of $a+b$ and $a^2+b^2$ is therefore $1$. If $a$ and $b$ are both odd, then $a^2+b^2\equiv 2\pmod{8}$, and therefore the highest power of $2$ that divides $a^2+b^2$ is $2^1$. So in that case $\gcd(a+b,a^2+b^2)=2$.

Last two digits of $2^{11212}(2^{11213}-1)$ What are the last two digits of the perfect numbers $2^{11212}(2^{11213}-1)$? I know that if $2^n-1$ is a prime, then $2^{n-1}(2^n-1)$ is a perfect number and that every even perfect number can be written in the form $2^n(2^n-1)$ where $2^n-1$ is prime. I'm not sure how to use this information though.

We want the remainder when the product is divided by $100$. Th remainder on division by $4$ is $0$, so all we need is the remainder on division by $25$. Note that $\varphi(25)=20$. So by Euler's Theorem, $2^{20}\equiv 1\pmod{25}$. It is easier to note that $2^{10}\equiv -1\pmod{25}$. It follows that $2^{11212}\equiv -4\pmod{25}$. The same idea shows that $2^{11213}-1\equiv -9\pmod{25}$. Thus our product is congruent to $36$, or equivalently $11$, modulo $25$. Now solve $x\equiv 0\pmod{4}$, $x\equiv 11\pmod{25}$. The solution is $36$ modulo $100$.

Question about Fermat's Theorem I'm trying to find $2^{25} \mod 21 $. By Fermat's theorem, $2^{20} \cong_{21} 1 $. Therefore, $2^{25} = 2^{20}2^{5} \cong_{21} 2^5 = 32 \cong_{21} 11 $. However, the answer in my book is $2$! What am I doing wrong? Also, I would like to ask what are the last two digits of $1 + 7^{162} + 5^{121} \times 3^{312} $ Thanks for your help.

By Euler's Theorem, we have $2^{12}\equiv 1\pmod{21}$. That is because $\varphi(21)=(2)(6)=12$. Thus $2^{25}=2^{12\cdot 2}\cdot 2^1\equiv 2\pmod{21}$. If we want to use Fermat's Theorem, we work separately modulo $3$ and modulo $7$. We have $2^2\equiv 1\pmod{3}$, and therefore $2^{25}=2^{2\cdot 12}\cdot 2^1\equiv 2\pmod{3}$. Similarly, $2^6\equiv 1\pmod{7}$ and therefore $2^{25}\equiv 2\pmod{7}$. It follows that $2^{25}\equiv 2\pmod{21}$. Added: For the last two digits of $1+7^{162}+(5^{121})(3^{312})$, it will be enough to evaluate modulo $4$ and modulo $25$. Modulo $4$: We have $7\equiv -1\pmod{4}$, so $7^{162}\equiv 1\pmod{4}$. Similarly, $5^{121}\equiv 1\pmod{4}$ and $3^{312}\equiv 1\pmod{4}$. Adding up, we get the sum is $\equiv 3\pmod{4}$. Modulo $25$: We don't have to worry about the messy last term. Note that $7^2\equiv -1\pmod{25}$, so $7^{160}\equiv 1\pmod{25}$. Thus $7^{162}\equiv -1\pmod{25}$. Thus our sum is congruent to $0$ modulo $25$. Finally, we want the multiple of $25$ between $0$ and $75$ which is congruent to $3$ modulo $4$. A quick scan shows the answer is $75$.

How to prove $(1+x)^n\geq 1+nx+\frac{n(n-1)}{2}x^2$ for all $x\geq 0$ and $n\geq 1$? I've got most of the inductive work done but I'm stuck near the very end. I'm not so great with using induction when inequalities are involved, so I have no idea how to get what I need... \begin{align} (1+x)^{k+1}&\geq (1+x)\left[1+kx+\frac{k(k-1)}{2}x^2\right]\\ &=1+kx+\frac{k(k-1)}{2}x^2+x+kx^2+\frac{k(k-1)}{2}x^3\\ &=1+(k+1)x+kx^2+\frac{k(k-1)}{2}x^2+\frac{k(k-1)}{2}x^3 \end{align} And here's where I have no clue how to continue. I thought of factoring out $kx^2$ from the remaining three terms, but I don't see how that can get me anywhere.

If $n=1$, it is trivial. Suppose it is true for $n$. We will show that this formula is true for $n+1$. $$ \begin{aligned} (1+x)^{n+1}&=(1+x)^n(1+x)\\ &\ge \left( 1+nx+\frac{n(n-1)}{2} x^2 \right)(1+x)\\ &= 1+nx+\frac{n(n-1)}{2}x^2 +x+nx^2 +\frac{n(n-1)}{2} x^3\\ &= 1+(n+1)x+\frac{n^2-n+2n}{2}x^2+\frac{n(n-1)}{2} x^3\\ &= 1+(n+1)x +\frac{n^2+n}{2}x^2+\frac{n(n-1)}{2} x^3\\ &\ge 1+(n+1)x +\frac{n(n+1)}{2}x^2\\ &= 1+(n+1)x +\frac{(n+1)(n+1-1)}{2}x^2 \end{aligned} $$ so we get desired result.

How to solve a recursive equation I have been given a task to solve the following recursive equation \begin{align*} a_1&=-2\\ a_2&= 12\\ a_n&= -4a_n{}_-{}_1-4a_n{}_-{}_2, \quad n \geq 3. \end{align*} Should I start by rewriting $a_n$ or is there some kind of approach to solve these? I tried rewriting it to a Quadratic Equation (English isn't my native language, sorry if this is incorrect). Is this the right approach, if so how do I continue? \begin{align*} a_n&= -4a_n{}_-{}_1-4a_n{}_-{}_2\\ x^2&= -4x-4\\ 0&= x^2 + 4x + 4 \end{align*}

$\displaystyle{\large a_{1} = -2\,\quad a_{2} = 12,\quad a_{n} = -4a_{n - 1} - 4a_{n - 2}\,,\quad n \geq 3}$. $$ \sum_{n = 3}^{\infty}a_{n}z^{n} = -4\sum_{n = 3}^{\infty}a_{n - 1}z^{n} - 4\sum_{n = 3}^{\infty}a_{n - 2}z^{n} = -4\sum_{n = 2}^{\infty}a_{n}z^{n + 1} - 4\sum_{n = 1}^{\infty}a_{n}z^{n + 2} $$ $$ \Psi\left(z\right) - a_{1}z - a_{2}z^{2} = -4z\left[\Psi\left(z\right) - a_{1}z\right] - 4z^{2}\Psi\left(z\right) $$ where $\displaystyle{\Psi\left(z\right) \equiv \sum_{n = 1}^{\infty}a_{n}z^{n}. \quad z \in {\mathbb C}.\quad \left\vert z\right\vert < {1 \over 2}}$. \begin{align} \Psi\left(z\right) &= {\left(4a_{1} +a_{2}\right)z^{2} + a_{1}z \over 4z^{2} + 4z + 1} = {4z^{2} - 2z \over 4z^{2} + 4z + 1} = {4z^{2} - 2z \over \left(1 + 2z\right)^{2}} = \left(z - 2z^{2}\right)\,{{\rm d} \over {\rm d} z}\left(1 \over 1 + 2z\right) \\[3mm]&= \left(z - 2z^{2}\right)\,{{\rm d} \over {\rm d} z}\sum_{n = 0}^{\infty}\left(-\right)^{n}2^{n}z^{n} = \left(z - 2z^{2}\right)\sum_{n = 1}^{\infty}\left(-\right)^{n}2^{n}nz^{n - 1} \\[3mm]&= \sum_{n = 1}^{\infty}\left(-1\right)^{n}2^{n}nz^{n} - \sum_{n = 1}^{\infty}\left(-1\right)^{n}2^{n + 1}nz^{n + 1} = \sum_{n = 1}^{\infty}\left(-1\right)^{n}2^{n}nz^{n} - \sum_{n = 2}^{\infty}\left(-1\right)^{n - 1}2^{n}\left(n - 1\right)z^{n} \\[3mm]&= -2z + \sum_{n = 2}^{\infty}\left(-1\right)^{n}2^{n}nz^{n} - \sum_{n = 2}^{\infty}\left(-1\right)^{n - 1}2^{n}\left(n - 1\right)z^{n} \\[3mm]&= -2z + \sum_{n = 2}^{\infty}\left(-1\right)^{n}\left(2n - 1\right)2^{n}z^{n} \end{align} $$ \Psi\left(z\right) = \sum_{n = 1}^{\infty}a_{n}z^{n} = \sum_{n = 1}^{\infty}\left(-1\right)^{n}\left(2n - 1\right)2^{n}z^{n} $$ $$ \begin{array}{|c|}\hline\\ \color{#ff0000}{\large\quad% a_{n} \color{#000000}{\large\ =\ } \left(-1\right)^{n}\left(2n - 1\right)2^{n}\,, \qquad\qquad n = 1, 2, 3, \ldots \quad} \\ \\ \hline \end{array} $$

If $\frac{\cos x}{\cos y}=\frac{a}{b}$ then $a\tan x +b\tan y$ equals If $$\frac{\cos x}{\cos y}=\frac{a}{b}$$ Then $$a \cdot\tan x +b \cdot\tan y$$ Equals to (options below): (a) $(a+b) \cot\frac{x+y}{2}$ (b) $(a+b)\tan\frac{x+y}{2}$ (c) $(a+b)(\tan\frac{x}{2} +\tan\frac{y}{2})$ (d) $(a+b)(\cot\frac{x}{2}+\cot\frac{y}{2})$ My approach : $$\frac{\cos x}{\cos y} = \frac{a}{b} $$ [ Using componendo and dividendo ] $$\frac{\cos x +\cos y}{\cos x -\cos y} = \frac{a+b}{a-b}$$ $$=\frac{2\cos\frac{x+y}{2}\cos\frac{x-y}{2}}{2\sin\frac{x+y}{2}\sin\frac{y-x}{2}}$$ I'm stuck, I'd aprecciate any suggestions. Thanks.

So, we have $$\frac a{\cos x}=\frac b{\cos y}=\frac{a+b}{\cos x+\cos y}$$ $$\implies a\tan x+b\tan y=\frac a{\cos x}\cdot\sin x+\frac b{\cos y}\cdot\sin y$$ Putting the values of $\displaystyle\frac a{\cos x},\frac b{\cos y}$ $$a\tan x+b\tan y=(\sin x+\sin y)\frac{(a+b)}{\cos x+\cos y}$$ Now, $\displaystyle \cos x+\cos y=2\cos\frac{x+y}2\cos\frac{x-y}2$ and $\displaystyle \sin x+\sin y=2\sin\frac{x+y}2\cos\frac{x-y}2$

Prove that $x^2 + xy + y^2 \ge 0$ by contradiction Using the fact that $x^2 + 2xy + y^2 = (x + y)^2 \ge 0$, show that the assumption $x^2 + xy + y^2 < 0$ leads to a contradiction... So do I start off with... "Assume that $x^2 + xy + y^2 <0$, then blah blah blah"? It seems true...because then I go $(x^2 + 2xy + y^2) - (x^2 + xy + y^2) \ge 0$. It becomes $2xy - xy \ge 0$, then $xy \ge 0$. How is this a contradiction? I think I'm missing some key point.

By completing square, $$x^2+xy+y^2 = x^2+2x\frac{y}{2}+\frac{y^2}{4} + \frac{3y^2}{4} = \left(x+\frac{y}{2}\right)^2+\frac{3y^2}{4}\ge 0$$

How to express a vector as a linear combination of others? I have 3 vectors, $(0,3,1,-1), (6,0,5,1), (4,-7,1,3)$, and using Gaussian elimination I found that they are linearly dependent. The next question is to express each vector as a linear combination of the other two. Different resources say just to use Gaussian elimination, but I just end up with a matrix in RREF. How can I find different vectors as a linear combination of others?

Let's look at Gaussian elimination: \begin{align} \begin{bmatrix} 0 & 6 & 4 \\ 3 & 0 & -7 \\ 1 & 5 & 1 \\ -1 & 1 & 3 \end{bmatrix} \xrightarrow{\text{swap row 1 and 3}}{}& \begin{bmatrix} 1 & 5 & 1 \\ 3 & 0 & -7 \\ 0 & 6 & 4 \\ -1 & 1 & 3 \end{bmatrix}\\ \xrightarrow{R_2-3R_1}{}& \begin{bmatrix} 1 & 5 & 1 \\ 0 & -15 & -10 \\ 0 & 6 & 4 \\ -1 & 1 & 3 \end{bmatrix}\\ \xrightarrow{R_4+R_1}{}& \begin{bmatrix} 1 & 5 & 1 \\ 0 & -15 & -10 \\ 0 & 6 & 4 \\ 0 & 6 & 4 \end{bmatrix}\\ \xrightarrow{-\frac{1}{15}R_2}{}& \begin{bmatrix} 1 & 5 & 1 \\ 0 & 1 & 2/3 \\ 0 & 6 & 4 \\ 0 & 6 & 4 \end{bmatrix}\\ \xrightarrow{R_3-6R_2}{}& \begin{bmatrix} 1 & 5 & 1 \\ 0 & 1 & 2/3 \\ 0 & 0 & 0 \\ 0 & 6 & 4 \end{bmatrix}\\ \xrightarrow{R_4-6R_2}{}& \begin{bmatrix} 1 & 5 & 1 \\ 0 & 1 & 2/3 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix}\\ \xrightarrow{R_1-5R_2}{}& \begin{bmatrix} 1 & 0 & -7/3 \\ 0 & 1 & 2/3 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix}\\ \end{align} If $v_1$, $v_2$ and $v_3$ are your vectors, this says that $$ v_3=-\frac{7}{3}v_1+\frac{2}{3}v_2 $$ because elementary row operations don't change linear relations between the columns.

Hard Integral $\frac{1}{(1+x^2+y^2+z^2)^2}$ Prove that $\displaystyle\int_{-\infty}^{\infty}\displaystyle\int_{-\infty}^{\infty}\displaystyle\int_{-\infty}^{\infty} \frac{1}{(1+x^2+y^2+z^2)^2}\, dx \, dy \, dz = \pi^2$ I tried substitution, trigonometric substitution, and partial fraction decomposition, but I can't solve this problem, I only know that $\frac{1}{(1+x^2+y^2+z^2)^2}$ is a even function :( then $$\int_{-\infty}^{\infty} \int_{-\infty}^{\infty} \int_{-\infty}^{\infty} \frac{1}{(1+x^2+y^2+z^2)^2}\, dx \, dy \, dz = $$ $$ 2 \int_{-\infty}^{\infty} \int_{-\infty}^{\infty} \int_{0}^{\infty} \frac{1}{(1+x^2+y^2+z^2)^2}\, dx \, dy \, dz $$

An alternative is to overkill it with some measure theory. Unfortunately I don't know the names of the theorems and objects used (not in my language and not in english). If someone does, please edit my answer as you see fit. Firstly note that $$\displaystyle\int \limits_{-\infty}^{\infty}\displaystyle\int \limits_{-\infty}^{\infty}\displaystyle\int \limits_{-\infty}^{\infty} \dfrac{1}{(1+x^2+y^2+z^2)^2}\mathrm dx \,\mathrm dy \,\mathrm dz=\iiint \limits_{\Bbb R^3\setminus\{0_{\Bbb R^3}\}}\dfrac{1}{(1+x^2+y^2+z^2)^2}\mathrm dx \,\mathrm dy \,\mathrm dz,$$ then use change of variables and something which translates to generalized polar coordinates to get $$\iiint \limits_{\Bbb R^3\setminus\{0_{\Bbb R^3}\}}\dfrac{1}{(1+x^2+y^2+z^2)^2}\mathrm dx \,\mathrm dy \,\mathrm dz=\iint \limits_{]0,+\infty[\times S_2} \dfrac{t^2}{\left(1+(tx)^2+(ty)^2+(tz)^2\right)^2}\mathrm dt \,\mathrm d\mu_{S_2}(x,y,z).$$ Now $$\displaystyle \begin{align} \iint \limits_{]0,+\infty[\times S_2} \dfrac{t^2\mathrm dt \,\mathrm d\mu_{S_2}(x,y,z)}{\left(1+(tx)^2+(ty)^2+(tz)^2\right)^2}&=\iint \limits_{]0,+\infty[\times S_2} \dfrac{t^2\mathrm dt \,\mathrm d\mu_{S_2}(x,y,z)}{\left(1+t^2(x^2+y^2+z)^2\right)^2}\\ &=\iint \limits_{]0,+\infty[\times S_2} \dfrac{t^2\mathrm dt \,\mathrm d\mu_{S_2}(x,y,z)}{\left(1+t^2\right)^2}\\ &=\underbrace{\mu _{S_2}(S_2)}_{\large 4\pi}\int \limits_{]0,+\infty[} \dfrac{t^2}{\left(1+t^2\right)^2}\mathrm dt\\ &=4\pi\cdot \dfrac \pi 4=\pi ^2\end{align}$$

Congruences doubt! What the rest of the division $2^{100}$ by $11$? $$2^5=32\equiv10\equiv-1\pmod{11}\\(2^5)^{20}=2^{100}\equiv-1^{20}\;\text{or}\; (-1)^{20}$$??

We have \begin{align*} 2^{10} &= 2^5 \cdot 2^5 &\equiv (-1) \cdot (-1) &= (-1)^2 \pmod{11}\\ &\vdots\\ 2^{100} &=2^5 \cdots 2^5 &\equiv (-1) \cdots (-1) &= (-1)^{20} \pmod{11} \end{align*}

What the rest of the division $1^6+2^6+...+100^6$ by $7$? What the rest of the division $1^6+2^6+...+100^6$ by $7$? $1^6\equiv1\pmod7\\2^6\equiv64\equiv1\pmod7\\3^6\equiv729\equiv1\pmod7$ Apparently all the leftovers are $one$, I thought of using Fermat's Little Theorem, however the $(7,7 k) = 7$, so you can not generalize, I think. help please.

The number of multiples of $7$ from $1$ to $100$ is $\left\lfloor\frac{100}{7}\right\rfloor = 14$ so... By Fermat's Little Theorem : $7\mid 1^6+2^6+…+100^6 - 86 \implies 1^6+2^6+…+100^6 \equiv 86 \equiv 2 \pmod7 $

Use the Chinese remainder theorem to find the general solution of $x \equiv a \pmod {2^3}, \; x \equiv b \pmod {3^2}, \; x \equiv c \pmod {11}$ Help! Midterm exam is coming, but i still unable to solve this simple problem using the Chinese remainder theorem. $$x \equiv a \pmod {2^3}, \quad x \equiv b \pmod {3^2}, \quad x \equiv c \pmod {11}.$$

From the condition of the equation we have: $$x \equiv a \pmod 8 \implies x = 8k + a$$ $$x \equiv b \pmod 9 \implies x = 9n + b$$ $$x \equiv c \pmod {11} \implies x = 11m + c$$ Now we have: $$8k+a=9n+b$$ $$8k+a\equiv b \pmod 9$$ $$8k\equiv b-a \pmod 9$$ Having actual values would be easier to get congruence relation for k modulo 9, but now we'll use: $$k \equiv \frac{b-a + 9s}{8} \pmod 9 \implies k = 9t + \frac{b-a+9s}{8}$$ Note that if we add 9s on the RHS the congruence relation won't change. Sowe add $9s$ in order to get an integer when we divide by 9. Now substitute back we have: $$x=8k+a = 8\left(9t + \frac{b-a}{8}\right) = 72t + b-a + 9s$$ Now using this relation for x we do the same thing for this realtion and $x = 11m + c$. It maybe clearer to you with example: $$x \equiv 3 \pmod 8 \implies x = 8k + 3$$ $$x \equiv 2 \pmod 9 \implies x = 9n + 2$$ $$x \equiv 5 \pmod {11} \implies x = 11m + 5$$ $$8k + 3 = 9n + 2$$ $$8k + 3 \equiv 2 \pmod 9$$ $$8k \equiv -1 \equiv 8 \pmod 9$$ $$k \equiv 1 \pmod 9 \implies k = 9t + 1$$ Now we substitute back: $$x = 8k + 3 = 8(9t+1) + 3 = 72t + 8 + 3 = 72t + 11$$ Now we repeat the same procedure: $$72t + 11 = 11m + 5$$ $$72t + 11 \equiv 5 \pmod {11}$$ $$72t \equiv 5 \equiv 720 \pmod {11}$$ $$t \equiv 10 \pmod {11} \implies t = 11s + 10$$ Now we substitue: $$x = 72t + 11 = 72(11s + 10) + 11 = 792s + 720 + 11 = 792s + 731$$ We have a congruence relation for $x$: $$x \equiv 731 \pmod{792}$$ You can see that $792=8\cdot 9 \cdot 11$, that's because all moduli are coprime, otherwise we would end up with the least common multiple of the moduli.

Induction proof that $n! > n^3$ for $n \ge 6$, and $\frac{(2n)!}{n! 2^n}$ is an integer for $n \ge 1$ Prove by induction that (a) $n! > n^3$ for every $n \ge 6$. (b) prove $\frac{(2n)!}{n!2^n}$ is an integer for every $n\geq 1$ I'm quite terrible with induction so any help would be appreciated.

(a) $n! > n^3$ for every $n \geq 6$ For the induction base, we simple have to show that $6! \geq 6^3$. Hence, $6 * 5 * 4 * (3 * 2) * 1 \geq 6*6*6$. Hence $6 * 6 * (5 * 4) \geq 6^3$. This is evidently true, as $5*4 = 20 > 6$. For the induction step, we have to show that $(n+1)! \geq (n+1)^3$ for $n\geq 6$. Now suppose $n! > n^3$ is true for all $n \geq 6$. Then $(n+1)! = (n+1)n! \geq (n+1)n^3$. Hence, we are done if $(n+1)n^3 \geq (n+1)^3$ for $n\geq 6$. Thus, we need to show that $n^4 + n^3 \geq n^3 + 3n^2 + 3n + 1$ for $n\geq 6$. Rearranging yields $n^4 - 3n^2 - 3n - 1 \geq 0$ for $n\geq 6$. The second derivative of this is $12n^2 - 6$, which is strictly convex for $n\geq 6$. Hence, we are through if both $f(n) := n^4 - 3n^2 - 3n - 1 \geq 0$ and $f'(n) = 4n^3 -6n - 3 \geq 0$ for $n=6$. This is trivial to check. (b) $\frac{(2n)!}{n! 2^n}$ is an integer for every $n\geq 1$. The induction base is trivial: simply insert $n=1$. For the induction step, we need to show that $\frac{(2(n+1))!}{(n+1)^2 2^{n+1}} = \frac{(2n+2)!}{(n+1)n! * 2 * 2^n} = \frac{(2n+2)(2n+1)}{2(n+1)} * \frac{(2n)!}{n! 2^n}$ is an integer for all $n\geq 1$. Now, by the induction hypothesis, the right multiplicand must be an integer. Hence, it suffices to show that the left multiplicand must be an integer as well. Since $(2n+2) = 2(n+1)$, this is obviously the case.

If $ac-bd=p$ and $ad+bc=0$, then $a^2+b^2\neq 1$ and $c^2+d^2\neq 1$? I'm trying to prove the following: Let $a,b,c,d\in\Bbb{Z}$ and $p$ be a prime integer. If $ac-bd=p$ and $ad+bc=0$, prove that $a^2+b^2\neq 1$ and $c^2+d^2\neq 1$. Actually I'm not even sure if this is correct. A proof or counter-example (in case this assertion is wrong) would be great. I got the following 3 results: * *$p^2=(a^2+b^2)(c^2+d^2)$ *$b(c^2+d^2)=-pd$ *$a(c^2+d^2)=pc$ *$c(a^2+b^2)=pa$ *$d(a^2+b^2)=-pb$ I'm at a loss as to how to proceed beyond this. Assuming $c^2+d^2=1$ or $a^2+b^2=1$ does not seem to cause any contradictions. Thanks in advance!

Consider $p^2 = p^2 + 0^2 = (ac-bd)^2 + (ad+bc)^2 = (a^2+b^2)(c^2+d^2)$. Then either one of $a^2+b^2$, $c^2+d^2$ equals $p^2$ and the other equals $1$ or both equal $p$. This follows from unique factorization of $\mathbb{Z}$. Considering the first case, suppose wlog $a^2+b^2=1$ and $a=0$. Then $c^2+d^2=p^2$. Now if both $c$ and $d$ are nonzero, then, since $b$ is also not equal to zero, $ad+bc$ cannot be zero, a contradiction. Thus suppose, again wlog, that $c=0$ and $d^2=p^2$. This leads to a counterexample, namely $a=0,b=1,c=0,d=-p$ or $a=0,b=-1,c=0,d=p$.

How prove this analysis function $a\le\frac{1}{2}$ let $$f(x)=\begin{cases} x\sin{\dfrac{1}{x}}&x\neq 0\\ 0&x=0 \end{cases}$$ show that:there exsit $M>0,(x^2+y^2\neq 0)$ , $$F(x,y)=\dfrac{f(x)-f(y)}{|x-y|^{a}}|\le M \Longleftrightarrow a\le\dfrac{1}{2}$$ My try: (1)if $a\le\dfrac{1}{2}$, then $$\dfrac{f(x)-f(y)}{|x-y|^a}=\dfrac{x\sin{\frac{1}{x}}-y\sin{\frac{1}{y}}}{|x-y|^a}$$ then How can prove $$|x\sin{\frac{1}{x}}-y\sin{\frac{1}{y}}|<M|x-y|^a,a\le\dfrac{1}{2}$$ By other hand: and if for any $x,y\in R$,and such $$ |x\sin{\frac{1}{x}}-y\sin{\frac{1}{y}}|<M|x-y|^a$$ then How prove must $a\le\dfrac{1}{2}$? I think this is nice problem,Thank you By the way:when I deal this problem, I find this nice equality $$|x\sin{\frac{1}{x}}-y\sin{\frac{1}{y}}|<2\sqrt{|x-y|}$$ But I can't prove ,Thank you

I think that this is just a partial answer or this is just an estimation for $a_0$ where $a\leq a_0$. $ a_n=\frac{1}{2n\pi + \frac{\pi}{2}},\ b_n = \frac{1}{2n\pi - \frac{\pi}{2}}$ $f(a_n)- f(b_n) = a_n +b_n = \frac{4n\pi}{4n^2\pi^2 - (\pi/2)^2} \approx \frac{1}{n\pi}$ and $|a_n - b_n|^{a_0} \approx |\frac{\pi}{4n^2\pi^2} |^{a_0}$ Intuitively we conclude that $a_0=1/2$.

In the field $\mathbb{Z}_7[x]/\langle x^4+x+1\rangle$, find the inverse of $f(x)=x^3+x+3$. In the field $\mathbb{Z}_7[x]/\langle x^4+x+1\rangle$, find the inverse of $f(x)=x^3+x+3$. I know how to find the inverses of elements within sets, rings, and fields. I know what to do if the field was just $\mathbb{Z}_7$, but the fact that the field is $\mathbb{Z}_7[x]/\langle x^4+x+1\rangle$ confuses me. I don't know where to start.

The computation here is the same in ${\mathbb Q}$ as in ${\mathbb Z}_7$. You look for a solution of the form $$ z=a+bx+cx^2+dx^3 \tag{1} $$ You then have $$ z(x^3+x+1)=dx^6+cx^5+(b+d)x^4+(a+c+3d)x^3+(b+3c)x^2+(a+3b)x+3a=Q(x) \tag{2} $$ Next, divide the result by $x^4+x+1$ : $$ Q(x)=(x^4+x+1)(dx^2+cx+b+d)+R(x) \tag{3} $$ where the remainder $R(x)$ equals $$ R(x)=(a+c+2d)x^3+(b+2c-d)x^2+(a+2b-c-d)x+(3a-b-d) \tag{4} $$ Then, solve the system $$ a+c+2d=b+2c-d=a+2b-c-d=0, \ \ 3a-b-d=1 \tag{5} $$ This will lead you to the solution $$ a=\frac{11}{47}, b=\frac{-8}{47}, c=\frac{1}{47}, d=\frac{-6}{47}, z=\frac{11-8x+x^2-6x^3}{47} \tag{6} $$

Problem finding in simple algebra It is given, $$x= \sqrt{3}+\sqrt{2}$$ How to find out the value of $$x^4-\frac{1}{x^4}$$/ The answer is given $40 \sqrt{6}$ but my answer was not in a square-root form I have done in thsi way: $$x+ \frac{1}{x}= 2 \sqrt{3}$$ Then, $$(x^2)^2-\left(\frac{1}{x^2}\right)^2= \left(x^2 + \frac{1}{x^2}\right)^2-2$$ But this way is not working. Where I am wrong?

The idea you're having to change it to terms of $x^2$ isn't bad, but it seems a little overfancy. (Maybe I overlooked some economy about it, but I haven't seen the benefit yet.) Why not just calculate it directly? (Hints follow:) $x^2=3+2+2\sqrt{6}=5+2\sqrt{6}$ $x^4=(5+2\sqrt{6})^2=25+24+20\sqrt{6}=49+20\sqrt{6}$ $\dfrac{1}{x^4}=\dfrac{1}{49+20\sqrt{6}}=\dfrac{49-20\sqrt{6}}{2401-2400}=49-20\sqrt{6}$ You can take it from here, I think.

Proving that $n!≤((n+1)/2)^n$ by induction I'm new to inequalities in mathematical induction and don't know how to proceed further. So far I was able to do this: $V(1): 1≤1 \text{ true}$ $V(n): n!≤((n+1)/2)^n$ $V(n+1): (n+1)!≤((n+2)/2)^{(n+1)}$ and I've got : $(((n+1)/2)^n)\cdot(n+1)≤((n+2)/2)^{(n+1)}$ $((n+1)^n)n(n+1)≤((n+2)^n)((n/2)+1)$

It is more easy to prove this inequality without induction. Really $$0 < i\cdot (n + 1 - i) = \left(\frac{n+1}2 + \frac{2i - n - 1}2\right)\left(\frac{n+1}2 - \frac{2i - n - 1}2\right) = \left(\frac{n+1}2\right)^2 - \left(\frac{2i - n - 1}2\right)^2 \le \left(\frac{n+1}2\right)^2.$$ Multiply this inequalities for all $i = 1, 2, \ldots, \left\lfloor\frac n2\right\rfloor$ and by $\frac{n+1}2 = \frac{n+1}2$ for odd $n$ to get $n! \le \left(\frac{n+1}2\right)^n$ as desired.

Finding minimum $\frac{x+y}{z}+\frac{x+z}{y}+\frac{y+z}{x}$ I would appreciate if somebody could help me with the following problem Q. Finding maximum minimum $$\frac{x+y}{z}+\frac{x+z}{y}+\frac{y+z}{x}(\text{where} ~~x,y,z>0)$$

Note that if $F(x,y,z) = \frac{x+y}{z}+\frac{x+z}{y} + \frac{y+z}{x}$, then $F(kx,ky,kz)=F(x,y,z),\ k>0$. So we will use Lagrange multiplier method. Let $g(x,y,z)=x+y+z$. Constraint is $x+y+z=1$. $$\nabla F = (\frac{1}{z}+\frac{1}{y} - \frac{y+z}{x^2},\frac{1}{z}+\frac{1}{x} - \frac{x+z}{y^2},\frac{1}{y}+\frac{1}{x} - \frac{x+y}{z^2} ) =\lambda \nabla g$$ So $$ \frac{x^2(z+y) -(z+y)^2}{x^2yz}=\frac{z^2(x+y) -(x+y)^2}{xyz^2}= \frac{y^2(z+x) -(z+x)^2}{xy^2z} =\lambda $$ $$ \frac{x^2(1-x) -(1-x)^2}{x^2yz}=\frac{z^2(1-z) -(1-z)^2}{xyz^2}= \frac{y^2(1-y) -(1-y)^2}{xy^2z} =\lambda $$ Note that $\lambda\neq 0$ by computation. Hence we have $$ (xz-xyz-1)(x-z)=(xy-xyz-1)(x-y)=(yz-xyz-1)(y-z)=0$$ $x=z\neq y$ implies that $2x^3-3x^2+x-1=0$. But it has only one solution larger than $1$. If $x,\ y,\ z$ are distinct, $xz=xy=yz$. Contradiction. So $x=y=z$. $F(1/3,1/3,1/3)=6$ is minimum.

$a+b+c=0;a^2+b^2+c^2=1$ then $a^4+b^4+c^4$ is equal to what? $a+b+c=0;a^2+b^2+c^2=1$ then $a^4+b^4+c^4$ is equal to what? I tried to solve this problem, and I get $a^4+b^4+c^4 = 2(a^2b^2 + 1)$ but I'm not sure if it's correct

$$a+b+c=0\Rightarrow c=-(a+b)$$ $$1=a^2+b^2+(a+b)^2=2a^2+2ab+2b^2=2(a^2+ab+b^2)\Rightarrow a^2+ab+b^2=\frac12$$ $$\begin{align*} a^4+b^4+c^4 &= a^4+b^4+(a^4+4a^3b+6a^2b^2+4ab^3+b^4)\\ &= 2(a^4+2a^3b+3a^2b^2+2ab^3+b^4)\\ &= 2(a^2(a^2+ab+b^2)+b^2(a^2+ab+b^2)+ab(a^2+ab+b^2))\\ &= a^2+ab+b^2=\frac12 \end{align*}$$

find $\lim_{n \to \infty} \frac{(n^2+1)(n^2+2) \cdots (n^2+n)}{(n^2-1)(n^2-2) \cdots (n^2-n)}$ Find $$\lim_{n \to \infty} \frac{(n^2+1)(n^2+2) \cdots (n^2+n)}{(n^2-1)(n^2-2) \cdots (n^2-n)}$$ I tried to apply the squeeze theorem, yet none of my attempts led me to the solution.

Let $f(n)$ be defined by \begin{align*} f(n) &= \frac{(n^2+1)(n^2+2)\cdots(n^2+n)}{(n^2-1)(n^2-2)\cdots(n^2-n)}\\ &= \frac{(1 + {1 \over n^2} )(1 +{2 \over n^2})\cdots(1 + {n \over n^2})}{(1 - {1 \over n^2} )(1 - {2 \over n^2})\cdots(1 - {n \over n^2})} \end{align*} Then $$\ln f(n) = \sum_{k=1}^n \ln\left(1 + {k \over n^2}\right) - \sum_{k=1}^n \ln\left(1 - {k \over n^2}\right)$$ Using the Taylor series for $\ln(1 + x)$, this can be rewritten as $$ \ln f(n) = \sum_{k=1}^n \left({k \over n^2} + O\left({k^2 \over n^4}\right)\right) - \left[\sum_{k=1}^n \left(-{k \over n^2} + O\left({k^2 \over n^4}\right)\right)\right]$$ Using the formulas for $\sum_{k=1}^n k$ and $\sum_{k=1}^n k^2$, the above says that $$ \ln f(n) = {n(n + 1) \over n^2} + O\left({1 \over n}\right)$$ Thus $$\lim_{n \to \infty} \ln f(n) = 1$$ We conclude that $$\lim_{n \to \infty} f(n) = e^1 = e$$

How find the sum $\sum_{n=1}^{\infty}\frac{\binom{4n-4}{n-1}}{2^{4n-3}(3n-2)}$ today,I see a amazing math problem: show that $$\sum_{n=1}^{\infty}\dfrac{\binom{4n-4}{n-1}}{2^{4n-3}(3n-2)}=\dfrac{\sqrt[3]{17+3\sqrt{33}}}{3}-\dfrac{2}{3\sqrt[3]{17+3\sqrt{33}}}-\dfrac{1}{3}$$ This problem is from here. But I consider sometimes,and I think it maybe use Taylor therom $$\arcsin{x}=\sum_{n=0}^{\infty}\dfrac{(2n-1)!!}{(2n+1)(2n)!!}x^{2n+1}$$ Thank you,Now is 24:00 in beijing time,so I must go bed.I hope someone can help.Thank you

The transformed series in Marin's post seems suspiciously like Taylor's expansion of Bring radical $$\begin{aligned}\sum_{n=1}^{\infty}\dfrac{\binom{4n-4}{n-1}}{2^{4n-3}(3n-2)} &= \sum_{n=0}^{\infty} \binom{4n}{n} \frac{1}{(3n + 1)2^{4n + 1}} \\ &= \frac{16^{1/3}}{2} \sum_{n=0}^{\infty} \binom{4n}{n} \frac{\left ( 16^{-1/3} \right )^{3n + 1}}{(3n + 1)} \\ &= \frac{1}{2^{1/3}}\sum_{n = 0}^{\infty} \frac{\left(16^{-1/3}\right)^n}{n!} \left [\frac{d^{n-1}}{dz^{n-1}} \left ( 1 + z \right )^{4n} \right ]_{z = 0}\end{aligned}$$ The last sum is the Lagrange-Burmann inversion formula applied to $\phi(z)=(1+z)^4$. It is straightforward now that the series (excluding the constant factor $1/2^{1/3}$) is one of the two real roots of the quartic $z^4 + z + 16^{-1/3}$. One can exclude $-1/2^{1/3}$ via invoking a numerical check, and thus $$\sum_{n=1}^{\infty}\dfrac{\binom{4n-4}{n-1}}{2^{4n-3}(3n-2)} = \dfrac{\sqrt[3]{17+3\sqrt{33}}}{3}-\dfrac{2}{3\sqrt[3]{17+3\sqrt{33}}}-\dfrac{1}{3} \blacksquare$$

Differentiation answer check $$f(x)=\arcsin \sqrt{\frac{x}{x+1}} + \arctan \sqrt{x} \mbox{.}$$ $$f'(x) = \frac{1}{\sqrt{1-\frac{x}{x+1}}} \cdot \frac{1}{2}\sqrt{\frac{x+1}{x}} \cdot -x^{-2} + \frac{1}{1+x} \cdot \frac{1}{2\sqrt{x}} = -\frac{\sqrt{x+1}}{2\sqrt{1-\frac{x}{x+1}}{\sqrt{x}}\frac{1}{x^2}} + \frac{1}{2\sqrt{x} + 2x}\mbox{.}$$ Is there anything else I can do with the answer?

HINT: Let $\displaystyle\arcsin\sqrt{\frac x{x+1}}=y$ $$\implies \sqrt{\frac x{x+1}}=\sin y\implies \frac x{x+1}=\sin^2y$$ $$\implies x=\tan^2y\implies \sqrt x=\tan y \implies y=\arctan \sqrt x$$ Alternatively, let $\displaystyle\arctan \sqrt x=u\implies \sqrt x=\tan u\implies x=\tan^2u$ $\displaystyle\implies \frac x{x+1}=\frac{\tan^2u}{1+\tan^2u}=\frac{\tan^2u}{\sec^2u}=\sin^2u$ $\displaystyle\implies \arcsin\sqrt{\frac x{x+1}}=\arcsin (\sin u)=u=\arctan \sqrt x$