hcju/proofwikiqa huggingface dataset:
Let $\left \langle {x_n} \right \rangle$ be any sequence of points of $S$ such that: : $\forall n \in \N_{>0}: x_n \ne c$ : $\displaystyle \lim_{n \mathop \to \infty} x_n = c$ By Limit of Function by Convergent Sequences: : $\displaystyle \lim_{n \mathop \to \infty} f \paren {x_n} = l$ By the Multiple Rule for Sequences: : $\displaystyle \lim_{n \mathop \to \infty} \paren {\lambda f \paren {x_n} } = \lambda l$ Applying Limit of Function by Convergent Sequences again: : $\displaystyle \lim_{x \mathop \to c} \paren {\lambda f \paren x} = \lambda l$ {{qed}} Category:Limits of Functions o0236k677eu03ljik97o08h3tyyemdx
Let $\left \langle {x_n} \right \rangle$ be any sequence of points of $S$ such that: : $\forall n \in \N^*: x_n \ne c$ : $\displaystyle \lim_{n \mathop \to \infty} x_n = c$ By Limit of Function by Convergent Sequences: : $\displaystyle \lim_{n \mathop \to \infty} f \left({x_n}\right) = l$ : $\displaystyle \lim_{n \mathop \to \infty} g \left({x_n}\right) = m$ By the Combined Sum Rule for Sequences: : $\displaystyle \lim_{n \mathop \to \infty} \left({\lambda f \left({x_n}\right) + \mu g \left({x_n}\right)}\right) = \lambda l + \mu m$ Applying Limit of Function by Convergent Sequences again, we get: : $\displaystyle \lim_{x \mathop \to c} \left({\lambda f \left({x}\right) + \mu g \left({x}\right)}\right) = \lambda l + \mu m$ {{qed}}
Let $\sequence {x_n}$ be any sequence of points of $S$ such that: :$\forall n \in \N: x_n \ne c$ :$\displaystyle \lim_{n \mathop \to \infty} x_n = c$ By Limit of Function by Convergent Sequences: :$\displaystyle \lim_{n \mathop \to \infty} \map f {x_n} = l$ :$\displaystyle \lim_{n \mathop \to \infty} \map g {x_n} = m$ By the Product Rule for Sequences: :$\displaystyle \lim_{n \mathop \to \infty} \ \paren {\map f {x_n} \map g {x_n} } = l m$ Applying Limit of Function by Convergent Sequences again, we get: :$\displaystyle \lim_{x \mathop \to c} \paren {\map f x \map g x} = l m$ {{qed}}
Let $\sequence {x_n}$ be any sequence of points of $S$ such that: :$\forall n \in \N_{>0}: x_n \ne c$ :$\displaystyle \lim_{n \mathop \to \infty} x_n = c$ By Limit of Function by Convergent Sequences: :$\displaystyle \lim_{n \mathop \to \infty} \map f {x_n} = l$ :$\displaystyle \lim_{n \mathop \to \infty} \map g {x_n} = m$ By the Quotient Rule for Sequences: :$\displaystyle \lim_{n \mathop \to \infty} \frac {\map f {x_n} } {\map g {x_n} } = \frac l m$ provided that $m \ne 0$. Applying Limit of Function by Convergent Sequences again, we get: :$\displaystyle \lim_{x \mathop \to c} \frac {\map f x} {\map g x} = \frac l m$ provided that $m \ne 0$. {{qed}}
By definition of continuous, we have that :$\forall c \in S: \displaystyle \lim_{x \mathop \to c} \map f x = \map f c$ :$\forall c \in S: \displaystyle \lim_{x \mathop \to c} \map g x = \map g c$ Let $f$ and $g$ tend to the following limits: :$\displaystyle \lim_{x \mathop \to c} \map f x = l$ :$\displaystyle \lim_{x \mathop \to c} \map g x = m$ From the Sum Rule for Limits of Functions, we have that: :$\displaystyle \lim_{x \mathop \to c} \paren {\map f x + \map g x} = l + m$ So, by definition of continuous again, we have that $f + g$ is continuous on $S$. {{qed}}
By definition of continuous: :$\forall c \in S: \displaystyle \lim_{x \mathop \to c} \map f x = \map f c$ :$\forall c \in S: \displaystyle \lim_{x \mathop \to c} \map g x = \map g c$ Let $f$ and $g$ tend to the following limits: :$\displaystyle \lim_{x \mathop \to c} \map f x = l$ :$\displaystyle \lim_{x \mathop \to c} \map g x = m$ From the Product Rule for Limits of Functions: :$\displaystyle \lim_{x \mathop \to c} \paren {\map f x \, \map g x} = l m$ So, by definition of continuous again, we have that $f g$ is continuous on $S$. {{qed}} {{link wanted|Review links for Definition:Continuous Mapping (Metric Spaces) and see whether there is something more directly appropriate for this level of analysis}}
By definition of continuous, we have that : $\forall c \in S: \displaystyle \lim_{x \mathop \to c} \map f x = \map f c$ Let $f$ tend to the following limit: :$\displaystyle \lim_{x \mathop \to c} \map f x = l$ From the Multiple Rule for Limits of Functions, we have that: :$\displaystyle \lim_{x \mathop \to c} \paren {\lambda \, \map f x} = \lambda l$ So, by definition of continuous again, we have that $\lambda f$ is continuous on $S$. {{qed}} Category:Continuous Functions rwn44x1jsraa15u99j1w77y4wcq43q5
By definition of continuous, we have that : $\forall c \in S: \displaystyle \lim_{x \mathop \to c} f \left({x}\right) = f \left({c}\right)$ : $\forall c \in S: \displaystyle \lim_{x \mathop \to c} g \left({x}\right) = g \left({c}\right)$ Let $f$ and $g$ tend to the following limits: :$\displaystyle \lim_{x \mathop \to c} f \left({x}\right) = l$ :$\displaystyle \lim_{x \mathop \to c} g \left({x}\right) = m$ From the Combined Sum Rule for Limits of Functions, we have that: :$\displaystyle \lim_{x \mathop \to c} \left({\lambda f \left({x}\right) + \mu g \left({x}\right)}\right) = \lambda l + \mu m$ So, by definition of continuous again, we have that $\lambda f + \mu g$ is continuous on $S$. {{qed}}
By definition of continuous, we have that :$\forall c \in S: \displaystyle \lim_{x \mathop \to c} \map f x = \map f c$ :$\forall c \in S: \displaystyle \lim_{x \mathop \to c} \map g x = \map g c$ Let $f$ and $g$ tend to the following limits: :$\displaystyle \lim_{x \mathop \to c} \map f x = l$ :$\displaystyle \lim_{x \mathop \to c} \map g x = m$ From the Quotient Rule for Limits of Functions, we have that: :$\displaystyle \lim_{x \mathop \to c} \frac {\map f x} {\map g x} = \frac l m$ wherever $m \ne 0$. So, by definition of continuous again, we have that $\dfrac f g$ is continuous on all points $x$ of $S$ where $\map g x \ne 0$. {{qed}}
Suppose there exists such a mapping $f$ of the unit ball to itself without fixed points. Since $\map f x \ne x$, the two points $x$ and $\map f x$ are distinct and there is a unique ray from $x$ to $\map f x$ on which they both lie. Call this line $L$ and let $\map h x = \partial B^n \cap L$. If $x \in \partial B^n$, then $\map h x = x$ and $h$ restricts to the identity on $\partial B^n$. Since $x$ is in the line segment between $\map f x$ and $\map h x$, one may write the vector $\map h x - \map f x$ as a multiple $t$ times the vector $x - \map f x$, where $t \ge 1$. Hence: :$\map h x = t x + \paren {1 - t} \map f x$ Since $f$ is smooth, the smoothness of $t$ with respect to $x$ implies the smoothness of $h$. Taking the dot product of both sides of this formula and noting that $\size {\map h x} = 1$: :$t^2 \size {x - \map f x}^2 + 2 t \map f x \cdot \paren {x - \map f x} + \size {\map f x}^2 - 1 = 0$ Applying the Quadratic Formula gives: :$\displaystyle t = \frac {\map f x \cdot \paren {\map f x - x} } {\size {x - \map f x}^2}$ an expression for $t$ in smooth terms of $x$. Hence $h$ is a smooth retract of a compact manifold onto its boundary, which contradicts the Retraction Theorem. {{qed}} {{Namedfor|Luitzen Egbertus Jan Brouwer|cat = Brouwer}} Category:Brouwer's Fixed Point Theorem q21bc3euefxs0nvew6ccoim903ke55l
When $n \ge 0$ we use the result for Natural Number Index. Now let $n \in \Z: n < 0$. Then let $m = -n$ and so $m > 0$. Thus $x^n = \dfrac 1 {x^m}$. {{begin-eqn}} {{eqn | l = \map D {x^n}       | r = \map D {\frac 1 {x^m} }       | c =  }} {{eqn | r = \frac {x^m \cdot 0 - 1 \cdot m x^{m - 1} } {x^{2 m} }       | c = Quotient Rule for Derivatives }} {{eqn | r = -m x^{-m - 1}       | c =  }} {{eqn | r = n x^{n - 1}       | c =  }} {{end-eqn}} {{qed}}
Let $n \in \Q$, such that $n = \dfrac p q$ where $p, q \in \Z, q \ne 0$. Then we have: {{begin-eqn}} {{eqn | l = \map D {x^n}       | r = \map D {x^{p/q} }       | c =  }} {{eqn | r = \map D {\paren {x^p}^{1/q} }       | c =  }} {{eqn | r = \frac 1 q \paren {x^p}^{1/q} x^{-p} p x^{p-1}       | c = Chain Rule for Derivatives }} {{eqn | r = \frac p q x^{\frac p q - 1}       | c = after some algebra }} {{eqn | r = n x^{n - 1}       | c =  }} {{end-eqn}} {{qed}}
The proof consists of a uniqueness and an existence part. Let $S'$ be a non-empty subset of $S$. === Existence === As $S$ is well-ordered by $\preceq$, we have by definition existence of a minimal elements $s$ of $S'$ (with respect to $\preceq$). === Uniqueness === Let $s, s'$ be minimal elements of $S'$.  By Well-Ordering is Total Ordering, we have at least one of $s \preceq s'$, $s'\preceq s$. Since both $s$ and $s'$ are minimal elements, we conclude that necessarily $s = s'$. {{qed}}
As $f$ is continuous, by the first part of the theorem, it has a primitive. Call it $F$. $\closedint a b$ can be divided into any number of closed subintervals of the form $\closedint {x_{k - 1} } {x_k}$ where: :$a = x_0 < x_1 \cdots < x_{k-1} < x_k = b$ Fix such a finite subdivision of the interval $\closedint a b$; call it $P$. Next, we observe the following telescoping sum identity: {{begin-eqn}} {{eqn | n = 1       | l = \sum_{i \mathop = 1}^k \paren {\map F {x_i} - \map F {x_{i - 1} } }       | r = \map F b - \map F a }} {{end-eqn}} Because $F' = f$, $F$ is differentiable. By Differentiable Function is Continuous, $F$ is also continuous. Therefore we can apply the Mean Value Theorem on $F$. It follows that in every closed subinterval $I_i = \closedint {x_{i - 1} } {x_i}$ there is some $c_i$ such that: :$\map {F'} {c_i} = \dfrac {\map F {x_i} - \map F {x_{i - 1} } } {x_i - x_{i - 1} }$  It follows that: {{begin-eqn}} {{eqn | l = \map F {x_i} - \map F {x_{i - 1} }       | r = \map {F'} {c_i} \paren {x_i - x_{i - 1} } }} {{eqn | n = 2       | l = \map F b - \map F z        | r = \sum_{i \mathop = 1}^k \map {F'} {c_i} \paren {x_i - x_{i - 1} }       | c = from $(1)$ }} {{eqn | r = \sum_{i \mathop = 1}^k \map f {c_i} \paren {x_i - x_{i - 1} }       | c = because $F' = f$ }} {{end-eqn}} From the definitions of supremum and infimum, we have for all $i$ (recall $I_i = \closedint {x_{i - 1} } {x_i}$): :$\displaystyle \inf_{x \mathop \in I_i} \map f x \le \map f {c_i} \le \sup_{x \mathop \in I_i} \map f x$ From the definitions of upper and lower sums, we conclude for any finite subdivision $P$: :$\displaystyle \map L P \le \sum_{i \mathop = 1}^k \map f {c_i} \paren {x_i - x_{i - 1} } \le \map U P$ Lastly, from the definition of a definite integral and from $(2)$, we conclude: :$\displaystyle \map F b - \map F a = \int_a^b \map f t \rd t$ {{qed}}
We use the definition of the natural logarithm as an integral: :$\displaystyle \ln x = \int_1^x \frac {\mathrm d t} t$ From Integral on Zero Interval: :$\displaystyle \ln 1 = \int_1^1 \frac {\mathrm d t} t = 0$ {{qed}}
We use the definition of the natural logarithm as the inverse of the exponential: :$\ln x = y \iff e^y = x$ Then: {{begin-eqn}} {{eqn | l = e^0       | r = 1       | c = Exponential of Zero }} {{eqn | ll= \leadstoandfrom       | l = \ln 1       | r = 0 }} {{end-eqn}} {{qed}}
We use the definition of the natural logarithm as the limit of a sequence: :$\displaystyle \ln x = \lim_{n \mathop \to \infty} n \left({\sqrt [n] x - 1}\right)$ Then: {{begin-eqn}} {{eqn | l = \ln 1       | r = \lim_{n \mathop \to \infty} n \left({\sqrt [n] 1 - 1}\right) }} {{eqn | r = \lim_{n \mathop \to \infty} n \left({0}\right) }} {{eqn | r = \lim_{n \mathop \to \infty} 0 }} {{eqn | r = 0 }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn | l = \ln x       | o = :=       | r = \int_1^x \dfrac 1 t \rd t       | c = {{Defof|Natural Logarithm|subdef = Positive Real|index = 1}} }} {{eqn | l = \frac \d {\d x} \ln x       | r = \frac \d {\d x} \int_1^x \dfrac 1 t \rd t }} {{eqn | r = \frac 1 x       | c = Fundamental Theorem of Calculus }} {{end-eqn}} {{qed}}
This proof assumes the definition of the natural logarithm as the inverse of the exponential function, where the exponential function is defined as the limit of a sequence: :$e^x := \displaystyle \lim_{n \to +\infty} \left({1 + \frac x n}\right)^n$ It also assumes the Laws of Logarithms. {{begin-eqn}} {{eqn | l = \map {\frac \d {\d x} } {\ln x}       | r = \lim_{\Delta x \mathop \to 0} \frac {\map \ln {x + \Delta x} - \ln x} {\Delta x}       | c = {{Defof|Derivative}} }} {{eqn | r = \lim_{\Delta x \mathop \to 0} \frac {\map \ln {\frac {x + \Delta x} x} } {\Delta x}       | c = Difference of Logarithms }} {{eqn | r = \lim_{\Delta x \mathop \to 0} \paren {\frac 1 {\Delta x} \centerdot \map \ln {1 + \frac {\Delta x} x} } }} {{eqn | n = 1       | r = \lim_{\Delta x \mathop \to 0} \paren {\map \ln {\paren {1 + \frac {\Delta x} x}^{1 / \Delta x} } }       | c = Natural Logarithm of Power }} {{end-eqn}} Define $u$ as: {{begin-eqn}} {{eqn | l = u       | r = \dfrac {\Delta x} x       | c =  }} {{eqn | ll= \leadsto       | l = \Delta x       | r = u x       | c =  }} {{eqn | ll= \leadsto       | l = \frac 1 {\Delta x}       | r = \frac 1 x \cdot \frac 1 u       | c =  }} {{end-eqn}} Hence: {{begin-eqn}} {{eqn | r = \lim_{u \mathop \to 0} \paren {\map \ln {\paren {1 + u}^{\frac 1 u \cdot \frac 1 x} } }       | c = substituting $u x$ for $\Delta x$ in $(1)$ }} {{eqn | r = \lim_{u \mathop \to 0} \paren {\frac 1 x \cdot \map \ln {1 + u}^{\frac 1 u} }       | c = Natural Logarithm of Power }} {{eqn | r = \frac 1 x \cdot \lim_{u \mathop \to 0} \paren {\map \ln {1 + u}^{\frac 1 u} }       | c = factoring out constants }} {{eqn | r = \frac 1 x \cdot \lim_{v \mathop \to +\infty} \paren {\map \ln {1 + \frac 1 v}^v}       | c = substituting $\dfrac 1 v$ for $u$ }} {{eqn | r = \frac 1 x \cdot \ln e^1       | c = Limit of Composite Function, Limit definition of $e^x$, Natural Logarithm Function is Continuous }} {{eqn | r = \frac 1 x       | c = Exponential of Natural Logarithm }} {{end-eqn}} {{qed}}
This proof assumes the definition of the natural logarithm as the inverse of the exponential function as defined by differential equation: :$y = \dfrac {\d y} {\d x}$ :$y = e^x \iff \ln y = x$ {{begin-eqn}} {{eqn | l = \frac {\d y} {\d x}       | r = y       | c = {{Defof|Exponential Function/Real|subdef = Differential Equation|Exponential Function}} }} {{eqn | l = \int \frac 1 y \rd y        | r = \int \rd x       | c = Separation of Variables }} {{eqn | r = x + C_0       | c = Integral of Constant where that constant is $1$ }} {{eqn | r = \ln y + C_0       | c = {{Defof|Natural Logarithm/Positive Real|index = 2|Natural Logarithm}}: $x = \ln y$ }} {{end-eqn}} The result follows from the definition of the antiderivative and the defined  initial condition: :$\tuple {x_0, y_0} = \tuple {0, 1}$ {{qed}}
This proof assumes the definition of the natural logarithm as the limit of a sequence of real functions. Let $\sequence {f_n}$ be the sequence of mappings $f_n: \R_{>0} \to \R$ defined as: :$\map {f_n} x = n \paren {\sqrt [n] x - 1}$ Fix $x_0 \in \R_{>0}$. Pick $k \in \N : x_0 \in J := \closedint {\dfrac 1 k} k$. From definition of bounded interval, $J$ is bounded. From Derivative of Nth Root and Combination Theorem for Sequences: :$\forall n \in \N : \forall x \in J : D_x \map {f_n} x = \dfrac {\sqrt [n] x} x$ In particular: :$\forall n: f_n$ is differentiable on $J$ From Defining Sequence of Natural Logarithm is Convergent, $\sequence {\map {f_n} {x_0} }$ is convergent. === Lemma === {{:Derivative of Natural Logarithm Function/Proof 4/Lemma}}{{qed|lemma}} From the lemma, $\sequence { {f_n}'}$ converges uniformly to $\dfrac 1 x$ on $J$. From Derivative of Uniformly Convergent Sequence of Differentiable Functions, $\map {f'} x = \dfrac 1 x$ on $J$ In particular: :$\map {f'} {x_0} = \dfrac 1 {x_0}$ Hence the result. {{qed}}
From Change of Base of Logarithm, we have: :$\log_e x = \dfrac {\log_a x} {\log_a e}$ from which: :$\log_a x = \log_a e \log_e x$ In this context, $\log_a e$ is a constant. The result then follows from Derivative of Natural Logarithm Function and Derivative of Constant Multiple. {{qed}} Category:Derivatives Category:Logarithms no3yjlbmxn4chs06xc44gyyr7w60b3w
From Derivative of Natural Logarithm Function $D \ln x = \dfrac 1 x$, which is strictly positive on $x > 0$. From Derivative of Monotone Function it follows that $\ln x$ is strictly increasing on $x > 0$. {{qed}}
From Natural Logarithm of 2 is Greater than One Half: :$\ln 2 \ge \dfrac 1 2$ From the definition of infinite limit at infinity, our assertion is: :$\forall M \in \R_{>0} : \exists N > 0 : x > N \implies \ln x > M$. As $x \to +\infty$, we will restrict our attention to sufficiently large $M$. From Logarithm is Strictly Increasing: :$\ln x$ is strictly increasing. So, for sufficiently large $M$: :$x > 2^{2 M} \implies \ln x > \ln 2^{2 M}$ From the Laws of Logarithms: {{begin-eqn}} {{eqn | l = \ln 2^{2 M}       | r = 2 M \ln 2 }} {{eqn | o = \ge        | r = 2 M \cdot \dfrac 1 2 }} {{eqn | r = M }} {{end-eqn}} Choosing $N = \ln 2^{2 M}$: :$\forall M \ge a: \exists N > 0: x > N \implies \ln x > M$ for some $a \in \R$. Hence the result, by the definition of infinite limit at infinity. {{qed}}
From the definition of the natural logarithm: {{begin-eqn}} {{eqn | l = \ln x       | r = \int_1^x \dfrac 1 t \ \mathrm d t }} {{end-eqn}} The result follows from Integral of Reciprocal is Divergent. {{qed}}
From the definition of natural logarithm: {{begin-eqn}} {{eqn | l = \ln x       | r = \int_1^x \dfrac 1 t \ \mathrm dt }} {{end-eqn}} The result follows from Integral of Reciprocal is Divergent. {{qed}}
From Derivative of Natural Logarithm Function: :$\dfrac \d {\d x} \ln x = \dfrac 1 x$ From the Power Rule for Derivatives: Integer Index: :$\dfrac {\d^2} {\d x^2} \ln x = \dfrac \d {\d x} \dfrac 1 x = -\dfrac 1 {x^2}$ {{qed}}
By definition, the exponential function is the inverse of the natural logarithm function. From Logarithm is Strictly Increasing, $\ln x$ is strictly increasing. The result follows from Inverse of Strictly Monotone Function. {{qed}}
For all $x \in \R$: {{begin-eqn}} {{eqn | l = D_x \exp x       | r = \exp x       | c = Derivative of Exponential Function }} {{eqn | o = >       | r = 0       | c = Exponential of Real Number is Strictly Positive }} {{end-eqn}} Hence the result, from Derivative of Monotone Function. {{qed}}
=== The exponential function approaches positive infinity as x approaches positive infinity === Let $M$ be a strictly positive real number. Let $N$ be $\ln M$. $N$ is real because $M > 0$. From Exponential is Strictly Increasing: :$\forall x: x > N \implies \exp x > \exp N = M$ Therefore: :$\forall M \in \R_{>0} : \exists N \in \R : \forall x > N : \exp x > M$ The result follows from the definition of infinite limit at infinity. {{qed}} === The exponential function approaches 0 as x approaches negative infinity === Let $\epsilon$ be a strictly positive real number. Let $c$ be $\ln \epsilon$. Let $x$ be any real number that smaller than $c$. From Exponential of Real Number is Strictly Positive: :$\exp x > 0 \implies \size {\exp x} = \exp x$ From Exponential is Strictly Increasing: :$\size {\exp x - 0} = \exp x < \exp c = \epsilon$ Therefore: :$\forall \epsilon \in \R_{>0} : \exists c \in \R : \forall x < c : \size {\exp x - 0} < \epsilon$ From the definition of limit at minus infinity, the result follows. {{qed}} === The exponential function has domain $\R$ and image $\openint 0 \to$ === {{questionable|the above uses this result}} We have that the Exponential is Strictly Increasing. From above, $\displaystyle \lim_{x \mathop \to \infty} \exp x = \infty$ From above, $\displaystyle \lim_{x \mathop \to -\infty} \exp x = 0$ Hence the result. {{qed}}
From the definition of the exponential function: :$e^y = x \iff \ln x = y$ Raising both sides of the equation $\ln x = y$ to the power of $e$: {{begin-eqn}} {{eqn | l = e^{\ln x}       | r = e^y       | c =  }} {{eqn | r = x       | c =  }} {{end-eqn}} {{qed}} Category:Exponential Function Category:Natural Logarithms 0fegiec28brg4h8iqx8wa7g1pjg3pzq
{{begin-eqn}} {{eqn | l = \paren {a b}^x       | r = \map \exp {x \map \ln {a b} }       | c = {{Defof|Power to Real Number}} }} {{eqn | r = \map \exp {x \ln a + x \ln b}       | c = Sum of Logarithms }} {{eqn | r = \map \exp {x \ln a} \map \exp {x \ln b}       | c = Exponential of Sum }} {{eqn | r = a^x b^x       | c = {{Defof|Power to Real Number}} }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn | l = a^{-x}       | r = \map \exp {-x \ln a}       | c = {{Defof|Power to Real Number}} }} {{eqn | r = \paren {\map \exp {x \ln a} }^{-1}       | c = Exponential of Product }} {{eqn | r = \frac 1 {\map \exp {x \ln a} }       | c =  }} {{eqn | r = \frac 1 {a^x}       | c = {{Defof|Power to Real Number}} }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn | l = \frac {a^x} {a^y}       | r = a^x \paren {\frac 1 {a^y} }       | c =  }} {{eqn | r = \paren {a^x} \paren {a^{-y} }       | c = Exponent Combination Laws: Negative Power }} {{eqn | r = a^{x - y}       | c = Product of Powers }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn | l = \paren {\frac a b}^x       | r = \map \exp {x \, \map \ln {\frac a b} }       | c = {{Defof|Power to Real Number}} }} {{eqn | r = \map \exp {x \ln a - x \ln b}       | c = Sum of Logarithms }} {{eqn | r = \frac {\map \exp {x \ln a} } {\map \exp {x \ln b} }       | c = Exponential of Sum }} {{eqn | r = \frac {a^x} {b^x}       | c = {{Defof|Power to Real Number}} }} {{end-eqn}} {{qed}} Category:Exponent Combination Laws ttlj4crc4wfbmmdthekbz8ku0htevke
{{begin-eqn}} {{eqn | l = \left({\frac a b}\right) ^{-x}       | r = \left({\frac 1 {\left({\frac a b}\right)} }\right)^x       | c = Exponent Combination Laws/Negative Power }} {{eqn | r = \left({\frac b a}\right)^x       | c =  }} {{end-eqn}} {{qed}} Category:Exponent Combination Laws 9kknb0tyyxmivkpeusz701e07egx70z
Recall the definition of the sine function: : $\displaystyle \sin x = \sum_{n \mathop = 0}^\infty \paren {-1}^n \frac {x^{2 n + 1} } {\paren {2 n + 1}!} = x - \frac {x^3} {3!} + \frac {x^5} {5!} - \cdots$ Thus $\sin x$ is expressible in the form of a power series. From Sine Function is Absolutely Convergent, we have that the interval of convergence of $\sin x$ is the whole of $\R$.  From Power Series is Differentiable on Interval of Convergence, it follows that $\sin x$ is continuous on the whole of $\R$. {{qed}} Category:Sine Function 5kqc8cp5vcnibppg3vt1lipqsuhcnsc
Recall the definition of the sine function: : $\displaystyle \sin x = \sum_{n \mathop = 0}^\infty \left({-1}\right)^n \frac {x^{2 n + 1} } {\left({2 n + 1}\right)!} = x - \frac {x^3} {3!} + \frac {x^5} {5!} - \cdots$ For: : $\displaystyle \sum_{n \mathop = 0}^\infty \left({-1}\right)^n \frac {x^{2 n + 1} } {\left({2 n + 1}\right)!}$ to be absolutely convergent we want: : $\displaystyle \sum_{n \mathop = 0}^\infty \left|{\left({-1}\right)^n \frac {x^{2 n + 1} } {\left({2 n + 1}\right)!} }\right| = \sum_{n \mathop = 0}^\infty \frac {\left|{x}\right|^{2 n + 1} } {\left({2 n + 1}\right)!}$ to be convergent. But: :$\displaystyle \sum_{n \mathop = 0}^\infty \frac {\left|{x}\right|^{2 n + 1} } {\left({2 n + 1}\right)!}$ is just the terms of: :$\displaystyle \sum_{n \mathop = 0}^\infty \frac {\left|{x}\right|^n}{n!}$ for odd $n$. Thus: : $\displaystyle \sum_{n \mathop = 0}^\infty \frac {\left|{x}\right|^{2 n + 1} } {\left({2 n + 1}\right)!} < \sum_{n \mathop = 0}^\infty \frac {\left|{x}\right|^n} {n!}$ But: : $\displaystyle \sum_{n \mathop = 0}^\infty \frac {\left|{x}\right|^n} {n!} = \exp \left|{x}\right|$ from the Taylor Series Expansion for Exponential Function of $\left|{x}\right|$, which converges for all $x \in \R$. The result follows from the Squeeze Theorem. {{qed}}
Recall the definition of the sine function: :$\displaystyle \sin x = \sum_{n \mathop = 0}^\infty \paren {-1}^n \frac {x^{2 n + 1} } {\paren {2 n + 1}!} = x - \frac {x^3} {3!} + \frac {x^5} {5!} - \cdots$ Thus: :$\displaystyle \sin 0 = 0 - \frac {0^3} {3!} + \frac {0^5} {5!} - \cdots = 0$ {{qed}}
Recall the definition of the sine function: {{begin-eqn}} {{eqn | l = \sin z       | r = \sum_{n \mathop = 0}^\infty \paren {-1}^n \frac {z^{2 n + 1} } {\paren {2 n + 1}!} }} {{eqn | r = z - \frac {z^3} {3!} + \frac {z^5} {5!} - \cdots }} {{end-eqn}} From Sign of Odd Power, we have that: :$\forall n \in \N: -\paren {z^{2 n + 1} } = \paren {-z}^{2 n + 1}$ The result follows directly. {{qed}}
{{begin-eqn}} {{eqn	| l = \map \sinh {-x} 	| r = -i \, \map \sin {-i x} 	| c = Hyperbolic Sine in terms of Sine }} {{eqn	| r = i \, \map \sin {i x} 	| c = Sine Function is Odd }} {{eqn	| r = -\sinh x 	| c = Hyperbolic Sine in terms of Sine }} {{end-eqn}} {{qed}}