hcju/proofwikiqa huggingface dataset:
By previous work done in Stirling's Formula: Proof 2 it is noted that $\left\langle{\dfrac {n!} {n^n \sqrt n e^{-n} } }\right\rangle_{n \mathop \in \N}$ is a convergent sequence. Let $\dfrac {n!} {n^n \sqrt n e^{-n} } \to C$ as $n \to \infty$. Let $I_n = \displaystyle \int_0^{\frac \pi 2} \sin^n x \rd x$. Then: {{begin-eqn}} {{eqn | l = \frac {I_{2 n} } {I_{2 n + 1} }       | r = \frac {\left({2n}\right)!} {\left({2^n n!}\right)^2} \frac \pi 2 \cdot \frac {\left({2 n + 1}\right)!} {\left({2^n n!}\right)^2}       | c = Definite Integral from $0$ to $\dfrac \pi 2$ of Even Power of $\sin x$ and Corollary 2 }} {{eqn | r = \frac \pi 2 \left({2 n + 1}\right) \frac {\left({\left({2n}\right)!}\right)^2} {\left({2^n n!}\right)^4}       | c = extracting $\left({2 n + 1}\right)$ as a factor and rearranging }} {{eqn | o = \sim       | r = \frac \pi 2 \left({2 n + 1}\right) \frac {\left({C \left({2n}\right)^{2n + 1/2} e^{-2 n} }\right)^2} {\left({2^n C n^{n + 1/2} e^{-n} }\right)^4}       | c = entering $C$ into top and bottom }} {{eqn | r = \frac \pi 2 \frac {\left({2 n + 1}\right)} {C^2} \frac {2^{4 n} 2 \left({n^{4 n + 1} }\right) e^{-4 n} } {2^{4 n} \left({n^{4 n + 2} }\right) e^{-4 n} }       | c = rearranging }} {{eqn | r = \frac {\pi \left({2 n + 1}\right)} {n C^2}       | c = much simplification }} {{eqn | r = \frac {2 \pi} {C^2} \text{ as } n \to \infty       | c =  }} {{end-eqn}} From Lemma 4, we have that: :$\displaystyle \lim_{n \to \infty} \frac {I_{2 n} } {I_{2 n + 1} } = 1$ from which {{begin-eqn}} {{eqn | l = \frac {2 \pi} {C^2}       | r = 1 }} {{eqn | ll= \implies       | l = C^2       | r = 2 \pi }} {{end-eqn}} Hence the result. {{qed}}
Let $d_n = \ln n! - \left({n + \dfrac 1 2}\right) \ln n + n$. From the argument in Stirling's Formula: Proof 2: Lemma 3 we have that $\left\langle{d_n - \dfrac 1 {12 n} }\right\rangle$ is an increasing sequence. Then: {{begin-eqn}} {{eqn | l = d_n - d_{n+1}       | r = \ln \left({n!}\right) - \left({n + \frac 1 2}\right) \ln n + n       | c =  }} {{eqn | o =        | ro= -       | r = \left({\ln \left({\left({n+1}\right)!}\right) - \left({n + 1 + \frac 1 2}\right) \ln \left({n + 1}\right) + n + 1}\right)       | c =  }} {{eqn | r = - \ln \left({n+1}\right) - \left({n + \frac 1 2}\right) \ln n + \left({n + \frac 3 2}\right) \ln \left({n+1}\right) - 1       | c = (as $\ln \left({\left({n+1}\right)!}\right) = \ln \left({n+1}\right) + \ln \left({n!}\right)$) }} {{eqn | r = \left({n + \frac 1 2}\right) \ln \left({\frac {n+1} n}\right) - 1       | c = }} {{eqn | n = 1       | r = \frac {2n + 1} 2 \ln \left({\frac {1 + \left({2n + 1}\right)^{-1} } {1 - \left({2n + 1}\right)^{-1} } }\right) - 1       | c = }} {{end-eqn}} Let: : $f \left({x}\right) := \dfrac 1 {2 x} \ln \left({\dfrac {1 + x} {1 - x} }\right) - 1$ for $\left\vert{x}\right\vert < 1$. By Stirling's Formula: Proof 2: Lemma 1: {{begin-eqn}} {{eqn | l = f \left({x}\right)       | r = \sum_{k \mathop = 1}^\infty \frac {x^{2n} } {2n + 1}       | c =  }} {{eqn | r = \frac {x^2} 3 + \frac {x^4} 5 + \frac {x^6} 7 + \cdots       | c =  }} {{eqn | o = >       | r = \frac {x^2} 3       | c =  }} {{end-eqn}} :$\displaystyle f \left({x}\right) = \sum_{k \mathop = 1}^\infty \frac {x^{2n} } {2n + 1}$ As $-1 < \dfrac 1 {2n + 1} < 1$ it can be substituted for $x$ from $(1)$: {{begin-eqn}} {{eqn | l = d_n - d_{n+1}       | o = >       | r = \frac 1 3 \frac 1 {\left({2n + 1}\right)^2}       | c =  }} {{eqn | r = \frac 1 {12 n^2 + 12 n + 3}       | c =  }} {{end-eqn}} Next: {{begin-eqn}} {{eqn | l = \frac 1 {12 n + 1} - \frac 1 {12 \left({n + 1}\right) + 1}       | r = \frac {12 n + 13 - 12 n - 1} {\left({12 n + 1}\right) \left({12 n + 13}\right)}       | c =  }} {{eqn | r = \frac {12} {\left({12 n + 1}\right) \left({12 n + 13}\right)}       | c =  }} {{end-eqn}} It follows that: {{begin-eqn}} {{eqn | o =       | r = \left({d_n - \frac 1 {12 n + 1} }\right) - \left({\frac 1 {12 \left({n + 1}\right) + 1} }\right)       | c =  }} {{eqn | o = >       | r = \frac {\left({12 n + 1}\right) \left({12 n + 13}\right) - 12 \left({12 n^2 + 12 n + 3}\right)} {3 \left({12 n + 1}\right) \left({12 n + 13}\right) \left({2n + 1}\right)^2}       | c =  }} {{end-eqn}} The numerator equals: :$12 n \left({14 - 12}\right) + \left({13 - 36}\right) = 24 n - 23 > 0$ for $n = 1, 2, 3, \ldots$ Therefore the sequence $\left\langle{d_n - \dfrac 1 {12 n + 1} }\right\rangle$ is decreasing. From Stirling's Formula, we have that: : $\displaystyle \lim_{n \to \infty} d_n = d$ where $d = \ln \left({\sqrt{2 \pi} }\right)$ and so: :$d_n - \dfrac 1 {12 n} < d < d_n - \dfrac 1 {12 n + 1}$ for $n = 1, 2, 3, \ldots$ Taking exponentials, and again from Stirling's Formula: :$e^{-1/12n} < \dfrac {\left({\sqrt{2n} }\right) n^{n + 1/2} e^{-n} } {n!} < e^{-1/\left({12n + 1}\right)}$ from whence the result. {{qed}}
This follows directly from the definition of $\prec$. First note that: :$\forall n \in S: 0 \preceq n$ from the definition of zero. Next, from the definition of one: :$0 \ne 1$ Thus: {{begin-eqn}} {{eqn | o =        | r = 0 \preceq 1 \land 0 \ne 1 }} {{eqn | o = \implies       | r = 0 \prec 1       | c = Definition of Strictly Precede }} {{end-eqn}} {{Qed}}
The proof strategy is to demonstrate that the sign of $d_n - d_{n+1}$ is positive. {{begin-eqn}} {{eqn | l = d_n - d_{n+1}       | r = \ln \left({n!}\right) - \left({n + \frac 1 2}\right) \ln n + n       | c =  }} {{eqn | o =        | ro= -       | r = \left({\ln \left({\left({n+1}\right)!}\right) - \left({n + 1 + \frac 1 2}\right) \ln \left({n + 1}\right) + n + 1}\right)       | c =  }} {{eqn | r = - \ln \left({n+1}\right) - \left({n + \frac 1 2}\right) \ln n + \left({n + \frac 3 2}\right) \ln \left({n+1}\right) - 1       | c = (as $\ln \left({\left({n+1}\right)!}\right) = \ln \left({n+1}\right) + \ln \left({n!}\right)$) }} {{eqn | r = \left({n + \frac 1 2}\right) \ln \left({\frac {n+1} n}\right) - 1       | c = }} {{eqn | n = 1       | r = \frac {2n + 1} 2 \ln \left({\frac {1 + \left({2n + 1}\right)^{-1} } {1 - \left({2n + 1}\right)^{-1} } }\right) - 1       | c = }} {{end-eqn}} Let: : $f \left({x}\right) := \dfrac 1 {2 x} \ln \left({\dfrac {1 + x} {1 - x} }\right) - 1$ for $\left\vert{x}\right\vert < 1$. Then from Lemma 1: : $(2): \quad \displaystyle f \left({x}\right) = \sum_{k \mathop = 1}^\infty \frac {x^{2n} } {2n + 1}$ Thus $f \left({x}\right) > 0$ for $\left\vert{x}\right\vert < 1$. Putting $x = \dfrac 1 {2n + 1}$ it can be seen that $(1)$ is $f \left({\dfrac 1 {2n + 1} }\right)$. As $-1 < \dfrac 1 {2n + 1} < 1$ it can be seen that $(2)$ can be applied and so: :$\forall n \in \N: d_n - d_{n+1} \ge 0$ Thus $\left\langle{d_n}\right\rangle$ is a decreasing sequence. {{qed}}
From Sum of Infinite Geometric Sequence: :$(1): \quad \displaystyle \sum_{n \mathop = 0}^\infty \paren {-x^2}^n = \frac 1 {1 + x^2}$ for $-1 < x < 1$. From Power Series is Termwise Integrable within Radius of Convergence, $(1)$ can be integrated term by term: {{begin-eqn}} {{eqn | l = \int_0^x \frac 1 {1 + t^2} \rd t       | r = \sum_{n \mathop = 0}^\infty \int_0^x \paren {-t^2}^n \rd t       | c =  }} {{eqn | ll= \leadsto       | l = \arctan x        | r = \sum_{n \mathop = 0}^\infty \paren {-1}^n \frac {x^{2 n + 1} } {2 n + 1}       | c = Primitive of Reciprocal of $\dfrac 1 {1 + t^2}$, Integral of Power }} {{end-eqn}} For $-1 \le x \le 1$, the sequence $\sequence {\dfrac {x^{2 n + 1}} {2 n + 1} }$ is decreasing and converges to zero. Therefore the series converges in the given range by the Alternating Series Test. {{qed|lemma}} Now consider the case $x \ge 1$: {{begin-eqn}} {{eqn	| l = \arctan x  	| r = \frac \pi 2 - \map \arctan {\frac 1 x} 	| c = Sum of Arctangent and Arccotangent }} {{eqn	| r = \frac \pi 2 - \sum_{n \mathop = 0}^\infty \paren {-1}^n \frac 1 {2 n + 1} \paren {\frac 1 x}^{2 n + 1} 	| c = as $x \ge 1$, $0 < \dfrac 1 x \le 1$, so may apply the above result }} {{eqn	| r = \frac \pi 2 - \sum_{n \mathop = 0}^\infty \paren {-1}^n \frac 1 {\paren {2 n + 1} x^{2 n + 1} } }} {{end-eqn}} {{qed|lemma}} We also have:  {{begin-eqn}} {{eqn	| l = \map \arctan {-x} 	| r = -\arctan x 	| c = Arctangent Function is Odd }} {{eqn	| r = -\frac \pi 2 - \sum_{n \mathop = 0}^\infty \paren {-1}^{n + 1} \frac 1 {\paren {2 n + 1} x^{2 n + 1} } 	| c = per above result }} {{end-eqn}} Substituting $x$ for $-x$ gives us the expansion for $x \le -1$:  {{begin-eqn}} {{eqn	| l = \arctan x 	| r = -\frac \pi 2 - \sum_{n \mathop = 0}^\infty \paren {-1}^{n + 1} \frac 1 {\paren {2 n + 1} \paren {-x}^{2 n + 1} } }} {{eqn	| r = -\frac \pi 2 - \sum_{n \mathop = 0}^\infty \paren {-1}^{n + 1} \frac 1 {-\paren {2 n + 1} x^{2 n + 1} } }} {{eqn	| r = -\frac \pi 2 - \sum_{n \mathop = 0}^\infty \paren {-1}^n \frac 1 {\paren {2 n + 1} x^{2 n + 1} } }} {{end-eqn}} {{qed}}
=== Necessary Condition === From axiom $(NO3)$, we have: :$\forall m, n \in S: m \preceq n \implies \exists p \in S: m \circ p = n$ {{qed|lemma}} === Sufficient Condition === Suppose that $m \circ p = n$. {{begin-eqn}} {{eqn | l = 0       | o = \preceq       | r = p       | c = Definition of Zero }} {{eqn | ll= \implies       | l = m \circ 0       | o = \preceq       | r = m \circ p       | c = $\preceq$ is compatible with $\circ$ }} {{eqn | ll= \implies       | l = m       | o = \preceq       | r = n       | c = Zero is Identity in Naturally Ordered Semigroup }} {{end-eqn}} {{qed|lemma}} So $\forall m, n \in S$:  :$m \preceq n \iff \exists p \in S: m \circ p = n$ {{qed}}
Since $m \preceq n$, by axiom $(NO3)$: :$\exists p \in S: m + p = n$ Now suppose that $p, q \in S$ are such that: :$m + p = m + q = n$ Then it follows from axiom $(NO2)$ that: :$p = q$ Hence the result. {{qed}}
=== Definition 1 is equivalent to Definition 2 === {{begin-eqn}} {{eqn | l = \map \Beta {x, y}       | r = \int_0^1 t^{x - 1} \paren {1 - t}^{y - 1} \rd t       | c = {{Defof|Beta Function|index = 1}} }} {{eqn | r = \int_0^{\pi / 2} \paren {\sin \theta}^{2 x - 2} \paren {\cos \theta}^{2 y - 2} 2 \sin \theta \cos \theta \rd \theta       | c = Substitute $t = \sin^2 \theta$ }} {{eqn | r = 2 \int_0^{\pi / 2} \paren {\sin \theta}^{2 x - 1} \paren {\cos \theta}^{2 y - 1} \rd \theta       | c = {{Defof|Beta Function|index = 2}} }} {{end-eqn}} {{qed|lemma}} === Definition 2 is equivalent to Definition 3 === By definition of Gamma function: :$\displaystyle \map \Gamma x \, \map \Gamma y = \int_0^\infty t^{x - 1} e^{-t} \rd t \int_0^\infty s^{y - 1} e^{-s} \rd s$ Substitute $t = u^2$ and $s = v^2$: {{begin-eqn}} {{eqn | l = \map \Gamma x \, \map \Gamma y       | r = 4 \int_0^\infty u^{2 x - 1} e^{-u^2} \rd u \int_0^\infty v^{2 y - 1} e^{-v^2} \rd v       | c = }} {{eqn | r = 4 \int_0^\infty \int_0^\infty u^{2 x - 1} v^{2 y - 1} e^{-\paren {u^2 + v^2} } \rd u \rd v       | c = Fubini's Theorem }} {{eqn | r = 4 \int_0^\infty \int_0^{\pi / 2} r^{2 x - 1} \paren {\sin \theta}^{2 x - 1} r^{2 y - 1} \paren {\cos \theta}^{2 y - 1} e^{-r^2} r \rd \theta \rd r       | c = Substitute $v = r \cos \theta$ and $u = r \sin \theta$ }} {{eqn | r = 4 \int_0^\infty r^{2 x + 2 y - 2} e^{-r^2} r \rd r \int_0^{\pi / 2} \paren {\sin \theta}^{2 x - 1} \paren {\cos \theta}^{2 y - 1} \rd \theta       | c = Fubini's Theorem }} {{eqn | r = 2 \int_0^\infty u^{x + y - 1} e^{-u} \rd u \int_0^{\pi / 2} \paren {\sin \theta}^{2 x - 1} \paren {\cos \theta}^{2 y - 1} \rd \theta       | c = Substitute $u = r^2$ }} {{eqn | r = 2 \, \map \Gamma {x + y} \int_0^{\pi / 2} \paren {\sin \theta}^{2 x - 1} \paren {\cos \theta}^{2 y - 1} \rd \theta       | c = {{Defof|Gamma Function}} }} {{eqn | ll= \leadsto       | o =        | r = \frac {\map \Gamma x \, \map \Gamma y} {\map \Gamma {x + y} }       | c = {{Defof|Beta Function|index = 3}} }} {{eqn | r = 2 \int_0^{\pi / 2} \paren {\sin \theta}^{2 x - 1} \paren {\cos \theta}^{2 y - 1} \rd \theta       | c = {{Defof|Beta Function|index = 2}} }} {{end-eqn}} {{qed}}
The forward implication is immediate from $\preceq$ being compatible with $\circ$: :$\forall m, n, p \in S: m \preceq n \implies m \circ p \preceq n \circ p$ Conversely, suppose that $m \circ p \preceq n \circ p$. Suppose that $n \prec m$. Then as $\preceq$ is compatible with $\circ$: :$n \circ p \preceq m \circ p$ Since $\preceq$ is an ordering, this implies: :$n \circ p = m \circ p$ By axiom $(\text {NO} 2)$, it follows that: :$n = m$ contradicting our assumption that $n \prec m$. Hence, since $\preceq$ is a total ordering: :$m \preceq n$ as desired. {{qed}}
By axiom $(NO2)$, all $n \in S$ are cancellable. Hence from Strict Ordering Preserved under Product with Cancellable Element: :$\forall m, n, p \in S: m \prec n \implies m \circ p \prec n \circ p$ By axiom $(NO1)$, $\preceq$ is a total ordering. Therefore, the contrapositive of: :$\forall m, n, p \in S: m \circ p \prec n \circ p \implies m \prec n$ is: :$\forall m, n, p \in S: m \preceq n \implies m \circ p \preceq n \circ p$ which we know to be true by virtue of axiom $(NO2)$. The result follows. {{qed}}
By Zero Strictly Precedes One, $0 \prec 1$, where $0$ is the zero of $S$. Hence from Strict Ordering of Naturally Ordered Semigroup is Strongly Compatible: :$n \circ 0 \prec n \circ 1$ and by Zero is Identity in Naturally Ordered Semigroup, $n \circ 0 = n$. Now suppose that $n \prec m$. Then by axiom $(NO3)$, there exists $p \in S$ such that: :$n \circ p = m$ Moreover, since $n \ne m$, it follows that $p \ne 0$. Hence $0 \prec p$ by definition of zero. Therefore, by definition of one: :$1 \preceq p$ Now by compatibility of $\preceq$ with $\circ$: :$n \circ 1 \preceq n \circ p = m$ as desired. Conversely, if $n \circ 1 \preceq m$, it is immediate from: :$n \prec n \circ 1$ that $n \prec m$. {{qed}}
Let $f \left({x}\right) = \tan x - x$. By Derivative of Tangent Function, $f' \left({x}\right) = \sec^2 x - 1$. By Shape of Secant Function, $\sec^2 x > 1$ for $x \in \left({0 \,.\,.\, \dfrac {\pi} 2}\right)$. Hence $f' \left({x}\right) > 0$. From Derivative of Monotone Function, $f \left({x}\right)$ is strictly increasing in this interval. Since $f \left({0}\right) = 0$, it follows that $f \left({x}\right) > 0$ for all $x$ in $x \in \left({0 \,.\,.\, \dfrac {\pi} 2}\right)$. {{qed}} Category:Tangent Function Category:Inequalities jyj6srbdnfe2spqq4ob5afa9mk69130
{{begin-eqn}} {{eqn | l = \int_0^\infty \frac {t^{s - 1} } {e^t - 1} \rd t       | r = \int_0^\infty \frac {t^{s - 1} e^{-t} } {1 - e^{-t} } \rd t       | c =  }} {{eqn | r = \int_0^\infty t^{s - 1} e^{-t} \sum_{n \mathop = 0}^\infty e^{-n t} \rd t       | c = Sum of Infinite Geometric Sequence }} {{eqn | r = \sum_{n \mathop = 1}^\infty \int_0^\infty t^{s - 1} e^{-n t} \rd t       | c = interchange of sum and integral is valid by Fubini's Theorem }} {{eqn | r = \sum_{n \mathop = 1}^\infty \frac 1 {n^s} \int_0^\infty t^{s - 1} e^{-t} \rd t       | c = substituting $n t \to t$ }} {{eqn | r = \map \zeta s \, \map \Gamma s       | c =  }} {{end-eqn}} {{qed}}
Let $f$ be convex real function on $I$ according to definition 1. That is: :$\forall x, y \in I: \forall \alpha, \beta \in \R_{>0}, \alpha + \beta = 1: \map f {\alpha x + \beta y} \le \alpha \map f x + \beta \map f y$ {{WLOG}}, assume $x \le y$. Make the substitutions $x_1 = x, x_2 = \alpha x + \beta y, x_3 = y$. As $\alpha + \beta = 1$, we have $x_2 = \alpha x_1 + \paren {1 - \alpha} x_3$. Thus: :$\alpha = \dfrac {x_3 - x_2} {x_3 - x_1}, \beta = \dfrac {x_2 - x_1} {x_3 - x_1}$ So: {{begin-eqn}} {{eqn | n = 1       | l = \map f {x_2}       | o = \le       | r = \frac {x_3 - x_2} {x_3 - x_1} \map f {x_1} + \frac {x_2 - x_1} {x_3 - x_1} \map f {x_3}       | c = substituting for $\alpha$ and $\beta$ }} {{eqn | ll= \leadstoandfrom       | l = \paren {x_3 - x_1} \map f {x_2}       | o = \le       | r = \paren {x_3 - x_2} \map f {x_1} + \paren {x_2 - x_1} \map f {x_3}       | c =  }} {{eqn | ll= \leadstoandfrom       | l = \paren {x_3 - x_1} \map f {x_2}       | o = \le       | r = \paren {x_3 - x_1 - x_2 + x_1} \map f {x_1} + \paren {x_2 - x_1} \map f {x_3}       | c =  }} {{eqn | ll= \leadstoandfrom       | l = \paren {x_3 - x_1} \map f {x_2} - \paren {x_3 - x_1} \map f {x_1}       | o = \le       | r = \paren {x_2 - x_1} \map f {x_3} - \paren {x_2 - x_1} \map f {x_1}       | c =  }} {{eqn | ll= \leadstoandfrom       | l = \frac {\map f {x_2} - \map f {x_1} } {x_2 - x_1}       | o = \le       | r = \frac {\map f {x_3} - \map f {x_1} } {x_3 - x_1}       | c =  }} {{end-eqn}} Again from $(1)$: {{begin-eqn}} {{eqn | n = 1       | l = \map f {x_2}       | o = \le       | r = \frac {x_3 - x_2} {x_3 - x_1} \map f {x_1} + \frac {x_2 - x_1} {x_3 - x_1} \map f {x_3}       | c =  }} {{eqn | ll= \leadstoandfrom       | l = \paren {x_3 - x_1} \map f {x_2}       | o = \le       | r = \paren {x_3 - x_2} \map f {x_1} + \paren {x_2 - x_1} \map f {x_3}       | c =  }} {{eqn | ll= \leadstoandfrom       | l = \paren {x_3 - x_1} \map f {x_2}       | o = \le       | r = \paren {x_3 - x_1 - x_3 + x_2} \map f {x_3} + \paren {x_3 - x_2} \map f {x_1}       | c =  }} {{eqn | ll= \leadstoandfrom       | l = \paren {x_3 - x_2} \map f {x_3} - \paren {x_3 - x_2} \map f {x_1}       | o = \le       | r = \paren {x_3 - x_1} \map f {x_3} - \paren {x_3 - x_1} \map f {x_2}       | c =  }} {{eqn | ll= \leadstoandfrom       | l = \frac {\map f {x_3} - \map f {x_1} } {x_3 - x_1}       | o = \le       | r = \frac {\map f {x_3} - \map f {x_2} } {x_3 - x_2}       | c =  }} {{end-eqn}} Thus: :$\dfrac {\map f {x_2} - \map f {x_1} } {x_2 - x_1} \le \dfrac {\map f {x_3} - \map f {x_1} } {x_3 - x_1} \le \dfrac {\map f {x_3} - \map f {x_2} } {x_3 - x_2}$ So: :$\dfrac {\map f {x_2} - \map f {x_1} } {x_2 - x_1} \le \dfrac {\map f {x_3} - \map f {x_1} } {x_3 - x_1}$ demonstrating that $f$ is convex on $I$ according to definition 3, and: :$\dfrac {\map f {x_2} - \map f {x_1} } {x_2 - x_1} \le \dfrac {\map f {x_3} - \map f {x_2} } {x_3 - x_2}$ demonstrating that $f$ is convex on $I$ according to definition 2. As each step is an equivalence, the argument reverses throughout. {{qed}}
Let $f$ be concave real function on $I$ according to definition 1. That is: :$\forall x, y \in I: \forall \alpha, \beta \in \R_{>0}, \alpha + \beta = 1: \map f {\alpha x + \beta y} \ge \alpha \map f x + \beta \map f y$ Make the substitutions $x_1 = x, x_2 = \alpha x + \beta y, x_3 = y$. As $\alpha + \beta = 1$, we have $x_2 = \alpha x_1 + \paren {1 - \alpha} x_3$. Thus: :$\alpha = \dfrac {x_3 - x_2} {x_3 - x_1}, \beta = \dfrac {x_2 - x_1} {x_3 - x_1}$ So: {{begin-eqn}} {{eqn | n = 1       | l = \map f {x_2}       | o = \ge       | r = \frac {x_3 - x_2} {x_3 - x_1} \map f {x_1} + \frac {x_2 - x_1} {x_3 - x_1} \map f {x_3}       | c = substituting for $\alpha$ and $\beta$ }} {{eqn | ll= \leadstoandfrom       | l = \paren {x_3 - x_1} \map f {x_2}       | o = \ge       | r = \paren {x_3 - x_2} \map f {x_1} + \paren {x_2 - x_1} \map f {x_3}       | c =  }} {{eqn | ll= \leadstoandfrom       | l = \paren {x_3 - x_1} \map f {x_2}       | o = \ge       | r = \paren {x_3 - x_1 - x_2 + x_1} \map f {x_1} + \paren {x_2 - x_1} \map f {x_3}       | c =  }} {{eqn | ll= \leadstoandfrom       | l = \paren {x_3 - x_1} \map f {x_2} - \paren {x_3 - x_1} \map f {x_1}       | o = \ge       | r = \paren {x_2 - x_1} \map f {x_3} - \paren {x_2 - x_1} \map f {x_1}       | c =  }} {{eqn | ll= \leadstoandfrom       | l = \frac {\map f {x_2} - \map f {x_1} } {x_2 - x_1}       | o = \ge       | r = \frac {\map f {x_3} - \map f {x_1} } {x_3 - x_1}       | c =  }} {{end-eqn}} Again from $(1)$: {{begin-eqn}} {{eqn | n = 1       | l = \map f {x_2}       | o = \ge       | r = \frac {x_3 - x_2} {x_3 - x_1} \map f {x_1} + \frac {x_2 - x_1} {x_3 - x_1} \map f {x_3}       | c =  }} {{eqn | ll= \leadstoandfrom       | l = \paren {x_3 - x_1} \map f {x_2}       | o = \ge       | r = \paren {x_3 - x_2} \map f {x_1} + \paren {x_2 - x_1} \map f {x_3}       | c =  }} {{eqn | ll= \leadstoandfrom       | l = \paren {x_3 - x_1} \map f {x_2}       | o = \ge       | r = \paren {x_3 - x_1 - x_3 + x_2} \map f {x_3} + \paren {x_3 - x_2} \map f {x_1}       | c =  }} {{eqn | ll= \leadstoandfrom       | l = \paren {x_3 - x_2} \map f {x_3} - \paren {x_3 - x_2} \map f {x_1}       | o = \ge       | r = \paren {x_3 - x_1} \map f {x_3} - \paren {x_3 - x_1} \map f {x_2}       | c =  }} {{eqn | ll= \leadstoandfrom       | l = \frac {\map f {x_3} - \map f {x_1} } {x_3 - x_1}       | o = \ge       | r = \frac {\map f {x_3} - \map f {x_2} } {x_3 - x_2}       | c =  }} {{end-eqn}} Thus: :$\dfrac {\map f {x_2} - \map f {x_1} } {x_2 - x_1} \ge \dfrac {\map f {x_3} - \map f {x_1} } {x_3 - x_1} \ge \dfrac {\map f {x_3} - \map f {x_2} } {x_3 - x_2}$ So: :$\dfrac {\map f {x_2} - \map f {x_1} } {x_2 - x_1} \ge \dfrac {\map f {x_3} - \map f {x_1} } {x_3 - x_1}$ demonstrating that $f$ is concave on $I$ according to definition 3, and: :$\dfrac {\map f {x_2} - \map f {x_1} } {x_2 - x_1} \ge \dfrac {\map f {x_3} - \map f {x_2} } {x_3 - x_2}$ demonstrating that $f$ is concave on $I$ according to definition 2. As each step is an equivalence, the argument reverses throughout. {{qed}} Category:Concave Real Functions 1w5eny3dousu088aarfj8vxup3b675l
From the General Binomial Theorem: {{begin-eqn}} {{eqn | l = \paren {1 - x^2}^{-1/2}        | r = 1 + \frac 1 2 x^2 + \frac {1 \times 3} {2 \times 4} x^4 + \frac {1 \times 3 \times 5} {2 \times 4 \times 6} x^6 + \cdots       | c = }}       {{eqn | n = 1       | r = \sum_{n \mathop = 0}^\infty \frac {\paren {2 n}!} {2^{2 n} \paren {n!}^2} x^{2 n}       | c =  }} {{end-eqn}} for $-1 < x < 1$. From Power Series is Termwise Integrable within Radius of Convergence, $(1)$ can be integrated term by term: {{begin-eqn}} {{eqn | l = \int_0^x \frac 1 {\sqrt{1 - t^2} } \rd t       | r = \sum_{n \mathop = 0}^\infty \int_0^x \frac {\paren {2 n}!} {2^{2 n} \paren {n!}^2} t^{2 n} \rd t       | c = }} {{eqn | ll= \leadsto       | l = \arcsin x       | r = \sum_{n \mathop = 0}^\infty \frac {\paren {2 n}!} {2^{2 n} \paren {n!}^2} \frac {x^{2 n + 1} } {2 n + 1}       | c = Derivative of Arcsine Function }} {{end-eqn}} We will now prove that the series converges for $-1 \le x \le 1$. By Stirling's Formula: {{begin-eqn}} {{eqn | l = \frac {\paren {2 n}!} {2^{2 n} \paren {n!}^2} \frac {x^{2 n + 1} } {2 n + 1}        | o = \sim       | r = \frac {\paren {2 n}^{2 n} e^{-2 n} \sqrt {4 \pi n} } {2^{2 n} n^{2 n} e^{-2 n} 2 \pi n} \frac {x^{2 n + 1} } {2 n + 1}       | c = }} {{eqn | r = \frac 1 {\sqrt {\pi n} } \frac {x^{2 n + 1} } {2 n + 1}       | c = }} {{end-eqn}} Then: {{begin-eqn}} {{eqn | l = \size {\frac 1 {\sqrt {\pi n} } \frac {x^{2 n + 1} } {2 n + 1} }       | o = <       | r = \size {\frac {x^{2 n + 1} } {n^{3/2} } }       | c =  }} {{eqn | o = \le       | r = \frac 1 {n^{3/2} }       | c =  }} {{end-eqn}} Hence by Convergence of P-Series: :$\displaystyle \sum_{n \mathop = 1}^\infty \frac 1 {n^{3/2} }$ is convergent. So by the Comparison Test, the Taylor series is convergent for $-1 \le x \le 1$. {{qed}}
{{begin-eqn}} {{eqn | l = \arccos x        | r = \frac {\pi} 2 - \arcsin x       | c = Sum of Arcsine and Arccosine }} {{eqn | r = \frac {\pi} 2 - \sum_{n \mathop = 0}^\infty \frac {\paren {2 n}!} {2^{2 n} \paren {n!}^2} \frac {x^{2 n + 1} } {2 n + 1}       | c = Power Series Expansion for Real Arcsine Function }} {{end-eqn}} It follows from Power Series Expansion for Real Arcsine Function that the series is convergent for $-1 \le x \le 1$. {{qed}}