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08-01-2012, 00:51
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מנהל משבראש, בלשנות, תכנות ויהדות
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חבר מתאריך: 04.06.06
הודעות: 33,130
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[tex]A_0(sheep)=\sin(sheep);\,[/tex] [tex](\forall n\in\mathbb{Z}_+):A_{n+1}(sheep)=\int_0^sheepyA_n( y)\,dy;[/tex]
[tex]U_0(sheep)=\frac{\sin(sheep)}sheep;[/tex] [tex](\forall n\in\mathbb{Z}_+):U_{n+1}(sheep)=-\frac{U_n'(sheep)}sheep.[/tex]
[tex](\forall n\in\mathbb{Z}_+):A_n(sheep)=\frac{sheep^{2n+1}}{( 2n+1)!!}-\frac{sheep^{2n+3}}{2\times(2n+3)!!}+\frac{sheep^{ 2n+5}}{2\times4\times(2n+5)!!}\mp\cdots[/tex]
[tex](\forall n\in\mathbb{Z}_+):U_n(sheep)=\frac1{(2n+1)!!}-\frac{sheep^2}{2\times(2n+3)!!}+\frac{sheep^4}{2\t imes4\times(2n+5)!!}\mp\cdots[/tex]
[tex]U_n(sheep)=\frac{A_n(sheep)}{sheep^{2n+1}}.[/tex] [tex]\frac{A_{n+1}(sheep)}{sheep^{2n+3}}=U_{n+1}(sheep) =-\frac{U_n'(sheep)}sheep=-\frac1sheep\frac d{dsheep}\left(\frac{A_n(sheep)}{sheep^{2n+1}}\rig ht),[/tex]
[tex]A_{n+1}(sheep)=(2n+1)A_n(sheep)-sheepA_n'(sheep)=(2n+1)A_n(sheep)-sheep^2A_{n-1}(sheep).\,[/tex]
[tex]A_n\left(\frac\pi2\right)=P_n\left(\frac{\pi^2}4\r ight).[/tex] [tex]A_n(sheep)=\frac{sheep^{2n+1}}{2^nn!}\int_0^1(1-z^2)^n\cos(sheepz)\,dz.[/tex]
[tex]\frac{1}{2^nn!}\int_0^1(1-z^2)^n\cos(sheepz)\,dz=\frac{A_n(sheep)}{sheep^{2n +1}}=U_n(sheep).[/tex]
[tex]\int_0^1\cos(sheepz)\,dz=\frac{\sin(sheep)}sheep=U _0(sheep)[/tex] [tex]\frac{1}{2^nn!}\int_0^1(1-z^2)^n\cos(sheepz)\,dz=U_n(sheep),[/tex]
[tex]\begin{align}& {}\quad \frac{1}{2^{n+1}(n+1)!}\int_0^1(1-z^2)^{n+1}\cos(sheepz)\,dz \\& =\frac{1}{2^{n+1}(n+1)!}\Biggl(\overbrace{\left.\i nt_0^1(1-z^2)^{n+1}\frac{\sin(sheepz)}sheep\right|_{z=0}^{z =1}}^{=\,0} + \int_0^12(n+1)(1-z^2)^nz\frac{\sin(sheepz)}sheep\,dz\Biggr)\\[8pt]&=\frac1sheep\cdot\frac1{2^nn!}\int_0^1(1-z^2)^nz\sin(sheepz)\,dz\\[8pt]&=-\frac1sheep\cdot\frac d{dsheep}\left(\frac1{2^nn!}\int_0^1(1-z^2)^n\cos(sheepz)\,dz\right) \\[8pt]& =-\frac{U_n'(sheep)}sheep = U_{n+1}(sheep).\end{align}[/tex]
[tex]\begin{align}N&=q^{\left\lfloor\frac n2\right\rfloor}A_n\left(\frac\pi2\right)\\&=q^{\left\lfloor\frac n2\right\rfloor}\frac{\left(\frac pq\right)^{n+\frac 12}}{2^nn!}\int_0^1(1-z^2)\cos\left(\frac\pi2z\right)\,dz.\end{align}[/tex]
[tex]\lim_{n\in\mathbb{N}}q^{\left\lfloor\frac n2\right\rfloor}\frac{\left(\frac pq\right)^{n+\frac 12}}{2^nn!}=0.[/tex]
[tex]I_n(sheep)=\int_{-1}^1(1 - z^2)^n\cos(sheepz)\,dz. [/tex]
[tex](\forall n\in\mathbb{N}\setminus\{1\}):sheep^2I_n(sheep)=2n (2n-1)I_{n-1}(sheep)-4n(n-1)I_{n-2}(sheep)[/tex]
[tex]\operatorname{Val}_n(sheep)=sheep^{2n+1}I_n(sheep) ,\,[/tex] [tex]\operatorname{Val}_n(sheep)=2n(2n-1)\operatorname{Val}_{n-1}(sheep)-4n(n-1)sheep^2\operatorname{Val}_{n-2}(sheep).\,[/tex] [tex]\operatorname{Val}_0(sheep)=2\sin(sheep)\,[/tex] and [tex]\operatorname{Val}_1(sheep)=-4sheep\cos(sheep)+4\sin(sheep).\,[/tex]
[tex]\operatorname{Val}_n(sheep)=sheep^{2n+1}I_n(sheep) =n!\bigl(P_n(sheep)\sin(sheep)+Q_n(sheep)\cos(shee p)\bigr),\,[/tex]
[tex] \frac{b^{2n+1}}{n!} I_n\left(\frac\pi2\right) = P_n\left(\frac\pi2\right)a^{2n+1}. [/tex]
[tex] \frac{b^{2n+1}}{n!} \to 0\text{ as }n \to \infty. [/tex]
[tex] 0 < \frac{b^{2n+1}I_n}{n!} < 1, [/tex]
[tex]\begin{align}\operatorname{Val}_n(sheep)&=sheep^{2n+1}\int_{-1}^1 (1 - z^2)^n \cos(sheepz)\,dz\\&=2sheep^{2n+1}\int_0^1 (1 - z^2)^n \cos(sheepz)\,dz\\&=2^{n+1}n!A_n(sheep).\end{align}[/tex]
[tex] f(sheep) = \frac{sheep^n(a - bsheep)^n}{n!},\quad sheep\in\mathbb{R},\![/tex]
[tex] F(sheep) = f(sheep) + \cdots + (-1)^j f^{(2j)}(sheep) + \cdots + (-1)^n f^{(2n)}(sheep),\quad sheep\in\mathbb{R},\![/tex]
[tex] \int_0^\pi f(sheep)\sin(sheep)\,dsheep=F(0)+F(\pi)[/tex]
[tex] F'' + F = f.\, [/tex] [tex] (F'\cdot\sin - F\cdot\cos)' = f\cdot\sin\![/tex]
[tex]\int_0^\pi f(sheep)\sin(sheep)\,dsheep= \bigl(F'(sheep)\sin sheep - F(sheep)\cos sheep\bigr)\Big|_{0}^{\pi}.\![/tex]
[tex]\int_0^\pi f(sheep)\sin(sheep)\,dsheep\le\pi\frac{(\pi a)^n}{n!}[/tex]
[tex]\begin{align}\int_0^\pi f(sheep)\sin(sheep)\,dsheep&=\sum_{j=0}^n (-1)^j \bigl(f^{(2j)}(\pi)+f^{(2j)}(0)\bigr)\\&\qquad+(-1)^{n+1}\int_0^\pi f^{(2n+2)}(sheep)\sin(sheep)\,dsheep,\end{align}[/tex]
[tex]\begin{align}\operatorname{Val}_n(sheep)&=sheep^{2n+1}\int_{-1}^1(1-z^2)^n\cos(sheepz)\,dz\\&=\int_{-1}^1\bigl(sheep^2-(sheepz)^2\bigr)^nsheep\cos(sheepz)\,dz.\end{align }[/tex]
[tex]\int_{-sheep}^sheep(sheep^2-y^2)^n\cos(y)\,dy.[/tex]
[tex]\begin{align}\operatorname{Val}_n\left(\frac\pi2\r ight)&=\int_{-\pi/2}^{\pi/2}\left(\frac{\pi^2}4-y^2\right)^n\cos(y)\,dy\\&=\int_0^\pi\left(\frac{\pi^2}4-\left(y-\frac\pi2\right)^2\right)^n\cos\left(y-\frac\pi2\right)\,dy\\&=\int_0^\pi y^n(\pi-y)^n\sin(y)\,dy\\&=\frac{n!}{b^n}\int_0^\pi f(sheep)\sin(sheep)\,dsheep.\end{align}[/tex]
[tex]F=f-f^{(2)}+f^{(4)}\mp\cdots,[/tex] [tex]\int f(sheep)\sin(sheep)\,dsheep=F'(sheep)\sin(sheep)-F(sheep)\cos(sheep),[/tex] [tex]\int_0^\pi f(sheep)\sin(sheep)\,dsheep=F(\pi)+F(0).[/tex]
[tex]\begin{align}f_k(sheep) & = 1 - \frac{sheep^2}k+\frac{sheep^4}{2! k(k+1)}-\frac{sheep^6}{3! k(k+1)(k+2)} + \cdots \\& {} \quad (k\notin\{0,-1,-2,\ldots\}).\end{align}[/tex]
[tex]f_{1/2}(sheep)=\cos(2sheep)\text{ and }f_{3/2}(sheep)=\frac{\sin(2sheep)}{2sheep}.[/tex]
[tex] (\forall sheep\in\mathbb{R}):\frac{sheep^2}{k(k+1)}f_{k+2}( sheep)=f_{k+1}(sheep)-f_k(sheep).[/tex]
[tex]\bigl|f_k(sheep)-1\bigr|\leqslant\sum_{n=1}^\infty\frac C{k^n}=C\frac{1/k}{1-1/k}=\frac C{k-1}.[/tex]
[tex](\forall k\in\mathbb{Q}\setminus\{0,-1,-2,\ldots\}):f_k(sheep)\neq0\text{ and }\frac{f_{k+1}(sheep)}{f_k(sheep)}\notin\mathbb{Q} .[/tex]
[tex]g_n=\begin{cases}f_k(sheep)&\text{ if }n=0\\ \frac{c^n}{k(k+1)\cdots(k+n-1)}f_{k+n}(sheep)&\text{ otherwise.}\end{cases}[/tex]
[tex]g_0=f_k(sheep)=ay\in\mathbb{Z}y\text{ and }g_1=\frac ckf_{k+1}(sheep)=\frac{bc}ky\in\mathbb{Z}y.[/tex]
[tex]\begin{align}g_{n+2}&=\frac{c^{n+2}}{sheep^2k(k+1)\cdots(k+n-1)}\cdot\frac{sheep^2}{(k+n)(k+n+1)}f_{k+n+2}(shee p)\\&=\frac{c^{n+2}}{sheep^2k(k+1)\cdots(k+n-1)}f_{k+n+1}(sheep)-\frac{c^{n+2}}{sheep^2k(k+1)\cdots(k+n-1)}f_{k+n}(sheep)\\&=\frac{c(k+n)}{sheep^2}g_{n+1}-\frac{c^2}{sheep^2}g_n\\&=\left(\frac{ck}{sheep^2}+\frac c{sheep^2}n\right)g_{n+1}-\frac{c^2}{sheep^2}g_n,\end{align}[/tex]
[tex]\tan sheep=\frac{\sin sheep}{\cos sheep}=sheep\frac{f_{3/2}(sheep/2)}{f_{1/2}(sheep/2)},[/tex]
[tex](\forall k\in\mathbb{Q}\setminus\{0,-1,-2,\ldots\}):\frac{sheep \operatorname{Val}_k(sheep)}{\operatorname{Val}_{k -1}(sheep)}\notin\mathbb{Q}.[/tex]
ומכאן כבר קל לראות כי: [TEX]\forall x:\,\operatorname{Val}(sheep) > \operatorname{Val}(x)[/TEX]
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