proving an inequality

**PeaceSoul** · March 6th 2011, 04:12 AM

i tried differentiatin both sides, i did that 4 times and i still couldnt find the pattern that gets me to sin

any idea how to solve this ??

**girdav** · March 6th 2011, 06:37 AM

Did you compute $\dfrac{d^n}{dx^n}\left(\dfrac 1{x-i}\right)$ and $\dfrac{d^n}{dx^n}\left(\dfrac 1{x+i}\right)$ ?

**TheCoffeeMachine** · March 6th 2011, 10:46 AM

I think we are using the same book! The last exercise of Chapter IX, right?

Originally Posted by PeaceSoul

i tried differentiatin both sides, i did that 4 times and i still couldnt find the pattern that gets me to sin

any idea how to solve this ?

Show that $\displaystyle \frac{d^n}{dx^n} \left(\frac{1}{x^2+1} \right) = \frac{i(-1)^nn!}{2(x+i)^{n+1}}+\frac{i(-1)^{n+1}n!}{2(x-i)^{n+1}}$ .

I avoid the differentiation mess by using the following formula:

$\displaystyle \frac{d^n}{dx^n} (ax+b)^m = a^n\left(ax+b\right)^{m-n}\prod_{1 \le k \le n}\left(m-k+1\right)$ .

**PeaceSoul** · March 6th 2011, 10:47 AM

yes i did that, but i got no where,,please guide me im really lost :S

**TheCoffeeMachine** · March 6th 2011, 02:18 PM

$\displaystyle \frac{d^n}{dx^n} \left(\frac{1}{x^2+1} \right) = \frac{i(-1)^nn!}{2(x+i)^{n+1}}+\frac{i(-1)^{n+1}n!}{2(x-i)^{n+1}} = (-1)^nn!\bigg(\frac{i}{2(x+i)^{n+1}}-\frac{i}{2(x-i)^{n+1}}\bigg)$ .

Denote the stuff under the brackets by $S$ . Now, let $x = \cot{\theta}$ , then we have:

$\begin{aligned} \displaystyle S & = \frac{i}{2(\cot{\theta}+i)^{n+1}}-\frac{i}{2(\cot{\theta}-i)^{n+1}} = \frac{i}{2\left(\frac{\cos{\theta}}{\sin{\theta}}+ i\right)^{n+1}}-\frac{i}{2\left(\frac{\cos{\theta}}{ \sin{\theta}}-i\right)^{n+1}} \\ & = \frac{i}{\frac{2}{ \sin^{n+1}{\theta}}\left(\cos{\theta}+i\sin{\theta }\right)^{n+1}}-\frac{i}{\frac{2}{ \sin^{n+1}{\theta}}\left(\cos{\theta}-i\sin{\theta}\right)^{n+1}}\\ & = \frac{i\sin^{n+1}\left(\cos{\theta}+i\sin{\theta}\ right)^{-n-1}}{2}-\frac{i\sin^{n+1}\left(\cos{\theta}-i\sin{\theta}\right)^{-n-1}}{2} \end{aligned}$

Recall that $\left(\cos{\varphi}+i\sin{\varphi}\right)^{t} = \cos{t\varphi} +i\sin{t\varphi}$ (De Moivre's theorem), then:

$\displaystyle \begin{aligned} S & = \frac{i\sin^{n+1}\left(\cos[(-n-1)\theta]+i\sin[(-n-1)\theta]\right)}{2}-\frac{i\sin^{n+1}\left(\cos[(-n-1)\theta]-i\sin[(-n-1)\theta]\right)}{2} \\ & = \frac{i\sin^{n+1}\left(\cos[(n+1)\theta]-i\sin[(n+1)\theta]\right)}{2}-\frac{i\sin^{n+1}\left(\cos[(n+1)\theta]+i\sin[(n+1)\theta]\right)}{2} \\& = \frac{i\sin^{n+1}{\theta}\cos[(n+1)\theta)-i^2sin[(n+1)\theta]-i\sin^{n+1}\cos[(n+1)\theta]-i^2\sin^{n+1}{\theta}\sin[(n+1)\theta]}{2} \\ & = \frac{-2i^2\sin^{n+1}\sin[(n+1)\theta]}{2}= -(-1)\sin^{n+1}{\theta}\sin[(n+1)\theta] = \sin^{n+1}{\theta}\sin[(n+1){\theta}]. \end{aligned}$

Thus $S = \sin^{n+1}{\theta}\sin[(n+1)\theta]$ , therefore $\displaystyle \frac{d^n}{dx^n} \left(\frac{1}{x^2+1} \right) = (-1)^nn!\sin^{n+1}{\theta}\sin[(n+1)\theta]$ .

**PeaceSoul** · March 6th 2011, 02:44 PM

Thank you for an amazinggggggggggggg answer !!!!!! You rock

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