How to prove?

**RCola** · December 17th 2010, 04:16 AM

1. { x= sin t ; y= sin kt} <=> (1-x)*[(d^2*y)/(dx^2)] - x*[(dy)/(dx)]+ k^2*y=0
?

And how to solve this limit? ->

2. Lim[x->Pi/2][tan x]^(2x-Pi) ?

**Danny** · December 17th 2010, 05:49 AM

I don't believe that

$x = \sin t, y = \sin kt$ is a solution to

$(1-x) \dfrac{d^2y}{dx^2} - x \dfrac{dy}{dx} + k^2 y = 0$

for general $k$ . It works for $k = 1$ though. Check both the equation and solution.

EDIT. I got it to work for

$(1-x^2) \dfrac{d^2y}{dx^2} - x \dfrac{dy}{dx} + k^2 y = 0$

Is this the equation?

**Danny** · December 17th 2010, 06:00 AM

Interesting variant to this thread. Introduce the new independent variable

$x = \sin t$

in the equation

$(1-x^2) \dfrac{d^2y}{dx^2} - x \dfrac{dy}{dx} + k^2 y = 0$

Spoiler:

**Prove It** · December 17th 2010, 06:53 AM

$\displaystyle \lim_{x \to \frac{\pi}{2}}(\tan{x})^{2x-\pi} = \lim_{x \to \frac{\pi}{2}}e^{\ln{[(\tan{x})^{2x-\pi}]}}$

$\displaystyle = \lim_{x \to \frac{\pi}{2}}e^{(2x-\pi)\ln{(\tan{x})}}$

$\displaystyle = e^{\lim_{x \to \frac{\pi}{2}}(2x-\pi)\ln{(\tan{x})}}$

$\displaystyle = e^{\lim_{x \to \frac{\pi}{2}}\frac{\ln{(\tan{x})}}{(2x-\pi)^{-1}}}$

$\displaystyle = e^{\lim_{x \to \frac{\pi}{2}}\frac{\frac{\sec^2{x}}{\tan{x}}}{-2(2x-\pi)^{-2}}}$ by L'Hospital's Rule

$\displaystyle = e^{-\frac{1}{2}\lim_{x \to \frac{\pi}{2}}\frac{(2x-\pi)^2\sec^2{x}}{\tan{x}}}$

$\displaystyle = e^{-\frac{1}{2}\lim_{x \to \frac{\pi}{2}}\frac{(2x-\pi)^2}{\sin{x}\cos{x}}}$

$\displaystyle = e^{-\frac{1}{2}\lim_{x \to \frac{\pi}{2}}\frac{(2x - \pi)^2}{\frac{1}{2}\sin{2x}}}$

$\displaystyle = e^{-\lim_{x \to \frac{\pi}{2}}\frac{(2x-\pi)^2}{\sin{2x}}}$

$\displaystyle = e^{-\lim_{x \to \frac{\pi}{2}}\frac{4(2x-\pi)}{2\cos{2x}}}$ by L'Hospital's Rule

$\displaystyle = e^{-2\lim_{x \to \frac{\pi}{2}}\frac{2x-\pi}{\cos{2x}}$

$\displaystyle = e^{-2\left(\frac{0}{-1}\right)}$

$\displaystyle = e^0$

$\displaystyle = 1$ .

PHEW!!!

**Soroban** · December 17th 2010, 07:02 AM

Hello, RCola!

$\displaystyle 2.\;\lim_{x\to\frac{\pi}{2}}(\tan x)^{2x-\pi}$

Let: $y \:=\<img src=$ \tan x)^{2x-\pi}" alt="y \:=\

\tan x)^{2x-\pi}" />

Take logs: . $\ln y \:=\:\ln(\tan x)^{2x-\pi} \;=\;(2x-\pi)\ln(\tan x)$

. . . . . . . . . . . . $=\;\dfrac{\ln(\tan x)}{(2x-\pi)^{-1}}\quad\Rightarrow\quad \dfrac{\infty}{\infty}$

Apply L'Hopital: . $\dfrac{\frac{\sec^2\!x}{\tan x}}{-(2x-\pi)^{-2}\cdot2} \;=\;\dfrac{\frac{1}{\cos^2\!x}\cdot\frac{\cos x}{\sin x}}{-2(2x-\pi)^{-2}}$

. . . . . . . . $=\;\dfrac{\frac{1}{\sin x\cos x}}{-2(2x-\pi)^{-2}} \;=\;\dfrac{\frac{2}{\sin2x}}{\frac{-2}{(2x-\pi)^2}} \;=\;-\dfrac{(2x-\pi)^2}{\sin2x}\quad\Rightarrow\quad \dfrac{0}{0}$

Apply l'Hopital: . $-\dfrac{2(2x-\pi)\cdot2}{2\cos2x} \;=\;\dfrac{2(\pi-2x)}{\cos 2x}$

. . Then: . $\displaystyle \lim_{x\to\frac{\pi}{2}} \frac{2(\pi-2x)}{\cos2x} \;=\;\frac{0}{-1} \;=\;0$

$\displaystyle \text{Since }\lim_{x\to\frac{\pi}{2}} \ln y \;=\;0,\:\text{ then: }\:\lim_{x\to\frac{\pi}{2}} y \;=\;e^0 \;=\;1$

. . $\displaystyle \text{Therefore: }\;\lim_{x\to\frac{\pi}{2}} (\tan x)^{2x-\pi} \;=\;1$

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

Edit: Ah, Prove It beat me to it! .And did an excellent job!
. . . .(Oh well, I did my work at "ground level" . . .)

**Prove It** · December 17th 2010, 07:03 AM

Originally Posted by Soroban

Hello, RCola!

Let: $y \:=\<img src=$ \tan x)^{2x-\pi}" alt="y \:=\

\tan x)^{2x-\pi}" />

Take logs: . $\ln y \:=\:\ln(\tan x)^{2x-\pi} \;=\;(2x-\pi)\ln(\tan x)$

. . . . . . . . . . . . $=\;\dfrac{\ln(\tan x)}{(2x-\pi)^{-1}}\quad\Rightarrow\quad \dfrac{\infty}{\infty}$

Apply L'Hopital: . $\dfrac{\frac{\sec^2\!x}{\tan x}}{-(2x-\pi)^{-2}\cdot2} \;=\;\dfrac{\frac{1}{\cos^2\!x}\cdot\frac{\cos x}{\sin x}}{-2(2x-\pi)^{-2}}$

. . . . . . . . $=\;\dfrac{\frac{1}{\sin x\cos x}}{-2(2x-\pi)^{-2}} \;=\;\dfrac{\frac{2}{\sin2x}}{\frac{-2}{(2x-\pi)^2}} \;=\;-\dfrac{(2x-\pi)^2}{\sin2x}\quad\Rightarrow\quad \dfrac{0}{0}$

Apply l'Hopital: . $-\dfrac{2(2x-\pi)\cdot2}{2\cos2x} \;=\;\dfrac{2(\pi-2x)}{\cos 2x}$

. . Then: . $\displaystyle \lim_{x\to\frac{\pi}{2}} \frac{2(\pi-2x)}{\cos2x} \;=\;\frac{0}{-1} \;=\;0$

$\displaystyle \text{Since }\lim_{x\to\frac{\pi}{2}} \ln y \;=\;0,\:\text{ then: }\:\lim_{x\to\frac{\pi}{2}} y \;=\;e^0 \;=\;1$

. . $\displaystyle \text{Therefore: }\;\lim_{x\to\frac{\pi}{2}} (\tan x)^{2x-\pi} \;=\;1$

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

Edit: Ah, Prove It beat me to it! .And did an excellent job!
. . . .(Oh well, I did my work at "ground level" . . .)

Well it's nice to know I didn't make any mistakes

**RCola** · December 17th 2010, 11:19 AM

Originally Posted by Danny

I don't believe that

$x = \sin t, y = \sin kt$ is a solution to

$(1-x) \dfrac{d^2y}{dx^2} - x \dfrac{dy}{dx} + k^2 y = 0$

for general $k$ . It works for $k = 1$ though. Check both the equation and solution.

EDIT. I got it to work for

$(1-x^2) \dfrac{d^2y}{dx^2} - x \dfrac{dy}{dx} + k^2 y = 0$

Is this the equation?

Originally Posted by Danny

Interesting variant to this thread. Introduce the new independent variable

$x = \sin t$

in the equation

$(1-x^2) \dfrac{d^2y}{dx^2} - x \dfrac{dy}{dx} + k^2 y = 0$

Spoiler:

\dfrac{dy}{dx} = smth...
This function is given in parametric form. Also there was given some definition, something like $\dfrac{d^2y}{dx^2} = some reference to t variable; and ;$ . So it need to be proven that taking derivatives up to 2nd power/ level will be equal to $(1-x^2) \dfrac{d^2y}{dx^2} - x \dfrac{dy}{dx} + k^2 y = 0$

Is there any tool to easily write math formulas between MATH quotes?

**Prove It** · December 17th 2010, 04:57 PM

Originally Posted by RCola

Is there any tool to easily write math formulas between MATH quotes?

Yes, improving your touch typing :P

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