1. ## How to prove?

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) ?

2. I don't believe that

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

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

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

EDIT. I got it to work for

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

Is this the equation?

3. Interesting variant to this thread. Introduce the new independent variable

$\displaystyle x = \sin t$

in the equation

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

Spoiler:
$\displaystyle \dfrac{d^2y}{dt^2} + k^2 y = 0$

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

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

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

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

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

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

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

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

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

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

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

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

$\displaystyle \displaystyle = e^0$

$\displaystyle \displaystyle = 1$.

PHEW!!!

5. Hello, RCola!

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

Let: $\displaystyle y \:=\\tan x)^{2x-\pi}$

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

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

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

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

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

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

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

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

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

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

6. Originally Posted by Soroban
Hello, RCola!

Let: $\displaystyle y \:=\\tan x)^{2x-\pi}$

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

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

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

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

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

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

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

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

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

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

7. Originally Posted by Danny
I don't believe that

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

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

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

EDIT. I got it to work for

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

Is this the equation?
Originally Posted by Danny
Interesting variant to this thread. Introduce the new independent variable

$\displaystyle x = \sin t$

in the equation

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

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

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

8. Originally Posted by RCola
Is there any tool to easily write math formulas between MATH quotes?
Yes, improving your touch typing :P