Interval of Convergence for a Series

calculusisfun
Messages
30
Reaction score
0

Homework Statement


Ok, so I don't need help with this part, I just got stuck at the following step when attempting to find the interval of convergence:

The Attempt at a Solution



I got here:

-4 < x^2 < 4

So, I need to solve this inequality. But can I? How can I take the square root of negative 4? And if this isn't possible to solve, what is the interval of convergence?
 
Physics news on Phys.org
The inequality is perfectly easy to solve. It's -2<x<2. For every real number x^2 is greater than -4, so you don't even care about that limit.
 
Last edited:
Thanks for the prompt response.

Okay, I understand what you're saying partially.

But to solve the inequality -4 < x^2 < 4, wouldn't you take the square root of both sides of the inequality to get the following:

root(-4) < x < root(4)

And root(-4) is an imaginary number is it not? Which would mean what for the interval of convergence? o.O
 
calculusisfun said:
Thanks for the prompt response.

Okay, I understand what you're saying partially.

But to solve the inequality -4 < x^2 < 4, wouldn't you take the square root of both sides of the inequality to get the following:

root(-4) < x < root(4)

And root(-4) is an imaginary number is it not? Which would mean what for the interval of convergence? o.O

Yes, sqrt(-4) is imaginary, but who cares? You aren't looking for imaginary solutions. You want real solutions. Just replace it with 0<=x^2<4.
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
Back
Top