Maclaurin Series used to find associated radius of convergence Q

badtwistoffate
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I have the Maclaurin series for cos (x), is their a way to find its radius of convergence from that?

ALSO
Is there a trick to find the shorter version of the power series for the Maclaurin series, I can never seem to find it so instead of the long series with each term but like E summation (the series)
 
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badtwistoffate said:
I have the Maclaurin series for cos (x), is their a way to find its radius of convergence from that?

you can try the ratio test.

badtwistoffate said:
ALSO
Is there a trick to find the shorter version of the power series for the Maclaurin series, I can never seem to find it so instead of the long series with each term but like E summation (the series)

Err, do you mean writing the series using the sigma notation instead of the first few terms followed by some ...? You want to look for patterns in the coefficients. No real trick, practice will help though.
 
shmoe said:
you can try the ratio test.
Err, do you mean writing the series using the sigma notation instead of the first few terms followed by some ...? You want to look for patterns in the coefficients. No real trick, practice will help though.

Yeah i tried the ratio test, but the radius of convergence it sayed in the big is infinity, how is that possible as it has to be n < 1?
 
badtwistoffate said:
Yeah i tried the ratio test, but the radius of convergence it sayed in the big is infinity, how is that possible as it has to be n < 1?

I don't understand what you're saying, what's "in the big" mean? What are you calling n that it has to be less than 1?
 
shmoe said:
I don't understand what you're saying, what's "in the big" mean? What are you calling n that it has to be less than 1?

Sorry I ment to say in the book the radius of convergence is infinity, how is that possible seeing the result of the ratio test gives you L and it has to be less then 1? how to you get that radius of infinity?
 
You should have found that for *any* value of x, the limit the ratio test gives is always less than 1, hence the series converges for all values of x and we say the radius of convergence is infinity.
 
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...
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