# Power Series of arcsin

1. Apr 1, 2012

### chromium1387

1. The problem statement, all variables and given/known data
a. Find a power series expansion for arcsin(x) centered at 0.
b. Find the radius of convergence and interval of convergence of the power series in a.
c. Choose an appropriate value of x to plug into the power series found in a. to find a series that converges to $\frac{$\pi$}{2}$.

2. Relevant equations
binomial theorem
ratio test

3. The attempt at a solution
a. Using the Binomial Theorem, I found a power series representation for $\frac{1}{\sqrt{1-x^2}}$ and integrated that to find a power series for arcsin. What I got was: x+$\sum\frac{1*3*5*...*(2n-1)x^{2n+1}}{(2n+1)(2^{n})(n!)}$
I'm fairly sure this is correct.
b. However, when I go and use the ratio test, I am a bit confused. I get it simplified down to :
|$\frac{x^{2}n}{(n+1)(2n-1)}$| $\rightarrow$ 0 as n $\rightarrow$ $\infty$
If I just left it here, R=$\infty$ and the interval of convergence would be (-$\infty$,$\infty$), correct?
c. And I have no idea how to do this one...

Last edited: Apr 1, 2012
2. Apr 1, 2012

### Poopsilon

The x in front of your sum doesn't matter for the ratio test since you are only concerned with the nth and the (n+1)th term.

As for part c, remember that arcsin is the inverse of sine, thus you want the value x such that sin(∏/2) = x.

3. Apr 1, 2012

### chromium1387

Okay, so does this mean that my radius and interval of convergence are correct?

This should be easy enough then. Thanks!

4. Apr 1, 2012

### Poopsilon

Well if this is over the reals then radius of convergence and interval of convergence are the same thing. I didn't explicitly check your calculation but as long as you were careful I'm sure its fine. Although I'm not sure the power series of arcsin outside of the interval [-1,1] even has any meaning, even if it is convergent on the whole real line, since sin() maps the entire real line to only values in that interval.

5. Apr 1, 2012

### chromium1387

Okay. Thank you!

6. Apr 1, 2012

### Dick

You should not have gotten radius of convergence equal infinity from your power series. You should have gotten 1. Your ratio test expression is wrong. Can you show how you got it?

7. Apr 1, 2012

### chromium1387

lim n$\rightarrow$$\infty$ |$\frac{1*3*5*...*(2n)x^{2n+2}}{2^{n+1}(n+1)!}$*$\frac{2^{n}n!}{1*3*5*...*(2n-1)x^{2n}}$|
=lim n$\rightarrow$$\infty$ |$\frac{n}{(n+1)(2n-1)}$|x$^{2}$
which, by l'hopital's rule is:
lim n$\rightarrow$$\infty$ |$\frac{1}{2n+2(n+1)-1}$|x$^{2}$
=0

8. Apr 1, 2012

### Dick

Well, that's wrong. How did you get more n's in the denominator than the numerator? You have a factorial or a double factorial in each. Check it again.

9. Apr 1, 2012

### chromium1387

i, uh, i'm not really sure..
i just cancelled things.
like the 1*3*5*... cancel.
the 2n stays.
the x^(2n+2) / x^2n make x^2.
the n!/(n+1)! make n+1.
the 2^n/2^n+1 just gives me 1/2. which cancels with the 2 in 2n.
so i got n/(n-1)(2n-1) all times x^2.
:s

10. Apr 1, 2012

### Dick

Ok, so why 2n-1 in the denominator? Didn't that cancel? There's at least another typo in there but that doesn't matter so much. You've got 1*3*5*...*(2n-1)*(2n+1) over 1*3*5*...*(2n-1).

11. Apr 1, 2012

### chromium1387

righttt! so then i'll just have n/n+1.

12. Apr 1, 2012

### Dick

Ok, so radius of convergence 1. Yes?

13. Apr 1, 2012

### chromium1387

yesyesyes! thank you!
crazy how overlooking one silly thing can mess things up!