Convergence of Infinite Sums of Trigonometric Functions: Finding the Range of x

[stukbv]

Homework Statement

Find what range of values of x the infinite sum of sin²ⁿ(x) and infinite sum 2nsin^2n-1(x) converge and find an expression for their sums, carefully justifying your answers.

The Attempt at a Solution

I used cauchys root testand basically got that the first sum converges for all x in the reals except
x= {(2n+1)pi/2 } for n=0,1,2,3...

Then i said that the radius of convergence was the same as the first since it is the derivative of the first.

I hope this is right so far, if so what does it mean by "find an expression for their sums"?

Thanks

 

[micromass]

Hi stukbv!

The Attempt at a Solution[/h2]
I used cauchys root testand basically got that the first sum converges for all x in the reals except
x= {(2n+1)pi/2 } for n=0,1,2,3...

Seems ok!

Then i said that the radius of convergence was the same as the first since it is the derivative of the first.

Hmm, but the second is not exactly the deravitive of the first is it? It looks a lot like it, but it isn't exactly that.

I hope this is right so far, if so what does it mean by "find an expression for their sums"?

It means to calculate the infinite sum. (Hint: the series is of a very special type).

 

[stukbv]

Does it mean to use taylors formula about x=0?
Also how would i deal with the second sum then, would i do cauchys root test again on it ?

 

[micromass]

Does it mean to use taylors formula about x=0?
Also how would i deal with the second sum then, would i do cauchys root test again on it ?

Let's wait a bit with the second series. Let's first do the first one. Think of geometric series...

 

[stukbv]

1/1-(sin(x))^2 ??

 

[micromass]

1/1-(sin(x))^2 ??

Indeed! Now, what do you get if you differentiate a geometric series?

 

[stukbv]

(cos(x))^2 + 2sinxcosx / (cos(x))^2

 

[micromass]

What?

I just asked you to calculate the derivative of

$$\sum_{n=0}^{+\infty}{x^{2n}}$$

what are those sines and cosines coming from??

 

[stukbv]

oh sorry i thought you meant differentiate my previous answer,
2x/(1-x^2)^2 ?

 

[micromass]

Yes, that's already fine!

Now, what if you differentiate the series itself? (i.e. not the sum)

 

[stukbv]

2nx^(2n-1)

 

[micromass]

Yes, now apply all that information to your second series...

 

[stukbv]

2sinx/(1-sin^2)^2 ? ahh i don't know :(

 

[micromass]

2sinx/(1-sin^2)^2 ? ahh i don't know :(

That's correct! See, it's not so hard

 

[stukbv]

ahh good thanks so much! does the second one also just converge for the same values as the first by the way?

 

[micromass]

ahh good thanks so much! does the second one also just converge for the same values as the first by the way?

I don't understand. You just calculated to where the second series converges?

 

[stukbv]

ah, so i just say it converges for all x such that sin^2(x) doesn't equal one.
Thanks!