# power series

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• May 16th 2009, 11:23 AM
Showcase_22
power series
Quote:

Find the range of values of values of x for which the series $\displaystyle \sum_{n=0}^{\infty} \sin^{2n} (x)$ and $\displaystyle \sum_{n=0}^{\infty} 2n \sin ^{2n-1}(x)$ converge, and find an expression for their sums, carefully justifying your answers.
I've calculated the radius of convergence for $\displaystyle \sum_{n =0}^{\infty} \sin^{2n} x$ in two different ways and I get two different answers.

The first way:

Let $\displaystyle a_n= \sin^{2n} x$.
Therefore $\displaystyle limsup_{n \rightarrow \infty} | \sin^{2n} x|^{\frac{1}{n} }=limsup_{n \rightarrow \infty} | \sin^2 x|=1$

Hence $\displaystyle R=\frac{1}{1}=1$.

Alternatively, using the ratio test:

$\displaystyle \left| \frac{a_{n+1}}{a_n} \right|=\left|\frac{\sin^{2n+2} x}{ \sin^{2n} x} \right|=\left|\sin^2 (x) \right|< 1 \Leftrightarrow \sin x<1 \Rightarrow x<\frac{\pi}{2}$

This tells me that $\displaystyle R=\frac{\pi}{2}$.
Which one is correct.

For $\displaystyle \sum_{n=0}^{\infty} 2n \sin^{2n-1} x$:

$\displaystyle \left| \frac{a_{n+1}}{a_n} \right|=\left| \frac{2(n+1) \sin^{2n+1} x}{2n \sin^{2n-1} x } \right|= \left| \frac{n+1}{n} \sin^2 x \right| \rightarrow |sin^2 x|<1 \Rightarrow \ x< \frac{\pi}{2}$

The main problem I have now is trying to find the expression for the sums. Can someone show me how to do that?
• May 16th 2009, 12:27 PM
putnam120
Well the first one is just a geometric series with common ration $\displaystyle \sin(x)$. So you need $\displaystyle |\sin(x)|<1\Longrightarrow -\frac{\pi}{2}<x<\frac{\pi}{2}$.

Additionally, the root test states that the series converges absolutely if $\displaystyle \limsup \sqrt[n]{|a_n|}<1$. So using what you have above, you will arrive at the same result.
• May 16th 2009, 02:19 PM
HallsofIvy
Quote:

Originally Posted by Showcase_22
I've calculated the radius of convergence for $\displaystyle \sum_{n =0}^{\infty} \sin^{2n} x$ in two different ways and I get two different answers.

The first way:

Let $\displaystyle a_n= \sin^{2n} x$.
Therefore $\displaystyle limsup_{n \rightarrow \infty} | \sin^{2n} x|^{\frac{1}{n} }=limsup_{n \rightarrow \infty} | \sin^2 x|=1$

Hence $\displaystyle R=\frac{1}{1}=1$.

Alternatively, using the ratio test:

$\displaystyle \left| \frac{a_{n+1}}{a_n} \right|=\left|\frac{\sin^{2n+2} x}{ \sin^{2n} x} \right|=\left|\sin^2 (x) \right|< 1 \Leftrightarrow \sin x<1 \Rightarrow x<\frac{\pi}{2}$

This tells me that $\displaystyle R=\frac{\pi}{2}$.
Which one is correct.

For $\displaystyle \sum_{n=0}^{\infty} 2n \sin^{2n-1} x$:

$\displaystyle \left| \frac{a_{n+1}}{a_n} \right|=\left| \frac{2(n+1) \sin^{2n+1} x}{2n \sin^{2n-1} x } \right|= \left| \frac{n+1}{n} \sin^2 x \right| \rightarrow |sin^2 x|<1 \Rightarrow \ x< \frac{\pi}{2}$

The main problem I have now is trying to find the expression for the sums. Can someone show me how to do that?

Both reduce, as you show, to $\displaystyle \left|sin^2(x)\right|< 1$!

How did you then decide that the first $\displaystyle \left|sin^2(x)\right|< 1$ tells you that |x|< 1 and the the second $\displaystyle \left|sin^2(x)\right|< 1$ that |x|< $\displaystyle \pi/2$?
• May 16th 2009, 04:54 PM
shawsend
$\displaystyle \sum_{n=0}^{\infty}2n \sin^{2n-1}(x)\equiv \frac{2}{a}\sum_{n=0}^{\infty} n(a^2)^n$

Now, what is:

$\displaystyle \sum_{n=0}^{\infty} n k^n$
• May 16th 2009, 07:28 PM
putnam120
Well for $\displaystyle |x|<1$ you have that $\displaystyle \sum_{n=0}^\infty x^n=\frac{1}{1-x}$. So doing the taking the derivative you have that $\displaystyle \sum_{n=0}^\infty nx^{n-1}=\frac{1}{(1-x)^2}$. From here you should see how to get to $\displaystyle \sum nx^n$
• May 17th 2009, 04:56 AM
Showcase_22
Quote:

Additionally, the root test states that the series converges absolutely if . So using what you have above, you will arrive at the same result.
That's one of my problems.

Is it true that $\displaystyle limsup_{n \rightarrow \infty} \sin^2 x =1$?

Therefore the radius of convergence R should be $\displaystyle R=\frac{1}{1}=1$.

However, when I use the ratio test (as seen in previous posts) I get $\displaystyle \frac{\pi}{2}$. It seems odd that i'm getting two different answers (Worried)
• May 17th 2009, 08:29 AM
putnam120
Quote:

Originally Posted by Showcase_22
That's one of my problems.

Is it true that $\displaystyle limsup_{n \rightarrow \infty} \sin^2 x =1$?

Therefore the radius of convergence R should be $\displaystyle R=\frac{1}{1}=1$.

However, when I use the ratio test (as seen in previous posts) I get $\displaystyle \frac{\pi}{2}$. It seems odd that i'm getting two different answers (Worried)

No, you are using $\displaystyle \limsup \sqrt[n]{|a_n|}=1$, this is different from $\displaystyle \limsup\sqrt[n]{|a_n|}<1$, that is where you problem is coming from.
• May 17th 2009, 11:34 AM
Showcase_22
Quote:

No, you are using , this is different from , that is where you problem is coming from.
Sorry but I can't see the difference (Worried)

On a lighter note, he's my working on the sums:

$\displaystyle \sum_{n=0}^{\infty}\sin^{2n} x=\frac{1}{1-\sin^2 x}=\frac{1}{\cos^2 x}=\sec^2 x$

$\displaystyle \sum_{n=0}^{\infty} 2n \sin^{2n-1} x=\frac{d}{dx} \left(\frac{1}{\cos^2 x}\right)=\frac{\cos^2 x (0)+2 \cos x \sin x}{\cos^4 x}=\frac{2 \sin x}{\cos^3 x}=2 \tan x \sec^2 x$

Finally, Shawsend i'm quite interested in solving what you've put. Unfortunately, what is the "a" in your $\displaystyle \sum_{n=0}^{\infty}2n \sin^{2n-1}(x)\equiv \frac{2}{a}\sum_{n=0}^{\infty} n(a^2)^n$?
• May 17th 2009, 01:59 PM
shawsend
Quote:

Originally Posted by Showcase_22

Finally, Shawsend i'm quite interested in solving what you've put. Unfortunately, what is the "a" in your $\displaystyle \sum_{n=0}^{\infty}2n \sin^{2n-1}(x)\equiv \frac{2}{a}\sum_{n=0}^{\infty} n(a^2)^n$?

$\displaystyle a=\sin(x)$
• May 17th 2009, 02:01 PM
putnam120
In what you are using there is an equal sign, $\displaystyle =$, however it should be replaces with less than, $\displaystyle <$.

I am sure that we would both agree that the following are different:
$\displaystyle x=1$ and $\displaystyle x<1$.
• May 17th 2009, 03:06 PM
shawsend
I should have written:

$\displaystyle \sum_{n=0}^{\infty}2n \sin^{2n-1}(x)= \frac{2}{a}\sum_{n=0}^{\infty} n(a^2)^n;\quad a=\sin(x)$
• May 18th 2009, 09:06 AM
Showcase_22
$\displaystyle \sum_{n=0}^{\infty} n k^n$

We know that $\displaystyle \sum_{n=0}^\infty x^n=\frac{1}{1-x}$ and $\displaystyle \sum_{n=0}^\infty nx^{n-1}=\frac{1}{(1-x)^2}$.

This implies that $\displaystyle \sum_{n=0}^{\infty} n k^n =\frac{k}{(k-1)^2}$
• May 18th 2009, 10:25 AM
putnam120
Well you might want to check the lower index on $\displaystyle \sum_{n=0}^\infty nx^{n-1}$, but that's the general idea.
• May 18th 2009, 12:39 PM
Showcase_22
why?

What's wrong with it?
• May 18th 2009, 02:39 PM
putnam120
The case where x=0 would cause some problems since how would you define $\displaystyle 0\cdot 0^{-1}$? But I suppose the same can be said about the original sum with its $\displaystyle 0^0$ term.
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