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Thread: power series

  1. #1
    Super Member Showcase_22's Avatar
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    power series

    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?
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  2. #2
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    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.
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  3. #3
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    Quote Originally Posted by Showcase_22 View Post
    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|< $\displaystyle \pi/2$?
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  4. #4
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    $\displaystyle \sum_{n=0}^{\infty}2n \sin^{2n-1}(x)\equiv \frac{2}{a}\sum_{n=0}^{\infty} n(a^2)^n$

    Now, what is:

    $\displaystyle \sum_{n=0}^{\infty} n k^n $
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    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 $\displaystyle \sum nx^n$
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  6. #6
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    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
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  7. #7
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    Quote Originally Posted by Showcase_22 View Post
    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
    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.
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  8. #8
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    Smile

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

    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
    $?
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  9. #9
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    Quote Originally Posted by Showcase_22 View Post

    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
    $?
    $\displaystyle a=\sin(x)$
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  10. #10
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    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 $\displaystyle x<1$.
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  11. #11
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    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)
    $
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  12. #12
    Super Member Showcase_22's Avatar
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    $\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}$
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  13. #13
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    Well you might want to check the lower index on $\displaystyle \sum_{n=0}^\infty nx^{n-1}$, but that's the general idea.
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  14. #14
    Super Member Showcase_22's Avatar
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    why?

    What's wrong with it?
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  15. #15
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    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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