power series of $left(frac{sin z}zright)^2$?












2












$begingroup$


So I need to find the power series of $left(frac{sin z}zright)^2$ around $0$, and find its radius of convergence.

Also the function is defined to be $1$ in $z=0$.
My guess about the radius is infinity, but I'm stuck in the power series.
This is what I've done so far:



$$left(frac{sin z}zright)^2 = left(frac{z-frac{z^3}{3!}+frac{z^5}{5!}-cdotsfrac{z^{2n+1}}{(2n+1)!}+cdots}zright)^2$$
$$=left(1-frac{z^2}{3!}+frac{z^4}{5!}-cdotsfrac{z^{2n}}{(2n+1)!}+cdotsright)^2$$
$$=left(sum_{n=0}^{infty}frac{(-1)^nz^{2n}}{(2n+1)!}right)^2$$



could really use some help. Thanks very much!






calculus sequences-and-series convergence







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  • 7




    $begingroup$
    Use $sin^2z=frac12(1-cos 2z)$.
    $endgroup$
    – Lord Shark the Unknown
    Dec 22 '18 at 17:55
















2












$begingroup$


So I need to find the power series of $left(frac{sin z}zright)^2$ around $0$, and find its radius of convergence.

Also the function is defined to be $1$ in $z=0$.
My guess about the radius is infinity, but I'm stuck in the power series.
This is what I've done so far:



$$left(frac{sin z}zright)^2 = left(frac{z-frac{z^3}{3!}+frac{z^5}{5!}-cdotsfrac{z^{2n+1}}{(2n+1)!}+cdots}zright)^2$$
$$=left(1-frac{z^2}{3!}+frac{z^4}{5!}-cdotsfrac{z^{2n}}{(2n+1)!}+cdotsright)^2$$
$$=left(sum_{n=0}^{infty}frac{(-1)^nz^{2n}}{(2n+1)!}right)^2$$



could really use some help. Thanks very much!






calculus sequences-and-series convergence







share|cite|improve this question











$endgroup$








  • 7




    $begingroup$
    Use $sin^2z=frac12(1-cos 2z)$.
    $endgroup$
    – Lord Shark the Unknown
    Dec 22 '18 at 17:55














2












2








2





$begingroup$


So I need to find the power series of $left(frac{sin z}zright)^2$ around $0$, and find its radius of convergence.

Also the function is defined to be $1$ in $z=0$.
My guess about the radius is infinity, but I'm stuck in the power series.
This is what I've done so far:



$$left(frac{sin z}zright)^2 = left(frac{z-frac{z^3}{3!}+frac{z^5}{5!}-cdotsfrac{z^{2n+1}}{(2n+1)!}+cdots}zright)^2$$
$$=left(1-frac{z^2}{3!}+frac{z^4}{5!}-cdotsfrac{z^{2n}}{(2n+1)!}+cdotsright)^2$$
$$=left(sum_{n=0}^{infty}frac{(-1)^nz^{2n}}{(2n+1)!}right)^2$$



could really use some help. Thanks very much!






calculus sequences-and-series convergence







share|cite|improve this question











$endgroup$




So I need to find the power series of $left(frac{sin z}zright)^2$ around $0$, and find its radius of convergence.

Also the function is defined to be $1$ in $z=0$.
My guess about the radius is infinity, but I'm stuck in the power series.
This is what I've done so far:



$$left(frac{sin z}zright)^2 = left(frac{z-frac{z^3}{3!}+frac{z^5}{5!}-cdotsfrac{z^{2n+1}}{(2n+1)!}+cdots}zright)^2$$
$$=left(1-frac{z^2}{3!}+frac{z^4}{5!}-cdotsfrac{z^{2n}}{(2n+1)!}+cdotsright)^2$$
$$=left(sum_{n=0}^{infty}frac{(-1)^nz^{2n}}{(2n+1)!}right)^2$$



could really use some help. Thanks very much!






calculus sequences-and-series convergence




calculus sequences-and-series convergence






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share|cite|improve this question













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share|cite|improve this question








edited Dec 22 '18 at 22:23









Somos

14.4k11336




14.4k11336










asked Dec 22 '18 at 17:54









RoeeRoee

374




374








  • 7




    $begingroup$
    Use $sin^2z=frac12(1-cos 2z)$.
    $endgroup$
    – Lord Shark the Unknown
    Dec 22 '18 at 17:55














  • 7




    $begingroup$
    Use $sin^2z=frac12(1-cos 2z)$.
    $endgroup$
    – Lord Shark the Unknown
    Dec 22 '18 at 17:55








7




7




$begingroup$
Use $sin^2z=frac12(1-cos 2z)$.
$endgroup$
– Lord Shark the Unknown
Dec 22 '18 at 17:55




$begingroup$
Use $sin^2z=frac12(1-cos 2z)$.
$endgroup$
– Lord Shark the Unknown
Dec 22 '18 at 17:55










1 Answer
1






active

oldest

votes


















3












$begingroup$

Using Lord Shark the Unknown's hint we can rewrite the problem in the following way



$$left(frac{sin z}zright)^2=frac1{2z^2}(1-cos 2z)$$



Utilizing the well known Taylor Series of the cosine function this further becomes



$$begin{align}
frac1{2z^2}(1-cos 2z)&=frac1{2z^2}left(1-sum_{n=0}^{infty}(-1)^nfrac{(2z)^{2n}}{(2n)!}right)\
&=frac1{2z^2}-2sum_{n=0}^{infty}(-1)^nfrac{(2z)^{2n-2}}{(2n)!}\
&=-2sum_{n=1}^{infty}(-1)^nfrac{(2z)^{2n-2}}{(2n)!}
end{align}$$




$$left(frac{sin z}zright)^2=2sum_{n=0}^{infty}(-1)^nfrac{4^n z^{2n}}{(2n+2)!}$$




Regarding the radius of convergence one could apply the ratio test to verfity that it is actually infinity.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    thank you very much for your answer! it helped me a lot. notice a small mistake in line 2, where you should multiply the series by (1/2) and not 2, not that I mind of course.
    $endgroup$
    – Roee
    Dec 22 '18 at 18:44










  • $begingroup$
    @Roee No problem. Actually it is right like this. Note that $$frac1{2z^2}sum_{n=0}^{infty}(-1)^nfrac{(2z)^{2n}}{(2n)!}=frac{color{red}{2}}{(2z)^2}sum_{n=0}^{infty}(-1)^nfrac{(2z)^{2n}}{(2n)!}=color{red}{2}sum_{n=0}^{infty}(-1)^nfrac{(2z)^{2n-2}}{(2n)!}$$ I the first place I have made a mistake within this step leading to the power series of $1/4(sin z/z)^2$ instead of the wanted for $(sin z/z)^2$.
    $endgroup$
    – mrtaurho
    Dec 22 '18 at 18:52













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1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









3












$begingroup$

Using Lord Shark the Unknown's hint we can rewrite the problem in the following way



$$left(frac{sin z}zright)^2=frac1{2z^2}(1-cos 2z)$$



Utilizing the well known Taylor Series of the cosine function this further becomes



$$begin{align}
frac1{2z^2}(1-cos 2z)&=frac1{2z^2}left(1-sum_{n=0}^{infty}(-1)^nfrac{(2z)^{2n}}{(2n)!}right)\
&=frac1{2z^2}-2sum_{n=0}^{infty}(-1)^nfrac{(2z)^{2n-2}}{(2n)!}\
&=-2sum_{n=1}^{infty}(-1)^nfrac{(2z)^{2n-2}}{(2n)!}
end{align}$$




$$left(frac{sin z}zright)^2=2sum_{n=0}^{infty}(-1)^nfrac{4^n z^{2n}}{(2n+2)!}$$




Regarding the radius of convergence one could apply the ratio test to verfity that it is actually infinity.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    thank you very much for your answer! it helped me a lot. notice a small mistake in line 2, where you should multiply the series by (1/2) and not 2, not that I mind of course.
    $endgroup$
    – Roee
    Dec 22 '18 at 18:44










  • $begingroup$
    @Roee No problem. Actually it is right like this. Note that $$frac1{2z^2}sum_{n=0}^{infty}(-1)^nfrac{(2z)^{2n}}{(2n)!}=frac{color{red}{2}}{(2z)^2}sum_{n=0}^{infty}(-1)^nfrac{(2z)^{2n}}{(2n)!}=color{red}{2}sum_{n=0}^{infty}(-1)^nfrac{(2z)^{2n-2}}{(2n)!}$$ I the first place I have made a mistake within this step leading to the power series of $1/4(sin z/z)^2$ instead of the wanted for $(sin z/z)^2$.
    $endgroup$
    – mrtaurho
    Dec 22 '18 at 18:52


















3












$begingroup$

Using Lord Shark the Unknown's hint we can rewrite the problem in the following way



$$left(frac{sin z}zright)^2=frac1{2z^2}(1-cos 2z)$$



Utilizing the well known Taylor Series of the cosine function this further becomes



$$begin{align}
frac1{2z^2}(1-cos 2z)&=frac1{2z^2}left(1-sum_{n=0}^{infty}(-1)^nfrac{(2z)^{2n}}{(2n)!}right)\
&=frac1{2z^2}-2sum_{n=0}^{infty}(-1)^nfrac{(2z)^{2n-2}}{(2n)!}\
&=-2sum_{n=1}^{infty}(-1)^nfrac{(2z)^{2n-2}}{(2n)!}
end{align}$$




$$left(frac{sin z}zright)^2=2sum_{n=0}^{infty}(-1)^nfrac{4^n z^{2n}}{(2n+2)!}$$




Regarding the radius of convergence one could apply the ratio test to verfity that it is actually infinity.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    thank you very much for your answer! it helped me a lot. notice a small mistake in line 2, where you should multiply the series by (1/2) and not 2, not that I mind of course.
    $endgroup$
    – Roee
    Dec 22 '18 at 18:44










  • $begingroup$
    @Roee No problem. Actually it is right like this. Note that $$frac1{2z^2}sum_{n=0}^{infty}(-1)^nfrac{(2z)^{2n}}{(2n)!}=frac{color{red}{2}}{(2z)^2}sum_{n=0}^{infty}(-1)^nfrac{(2z)^{2n}}{(2n)!}=color{red}{2}sum_{n=0}^{infty}(-1)^nfrac{(2z)^{2n-2}}{(2n)!}$$ I the first place I have made a mistake within this step leading to the power series of $1/4(sin z/z)^2$ instead of the wanted for $(sin z/z)^2$.
    $endgroup$
    – mrtaurho
    Dec 22 '18 at 18:52
















3












3








3





$begingroup$

Using Lord Shark the Unknown's hint we can rewrite the problem in the following way



$$left(frac{sin z}zright)^2=frac1{2z^2}(1-cos 2z)$$



Utilizing the well known Taylor Series of the cosine function this further becomes



$$begin{align}
frac1{2z^2}(1-cos 2z)&=frac1{2z^2}left(1-sum_{n=0}^{infty}(-1)^nfrac{(2z)^{2n}}{(2n)!}right)\
&=frac1{2z^2}-2sum_{n=0}^{infty}(-1)^nfrac{(2z)^{2n-2}}{(2n)!}\
&=-2sum_{n=1}^{infty}(-1)^nfrac{(2z)^{2n-2}}{(2n)!}
end{align}$$




$$left(frac{sin z}zright)^2=2sum_{n=0}^{infty}(-1)^nfrac{4^n z^{2n}}{(2n+2)!}$$




Regarding the radius of convergence one could apply the ratio test to verfity that it is actually infinity.






share|cite|improve this answer











$endgroup$



Using Lord Shark the Unknown's hint we can rewrite the problem in the following way



$$left(frac{sin z}zright)^2=frac1{2z^2}(1-cos 2z)$$



Utilizing the well known Taylor Series of the cosine function this further becomes



$$begin{align}
frac1{2z^2}(1-cos 2z)&=frac1{2z^2}left(1-sum_{n=0}^{infty}(-1)^nfrac{(2z)^{2n}}{(2n)!}right)\
&=frac1{2z^2}-2sum_{n=0}^{infty}(-1)^nfrac{(2z)^{2n-2}}{(2n)!}\
&=-2sum_{n=1}^{infty}(-1)^nfrac{(2z)^{2n-2}}{(2n)!}
end{align}$$




$$left(frac{sin z}zright)^2=2sum_{n=0}^{infty}(-1)^nfrac{4^n z^{2n}}{(2n+2)!}$$




Regarding the radius of convergence one could apply the ratio test to verfity that it is actually infinity.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Dec 22 '18 at 18:21

























answered Dec 22 '18 at 18:15









mrtaurhomrtaurho

5,74551540




5,74551540












  • $begingroup$
    thank you very much for your answer! it helped me a lot. notice a small mistake in line 2, where you should multiply the series by (1/2) and not 2, not that I mind of course.
    $endgroup$
    – Roee
    Dec 22 '18 at 18:44










  • $begingroup$
    @Roee No problem. Actually it is right like this. Note that $$frac1{2z^2}sum_{n=0}^{infty}(-1)^nfrac{(2z)^{2n}}{(2n)!}=frac{color{red}{2}}{(2z)^2}sum_{n=0}^{infty}(-1)^nfrac{(2z)^{2n}}{(2n)!}=color{red}{2}sum_{n=0}^{infty}(-1)^nfrac{(2z)^{2n-2}}{(2n)!}$$ I the first place I have made a mistake within this step leading to the power series of $1/4(sin z/z)^2$ instead of the wanted for $(sin z/z)^2$.
    $endgroup$
    – mrtaurho
    Dec 22 '18 at 18:52




















  • $begingroup$
    thank you very much for your answer! it helped me a lot. notice a small mistake in line 2, where you should multiply the series by (1/2) and not 2, not that I mind of course.
    $endgroup$
    – Roee
    Dec 22 '18 at 18:44










  • $begingroup$
    @Roee No problem. Actually it is right like this. Note that $$frac1{2z^2}sum_{n=0}^{infty}(-1)^nfrac{(2z)^{2n}}{(2n)!}=frac{color{red}{2}}{(2z)^2}sum_{n=0}^{infty}(-1)^nfrac{(2z)^{2n}}{(2n)!}=color{red}{2}sum_{n=0}^{infty}(-1)^nfrac{(2z)^{2n-2}}{(2n)!}$$ I the first place I have made a mistake within this step leading to the power series of $1/4(sin z/z)^2$ instead of the wanted for $(sin z/z)^2$.
    $endgroup$
    – mrtaurho
    Dec 22 '18 at 18:52


















$begingroup$
thank you very much for your answer! it helped me a lot. notice a small mistake in line 2, where you should multiply the series by (1/2) and not 2, not that I mind of course.
$endgroup$
– Roee
Dec 22 '18 at 18:44




$begingroup$
thank you very much for your answer! it helped me a lot. notice a small mistake in line 2, where you should multiply the series by (1/2) and not 2, not that I mind of course.
$endgroup$
– Roee
Dec 22 '18 at 18:44












$begingroup$
@Roee No problem. Actually it is right like this. Note that $$frac1{2z^2}sum_{n=0}^{infty}(-1)^nfrac{(2z)^{2n}}{(2n)!}=frac{color{red}{2}}{(2z)^2}sum_{n=0}^{infty}(-1)^nfrac{(2z)^{2n}}{(2n)!}=color{red}{2}sum_{n=0}^{infty}(-1)^nfrac{(2z)^{2n-2}}{(2n)!}$$ I the first place I have made a mistake within this step leading to the power series of $1/4(sin z/z)^2$ instead of the wanted for $(sin z/z)^2$.
$endgroup$
– mrtaurho
Dec 22 '18 at 18:52






$begingroup$
@Roee No problem. Actually it is right like this. Note that $$frac1{2z^2}sum_{n=0}^{infty}(-1)^nfrac{(2z)^{2n}}{(2n)!}=frac{color{red}{2}}{(2z)^2}sum_{n=0}^{infty}(-1)^nfrac{(2z)^{2n}}{(2n)!}=color{red}{2}sum_{n=0}^{infty}(-1)^nfrac{(2z)^{2n-2}}{(2n)!}$$ I the first place I have made a mistake within this step leading to the power series of $1/4(sin z/z)^2$ instead of the wanted for $(sin z/z)^2$.
$endgroup$
– mrtaurho
Dec 22 '18 at 18:52




















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