How do you prove this complex summation?

[Kulla_9289]

Hi,
I have this complex summation problem. I tried my best to get to the answer but halfway through I am stuck. Can anyone please help me out?

Thanks

 

[Dr.Peterson]

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First, what do you mean by this?

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Did you mean to say the imaginary part, perhaps?

Second, have noticed the relationship between your denominator and theirs? That suggests that you are going in the right direction.

 

[Kulla_9289]

Did you mean to say the imaginary part, perhaps?

Yes.

Second, have noticed the relationship between your denominator and theirs? That suggests that you are going in the right direction.

Maybe. But after this I am lost.

 

[Dr.Peterson]

Yes.

Do you have a notation for indicating that? Writing i times something is not the same.

Maybe. But after this I am lost.

Give it a try. What is the imaginary part of your last line?

 

[Kulla_9289]

Ok I tried. I got up to this. Not sure how they got 1025

 

[Dr.Peterson]

Check your work in the numerator; I suggest showing more steps to make errors more visible.

 

[logistic_guy]

Here is my approach of how to solve this problem.

\(\displaystyle \sum_{n=1}^{N}2^{-n}z^n = \sum_{n=1}^{N}\left(\frac{z}{2}\right)^n = \frac{\frac{z}{2}\left(1 - \left(\frac{z}{2}\right)^N\right)}{1 - \frac{z}{2}}\)

This result comes from the fact that the sum is a geometric series.

Now we have this sum:

\(\displaystyle \sum_{n=1}^{10}2^{-n}\sin\left(\frac{n\pi}{10}\right)\)

which is the imaginary part of:

\(\displaystyle \sum_{n=1}^{10}2^{-n}e^{in\pi/10}\)

Since \(\displaystyle e^{in\pi/10} = \cos\left(\frac{n\pi}{10}\right) + i\sin\left(\frac{n\pi}{10}\right)\),

we can solve the sum that contains the exponential and then extract the imaginary part of it.

\(\displaystyle \sum_{n=1}^{10}2^{-n}e^{in\pi/10} = \sum_{n=1}^{10}\left(\frac{e^{i\pi/10}}{2}\right)^n = \frac{\frac{e^{i\pi/10}}{2}\left(1 - \left(\frac{e^{i\pi/10}}{2}\right)^{10}\right)}{1 - \frac{e^{i\pi/10}}{2}}\)


\(\displaystyle = \frac{\frac{e^{i\pi/10}}{2}\left(1 - \left(\frac{e^{i\pi}}{2^{10}}\right)\right)}{1 - \frac{e^{i\pi/10}}{2}} = \frac{\frac{e^{i\pi/10}}{2}\left(\frac{2^{10}}{2^{10}} - \left(\frac{e^{i\pi}}{2^{10}}\right)\right)}{\frac{2}{2} - \frac{e^{i\pi/10}}{2}}= \frac{e^{i\pi/10}\left(2^{10} - e^{i\pi}\right)}{2^{10}(2 - e^{i\pi/10})}\)

We know that \(\displaystyle e^{i\pi} = -1\), then

\(\displaystyle \sum_{n=1}^{10}2^{-n}e^{in\pi/10} = \frac{e^{i\pi/10}\left(2^{10} + 1\right)}{2^{10}(2 - e^{i\pi/10})} = \frac{(2^{10} + 1)(\cos \pi/10 + i\sin \pi/10)}{2^{10}(2 - \cos \pi/10 - i\sin \pi/10)}\)

Multiply the numerator and the denominator by \(\displaystyle (2 - \cos \pi/10 + i\sin \pi/10)\)

\(\displaystyle \color{red} \bold{numerator:}\)\(\displaystyle (\cos \pi/10 + i\sin \pi/10)(2 - \cos \pi/10 + i\sin \pi/10) = 2\cos\left(\frac{\pi}{10}\right) - 1 + 2i\sin\left(\frac{\pi}{10}\right)\)
\(\displaystyle \color{blue} \bold{denominator:}\)\(\displaystyle (2 - \cos \pi/10 - i\sin \pi/10)(2 - \cos \pi/10 + i\sin \pi/10) = 5 - 4\cos\left(\frac{10}{\pi}\right)\)

This gives:

\(\displaystyle \sum_{n=1}^{10}2^{-n}e^{in\pi/10} = \frac{(2^{10} + 1)\bigg[2\cos\left(\frac{\pi}{10}\right) - 1 + 2i\sin\left(\frac{\pi}{10}\right)\bigg]}{2^{10}\bigg[5 - 4\cos\left(\frac{10}{\pi}\right)\bigg]}\)

Take the imaginary part of this result.

\(\displaystyle \sum_{n=1}^{10}2^{-n}\sin\left(\frac{n\pi}{10}\right) = \frac{(2^{10} + 1)\bigg[2\sin\left(\frac{\pi}{10}\right)\bigg]}{2^{10}\bigg[5 - 4\cos\left(\frac{10}{\pi}\right)\bigg]} = \frac{1025\bigg[2\sin\left(\frac{\pi}{10}\right)\bigg]}{1024\bigg[5 - 4\cos\left(\frac{10}{\pi}\right)\bigg]} = \frac{1025\bigg[\sin\left(\frac{\pi}{10}\right)\bigg]}{512\bigg[5 - 4\cos\left(\frac{10}{\pi}\right)\bigg]}\)

\(\displaystyle \textcolor{green}{\textbf{Finally.}}\)

\(\displaystyle \sum_{n=1}^{10}2^{-n}\sin\left(\frac{n\pi}{10}\right) = \frac{1025\sin\left(\frac{\pi}{10}\right)}{2560 - 2048\cos\left(\frac{10}{\pi}\right)}\)

 

[logistic_guy]

\(\displaystyle \color{blue} \bold{denominator:}\)\(\displaystyle (2 - \cos \pi/10 - i\sin \pi/10)(2 - \cos \pi/10 + i\sin \pi/10) = 5 - 4\cos\left(\frac{10}{\pi}\right)\)

Not here that I unintentionally wrote \(\displaystyle \cos\left(\frac{10}{\pi}\right)\). It should be \(\displaystyle \cos\left(\frac{\pi}{10}\right)\) instead.