Sum to infinity (complex numbers)

[TsAmE]

Find the sum of the geometric series: \(\displaystyle \sum_{k = 0}^{\infty}\frac{e^{ki\theta}}{2^k}\)

Attempt:

\(\displaystyle S_\infty\)
\(\displaystyle = \frac{a}{1 - r}\)
\(\displaystyle =\frac{1}{1 - \frac{1}{2}e^{i\theta}}\)

Would this be right?

 

[Deleted member 4993]

Find the sum of the geometric series: \(\displaystyle \sum_{k = 0}^{\infty}\frac{e^{ki\theta}}{2^k}\)

Attempt:

\(\displaystyle S_\infty\)
\(\displaystyle = \frac{a}{1 - r}\)
\(\displaystyle =\frac{1}{1 - \frac{1}{2}e^{i\theta}}\) ............. Correct

Would this be right?

 

[TsAmE]

Oh ok. Im am stuck with the question which follows:

By considering the real and imaginary parts of the series in the previous question, find the sums of:

\(\displaystyle \sum_{k=0}^{\infty}\frac{cos(k\theta)}{2^k}\)

and

\(\displaystyle \sum_{k=1}^{\infty}\frac{sin(k\theta)}{2^k}\)

I know the "a" values, but I am not sure how to calculate r, but I know that each expression differs by \(\displaystyle \frac{1}{2}\)and an extra\(\displaystyle \theta\)

 

[Deleted member 4993]

Oh ok. Im am stuck with the question which follows:

By considering the real and imaginary parts of the series in the previous question, find the sums of:

\(\displaystyle \sum_{k=0}^{\infty}\frac{cos(k\theta)}{2^k}\)

and

\(\displaystyle \sum_{k=1}^{\infty}\frac{sin(k\theta)}{2^k}\)

I know the "a" values, but I am not sure how to calculate r, but I know that each expression differs by \(\displaystyle \frac{1}{2}\)and an extra\(\displaystyle \theta\)

\(\displaystyle \ [real] \ \sum_{k=0}^{\infty}\frac{e^{i(k\theta)}}{2^k} \ = \ \sum_{k=0}^{\infty}\frac{cos(k\theta)}{2^k}\)

\(\displaystyle \ [imaginary] \ \sum_{k=0}^{\infty}\frac{e^{i(k\theta)}}{2^k} \ = \ \sum_{k=0}^{\infty}\frac{sin(k\theta)}{2^k}\)

 

[TsAmE]

\(\displaystyle \ [real] \ \sum_{k=0}^{\infty}\frac{e^{i(k\theta)}}{2^k} \ = \ \sum_{k=0}^{\infty}\frac{cos(k\theta)}{2^k}\)

\(\displaystyle \ [imaginary] \ \sum_{k=0}^{\infty}\frac{e^{i(k\theta)}}{2^k} \ = \ \sum_{k=0}^{\infty}\frac{sin(k\theta)}{2^k}\)

I am aware of this and got the following:

\(\displaystyle \ \sum_{k=0}^{\infty}\frac{cos(k\theta)}{2^k} = 1 + \frac{1}{2}cos\theta + \frac{1}{4}cos(2\theta) + \frac{1}{8}cos(3\theta)...\)
\(\displaystyle \ \sum_{k=0}^{\infty}\frac{sin(k\theta)}{2^k} = \frac{1}{2}sin\theta + \frac{1}{4}sin(2\theta) + \frac{1}{8}sin(3\theta)...\)

but by using \(\displaystyle S_\infty = \frac{a}{1 - r}\), I dont know what r is in both cases

 

[galactus]

Since you now know that \(\displaystyle \sum_{k=0}^{\infty}\frac{e^{ki{\theta}}}{2^{k}}=\frac{2}{2-e^{i{\theta}}}\),

you can use the fact that \(\displaystyle e^{ik{\theta}}=cos(k{\theta})+isin(k{\theta})\) and perhaps:

\(\displaystyle \frac{e^{ik{\theta}}}{2}+\frac{e^{-ik{\theta}}}{2}=cos(k{\theta})\)

to find the sums.

 

[TsAmE]

Using what you told me, here is what I have done:

\(\displaystyle \sum_{k = 0}^{\infty} \frac{cos(k\theta)}{2^k}\)
\(\displaystyle =\sum_{k = 0}^{\infty} \frac{e^{i\theta} + e^{-i\theta}}{2} \div 2^k\)
\(\displaystyle =\sum_{k = 0}^{\infty} \frac{e^{i\theta} + e^{-i\theta}}{2^{k + 1}} = \frac{e^{i\theta} + e^{-i\theta}}{2} + \frac{e^{i\theta} + e^{-i\theta}}{4} + \frac{e^{i\theta} + e^{-i\theta}}{8}\)

\(\displaystyle s_\infty\)
\(\displaystyle = \frac{a}{1 - r}\)
\(\displaystyle = \frac{e^{i\theta} + e^{-i\theta}}{2} \div 1 - \frac{1}{2}\)
\(\displaystyle =e^{i\theta} + e^{-i\theta}\)

but this definitely isnt right

 

[lookagain]

Using what you told me, here is what I have done:

\(\displaystyle \sum_{k = 0}^{\infty} \frac{cos(k\theta)}{2^k}\)

\(\displaystyle =\sum_{k = 0}^{\infty} \frac{e^{i\theta} + e^{-i\theta}}{2} \div 2^k\)

\(\displaystyle =\sum_{k = 0}^{\infty} \frac{e^{i\theta} + e^{-i\theta}}{2^{k + 1}} = \frac{e^{i\theta} + e^{-i\theta}}{2} + \frac{e^{i\theta} + e^{-i\theta}}{4} + \frac{e^{i\theta} + e^{-i\theta}}{8}\)

\(\displaystyle s_\infty\)

\(\displaystyle = \frac{a}{1 - r}\)

\(\displaystyle = \frac{e^{i\theta} + e^{-i\theta}}{2} \div (1 - \frac{1}{2})\)

\(\displaystyle =e^{i\theta} + e^{-i\theta}\)

but this definitely isnt right

TsAmE,

in all of your posts, put in vertical lines of space to keep your expressions
and equations (particularly when they're in Latex) from jamming into each
other as shown amended in the above quote box.

Also, the expression \(\displaystyle 1 - \frac{1}{2}\) must be inside grouping symbols
as shown amended in the above quote box.

 

[Deleted member 4993]

Find the sum of the geometric series: \(\displaystyle \sum_{k = 0}^{\infty}\frac{e^{ki\theta}}{2^k}\)

Attempt:

\(\displaystyle S_\infty\)
\(\displaystyle = \frac{a}{1 - r}\)
\(\displaystyle =\frac{1}{1 - \frac{1}{2}e^{i\theta}}\)

Would this be right?

\(\displaystyle \sum_{k = 0}^{\infty}\frac{e^{ki\theta}}{2^k}\ = \ \frac{1}{1 - \frac{1}{2}e^{i\theta}}\)

\(\displaystyle \sum_{k = 0}^{\infty}\frac{cos(k\theta)}{2^k}\ + i \sum_{k = 0}^{\infty}\frac{sin(k\theta)}{2^k} \ = \ \frac{1}{1 - \frac{1}{2}[cos(\theta) + i sin(\theta)]}}\)

Separate the real and imaginary parts of the right-hand-side (RHS) of the equation above.

Equate the real part of the LHS with the real part of the RHS and equate the imaginary part of the LHS with the imaginary part of the RHS.

And you are done.......

 

[galactus]

SK's way is more efficient and easier, but what I was getting at is since:

\(\displaystyle \sum_{k=0}^{\infty}\frac{e^{i{\theta}k}}{2^{k}}=\frac{2}{2-e^{i{\theta}}}\)

then, \(\displaystyle \sum_{k=0}^{\infty}\frac{e^{-i{\theta}k}}{2^{k}}=\frac{2e^{i{\theta}}}{2e^{i{\theta}}-1}\)

Now, from the identity, we get:

\(\displaystyle \sum_{k=0}^{\infty}\frac{cos{\theta}k}{2^{k}}=\frac{1}{2}\left(\frac{2}{2-e^{i{\theta}}}\right)+\frac{1}{2}\left(\frac{2e^{i{\theta}}}{2e^{i{\theta}}-1}\right)=\frac{2(cos{\theta}-2)}{4cos{\theta}-5}\)

The sum with sin can then be found by just a little algebra.

 

[TsAmE]

Thanks. I got the following:

\(\displaystyle \sum_{k=0}^{\infty}\frac{cos(k\theta)}{2^k}\)

\(\displaystyle = Re(\frac{1}{1 - \frac{1}{2}(cos\theta - isin\theta)})\)

\(\displaystyle = \frac{2}{2 - cos\theta}\)

but the correct answer was

\(\displaystyle \frac{4 - 2cos\theta}{5 - 4cos\theta}\)

so I tried to get it into this form:

\(\displaystyle = \frac{4 - 2cos\theta}{4 - 4cos\theta + cos^2\theta}\)

but didnt exactly work out

and for:

\(\displaystyle \sum_{k=1}^{\infty}\frac{sin(k\theta)}{2^k}\)

\(\displaystyle = Im(\frac{1}{1 - \frac{1}{2}(cos\theta - isin\theta)})\)

= \(\displaystyle \frac{2}{2 + sin\theta}\), same reasoning (correct answer: \(\displaystyle \frac{2sin\theta}{5 - 4cos\theta}\))

 

[Deleted member 4993]

Thanks. I got the following:

\(\displaystyle \sum_{k=0}^{\infty}\frac{cos(k\theta)}{2^k}\)

\(\displaystyle = Re(\frac{1}{1 - \frac{1}{2}(cos\theta - isin\theta)})\)

\(\displaystyle = \frac{2}{2 - cos\theta}\)..... Certainly not

RE[1/(a+ib)] = RE[(a-ib)/(a[sup:31dx720s]2[/sup:31dx720s] + b[sup:31dx720s]2[/sup:31dx720s])] = a/(a[sup:31dx720s]2[/sup:31dx720s] + b[sup:31dx720s]2[/sup:31dx720s])

If you do your algebra correctly - you'll get correct answer.

but the correct answer was

\(\displaystyle \frac{4 - 2cos\theta}{5 - 4cos\theta}\)

so I tried to get it into this form:

\(\displaystyle = \frac{4 - 2cos\theta}{4 - 4cos\theta + cos^2\theta}\)

but didnt exactly work out

and for:

\(\displaystyle \sum_{k=1}^{\infty}\frac{sin(k\theta)}{2^k}\)

\(\displaystyle = Im(\frac{1}{1 - \frac{1}{2}(cos\theta - isin\theta)})\)

= \(\displaystyle \frac{2}{2 + sin\theta}\), same reasoning (correct answer: \(\displaystyle \frac{2sin\theta}{5 - 4cos\theta}\))

 

[TsAmE]

RE[1/(a+ib)] = RE[(a-ib)/(a[sup:20hd9i28]2[/sup:20hd9i28] + b[sup:20hd9i28]2[/sup:20hd9i28])] = a/(a[sup:20hd9i28]2[/sup:20hd9i28] + b[sup:20hd9i28]2[/sup:20hd9i28])

I understand this, but in the context of this question for getting the sums of the cos's:

\(\displaystyle Re(\frac{2}{2 - cos\theta - isin\theta})\), has an extra term in the denominator (the 2), and so isnt in the same form as what you said above, as it has 3 terms in the denominator and not 2. I tried mulitplying this fraction by the conjugate of the denominator, but it didnt really work out.

 

[Deleted member 4993]

RE[1/(a+ib)] = RE[(a-ib)/(a[sup:3gfdqjah]2[/sup:3gfdqjah] + b[sup:3gfdqjah]2[/sup:3gfdqjah])] = a/(a[sup:3gfdqjah]2[/sup:3gfdqjah] + b[sup:3gfdqjah]2[/sup:3gfdqjah])

I understand this, but in the context of this question for getting the sums of the cos's:

\(\displaystyle Re(\frac{2}{2 - cos\theta - isin\theta})\), has an extra term in the denominator (the 2), and so isnt in the same form as what you said above, as it has 3 terms in the denominator and not 2. I tried mulitplying this fraction by the conjugate of the denominator, but it didnt really work out.

It is exactly the same as I indicated before:

a = 2 - cos(?) ? a[sup:3gfdqjah]2[/sup:3gfdqjah] = 4 - 4cos(?) + cos[sup:3gfdqjah]2[/sup:3gfdqjah](?)

b = - sin(?) ? b[sup:3gfdqjah]2[/sup:3gfdqjah] = sin[sup:3gfdqjah]2[/sup:3gfdqjah](?)

\(\displaystyle Re\left (\frac{2}{2 - cos\theta - isin\theta}\right )\)

\(\displaystyle = \ Re\left (\frac{2[(2-cos\theta) + isin\theta}{(2 - cos\theta)^2 + (sin\theta)^2}\right )\)

\(\displaystyle = \ \frac{2(2-cos\theta)}{(2 - cos\theta)^2 + (sin\theta)^2}\)

 

[TsAmE]

Oh ok thanks. For the sin part I tried:

\(\displaystyle Im(\frac{4 - 2cos\theta + 2isin\theta}{5 - 4cos\theta})\)

\(\displaystyle =2sin\theta\)

but the answer was \(\displaystyle \frac{2sin\theta}{5 - 4cos\theta}\)

Why is it that the denominator is included, even though it not an imaginary part?

 

[Deleted member 4993]

Oh ok thanks. For the sin part I tried:

\(\displaystyle Im(\frac{4 - 2cos\theta + 2isin\theta}{5 - 4cos\theta})\)

\(\displaystyle =2sin\theta\)

but the answer was \(\displaystyle \frac{2sin\theta}{5 - 4cos\theta}\)

Why is it that the denominator is included, even though it not an imaginary part?

because

\(\displaystyle \frac{ax + by}{c + d} \ = \ \frac{ax}{c+d} \ + \ \frac{by}{c+d}\)

 

[TsAmE]

Thanks a lot for the help, I learnt a lot.