HELP: A complex valued power series

[Hummingbird25]

Hi All,

I have this here power series which is complex valued

$$\frac{2n+1}{2^n} i^n = \frac{4}{25} + \frac{22}{25}i$$

My task is to prove that this power series has the above mentioned sum.

To do this I separate the sum into two sums

$$S_0 + S_1 = (i/2)^n + 2* (i/2)*n*(i/2)^{(n-1)}$$

the first is easy since can use

$$S_0 = \frac{1}{1-(i/2)} = 4/5+2/5i$$

The second is quite hard to compute. I'm to told that I need to differentiate, by using the sum

$$\sum _{n=0} ^{\infty} x^n = \frac{i}{2}$$

then by differentiating the sum I get

$$f'(x) = \sum_{n=0} ^{\infty} n x^{n-1}$$

In my textbook it says that x^n converge to 1/(1-x)

Then since |i/2| < 1, I differentiate both sides of the equation.

$$\frac{1}{(1-(i/2))^2} = \sum_{n=0} ^{\infty} n (i/2)^{n-1}$$

I multiply by i/2 on both sides of the equation to get.

$$\frac{i/2 * 1}{(1-(i/2))^2} =\sum_{n=0} ^{\infty}(i/2) n (i/2)^{n-1}$$

then since |i/2| < 1, then I multiply by a factor of two on both sides of the equation

$$2\frac{i/2 * 1}{(1-(i/2))^2} = \sum_{n=0} ^{\infty}2 (i/2) n (i/2)^{n-1}$$

$$-16/25 + 12/25i = \sum_{n=0} ^{\infty}2 (i/2) n (i/2)^{n-1}$$

Finally $$\frac{4}{5}+ \frac{2}{5i}+ (\frac{-16}{25} + \frac{12}{25i}) = \frac{4}{25} + \frac{22}{25}i$$

I have the feeling that I'm doing something wrong?

Sincerely

Hummingbird25

 

[Curious3141]

Hi All,

I have this here power series which is complex valued

$$\frac{2n+1}{2^n} i^n = \frac{4}{25} + \frac{22}{25}i$$

My task is to prove that this power series has the above mentioned sum.

To do this I separate the sum into two sums

$$S_0 + S_1 = (i/2)^n + 2* (i/2)*n*(i/2)^{(n-1)}$$

the first is easy since can use

$$S_0 = \frac{1}{1-(i/2)} = 4/5+2/5i$$

The second is quite hard to compute. I'm to told that I need to differentiate, by using the sum

$$\sum _{n=0} ^{\infty} x^n = \frac{i}{2}$$

then by differentiating the sum I get

$$f'(x) = \sum_{n=0} ^{\infty} n x^{n-1}$$

In my textbook it says that x^n converge to 1/(1-x)

Then since |i/2| < 1, I differentiate both sides of the equation.

$$\frac{1}{(1-(i/2))^2} = \sum_{n=0} ^{\infty} n (i/2)^{n-1}$$

I multiply by i/2 on both sides of the equation to get.

$$\frac{i/2 * 1}{(1-(i/2))^2} =\sum_{n=0} ^{\infty}(i/2) n (i/2)^{n-1}$$

then since |i/2| < 1, then I multiply by a factor of two on both sides of the equation

$$2\frac{i/2 * 1}{(1-(i/2))^2} = \sum_{n=0} ^{\infty}2 (i/2) n (i/2)^{n-1}$$

$$-16/25 + 12/25i = \sum_{n=0} ^{\infty}2 (i/2) n (i/2)^{n-1}$$

Finally $$\frac{4}{5}+ \frac{2}{5i}+ (\frac{-16}{25} + \frac{12}{25i}) = \frac{4}{25} + \frac{22}{25}i$$

I have the feeling that I'm doing something wrong?

Sincerely

Hummingbird25

Sorry, I don't really have the time to really scrutinise the working you gave. There does seem to be a little typo in the LaTex, but it seems generally OK.

There's a much easier way than differentiation to work out sums of the form

$$S = \sum{nx^{-n}}$$ or variants on the theme.

Just multiply by x to get

$$Sx = \sum{nx^{1-n}}$$

and compare like powers of Sx to S term by term.

You will find that $$Sx - S = S(1-x)$$ will amount to a very easy geometric progression and you can just divide that by (1-x) to find S.

Much easier than differentiation, IMHO.