1. ## Math133

Hey would anyone be able to help me with these 2 q's:

For this one I don't know how to do the series with 1/n

Capture.JPG

For this one I just dont understand the binomial form in general. Oops find the tayler series for it

Capture.JPG

Any help would be greatly appreciated!

2. ## Re: Math133

Originally Posted by BenHowe
Hey would anyone be able to help me with these 2 q's:

For this one I don't know how to do the series with 1/n

Capture.JPG

For this one I just dont understand the binomial form in general. Oops find the tayler series for it

Capture.JPG

Any help would be greatly appreciated!
$\noindent For the first one, split up the series and use \sum_{n=1}^{\infty}z^n = \frac{1}{1-z}-1 and \sum_{n=1}^{\infty}\frac{1}{n}z^{n} = -\log{(1-z)}, noting the intervals of convergence.$

3. ## Re: Math133

Why is it -log(1-z)?

4. ## Re: Math133

Originally Posted by BenHowe
Why is it -log(1-z)?
It's one of the standard Taylor series. You can obtain it by integrating the geometric series.

5. ## Re: Math133

$x^{-1/2}=\frac{1}{3}\left(\frac{x-9}{9}+1\right)^{-1/2}=\frac{1}{3}\sum_{k=0}^\infty 3^{-2k}\binom{-1/2}{k}(x-9)^k.$

You can also express the binomial coefficient in terms of factorials and powers of 2 if you like, but I leave that to you.

6. ## Re: Math133

In case you haven't seen fractional binomial coefficients before:

$\binom{\alpha}{k}:=\frac{\alpha(\alpha-1)\ldots(\alpha-k+1)}{k!}.$

7. ## Re: Math133

Originally Posted by InteGrand
It's one of the standard Taylor series. You can obtain it by integrating the geometric series.
So you can just the sum be J or whatever.

$J=\sum_{n=0}^{\infty} \frac{x^{n}}{n}\\ {J}^{\prime}=\sum_{n=1}^{\infty} x^{n-1}\\ J^{\prime}\\ \text{This is effectively the normal geom series now which you show by subbing another variable. So}\\ J^{\prime}=\frac{1}{1-x}\\ \int J^{\prime} = \int \frac{1}{1-x} dx\\ J=-ln|1-x| \text{The constant term goes to 0}$

One question tho, do you get the right answer if multiply by x in the final step rather than shifting the sum so to speak i.e. integrand becomes 1/x(1-x)

8. ## Re: Math133

Originally Posted by seanieg89
In case you haven't seen fractional binomial coefficients before:

$\binom{\alpha}{k}:=\frac{\alpha(\alpha-1)\ldots(\alpha-k+1)}{k!}.$
Yeah thanks heaps. I've done a lot of binomial stuff this semester in another unit, it just wasnt making sense. But just to clarify you can either do the general computation for tayler/macluaran series or you can fiddle with a function until it reduces a well-known result?

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