1. ## Series ;)

Ok so i have these 2 problems. The first i can get the answer in the book except for the $ln(5)$ part. The Second i'm not sure how to even start. So if you could help me figure out the first and just get me started on the second that would be great ^.^

Directions: Find the power series representation for the function and determine the radius of convergence.

1. $f(x)=ln(5-x)$

Answer: $ln(5)-\sum^{\infty}_{n=1}\frac{x^n}{n5^n}$

2. $f(x)=\frac{x^3}{(x-2)^2}$

Answer: $\sum^{\infty}_{n=3}\frac{n-2}{2^{n-1}}x^n$

Thank you

2. Originally Posted by mortalapeman
Ok so i have these 2 problems. The first i can get the answer in the book except for the $ln(5)$ part.
1. $f(x)=ln(5-x)$

Answer: $ln(5)-\sum^{\infty}_{n=1}\frac{x^n}{n5^n}$

Well, to get the power series representation, you first have to take a derivative. Then, at the end, you have to take an integral, and when you take an indefinite integral, you have to add "+C" to it to represent any constant that the derivative could have knocked out. Then to solve for "C," you set x equal to 0, so you get..

ln(5-0) = 0 + C
ln(5) = C

I imagine that's what they're getting it from

3. 1. Note that: $\ln (5-x) = - \int \frac{1}{5-x} \ dx$

So you if you can find the power series for: $f(x) = \frac{1}{5-x} = \frac{1}{5} \cdot \frac{1}{1 - \frac{x}{5}}$

then you can integrate each term to get the power series for $\ln (5-x)$.

The $\ln 5$ comes from the fact that you get the constant of integration after integrating and you would need to solve for it. This can be done by simply plugging in $x = 0$ to your series representation of $\ln (5-x)$

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2. Find the power series of $\frac{1}{(x-1)^2}$ first and multiply your result by $x^3$.

Note that: $\frac{d}{dx} \frac{1}{1-x} = \frac{1}{(x-1)^2}$

The power series of $\frac{1}{1-x}$ is a standard one. So if you differentiate each term, you'll get the power series of $\frac{1}{(x-1)^2}$.