Coefficients of a Taylor Series

[lxman]

Homework Statement

The function f(x)=ln(10-x) is represented as a power series:
\sum^{\infty}_{n=0}a_{n}x^{n}
Find the first few coefficients in the power series. Hint: First find the power series for the derivative of .

The Attempt at a Solution

Okay, start seems fairly straightforward:

f'(x)=\frac{1}{10-x}

I factor out \frac{1}{10} to arrive at:

f'(x)=\frac{1}{10}*\frac{1}{1-\frac{x}{10}}

I then arrive at the geometric series:

\sum^{\infty}_{n=0}\frac{1}{10}*\frac{x^{n}}{10^{n}}

Things begin to get a bit fuzzy for me from here. Next, I need to integrate WRT x to arrive at a solution for the original f(x). I believe this would result in:

\sum^{\infty}_{n=0}\frac{1}{10}*\frac{x^{2n}}{2(10^{n+1})}

Am I correct to this point, and where do I go from here?

 

[micromass]

Homework Statement

The function f(x)=ln(10-x) is represented as a power series:
\sum^{\infty}_{n=0}a_{n}x^{n}
Find the first few coefficients in the power series. Hint: First find the power series for the derivative of .

The Attempt at a Solution

Okay, start seems fairly straightforward:

f'(x)=\frac{1}{10-x}

I factor out \frac{1}{10} to arrive at:

f'(x)=\frac{1}{10}*\frac{1}{1-\frac{x}{10}}

I then arrive at the geometric series:

\sum^{\infty}_{n=0}\frac{1}{10}*\frac{x^{n}}{10^{n}}

Good so far, but

Things begin to get a bit fuzzy for me from here. Next, I need to integrate WRT x to arrive at a solution for the original f(x). I believe this would result in:

\sum^{\infty}_{n=0}\frac{1}{10}*\frac{x^{2n}}{2(10^{n+1})}

Am I correct to this point, and where do I go from here?

I'm quite lost at how you found this integral. The integral of

\sum_{n=0}^{+\infty}{a_nx^n}

is

C+\sum_{n=0}^{+\infty}{\frac{a_n}{n+1}x^{n+1}}

So I don't quite see how you got that 2n in the exponent...

 

[lxman]

Thank you for the reply.

I'm quite lost at how you found this integral.

I am sure your misgivings are justified. I am viewing this as an integration problem. Therefore I conclude that I need to obtain:

\int\frac{1}{10}*\frac{x^{n}}{10^{n}}dx

Using standard integration techniques I perform the following:

\frac{1}{10}\int\frac{x^{n}}{10^{n}}dx
=\frac{1}{10}\ \frac{x^{2n}}{2(10^{n+1})}

Have I erred at this step?

 

[lxman]

Okay, writing that out and looking back over it, I see an error.

x^{2}\ x^{n} should be x^{2+n}.

Correcting that mistake, I now arrive at:

\sum^{+\infty}_{n=0}\frac{1}{10}\ \frac{x^{2+n}}{2(10^{n+1})}

How about now?

 

[micromass]

No, that's also not good I'm confused at how you obtained that 2...

What is \int{x^ndx}??

 

[eczeno]

Thank you for the reply.
I am sure your misgivings are justified. I am viewing this as an integration problem. Therefore I conclude that I need to obtain:

\int\frac{1}{10}*\frac{x^{n}}{10^{n}}dx

Using standard integration techniques I perform the following:

\frac{1}{10}\int\frac{x^{n}}{10^{n}}dx
=\frac{1}{10}\ \frac{x^{2n}}{2(10^{n+1})}

Have I erred at this step?

this last step is incorrect. first, i would factor out the \frac{1}{10^{n}} so we get:

\frac{1}{10^{n+1}}\int{x^{n}}dx

which integrates to:

\frac{1}{10^{n+1}}\frac{x^{n+1}}{n+1}+C

 

[lxman]

OK, now that I see it, I understand. I was taking n to be a constant and performing:

\int\ x\ dx

which gives me:

\frac{x^{2}}{2}

and then trying to lump the n's back in. My mistake there.

So, taking:

\frac{1}{10^{n+1}}\ \frac{x^{n+1}}{n+1}+C

my first few terms would then be:

\frac{x}{10}+C

\frac{x^{2}}{200}+C

\frac{x^{3}}{3000}+C

\frac{x^{4}}{40000}+C

\frac{x^{5}}{500000}+C

?

 

[eczeno]

seems correct. you don't need a +C for each term though, you can combine them all to one constant.

cheers