# What are the first few coefficients for this series?

## Homework Statement

The function $f(x)=ln(2−x)$ is represented as a power series in the form:

$\sum C_{n}x^{n},n,0,inf$

Find the first five coefficients in the power series.

Basically, this is a problem from my online homework. I did a lot of work to get to my answers, and I feel sort of tired to type up all of it so I'll just show my final result (hopefully you guys don't mind). And if need be to know, I first differentiated f(x), then wrote down the power series for it, then integrated the power series to obtain the power series for the original f(x).

Afterwards, I expanded this power series for the first five terms and wrote down the coefficients:

$(-1/2), (-1/8), (-1/24), (-1/64), (-1/160)$

I then inputted these numbers into the boxes specified on the web page and I received a message saying that they were wrong. I've checked my work twice, my TI89, and even Wolfram Alpha; I get the same answers from these sources, so can someone tell me what is wrong? Also if someone does have a solution, you do not have to tell me what the actual answers are, just notify me with what I did wrong.

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## Answers and Replies

vela
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What about the constant term?

There was no constant term; after integrating the whole power series, I obtained x^(n+1)/(n+1) times a number. If n = 0 as the first term, then you would get x^(0+1)/(0+1), which is equal to just x.

Edit: Also if you are interested to know, this was the power series I derived:

$\sum\frac{-1}{2}\frac{1}{2^n}\frac{x^{n+1}}{n+1},n,0,∞$

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Ray Vickson
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There was no constant term; after integrating the whole power series, I obtained x^(n+1)/(n+1) times a number. If n = 0 as the first term, then you would get x^(0+1)/(0+1), which is equal to just x.

Edit: Also if you are interested to know, this was the power series I derived:

$\sum\frac{-1}{2}\frac{1}{2^n}\frac{x^{n+1}}{n+1},n,0,∞$

There must be a constant term: what is f(0)? Is it zero? By the way, is your notation supposed to be $\sum_{n=0}^\infty C_n x^n$?

RGV

I don't think that's the case because if the derivative of the function is $f ' (x)=\frac{-1}{2-x}$, then the power series representation would be:

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

If I wrote out the first five terms of this series, I would get:

$(\frac{-1}{2})-(\frac{1}{4})x-(\frac{1}{8})x^2-(\frac{1}{16})x^3-(\frac{1}{32})x^4$

Next I would have to integrate these terms to get a partial representation for $f(x) = ln(2-x)$:

$\int (\frac{-1}{2})-(\frac{1}{4})x-(\frac{1}{8})x^2-(\frac{1}{16})x^3-(\frac{1}{32})x^4 dx$
$= (\frac{-1}{2})x-(\frac{1}{8})x^2-(\frac{1}{24})x^3-(\frac{1}{64})x^4-(\frac{1}{160})x^5$

So as seen from above, the coefficients would be as I specified in my first post?

EDIT: I tried inputting 0, -1/2, -1/8, -1/24, -1/64 and I got everything except for 0 right; so now I know there must be something before -1/2? But if there is something before -1/2, how would I find it?

EDIT2: I tested ln(2) and found that to be the replacement for 0, but can someone explain to me how the first term is ln(2)?

EDIT3: Never mind, I found out why it was ln(2); the C that is resulted from integration had to have been solved, I totally neglected that so thank you guys for pointing that out.

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HallsofIvy
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When you integrated the series term by term, did you include the constant of integration?

Okay so here's another similar problem: I first differentiated the function $f(x)=ln(1-x^2)$ and wrote down the power series for it:

$\sum\stackrel{∞}{n=0}(-2x)(x^2)^n$

Then I integrated the power series to get a series for the original function:

$\sum\stackrel{∞}{n=0}(-2)\frac{x^{2n+2}}{2n+2}$

There was suppose to be a constant of integration but I found it to be zero, so after expanding the series for five terms, I found the coefficients to be:

$0, -1, -1/2, -1/3, -1/4$

After inputting these numbers, as seen by the above picture, the web page said they were wrong; can someone tell me why?

Dick
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Okay so here's another similar problem:
View attachment 40518

I first differentiated the function $f(x)=ln(1-x^2)$ and wrote down the power series for it:

$\sum\stackrel{∞}{n=0}(-2x)(x^2)^n$

Then I integrated the power series to get a series for the original function:

$\sum\stackrel{∞}{n=0}(-2)\frac{x^{2n+2}}{2n+2}$

There was suppose to be a constant of integration but I found it to be zero, so after expanding the series for five terms, I found the coefficients to be:

$0, -1, -1/2, -1/3, -1/4$

After inputting these numbers, as seen by the above picture, the web page said they were wrong; can someone tell me why?

I think the homework is expecting c1 to be the coefficient of x in the series. That's zero. The first nonzero term, -1, is the coefficient of x^2 in the series. That should be c2.

I tried what you said and it turns out c1 is indeed zero, meaning c0 is zero, c1 is zero, and c2 is -1. However c3 and c4 are still wrong; I used c3 = -1/2 and c4 = -1/3, and the web page said they were wrong?

Dick
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I tried what you said and it turns out c1 is indeed zero, meaning c0 is zero, c1 is zero, and c2 is -1. However c3 and c4 are still wrong; I used c3 = -1/2 and c4 = -1/3, and the web page said they were wrong?

You aren't thinking about the powers of x in your series. It's -x^2-x^4/2-x^6/3-... isn't it. c3 should be the coefficient of the power of x^3. Etc.

Oh well that works, c3 is actually zero, and c4 is -1/2, but what I don't get is why these numbers don't correspond to the exponent of x. For example, x^3 corresponds to the coefficient of -1/2, but instead x^3 corresponds to the coefficient of zero?

Dick
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Oh well that works, c3 is actually zero, and c4 is -1/2, but what I don't get is why these numbers don't correspond to the exponent of x. For example, x^3 corresponds to the coefficient of -1/2, but instead x^3 corresponds to the coefficient of zero?

The attachment says the series is c_n*x^n. So c1 is the coefficient of x, c2 is the coefficient of x^2, etc etc. The odd powers are missing from your series, so all of the odd coefficients should be 0. Right?

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Yes that's true, but the terms that are suppose to be missing are the even terms; expanding the series produces exponents of odd numbers, that's why I'm confused as to why the even numbers are showing and the odd numbers are missing.
$\sum\stackrel{∞}{n=0}(-2)\frac{x^{2n+2}}{2n+2}$

Dick
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Homework Helper
Yes that's true, but the terms that are suppose to be missing are the even terms; expanding the series produces exponents of odd numbers, that's why I'm confused as to why the even numbers are showing and the odd numbers are missing.
$\sum\stackrel{∞}{n=0}(-2)\frac{x^{2n+2}}{2n+2}$

2n+2 is even, isn't it? That's the exponent. What do you mean 'exponents of odd numbers'?

Whoa my bad, my scratch paper was conglomerated with numerous revisions of different series, making me look at the wrong one. :uhh: