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In this example problem in my textbook:

"Find a power series representation for ln(1-x) and its radius of convergence."

They integrate both sides:

-ln(1-x) = integral (1/1-x)dx which comes out to be SUM x^n/n + C.

They solve for C, C=0. This is where I get stuck. They proceed to show what the series looks like when C=0, and show:

ln(1-x) = -x - x^2/2 - x^3/x - ... = -SUM x^n/n |x| < 1

How come these are all negative terms?

They then say: "Notice what happens if we put x = 1/2 in the result of Example 6. Since ln(1/2) = -ln2 we see that: "

ln2 = 1/2 + 1/8 + 1/24 + 1/64 + ... = SUM 1/(n2^n)

Why was x = 1/2 put back into -ln(1-x)? Why not ln(1-x)? Now why are they all positive?

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You're missing a minus sign in front of the integral. –  Jonas Meyer Feb 2 '11 at 1:14

1 Answer 1

up vote 3 down vote accepted

$\ln(1-x) \lt 0$ for $0 \lt x \lt 1$. So you expect negative terms.

If you have a series $f(x) = \sum a_n x^n$ valid for $|x| \lt 1$, then we have that $-f(x) = g(x) = \sum (-a_n) x^n$ is also valid for $|x| \lt 1$.

Does that help?

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What about plugging back in 1/2? If x = 1/2, then it's ln(1/2) = -ln2. What happened to the negative sign in front of ln2? It seems like they plugged it back into -ln(1-x) and not ln(1-x). If they did, why -ln(1-x)? Aren't we looking for ln(1-x)? –  ShrimpCrackers Feb 2 '11 at 1:11
    
@Shrimp: $\ln a = - \ln (1/a)$ So to get $\ln 2$, you need $- \ln (1/2)$. Hence you need $-\ln(1-x)$ for $x = 1/2$. And like I said, if you have a valid power series, you can multiply it by $-1$ and it is still valid. –  Aryabhata Feb 2 '11 at 1:16
    
Thanks Moron. Makes sense now. Funny name, but it doesn't match you. –  ShrimpCrackers Feb 2 '11 at 1:21

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