Taylor Series for ln(1-3x) about x = 0 | Homework Question

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SUMMARY

The Taylor Series for the function f(x) = ln(1-3x) about x = 0 is derived using the formula ln(1+x) = Σ (-1)^(n+1) x^n / n. By substituting -3x for x, the series becomes ln(1-3x) = Σ (-1)^(n+2) (3x)^n / n. This transformation correctly incorporates the factor of 3 into the series, confirming the validity of the approach. The first few terms of the series can be computed to verify the result.

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Homework Statement


Determine the Taylor Series for f(x) = ln(1-3x) about x = 0

Homework Equations



ln(1+x) = \sum\fract(-1)^n^+^1 x^n /{n}

The Attempt at a Solution



ln(1-3x) = ln(1+(-3x))

ln(1+(-3x)) = \sum\fract(-1)^n^+^2 x^3^n /{n}

Is that right?
 
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The -1 is in the right place, but I'm not sure why the 3 migrated to the exponent.
 
So is it:
<br /> \sum\fract(-1)^n^+^2 3x^n /{n}<br />
 
You check it yourself by computing the first couple of terms in the Taylor series.
 

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