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

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Homework Help Overview

The discussion revolves around finding the Taylor Series for the function f(x) = ln(1-3x) centered at x = 0. Participants are examining the application of the Taylor series expansion for logarithmic functions.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • The original poster attempts to apply the Taylor series for ln(1+x) to ln(1-3x) by substituting -3x. Some participants question the correctness of the exponent manipulation and the placement of the coefficient 3.

Discussion Status

The discussion is ongoing, with participants providing feedback on the original poster's attempt and suggesting verification through computation of initial terms. There is no explicit consensus yet on the correct formulation of the series.

Contextual Notes

Participants are navigating the nuances of series expansion and the implications of substituting values within the logarithmic function. There may be assumptions regarding familiarity with series notation and convergence that are not explicitly stated.

ganondorf29
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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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