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What is the Maclaurin series for #(1-x)ln(1-x)#?

Answer 1

Ezra Adams

# -x + 1/2x^2 + 1/6x^3 + 1/12x^4 #

Start with the known Maclaurin series for #ln(1-x)#, which is:

#ln(1-x) = -x-1/2x^2-1/3x^3-1/4x^4 -1/5x^5- ... #

Then we can just use algebra to multiply this series by #(1-x)#, hence:

# f(x) = (1-x)ln(1-x) # # " " = (1-x){-x-1/2x^2-1/3x^3-1/4x^4 - 1/5x^5... } #

# " " = (1){-x-1/2x^2-1/3x^3-1/4x^4 - 1/5x^5 ... } # # " " - x{-x-1/2x^2-1/3x^3-1/4x^4 - 1/5x^5... }#

# " " = -x-1/2x^2-1/3x^3-1/4x^4 - 1/5x^5... + # # " " x^2+1/2x^3+1/3x^4+ 1/4x^5... #

# " " = -x + 1/2x^2 + 1/6x^3 + 1/12x^4 + 1/20x^5... #

So the required polynomial of degree #4# is

# -x + 1/2x^2 + 1/6x^3 + 1/12x^4 #

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Answer 2

Paisley Johnson

The Maclaurin series for (1-x)ln(1-x) is:

∑[n=1 to ∞] (-1)^(n+1) * (x^n / n^2)

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.