Solve and give correct answer for this taylor series of ln(1+x)?

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Thomas White · Accepted Answer

Compute the taylor series of ln(1+x)I&#039;ve first computed derivatives (up to the 4th) of ln(1+x)f′(x)=11+xf′′(x)=−1(1+x)2f′′′(x)=2(1+x)3f′′′′(x)=−6(1+x)4Therefore the series:ln⁡(1+x)=f(a)+11+ax−a1!−1(1+a)2(x−a)22!+2(1+a)3(x−a)33!−6(1+a)4(x−a)44!+…But this doesn&#039;t seem to be correct. Can anyone please explain why this doesn&#039;t work?The supposed correct answer is:Asln⁡(1+x)=∫(11+x)dxln⁡(1+x)=∑k=0∞∫(−x)kdx

John Koga · Accepted Answer

You got the general expansion about x=a. Here we are intended to take a=0. That is, we are finding the Maclaurin series of ln(1+x).
That will simplify your expression considerably. Note also that

\frac{(n - 1)!}{n!} = \frac{1}{n} .

The approach in the suggested solution also works. We note that

\frac{1}{1 + t} = 1 - t + t^{2} - t^{3} + \dots (1)

if

| t | < 1

(infinite geometric series). Then we note that

\ln (1 + x) = \int_{0}^{x} \frac{1}{1 + t} d t .

Then we integrate the right-hand side of (1) term by term. We get

\ln (1 + x) = x - \frac{x^{2}}{2} + \frac{x^{3}}{3} - \frac{x^{4}}{4} + \dots,

precisely the same thing as what one gets by putting a=0 in your expression.

nick1337 · Accepted Answer

Note that 11+x=∑n≥0(−1)nxn Integrating both sides gives you ln(1+x)=∑n≥0(−1)nxn+1n+1 =x−x22+x33... Alternatively, f(1)(x)=(1+x)−1⇒f(1)(0)=1 f(2)(x)=−(1+x)−2⇒f(2)(0)=−1 f(3)(x)=2(1+x)−3⇒f(3)(0)=2 f(4)(x)=−6(1+x)−4⇒f(4)(0)=−6 We deduce that f(n)(0)=(−1)n−1(n−1)! Hence, ln(1+x)=∑n≥1f(n)(0)n!xn =∑n≥1(−1)n−1(n−1)!n!xn =∑n≥1(−1)n−1nxn =∑n≥1(−1)nn+1xn+1 =x−x22x33−... Edit: To derive a closed for for the geometric series, let S=1−x+x2−x3+... xS=x−x2+x3−x4... S+xS=1 S=11+x To prove in the other direction, use the binomial theorem or simply compute the series about 0 manually.

Solve and give correct answer for this taylor series of

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