 Pamela Meyer

2021-12-21

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

Compute the taylor series of ln(1+x)
I've first computed derivatives (up to the 4th) of ln(1+x)
$f\prime \left(x\right)=\frac{1}{1+x}$
$f\prime \prime \left(x\right)=\frac{-1}{{\left(1+x\right)}^{2}}$
$f\prime \prime \prime \left(x\right)=\frac{2}{{\left(1+x\right)}^{3}}$
$f\prime \prime \prime \prime \left(x\right)=\frac{-6}{{\left(1+x\right)}^{4}}$
Therefore the series:
$\mathrm{ln}\left(1+x\right)=f\left(a\right)+\frac{1}{1+a}\frac{x-a}{1!}-\frac{1}{{\left(1+a\right)}^{2}}\frac{{\left(x-a\right)}^{2}}{2!}+\frac{2}{{\left(1+a\right)}^{3}}\frac{{\left(x-a\right)}^{3}}{3!}-\frac{6}{{\left(1+a\right)}^{4}}\frac{{\left(x-a\right)}^{4}}{4!}+\dots$
But this doesn't seem to be correct. Can anyone please explain why this doesn't work?
$As\mathrm{ln}\left(1+x\right)=\int \left(\frac{1}{1+x}\right)dx$
$\mathrm{ln}\left(1+x\right)=\sum _{k=0}^{\mathrm{\infty }}\int {\left(-x\right)}^{k}dx$ John Koga

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 $\frac{\left(n-1\right)!}{n!}=\frac{1}{n}.$
The approach in the suggested solution also works. We note that
$\frac{1}{1+t}=1-t+{t}^{2}-{t}^{3}+\cdots \left(1\right)$
if $|t|<1$ (infinite geometric series). Then we note that
$\mathrm{ln}\left(1+x\right)={\int }_{0}^{x}\frac{1}{1+t}dt.$
Then we integrate the right-hand side of (1) term by term. We get
$\mathrm{ln}\left(1+x\right)=x-\frac{{x}^{2}}{2}+\frac{{x}^{3}}{3}-\frac{{x}^{4}}{4}+\cdots ,$
precisely the same thing as what one gets by putting a=0 in your expression. nick1337

Note that
$\frac{1}{1+x}=\sum _{n\ge 0}\left(-1{\right)}^{n}{x}^{n}$
Integrating both sides gives you
$ln\left(1+x\right)=\sum _{n\ge 0}\frac{\left(-1{\right)}^{n}{x}^{n+1}}{n+1}$
$=x-\frac{{x}^{2}}{2}+\frac{{x}^{3}}{3}...$
Alternatively,
${f}^{\left(1\right)}\left(x\right)=\left(1+x{\right)}^{-1}⇒{f}^{\left(1\right)}\left(0\right)=1$
${f}^{\left(2\right)}\left(x\right)=-\left(1+x{\right)}^{-2}⇒{f}^{\left(2\right)}\left(0\right)=-1$
${f}^{\left(3\right)}\left(x\right)=2\left(1+x{\right)}^{-3}⇒{f}^{\left(3\right)}\left(0\right)=2$
${f}^{\left(4\right)}\left(x\right)=-6\left(1+x{\right)}^{-4}⇒{f}^{\left(4\right)}\left(0\right)=-6$
We deduce that
${f}^{\left(n\right)}\left(0\right)=\left(-1{\right)}^{n-1}\left(n-1\right)!$
Hence,
$ln\left(1+x\right)=\sum _{n\ge 1}\frac{{f}^{\left(n\right)\left(0\right)}}{n!}{x}^{n}$
$=\sum _{n\ge 1}\frac{\left(-1{\right)}^{n-1}\left(n-1\right)!}{n!}{x}^{n}$
$=\sum _{n\ge 1}\frac{\left(-1{\right)}^{n-1}}{n}{x}^{n}$
$=\sum _{n\ge 1}\frac{\left(-1{\right)}^{n}}{n+1}{x}^{n+1}$
$=x-\frac{{x}^{2}}{2}\frac{{x}^{3}}{3}-...$
Edit: To derive a closed for for the geometric series, let
$S=1-x+{x}^{2}-{x}^{3}+...$
$xS=x-{x}^{2}+{x}^{3}-{x}^{4}...$
$S+xS=1$
$S=\frac{1}{1+x}$
To prove in the other direction, use the binomial theorem or simply compute the series about 0 manually.

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