Pamela Meyer

2021-12-21

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

Thomas White

Beginner2021-12-22Added 40 answers

Compute the taylor series of ln(1+x)

I've first computed derivatives (up to the 4th) of ln(1+x)

$f\prime \left(x\right)=\frac{1}{1+x}$

$f\prime \prime \left(x\right)=\frac{-1}{{(1+x)}^{2}}$

$f\prime \prime \prime \left(x\right)=\frac{2}{{(1+x)}^{3}}$

$f\prime \prime \prime \prime \left(x\right)=\frac{-6}{{(1+x)}^{4}}$

Therefore the series:

$\mathrm{ln}(1+x)=f\left(a\right)+\frac{1}{1+a}\frac{x-a}{1!}-\frac{1}{{(1+a)}^{2}}\frac{{(x-a)}^{2}}{2!}+\frac{2}{{(1+a)}^{3}}\frac{{(x-a)}^{3}}{3!}-\frac{6}{{(1+a)}^{4}}\frac{{(x-a)}^{4}}{4!}+\dots$

But this doesn't seem to be correct. Can anyone please explain why this doesn't work?

The supposed correct answer is:

$As\mathrm{ln}(1+x)=\int \left(\frac{1}{1+x}\right)dx$

$\mathrm{ln}(1+x)=\sum _{k=0}^{\mathrm{\infty}}\int {(-x)}^{k}dx$

I've first computed derivatives (up to the 4th) of ln(1+x)

Therefore the series:

But this doesn't seem to be correct. Can anyone please explain why this doesn't work?

The supposed correct answer is:

John Koga

Beginner2021-12-23Added 33 answers

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

$\frac{1}{1+t}=1-t+{t}^{2}-{t}^{3}+\cdots \left(1\right)$

if$\left|t\right|<1$ (infinite geometric series). Then we note that

$\mathrm{ln}(1+x)={\int}_{0}^{x}\frac{1}{1+t}dt.$

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

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

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

That will simplify your expression considerably. Note also that

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

if

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

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

nick1337

Expert2021-12-27Added 777 answers

Note that

Integrating both sides gives you

Alternatively,

We deduce that

Hence,

Edit: To derive a closed for for the geometric series, let

To prove in the other direction, use the binomial theorem or simply compute the series about 0 manually.

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