Is there any expansion for $\mathrm{log}(1+x)$ when $x>1$?

Is there any expansion for $\mathrm{log}(1+x)$ when $x>1$?

Is there any expansion for $\mathrm{log}(1+x)$ when $x>1$?

Trace Glass
2022-10-24
Answered

Is there any expansion for $\mathrm{log}(1+x)$ when $x>1$?

Is there any expansion for $\mathrm{log}(1+x)$ when $x>1$?

Is there any expansion for $\mathrm{log}(1+x)$ when $x>1$?

You can still ask an expert for help

bargeolonakc

Answered 2022-10-25
Author has **16** answers

You can expand the function $\mathrm{log}(1+x)$ around any point at which it is defined. This means there exists an expansion of $\mathrm{log}(1+x)$ around the point $x=2$, for example, however it will be of the form

$\mathrm{log}(1+x)=\sum _{i=1}^{\mathrm{\infty}}{a}_{i}(x-2{)}^{i},$

and the expansion will be valid for $x\in (-1,5)$

However, I imagine you want the expansion to of the form

$\mathrm{log}(1+x)=\sum _{i=1}^{\mathrm{\infty}}{a}_{i}{x}^{i},$

for which you will have a problem. The problem is that any power series

$\sum _{i=1}^{\mathrm{\infty}}{a}_{i}(x-{x}_{0}{)}^{i}$

will converge on a interval symmetric around ${x}_{0}$ (meaning an interval of the type $({x}_{0}-\delta ,{x}_{0}+\delta )$). This means that if the series expansion for $\mathrm{log}(1+x)$ will converge for $x>1$, it will also converge for some $x<-1$ which is impossible.

$\mathrm{log}(1+x)=\sum _{i=1}^{\mathrm{\infty}}{a}_{i}(x-2{)}^{i},$

and the expansion will be valid for $x\in (-1,5)$

However, I imagine you want the expansion to of the form

$\mathrm{log}(1+x)=\sum _{i=1}^{\mathrm{\infty}}{a}_{i}{x}^{i},$

for which you will have a problem. The problem is that any power series

$\sum _{i=1}^{\mathrm{\infty}}{a}_{i}(x-{x}_{0}{)}^{i}$

will converge on a interval symmetric around ${x}_{0}$ (meaning an interval of the type $({x}_{0}-\delta ,{x}_{0}+\delta )$). This means that if the series expansion for $\mathrm{log}(1+x)$ will converge for $x>1$, it will also converge for some $x<-1$ which is impossible.

Deja Bradshaw

Answered 2022-10-26
Author has **4** answers

You are looking for Laurent series for $|x|>1$

$\mathrm{ln}(1+x)=\mathrm{ln}(x(1+1/x))=\mathrm{ln}(x)+\mathrm{ln}(1+1/x)=\mathrm{ln}(x)+\sum _{k=1}^{\mathrm{\infty}}\frac{(-1{)}^{k}}{k}{x}^{-k}$

with the condition of convergence $|x|>1$

$\mathrm{ln}(1+x)=\mathrm{ln}(x(1+1/x))=\mathrm{ln}(x)+\mathrm{ln}(1+1/x)=\mathrm{ln}(x)+\sum _{k=1}^{\mathrm{\infty}}\frac{(-1{)}^{k}}{k}{x}^{-k}$

with the condition of convergence $|x|>1$

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Assume $x>0$ and let

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$u(1)=4.$

I started off by doing some algebra to get:

$\frac{1}{{u}^{2}}du=\frac{1}{{x}^{2}+x}dx.$

I then took the partial fraction of the right side of the equation:

$\frac{1}{{u}^{2}}du=(\frac{1}{x}-\frac{1}{x+1}).$

I then took the integral of both sides:

$-\frac{1}{u}=\mathrm{log}x-\mathrm{log}(x+1)+C.$

From here I don't know what to do because we are solving for $u(x)$ and I'm not sure how to get that from $-\frac{1}{u}$

Assume $x>0$ and let

$x(x+1)\frac{du}{dx}={u}^{2},$

$u(1)=4.$

I started off by doing some algebra to get:

$\frac{1}{{u}^{2}}du=\frac{1}{{x}^{2}+x}dx.$

I then took the partial fraction of the right side of the equation:

$\frac{1}{{u}^{2}}du=(\frac{1}{x}-\frac{1}{x+1}).$

I then took the integral of both sides:

$-\frac{1}{u}=\mathrm{log}x-\mathrm{log}(x+1)+C.$

From here I don't know what to do because we are solving for $u(x)$ and I'm not sure how to get that from $-\frac{1}{u}$

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I moved only with the fact that $4={\mathrm{log}}_{2}16$. I have no idea what to do next.

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$$\sum _{n=1}^{\mathrm{\infty}}\mathrm{ln}(n){z}^{n}$$

Find the convergence radius and a the function f to which the series converges.

I have easily found that $R=1$ is the convergence radius, however I can not find the function. I was trying to found an elemental function with this power series expantion, but I have failed. Anyone knows such function and how to prove the convergence?