taylor series of $\ln(1+x)$?

Question

Compute the taylor series of $\ln(1+x)$

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

$f^{'}(x)$ = $\frac{1}{1+x}$
$f^{''}(x) = \frac{-1}{(1+x)^2}$
$f^{'''}(x) = \frac{2}{(1+x)^3}$
$f^{''''}(x) = \frac{-6}{(1+x)^4}$
Therefore the series:
$\ln(1+x) = f(a) + \frac{1}{1+a}\frac{x-a}{1!} - \frac{1}{(1+a)^2}\frac{(x-a)^2}{2!} + \frac{2}{(1+a)^3}\frac{(x-a)^3}{3!} - \frac{6}{(1+a)^4}\frac{(x-a)^4}{4!} + ...$

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

The supposed correct answers are: $$\ln(1+x) = \int \left(\frac{1}{1+x}\right)dx$$ $$\ln(1+x) = \sum_{k=0}^{\infty} \int (-x)^k dx$$

Answer 1

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 $$\frac{1}{1+t}=1-t+t^2-t^3+\cdots\tag{1}$$ if $|t|\lt 1$ (infinite geometric series). Then we note that $$\ln(1+x)=\int_0^x \frac{1}{1+t}\,dt.$$ Then we integrate the right-hand side of (1) term by term. We get $$\ln(1+x) = x-\frac{x^2}{2}+\frac{x^3}{3}-\frac{x^4}{4}+\cdots,$$ precisely the same thing as what one gets by putting $a=0$ in your expression.

Answer 2

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

Answer 3

Here is another method (not efficient) :

We use the fact that for all $x \in ]-1,1[$ , $\frac{1}{1+x}=\sum \limits_{n\ge 0} (-1)^n x^n$.

Then for all $x \in ]-1,1[$, we want to prove that : $\ln(1+x) =\sum \limits_{n\ge 0} (-1)^n \frac{x^{n+1}}{n+1}$.

Notice that for all $x \in[0,1[$, we have $\ln(1+x)= \int\limits_{0}^{x} \frac{1}{1+t}\mathrm{d}t$ and for all $x \in]-1,0]$, we have $\ln(1+x)= -\int\limits_{x}^{0} \frac{1}{1+t}\mathrm{d}t$. (Note that the function $t \mapsto \pm \frac{1}{1+t}$ is continuous on the compacts $[0,x]$ and $[x,0]$).

Now we can focus on the case where $x\in [0,1[$. The other case will be similar...

As the function $t \mapsto \frac{1}{1+t}\mathbb{1}_{[0,x]}(t)$ is positive and Lebesgue-measurable on $[0,1[$ we can write $\ln(1+x)= \int\limits_{0}^{1} \frac{1}{1+t}\mathbb{1}_{[0,x]}(t)\mathrm{d}t$.

Then $\ln(1+x)= \int\limits_{0}^{1} \sum \limits_{n\ge 0} (-1)^n t^n\mathbb{1}_{[0,x]}(t)\mathrm{d}t$ and we introduce for all $n\ge 0$ and for all $t\in [0,1[$ : $S_n(t,x)=\sum \limits_{k=0}^{n} (-1)^k t^k\mathbb{1}_{[0,x]}(t)$.

So $\ln(1+x)= \int\limits_{0}^{1} \sum \limits_{n\ge 0} (-1)^n t^n\mathbb{1}_{[0,x]}(t)\mathrm{d}t =\int\limits_{0}^{1} \lim\limits_{n\to+\infty} S_n(t,x)\mathrm{d}t$.

Then for all $n\ge 0$, the sequence of partial sums $S_n$ is Lebesgue-measurable on $[0,1[$ and for each $t\in[0,1[$ point-wise convergent to $S =\sum \limits_{n\ge 0} (-1)^n t^n\mathbb{1}_{[0,x]}(t)=\frac{1}{1+t}\mathbb{1}_{[0,x]}(t)$.

Moreover for all $n\ge 0$ and $t\in [0,1[$, we have $\vert S_n(t,x)\vert \le \sum \limits_{k= 0}^{n}t^k\mathbb{1}_{[0,x]}(t) \le \lim\limits_{n\to +\infty}\sum \limits_{k= 0}^{n}t^k\mathbb{1}_{[0,x]}(t)$. Because for all $k\ge 0$, the functions $t\mapsto t^{k}\mathbb{1}_{[0,x]}$ form a positive sequence of functions on $[0,1[$. That's why $\vert S_n(t,x)\vert \le \frac{1}{1-t}\mathbb{1}_{[0,x]}(t)$.

The function $t\mapsto \frac{1}{1-t}\mathbb{1}_{[0,x]}(t)$ is positive and Lebesgue-measurable and even Lebesgue-integrable on $[0,1[$ because $\int \limits_{0}^{1} =\frac{1}{1-t}\mathbb{1}_{[0,x]}(t)\mathrm{dt} = \int \limits_{0}^{x}\frac{1}{1-t}\mathrm{d}t = \ln(1-x)<+\infty$

Then using the dominated convergence theorem we can write that :

$\ln(1+x)=\int\limits_{0}^{1} \lim\limits_{n\to+\infty} S_n(t,x)\mathrm{d}t = \lim\limits_{n\to+\infty} \int \limits_{0}^{1} S_n(t,x)\mathrm{d}t = \lim\limits_{n\to+\infty} \int\limits_{0}^{1}\sum \limits_{k=0}^{n} (-1)^k t^k\mathbb{1}_{[0,x]}(t)\mathrm{d}t$

$= \lim\limits_{n\to+\infty} \sum \limits_{k=0}^{n} (-1)^k \int\limits_{0}^{1} t^k\mathbb{1}_{[0,x]}(t)\mathrm{d}t = \lim\limits_{n\to+\infty} \sum \limits_{k=0}^{n} (-1)^k \int\limits_{0}^{x} t^k\mathrm{d}t = \lim\limits_{n\to+\infty} \sum \limits_{k=0}^{n} (-1)^k \frac{x^{k+1}}{k+1} = \sum \limits_{k=0}^{+\infty} (-1)^k \frac{x^{k+1}}{k+1}$.

With the same reasoning we can deduce the same result for $x\in ]-1,0]$.

Finally for all $x\in ]-1,1[$, we have $\ln(1+x) =\sum \limits_{n\ge 0} (-1)^n \frac{x^{n+1}}{n+1}$.

NB : With properties on power series it takes four lines...

Answer 4

Start with the Taylor series of the derivative $$\frac 1{1+x}=\frac 1{(1+a)+(x-a)}=\frac 1{1+a}\,\,\frac 1{ 1+\frac {x-a}{1+a}}$$ Let $t=\frac {x-a}{1+a}$ $$\frac 1{1+t}=\sum_{n=0}^\infty (-1)^n t^n$$ So $$\frac 1{1+x}=\frac 1{1+a}\,\,\sum_{n=0}^\infty (-1)^n \left(\frac{x-a}{a+1}\right)^n$$ Now, integrate $$\log(1+x)=\frac 1{1+a}\,\,\sum_{n=0}^\infty (-1)^n \int\left(\frac{x-a}{a+1}\right)^n \,dx=\frac 1{1+a}\,\,\sum_{n=0}^\infty (-1)^n \frac{1+a}{n+1} \left(\frac{x-a}{a+1}\right)^{n+1}+C$$ Simplify $$\log(1+x)=\sum_{n=0}^\infty (-1)^n \frac{1}{n+1} \left(\frac{x-a}{a+1}\right)^{n+1}+C$$ and using $x=a$, then $C=\log(1+a)$