# Power series solution for Log(1+x)

#### ssh

##### New member
Show that,

$\log(1+x)=x-\frac{x^2}{2}+\frac{x^3}{3}+\cdots$

Last edited by a moderator:

#### Fernando Revilla

##### Well-known member
MHB Math Helper
Show that log(1+x) = x - x2\2 + x3​\3................
Use that $\dfrac{1}{1+x}=1-x+x^2-x^3+\ldots\quad (|x|<1)$ and take into account that all power series can be integrated term by term on an interval lying inside the interval of convergence.

#### ssh

##### New member
Can we write this as a Taylor's series as f(x) = Log(1+x), then f'(x)=1\1+x so on.

#### Fernando Revilla

##### Well-known member
MHB Math Helper
Can we write this as a Taylor's series as f(x) = Log(1+x), then f'(x)=1\1+x so on.
Of course you can. But using that method you only obtain the Taylor series of $f(x)=\log (1+x)$, that is $\displaystyle\sum_{n=1}^{\infty}(-1)^{n-1}\dfrac{x^n}{n}$. To prove that $\log (1+x)=\displaystyle\sum_{n=1}^{\infty}(-1)^{n-1}\dfrac{x^n}{n}$ in $(-1,1)$ (also in $x=1$) you need to verify that the remainder of the Taylor series converges to $0$.

#### Prove It

##### Well-known member
MHB Math Helper
Show that,

$\log(1+x)=x-\frac{x^2}{2}+\frac{x^3}{3}+\cdots$
If you want to use the long method, remember that a Maclaurin series for a function is given by \displaystyle \begin{align*} f(x) = \sum_{n = 0}^{\infty} \frac{f^{(n)}(0)}{n!}x^n \end{align*}

So evaluating the derivatives gives

\displaystyle \begin{align*} f(x) &= \ln{(1 + x)} \\ f(0) &= 0 \\ f'(x) &= \frac{1}{1 + x} \\ f'(0) &= 1 \\ f''(x) &= -\frac{1}{(1 + x)^2} \\ f''(0) &= -1 \\ f'''(x) &= \frac{2}{(1 +x)^3} \\ f'''(0) &= 2 \\ f^{(4)}(x) &= -\frac{3!}{(1 + x)^4} \\ f^{(4)}(0) &= -3! \\ f^{(5)}(x) &= \frac{4!}{(1 + x)^5} \\ f^{(5)}(0) &= 4! \\ \vdots \end{align*}

So substituting these in gives

\displaystyle \begin{align*} \ln{(1 + x)} &= \frac{0}{0!}x^0 + \frac{1}{1!}x^1 + \frac{(-1)}{2!}x^2 + \frac{2}{3!}x^3 - \frac{3!}{4!}x^4 + \frac{4!}{5!}x^5 - \dots + \dots \\ &= x - \frac{x^2}{2} + \frac{x^3}{3} - \frac{x^4}{4} + \frac{x^5}{5} - \dots + \dots \end{align*}

#### Fernando Revilla

##### Well-known member
MHB Math Helper
Now, we have to prove the Taylor's remainder $$R_n(x)$$:

$$R_n(x)=\dfrac{f^{(n+1)}(\xi)}{(n+1)!}x^n=\dfrac{(-1)^n}{(1+\xi)^{n+1}}\dfrac{x^{n+1}}{(n+1)!}\quad (\xi \mbox{ between }0\mbox{ and }x)$$

has limit $$0$$ for $$x\in (-1,1)$$ as $$n\to \infty$$. For that reason is better to use the series expansion of $$1/(1+x)$$.

#### Saknussemm

##### New member
Taking the derivative of the MacLaurin series gives you
$1 -x +x^2 - x^3 + x^4 + \ldots$
Since this is a geometric series with ratio $-x$, it equals $\frac{1}{1 + x}$ when x is in $(-1, 1)$.
This shows the expression $\ln(1+x)$ and its MacLaurin expansion to have the same derivative over $(-1, 1)$, which means they are equal within a constant. And, since they are equal at $x=0$, this constant is zero.

If my reasoning is correct, this is simpler than proving the limit of the Taylor remainder.

#### HallsofIvy

##### Well-known member
MHB Math Helper
Taking the derivative of the MacLaurin series gives you
$1 -x +x^2 - x^3 + x^4 + \ldots$
Since this is a geometric series with ratio $-x$, it equals $\frac{1}{1 + x}$ when x is in $(-1, 1)$.
This shows the expression $\ln(1+x)$ and its MacLaurin expansion to have the same derivative over $(-1, 1)$, which means they are equal within a constant. And, since they are equal at $x=0$, this constant is zero.

If my reasoning is correct, this is simpler than proving the limit of the Taylor remainder.
It is, however, the essentially the same as Fernando Revilla's suggestion in the first response to this thread.

#### Saknussemm

##### New member
It is, however, the essentially the same as Fernando Revilla's suggestion in the first response to this thread.
The logic and context are not the same, as I was answering the question whether you can use the Taylor series. In general, you prove the validity of the Taylor expansion over a given interval by proving the Taylor reminder tends to zero as n goes to infinity. But here, because we can bring out a geometric series, as Fernando Revilla does in the initial answer, we have a simpler alternative. It is actually an exceptional case that is worth noting.