# Power series solution for Log(1+x)

• MHB
• ssh
In summary, the given conversation discusses different methods of proving the equation $\log(1+x)=x-\frac{x^2}{2}+\frac{x^3}{3}+\cdots$ and concludes that it can be proven by using the Maclaurin series expansion of $1/(1+x)$. This simplifies the proof as it shows the two expressions have the same derivative, making the constant term in the Maclaurin series equal to zero. This is a unique case that is worth noting.

#### ssh

Show that,

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

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ssh said:
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.

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

ssh said:
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 $0$.

ssh said:
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*}

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 $$1/(1+x)$$.

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.

Saknussemm said:
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.

HallsofIvy said:
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.