Logarithmic formula

Question

By making as few assumptions as possible, prove that

$$\log \left({{1+x} \over {1-x}} \right)=2 \left(\frac{x}{1}+\frac{x^3}{3}+\frac{x^5}{5}+\cdots\right)$$

for $|x|<1$.

Answer 1

HINT:

$$\log\left(\frac{1+x}{1-x}\right)=\log(1+x)-\log(1-x)$$

And integrate the following:

$$\frac1{1-r}=1+r+r^2+r^3+\dots\forall\ |r|<1$$

Edit:

If you think the geometric series is for granted, see that

$$\frac{1-r^N}{1-r}=1+r+r^2+r^3+\dots+r^{N-1}$$

and take $N\to\infty$

Answer 2

Use Taylor expansion for $\ln(1+x)$, it is $\ln(1+x)=x-1/2x^2+1/3x^3-1/4x^4+... $, so similarly you can know $\ln(1-x)=-x+1/2x-1/3x^3-1/4x^4+...$.

Now let look at $\ln(\frac{1+x}{1-x})=\ln(1+x)-\ln(1-x)=2/(x+1/3x^3+1/5x^5+...)$

Answer 3

Note that for $|x|<1$, we can write

$$\begin{align} \log\left(\frac{1+x}{1-x}\right)&=\int_0^{x}\color{#034da3}{\left(\frac{1}{1+u}+\frac{1}{1-u}\right)}\,du\\\\ &=\int_0^x \color{#034da3}{\frac{2}{1-u^2}}\,du\\\\ &=\int_0^x \overbrace{\color{#034da3}{2\sum_{n=0}^\infty u^{2n}}}^{\color{#034da3}{\text{Geometric Series}}}\,du\\\\ &=2\sum_{n=0}^\infty \frac{x^{2n+1}}{2n+1}\\\\ &=2\sum_{n=1}^\infty\frac{x^{2n-1}}{2n-1} \end{align}$$

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Alternatively, for $|x|<1$, we can write

$$\begin{align} \log\left(\frac{1+x}{1-x}\right)&=\int_0^{x}\color{#034da3}{\left(\frac{1}{1+u}+\frac{1}{1-u}\right)}\,du\\\\ &=\int_0^x \overbrace{\color{#034da3}{\sum_{n=0}^\infty (1+(-1)^n)u^n}}^{\color{#034da3}{\text{Geometric Series of the integrand}}}\,dx\\\\ &=\underbrace{\sum_{n=0}^\infty\frac{(1+(-1)^n)x^{n+1}}{n+1}}_{\text{After integrating the power series term-by-term}}\\\\ &=2\sum_{n=0}^\infty \frac{x^{2n+1}}{2n+1}\\\\ \end{align}$$

Answer 4

From the non-standard definition of the logarithm [proved here] (Trying to show that $\ln(x) = \lim_{n\to\infty} n(x^{1/n} -1)$)

$$\log(t):=\lim_{n\to\infty}n(t^{1/n}-1)$$

after substituting $t=1+x$, we get

$$\log(1+x):=\lim_{n\to\infty}n((1+x)^{1/n}-1).$$

Then using the generalized binomial theorem and the falling factorial notation,

$$\log(1+x)=\lim_{n\to\infty}n\left(\sum_{k=0}^\infty\left(\frac1n\right)_{k}\frac{x^k}{k!}-1\right)=\lim_{n\to\infty}n\sum_{k=1}^\infty\left(\frac1n\right)_{k}\frac{x^k}{k!}.$$

As the coefficient of the $k^{th}$ power is

$$n\left(\frac1n\right)_k=n\frac1n\left(\frac1n-1\right)\cdots\left(\frac1n-k+1\right)=\left(\frac1n-1\right)\cdots\left(\frac1n-k+1\right),$$

its limit is clearly $(-1)(-2)\cdots(-k+1)=(-1)^{k-1}(k-1)!$

Then

$$\log(1+x)=\sum_{k=1}^\infty(-1)^{k-1}\frac{x^k}{k}$$ and

$$\log(1+x)-\log(1-x)=2\sum_{k=1}^\infty\frac{x^{2k-1}}{2k-1}.$$

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For complete rigor, one needs to justify that the limit of the sum is the sum of the limits.