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How to expand $f(x)=\ln\left( x+\sqrt{1+x^2} \right)$ into series at $x_0=0$? I've tried using Taylor's formula but counting consecutive derivatives was inconvenient and I couldn't find the general formula.

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Hint: Apply Newton's generalised binomial theorem to $f'(x) = \frac{1}{\sqrt{1+x^2}}$, and then $f(x) = f(0) + \int_0^x f'(t)dt$, where $x\in(-1,1)$. –  Eastsun Jun 14 '12 at 18:51
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up vote 5 down vote accepted

This is a somewhat laborious expansion, but it has a nice closed formula.

We begin with

$$f(x) = \frac{1}{\sqrt{1+x^2}}$$

and note that you function $F$ is such that $F'(x) = f(x)$. We can make use of the Generalized Binomial Theorem, namely

$$(1+\mathrm x)^\alpha = \sum_{n=0}^\infty {\alpha \choose n}\mathrm x^n$$

In your case, set $\mathrm x= x^2$ and $\alpha = -\dfrac{1}{2}$.

$$(1+x^2)^{-1/2}=\frac{1}{\sqrt{1+x^2}}=\sum_{n=0}^\infty {-1/2 \choose n}x^{2n}$$

It is clear the most important calculation will be that of $$ c_n={-1/2 \choose n}$$

Writing this explicitly, it gives

$$\eqalign{ & {c_n} = \frac{{\left( { - \frac{1}{2}} \right)\left( { - \frac{1}{2} - 1} \right)\left( { - \frac{1}{2} - 2} \right) \cdots \left( { - \frac{1}{2} - n + 1} \right)}}{{n!}} \cr & {c_n} = \frac{1}{{n!}}\prod\limits_{k = 0}^{n - 1} {\left( { - \frac{1}{2} - k} \right)} \cr & {c_n} = \frac{1}{{n!}}{\left( { - 1} \right)^n}\prod\limits_{k = 0}^{n - 1} {\left( {\frac{1}{2} + k} \right)} \cr & {c_n} = \frac{1}{{n!}}{\left( { - 1} \right)^n}\prod\limits_{k = 0}^{n - 1} {\left( {\frac{{2k + 1}}{2}} \right)} \cr & {c_n} = \frac{1}{{n!}}{\left( { - \frac{1}{2}} \right)^n}\prod\limits_{k = 0}^{n - 1} {\left( {2k + 1} \right)} \cr} $$

Note that for $n=1$ the product is empty, thati is, it is $1$.

If we write the product explicitly, we get

$$\prod\limits_{k = 0}^{n - 1} {\left( {2k + 1} \right)} = 1 \cdot 3 \cdots \left( {2n - 3} \right)\left( {2n - 1} \right)$$

We can "complete" it by adjoining the even numers:

$$\prod\limits_{k = 0}^{n - 1} {\left( {2k + 1} \right)} = {2^n}n!\frac{{1 \cdot 3 \cdots \left( {2n - 3} \right)\left( {2n - 1} \right)}}{{{2^n}n!}} = \frac{{1 \cdot 2 \cdot 3 \cdot 4 \cdots \left( {2n - 3} \right)\left( {2n - 2} \right)\left( {2n - 1} \right)2n}}{{{2^n}n!}}$$

and get

$$\prod\limits_{k = 0}^{n - 1} {\left( {2k + 1} \right)} = \frac{{\left( {2n} \right)!}}{{{2^n}n!}} = \frac{1}{{{2^n}}}\frac{{\left( {2n} \right)!}}{{\left( {2n - n} \right)!}}$$

so that

$${c_n} = {\left( { - 1} \right)^n}\frac{1}{{{2^n}n!}}\prod\limits_{k = 0}^{n - 1} {\left( {2k + 1} \right)} = {\left( { - 1} \right)^n}\frac{1}{{{4^n}}}\frac{{\left( {2n} \right)!}}{{n!\left( {2n - n} \right)!}} = {\left( { - 1} \right)^n}\frac{1}{{{4^n}}}{2n \choose n}$$

Thus we have that

$$\frac{1}{{\sqrt {1 + {x^2}} }} = \sum\limits_{n = 0}^\infty {{c_n}} {x^{2n}} = \sum\limits_{n = 0}^\infty {{{\left( { - \frac{1}{4}} \right)}^n}}{2n \choose n} {x^{2n}}$$

Thus, from our previous considerations, we get

$$\log \left( {x + \sqrt {1 + {x^2}} } \right) = \int\limits_0^x {\frac{{dt}}{{\sqrt {1 + {t^2}} }}} = \sum\limits_{n = 0}^\infty {{{\left( { - \frac{1}{4}} \right)}^n}}{2n \choose n} \frac{{{x^{2n + 1}}}}{{2n + 1}}$$

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awesome answer! than you so much :-) –  xan Jun 14 '12 at 19:57
    
@xan Glad to help! –  Pedro Tamaroff Jun 14 '12 at 21:59
    
I think the above development is valid only for $\,|x|<1\,$ ... –  DonAntonio Jun 15 '12 at 2:40
    
@DonAntonio Yes, it is, since the binomial expansion is valid in $|x| < 1$. What is your point/preoccupation? –  Pedro Tamaroff Jun 15 '12 at 2:44
    
@Peter Well, I didn't look at the binomial expansion but just at the final form of the MacClaurin series. Interesting that this power series's convergence radius is so limited whereas the one for the function $\,\sinh x\,$ isn't. That's all. –  DonAntonio Jun 15 '12 at 3:02
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Check that your function is the inverse hyperbolic sine: $$f(x)=\operatorname{arsinh}x\text{, where }\sinh x=\frac{e^x-e^{-x}}{2}$$ and you'll be able to get the derivative using the theorem for the inverse function...:)

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