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.
Expanding logarithm into series
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4Hint: 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 $x\in(-1,1)$. – 2012-06-14
2 Answers
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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0Oh, ok. I know the Laurent series for the function when \,|x|>1\, but didn't think of considering it a complex function but a real one, @Peter. Thanks. Oh, and +1 , of course. – 2012-06-15
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...:)