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The answer is $$x-\frac{ 1}{2}\frac{ x^3}{3}+\frac{ 1\cdot 3}{2\cdot 4}\frac{ x^5}{5}-\frac{ 1\cdot 3\cdot 5}{2\cdot 4\cdot 6}\frac{ x^7}{7}+\cdots$$ But I can't see how. Unfortunately, "how" can't be using Taylor's formula, because that isn't introduced until the next section. (Simmons' Calculus)

The hint given with the problem is to integrate another series. This series must be $$1-\frac{ 1}{2} x^2+\frac{ 1\cdot 3}{2\cdot 4} x^4-\frac{ 1\cdot 3\cdot 5}{2\cdot 4\cdot 6} x^6+\cdots$$

But where does it come from?
[How would I know to integrate that series, if it weren't for the hint?]

The derivative of $\ln(x+\sqrt{1+x^2})$ is $\displaystyle \frac{ 1}{\sqrt{1+x^2}}$,

but I can't see how to get from $\displaystyle \frac{ 1}{\sqrt{1+x^2}}$ to that series.

The derivative can be written as $\displaystyle \frac{ 1}{x+\sqrt{1+x^2}}\left(1+\frac{x}{\sqrt{1+x^2}}\right)$, but that looks even worse.

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As TonyK suggested, first simplify your expression of the derivative (try to write the parenthesis as a single fraction and... bingo!). –  Did May 31 '11 at 12:47

3 Answers 3

You can use the fact, as you write, that the derivative of such function is $$y' = \frac{1}{\sqrt{1+x^2}}$$

But from the binomial theorem you have that

$$\frac{1}{\sqrt{1+x^2}} = \sum_{k=0}^{\infty} {-\frac{1}{2} \choose k} x^{2k}$$

You can check by yourself that

$${-\frac{1}{2} \choose k} ={\left( { - 1} \right)^k}\frac{{\left( {2k - 1} \right)!!}}{{\left( {2k} \right)!!}}$$

using induction over the following expression

$$\frac{1}{{k!}}\left( { - \frac{1}{2}} \right)\left( { - \frac{3}{2}} \right)\left( { - \frac{5}{2}} \right) \cdots \left( {\frac{1}{2} - k} \right)$$

Where I simply replaced the value of $n$ in $$\frac{{n\left( {n - 1} \right)\left( {n - 2} \right) \cdots \left( {n - k + 1} \right)}}{{k!}}$$

As a consequence you get

$$\frac{1}{\sqrt{1+x^2}} = \sum_{k=0}^{\infty} {\left( { - 1} \right)^k}\frac{{\left( {2k - 1} \right)!!}}{{\left( {2k} \right)!!}} x^{2k}$$

Upon integration you have

$$ \log\left({x+\sqrt{1+x^2}}\right) = \sum_{k=0}^{\infty} {\left( { - 1} \right)^k}\frac{{\left( {2k - 1} \right)!!}}{{\left( {2k} \right)!!}} \frac{x^{2k+1}}{2k+1} = x-\frac{ 1}{2}\frac{ x^3}{3}+\frac{ 1\cdot 3}{2\cdot 4}\frac{ x^5}{5}-\frac{ 1\cdot 3\cdot 5}{2\cdot 4\cdot 6}\frac{ x^7}{7}+\cdots$$

Remember that by definition $0!! = 1$ and $(-1)!! = 1$

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The first part of the derivate you calculated can be transformed to $$\frac{ 1}{x+\sqrt{1+x^2}} = \frac{ 1}{x+\sqrt{1+x^2}} \frac{ x-\sqrt{1+x^2}}{x-\sqrt{1+x^2}} = \sqrt{1+x^2}-x$$ and therefore the derivate is $$\sqrt{x^2+1} - \frac{x^2}{\sqrt{x^2+1}}$$ The term $\frac{x}{\sqrt{x^2+1}}$ is the derivate of $\sqrt{x^2+1}$.

The expansion of $$\sqrt{1+x}=\sum_{n=0}^{\infty}{\frac{(-1)^n(2n)!}{(1-2n)(n!)^2 4^n}x^n}$$

You can try to guess the coeffizients of the series:

Let $w(x)=\sqrt{x+1}$ Then $$(x+1)^{\frac{3}{2}}=w(x)(1+x)$$ Differentiating this equation gives $$\frac{3}{2}w(x)=w(x)+(1+x)w'(x)$$ $$w(x)=2(x+1)w'(x)$$ If $p(x)=\sum_{n=0}^{\infty}a_n x^n$ then we get $$a_n=2(na_n+(n+1)a_{n+1})$$ and $$a_{n+1}=(-1)\frac{2n+1}{2(n+1)}a_n$$ if a powerseries $p(x)$ should fullfill this equation. From this we get $$a_n=\frac{(-1)^n(2n)!}{(1-2n)(n!)^2 4^n}$$ by induction. So if w(x) there is a power series expansion of w(x) these are the coefficients. Both w(x) and the power series p(x) satisfy the equation $$\frac{f'(x)}{f(x)}=\frac{1}{x+1}$$ and therefore $$(\log{(f(x)})'=\frac{1}{x+1}$$

Therefore $$(\log{(w(x)})'=(\log{(p(x)})'=\frac{1}{x+1}$$ $$\log{(w(x))}=\log{(p(x))}+C$$ $$w(x)=p(X)*D$$ where $D=e^C$. From $w(0)=p(0)=1$ we conclude that $D=1$ and $w(x)=p(x)$.

We have now proved that this is the power series expansion of $\sqrt{x+1}$.

Substitute $x$ by $x^2$ in $\sqrt{x+1}$ and you get $$\sqrt{1+x^2}=\sum_{n=0}^{\infty}{\frac{(-1)^n(2n)!}{(1-2n)(n!)^2 4^n}x^{2n}}$$ differentiate the equation and get $$\frac{x}{\sqrt{1+x^2}}=\sum_{n=1}^{\infty}{\frac{(-1)^n(2n)!(2n)}{(1-2n)(n!)^2 4^n}x^{2n-1}}$$ and therefore $$\sqrt{x^2+1} - x*\frac{x}{\sqrt(1+x^2)}=\sum_{n=0}^{\infty}{\frac{(-1)^n(2n)!}{(n!)^2 4^n}x^{2n}}$$ This is the series of your hint.

Integration now gives $$C+\sum_{n=0}^{\infty}{\frac{(-1)^n(2n)!}{(n!)^2 4^n (2n+1)}x^{2n+1}}$$ For x=0 the function is 0, therefore C=0

If you calculate $x$ from $y=\ln{(x+\sqrt{x^2+1})}$ you get $$x=\frac{(e^y)^2-1}{2e^y}=\frac{e^y-e^{-y}}{2}=\sinh(y)$$ Therefore your function is the inverse of the sinh, the arsinh.

You can chek your results at http://en.wikipedia.org/wiki/List_of_mathematical_series

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@stuart: now the proof is "Taylor series" free. I am not sure about your prerequisites but i assume that of some basic facts about integration/differentiation/powerseries can be used. –  miracle173 Jun 2 '11 at 21:40
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I honestly think this solution is very clever but unecessarily difficult. –  Pedro Tamaroff Feb 5 '12 at 1:44

Do you know how to expand $1/\sqrt{1+x^2}$? See for instance this Wikipedia article.

Updated to add: You don't need calculus to derive this series: $$\frac{1}{\sqrt{1-y}} = 1+\frac{ 1}{2} y+\frac{ 1\cdot 3}{2\cdot 4} y^2+\frac{ 1\cdot 3\cdot 5}{2\cdot 4\cdot 6} y^3+\cdots$$ Just square the right-hand side, and you will get (miraculously) $$1 + y + y^2 + y^3 +\cdots$$ which you recognise as $1/(1-y)$ (right?). Now put $y = -x^2$ to obtain your expression for $1/\sqrt{1+x^2}$.

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@TonkyK, Thank you very much for your answer. Thank you as well, @miracle173. I found the expansion you suggested that I use in my book. It's derived as a result of Taylor's formula in a section that I haven't gotten to yet. Is there some other way to derive the binomial series or is there some way to solve this problem without it? Thanks again. –  stuart May 31 '11 at 14:13
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@stuart: You don't need Taylor's formula. See my updated answer. –  TonyK May 31 '11 at 15:16
    
thanks again. This clears things up to the point where I understand the solution, but I'm still uneasy with it. Squaring a power series isn't something that's been discussed in the book yet. It (multiplying power series, that is) is in an optional section about two sections further from where this problem was found. Can you conceive of any other way to solve this problem? –  stuart May 31 '11 at 18:20
    
@stuart: I'm afraid it doesn't get any simpler than this. –  TonyK Jun 1 '11 at 11:40
    
@TonyK Athough it is true squaring the RHS $gives 1+y+y^2+\cdots$ it is still imperative to show where the RHS came from. –  Pedro Tamaroff Feb 5 '12 at 1:49

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