3

$\sqrt{1+x^2}\cdot\ln (x+\sqrt{1+x^2}) = x+\frac{x^3}{3} - \frac{2}{3}\frac{x^5}{5}+ \frac{2}{3}\frac{4}{5}\frac{x^7}{7}-...$

I stuck with performing right part. Where are these coefficients from?

Desh
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  • The expression on the right hand side of the equation is the product of the series around $x=0$ of the two functions on the left hand side of the equation. – MasB Oct 31 '16 at 18:13
  • $ \ln(x+\sqrt{1+x^2})=\int \frac{dx}{\sqrt{1+x^2}}$-may be this may help – vidyarthi Oct 31 '16 at 18:18
  • I recalculated it and confirmed that the coefficients are correct as shown in the question (I has an error before). They are obtained by Taylor expansion at $x=0$. – am301 Oct 31 '16 at 18:59
  • Both Maple and Wolfram give the equality given by the OP. I believe your series might be in error. – MasB Oct 31 '16 at 19:49

1 Answers1

1

We are looking for the Taylor series of $$f(x)=\sqrt{1+x^2}\int\frac{dx}{\sqrt{1+x^2}}\tag{1}$$ that is clearly related with the Taylor series of $$ g(x) = \sqrt{1-x^2}\int\frac{dx}{\sqrt{1-x^2}} = \sqrt{1-x^2}\arcsin(x).\tag{2}$$ We may now invoke some heavy artillery. For instance, the following identity: $$ \arcsin^2(x) = \sum_{n\geq 1}\frac{(4x^2)^n}{2n^2\binom{2n}{n}}\tag{3}$$ leading to: $$ \frac{\arcsin x}{\sqrt{1-x^2}}=\sum_{n\geq 1}\frac{4^n}{(2n)\binom{2n}{n}} x^{2n-1} \tag{4} $$ and to: $$ x-\sqrt{1-x^2}\arcsin(x) = \int\frac{x\arcsin(x)}{\sqrt{1-x^2}}\,dx = \sum_{n\geq 1}\frac{4^n}{(2n)(2n+1)\binom{2n}{n}}x^{2n+1}\tag{5} $$ from which: $$ g(x)=x-\sum_{n\geq 1}\frac{4^n}{(2n)(2n+1)\binom{2n}{n}}x^{2n+1}\tag{6}$$ and $$\boxed{ f(x)=x-\sum_{n\geq 1}\frac{4^n(-1)^n}{(2n)(2n+1)\binom{2n}{n}}x^{2n+1}}\tag{7}$$ Two distinct prof of $(3)$ (one proof through Bonnet's recursion formula and Legendre polynomials, a second proof through the residue theorem) can be found in my course notes, at pages $18$ and $19$. A little warning: they are in Italian.

Community
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Jack D'Aurizio
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