I'm trying to find a recurrence relation for the coefficients for the Maclaurin series for $\tan(x)$ by substituting $y=\sum_{k=0}^{\infty}C_{2k+1}x^{2k+1}$ into the differential equation $y'=1+y^2$. This is because $\tan(x)$ is the solution to the initial value problem for the aforementioned DE with the initial condition $y(0)=0$; this is also where the form $\sum_{k=0}^{\infty}C_{2k+1}x^{2k+1}$ comes from (the fact that $\tan(x)$ is an odd function and that $y(0)=0$ which implies $C_0=0$). But I have no clue how to work "around" the expression $y^2=\big(\sum_{n=1}^{\infty}C_{2k+1}x^{2k+1}\big)^2$. How can I find a recurrence relation with an infinite squared summation? Any help is appreciated, thank you.
Squaring an arbitrary summation?
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sequences-and-series
ordinary-differential-equations
power-series
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0The coefficients will get convolved i.e. $(\sum_{k\ge0}C_kx^k)^2=\sum_{k\ge0}(\sum_{i=0}^kC_iC_{k-i})x^k$. – 2012-04-30
1 Answers
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Take the Cauchy product:
$\begin{align*} \left(\sum_{k\ge 0}C_{2k+1}x^{2k+1}\right)^2&=x^2\left(\sum_{k\ge 0}C_{2k+1}x^{2k}\right)^2\\\\ &=x^2\sum_{k\ge 0}D_{2k}x^{2k}\;, \end{align*}$
where $D_{2k}=\sum_{i=0}^kC_{2i+1}C_{2(k-i)+1}\;.$ Thus, the differential equation becomes
$\begin{align*} \sum_{k\ge 0}C_{2k+1}(2k+1)x^{2k}&=1+x^2\sum_{k\ge 0}\sum_{i=0}^kC_{2i+1}C_{2(k-i)+1}x^{2k}\\ &=1+\sum_{k\ge 1}\sum_{i=0}^{k-1}C_{2i+1}C_{2(k-i)-1}x^{2k}\;, \end{align*}$
and we have $C_1=1$ and $C_{2k+1}=\frac1{2k+1}\sum_{i=0}^{k-1}C_{2i+1}C_{2(k-i)-1}$ for $k\ge 1$.
E.g.,
$\begin{align*} C_3&=\frac13 C_1^2=\frac13\;,\\ C_5&=\frac15(2C_1C_3)=\frac2{15}\;,\text{ and}\\ C_7&=\frac17(2C_1C_5+C_3^2)=\frac{17}{315}\;. \end{align*}$
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0@Hautdesert: Glad I could help! – 2012-04-30