# Squaring an arbitrary summation?

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.

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 The coefficients will get convolved i.e. $(\sum_{k\ge0}C_kx^k)^2=\sum_{k\ge0}(\sum_{i=0}^kC_iC_{k-i})x^k$. – sai Apr 30 '12 at 0:11

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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 In other contexts, the coefficients of the squared function would be what is termed a self-convolution, or autoconvolution. – J. M. Apr 30 '12 at 0:41 Say we have $\big(\sum_{i=0}^{n}a_i\big)\big(\sum_{j=0}^{n}b_j\big)$ $=a_0\big(\sum_{j=0}^{n}\big)+a_1\big(\sum_{j=0}^{n}\big)+...+a_n\big(\sum_{j=0}‌​^{n}\big)$ which equals $\sum_{i=0}^{n}\sum_{j=0}^{n}a_ib_j$. Why do we shift the index of the second summation? – Hautdesert Apr 30 '12 at 1:39 @Hautdesert: I don’t understand. This example is very different from the product of two series (or for that matter two polynomials), and I don’t know what index you think is being shifted. – Brian M. Scott Apr 30 '12 at 1:44 @BrianM.Scott: Is the reasoning in my previous comment not a generalization of the product of two arbitrary polynomials? I thought we could just set $a_i=c_ix^i$ and derive the Cauchy product. – Hautdesert Apr 30 '12 at 1:49 @Hautdesert: Not if you want to group the product terms according to powers of $x$ so as to be able to match coefficients in $y'$ and $1+y^2$. That’s why the coefficient $D_{2k+1}$ of $x^{2k}$ above is itself a sum of products. Similarly, in $(a+bx+cx^2)(d+ex+fx^2)$ the coefficient of $x^3$ is $bf+cd$, since there are two $x^3$ terms. – Brian M. Scott Apr 30 '12 at 1:55