# Power Series with complicated Algebra

I need to find the Power Series for $\frac{ac^2}{\sqrt{1-\frac{b^2}{c^2}}}$. I said let $\frac{b^2}{c^2}=x$. Therefore we have $\frac{ac^2}{\sqrt{1-x}}$. Well we know the Power Series for $\frac{1}{1-x}$. Therefore we substitute and get
$$ac^2(1+x+x^2+x^3+x^4+\cdots)^{0.5}$$ If I substitute back in for $x$ though I get stuck on the next step. Any tips or other ways would be much appreciated.

• Do you know the power series for $(1-x)^{-1/2}$? If not, can you calculate it using the binomial theorem? – Mathmo123 Sep 8 '15 at 23:26
• No. We are not supposed to figure it out with the binomial theorem, but I would be interested if you had a link I could follow. My other idea was to put the ac^2 back into the square root and solve it that way to cross off some things perhaps. – Jack Armstrong Sep 8 '15 at 23:28
• "The power series for $\frac{ac^2}{\sqrt{1-\frac{b^2}{c^2}}}$" could mean a power series for $a\mapsto \frac{ac^2}{\sqrt{1-\frac{b^2}{c^2}}}$, or a power series for $b\mapsto \frac{ac^2}{\sqrt{1-\frac{b^2}{c^2}}}$, or a power series for $c\mapsto \frac{ac^2}{\sqrt{1-\frac{b^2}{c^2}}}$, or even a power series for $b^2/c^2 \mapsto \frac{ac^2}{\sqrt{1-\frac{b^2}{c^2}}}$ (which is how you've construed it), or a power series for $b^2\mapsto \frac{ac^2}{\sqrt{1-\frac{b^2}{c^2}}}$, etc. So the question as written is somewhat ambiguous. ${}\qquad{}$ – Michael Hardy Sep 9 '15 at 0:11

## 1 Answer

Suppose $f(x) =\frac{1}{\sqrt{1-x}}$. Then $f^2(x) =\frac1{1-x} =\sum_{n=0}^{\infty} x^{n}$.

If $f(x) =\sum_{n=0}^{\infty} a_n x^n$,

$\begin{array}\\ f^2(x) &=\sum_{n=0}^{\infty} a_n x^n \sum_{m=0}^{\infty} a_m x^m\\ &=\sum_{n=0}^{\infty} \sum_{m=0}^{\infty} a_n a_m x^{n+m}\\ &=\sum_{k=0}^{\infty} \sum_{j=0}^{k} a_j a_{k-j} x^{k}\\ &=\sum_{k=0}^{\infty} x^k \sum_{j=0}^{k} a_j a_{k-j}\\ \end{array}$

Therefore $1 =\sum_{j=0}^{k} a_j a_{k-j}$.

If $k=0$, $1 = a_0^2$ so $a_0 = 1$ or $-1$. I will choose $a_0 = 1$; the other choice will reverse all the signs.

If $k \ge 1$, $1 =\sum_{j=0}^{k} a_j a_{k-j} =2a_0a_k +\sum_{j=1}^{k-1} a_j a_{k-j} =2a_k +\sum_{j=1}^{k-1} a_j a_{k-j}$ or $a_k =\frac12(1-\sum_{j=1}^{k-1} a_j a_{k-j})$.

For $k=1$, $a_1 = \frac12$.

For $k=2$, $a_2 =\frac12(1-a_1^2) =\frac12(1-\frac14) =\frac{3}{8}$.

As a check so far, $(1+x/2+3x^2/8)^2 =1+x(1/2+1/2)+x^2(1/4+2(3/8)) + (...)x^3 =1+x+x^2+(...)x^3$.

For $k=3$, $a_3 =\frac12(1-a_1a_2-a_2a_1) =\frac12(1-2a_1a_2) =\frac12(1-2\frac12 \frac{3}{8}) =\frac{5}{16}$.

And so on.