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.

  • $\begingroup$ Do you know the power series for $(1-x)^{-1/2}$? If not, can you calculate it using the binomial theorem? $\endgroup$ – Mathmo123 Sep 8 '15 at 23:26
  • $\begingroup$ 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. $\endgroup$ – Jack Armstrong Sep 8 '15 at 23:28
  • 1
    $\begingroup$ "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. ${}\qquad{}$ $\endgroup$ – Michael Hardy Sep 9 '15 at 0:11

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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.