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Is there a way to simplify the expression:

$D = (f_1(\omega)+f_2(\omega))^n-(f_1(\omega)-f_2(\omega))^n$

where $n$ is a positive integer.

In this particular problem:

$f_1(\omega)=-\omega^2+2$

$f_2(\omega)=\omega \sqrt{\omega^2-4}$

Expanding $D$ for some values of $n$:

$n=1$: $\sqrt{{\omega}^{2}-4}(2 \omega)$

$n=2$: $\sqrt{{\omega}^{2}-4}(-4\omega^3 + 8\omega)$

$n=3$: $\sqrt{{\omega}^{2}-4}\left( 8\,\omega^{5}-32\,\omega^{3}+24\,\omega\right) $

$n=4$: $\sqrt{{\omega}^{2}-4}\left(-16\omega^7+96\omega^5-160\omega^3+64\omega\right)$

$n=5$: $\sqrt{{\omega}^{2}-4}\left(32\omega^9-256\omega^7+672\omega^5-640\omega^3+160\omega\right)$

Paulo J. P. Gonçalves
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1 Answers1

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By the binomial theorem $$ (f_1+f_2)^n -(f_1-f_2)^n = \sum_{k=0}^{n}\binom{n}{k}f_1^{n-k}f_2^k -\sum_{k=0}^{n}\binom{n}{k}f_1^{n-k}(-f_2)^k. $$ Notice that since we have $(-f_2)^k$ the second sum is alternating, so every other term cancels, leaving $$ 2\sum_{j=0}^{\lfloor\frac{n}{2}\rfloor}\binom{n}{2j+1}f_1^{n-(2j+1)}f_2^{2j+1}=2\left( \binom{n}{1}f_1^{n-1}f_2 + \binom{n}{3}f_1^{n-3}f_2^3 + \dots \right) $$ where $\lfloor\ \rfloor$ is the floor function.

M_B
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  • Many thanks, binomial theorem seems a good way to re-write the problem. But the binomial expression seems more complicated than the original expression. Any idea how to define the binomial as single expression to the power $n$. – Paulo J. P. Gonçalves Apr 08 '18 at 21:43
  • I don't believe such a formula exists in general, but I could very well be wrong. – M_B Apr 08 '18 at 21:47