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Verify a summation formula for sin3 x

by RoRi

Assume that \sin^3 x has a power-series representation in terms of powers of x, verify that it has the form

    \[ \sin^3 x = \sum_{n=1}^{\infty} (-1)^{n+1} \frac{3^{2n}-1}{(2n+1)!} x^{2n+1} \]

for all x \in \mathbb{R}.

First, we recall the triple angle identity for the sine,

    \[ \sin (3x) = 3 \sin x - 4 \sin^3 x \quad \implies \quad \sin^3 x = \frac{3}{4} \sin x - \frac{1}{4} \sin (3x). \]

Since we know the expansion for \sin x is

    \[ \sin x = \sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{(2n-1)!} x^{2n-1}, \]

we then have

    \begin{align*} \sin^3 x &= \frac{3}{4} \sin x - \frac{1}{4} \sin (3x) \\[9pt] &= \frac{3}{4} \left( \sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{(2n-1)!} x^{2n-1} \right) - \frac{1}{4} \left( \sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{(2n-1)!} (3x)^{2n-1} \right) \\[9pt] &= \frac{3}{4} \left( \sum_{n=1}^{\infty} \left( \frac{(-1)^{n+1}}{(2n-1)!} x^{2n-1} \right) \right) - \frac{3}{4} \left( \sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{(2n-1)!} 3^{2n-2}x^{2n-1} \right) \\[9pt] &= \frac{3}{4} \sum_{n=1}^{\infty} \left( \frac{x^{2n-1}}{(2n-1)!} \cdot \big( (-1)^{n+1} - (-1)^{n+1} 3^{n-2} \big) \right) \\[9pt] &= \frac{3}{4} \sum_{n=1}^{\infty} \frac{x^{2n-1}}{(2n-1)!} (-1)^{n+1}\big( 1 - 3^{2n-2} \big) \\[9pt] &= \frac{3}{4} \sum_{n=1}^{\infty} \frac{x^{2n-1}}{(2n-1)!} (-1)^n \big( 3^{2n-2} - 1 \big) \\[9pt] &= \frac{3}{4} \sum_{n=2}^{\infty} \frac{x^{2n-1}}{(2n-1)!} (-1)^n \big( 3^{2n-2} - 1 \big) \\[9pt] &= \frac{3}{4} \sum_{n=1}^{\infty} (-1)^{n+1} \frac{3^{2n} - 1}{(2n+1)!} x^{2n+1}. \end{align*}

In the last two steps, we moved the sum from n=1 to infinity to n=2 to infinity since the n=1 term was 0, and in the final step we reindexed the sum. This was the requested identity.

2 comments

  1. Anonymous says:

    There is an error in your solution when you divide the second series by 3, Therefore,instead of writing (3x)^{2n-2} it should be (3)^{2n-2}(x)^{2n-1}.

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