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Determine the 98th coefficient in the power series of sin (2x + (1/4)π)

Consider the power series expansion

    \[ \sin \left( 2x + \frac{1}{4} \pi \right) = \sum_{n=0}^{\infty} a_n x^n. \]

Find the coefficient a_{98} in this expansion.


First, we recall the identity for the sine of a sum,

    \[ \sin (x+y) = \sin x \cos y + \sin y \cos x. \]

So we have

    \begin{align*}  \sin \left( 2x + \frac{1}{4} \pi \right) &= \sin (2x) \cos \left( \frac{\pi}{4} \right) + \sin \left( \frac{\pi}{4} \right) \cos (2x) \\[9pt]  &= \frac{\sqrt{2}}{2} \left( \sin (2x) + \cos (2x) \right). \end{align*}

Then, we know the expansions of \sin x and \cos x are

    \begin{align*}  \sin x &= \sum_{n=0}^{\infty} \frac{(-1)^n}{(2n+1)!} x^{2n+1} \\[9pt]  \cos x &= \sum_{n=0}^{\infty} \frac{(-1)^n}{(2n)!} x^{2n}. \end{align*}

Therefore, we have

    \begin{align*}  \sin \left( 2x + \frac{\pi}{4} \right) &= \frac{\sqrt{2}}{2} \left( \sin (2x) + \cos (2x) \right) \\[9pt]  &= \frac{\sqrt{2}}{2} \left( \sum_{n=0}^{\infty} \frac{(-1)^n (2x)^{2n+1}}{(2n+1)!} + \sum_{n=0}^{\infty} \frac{(-1)^n (2x)^{2n}}{(2n)!}\right) \\[9pt]  &= \frac{\sqrt{2}}{2} \left( \sum_{n=0}^{\infty} (-1)^n \left( \frac{2^{2n+1}}{(2n+1)!} x^{2n+1} + \frac{2^{2n}}{(2n)!} x^{2n} \right)\right) \\[9pt]  &= \frac{\sqrt{2}}{2} \left( \sum_{n=0}^{\infty} \left(\frac{2^n}{n!} \right)(-1)^{\left \lfloor \frac{n}{2} \right \rfloor} x^n \right). \end{align*}

(Where \left \lfloor \frac{n}{2} \right \rfloor denotes the greatest integer less than or equal to \frac{n}{2}.) So, the 98th coefficient is given by

    \[ a_{98} = \frac{2^{98}}{\sqrt{2} \cdot (98)!} (-1)^{49} = -\frac{\sqrt{2} 2^{97}}{98!}. \]

3 comments

    • RoRi says:

      The last step before the calculation of a_{98} (the one where we combine the even and odd terms) or the actual computation of a_{98}?

    • RoRi says:

      Assuming it is the combining the even and odd terms. We start with

          \[ \frac{\sqrt{2}}{2} \left( \sum_{n=0}^{\infty} (-1)^n \left( \frac{2^{2n+1}}{(2n+1)!} x^{2n+1} + \frac{2^{2n}}{(2n)!} x^{2n} \right) \right). \]

      For each n in the sum we get two terms, one with even powers and one with odd powers, and they are either both positive or both negative. So, we’re actually getting every term \frac{2^n}{n!}, but instead of alternating with each term, it is two positive terms and then two negative terms. That’s what the (-1)^{\lfloor \frac{n}{2} \rfloor} does. Looking at the sum we combine them into

          \begin{align*}  \frac{\sqrt{2}}{2} \sum_{n=0}^{\infty} (-1)^{\lfloor \frac{n}{2} \rfloor} \frac{2^n}{n!} x^n &= \frac{\sqrt{2}}{2} \left( \frac{2^0}{0!} x^0  + \frac{2^1}{1!} x - \frac{2^2}{2!}x^2 - \frac{2^3}{3!}x^3 + \cdots \right)\\[9pt]  &= \frac{\sqrt{2}}{2} \left( \sum_{n=0}^{\infty} (-1)^n \left( \frac{2^{2n+1}}{(2n+1)!}x^{2n+1} + \frac{2^{2n}}{(2n)!}x^{2n} \right) \right). \end{align*}

      So, I think it does work as claimed.

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