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Compute the coefficients of a power series with coefficients satisfying a given identity

Consider the power series

    \[ f(x) = \sum_{n=0}^{\infty} a_n x^n \]

whose coefficients satisfy the identity

    \[ \cos x = \sum_{n=0}^{\infty} a_n (n+2)x^n. \]

Compute the coefficients a_5, a_6 and determine the value of f(\pi).


First, we know the power-series expansion of \cos x is given by

    \[ \cos x = \sum_{n=0}^{\infty} 1 - \frac{1}{2}x^2 + \frac{1}{24} x^4 - \frac{1}{720} x^6 + \cdots. \]

So, we have

    \[ 1 - \frac{1}{2} x^2 + \frac{1}{24}x^4 - \frac{1}{720} x^6 + \cdots = \sum_{n=0}^{\infty} a_n (n+2)x^n. \]

Equating like powers of x we can compute the coefficients a_5 and a_6,

    \begin{align*}  a_5 &= 0 \\[9pt]   8 a_6 &= \frac{-1}{6!} & \implies && a_6 &= \frac{-7}{8!}. \end{align*}

Then to compute the value of f(\pi) we solve the differential equation

    \begin{align*}  \cos x &= \sum_{n=0}^{\infty} a_n (n+2) x^n \\[9pt]  &= \sum_{n=0}^{\infty}n a_n x^n + 2 \sum_{n=0}^{\infty} a_n x^n \\[9pt]  &= \sum_{n=1}^{\infty} n a_n x^n + 2 \sum_{n=0}^{\infty} a_n x^n \\[9pt]  &= xy' + 2y \end{align*}

If x = 0 we have

    \[ \cos x = \sum_{n=0}^{\infty} a_n (n+2) x^n \quad \implies \quad 1 = 2a_0 \quad \implies \quad a_0 = \frac{1}{2}. \]

Therefore, f(0) = \frac{1}{2}. If x \neq 0 then we can divide the above differential equation by x to obtain the first order linear differential equation

    \[ y' + \frac{2}{x} y = \frac{\cos x}{x}. \]

Now, we can solve this differential equation as follows,

    \begin{align*}  && y' + \frac{2}{x} y &= \frac{\cos x}{x} \\[9pt]  \implies && x^2 y' + 2x y &= x \cos x \\[9pt]  \implies && x^2 y' + (x^2)' y &= x \cos x \\[9pt]  \implies && \frac{d}{dx} \left( x^2 y \right) &= x \cos x &(\text{prod. rule})\\[9pt]  \implies && \int \frac{d}{dx} \left( x^2 y \right) \, dx &= \int x \cos x \, dx \\[9pt]  \implies && x^2 y &= \cos x + x \sin x + C \\[9pt]  \implies && y &= \frac{\cos x}{x^2} + \frac{\sin x}{x} + \frac{C}{x^2}. \end{align*}

Where C is an arbitrary constant.

(Incomplete. Judging by the answer in the back of the book, Apostol computes this constant as C = -1. I’m not sure how to get that though. I think we need some kind of initial condition to determine the constant, and so get a unique solution for f(x). Maybe we can assume this must be continuous at 0 and then take a limit as x \to 0? I do think that would get us to C = -1, but I don’t know why can assume f is continuous at 0. Leave a comment if you have any suggestions.)

One comment

  1. tom says:

    How about using the interval -π π, then that f(x) seems to be even. Some jump discontinuities at the endpoints but still integrable.

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