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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.)


  1. Anonymous says:

    Do it instead by comparing with the terms in the expansion of cos(x). Then the solution is straightforward, without the constant C cropping up.

  2. Artem says:

    Ok guys, I got this due to your comments and Apostol answer.
    So, by theorem 11.8, if the function is represented by a power series in the interval (a-r, a+r), then it is continuous. However, we are automatically given the condition that f(x) power series exists! Moreover, the expansion is around 0 (since all terms are x^n), so no matter what radius is, a = 0 is always in the interval of convergence. Thus, the series f(x) MUST be continuous at x = 0.

    Now, since our answer from the differential equation clearly does not allow for a 0 substitution, we understand that this is a discontinuous function with removable discontinuity! In short, one point has been removed from it. Since Apostol assumed the series exists, it means the point has been inserted, and it should equal to the limit of the function at 0.

    Now, from the power series we know that f(0) = 1/2. We just examine the limit: \lim_{x \rightarrow 0} f(x) = \lim_{x \rightarrow 0} \frac{sin x}{x} + \lim_{x \rightarrow 0} \frac{cos x + C}{x^2}. Here we used the rule that the limit of sums is the sum of limits.

    But you may notice, that the limit with C exists ONLY when the nominator is 0 (otherwise we cannot apply L’Hopital rule!). So, at 0 the only possible option to choose is that C = -1, then \lim_{x \rightarrow 0} \frac{cos x + C}{x^2} = \lim_{x \rightarrow 0} \frac{cos x - 1}{x^2} = \lim_{x \rightarrow 0} \frac{-sin x}{2x} = \lim_{x \rightarrow 0} \frac{-cos x}{2} = -1/2! This is exactly what we should get, since lim_{x \rightarrow 0} \frac{sin x}{x} = 1.

    Thus, we have arrived at the solution in Apostol, and verified it!

  3. 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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