power series sine

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power series expansions for sine

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Question statement

Find the power series expansion of $\sin x$ by cnsidering the following.

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Explain why $\sin x$ is not a polynomial (Hint: Think about how many zeros $sin x$ has).

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a) We know that $\sin x$ has infitely many zeros. On the other hand a polynomial of degree $n$ can have at most $n$ roots. Hence $\sin x$ cannot be a polynomial. (If we try to express $sin x$ as a polynomial, we see that the degree of the polynomial must be infinite. In other words, we need a power series!)

b) Suppose we have a polynomial

$y(x) = a_0 + a_1x + a_2x^2 + a_3x^3 + \ldots + a_nx^n$

which is a solution to the given differential equation.

Then, since $y^{\prime\prime} = - y$ we have

$2a_2 + 6a_3x +12a_4x^2+\ldots + n(n-1)a_nx^{n-2} = -a_0 - a_1x - a_2x^2 - a_3x^3 - \ldots - a_nx^n .$

Then

$2a_2= -a_0=0$ (since y(0)=0), so $a_2=0$
$6a_3 = -a_1 = -1$ (since y^prime(0)=1), so $a_3 = \frac{-1}{3!}$ .
$12a_4 = -a_2 = 0$ , so $a_4=0$ .
$20a_5 = -a_3 = \frac{1}{5!}$ .
In general $a_n=0$ if $n$ is odd, and if $n$ is odd, $a_n=\frac{-1}{n!}$ if $n$ or $a_n=\frac{1}{n!}$ alternatingly

But observe that the degrees of $y^{\prime\prime}$ and $-y$ does not match. So, infact $y$ should be a power series:

$y(x) = a_0 + a_1 x + a_2 x^2 + a_3 x^3 + \ldots + a_nx^n+\ldots$

with the coefficents computed as above. Discarding the zero terms we can rewrite it as

$y(x) = -1+ -\frac{1}{3!} x^3+ \frac{1}{5!} x^5 -\frac{1}{7!}x^7+\frac{1}{9!}x^9 -\frac{1}{11!}x^{11} + \ldots.$

c) We know that $sin x$ is the only solution to the differential equation in part b) (why it is the only one?). So we must have

$\sin x =-1+ -\frac{1}{3!} x^3+ \frac{1}{5!} x^5 -\frac{1}{7!}x^7+\frac{1}{9!}x^9 -\frac{1}{11!}x^{11} + + \ldots.$

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Series Homework 2 (power series) by Ugur Efem in Ugur's workspace.

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Ugur Efem 2 years, 5 months ago

Ugur Efem 2 years, 5 months ago

Ugur Efem 2 years, 5 months ago

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