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power series expansions for sine

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

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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!)

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b) Suppose we have a polynomial 

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

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which is a solution to the given differential equation. 

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

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\\[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 .\\]

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Then

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

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

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

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\\[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.\\]

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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 

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

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Give an \"infinite polynomial\" solution to 

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\\[\\frac{d^2 y}{dx^2} = -y; \\mbox{ and } y(0)=0, y^\\prime(0) = 1.\\]

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Hence express $\\sin x$ as an \"infinite polynomial\".

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