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Math Help - Infinitely differentiable

  1. #1
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    Infinitely differentiable

    Suppose f is infinitely differentiable on \mathbb{R}.

    (a) If f''(x) + f(x) = 0, f(0) = 0, and f'(0) = 1, show that f(x) = sin x.
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  2. #2
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    Well, if f(x) = \sin{x}, then f'(x) = \cos{x} and f''(x) = -\sin{x}.
    So f''(x)+f(x) = -\sin{x}+\sin{x} = 0 and f'(0) = \cos(0) = 1.
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  3. #3
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    This is the (one) definition of \sin. It is clear that f^{(2n+1)}(0)=(-1)^n, and all other derivatives of f vanish. From this, we can construct a Taylor-like series definition of f, and see that it fits with the usual Taylor series of \sin:

    f(x)=\displaystyle\sum_{n=0}^\infty\frac{(-1)^n}{(2n+1)!}x^{2n+1}.

    You have to be careful in using this statement, which may not preclude the analyticity of f.
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  4. #4
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    Note that The CoffeeMachine's response shows that if f(x)= sin(x), then f satisfies the given conditions. To show the converse, which is the actual problem, you must also show uniqueness for solutions to such a differential equation. Once again, because the Original Poster shows no attempt at a solution, we do not know what he/she has available to do that.
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