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Thread: Infinitely differentiable

  1. #1
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    Infinitely differentiable

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

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

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

    You have to be careful in using this statement, which may not preclude the analyticity of $\displaystyle f$.
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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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