Suppose f(x) - f''(x) = 0 for all x, also f(0) = f'(0) = 0, use Taylor polynomials to show that f(x) = 0 for all x
A point in the right direction please? I've been trying to solve this for ages
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Anonymous2010-12-03 5:34
First show that f(x) - f''(x) = 0 (i.e. f(x) = f''(x)) implies that both f^{(2k)}(x) = f(x) and f^{(2k+1)}(x) = f'(x) for any nonnegative integer k (i.e. every even derivative is f(x) and every odd derivative is f'(x)). Then consider the Taylor series of f centered at zero.
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Anonymous2010-12-03 14:31
thanks anon
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Anonymous2010-12-08 15:43
\frac{a}{b}
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Anonymous2010-12-08 18:07
Its probably much easier than that. The only function which satisfies f(x) = f'(x), \forall x is
f(x) = c exp ^ {x}, where c is some constant.
The only way f(0) = 0 is when c = 0. Which means
f(x) = 0, \forall x. So is f'(x).
Name:
Anonymous2010-12-08 18:09
[math] \frac{d}{dx} f(x) = f(x) \rightarrow f(x) = c e^{x}