are stupid. The proof will almost undoubtedly be of the form I described.
>>9 The theorem that guarantee's uniqueness of solutions will almost undoubtedly be harder to prove than the first result.
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Anonymous2007-12-17 18:16
It is very simple. If f(x)=f'(x), then f'(x)/f(x)=1. Since f'(x)/f(x)-1=0, then d(ln(f(x))-x)/dx must also equal to 0 since it is equal to f'(x)/f(x)-1. Since it can easily be proven that only constant functions have 0 derivatives everywhere, then ln(f(x))-x must be a constant function C. The rest easily follows.
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Anonymous2007-12-17 18:19
I also forgot to mention that if f(x) is 0 anywhere, then the proof obviously does not work. You have to take the case separately in order to prove that if f(x) is 0 anywhere with f'(x) also equaling 0, then f(x) must be equal to the constant function f(x)=0.
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Anonymous2007-12-17 21:42
My calc 2 class basically started with d/dx f(x) = f(x) then did some shit with some properties and showed that the solution was e^(x). So yeah its just e^(x).
If you only care about magnitudes and allow complex numbers, e^(sqrt(-1)*t) would qualify. Its magnitude is 1 everywhere (just changed direction in the complex plane) and its derivative acts the same, just 90 degrees off.
Right. Assuming you ignore 2*e^x, 3*e^x, pi*e^x, and so on.
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Anonymous2007-12-17 23:32
>>13
e^(sqrt(-1)*t) does not work. Its derivative is i*e^(it) which is NOT equal to i*e^(it)
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Anonymous2007-12-17 23:40
>>15
I made this post, and I was not paying attention to >>13 about magnitudes, so it was a mistake. However, an important thing is wording "functions from R to R" in the OP which clearly excluded complex functions or their absolute values. If, however, their absolute values were to be included, then it would be the same as the regular real function solutions.
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Anonymous2007-12-18 9:23
someone please diff(e^(it) / (cost + isint)) step by step for me?
I tried to do it and it is NOT 1
or maybe I made a mistake somewhere, can someone do it?