Here’s the problem :
Prove : If f(x) = f '(x) then f(x) = Ce[sup]x[/sup].
It seems pretty simple, but I’m not sure if I can do it properly. Perhaps because it looks so easy, and similar to things I’ve seen before, that I’m making it hard.
Bearing that idea out, my approach is something out of Diff. Eq., though I had the feeling it wasn’t rigorous enough; anyway I’m not doing it right since I can’t solve it, but I can’t think of another approach.
I should note that in this text, ln x is defined as Int(1,x): 1/u du
for positive x.
e[sup]x[/sup]is the inverse of ln x (after e is defined using ln e = 1).
Continuing, after substituting :
f(0)*Int(1,f(x)/f(0)): 1/u du = Int(0,x): dt.
f(0)*ln( f(x)/f(0) ) = x [ Now I’ve assumed that f(x)/f(0) > 0.]
I’m pretty much stuck there; if I didn’t have that f(0) outside I could exp() both sides and get the answer, but I can’t get rid of it. I can’t see how to get ln x or e[sup]x[/sup] involved in another way either.
I’m not sure why you think he needs to assume an existence and uniqueness theorem, ultrafilter. He’s just solving the given equation explicitly by integration.
panamajack: you may be interested to note that you’re only making one assumption, not two. If you’re assuming that f(x)!=0, then since f is differentiable (and hence continuous), f(x) is either always positive or always negative. In either case, f(x)/f(0) has to be positive.
There is a simple trick. If f’(x) - f(x) = 0, multiply both sides by
exp(-x) (that’s e to the power -x) to get f’(x)exp(-x) - f(x)exp(-x) = 0. This is, by elementary calculus, d/dx(f(x)exp(-x) = 0. Now it is elementary that the only function whose derivative is 0, is constant. So multiply f(x)exp(-x) by exp(x) since exp(-x)exp(x) = exp(x-x) = exp(0) = 1 so f(x) = c exp(x). This is the only argument I know that doesn’t require analyzing the possible division by 0. The multiplier used is called an integrating factor.
bugg - not quite rigourous enough for my usage, but that was the basic approach I was using.
The specific problem is that I can’t do the step in between these two :
=> dy/y = dx
=> ln(y) = x + c1
with my current definition of ln(x) (I don’t have indefinite integrals either). Though by “can’t” and “don’t” I really mean “don’t want to use”, but that might be the only way to continue with this approach.
I like Hari Seldon’s proof, as it has no odd assumotions and is as simple as it should be, even if it has a ‘tricky’ air to it.
Good point about my assumption(s), Orbifold.
FriendRob, are you sure? I need to divide by f(0) to get the lower bound to be 1, which means I need to multiply by f(0) to cancel that term in substitution. Unless I’m really missing something.
Thanks to all the rest[sup]*[/sup], too.
*In the second season this’ll get changed to “ultrafilter and Punoqllads”, so don’t worry.
In order to do this, you have to use the fact that d/dx(e[sup]x[/sup]) = e[sup]x[/sup]. This may or may not be okay, depending on what you have to work with.
Another way is to use the substitution f(x)=g(x)e[sup]x[/sup]. This makes it quite easy (can you assume the product rule? And that de[sup]x[/sup]/dx=e[sup]x[/sup])