Moderators: gmalivuk, Prelates, Moderators General
coffeesneeze wrote:It seemed natural to try this by contradiction. Assume there exists a>0 for which f'(a) is less than or equal to exp(a). Define g(x)=f(x)-exp(x). We have g(0) is greater than or equal to 0 and g(a) is less than or equal to 0. So on (0,a) g is decreasing. By MVT there must be some b in (0,a) where g'(b)<0. And from here g(b)<g'(b)<0. So for some b in this interval f(b)<exp(b).
Where the crap is my contradiction though?
ameretrifle wrote:Magic space feudalism is therefore a viable idea.
letterX wrote:coffeesneeze wrote:It seemed natural to try this by contradiction. Assume there exists a>0 for which f'(a) is less than or equal to exp(a). Define g(x)=f(x)-exp(x). We have g(0) is greater than or equal to 0 and g(a) is less than or equal to 0. So on (0,a) g is decreasing. By MVT there must be some b in (0,a) where g'(b)<0. And from here g(b)<g'(b)<0. So for some b in this interval f(b)<exp(b).
Where the crap is my contradiction though?
This is nearly the entire proof. All you have to do to make it work is choose 'a' a little more carefully...
letterX wrote:definite integrals of functions positive everywhere but one point?
coffeesneeze wrote:letterX wrote:I think what you are suggesting is to take the least a for which f'(a) is less than or equal to exp(a). Can I simply assume that there is a least a with this property? It seems like I'd have to do some more work to show that there actually is a least a.
coffeesneeze wrote:jestingrabbit wrote:definite integrals of functions positive everywhere but one point?
What I'm assuming for contradiction gets me that there is at least one point, not just one. But a definite integral of an everywhere positive (except at one point) function would be... positive... /shrug.
ameretrifle wrote:Magic space feudalism is therefore a viable idea.
jestingrabbit wrote:What can you tell me about integrals of derivatives?
jestingrabbit wrote:I would try to make f(a) - exp(a) into an integral from 0 to a... of something (assuming that f(0) = 1).
skeptical scientist wrote:There's a really nice, really easy way to do this that nobody has thought of yet.Spoiler:
To be completely rigorous, you should prove that f(x)>0 first, but this should be easy for you.
Talith wrote:I was under the impression that the derivative of any differentiable function on the reals had a well defined Riemann integral on any interval.
PM 2Ring wrote:Hey, I thought of that approach.
I didn't post though, because I couldn't think of a simple way of proving f(x)>0, apart from using a Taylor's expansion, and that seems like overkill for this problem.
skeptical scientist wrote:Talith wrote:I was under the impression that the derivative of any differentiable function on the reals had a well defined Riemann integral on any interval.
Nope. Volterra's function is a counterexample.
ameretrifle wrote:Magic space feudalism is therefore a viable idea.
ameretrifle wrote:Magic space feudalism is therefore a viable idea.
coffeesneeze wrote:Is there anything from stopping me from starting with define g(x)=f'(x)/f(x), Note g(x) is an antiderivative for h(x)=ln(f(x))?
Then I could avoid invoking whatever that evil Volterra function is.
ameretrifle wrote:Magic space feudalism is therefore a viable idea.
jestingrabbit wrote:The Volterra function is also an antiderivative of its derivative, its just not unique. You need uniqueness, and you can't get that.
coffeesneeze wrote:Define g(x)=\ln{(f(x))}.
It's derivative is this:g'(x)=\frac{f'(x)}{f(x)}>1.So this\int_0^x g'(t)\ dt>\int_0^x1\ dx = x.
...
The problem being I can't assume g'(x) is integrable?
jestingrabbit wrote:The Volterra function is also an antiderivative of its derivative, its just not unique. You need uniqueness, and you can't get that.
The 62-foot tall statue of Jesus constructed out of styrofoam, wood and fiberglass resin caught on fire after the right hand of the statue was struck by lightning.
meatyochre wrote:And yea, verily the forums crowd spake: "Teehee!"
Dark Avorian wrote:I guess I assumed continuity of the derivative...so YMMV
The 62-foot tall statue of Jesus constructed out of styrofoam, wood and fiberglass resin caught on fire after the right hand of the statue was struck by lightning.
meatyochre wrote:And yea, verily the forums crowd spake: "Teehee!"
Dark Avorian wrote:mean value theorem relies on continuity of derivative, no?
Oh right, ignore me, differentiability implies that no jump discontinuities appear in the derivative.
The 62-foot tall statue of Jesus constructed out of styrofoam, wood and fiberglass resin caught on fire after the right hand of the statue was struck by lightning.
meatyochre wrote:And yea, verily the forums crowd spake: "Teehee!"
lalop wrote:According to its wiki page, the Henstock-Kurzweil integral of an everywhere-existing derivative always exists (thus presumably allowing integration as a proof method without additional hypotheses). I was wondering if anyone was familiar enough to confirm or deny.
Lemma: Let f be differentiable on R, f(0)≥1, and f'(x)>f(x) for all x in R. Then f(x)>0 for x≥0.
Proof: Suppose f(x)≤0 for some x≥0, and let b=inf{x : x≥0 and f(x)≤0}. Then b≥0 and, by continuity, f(b)=0, so b>0. By choice of b, f(x)>0 on [0,b). We know f is continuous on [0,b] and differentiable on (0,b), so by the mean value theorem, there is some c in (0,b) such that f'(c)=(f(b)-f(0))/(b-0)≤0. But then f(c)>0≥f'(x), contradicting f'(x)>f(x) for all x in R.
Theorem: Let f be differentiable on R, f(0)≥1, and f'(x)>f(x) for all x in R. Then f(x)>e^{x} for x>0.
Proof: By the lemma, the function g(x)=ln f(x) is well-defined for x≥0, and g'(x)=f'(x)/f(x)>1 for x≥0. By the mean value theorem, for all x>0 there is some c in (0,x) (depending on x) such that g(x)=g(0)+x*g'(c). Since g(0)≥0 and g'(c)>1, we have g(x)>x, whence f(x)=e^{g(x)}>e^{x}.
PM 2Ring wrote:I think I get it, SS. But where did that a come from in there is some c in (a,b). I guess it's just a typo, and that the interval should be (0, b).
skeptical scientist wrote:Lemma: Let f be differentiable on R, f(0)≥1, and f'(x)>f(x) for all x in R. Then f(x)>0 for x≥0.
Proof: Suppose f(x)≤0 for some x≥0, and let b=inf{x : x≥0 and f(x)≤0}. Then b≥0 and, by continuity, f(b)=0, so b>0. By choice of b, f(x)>0 on [0,b). We know f is continuous on [0,b] and differentiable on (0,b), so by the mean value theorem, there is some c in (0,b) such that f'(c)=(f(b)-f(0))/(b-0)≤0. But then f(c)>0≥f'(x), contradicting f'(x)>f(x) for all x in R.
letterX wrote:What if your h's happen to sum to some finite value? Then you only get that f(x) > 0 for all x up to whatever the h's happen to sum up to.
skeptical scientist wrote:I think the mean and intermediate value theorems are constructively false, so I doubt you'll find a purely constructive proof.
PM 2Ring wrote:skeptical scientist wrote:I think the mean and intermediate value theorems are constructively false, so I doubt you'll find a purely constructive proof.
Ah, well. It's a little bit annoying that seemingly straight-forward and intuitive stuff like this can't be proved constructively.