Unbounded function

Muggleton · 2009-01-21 03:14:21

Jai Ganesh · 2009-01-21 03:51:34

Does this thread fit in here?
If this is a question you're posing to the visitors as a puzzle/challenging problem, Puzzles or Exercises would be the ideal place. If this is a problem for which you require help in solving, this is the right place.
Muggleton, I am telling this because I understand, being a relative new entrant to the forum, you may be unaware of this.

Muggleton · 2009-01-21 03:56:56

I need help with finding a rigorous proof of the above.

mathsyperson · 2009-01-21 06:27:33

I've probably missed some rigour here and there, but I think this is correct.

Let f'(0) = k. k is by definition a positive number.
Since f''(x) is positive everywhere, x>0 implies f'(x) > k.

Therefore f(x) > f(0)+kx, for all x>0, since the RHS has a gradient of k everywhere.

f(0) + kx --> ∞ as x --> ∞, so f(x) must tend to ∞ as well.

Muggleton · 2009-01-21 09:23:27

Thank you, mathsyperson.

When I said rigorous, I was thinking of first principles. So

would mean

But your proof is good enough. I'm familiar with the results you used; those results are based on first principles or can be traced back to first principles.

mathsyperson · 2009-01-22 08:37:50

I *think* this result can be generalised so that instead of f' and f'', you have f[sup](m)[/sup] and f[sup](n)[/sup], where m and n are respectively odd and even.
You'd also need f to be differentiable max(m,n) times rather than just twice.

Haven't worked out a solid proof yet though.

mathsmypassion · 2009-01-22 09:32:03

For a rigouros proof it may work proof by contradiction (reductio ad absurdum). I have a feeling that will do it.

Muggleton · 2009-01-22 11:31:37

mathsyperson wrote:

I *think* this result can be generalised so that instead of f' and f'', you have f[sup](m)[/sup] and f[sup](n)[/sup], where m and n are respectively odd and even.
You'd also need f to be differentiable max(m,n) times rather than just twice.
Haven't worked out a solid proof yet though.

Intuitively, f', f'' > 0 means the function is convex and strictly increasing; such a curve should intuitively go to infinity as x goes to infinity. I just wanted a concrete analytical proof of what was intuitive to me.

In fact f' and f'' don't have to be positive on the whole real line, they just have to be positive on the interval [a,∞) for some real number a. This can even be easily shown - just replace 0 by a in the your proof above.

Last edited by Muggleton (2009-01-22 11:34:24)

Ricky · 2009-01-23 08:53:25

You almost never do these proofs in terms of epsilons and deltas. You want to be using the tools that you have made (IVT, MVT, f' > 0 means f is monotone increasing...) from those epsilons and deltas to analyze such a function.

mathsyperson, as soon as you prove that f'(x) > g'(x) for all x > 0 and f(0) = g(0) implies that f(x) > g(x) for all x > 0, then that proof will become rigorous. But I believe that is the crucial last step that you have left unjustified.

With these problems, it's always easiest to throw in an auxiliary function and work with that. mathsyperson is right on the nose when he talks about the tangent line at the point f(0).

Set g(x) = f(x) - xf'(0), with x > 0. Computing the derivative of g:

Since f'' > 0, f' is monotone increasing. This is a common fact proven in undergrad analysis, and is typically a problem on its own. This is why I would not expect a proof of it here. Since x > 0, we ahve that f'(x) - f'(0) > 0 (by monotone). Thus, g'(x) > 0 for all x > 0.

g(0) = f(0) - 0f'(0) = f(0). As g is monotone, g(x) = f(x) - xf'(0) > g(0) = f(0). Thus, we conclude that f(x) > f(0) + xf'(0) for all x. Therefore, as the RHS is unbounded, so is the LHS. QED.

To prove it for points where your interval is [a, infinity) that f' and f'' are positive, Argue that h(x) = g(x-a) has these same properties.

JaneFairfax · 2010-05-11 22:35:03

Muggleton wrote:

When I said rigorous, I was thinking of first principles. So
would mean

I suppose we could do that if we stated the problem a bit differently:

I have succeeded in proving from first principles that

for such a function, which may, or may not, be an intermediate result that is required.

By the way, last night, while thinking in my sleep (yes, thinking in my sleep ) about uniform continuity of real-valued functions, I remembered this thread, and then I conjectured if a function whose first and second derivatives are positive cannot be uniformly continuous. I might come up with a proof soon.

BO · 2010-05-16 18:16:02

f'(x)-ε<[f(x+h)-f(x)]/h<f'(x)+ε
f(x+h)>f(x)+f'(x)-ε
Since f'(x) is always +ve, f is a monotone increasing function.
Similarly f"(x) is a monotone increasing function.
Now, it is easy to prove that f(x+c)-f(x)=c f'(x+t), where 0<t<c, for any c and x.
We have f'(a)> (f"(a)-ε)*k+f'(a-k)=m for a point a in R.
Then, when x>a, f(x+nc)>ncm+f(x), which is unbounded above. Hence proved.

Jane Fairfax, consider the function f(x)=x^3 which is differentiable at every point of R. f'(x) is positive and differentiable at every point and so is f"(x), but f'(x) does not tend to 0 as x tends to infinity.

Last edited by BO (2010-05-16 18:25:42)

JaneFairfax · 2010-05-16 20:28:07

BO wrote:

Jane Fairfax, consider the function f(x)=x^3 which is differentiable at every point of R. f'(x) is positive and differentiable at every point and so is f"(x), but f'(x) does not tend to 0 as x tends to infinity.

Sorry, whats your point? The cubic function is not bounded above.

Last edited by JaneFairfax (2010-05-16 20:31:34)

BO · 2010-05-17 06:03:14

Oh, extremely sorry, I didn't notice that you were not talking about the topic problem.

BO · 2010-05-17 06:22:00

JaneFairfax wrote:

I suppose we could do that if we stated the problem a bit differently:

I have succeeded in proving from first principles that
for such a function, which may, or may not, be an intermediate result that is required.
By the way, last night, while thinking in my sleep (yes, thinking in my sleep ) about uniform continuity of real-valued functions, I remembered this thread, and then I conjectured if a function whose first and second derivatives are positive cannot be uniformly continuous. I might come up with a proof soon.

is non-negative for all values of x. Now if is positive for all values of x, then is a monotone increasing function, which in turn implies that is unbounded above. so there must be some for which

Last edited by BO (2010-05-17 06:23:17)

Math Is Fun Forum

#1 2009-01-21 03:14:21

Unbounded function

#2 2009-01-21 03:51:34

Re: Unbounded function

#3 2009-01-21 03:56:56

Re: Unbounded function

#4 2009-01-21 06:27:33

Re: Unbounded function

#5 2009-01-21 09:23:27

Re: Unbounded function

#6 2009-01-22 08:37:50

Re: Unbounded function

#7 2009-01-22 09:32:03

Re: Unbounded function

#8 2009-01-22 11:31:37

Re: Unbounded function

#9 2009-01-23 08:53:25

Re: Unbounded function

#10 2010-05-11 22:35:03

Re: Unbounded function

#11 2010-05-16 18:16:02

Re: Unbounded function

#12 2010-05-16 20:28:07

Re: Unbounded function

#13 2010-05-17 06:03:14

Re: Unbounded function

#14 2010-05-17 06:22:00

Re: Unbounded function

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