Dear friends, I need with the following problem.
Let for .
Then, I need to prove that for all and all .
Any help due to this direction would be very appreciated.
Dear friends, I need with the following problem.
Let for .
Then, I need to prove that for all and all .
Any help due to this direction would be very appreciated.
This looks like a difficult and interesting problem. Obviously f(z), being an exponential, is always positive (for z>0, which I'll assume throughout). Also, , which is always negative.
The second derivative is given by . This is always positive, because the function is always positive (g(0)=0, and , which is positive, so g is an increasing function and therefore positive).
So even for n=2 the result is not altogether easy, and for n≥3 it gets very messy. Clearly some new idea is needed to deal with the general case.
It looks hard when we do as you have shown.
With the additionional assumption for all (or equivalently ), we can use the Taylor expansions, i.e.,
....for all
and
....for all .
Hence, for all , we have
(we may reverse the order of the sum and the integral since the series converge uniformly)
I have just jammed here.
There should be a way of rewriting this as follows:
....for .
Is there a formula for multiplication of power series for more than two?
I think by using that formula it will be possible to obtain the latter formula...
By the way, it is also okay for me to prove the claim on .
(Continued). hmm, I have done something but seems more complicated now.
For all , we have
....(apply Cauchy's multiplication formula for times)
.......'
.......' ....(set and , then reverse the order of the sums)
.......'
Any ideas at this point are welcome, thanks!
Here are a few more thoughts on this intriguing problem. First,
(reversing the order of integration in the double integral). Therefore
.
Next, let A denote the set of all infinitely-differentiable function on the positive reals whose derivatives alternate in sign: -1)^kf^{(k)}(x)>0\text{ for all }x>0\text{ and all }k\geqslant0\}" alt="A := \{f\in C^\infty(\mathbb{R}_{>0})-1)^kf^{(k)}(x)>0\text{ for all }x>0\text{ and all }k\geqslant0\}" />.
Proposition. If then .
Outline proof: The n'th derivative of f is of the form , where . Also, you can see by differentiating that . It is then straightforward to prove by induction (using the condition ) that the functions alternate in sign, with for all x>0 and all n.
It follows from that Proposition and the previous remarks that if the function is in A then so is the function .
I suspect that the function u(z) is indeed in A, but I haven't been able to prove that. However, this seems like a step in the right direction, because u(z) is a considerably simpler-looking function than f(z).
One final thought: functions whose derivatives alternate in sign seem to occur in Boltzmann's approach to the theory of entropy. I wonder if that is where this problem originated?
Remark. This fails to hold for , i.e., .
See the link the above discussion is mentioned to be a typo: http://www.mathhelpforum.com/math-he...594-post9.html
I now think that the proposition belongs to you (Opalg).
Just modified the proposition as follows.
Proposition 1. Let . Then for all , where
....on .
As Opalg mentioned that it is straight forward to prove Proposition 1 but it is a little bit tricky since the derivative of the preceding term appears in the recursion. But the following modification, which implies the one above, can be proved.
Proposition 2. Let , then for all .
Below, I would like to give the proof for those being interested with this problem (with the hope that myself and Opalg are not the only ones ).
Proof. Clearly, . Suppose now that for some , then for any , we have
.....................
.....................
.....................
which proves .
The proof of the proposition is now complete.
Corollary. If , then , where for .
Proof. for all .
The last corollary is now applicable for the original problem.
For the rest of the proof see: http://www.mathhelpforum.com/math-he...495-post8.html
Okay, after a second sleepless night I think I have cracked this problem.
According to my previous comment, we need to show that the function is in the class , namely for all x>0 and all n.
By Leibniz's formula, . But , and (for ) . Therefore
So we need to show that the function is positive for all x>0. But , so it will be sufficient to show that . Now , so the problem reduces to showing that . But that sum is less than the infinite sum , which is a geometric series with sum .
That completes the proof that .
Functions that alternate in sign with respect to its derivative are called as completely monotonic functions.
I came a cross with the reference [1] by chance, and saw that what we have done here is known from [1, Theorem 2].
References
[1] Miller and Samko, Completely monotonic functions, Integr. Transf. and Spec. Funct., vol. 12, no. 4, pp. 389--402, 2001.
http://w3.ualg.pt/~ssamko/dpapers/fi..._monotonic.pdf