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In one book the author defines the following value: $$ F(m) = \sup\limits_{v\geq 1}\frac{v^m}{e^v}\int\limits_1^v\frac{e^u}{u^m}\mathrm du $$ for $m\in [3,4]$. Further he puts an upper bound for this value: $$ F(m)\leq 1+\frac m2+\frac m2(m+1)^{m+1}e^{-m}, $$ so e.g. for $m = 3$ on has $F(M)\leq 21.67$. This bound seems to be too rough: I do not know the actual value for $F(3)$ but at lease Mathematica provides the answer $\approx3.58$.

It's clear that I cannot rely upon this approximation and I would like to find more nice bounds for $F(m)$ especially for $m\in [3,4]$.

If one define $$ f(v,m) = \frac{v^m}{e^v}\int\limits_1^v\frac{e^u}{u^m}\mathrm du $$ then $$ f'(v,m) = 1+\frac{v^{m-1}(m-v)}{e^v}\int\limits_1^v\frac{e^u}{u^m}\mathrm du $$ where the deirvative is taken w.r.t. $v$. Since we are interested only in the case $v\geq 1$ then for $v\in[1,m]$ clearly $f'\geq 1>0$. So there are two questions:

  1. if there $v^*$ such that $f'(v^*,m) = 0$?

  2. if it is unique?

These questions are strongly related to the behaviour of $$ \frac{v^{m-1}(m-v)}{e^v}\int\limits_1^v\frac{e^u}{u^m}\mathrm du $$ which I cannot understand well. E.g. calculation of $f''$ does not help much to me (again, the derivative is w.r.t $v$).

Edited: thanks to Willie Wong, I've fixed the bound for $F(3)\leq 21.67$. This bound is still to rough since I need to operate thereafter with values like $\exp(F(m))$ where the difference between numbers $3.58$ and $21.67$ is significant.

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$1 + 3/2 + 3/2 * 4^4 * e^{-3} \sim 21.67$. I am not sure where you got the 117 from... –  Willie Wong Jul 18 '11 at 16:26
    
@Willie Wong: thank you very much for this remark. I calculated it at the office, so I'll check tomorrow where is the mistake. Anyway, $21.67$ is remarkably bigger than $3.58$ - an I put the motivation why in the question. –  Ilya Jul 18 '11 at 18:07
    
I wonder if on the RHS of your definition of $F(m)$ the $m$s should be $m+1$ instead. See below where I recover a naive bound which look just like the bound that your book cites, except that $m$ is replaced by $m-1$. Can you double check what the value for $F(4)$ should be? –  Willie Wong Jul 18 '11 at 18:47

1 Answer 1

up vote 2 down vote accepted

Some random thoughts.

  • Rewrite the equation for $f'$ as $$ f'(v,m) = 1 + \left(\frac{m}{v} - 1\right) f(v,m) $$ which implies that $f' = 0 \iff f = \frac{1}{1 - m/v}$. Observe that $f$ by definition is non-negative, and $f(1,m) = 0$. This implies that if $f' \neq 0$ anywhere, $f$ must be strictly increasing. But $(1-m/v)^{-1}$ is strictly decreasing in $v > m$, so if $f'\neq 0$ anywhere, we must have that $f \leq \inf_{v > m} (1-m/v)^{-1} = 1$.
  • Suppose $f'(v^*,m) = 0$. Then $$ f''(v^*,m) = \left(\frac{m}{v^*} - 1 \right) f'(v^*,m) - \frac{m}{(v^*)^2} f(v^*,m) < 0 $$ hence any critical point of $f$ must be a maximum. This implies that the maximum, when attained at $v^* < \infty$, is unique. (And in particular answers your second question.)

Now, for $v > m$, $f'(v,m) < 1$. So we must have $$ f(v^*,m) - f(m,m) \leq (v^*-m) = m\left( f(v^*,m)-1\right)^{-1}$$ which can give a crude estimate of $f(v^*,m)$ using the quadratic formula $$ f(v^*,m) \leq 1 + f(m,m) $$

What remains is to get a good estimate on $f(m,m)$. Observe that for $u < m$, $e^u/u^m$ is decreasing and convex. Just by estimating the integral via a trapezoid, you have that

$$ f(m,m) \leq \frac{m^m}{e^m} \frac{(m-1)}{2} \left( e + \frac{e^m}{m^m}\right) = \frac{m-1}{2} \left( 1 + m^m e^{1-m}\right) $$

which gives that, for $m = 3$

$$ f(v^*,3) \leq 1 + \frac{2}{2} (1 + 3^3 e^{-2}) < 5.66 $$

and for $m = 4$

$$ f(v^*,4) \leq 1 + \frac{3}{2} (1 + 4^4 e^{-3}) < 21.7 $$

which are pretty reasonable estimates, I think.

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author provided bounds on $F(m)$ which I wrote but then he only use $F(m+1)$. So when I was doing calculations in Mathematica for this bounds, I calculated bounds for $m=4$ as you mentioned. On the other hand, your bounds are really his bounds for $m\to m-1$, but in the proof of his bounds he proved them exactly for $m$, so there seems not to be a mistake and you improved his bounds. –  Ilya Jul 19 '11 at 8:41
    
thank you very much for the answer. Based on the uniqueness of the maximum I can find it numerically provided that $f'(v_1,m)>0$ and $f'(v_2,m)<0$. Then $v_1$ and $v_2$ I can just find with Mathematica by asking it to solve an equation $f'=0$ and taking some value on the right and left of the approximated root. By the way, *But $(1-m/v)^{-1}$ is strictly decreasing in $v > m$, so if $f'\neq 0$ anywhere, we must have that $f \leq \inf_{v > m} (1-m/v)^{-1} = 1$. * Why does this implication hold? –  Ilya Jul 19 '11 at 8:56
    
If $f'\neq 0$, $f'$ is signed. Since it starts out positive, it remains positive. So $f$ is increasing. But $f'$ will equal zero if the graph of $f$ and $(1-m/v)^{-1}$ intersect. If a strictly increasing function doesn't intersect a strictly decreasing function, and if at one point the increasing function is below the decreasing function $\implies$ the increasing function must be always smaller than the smallest possible value for the decreasing function: this is just an application of the intermediate value theorem. (Assuming continuity of course.) –  Willie Wong Jul 19 '11 at 9:49

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