Nice problem Which is bigger $e^a$ and $a^3$

Question

let $$a=\sum_{n=0}^{\infty}\dfrac{\left(\dfrac{n+1}{3}\right)^n}{(n+1)!}$$

My Question:

Which is bigger $e^a$ and $a^3$

I guess

$$e^a=a^3$$

but I can't prove it,and I think this is nice problem.Thank you evryone

This problem from this http://www.math.org.cn/forum.php?mod=viewthread&tid=3577&extra=&page=1

why would the e-power and the exponential term be equal to each other? — imranfat, Sep 10 '13 at 02:21
It is highly likely these are equal... Numeric methods give identical approximations (I'm pretty sure those are approximations, not exact answers): cubic and exp — apnorton, Sep 10 '13 at 02:31

score 6 · Accepted Answer · edited Apr 13 '17 at 12:21

We can "avoid" the Lambert $W$ function by skipping directly to its series representation. We'll call upon the Lagrange inversion formula, which can be found in my answer here.

Suppose we're looking for the smallest positive root of the equation $e^x = x^3$, which is also the smallest positive root of the equation

$$ \frac{x}{e^{x/3}} = 1. $$

Let's call this root $a$. The Lagrange inversion formula yields the desired series representation,

$$ \begin{align} a &= \sum_{n=1}^{\infty} \frac{1}{n!} \left\{\left(\frac{d}{dx}\right)^{n-1} e^{nx/3} \right\}_{x=0} \\ &= \sum_{n=1}^{\infty} \frac{1}{n!} \left\{\left(\frac{n}{3}\right)^{n-1} e^{nx/3} \right\}_{x=0} \\ &= \sum_{n=1}^{\infty} \frac{1}{n!} \left(\frac{n}{3}\right)^{n-1} \\ &= \sum_{n=0}^{\infty} \frac{1}{(n+1)!} \left(\frac{n+1}{3}\right)^{n}. \end{align} $$

score 4 · Answer 2 · answered Sep 10 '13 at 02:32

4

The solution to $e^a=a^3$ is given by the Lambert W function:

$a=-3 \text{W} \left( - \frac{1}{3} \right)$

Now, take a look at the summation definition of the Lambert W function. If you change your limits so they start at $n=1$ instead of $n=0$, then it is easy to see that your summation is equal to the value of $a$ given above.

answered Sep 10 '13 at 02:32

Ryan

1,389

can you except use $Lambert W function$ methods? Thank you – math110 Sep 10 '13 at 02:37
Sorry, I don't understand. Do you mean to say that you want a proof without using the Lambert W function? – Ryan Sep 10 '13 at 03:53
Yes,That's my meaning. – math110 Sep 10 '13 at 05:02
The only other way is to find solutions to $\frac{a}{\text{ln }a} = 3$. It is very difficult to do this without the Lambert W function. I wouldn't even know where to start. – Ryan Sep 10 '13 at 05:09

score -2 · Answer 3 · answered Sep 10 '13 at 02:25

-2

well if e^a and a^3 are graphed the results are a=1.857 and a=4.5364 lim a=>-inf e^a is 0 and lim a=>-inf a^3 is -inf lim a=>inf e^a/a^3 lim a=>inf e^a/3a^2 lim a=>inf e^a/6a lim a>inf e^a/6 is inf by L'hopital rule so e^a will be greater than a^3 from interval (-inf,1.857)u(4.53864,inf)

answered Sep 10 '13 at 02:25

Michael Toth

45
4

No,I don't think you – math110 Sep 10 '13 at 02:28
$e^a$ and $a^3$ are numbers. It doesn't make since for $e^a$ to be greater than $a^3$ on an interval--what variable ranges over the interval? $e$ is a constant, $a$ is a constant (if the series converges, but Wolfram says it does), and $3$ is a constant. The answer is either $>$ or $<$. There are no intervals in this problem. – apnorton Sep 10 '13 at 02:28

Nice problem Which is bigger $e^a$ and $a^3$

3 Answers3