Re: e^x > x^a , a > 1

Xan <xancorreu@xxxxxxxxx> wrote:
On Nov 27, 10:19 pm, David W. Cantrell <DWCantr...@xxxxxxxxxxx> wrote:
Xan <xancor...@xxxxxxxxx> wrote:
On Nov 27, 5:55 pm, David W. Cantrell <DWCantr...@xxxxxxxxxxx> wrote:
Xan <xancor...@xxxxxxxxx> wrote:
On Nov 26, 11:38 pm, David W. Cantrell <DWCantr...@xxxxxxxxxxx>
wrote:
Xan <xancor...@xxxxxxxxx> wrote:
On Nov 26, 5:32 pm, David W. Cantrell
<DWCantr...@xxxxxxxxxxx> wrote:
hagman <goo...@xxxxxxxxxxxxx> wrote:
On 26 Nov., 16:43, Xan <xancor...@xxxxxxxxx> wrote:
Hi,

I have this problem:
For all a > 1, fixed, find x_0(a) such that for all
x > x_0: e^x > x^a

Obviously, x_0 exists, and now I think I could prove
that we could take x_0 = e^(2a^2)
But I need some "little" bound. What's the lowest
lowest bound of x_0 we take?

It is of course the greatestest solution of e^x = x^a
provided such a solution exists (note that e.g. e^x > x^1
for all x in R).
Also note that x_0 may even be negative: e^x > x^2 for
all x > -0.7.

True. But to have a sharp inequality, e^x > x^2 for x > -2
W_0(1/2) where W_0 denotes the principal (or 0) branch of
the Lambert W function. (For information of the Lambert W
function, see for example
<http://mathworld.wolfram.com/LambertW-Function.html>.)

However, I suspect that Xan may actually have intended to
restrict x to positive values and to have a >= e, despite
what was said originally. With that assumption, the largest
solution of e^x = x^a is given by

x = -a W_{-1}(-1/a)

where W_{-1} denotes the nonprincipal real (or -1) branch
of the Lambert W function.

Yes. I assume that x > 0 and a > 1 both real numbers.

Is there any bound for that nonprincipal real (or -1)
branch....?

Sure, but I'm not really sure what you want.

You asked for the lowest bound for x0 and I gave it precisely.

But you need something cruder, not involving the Lambert W
function. If so, how about the following?

e^x > x^a for x > 2 a log(a)

David

Yes, I wanted a more "crude" thing.
I amazing for this lower bound: how do you get x_0 = 2 a log (a)
?

I got that crude lower bound by thinking about the lowest bound,
-a W_{-1}(-1/a), and the behavior of the Lambert W function.

How can you get that. I saw your link (mathworld) and I can't get a
hint!
I don't see that x^a = e^x iff the solution of -a W_{-1}(-1/a)

That's not exactly what I said. It's certainly not "iff", for example.
Consider the equation x^4 = e^x. It has three real solutions. The
smaller two are given using the principal (or 0) branch of the Lambert
W function:

-4 W_0(1/4) = -0.81555... and -4 W_0(-1/4) = 1.4296...

Only the largest solution, which is the one pertinent to what you
wanted, is given by

-4 W_{-1}(-1/4) = 8.6131...

Can you teach some class about that?

It's easy to show how to solve x^a = e^x using the Lambert W function.
Recall that it's merely the inverse of f(t) = t e^t. Thus, to solve our
equation via the Lambert W function, we need to change the equation
into the form t e^t = c and so the solutions will be given by t = W(c).

So here's the basic process, assuming that x > 0:

x^a = e^x

x = e^(x/a)

x e^(-x/a) = 1

-x/a e^(-x/a) = -1/a

Now, of course, -x/a is like t and -1/a is like c, mentioned above, and
so we get

-x/a = W(-1/a)

x = -a W(-1/a)

For a > e, the above gives two positive solutions for x, using the two
real branches, called 0 and -1, of the Lambert W function. The larger
of those two solutions, which is what is pertinent to your problem, is
obtained using the -1 branch.

So, you have used W(x) as complex function, isn't?

I've only used portions of the 0 and -1 branches where W is real.

How can you proof this: that for a > e, the larger real positive
solution is in -1 branch?

For -1/e <= t < 0, -1 <= W_0(t) < 0 while W_{-1}(t) <= -1.

Take the graph of f(x) = x*e^x. Flip it about the line y=x to get a graph
of the inverse. That shows the real parts of both the 0 and -1 branches
of W, meeting at the point (-1/e, -1).

The most important, how to calculate W(-1/4)?

It can be approximated using series or iterative methods. Asking for
W(-1/4), both branches, is the same as asking for the two real solutions of
x*e^x = -1/4. Do you know a way to approximate those solutions?

Can you put a sketch of the proof of that? Is it tecnically
dificult (more than basic analysis)?

Perhaps someone else can provide a sketch without reference to the
Lambert W function. Or perhaps someone else could find a nicer
lower bound.

Well, and a demostration with Lambert function? Can you provide me
your proof? Is it extended if we consider only Lambert function
restriction to real numbers (non-complex)?

I was using only real numbers, but I haven't written out a proof that

e^x > x^a for x > 2 a log(a).

Establishing that crude bound actually seems messier, at least for me,
that getting the sharp bound in terms of the Lambert W function.

BTW, if a bound in the form k a log(a) is useful to you, perhaps I
should note that k = 2 is not best possible. The best possible bound in
that form has

k = e/(e - 1) = 1.58...

David

You impress me. I don't know how you calculate this. Can you show me.

I used a computer algebra system in the calculation. But it was basically
just a matter of finding the value k such that the curves
y = -a W_{-1}(-1/a) and y = k a log(a) are tangent to each other.

To help you visualize, the graph
<http://img122.imageshack.us/img122/5548/xanht3.gif>
shows y = -a W_{-1}(-1/a) in blue, y = e/(e - 1) a log(a) in red, and the
cruder bound y = 2 a log(a) in black, beginning at a = e.

David
.

Relevant Pages

  • Re: e^x > x^a , a > 1
    ... (For information of the Lambert W function, ... restrict x to positive values and to have a>= e, ... Only the largest solution, which is the one pertinent to what you wanted, ... So here's the basic process, ...
    (sci.math)
  • Re: e^x > x^a , a > 1
    ... (For information of the Lambert W ... Flip it about the line y=x to get a graph ... Okay. ... I suppose you use Taylor series to do that. ...
    (sci.math)
  • Re: e^x > x^a , a > 1
    ... (For information of the Lambert W function, ... I got that crude lower bound by thinking about the lowest bound, ... Only the largest solution, which is the one pertinent to what you wanted, ... So here's the basic process, ...
    (sci.math)
Quantcast