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Proof that a function is of exponential order 
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#1
Jul1911, 03:23 AM

P: 44

Hi,
I'm being asked to test whether a function is of exponential order, i.e. whether abs( f(x) ) <= M*exp(a * t), for all t >= T (which is finite). The function is x * ln( x ). Now, I have the solution right here, so I know how to solve it. However, I did it a different way and wanted to ask if it is valid as well. Here it goes: First, observe that ln( x ) < x for large x, because ln( x ) grows that a slower rate than x. Therefore, abs(x * ln( x )) <= x^2. If we can show that x^2 is of exponential order, then we have also shown that x * ln( x ) is of exponential order. x^2 = exp( 2 * ln( x )), which is clearly of exponential order. I think the problem with this approach is the step where I show that ln( x ) < x for large x. I know that that is right by intuition and by looking at the graphs, but I don't think that I've proved it properly. I'm sure it is not enough to prove that one function grows slower than the other (i.e. that one derivative is always smaller than the other after a certain value). How do I proof this in a rigorous way? Thanks, Alex EDIT: How about the following: after x = 1, ln( x ) grows slower than x. But since ln( 1 ) < 1, it follows that after x=1 ln( x ) will always be less than x (probably even before that). Is that good enough? 


#2
Jul1911, 04:25 AM

Sci Advisor
P: 1,742

Take the function f(x) = xlog(x) for x >= 2. the derivative is 11/x, which is positive. If f(x) = 0 for any x >= 2, rolles theorem says that f'(c) = 0 for some c between 2 and x. This is impossible, and since e.g. f(e) = e1 > 0, the intermediate value theorem implies that f(x) must always be positive, and hence that x > log(x) for all x >= 2.
I used 2 and e as arbitrary constants a and b such that 1<a and 1<b to ensure that the derivative is positive, and without having to prove that 2 > log(2). 


#3
Jul1911, 03:15 PM

HW Helper
Thanks
PF Gold
P: 7,591

Or a direct way:
[tex]\frac 1 t < 1 \hbox{ if } t>1[/tex] [tex]\int_1^x\frac 1 t\,dt < \int_1^x 1\,dt[/tex] [tex]\ln(x)\ln(1) < x1[/tex] [tex]\ln(x)<x1<x[/tex] 


#4
Jul1911, 03:46 PM

Math
Emeritus
Sci Advisor
Thanks
PF Gold
P: 39,363

Proof that a function is of exponential order
In fact, you have shown that xln(x) increases more slowly than [itex]x^2[/itex] which is NOT of exponential order itself. 


#5
Jul1911, 04:33 PM

HW Helper
Thanks
PF Gold
P: 7,591

x^{2}<e^{x} for (even not very) large x. 


#6
Jul1911, 07:52 PM

Math
Emeritus
Sci Advisor
Thanks
PF Gold
P: 39,363

Confusion as to exactly what "of exponential order" meant does the function increase at the same rate as [itex]Me^{kx}[/itex] for some M and k or just not as fast. I see now that we are talking about the second here.



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