- #1

awelex

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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?

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?

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