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Inertigratus Feb26-12 05:11 AM

Integral tail and error
 
Hi,

I have an integral from 0 to ∞ and divided into two integrals, one from 0 to C and one from C to ∞.
I'm trying to simplify the second integral so that I can solve for C to make the integral less than ε.
[itex]\int\frac{1-e^{-x^{3}}}{x^{2}} = -xf(x) + \int 3xe^{-x^{3}} < \int 3xe^{-x} = -3xe^{-x} + \int 3e^{-x} < \int 3e^{-x} = -3e^{-x} < ε[/itex]
For ε = 1e^-6 and limits from C to ∞ this gives me: [itex]3e^{-C} < 10^{-6} \Rightarrow C ≥ ln(3) + 6ln(10) ≈ 15[/itex].
However, Wolfram gives me that the value of the integral from 15 to ∞ is ≈0.0667 which is far from 1e^-6.

Is my inequality wrong?

JJacquelin Feb26-12 08:52 AM

Re: Integral tail and error
 
Quote:

Quote by Inertigratus (Post 3784866)
Hi,

I have an integral from 0 to ∞ and divided into two integrals, one from 0 to C and one from C to ∞.
I'm trying to simplify the second integral so that I can solve for C to make the integral less than ε.
[itex]\int\frac{1-e^{-x^{3}}}{x^{2}} = -xf(x) + \int 3xe^{-x^{3}} < \int 3xe^{-x} = -3xe^{-x} + \int 3e^{-x} < \int 3e^{-x} = -3e^{-x} < ε[/itex]
For ε = 1e^-6 and limits from C to ∞ this gives me: [itex]3e^{-C} < 10^{-6} \Rightarrow C ≥ ln(3) + 6ln(10) ≈ 15[/itex].
However, Wolfram gives me that the value of the integral from 15 to ∞ is ≈0.0667 which is far from 1e^-6.

Is my inequality wrong?

Hi !

Difficult to answer if the term f(x) is not defined in the equation.
And where the int 3x exp(-3 x^3) is it comming from ?

Inertigratus Feb26-12 10:11 AM

Re: Integral tail and error
 
Quote:

Quote by JJacquelin (Post 3785088)
Hi !

Difficult to answer if the term f(x) is not defined in the equation.
And where the int 3x exp(-3 x^3) is it comming from ?

f(x) is the whole thing in the first integral and the rest is integration by parts.
The primitive function of x^-2 is -x^-1 then I use integration by parts on exp(-x^3)*x^-2 by letting h(x) = exp(-x^3) and g'(x) = x^-2 which gives h(x)g(x) - ∫h'(x)g(x)dx.
h(x)g(x)*(-x^-1) is the term that I refer to as -xf(x).

JJacquelin Feb26-12 12:59 PM

Re: Integral tail and error
 
OK. It's clear.
3x exp(-x^3) < 3x exp(-x) is true, but the two functions are too much different.
For example, case x=2.5 : 3x exp(-x^3)=0.00000123 while 3x exp(-x)=0.616
So, it is not surpristng that the result of the respective integrals be very far from one to the other.
I suggest 3x exp(-x^3) < 3x exp(-x^3)
Integral (from x=C to infinity) of 3xexp(-x^3) dx = exp(-C^3)= 10^(-6)
Leading to C= 2.4
Integral (from x=2.4 to infinity) of 3x exp(-x^3) dx = 4*10^(-7) approximately

Inertigratus Feb26-12 01:26 PM

Re: Integral tail and error
 
Quote:

Quote by JJacquelin (Post 3785481)
OK. It's clear.
3x exp(-x^3) < 3x exp(-x) is true, but the two functions are too much different.
For example, case x=2.5 : 3x exp(-x^3)=0.00000123 while 3x exp(-x)=0.616
So, it is not surpristng that the result of the respective integrals be very far from one to the other.
I suggest 3x exp(-x^3) < 3x exp(-x^3)
Integral (from x=C to infinity) of 3xexp(-x^3) dx = exp(-C^3)= 10^(-6)
Leading to C= 2.4
Integral (from x=2.4 to infinity) of 3x exp(-x^3) dx = 4*10^(-7) approximately

I thought it would be good to be far off since if the approximation (the far off function) is supposed to be less then ε, then the original integral should be much less then ε, right?
Good suggestion, didn't think of it... remember though that the original function, f(x) is (1 - exp(-x^3))/x^2. The integral of that one from 2.4 to infinity is approx. 0.417 according to WolframAlpha. But that's what I don't understand, how can the integral of the original function be larger then that of the approximated function (which is larger)?
0.417 is much larger then 10^-7?
http://goo.gl/1TcPa (you have to put a parenthesis around 1 - exp(...))

JJacquelin Feb26-12 02:57 PM

Re: Integral tail and error
 
Quote:

remember though that the original function, f(x) is (1 - exp(-x^3))/x^2. The integral of that one from 2.4 to infinity is approx. 0.417 according to WolframAlpha. But that's what I don't understand, how can the integral of the original function be larger then that of the approximated function (which is larger)?
0.417 is much larger then 10^-7?
No,what you call "the original function" is no longer (1 - exp(-x^3))/x^2 because you took apart the term 1/x.
The tail of this term 1/x is :
Integal (from x=2.4 to infinity) of dx/x = 1/2.4 = 0.417
With C=2.4, the tail of the term -exp(-x^3)/x is what was said in my previous post : less than 10^-7.

If you consider the whole function (1 - exp(-x^3))/x^2 , it is completely different : You must not integrate 1/x separately.
Start with (1 - exp(-x^3))/x^2 < 1/x
Integral (x=C to infinity) of 1/x = 1/C = 10^-6
C= 10^6
Integral (from x=10^6 to infinity) of ((1 - exp(-x^3))/x^2)dx = 10^-6 approximately

Inertigratus Feb26-12 11:28 PM

Re: Integral tail and error
 
Quote:

Quote by JJacquelin (Post 3785679)
No,what you call "the original function" is no longer (1 - exp(-x^3))/x^2 because you took apart the term 1/x.
The tail of this term 1/x is :
Integal (from x=2.4 to infinity) of dx/x = 1/2.4 = 0.417
With C=2.4, the tail of the term -exp(-x^3)/x is what was said in my previous post : less than 10^-7.

If you consider the whole function (1 - exp(-x^3))/x^2 , it is completely different : You must not integrate 1/x separately.
Start with (1 - exp(-x^3))/x^2 < 1/x
Integral (x=C to infinity) of 1/x = 1/C = 10^-6
C= 10^6
Integral (from x=10^6 to infinity) of ((1 - exp(-x^3))/x^2)dx = 10^-6 approximately

Oh, I see. I partly understand but I'm still a bit confused, isn't the sum of the integrals of 1/x2 and -exp(-x^3)/x2 not the same as the integral of (1-exp(-x^3))/x2 ? Which means I should be able to calculate them separately and add them up?
Thanks!

JJacquelin Feb27-12 12:41 AM

Re: Integral tail and error
 
Quote:

Quote by Inertigratus (Post 3786412)
Oh, I see. I partly understand but I'm still a bit confused, isn't the sum of the integrals of 1/x2 and -exp(-x^3)/x2 not the same as the integral of (1-exp(-x^3))/x2 ? Which means I should be able to calculate them separately and add them up?
Thanks!

First : If you consider the tail of (1-exp(-x^3))/x
Of course, you are able to calculate separately the integrals and add them up.
But, also you are able to calculate separately the tail of each of them and add them up.
Not only one of the two tails has to be < 10-6 , but the sum of the two.

Second : On another viewpoint, you could say : I know exactly the integral of 1/x, which is -1/x. So there is no error on this part. As a consequence, I am interested only about the error on the integral of (-exp(-x^3))/x. It is completely different : The tail considered is only the tail of the integral of (-exp(-x^3))/x. As already shown, C=2.4 and this is to be tested NOT with the whole function (1-exp(-x^3))/x but on the part (-exp(-x^3))/x .The integral (from x=2.4 to infinity) of (-exp(-x^3))/x is < 10^-6 as expected.

Inertigratus Feb27-12 01:12 AM

Re: Integral tail and error
 
Quote:

Quote by JJacquelin (Post 3786483)
First : If you consider the tail of (1-exp(-x^3))/x
Of course, you are able to calculate separately the integrals and add them up.
But, also you are able to calculate separately the tail of each of them and add them up.
Not only one of the two tails has to be < 10-6 , but the sum of the two.

Second : On another viewpoint, you could say : I know exactly the integral of 1/x, which is -1/x. So there is no error on this part. As a consequence, I am interested only about the error on the integral of (-exp(-x^3))/x. It is completely different : The tail considered is only the tail of the integral of (-exp(-x^3))/x. As already shown, C=2.4 and this is to be tested NOT with the whole function (1-exp(-x^3))/x but on the part (-exp(-x^3))/x .The integral (from x=2.4 to infinity) of (-exp(-x^3))/x is < 10^-6 as expected.

I think I understand now, thanks again!


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