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Re: value of a function at y(0) given the definite integral [message #27548 is a reply to message #27378]

Fri, 26 October 2001 15:24

a
Messages: 4
Registered: July 1997

Junior Member

Hmmm...

;find integral of known fn x^2 over interval [0,1]
x = findgen(11)/10.
print, int_tabulated( x, x^2 )
0.333406

Here, we know
1) integral_aTOb = 1/3 (where a=0 & b=1)
2) y(0) = 0^2 = 0
3) y(1) = 1^2 = 1

Using JD's stuff, let's find A for this plain jane case

A integral_0to1(x^2 dx) = 1/3
A * (1^3/3 - 0^3/3) = 1/3
A = 1

Of course, it should be 1 because that's the normalization constant we
assumed when we calculated using int_tabulated( x, x^2 ) above. But, how do
you relate A to the value of y(0)? We know y(0) in this case is equal to 0,
not 1.

I think the problem is that his integral_aTOb is not = 1/3, but something
else, like 10. Therefore the y(0) must not be 0 but must be greater than
this, like 2.7234. I think he's looking for the *offset* of y(0) that would
make the area under the curve x^2 = 10 (or whatever). Wouldn't this be an
addition y(x)+A, not multiplication A*y(x)?

Under JD's solution for the example here of integral_aTOb(x^2) = 10:

A * integral_0to1(x^2 dx) = 10
A * (1^3/3 - 0^3/3) = 10
A * 1/3 = 10
A = 30

Again, I don't see it how this value relates to the value of y at x=0?

Sorry, just getting curious/
john

"JD Smith" <jdsmith@astro.cornell.edu> wrote in message
news:3BD9C768.8F4A8721@astro.cornell.edu...
>
> integral(Ay(x) dx)=A integral(y(x) dx)
>
> will guide the way.
>
> JD

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[Message index]

		value of a function at y(0) given the definite integral By: aqueous0123 on Mon, 22 October 2001 13:41
		Re: value of a function at y(0) given the definite integral By: a on Fri, 26 October 2001 15:24
		Re: value of a function at y(0) given the definite integral By: John-David T. Smith on Fri, 26 October 2001 13:28
		Re: value of a function at y(0) given the definite integral By: aqueous0123 on Fri, 26 October 2001 12:55

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