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        riddles >> putnam exam (pure math) >> 1987 B1 Not making sense
        
(Message started by: gordysc on Oct 25^th, 2005, 12:45pm)

Title: 1987 B1 Not making sense
Post by gordysc on Oct 25^th, 2005, 12:45pm

Ok, a group of us here are working on B1 from 1987:
Integrate (ln(9-x)^1/2)/((ln(9-x)^1/2)+(ln(3+x)^1/2))dx from 2 to 4. Now, the archives answer is to substitute x=6-t in and then add the integrals together to get 1. But, when we substitute x=6-t, we get -(ln(3+t)^1/2)/((ln(9-t)^1/2)+(ln(3+t)^1/2))dt
So how are we suppose to add these two different variables together? It just isn't clicking. Any feedback would be helpful and very much appreciated!

Title: Re: 1987 B1 Not making sense
Post by towr on Oct 25^th, 2005, 3:01pm

INT _{2 to 4} (ln(9-x)^1/2)/((ln(9-x)^1/2)+(ln(3+x)^1/2)) dx
== { sub x = 6-t }
INT _{4 to 2} (ln(3+t)^1/2)/((ln(3+t)^1/2)+(ln(9-t)^1/2)) d (6-t)
==
INT _{2 to 4} (ln(3+t)^1/2)/((ln(3+t)^1/2)+(ln(9-t)^1/2)) dt
== { sub t = x }
INT _{2 to 4} (ln(3+x)^1/2)/((ln(3+x)^1/2)+(ln(9-x)^1/2)) dx

Add first and last, to get twice the value you want

INT _{2 to 4} (ln(9-x)^1/2)/((ln(9-x)^1/2)+(ln(3+x)^1/2)) dx
+ INT _{2 to 4} (ln(3+x)^1/2)/((ln(3+x)^1/2)+(ln(9-x)^1/2)) dt
== { move addition inside integration }
INT _{2 to 4} (ln(9-x)^1/2 + (ln(3+x)^1/2) /((ln(9-x)^1/2)+(ln(3+x)^1/2)) dx
== {simplify}
INT _{2 to 4} 1 dx = 2
which is twice our integral at the start, which is thus 1.

Title: Re: 1987 B1 Not making sense
Post by Luke Gordon on Oct 25^th, 2005, 6:54pm

Ok, this is where we get confused. We state that x=6-t, but then you substitute x back into the equation again....x!=6-x We understand everything else. It's just that x has already been defined as equaling 6-t...

Title: Re: 1987 B1 Not making sense
Post by towr on Oct 26^th, 2005, 1:07am

It's not the same x. You could have done the substitution x -> 6-x, in one go. But that would be more confusing (and easier to make mistakes with).

The thing is, the x is a bounded variable, it doesn't exist outside the scope of integration. So it can't carry meaning over the equality signs either.

[sum]^N_x=1 x = N (N+1)/2
{sub x -> y}
[sum]^N_y=1 y = N (N+1)/2
{sub y -> z-20}
[sum]^N+20_z=21 (z-20) = N (N+1)/2
{sub z -> x}
[sum]^N+20_x=21 (x-20) = N (N+1)/2
{sub x -> x+20}
[sum]^N_x=1 x = N (N+1)/2

The same thing keeps happening; 1+2+3+4+...+N And that's what's important, not the names of the variables involved (which is why you can substitute).

Title: Re: 1987 B1 Not making sense
Post by Luke Gordon on Oct 26^th, 2005, 11:01am

Ok, that makes some sense. So then, we can say x=x^2 or anything we want to for a bounded variable?

Title: Re: 1987 B1 Not making sense
Post by towr on Oct 26^th, 2005, 1:51pm

on 10/26/05 at 11:01:04, Luke Gordon wrote:

Ok, that makes some sense. So then, we can say x=x^2 or anything we want to for a bounded variable?

Yes, as long as both expressions are equivalent after substitution it doesn't matter what the substitution is (or what the names of the variables in it are).
It's really just to make an expression more readable, or easier to manipulate. And naturally you have to be mindful of additional constarints; if x is in the range [-1,1], you can't substitute x with x², because it can't cover the same range.
I think the substitution should be reversable (for the range the variables are used in)

Title: Re: 1987 B1 Not making sense
Post by Icarus on Oct 26^th, 2005, 2:47pm

The variable in a definite integral is a dummy variable. That means it is only there as a placeholder. The integral itself does not depend in any way on a value for this variable.

So [int]_a^b f(x) dx = [int]_a^b f(t) dt = [int]_a^b f(y) dy = ..., entirely independent of any other meanings given to the variables x, t, y, etc, in the calculation. The value of the integral depends only on the values of a and b, and the definition of the function f.

Title: Re: 1987 B1 Not making sense
Post by towr on Oct 26^th, 2005, 2:52pm

On thing you do have to be carefull of is that the new variable wasn't already present in the expression being integrated.

[int]_a^b f(x,y) dx is not [int]_a^b f(y,y) dy