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[EarthStormFire]

There exsist an f(x) and a g(x), such that the int(f(x)*g(x)dx) between (a, b ) is equal to 0. Then f(x) and g(x) form an orthogonal set between a and b.


Now my question is prove that if the first statement is true, then second sentence must also be true.


Note. int = integral

^ That is what I want to prove

 

[rastex]

What's the definition of an orthogonal set?

 

[EarthStormFire]

It means the equations are perpendicular at all points of intersection between a and b.

 

[Suranga3]

It means the equations are perpendicular at all points of intersection between a and b.

For functions to be perpendicular don't their slopes have to negative reciprocals of each other? I'm sure that has something to do with it.

 

[rastex]

cool... ya, this stuff is beyond me, but what about integrating both sides? Would that get anywhere?

 

[EarthStormFire]

This actually isn't HW. I was told this, and was wondered if there was a proof for it. I know it is true. However the proof might actually be even well beyond what is thought in college.

 

[fart]

my calculus is pretty rusty, but uh, AB = 0 iff A = 0 or B = 0

 

[EarthStormFire]

Damn I made a mistake. There should only be 1 integral not 2. It should be int(f(x)*g(x)dx) from a to b.

 

[EarthStormFire]

I have corrected the problem. Maybe now someone can help me solve it.

 

[marsomega]

This is definately cal 3 stuff. There are a set of theorems you can use to prove. Vector projections or something similar can help you. Then again, this shouldn't be to hard using simple vector work I think. Then again I'd have to brush up on my cal 3. That and I can't help but think convolution everytime I see "*". Damn upper level maths f'ed me up.

I'll look into it.

 

[EarthStormFire]

This is definately cal 3 stuff. There are a set of theorems you can use to prove. Vector projections or something similar can help you. Then again, this shouldn't be to hard using simple vector work I think. Then again I'd have to brush up on my cal 3. That and I can't help but think convolution everytime I see "*". Damn upper level maths f'ed me up.

I'll look into it.

It is multiplication, I know it looks like the convolution symbol, but it not. I really need to brush up on my calc too, so I dont forget it compleatly.

 

[CajoleJuice]

Ask me in a few years.