- #1

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[tex]\int^b_a f(x)g(x)dx=0[/tex]

But what if we have functions of two variables ##f(x,y)## and ##g(x,y)## that are orthogonal on the interval ##[a,b]##. Is there some definitions

[tex]\int^b_a f(x,z)g(z,y)dz=0[/tex]?

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- Thread starter LagrangeEuler
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- #1

- 661

- 16

[tex]\int^b_a f(x)g(x)dx=0[/tex]

But what if we have functions of two variables ##f(x,y)## and ##g(x,y)## that are orthogonal on the interval ##[a,b]##. Is there some definitions

[tex]\int^b_a f(x,z)g(z,y)dz=0[/tex]?

- #2

FactChecker

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Yes. The definition of "orthogonal" is very general. You can do integrals on both variables. As long as there is an "inner product" defined, the definition of "orthogonal" is also defined. And it is not hard to show that the double integral would have the properties of an inner product. (see http://mathworld.wolfram.com/InnerProduct.html )

Last edited:

- #3

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But what if we have functions of two variables ##f(x,y)## and ##g(x,y)## that are orthogonal on the interval ##[a,b]##. Is there some definitions

[tex]\int^b_a f(x,z)g(z,y)dz=0[/tex]?

The integral you've written is not what you seem to mean. First, it shouldn't be ##dz##, it should be ##dxdy## or ##dA## for area. Also, it shouldn't be over a single interval the way you have it written. It should be over a set in the plane. That set could be a box defined by two intervals, one over ##x## and the other over ##y##, or it could be any other arbitrarily shaped set, as long as it's measurable.

- #4

Ssnow

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Ssnow

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