- #1

ainster31

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But what does it mean intuitively, analytically, or in terms of graphs?

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- Thread starter ainster31
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- #1

ainster31

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- 1

But what does it mean intuitively, analytically, or in terms of graphs?

- #2

arildno

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"So what", you may ask:

It can essentially be regarded as a generalization of the following situation:

If a(x) and b(x) are orthogonal functions over some interval I, it means that whenever on I a(x) is "active" (that is, from the most part DIFFERENT from zero), b(x) is "passive" (i.e, for the most part equal to zero), and vice versa.

So, the PRODUCT of a(x) and b(x) will be practically "passive" everywhere, rather than that a and b reinforce/twist the effects of each other.

- #3

HallsofIvy

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It can be generalized to any "inner product space" with orthogonal defined as "the inner product" (a generalization of dot product in R

One can show, for example, that [itex]\int_0^{2\pi} sin(nx)sin(mx)dx= 0[/itex] and [itex]\int_0^{2\pi} cos(nx)cos(mx)dx[/itex], as long as [itex]m\ne n[/itex] and that [itex]\int_0^{2\pi} sin(nx)cos(mx)= 0[/itex] for all m and n. Thus, the set of functions {sin(nx), cos(nx)} for for an "orthogonal set" of functions.

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