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What does it mean if two functions are orthogonal?

by ainster31
Tags: functions, orthogonal
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Oct16-13, 03:21 PM
P: 154
I already know the definition:

But what does it mean intuitively, analytically, or in terms of graphs?
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Oct16-13, 03:48 PM
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PF Gold
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In its simplest form, it means that the integral of product function of the two functions over a specified interval is zero.

"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.
Oct16-13, 07:44 PM
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PF Gold
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"Orthogonal", of course, comes from geometry meaning "perpendicular". One property of that is that if two vectors are perpendicular their dot product is 0.

It can be generalized to any "inner product space" with orthogonal defined as "the inner product" (a generalization of dot product in Rn). For spaces of functions, such as "[itex]L_2[a, b][/itex]", the set of all function that are "square integrable" on interval [0, 1], we can define the inner product to be [itex]\int_a^b f(x)g(x) dx[/itex] (or complex conjugate of g for complex valued functions). Two such functions, f and g, are said to be "orthogonal" if [itex]\int_a^b f(x)g(x)dx= 0[/itex].

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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