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I have been reading some articles and I see that there is an analogue of the dot product for functions in the form of an integral. However, I am confused by the fact that there seems to be 2 forms:

  1. $\int f_1(x)f_2(x)dx$
  2. $\int w(x)f_1(x)f_2(x)dx$ where $w(x)$ is called the weight function

What is going on? Perhaps the 1st case is a special case of the second where the weight function equals 1? When do you need the weight function?

Thanks.

  • 0
    If there's no context, what are we talking about, then? ;)2011-09-26

1 Answers 1

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Expanded summary of comments:

  • Yes, the first inner product is a special case of the second, with $w\equiv 1$
  • There are multiple reasons to consider spaces with weighted inner product (called weighted $L^2$ spaces):
    1. Polynomials are not square integrable on unbounded intervals $I$ such as $\mathbb R$ or $[0,\infty)$. If one wishes to have an orthogonal basis of polynomials on $L^2(I)$, a weight must be used. Two popular weights are $\exp(-x^2)$ and $\exp(-x)$.
    2. Even on a bounded interval, polynomials with interesting properties (such as Chebyshev polynomials $T_n$ on $[-1,1]$) happen to be orthogonal with a weight different from $1$.
    3. Eigenfunctions of a differential equation with nonconstant coefficients tend to be orthogonal with respect to weights related to the coefficients.

Is there any meaning to the question: "Given 2 functions $f(x)$ and $g(x)$, determine whether they are orthogonal." (with no additional information)?

Without any context, this is an unacceptably vague question. If I had to guess, I'd say that the inner product $\int_D fg$ should be used, where $D$ is the intersection of domains of $f$ and $g$.