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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.

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    See Examples 4 and 5. http://tutorial.math.lamar.edu/Classes/LinAlg/InnerProductSpaces.aspx2011-09-26
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    Given $n$ positive numbers $a_1, \ldots, a_n$, did you know that $(x,y) =_{def} \sum_{i=1}^n a_i x_i y_i$ is also an inner product (where $x,y \in \mathbb R^n$)? You can think of the second form as an analogue of this.2011-09-26
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    @M.B.: Thanks. The thing is I have read that when determining the orthogonality of eigenfunctions, the weight function ***must*** be included in the integral -- 2nd form. But I don't understand why and when the weight function should be necessary...2011-09-26
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    @SrivatsanNarayanan: Ah, interesting. No, I didn't know that. Thanks! There is still a problem though, as I said in my comment addressed to M.B., if I want to test the orthogonality of 2 eigenfunctions say of the forms $\sin(nx)$ and $\sin(mx)$ then would I need a weight function? I saw some demonstrations that just do the 1st form and some the 2nd... :-S2011-09-26
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    "Perhaps the 1st case is a special case of the second where the weight function equals 1?" - yes. In general, one considers a [measure](http://mathworld.wolfram.com/L2-InnerProduct.html) first and foremost, and then the inner product that goes along with it. The weight function accounts for the measure...2011-09-26
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    Orthogonality is determined by the inner product. If the inner product is given with a weight, then that weight should be used to determine orthogonality. Usually the weight is $1$, but there are cases where it is not, but those cases should hopefully be clear by context.2011-09-26
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    @J.M.: Thanks. How would I know what *measure* to use?2011-09-26
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    For instance, one might consider the Chebyshev measure $\frac{\mathrm dx}{\sqrt{1-x^2}}$ over $[-1,1]$ when one wants to emphasize the ends of the interval, or the Laguerre measure $\exp(-x)\mathrm dx$ when one is considering semi-infinite intervals...2011-09-26
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    @robjohn: Thanks. Say I have a differential equation and I found a bunch of eigenfunctions for it. Then how would I know what its weight is to test for orthogonality?2011-09-26
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    In general, the application usually *cries out* for the appropriate measure... ;)2011-09-26
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    Unless the problem explicitly mentions a weight for the inner product, I would use $1$. If the problem asks for solutions in $L^2(\omega)$ or $L^2(\omega\;\mathrm{d}x)$ for some weight $\omega$, the weight would be $\omega$. The weight for $L^2[a,b]$ would be the characteristic function of $[a,b]$.2011-09-26
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    @Confuserlearner: As regards differential equations, the standard situation where orthogonality plays a role is that of a Sturm-Liouville boundary value problem. If the Sturm-Liouville eigenvalue equation is written in the form $-\frac{d}{dx}\left[p(x)\frac{dy}{ dx}\right]+q(x)y=\lambda w(x)y$, the weight is $w(x)$. It is then a theorem that (under appropriate conditions) the eigenfunctions are orthogonal with respect to this weight.2011-09-26
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    @RobertIsrael: Thanks. Is there any meaning to the question: "Given 2 functions $f(x)$ and $g(x)$, determine whether they are orthogonal." (with no additional information)?2011-09-26
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    @Confused: depends on the context. For instance, when you mentioned "if I want to test the orthogonality of 2 eigenfunctions say of the forms $\sin(nx)$ and $\sin(mx)$", that context (particle in a box?) has $w(x)=1$...2011-09-26
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    @J.M.: Thanks. That example was just conjured up randomly... If there isn't a specified context, is the question meaningful at all?2011-09-26
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    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 $g$.