Here is the question: find the projection of sin(x) onto x^3 on the interval [-π,π], where the inner product is integral (-π,π)f(x)g(x)dx I computed this as zero. since ∫(-π,π)x^3sin(x)dx=<sin(x),x^3>=0. Is that correct?
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Originally Posted by Chris11 Here is the question: find the projection of sin(x) onto x^3 on the interval [-π,π], where the inner product is integral (-π,π)f(x)g(x)dx I computed this as zero. since ∫(-π,π)x^3sin(x)dx=<sin(x),x^3>=0. Is that correct? Why do you think it is zero? Can you show us your computation?
projection(sin(x) on x^3)= (<sin(x),x^3>/<x^3,x^3>)x^3. That's the computation that I ran. For the top inner product, since (x^3)sin(x) is odd, it's integral over -π to π is zero. Hence, the projection is zero.
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