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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- April 27th 2010, 01:42 PMChris11Please check this.
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? - April 27th 2010, 05:34 PMIsomorphism
- April 27th 2010, 06:48 PMChris11
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.