Least squares and integration problem

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SUMMARY

The discussion focuses on the least squares approximation in the vector space C[-1,1] with an inner product defined by the integral from -1 to 1 of f(x)g(x) dx. The orthonormal set of vectors u1(x) = 1/(2^.5) and u2(x) = (6^.5/2)x is established, confirming their inner products are zero and their lengths are one. The best least squares approximation to h(x) = x^(1/3) + x^(2/3) is expressed as p(x) = (c1)(u1(x)) + (c2)(u2(x)), where ci is determined by evaluating the integral of (ui(x))h(x). The evaluation at x = -1 does not yield complex numbers, as the cube root of -1 is -1, leading to a final value of (1/√2)(6/5).

PREREQUISITES
  • Understanding of inner product spaces in functional analysis
  • Familiarity with least squares approximation techniques
  • Knowledge of integration over defined intervals
  • Proficiency in evaluating integrals involving polynomial and root functions
NEXT STEPS
  • Study the properties of inner product spaces in C[-1,1]
  • Learn about the Gram-Schmidt process for orthonormalization
  • Explore advanced techniques in least squares approximation
  • Investigate the evaluation of integrals involving piecewise functions
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Mathematicians, students in advanced calculus or functional analysis, and anyone involved in numerical methods or approximation theory will benefit from this discussion.

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Question states
Consider the vector space C[-1,1] with an inner product defined by
<f,g> = the integral from 1 to -1 of f(x)g(x) dx
a)
Show that
u1(x)= 1/(2^.5) u2(x)= ((6^.5)/2)x
form an orthonormal set of vectors
b)
Use the result from a) to find the best least squates approximation to
h(x)= x^(1/3) + x^(2/3)
by a linear function.
For part a) I have shown that u1 and u2 each have an inner product of zero and a length of one.
I've been trying to find a solution to b) in the form
p(x) = (c1)(u1(x)) + (c2)(u2(x))
where ci = the integral from -1 to 1 of (ui(x))h(x)
evaluating this integral for c1 produces
(1/(2^.5))[(3/4)x^(4/3) + (3/5)x^(5/3)] evaluated from -1 to 1
Now I've finally got to the problem
Putting -1 in would produce complex numbers and I don't know how to proceed.
I'm not sure weather the problem here is that I'm not aproaching the least squares problem correctly or weather I'm not approaching the integral correctly. Any help would be appreciated, thanks.
 
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Putting -1 in for x? That doesn't produce complex numbers. The cube root of -1 is -1. Its value at x= 1 is [itex]\frac{1}{\sqrt{2}}\left(\frac{3}{4}+\frac{3}{5}\right)[/itex] and its value at x= -1 is [itex]\frac{1}{\sqrt{2}}\left(\frac{3}{4}-\frac{3}{5}\right)[/itex] Subtracting gives just [itex]\frac{1}{\sqrt{2}}\frac{6}{5}[/itex].
 

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