Solving for the Unique Vector h(x) in the Polynomial Space of Degree 2

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SUMMARY

The discussion focuses on finding the unique vector h(x) in the polynomial space of degree 2, defined by the inner product =int(f*h) from 0 to 1, and the linear transformation g(f)=f(0)+f'(1). The theorem states that for a finite dimensional inner product space V, there exists a unique vector y such that g(x)=. The solution involves expressing h(x) as a polynomial h(x)=h0+h1*x+h2*x^2 and determining coefficients by substituting specific polynomial forms for f(x) to derive equations for the h's.

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Homework Statement


For the given inner product space V=polynomial space of degree 2. Find a the unique vector h(x) such that <f,h>=int(f*h) from 0 to 1. and g(f)=f(0)+f'(1).



Homework Equations


Theorem: let V be a finite dimensional inner product space over F and let g:V->F be a linear transformation. Then there exists an unique vector, y in V such that g(x)=<x,y>



The Attempt at a Solution


well, <f,h>=g(f)

=> <f,h>=f(0)+f'(1)
=>int(f*h) from 0 to 1=f(0)+f'(1)

I have no clue how to solve this final step
 
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Be explicit. Let h(x)=h0+h1*x+h2*x^2, similarly for f(x). Now explicitly work out <f,g>=f(0)+f'(0). Hint: if the equation must be true for ALL f, then it must be true for say, f0=1, f1=0 and f2=0. Try some other choices for the coefficients of f, until you get enough equations to solve for the h's.
 

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