Straight through the curves

Algebra Level 3

Let V = P n ( R ) V = P_{n} \left(\mathbb{R}\right) be an inner product space of all polynomials with real coefficients and degree not greater than n n , and the inner product defined as f ( x ) , g ( x ) = 1 1 f ( t ) g ( t ) d t f , g V \langle f(x) , g(x)\rangle = \int_{-1}^{1}f(t)g(t)dt \: \forall f,g \in V The orthogonal projection of f P 3 ( R ) f \in P_{3}\left(\mathbb{R}\right) , f ( x ) = x 3 f(x)=x^3 on P 2 ( R ) P_{2}\left(\mathbb{R}\right) is a line. Find the slope of this line.


The answer is 0.6.

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1 solution

Otto Bretscher
Oct 12, 2018

x 3 a x x^3-ax is orthogonal to 1 1 and x 2 x^2 ; we want it to be orthogonal to x x as well. Now x , x 3 a x = 6 10 a 15 = 0 \langle x,x^3-ax\rangle=\frac{6-10a}{15}=0 when a = 6 10 a=\boxed{ \frac{6}{10}} .

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