Minimize integral using orthogonal basis

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The discussion revolves around verifying the least squares approximation using an orthogonal basis derived from the Gram-Schmidt process. The user initially finds a discrepancy between their calculated projection and the answer provided in the textbook. They correctly identify the polynomials used for orthogonality but mistakenly refer to one of them, leading to confusion. After realizing the error in the basis selection, they express frustration over the timing of this realization. The conversation highlights the importance of careful notation and the verification of each step in mathematical calculations.
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Homework Statement
Determine the polynomial ##p## of degree at most ##1## that minimizes ##\int_0^2|e^x-p(x)|^2 dx##. (Hint: first find an orthogonal basis for a suitably chosen space of polynomials of degree ##1##.)
Relevant Equations
##L^2## norm, inner product, least squares approximation.
I'm posting to inquire about a possible typo in the given answer in the back of the book, or if maybe I did something wrong, because my answer does not agree with the one stated.

So the exercise is about finding the least squares approximation. The norm is the ##L^2## norm and the corresponding inner product is $$\langle f, g\rangle=\int_0^2 f(x)g(x)dx.$$ I choose the polynomials ##v_1=1## and ##v_2=x## as a basis and make them orthogonal according to Gram-Schmidt, i.e. I find that ##u_1=1## and ##u_2=x-1## are orthogonal to each other under the inner product above. Put ##u=e^x##, then the orthogonal projection of ##u## on the subspace spanned by ##u_1,u_2## is given by $$P(u)=\frac{\langle u, u_1\rangle}{\langle u_1,u_1\rangle}u_1+\frac{\langle u, u_2\rangle}{\langle u_2,u_2\rangle}u_2.$$ Now, \begin{align}\langle u_1,u_1\rangle&=\int_0^2 dx=2 \nonumber \\ \langle u_2,u_2\rangle&=\int_0^2(x-1)^2dx=\int_0^2(x^2-2x+1)dx=\frac23, \nonumber\end{align}and\begin{align}
\langle u,u_1\rangle&=\int_0^2 e^xdx=e^2-1 \nonumber \\
\langle u,u_2\rangle&=\int_0^2(x-1)e^x dx=2. \nonumber
\end{align}
Plugging these into the equation for ##P(u)##, gives $$P(u)=3x+\frac12(e^2-1),$$but the answer given is ##3x+\frac12(e^2-7)##. Is there a way to check one's answer to know if you got the right one?
 
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Sorry, I just realized now I forgot to use ##u_2=x-1## instead of ##v_1=x##.:doh:
 
psie said:
Sorry, I just realized now I forgot to use ##u_2=x-1## instead of ##v_1=x##.:doh:
... and you tell me this now that I have literally checked every single integral and step and was right about to answer where your typo was ... :devil: :biggrin:
 

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