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In the image below, the first equation asks for the blue triangle to be right. The second equation asks for the pink triangle to be right. Since u and v are on the sides of both these triangles, this is all equivalent to $u \bot v$.

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For any subspace $V$ we have $V^\perp = \overline{V}^\perp$. Passing to the closure may simplify the problem. In (b) the closure of even polynomials is the space of all even $L^2$ functions. Indeed, since polynomials are dense in $L^2$, given an even function $f$ we can find a polynomial $p$ such that $\|f-p\|_{L^2}$ is small. Then $q(x)=(p(x)+p(-x))/2$ is ...

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If I correctly understood your question, you want to find a ortonormal basis of $V$, where $V=Span\{1,x\}$ w.r.t the given scalar product. In other words, you want to begin by correcting through normalization the fact that $$\langle 1, x\rangle=\int_0^1 x dx=\frac{1}{2}$$. If you search for an orthonormal basis, then you need to introduce at first ...

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You don't need to use the standard basis, you already have a basis of $\{1,x\}$ for the space you're interested in, namely $Span(\{1,x\})$. Use Gram-Schmidt on those vectors directly. As you point out $\langle 1,1\rangle=1$ already, so you have one orthonormal basis element already. For the other one, take $x-\frac{\langle x,1\rangle}{\langle ... 6 David Speyer successfully analyzed the asymptotics of$n\to \infty$. But there's more: even for a fixed$n$, the function$(1-x^2)^{1/4}P_n(x)$has nearly equal extremal values (it successfully pretends to be a Chebyshev polynomial). Here is$n=9$, not a large number: Encyclopedia of Math conveniently states the relevant estimate in terms of the ... 9 I claim that that curve is $$y=\pm \frac{\sqrt{2/\pi}}{ \sqrt[4]{1-x^2}}.$$ This argument will not be rigorous, and will cite a source I haven't fully understood. Take a look at Whittaker and Watson, A course in Modern Analysis, p. 316. They write: $$P_n(\cos \theta) = \frac{4}{\pi} \frac{2 \cdot 4 \cdots (2n)}{3\cdot 5 \cdots (2n+1)} \left( ... 5 All the local maxima do not sit on the same "nice curve". Under Wikipedia's normalisation with \|P_n\|^2 = \frac{2}{2n+1} we can use Bonnet's recursion formula to get that$$ (n+1)P_{n+1}(0) = - n P_{n-1}(0) $$Since your "normalised" \sqrt{\frac{2}{2n+1}}\tilde{P}_n = P_n we get that under your normalization$$ \frac{n+1}{n} ... 2 Completeness of an orthogonal sequence of functions is a bit tricky on unbounded intervals, while it is relatively straightforward on bounded intervals. In the case of Laguerre and Hermite polynomials, there is a nice trick due to von Neumann that allows the reduction to bounded intervals. There seems to be a bit of confusion about the interval in the ... 2 Very roughly speaking: Laguerre polynomials look like the family of trigonometric functions ($\sin nx$or$\cos nx$) in the region where the weight is concentrated. That is, they have moderate size and slowly increasing oscillation with$n$. Outside of this region, they look pretty much like any random collection of polynomials. When$\alpha=0$, the weight ... 2 First make a table of your data. $$\begin {array}{r|r|r} i&x_i&f(x_i)\\ \hline 1&10&0\\2&10.2&0.004\\3&10.4&0.016\\4&10.6&0.036\\5&10.8&0.064\\6&11&0.1 \end {array}$$ Then, as the instructions say,$\phi_0=1$and$\phi_1=x-\frac {(x\phi_0,\phi_0)}{\phi_0,\phi_0}=x-\frac {(x,1)}{(1,1)}\$ The terms in ...

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