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A family of functions is known as $\left(\varphi_{0}, \varphi_{1}, \cdots, \varphi_{n}\right)$.

I'd like to know how to express their inner product conveniently as follows:

$$\left(\begin{array}{cccc} \left(\varphi_{0}, \varphi_{0}\right) & \left(\varphi_{0}, \varphi_{1}\right) & \cdots & \left(\varphi_{0}, \varphi_{n}\right) \\ \left(\varphi_{1}, \varphi_{0}\right) & \left(\varphi_{1}, \varphi_{1}\right) & \cdots & \left(\varphi_{1}, \varphi_{n}\right) \\ \vdots & \vdots & & \vdots \\ \left(\varphi_{n}, \varphi_{0}\right) & \left(\varphi_{n}, \varphi_{1}\right) & \cdots & \left(\varphi_{n}, \varphi_{n}\right) \end{array}\right)$$

Where $(f(x), g(x))$ is the inner product: $(f(x), g(x))=\int_{a}^{b} f(x) g(x) \mathrm{d} x$

We can take $\{1,x,x^2,x^3,x^4\}$ and $\{a=-1,b=1\}$ as an example to realize the above requirements.

Outer[Integrate[#1*#2, {x, -1, 1}] &, {1, x, x^2, x^3}, {1, x, x^2, x^3}]

I wonder if there are any other ways to achieve this?

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    $\begingroup$ Just use Outer? Or am I missing something? With[{fns = Array[\[Phi], 3, 0]}, Outer[Integrate[#1[x]*#2[x], {x, a, b}] &, fns, fns]] - in your concrete case With[{fns = Array[Power[x, #] &, 5, 0]}, Outer[Integrate[#1*#2, {x, -1, 1}] &, fns, fns]] giving result {{2, 0, 2/3, 0, 2/5}, {0, 2/3, 0, 2/5, 0}, {2/3, 0, 2/5, 0, 2/7}, {0, 2/5, 0, 2/7, 0}, {2/5, 0, 2/7, 0, 2/9}} $\endgroup$ Commented Sep 15, 2020 at 1:03

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Since you're dealing with inner products, you can take advantage of the symmetry that $\left<f(x), g(x)\right> = \left<g(x), f(x)\right>$ to only compute the $n(n+1)/2$ upper triangular elements instead of the total $n^2$.

So we can use SymmetrizedArray as

ipmatrix[ϕ_, n_, {a_, b_} /; a <= b] := SymmetrizedArray[ {j_, k_} :> Integrate[ϕ[x, j]*ϕ[x, k], {x, a, b}], {n, n}, Symmetric ]

To yield your example of

Normal@ipmatrix[#1^(#2 - 1) &, 5, {-1, 1}]

{{2, 0, 2/3, 0, 2/5}, {0, 2/3, 0, 2/5, 0}, {2/3, 0, 2/5, 0, 2/7}, {0, 2/5, 0, 2/7, 0}, {2/5, 0, 2/7, 0, 2/9}}

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The results of the two methods are the same $\color{Gray} {\text{(2001 武汉 岩石 数值分析 4)}} $:

ClearAll["Global`*"] f[x_] := x^3 L[a_, b_, c_] := Integrate[(f[x] - (a*x^2 + b*x + c))^2, {x, 0, 1}] StagnationPoints = Solve[{D[L[a, b, c], a] == 0, D[L[a, b, c], b] == 0, D[L[a, b, c], c] == 0}] Print[Style["The best quadratic approximation polynomial is：", Red, Bold], a*x^2 + b*x + c /. Flatten[StagnationPoints]] LinearSolve[ Outer[Integrate[#1*#2, {x, 0, 1}] &, {1, x, x^2}, {1, x, x^2}], Map[Integrate[#1*x^3, {x, 0, 1}] &, {1, x, x^2}]].{1, x, x^2}
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    $\begingroup$ （-1） This isn't an answer to your question at all, why do you post it here? $\endgroup$ Commented Sep 15, 2020 at 11:27

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