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In an article about Legendre Polynomials, I encountered the following simplification.

\begin{align} (something)\dots&=\int_{-1}^{1} \int_{-1}^{1} \left[\sum_{i=n+1}^{\infty} \sum_{j=n+1}^{\infty} a_{ij}\, p_{i}(x) \,p_{j}(y) \right]^2\, dx\, dy \qquad (1)\\ &=\int_{-1}^{1} \int_{-1}^{1} \sum_{i=n+1}^{\infty} \sum_{j=n+1}^{\infty} \left[a_{ij}^2\, p_{i}^2(x) \,p_{j}^2(y) \right]\, dx\, dy \, \qquad (2) \end{align} where $p_i(x)$ and $p_j(y)$ are Legendre polynomials.

Could you please explain how we can simplify equation $(1)$ to equation $(2)$? As you know infinite sums do not have many properties that finite sums have. By the way, the article can be found here. The above equations are on page 88.

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  • $\begingroup$ The second integral is the result of multiplying the four series of the first integral then using orthogonality. $\endgroup$
    – Leucippus
    Oct 17, 2022 at 18:00
  • $\begingroup$ Dear @Leucippus , I am sorry. I could not understand what you mean. Could you be more clear? $\endgroup$
    – feza
    Oct 17, 2022 at 19:00

1 Answer 1

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The use of $$ \int_{-1}^{1} P_{n}(x) \, P_{m}(x) \, dx = \frac{2}{2 n + 1} \, \delta_{n,m}$$ will be made as follows: \begin{align} I &= \int_{-1}^{1} \int_{-1}^{1} \left[\sum_{i=n+1}^{\infty} \sum_{j=n+1}^{\infty} b_{i,j} \, P_{i}(x) \, P_{j}(y) \right]^2 \, dx\, dy \\ &= \sum_{i} \sum_{j} \sum_{k} \sum_{m} b_{i,j} \, b_{k,m} \, \int_{-1}^{1} \int_{-1}^{1} P_{i}(x) \, P_{k}(x) \, P_{j}(y) \, P_{m}(y) \, dx \, dy \\ &= \sum_{i} \sum_{j} \sum_{k} \sum_{m} b_{i,j} \, b_{k,m} \, \int_{-1}^{1} P_{i}(x) \, P_{k}(x) \, dx \times \, \int_{-1}^{1} P_{j}(y) \, P_{m}(y) \, dy \\ &= \sum_{i} \sum_{j} \sum_{k} \sum_{m} \frac{4 \, b_{i,j} \, b_{k,m}}{(2 i+1)(2 j +1)} \, \delta_{i,k} \, \delta_{j,m} \\ &= \sum_{i} \sum_{j} \frac{4 \, b_{i,j}^{2}}{(2 i+1)(2 j+1)} \end{align} Now, since $$ \int_{-1}^{1} P_{n}^{2}(x) \, dx = \frac{2}{2 n + 1} $$ then \begin{align} I &= \sum_{i} \sum_{j} \frac{4 \, b_{i,j}^{2}}{(2 i+1)(2 j+1)} \\ &= \sum_{i} \sum_{j} b_{i,j}^{2} \, \int_{-1}^{1} \int_{-1}^{1} P_{i}^{2}(x) \, P_{j}^{2}(x) \, dx \\ &= \int_{-1}^{1} \int_{-1}^{1} \sum_{i=n+1}^{\infty} \sum_{j=n+1}^{\infty} \left( b_{i,j}^2 \, P_{i}^2(x) \, P_{j}^2(y) \right) \, dx\, dy. \end{align} This gives $$ \int_{-1}^{1} \int_{-1}^{1} \left[\sum_{i=n+1}^{\infty} \sum_{j=n+1}^{\infty} b_{i,j} \, P_{i}(x) \, P_{j}(y) \right]^2 \, dx\, dy = \int_{-1}^{1} \int_{-1}^{1} \sum_{i=n+1}^{\infty} \sum_{j=n+1}^{\infty} \left( b_{i,j}^2 \, P_{i}^2(x) \, P_{j}^2(y) \right) \, dx\, dy $$

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  • $\begingroup$ Dear @Leucippus , Great explanation. Thank you. $\endgroup$
    – feza
    Oct 19, 2022 at 0:10

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