Answer on Question #85176 – Math – Differential Equations
Question
Expand the function f ( x ) = P 3 ′ ( x ) f(x) = P_3'(x) f ( x ) = P 3 ′ ( x )
∑ k = 0 ∞ A k P k ( x ) , where P 3 ( x ) = 5 x 3 − 3 x 2 . \sum_{k=0}^{\infty} A_k P_k(x), \text{ where } P_3(x) = \frac{5x^3 - 3x}{2}. k = 0 ∑ ∞ A k P k ( x ) , where P 3 ( x ) = 2 5 x 3 − 3 x .
The question is based on Legendre polynomials.
Solution
If P 3 ( x ) = 5 x 3 − 3 x 2 P_3(x) = \frac{5x^3 - 3x}{2} P 3 ( x ) = 2 5 x 3 − 3 x f ( x ) = 3 ( 5 x 2 − 1 ) 2 f(x) = \frac{3(5x^2 - 1)}{2} f ( x ) = 2 3 ( 5 x 2 − 1 )
We give the function f ( x ) f(x) f ( x ) − 1 ≤ x ≤ 1 -1 \leq x \leq 1 − 1 ≤ x ≤ 1
f ( x ) = ∑ k = 0 ∞ A k P k ( x ) f(x) = \sum_{k=0}^{\infty} A_k P_k(x) f ( x ) = ∑ k = 0 ∞ A k P k ( x ) P 0 = 1 , P 1 = x , P 2 = 3 x − 1 2 , … , P k = 1 2 k k ! d k ( x 2 − 1 ) k d x k P_0 = 1, P_1 = x, P_2 = \frac{3x - 1}{2}, \ldots, P_k = \frac{1}{2^k k!} \frac{d^{k(x^2 - 1)^k}}{dx^k} P 0 = 1 , P 1 = x , P 2 = 2 3 x − 1 , … , P k = 2 k k ! 1 d x k d k ( x 2 − 1 ) k A k = 2 k + 1 2 ∫ − 1 1 f ( x ) P k ( x ) d x A_k = \frac{2k + 1}{2} \int_{-1}^{1} f(x) P_k(x) dx A k = 2 2 k + 1 ∫ − 1 1 f ( x ) P k ( x ) d x
In our case A k = 0 A_k = 0 A k = 0 k = 2 m + 1 , m = 0 , 1 , 2 , 3 , … k = 2m + 1, m = 0,1,2,3,\ldots k = 2 m + 1 , m = 0 , 1 , 2 , 3 , … f ( x ) = 3 ( 5 x 2 − 1 ) 2 f(x) = \frac{3(5x^2 - 1)}{2} f ( x ) = 2 3 ( 5 x 2 − 1 )
A 0 = 1 2 ∫ − 1 1 ( 3 ( 5 x 2 − 1 ) 2 × 1 ) d x = ∫ 0 1 ( 3 ( 5 x 2 − 1 ) 2 ) d x = ( 5 x 3 2 − 3 x 2 ) ∣ 0 1 = 1 , A_0 = \frac{1}{2} \int_{-1}^{1} \left( \frac{3(5x^2 - 1)}{2} \times 1 \right) dx = \int_{0}^{1} \left( \frac{3(5x^2 - 1)}{2} \right) dx = \left( \frac{5x^3}{2} - \frac{3x}{2} \right) \Big|_{0}^{1} = 1, A 0 = 2 1 ∫ − 1 1 ( 2 3 ( 5 x 2 − 1 ) × 1 ) d x = ∫ 0 1 ( 2 3 ( 5 x 2 − 1 ) ) d x = ( 2 5 x 3 − 2 3 x ) ∣ ∣ 0 1 = 1 , A 2 = 5 ∫ 0 1 3 ( 5 x 2 − 1 ) 2 × 3 x 2 − 1 2 d x = 15 4 ∫ 0 1 ( 15 x 4 − 5 x 2 − 3 x 2 + 1 ) d x = 15 4 ( 3 x 5 − 8 x 3 3 + x ) ∣ 0 1 = 5 , A_2 = 5 \int_{0}^{1} \frac{3(5x^2 - 1)}{2} \times \frac{3x^2 - 1}{2} dx = \frac{15}{4} \int_{0}^{1} (15x^4 - 5x^2 - 3x^2 + 1) dx = \frac{15}{4} \left( 3x^5 - \frac{8x^3}{3} + x \right) \Big|_{0}^{1} = 5, A 2 = 5 ∫ 0 1 2 3 ( 5 x 2 − 1 ) × 2 3 x 2 − 1 d x = 4 15 ∫ 0 1 ( 15 x 4 − 5 x 2 − 3 x 2 + 1 ) d x = 4 15 ( 3 x 5 − 3 8 x 3 + x ) ∣ ∣ 0 1 = 5 , A 4 = 9 ∫ 0 1 3 ( 5 x 2 − 1 ) 2 × 35 x 4 − 30 x 2 + 3 8 d x = 9 ( 75 x 7 16 − 111 x 5 16 + 15 x 3 16 + 9 x 16 ) ∣ 0 1 = − 27 4 , A_4 = 9 \int_{0}^{1} \frac{3(5x^2 - 1)}{2} \times \frac{35x^4 - 30x^2 + 3}{8} dx = 9 \left( 75 \frac{x^7}{16} - 111 \frac{x^5}{16} + 15 \frac{x^3}{16} + 9 \frac{x}{16} \right) \Big|_{0}^{1} = -\frac{27}{4}, A 4 = 9 ∫ 0 1 2 3 ( 5 x 2 − 1 ) × 8 35 x 4 − 30 x 2 + 3 d x = 9 ( 75 16 x 7 − 111 16 x 5 + 15 16 x 3 + 9 16 x ) ∣ ∣ 0 1 = − 4 27 , … , \ldots, … , A 2 k = 3 ( 4 k + 1 ) 2 2 k + 1 ( 2 k ) ! ∫ 0 1 ( 5 x 2 − 1 ) × d 2 k ( x 2 − 1 ) 2 k d x 2 k d x . A_{2k} = \frac{3(4k + 1)}{2^{2k + 1}(2k)!} \int_{0}^{1} (5x^2 - 1) \times \frac{d^{2k(x^2 - 1)^{2k}}}{dx^{2k}} dx. A 2 k = 2 2 k + 1 ( 2 k )! 3 ( 4 k + 1 ) ∫ 0 1 ( 5 x 2 − 1 ) × d x 2 k d 2 k ( x 2 − 1 ) 2 k d x .
Then
3 ( 5 x 2 − 1 ) 2 = 1 + 5 ( 3 x 2 − 1 ) 2 − 27 ( 35 x 4 − 30 x 2 + 3 ) 32 + … = ∑ 2 k = 0 ∞ 3 ( 4 k + 1 ) 2 4 k + 1 ( ( 2 k ) ! ) 2 × d 2 k ( x 2 − 1 ) 2 k d x 2 k ∫ 0 1 ( 5 x 2 − 1 ) × d 2 k ( x 2 − 1 ) 2 k d x 2 k d x . \frac{3(5x^2 - 1)}{2} = 1 + \frac{5(3x^2 - 1)}{2} - \frac{27(35x^4 - 30x^2 + 3)}{32} + \ldots = \sum_{2k=0}^{\infty} \frac{3(4k + 1)}{2^{4k + 1}((2k)!)^2} \times \frac{d^{2k(x^2 - 1)^{2k}}}{dx^{2k}} \int_{0}^{1} (5x^2 - 1) \times \frac{d^{2k(x^2 - 1)^{2k}}}{dx^{2k}} dx. 2 3 ( 5 x 2 − 1 ) = 1 + 2 5 ( 3 x 2 − 1 ) − 32 27 ( 35 x 4 − 30 x 2 + 3 ) + … = 2 k = 0 ∑ ∞ 2 4 k + 1 (( 2 k )! ) 2 3 ( 4 k + 1 ) × d x 2 k d 2 k ( x 2 − 1 ) 2 k ∫ 0 1 ( 5 x 2 − 1 ) × d x 2 k d 2 k ( x 2 − 1 ) 2 k d x . Answer: 3 ( 5 x 2 − 1 ) 2 = ∑ 2 k = 0 ∞ 3 ( 4 k + 1 ) 2 4 k + 1 ( ( 2 k ) ! ) 2 × d 2 k ( x 2 − 1 ) 2 k d x 2 k ∫ 0 1 ( 5 x 2 − 1 ) × d 2 k ( x 2 − 1 ) 2 k d x 2 k d x . \text{Answer: } \frac{3(5x^2 - 1)}{2} = \sum_{2k=0}^{\infty} \frac{3(4k + 1)}{2^{4k + 1}((2k)!)^2} \times \frac{d^{2k(x^2 - 1)^{2k}}}{dx^{2k}} \int_{0}^{1} (5x^2 - 1) \times \frac{d^{2k(x^2 - 1)^{2k}}}{dx^{2k}} dx. Answer: 2 3 ( 5 x 2 − 1 ) = 2 k = 0 ∑ ∞ 2 4 k + 1 (( 2 k )! ) 2 3 ( 4 k + 1 ) × d x 2 k d 2 k ( x 2 − 1 ) 2 k ∫ 0 1 ( 5 x 2 − 1 ) × d x 2 k d 2 k ( x 2 − 1 ) 2 k d x .
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