User:Egm6321.f10.team2.patterson/Homework 6

= Problem 9: Verification that the Legendre Polynomials are Solutions to the Legendre Equation =

Given
The following equations are Legendre Polynomials, Pn(x), where n=0,1,2,3,4:


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$$ \displaystyle {{P}_{0}}\left( x \right)=1 $$ (eq 9.1)
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$$ \displaystyle {{P}_{1}}\left( x \right)=x $$ (eq 9.2)
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$$ \displaystyle {{P}_{2}}\left( x \right)=\frac{1}{2}\left(3x^{2}-1\right) $$ (eq 9.3)
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$$ \displaystyle {{P}_{3}}\left( x \right)=\frac{1}{2}\left(5x^{3}-3x\right) $$ (eq 9.4)
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$$ \displaystyle {{P}_{4}}\left( x \right)=\frac{1}{8}\left(35x^{4}-30x^{2}+3\right) $$    (eq 9.5)
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Verify that Equations 9.1, 9.2, 9.3, 9.4, and 9.5 are all solutions of the following Legendre Equation:
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$$ \displaystyle \left(1-x^2 \right)y'' -2x y' + n\left(n+1\right)y=0 $$ (eq 9.6)
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Solution
For each equation the following calculations are made:
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$$ \displaystyle y={{P}_{n}}\left( x \right) $$
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$$ \displaystyle y'=\frac{d}{dx}[{{P}_{n}}\left( x \right)] $$
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$$ \displaystyle y''=\frac{d}{dx}[\frac{d}{dx}[{{P}_{n}}\left( x \right)]] $$ Beginning with P0(x) {where n=0}, y=1 and y' and y" are both 0. Plugging these values into Equation 9.6 verifies it as a solution to the Legendre Equation as shown:
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$$ \displaystyle \cancel{\left(1-x^2 \right)[0]}^0 \cancel{-2x [0]}^0 + \cancel{(0)\left((0)+1\right)[1]}^0=0 $$ (eq 9.7)
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For P1(x) {where n=1}, y=x, y'=1 and y"=0. Plugging these values into Equation 9.6 verifies it as a solution to the Legendre Equation as shown:
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$$ \displaystyle \cancel{\left(1-x^2 \right)[0]}^0 -2x [1]+(1)\left((1)+1\right)[x]=0 $$ (eq 9.8)
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$$ \displaystyle \cancel{(-2+2)x}^0=0 $$ (eq 9.9) For P2(x) {where n=2}, y=1/2(3x2-1), y'=3x and y"=3. Plugging these values into Equation 9.6 verifies it as a solution to the Legendre Equation as shown:
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$$ \displaystyle \left(1-x^2 \right)[3] -2x [3x]+(2)\left((2)+1\right)[\frac{1}{2}\left(3x^2-1\right)]=0 $$ (eq 9.10)
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$$ \displaystyle \left(3-3x^2\right)-(6x^2)+(9x^2-3)=0 $$ (eq 9.11)
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$$ \displaystyle \cancel{(-3-6+9)x^2}^0+\cancel{(3-3)}^0=0 $$ (eq 9.12)
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For P3(x) {where n=3}, y=1/2(5x3-3x), y'=1/2(15x2-3) and y"=15x. Plugging these values into Equation 9.6 verifies it as a solution to the Legendre Equation as shown:
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$$ \displaystyle \left(1-x^2 \right)[15x] -2x \frac{1}{2}\left(15x^2-3\right)+(3)\left((3)+1\right)[\frac{1}{2}\left(5x^3-3x\right)]=0 $$ (eq 9.13)
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$$ \displaystyle \left(15x-15x^3\right)-(15x^3-3x)+(30x^3-18x)=0 $$ (eq 9.14)
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$$ \displaystyle \cancel{(-15-15+30)x^3}^0+\cancel{(15+3-18)x}^0=0 $$ (eq 9.15) For P4(x) {where n=4}, y=1/8(35x4-30x2+3), y'=1/8(140x3-60x) and y"=1/8(420x2-60). Plugging these values into Equation 9.6 verifies it as a solution to the Legendre Equation as shown:
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$$ \displaystyle \left(1-x^2 \right)[\frac{1}{8}\left(420x^2-60\right)] -2x \frac{1}{8}\left(140x^3-60x\right)+(4)\left((4)+1\right)[\frac{1}{8}\left(35x^4-30x^2+3\right)]=0 $$ (eq 9.16)
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$$ \displaystyle \left(-\frac{420}{8}x^4+60x^2-\frac{60}{8})\right)+(-35x^4+15x^2)+(\frac{175}{2}x^4-75x+\frac{15}{2})=0 $$ (eq 9.17)
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$$ \displaystyle \cancel{(-\frac{420}{8}-35+\frac{175}{2})x^4}^0+\cancel{(60+15-75)x^2}^0+\cancel{(-\frac{60}{8}+\frac{15}{2})}^0=0 $$ (eq 9.18)
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