User:Egm6322.s12.team2.Xia/RP8.4

=R*8.4 -Legendre Function =

Given
The following pairs: $$\displaystyle \{P_{0},\, Q_{0}\},\,\{P_{1},\! Q_{1}\},\cdots\,,\{P_{4},\, Q_{4}\}$$ (8.4.1)

where $$\displaystyle \{P_{i},\, Q_{i}\} $$ are Legendre polynomials.

Find
1.Plot the 8.4.1 in separate figures. 2.Observe even-ness and odd-ness of 8.4.1 and guess the value of the scalar products:

$$\displaystyle =\int_{\mu=-1}^{\mu=+1}P_{i}(\mu)Q_{i}(\mu)du=?$$

Solution
1.The Legendre polynomial is as follows: $$\displaystyle {P}_{n}(x)=\sum^{m}_{r=0}(-1)^r \frac{(2n-2r)!x^{n-2r}}{2^{n}r!}(n-r)!(n-2r)! $$ where $$\displaystyle m = \frac{n}{2}$$. $$\displaystyle {P}_{0}(x) = 1 $$

$$\displaystyle {P}_{1}(x) = x $$

$$\displaystyle {P}_{2}(x) = \frac{1}{2}(3x^2-1) $$

$$\displaystyle {P}_{3}(x) = \frac{1}{2}(5x^3-3x) $$

$$\displaystyle {P}_{4}(x) = \frac{35\,{x}^{4}}{8}-\frac{15\,{x}^{2}}{4}+\frac{3}{8} $$ and $$\displaystyle {Q}_{0}(x) = \frac{1}{2}\mathrm{log}\left(\frac{1+x}{1-x}\right) $$

$$\displaystyle {Q}_{1}(x) = \frac{1}{2}x \mathrm{log}\left(\frac{1+x}{1-x}\right)-1 $$

$$\displaystyle {Q}_{2}(x) = \frac{1}{4}(3x^2-1)\mathrm{log}\left(\frac{1+x}{1-x}\right)-\frac{3}{2}x $$

$$\displaystyle {Q}_{3}(x) = \frac{1}{4}(5x^3-3x)\mathrm{log}\left(\frac{1+x}{1-x}\right)-\frac{5}{2}x^2+\frac{2}{3} $$

$$\displaystyle {Q}_{4}(x) = \frac{35\,{x}^{4}\,\mathrm{log}\left( -\frac{x}{x-1}-\frac{1}{x-1}\right) }{16}-\frac{15\,{x}^{2}\,\mathrm{log}\left( -\frac{x}{x-1}-\frac{1}{x-1}\right) }{8}+\frac{3\,\mathrm{log}\left( -\frac{x}{x-1}-\frac{1}{x-1}\right) }{16}-\frac{35\,{x}^{3}}{8}+\frac{55\,x}{24} $$ Using Maxima with GNUplot we get the following figures: 2. It is easy to find that $$\displaystyle {P}_{0}(x) = 1={P}_{0}(-x) $$

$$\displaystyle {P}_{1}(x) = x= -{P}_{1}(-x) $$

$$\displaystyle {P}_{2}(x) = \frac{1}{2}(3x^2-1)= {P}_{2}(-x)$$

$$\displaystyle {P}_{3}(x) = \frac{1}{2}(5x^3-3x)= -{P}_{2}(-x) $$ Thus, we are able to guess that $$\displaystyle {P}_{2n}(x)={P}_{2n}(-x) $$ are even-ness while $$\displaystyle {P}_{2n+1}(x)=-{P}_{2n+1}(-x) $$ are odd-ness. Using the same method, we will find that $$\displaystyle {Q}_{2n}(x)=-{Q}_{2n}(-x) $$ are odd-ness while $$\displaystyle {Q}_{2n+1}(x)={Q}_{2n+1}(-x) $$ are even-ness. From the above, we can easily find that $$\displaystyle {P}_{i}(x){Q}_{i}(x) $$ are odd-ness, Since the integral of smooth odd functions from -a to a are zero. That's to say $$\displaystyle =\int_{\mu=-1}^{\mu=+1}P_{i}(\mu)Q_{i}(\mu)du= 0$$