Answer to Question #102656 in Calculus for BIVEK SAH

Question

Expand the function
f(x)=x^4-1 in a series of the form

(infinity)sum(k=0)=AkPk(x)

Answer 1

$f(x)=x^4-1$

$f(x)=a_0P_0(x)+a_1P_1(x)+a_2P_2(x)+... \ ,\qquad x\in[-1,1]$

Multiplying both sides by $P_m(x)$ and integrating over the interval $[-1,1]$ , we have:

$a_m=\frac{1}2{}(2m+1)\int \limits _{-1}^{1} f(x)P_m(x)dx$

Then we can use Rodrigues formula for Legendre polynomials :

$a_m=(-1)^m \frac{2m+1}{2^{m+1}m!} \int\limits_{-1}^{1}f(x)\frac{d^m(1-x^2)^m}{dx^m} dx$

Integrating by parts several times, we have:

$a_m=\frac{2m+1}{2^{m+1}m!}\int\limits_{-1}^{1} \frac{d^mf(x)}{dx^m}(1-x^2)^mdx$

Now we can calculate coefficients $a_m:$

$a_0=\frac{1}{2}\int\limits_{-1}^{1}(x^4-1)dx=\frac{1}{2}(\frac{x^5}{5}-x)|_{-1}^{1}=-\frac{4}{5}$

$a_1=\frac{3}{4}\int\limits_{-1}^{1}4x^3 (1-x^2)dx= 3\int\limits_{-1}^{1}(x^3 -x^5)dx=3(\frac{x^4}{4}-\frac{x^6}{6})\vert_{-1}^{1}=0$

$a_2=\frac{5}{8\times 2!}\int\limits_{-1}^{1}12x^2(1-x^2)^2dx= \frac{15}{4}\int\limits_{-1}^{1}(x^2 -2x^4+x^6)dx=$

$\frac{15}{4}(\frac{x^3}{3}-\frac{2x^5}{5}+\frac{x^7}{7})|_{-1}^{1}=\frac{4}{7}$

$a_3=\frac{7}{16\times 3!}\int\limits_{-1}^{1}24x (1-x^2)^3dx= \frac{7}{4}\int\limits_{-1}^{1}(x-3x^3+3x^5-x^7)dx=$

$=\frac{7}{4}(\frac{x^2}{2}-3\frac{x^4}{4}+3\frac{x^6}{6}-\frac{x^8}{8})\vert_{-1}^{1}=0$

$a_4=\frac{9}{32\times 4!}\int\limits_{-1}^{1}24(1-x^2)^4dx= \frac{9}{32}\int\limits_{-1}^{1}(1-4x^2+6x^4–4x^6+x^8)dx=$

$=\frac{9}{32}(x-\frac{4x^3}{3}+\frac{6x^5}{5}-\frac{4x^7}{7}+\frac{x^9}{9})|_{-1}^{1}=\frac{8}{35}$

$a_m=\frac{2m+1}{2^{m+1}m!}\int\limits_{-1}^{1}0\times (1-x^2)^mdx=0, \;\; m\geq5$

$-\frac{4}{5}P_0(x)+\frac{4}{7}P_2(x)+\frac{8}{35}P_4(x)= -\frac{4}{5}\times 1 +\frac{4}{7}\times \frac{1}{2}(3x^2-1)+\frac{8}{35}\times \frac{35x^4-30x^2+3}{8}=$

$=-\frac{4}{5}+\frac{6}{7}x^2-\frac{2}{7}+x^4-\frac{6}{7}x^2+\frac{3}{35}=x^4-1$ .

Answer: $f(x)=-\frac{4}{5}P_0(x)+\frac{4}{7}P_2(x)+\frac{8}{35}P_4(x).$