Question

Help me

Answer 1

Expand \(x^{3}-3x^{2} + x \) into \(\sum_{k=0}^{\infty} A_{k} P_{k}^{m} (x)\)

Answer 2

$$f(x)=x^3-3x^2+x\\A_k=\frac{2k+1}2\int_{-1}^1 P_k(x)f(x)\,dx$$Evaluate for some \(k\):$$P_0(x)=1\\A_0=\frac12\int_{-1}^1f(x)\,dx=\frac12\left[\frac14x^4-x^3+\frac12x^2\right]_{-1}^1=-1\\P_1(x)=x\\A_1=\frac32\int_{-1}^1xf(x)\,dx=\frac32\left[\frac15x^5-\frac34x^4+\frac13x^3\right]_{-1}^1=\frac85\\P_2(x)=\frac12(3x^2-1)\\A_2=\frac52\int\frac12(3x^2-1)f(x)\ dx=-2\\P_3(x)=\frac12(5x^3-3x)\\A_3=\frac72\int\frac12(5x^3-3x)f(x)\,dx=\frac25$$... and were done, since \(x^3-3x^2+x\) is a polynomial of degree \(3\). Our corresponding Legendre expansion is then:$$f(x)=P_0(x)+\frac85P_1(x)+-2P_2(x)+\frac25P_3(x)$$

Answer 3

oops \(f(x)=-P_0(x)+\dfrac85P_1(x)-2P_2(x)+\dfrac25P_3(x)\)