Question #0ffa1

2 Answers
Dec 22, 2017

A bit of experimenting with linear polynomials of the form p(y)=ay+b reveals that p(y)=3/2 y is one example that works.

Explanation:

To start with, you could hope that a linear polynomial p(y)=ay+b will work.

The condition that int_{-1}^{1}p(y)dy=0 implies that
1/2 ay^{2}+by|_{-1}^{1}=0, or 2b=0, so b=0.

Since yp(y)=ay^{2}+by, the condition that int_{-1}^{1}yp(y)dy=1 implies that 1/3 ay^{3}+1/2 by^{2}|_{-1}^{1}=1. This leads to 2/3 a = 1 so that a=3/2.

Dec 22, 2017

See below.

Explanation:

Calling p_n(x)=sum_(k=0)^n a_k x^k we have

int_-1^1 sum_(k=0)^n a_k x^k dx = sum_(k=0)^n a_k/(k+1) (1-(-1)^(k+1))=0 or

sum_(k=0)^floor(n/2) a_(2k)/(2k+1) = 0

and

int_-1^1 x p_n(x)dx = int_-1^1 sum_(k=0)^n a_k x^(k+1) dx = sum_(k=0)^n a_k/(k+2) (1-(-1)^(k+2))=1 or

sum_(k=0)^floor(n/2) a_(2k-1)/(2k+1) = 1/2

Then, the polynomials obeying the conditions

{(sum_(k=0)^floor(n/2) a_(2k)/(2k+1) = 0),(sum_(k=0)^floor(n/2) a_(2k-1)/(2k+1) = 1/2):}

are solutions.