Question

Question #0ffa1

Answer 1

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`.

Answer 2

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.