I need some guidance for the following:
For the polynomial q(x) = (x − 1)(x − 2)(x − 3)(x − 4), with degree 4, describe the basis for P3 that partial fractions asserts we have, and demonstrate that the collection of polynomials are linearly independent; that is, that a1p1(x) + a2p2(x) + a3p3(x) + a4p4(x) = 0 implies that a1 = a2 = a3 = a4 = 0. Explain why this implies that your collection of polynomials is therefore a basis for P3.
I know that Pn has the “standard” basis consisting of the polynomials (with one term, usually called monomials)
{1, x, x^2, . . . , x^n}
because they span Pn - the definition, esssentially, of polynomials as the functions
p(x) = a0^xn+a1^xn−1+· · ·+an−1^x+a^n
express polynomials as linear combinations of monomials. The monomials are also linearly independent:
If p(x) = a0x n+a1x n−1+· · ·+an−1x+an = 0, then plugging in x = 0 gives p(0) = an = 0, so p(x) = a0x n + a1x n−1 + · · · + an−1x = x[a0x n−1+a1x n−2+· · ·+an−1] = 0, so a0x n−1+a1x n−2+· · ·+an−1 = 0, as well. Repeating this argument (plugging in x = 0) will repeatedly show another coefficient is equal to zero, resulting in an = an−1 = · · · = a1 = a0 = 0, establishing linear independence.
Start with $p(x)=a_0+a_1x+a_2x^2+a_3x^3$ and use the partial fractions setup $$\frac{p(x)}{q(x)}=\frac{b_1}{x-1}+\frac{b_2}{x-2}+\frac{b_3}{x-3}+\frac{b_4}{x-4}.$$ Now as usual multiply both sides by $q(x)=(x-1)(x-2)(x-3)(x-4)$ and you'll get each $b_k$ multiplied by the product of three of the four factors $x-i$ on the right, and on the left you have the polynomial $p(x)$ with its four coefficients (up to third power). By equating coefficients you can get the values of the $b_i$ in terms of the $a_i.$ It would seem the basis referred to is just that obtained by multiplying (expanding out) each product of three of the four terms $x-i$, because then by use of the multipliers $b_i$ you get any cubic $p(x)$ as a linear combination of those four cubics.