How to prove this sequence of functions converge to a polynomial?

Question

Let (P_n) be a sequence of polynomials with real coefficients for which there exists a positive integer m such that degree(P_n) <= m for every n. Show that, if (P_n) converges for m +1 pairwise distinct elements of R, then (P_n) converges on R to a polynomial P. (Hint: Consider Lagrange Interpolation Formula)

Show that this conclusion may fail if the sequence (degree(P_n)) is unbounded.

Actually, if degree(P_n) is unbounded, then, even if P_n converges on R to a function f, f doesn't need to be a polynomial.

Answer 1

(What’s “pairwise distinct”? Isn’t that the same thing as “distinct”? Oh well…)

It sounds like you already have a handle on the counterexample, so here’s a proof of the first part.
Suppose P_n(x_j) -> y_j as n->oo for j = 0,1,2,…,m. Define

pi(x) = (x - x_0) (x – x_1) (x – x_2)…(x – x_m)

and let pi’(x_j) denote the same product but omitting the factor (x – x_j).

Define L_j(x) = pi(x) / ((x – x_j) pi’(x_j)).

The Lagrange interpolating polynomial for P_n, given its values at x_0, x_1,…,x_m is then

Sum{L_j(x) P_n(x_j); j= 0,1,…,m}.

Since P_n is a polynomial of degree at most m, it coincides with its interpolating polynomial, i.e.
P_n(x) = Sum{L_j(x) P_n(x_j); j= 0,1,…,m} for all x.

Taking limits, we see
lim P_n(x) = lim Sum{L_j(x) P_n(x_j); j= 0,1,…,m} = Sum{L_j(x) y_j; j=0,1,…,m}
as n-> oo, which is a polynomial.

Answer 2

Nealjking's answer is fairly stable. I even have no longer something exciting to function. I purely want to point out that, without pointing out it, he implicitly used the so observed as Lagrange Interpolation formula, which shows his matrix Z is definitely invertible. As for section 2, properly this is no longer a trivial difficulty. i will think of greater approximately it

Answer 3

tee-hee-hee