Let c0, c1,..., cn be distinct points in F. Consider the polynomials p0,...,Pn P

Question

Let c0, c1,..., cn be distinct points in F. Consider the polynomials p0,...,Pn Pn(F) defined by pk(x) = (x - c0)(x - c1) ... (x - ck-1)(x-ck+1) ... (x - cn) / (ck-c0)(ck - c1) ... (ck - ck-1)(ck - ck+1) ... (ck - cn). Show that (po, ...Pn) is linearly independent. Hint: show that pk(cj) = Show that (p0, ..., pn) is a basis for Pn(F) Prove that given any values y0,...,yn F, there is a unique polynomial in p Pn(F) such that p(ci) = yi for i = 0,..., n. Find the real cubic that passes through the points: (0, 1), (1, 17), (2,-25), (3,pi) [Don't do this by solving a linear system. Use parts (a)-(c).]

Explanation / Answer

Actually these are the lagranges polynomials, but whatever :-)

(a)

i) pk(ck) = 1 since denumerator and numerator are the same for x=ck
ii) pk(cj) = 0 since j!=k then one of the (cj-cl) for l=1..n and l!=k will be zero

We note in general delta_{ij} (kronecker symbol) = 1 of i=j and 0 otherwise), i'll use it now

Now suppsose a0p0 + ... + anpn = 0

Let j in 0..n, by applying cj to this expression :

a0p0(cj) + ... + anpn(cj) = 0 => ak = 0 (since pk(cj)=delta_{kj})

So (p0,...,pn) is linearly independent.


(b) We have (n+1) set of linearly independent vector, which is a subset of Pn(F) , which has dimension n+1.
So by theorem (p0,...,pn) is a basis for Pn(F)


(c)

Suppose y0,...,yn in F

Write p(x) = sum(k=0..n) yk pk (which is the vector (y0,...,yn) is the basis (p0,...,pn))

Notice that p(ci) = y0p0(cj) + .. + ynpn(cj) = yj ( since pi(cj)=delta_{ij} )

So we have found a polynom such that p(ci)=yi

Now suppose we have found another polnyom q(x) such that q(ci)=yi

Decompose q(x) = sum(k=0..n) ak pk in the basis (p0,...,pn)

Again q(ci) = ai since pi(cj) = delta_{ij}

So q(x) = sum(k=0..n) q(ci)pk = sum(k=0..n) yi pk = p(x)

Hence the polynom is unique.


(d)


p0 = (x-1)(x-2)(x-3)/(0-1)(0-2)(0-3) = -1/6(x-1)(x-2)(x-3)   ( p0(0)=1 and p0(1)=p0(2)=p0(3)=0)
p1 = x(x-2)(x-3)/((1-0)(1-2)(1-3)) = 1/2 x(x-2)(x-3)         (p1(1)=1 and p1(0)=p1(2)=p1(3)=0)
p2 = x(x-1)(x-3)/((2-0)(2-1)(2-3)) = -1/2x(x-1)(x-3)         (p2(2)=1 and p2(0)=p2(1)=p2(3)=0)
p3 = x(x-1)(x-2)/((3-0)(3-1)(3-2)) = x(x-1)(x-2)/6           (p3(3)=1 and p3(0)=p3(1)=p3(2)=0)


So p(x) = 1*p0 + 17*p1 -25*p2 + Pi*p3

p(x) = -1/6(x-1)(x-2)(x-3) + 17/2 x(x-2)(x-3) +25/2x(x-1)(x-3) + Pi*x(x-1)(x-2)/6

No need to expand it, but if you need :
p(x) = (pi+125)/6x^3 -(183+Pi)/2x^2 + (Pi+260)/3 x + 1

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