Prove the following statement:
Let f be a polynomial, which can be written in the form
fix) = a(n)X^(n) + a(n-1)X^(n-1) + • • • + a0
and also in the form
fix) = b(n)X^(n) + b(n-1)X^(n-1) + • • • + b0
Prove that a(i)=b(i) for all i=0,1,2,...,n-1,n
3. The Attempt at a Solution [/B]
0 = f(x) - f(x) = (a(n)-b(n))X^(n) + (a(n)-b(n))X^(n) + ... + a0 - b0
Need to prove that Di = (a(i)-b(i)) = 0 for i = 0,1,2,3,4,...n
But I do not know how.
This is the answer from book, using the method of contradiction:
Suppose that there exists some index i such that Di does not equal 0.
Let m be the largest of these indices, so that we can write
0 = D(m)X^(m) + ... + D0 for all x and D(m) does not equal 0. This contradicts Theorem 2. Therefore we conclude that Di = 0 for all i = 1, . . . , n, thus proving the theorem.
Theorem 2 is as stated:
Let f be a polynomial. Let a0,a1,...,a(n-1),a(n) be numbers such that a(n) does not equal 0, and such that we have: f(x) = a0 + a1X + ... + a(n)X^(N) for all x. Then f has at most n roots.
I don't understand the contradiction between theorem 2 and the statement regarding D(m). Could someone explain?