Four proof that exercises about taylor polynomials

[colt]

Four "proof that" exercises about taylor polynomials

Homework Statement
Definition: A function f is called C^n if f n times derivable and if the n-time derivable f^(n) is continuos. If is from class C^n then its called f ε C^n

Exercise 1) Be f ε C^n in the interval [a,x]. Be P a polynomial of degree n such that P^(k) (a) = f^(k) (a), for all k = 0,1,2..n. Proof that lim (x->a) [f(x) - P(x)] / (x-a)^n = 0.

Suggestion: apply L'Hopital rule n times then uses the continuity

The Attempt at a Solution

Well, deriving n times using L'Hopital I come with: lim (x->a) [f´(n) (x) - P´(n) (x)] / n*(n-1)*(n-2)*... (2)*(1)*(x-a)^(n-n) = 0

-> lim(x->a) [f´(n) (x) - P´(n) (x)] / n! = 0
Then I don't know how to proceed


Homework Statement
2)The following exercise is a modification from the first that doesn't demand the continuity of f'(n), but limits the choice of the P polynomial. Be a ε R e be f a function such that f'(a),...f'(n) (a) all exist. Be P a polynomial of degree n such that P^(k) (a) = f^(k) (a), for all k = 0,1,2..n., like before, but now under an extra condition:
P(x) = [f'(n) (a) * (x-a)^n] / n! + [f'(n-1) (a)*(x-a)^(n-1)] / (n-1)! +... (inferior order terms)

So: lim (x->a) [f(x) - P (x)] / (x-a)^n = 0
Sugestion applys the L'Hopital rule n-1 times and then use the properties of P

The Attempt at a Solution

P(x) has too many terms, so I can't imagine what it will look alike after I derivate each one N-1 times. Plus in the first
one I was thinking in derive n times not n-1


Homework Statement
3)a)Be P and Q two polynomials em (x-a) of degree <=n. Suppose that P and Q are equals until order n in a.
Be R:=P-Q. Proof that lim (x->a) R(x)/(x-a) = 0 for all k=0,1,2... Note that R(x)/(x-a)^k = R(x)*(x-a)^(n-k)/(x-a)^n

The Attempt at a Solution

No clue about how to do it or the tip. And := means approximately?

Homework Statement
3)b)Realize that P=Q
Suggestion: All is necessary is to prove that R=0. To do so, if R(x) = b0 + b1*(x-a)+...+bn*(x-a)^n use the previous item to prove that b0 = 0. So should be R(x) = b1 (x-a) + ...+bn*(x-a)^n and the previous can be used once more to demonstrate now that b1=0 and keeping so until bn.

The Attempt at a Solution

Depends on the previous item

 

[kurt.physics]

, but I think that by proving that R=0, then each coefficient of the polynomial must be 0. And to do so, I have to prove that R(x)/(x-a)^k = 0?