Proof of Taylor's formula for polynomials

Join the discussion
Ask a follow-up here, or get your own question answered by working scientists, mathematicians and engineers — people, not an autocomplete.
Real named experts · corrections over time · the nuance an AI answer skips
1 reply · 3K views
JG89
Messages
724
Reaction score
1
My book reads as follows:

"This is an entirely elementary algebraic formula concerning a polynomial in x or order n, say
[tex]f(x) = a_0 + a_1x + a_2x^2 + ... + a_nx^n[/tex].

If we replace x by a + h = b and expand each term in powers of h, there results immediately a representation of the form [tex]f(a+h) = c_0 = c_1h + c_2h^2 + ... + c_nh^n[/tex].

Taylor's formula is the relation: [tex]c_v = \frac{1}{v!}f^v(a)[/tex], for the coefficients c_v in terms of f and its derivatives at x = a. To prove this fact we consider the quantity h = b - a as the independent variable, and apply the chain rule which shows that differentiation with respect to h is the same as differentiation with respect to b = a + h."

I don't get how f'(h) = f'(a + h)?
 
Last edited:
Physics news on Phys.org
I think it's saying that if you have f'(h) and want f'(h+a), just substitute in h+a where you see h to get f'(h+a). This works because [tex]\frac{d}{dh} h+a = 1[/tex]