# Proof of Taylor's formula for polynomials

1. Apr 12, 2009

### JG89

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

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

Taylor's formula is the relation: $$c_v = \frac{1}{v!}f^v(a)$$, 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: Apr 12, 2009
2. Apr 12, 2009

### qntty

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 $$\frac{d}{dh} h+a = 1$$