6.5 Taylor's Theorem 333
Because f and Tn(x) have the same value and first n derivatives at a we
would expect the graphs of f(x) and Tn(x) to conform closely for values of
x near a. If this is not enough to convince you that the relationship between
a function and its nth Taylor polynomial is a very special one, the following
theorem may be enough.
Theorem 6.5.8 The nth Taylor polynomial Tn(x) is the unique nth degree
polynomial in powers of (x - a) with the properties identified in Theorem 6.5. 7.
n
That is, if p(x) =I: ak(x-al has the property that f(a) = p(a), f'(a) = p'(a),
k=O
f"(a) = p"(a), · · ·, f(n)(a) = p(n)(a), then the coefficients in p(x) are identical
f(k)(a)
to the coefficients inTn(x): i.e., Vk=0, 1,··· , n , ak = -k-,-.
Proof. Suppose that
g(x) = ao + a1(X - a)+ a2(X - a)^2 + · · · + an(X - ar, and
h(x) = bo + b1(x - a)+ b2(x - a)^2 + · · · + bn(x - a)n
are polynomials such that
g(a) = h(a), g'(a) = h'(a), g"(a) = h"(a), · · ·, g(n)(a) = h(nl(a).
Then,
g(a) = h(a) ==? ao + 0 + · · · + 0 = bo + 0 + · · · + 0. Thus,
ao = bo.
Differentiating g(x) and h(x), we have
g'(x) = a1 + 2a2(x - a)+ 3a3(x - a)^2 + · · · + nan(x - a)n-l;
h'(x) = b1 + 2b2(x - a)+ 3b3(x - a)^2 + · · · + nbn(x - a)n-^1.
Then g'(a) = h'(a) ==? a 1 + 0 + · · · + 0 = b1 + 0 + · · · + 0. Thus,
a1 = b1.
Differentiating again, we have
g"(x) = 2a2 + 2 · 3a3(X - a)+ 3 · 4a3(X - a)^2 + · · · + n(n - l)an(X - ar-^2 ;
h"(x) = 2b2 + 2 · 3b3(x - a)+ 3 · 4b3(x - a)^2 + · · · + n(n - l)bn(x - a)n-^2.
Then g"(a) = h"(a) ==? 2a 2 + 0 + · · · + 0 = 2b2 + 0 + · · · + 0. Thus,
a2 = b2.
Continuing in this way, we obtain a3 = b3, a4 = b4, · · ·, an = bn. That is ,
the coefficients of g and hare identical. Hence, there can be only one nth degree
polynomial in powers of (x -a) with the properties identified in Theorem 6.5.7.
To study the difference between f and its nth degree Taylor polynomial
about a, we introduce a notation for this difference.
