Taylor's Theorem

Suppose we're working with a function f that has derivatives up to the n-th order on an interval about 0.
We can approximate f(x) near x=0 by a polynomial Pn(x) of degree n:

• For n = 0, the best constant approximation near 0 is

 P0(x) = f(0)
which matches f at 0.

• For n = 1, the best linear approximation near 0 is

 P1(x) = f(0)+f ′(0)x.

Note that P1 matches f at 0 and P1 ′ matches f ′ at 0.

• For n = 2, the best quadratic approximation near 0 is

 P2(x) = f(0)+f ′(0)x+ f ′ ′(0)2! x2.

Note that P2, P2 ′, and P2  ′ match f, f ′, and f  ′, respectively, at 0.

Continuing this process,

 Pn(x) = f(0)+f ′(0)x+ f ′ ′(0)2! x2+…+ f(n)(0)n! xn.

This is the Taylor polynomial of degree n about 0. More generally, if f has n derivatives at x = a, the Taylor polynomial of degree n about a is

 n Σk = 0 f(k)(a)k! (x-a)k = f(a) + f ′(a)(x-a) + f ′ ′(a)2! (x-a)2 + … + f(n)(a)n! (x-a)n.

The Taylor polynomial of degree n about 0 is also called the Maclaurin polynomial of degree n.

Taylor's Theorem

Suppose f has n+1 derivatives on an open interval containing a. Then for each x in the interval,

 f(x) = n Σk = 0 f(k)(a)k! (x-a)k +Rn+1(x)
where the error term Rn+1(x) satisfies Rn+1(x) = (f(n+1)(c) / (n+1)!) (x-a)n+1 for some c strictly between a and x.
[basta che f(n) sia continua e che f(n+1) esista tra c ed x, non anche in c e in x]

This form for the error Rn+1(x), derived in 1797 by Joseph Lagrange, is called the Lagrange formula for the remainder. The infinite Taylor series converges pointwise [puntualmente, punto per punto] to f,

 f(x) = ∞ Σk = 0 f(k)(a)k! (x-a)k,
in I ∋ a   if and only if  for all x in I  limn→∞ Rn(x) = 0.

Nota. La serie di Taylor è stata illustrata da Taylor in un suo libro del 1715, ma era già stata introdotta qualche anno prima. La serie di MacLaurin fu invece studiata da Taylor. MacLaurin ottene invece risultati significativi in ambito geometrico. …

Examples of Taylor Series about 0

1. For f(x) = ex,

 f(k)(x) = ex    =>    f(k)(0) = 1.

So

ex =
 1 + x + x22! + x33! + …
=
 ∞Σk = 0 xkk!
which converges for all x since

 limn→∞ Rn(x) = limn→∞ ecx(n+1) / (n+1)! = 0
for all c between 0 and x.

2. For f(x) = ln(1+x),

 f ′(x) = 1/(1+x)
 f ′ ′(x) = -1/(1+x)2
 f ′ ′ ′(x) = 2/(1+x)3
 f(4)(x) = (-3·2)/(1+x)4
 ···

=>
 f(0) = 0
 f ′(0) = 1
 f ′ ′(0) = -1
 f ′ ′ ′(0) = 2
 f(4)(0) = -6

So

 ln(1+x)
=
 x - x22 + x33 - x44 + …
=
 ∞Σk = 0 (-1)k xk+1k+1
which converges only for -1 < x ≤ 1.

3. For  f(x) = sin(x),

f '(x) = cos(x),  f "(x) = −sin(x),  f(3)(x) = −cos(x),  f(4)(x) = sin(x), ... so:
sin(x) = x − x3/3! + x5/5! − x7/7! + ... =
k = 0...∞ (−1)k x2k+1 / (2k+1)!
and for  f(x) = cos(x),
f '(x) = −sin(x),  f "(x) = −cos(x),  f(3)(x) = sin(x),  f(4)(x) = cos(x), ... so:
cos(x) = 1 − x2/2! + x4/4! − x6/6! + ... =
k = 0...∞ (−1)k x2k / (2k)!
which converge for all x.

The Taylor Series in (x-a) is the unique power series in (x-a) converging to f(x) on an interval containing a. For this reason,

• By Example 1,

 e-2x = 1 - 2x + 2x2 - 43 x3+…
where we have substituted -2x for x.

• By Example 2, since d/dx[ln(1+x)] = 1/(1+x), we can differentiate the Taylor series for ln(1+x) to obtain
 11+x = 1-x+x2-x3+….

Substituting -x for x  (riabbiamo - vedi Infinite Series Convergence):

 11-x = 1+x+x2+x3+….

Clicca per altri esempi ed applicazioni (per calcolo di approssimazioni e di limiti)

In the Exploration, compare the graphs of various functions with their first Taylor polynomials about x = 0.

Exploration

Key Concept [index]

Taylor's Theorem

Suppose f has n+1 continuous derivatives on an open interval containing a. Then for each x in the interval,

 f(x) = n Σk = 0 f(k)(a)k! (x-a)k +Rn+1(x)
where the error term Rn+1(x) satisfies Rn+1(x) = [f(n+1)(c) / (n+1)!](x-a)n+1 for some c between a and x.