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
Σ
k = 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.

The Taylor polynomial approximates f(x) near a.
Abbiamo che
• se p(x) è il polinomio di Taylor di ordine n di f(x):
  f(x) = p(x) + R(x) con R(x) = o((x-a)n)
ossia quanto "resta" da aggiungere a p(x) per ottenere f(x) è, per x → a, un infinitesimo di ordine superiore rispetto a (x-a)n
e che
• se esiste f(n+1)(a),  R(x) = O((x-a)n+1)
ossia quanto "resta" da aggiungere a p(x) per ottenere f(x) è, per x → a, un infinitesimo dello stesso ordine o di ordine superiore rispetto a (x-a)n+1
[clicca qui per approfondimenti]
  Taylor's Theorem gives bounds for the error in this approximation.

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

Examples of Taylor Series about 0

  1. For f(x) = ex,

    f(k)(x) = ex    =>    f(k)(0) = 1.

    So

    ex =
    1 + x + x2
    2!
    		 + x3
    3!
    		 + …
    =
    ∞
    Σ
    k = 0 
    		 xk
    k!
    which converges for all x since

    lim
    n→∞    		 Rn(x)

    		 =

    		lim
    n→∞    		 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 - x2
    2
    		 + x3
    3
    		 - x4
    4
    		 + …
    =
    ∞
    Σ
    k = 0 
    		 (-1)k xk+1
    k+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 - 4
    3
    		 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
    1
    1+x
    		 = 1-x+x2-x3+….

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

    1
    1-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
Σ
k = 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.