Post on 17-Jan-2017
Taylor Expansions
Taylor Expansions In this section we find the polynomials and series expansions in (x – a) of infinitely differentiable functions at x = a.
Taylor Expansions In this section we find the polynomials and series expansions in (x – a) of infinitely differentiable functions at x = a. They are called Taylor expansions.
Taylor Expansions In this section we find the polynomials and series expansions in (x – a) of infinitely differentiable functions at x = a. They are called Taylor expansions.
Given a function f(x) that is infinitely differentiable at x = a, we define the n'th Taylor polynomial of f(x) as:
f(1)(a)(x – a) += f(a) +pn(x)1!
f(2)(a)(x – a)2+ ..2! + f(n)(a)(x – a)n
n!
Taylor Expansions In this section we find the polynomials and series expansions in (x – a) of infinitely differentiable functions at x = a. They are called Taylor expansions.
Given a function f(x) that is infinitely differentiable at x = a, we define the n'th Taylor polynomial of f(x) as:
f(1)(a)(x – a) += f(a) +pn(x)1!
pn(x) = Σk=0
n (x – a)k
k! f(k)(a)
f(2)(a)(x – a)2+ ..2! + f(n)(a)(x – a)n
n!or
Taylor Expansions In this section we find the polynomials and series expansions in (x – a) of infinitely differentiable functions at x = a. They are called Taylor expansions.
Given a function f(x) that is infinitely differentiable at x = a, we define the n'th Taylor polynomial of f(x) as:
f(1)(a)(x – a) += f(a) +pn(x)1!
pn(x) = Σk=0
n (x – a)k
k! f(k)(a)
f(2)(a)(x – a)2+ ..2! + f(n)(a)(x – a)n
n!or
TheTaylor series of f(x) is defined as:
P(x) = Σk=0
(x – a)k
k! f(k)(a)∞
Example: A. Find the Taylor-expansion of f(x) = 1 + x + x2 around x = 1.
Taylor Expansions
Example: A. Find the Taylor-expansion of f(x) = 1 + x + x2 around x = 1.We need the derivatives of f(x):
Taylor Expansions
Example: A. Find the Taylor-expansion of f(x) = 1 + x + x2 around x = 1.We need the derivatives of f(x):f(1) = 1 + 1 + 1 f(1) = 3
Taylor Expansions
Example: A. Find the Taylor-expansion of f(x) = 1 + x + x2 around x = 1.We need the derivatives of f(x):
(1)
f(1) = 1 + 1 + 1 f(1) = 3
Taylor Expansions
f (x) = 1 + 2x
Example: A. Find the Taylor-expansion of f(x) = 1 + x + x2 around x = 1.We need the derivatives of f(x):
(1) (1)
f(1) = 1 + 1 + 1 f(1) = 3
Taylor Expansions
f (x) = 1 + 2x f (1) = 3
Example: A. Find the Taylor-expansion of f(x) = 1 + x + x2 around x = 1.We need the derivatives of f(x):
(1) (1)
f(1) = 1 + 1 + 1 f(1) = 3
Taylor Expansions
f (x) = 1 + 2x f (1) = 3 (2) (2)f (x) = 2 f (1) = 2
Example: A. Find the Taylor-expansion of f(x) = 1 + x + x2 around x = 1.We need the derivatives of f(x):
(1) (1)
f(1) = 1 + 1 + 1 f(1) = 3
Taylor Expansions
f (x) = 1 + 2x f (1) = 3 (2) (2)f (x) = 2 f (1) = 2 (n)f (1) = 0 for n > 3.
Example: A. Find the Taylor-expansion of f(x) = 1 + x + x2 around x = 1.We need the derivatives of f(x):
(1) (1)
f(1) = 1 + 1 + 1 f(1) = 3
Taylor Expansions
f (x) = 1 + 2x f (1) = 3 (2) (2)f (x) = 2 f (1) = 2 (n)f (1) = 0 for n > 3. Hence,
= f(1) = 3p0(x)
Example: A. Find the Taylor-expansion of f(x) = 1 + x + x2 around x = 1.We need the derivatives of f(x):
(1) (1)
f(1) = 1 + 1 + 1 f(1) = 3
Taylor Expansions
f (x) = 1 + 2x f (1) = 3 (2) (2)f (x) = 2 f (1) = 2 (n)f (1) = 0 for n > 3. Hence,
= f(1) = 3p0(x)f(1)(1)(x – 1)= f(1) +p1(x)
1!
Example: A. Find the Taylor-expansion of f(x) = 1 + x + x2 around x = 1.We need the derivatives of f(x):
(1) (1)
f(1) = 1 + 1 + 1 f(1) = 3
Taylor Expansions
f (x) = 1 + 2x f (1) = 3 (2) (2)f (x) = 2 f (1) = 2 (n)f (1) = 0 for n > 3. Hence,
= f(1) = 3p0(x)f(1)(1)(x – 1)= f(1) +p1(x)
1! = 3(x – 1)3 +1!
Example: A. Find the Taylor-expansion of f(x) = 1 + x + x2 around x = 1.We need the derivatives of f(x):
(1) (1)
f(1) = 1 + 1 + 1 f(1) = 3
Taylor Expansions
f (x) = 1 + 2x f (1) = 3 (2) (2)f (x) = 2 f (1) = 2 (n)f (1) = 0 for n > 3. Hence,
= f(1) = 3p0(x)f(1)(1)(x – 1)= f(1) +p1(x)
1! = 3(x – 1)3 +1!
(= 3x)
Example: A. Find the Taylor-expansion of f(x) = 1 + x + x2 around x = 1.We need the derivatives of f(x):
(1) (1)
f(1) = 1 + 1 + 1 f(1) = 3
Taylor Expansions
f (x) = 1 + 2x f (1) = 3 (2) (2)f (x) = 2 f (1) = 2 (n)f (1) = 0 for n > 3. Hence,
= f(1) = 3p0(x)f(1)(1)(x – 1)= f(1) +p1(x)
1! = 3(x – 1)3 +1!
f(1)(1)(x – 1) += f(1) +p2(x)1!
f(2)(1)(x – 1)2
2!
(= 3x)
Example: A. Find the Taylor-expansion of f(x) = 1 + x + x2 around x = 1.We need the derivatives of f(x):
(1) (1)
f(1) = 1 + 1 + 1 f(1) = 3
Taylor Expansions
f (x) = 1 + 2x f (1) = 3 (2) (2)f (x) = 2 f (1) = 2 (n)f (1) = 0 for n > 3. Hence,
= f(1) = 3p0(x)f(1)(1)(x – 1)= f(1) +p1(x)
1! = 3(x – 1)3 +1!
f(1)(1)(x – 1) += f(1) +p2(x)1!
f(2)(1)(x – 1)2
2!= 3(x – 1)3 +
1! + 2(x – 1)2
2!
(= 3x)
Example: A. Find the Taylor-expansion of f(x) = 1 + x + x2 around x = 1.We need the derivatives of f(x):
(1) (1)
f(1) = 1 + 1 + 1 f(1) = 3
Taylor Expansions
f (x) = 1 + 2x f (1) = 3 (2) (2)f (x) = 2 f (1) = 2 (n)f (1) = 0 for n > 3. Hence,
= f(1) = 3p0(x)f(1)(1)(x – 1)= f(1) +p1(x)
1! = 3(x – 1)3 +1!
f(1)(1)(x – 1) += f(1) +p2(x)1!
f(2)(1)(x – 1)2
2!= 3(x – 1)3 +
1! + 2(x – 1)2
2! = 1 + x + x2
(= 3x)
Example: A. Find the Taylor-expansion of f(x) = 1 + x + x2 around x = 1.We need the derivatives of f(x):
(1) (1)
f(1) = 1 + 1 + 1 f(1) = 3
Taylor Expansions
f (x) = 1 + 2x f (1) = 3 (2) (2)f (x) = 2 f (1) = 2 (n)f (1) = 0 for n > 3. Hence,
= f(1) = 3p0(x)f(1)(1)(x – 1)= f(1) +p1(x)
1! = 3(x – 1)3 +1!
f(1)(1)(x – 1) += f(1) +p2(x)1!
f(2)(1)(x – 1)2
2!= 3(x – 1)3 +
1! + 2(x – 1)2
2! = 1 + x + x2
= 1 + x + x2 for n > 2.pn(x)
(= 3x)
Taylor Expansions
y=1+x+x2
(1, 3)
Graphs of Taylor polynomials at x = 1
Taylor Expansions
y=1+x+x2
y=3(1, 3)
Graphs of Taylor polynomials at x = 1
Taylor Expansions
y=1+x+x2
y=3
y=3x
(1, 3)
Graphs of Taylor polynomials at x = 1
Taylor Expansions
y=1+x+x2
y=3
y=3x
Graphs of Taylor polynomials at x = 1
(1, 3)
Example: B. Find the Taylor-expansion of f(x) = cos(x)
around x =
Taylor Expansions
π2
Example: B. Find the Taylor-expansion of f(x) = cos(x)
around x =
Taylor Expansions
π2
cos(x)
-sin(x)
-cos(x)
sin(x)
Example: B. Find the Taylor-expansion of f(x) = cos(x)
around x =
Taylor Expansions
π2
cos(x)
-sin(x)
-cos(x)
sin(x)
At x = π2 , we get the sequence of coefficients 0, 1, 0, -1, 0, 1, 0, -1, …
Example: B. Find the Taylor-expansion of f(x) = cos(x)
around x =
Taylor Expansions
π2
cos(x)
-sin(x)
-cos(x)
sin(x)
At x = π2 , we get the sequence of coefficients 0, 1, 0, -1, 0, 1, 0, -1, … So the Taylor expansions is
1(x – π/2) – = 0 +P(x) 1!(x – π/2)3
+ 03!+ 0
1(x – π/2)5+
5!– + 0
1(x – π/2)7
7!..
Example: B. Find the Taylor-expansion of f(x) = cos(x)
around x =
Taylor Expansions
π2
cos(x)
-sin(x)
-cos(x)
sin(x)
At x = π2 , we get the sequence of coefficients 0, 1, 0, -1, 0, 1, 0, -1, … So the Taylor expansions is
1(x – π/2) – = 0 +P(x) 1!(x – π/2)3
+ 03!+ 0
1(x – π/2)5+
5!– + 0
1(x – π/2)7
7!..
P(x)= Σ(-1)n+1(x – π/2)2n+1
(2n + 1)!n=0
n =∞
We can't expand Ln(x) at x = 0 but we can expand it at x = 1. Taylor Expansions
We can't expand Ln(x) at x = 0 but we can expand it at x = 1. Taylor Expansions
Example: Expand Ln(x) at x = 1.
We can't expand Ln(x) at x = 0 but we can expand it at x = 1. Taylor Expansions
y(1) = x-1, Example: Expand Ln(x) at x = 1.
We can't expand Ln(x) at x = 0 but we can expand it at x = 1. Taylor Expansions
y(1) = x-1, y(2) = -x-2, Example: Expand Ln(x) at x = 1.
We can't expand Ln(x) at x = 0 but we can expand it at x = 1. Taylor Expansions
y(1) = x-1, y(2) = -x-2, y(3) = 2x-3, Example: Expand Ln(x) at x = 1.
We can't expand Ln(x) at x = 0 but we can expand it at x = 1. Taylor Expansions
y(1) = x-1, y(2) = -x-2, y(3) = 2x-3, y(4) = -3*2x-4, Example: Expand Ln(x) at x = 1.
We can't expand Ln(x) at x = 0 but we can expand it at x = 1. Taylor Expansions
y(1) = x-1, y(2) = -x-2, y(3) = 2x-3, y(4) = -3*2x-4, y(5) = 4!x-5, Example: Expand Ln(x) at x = 1.
We can't expand Ln(x) at x = 0 but we can expand it at x = 1. Taylor Expansions
y(1) = x-1, y(2) = -x-2, y(3) = 2x-3, y(4) = -3*2x-4, y(5) = 4!x-5, we get the general formulay(n)=(-1)n-1(n – 1)! x-n for n = 1, 2, 3 ..
Example: Expand Ln(x) at x = 1.
We can't expand Ln(x) at x = 0 but we can expand it at x = 1. Taylor Expansions
y(1) = x-1, y(2) = -x-2, y(3) = 2x-3, y(4) = -3*2x-4, y(5) = 4!x-5, we get the general formulay(n)=(-1)n-1(n – 1)! x-n for n = 1, 2, 3 ..
Example: Expand Ln(x) at x = 1.
We can't expand Ln(x) at x = 0 but we can expand it at x = 1. Taylor Expansions
y(1) = x-1, y(2) = -x-2, y(3) = 2x-3, y(4) = -3*2x-4, y(5) = 4!x-5, we get the general formulay(n)=(-1)n-1(n – 1)! x-n for n = 1, 2, 3 ..
Example: Expand Ln(x) at x = 1.
We can't expand Ln(x) at x = 0 but we can expand it at x = 1. Taylor Expansions
y(1) = x-1, y(2) = -x-2, y(3) = 2x-3, y(4) = -3*2x-4, y(5) = 4!x-5, we get the general formulay(n)=(-1)n-1(n – 1)! x-n for n = 1, 2, 3 ..
Example: Expand Ln(x) at x = 1.
We can't expand Ln(x) at x = 0 but we can expand it at x = 1. Taylor Expansions
At x = 1, we get
y(1) = x-1, y(2) = -x-2, y(3) = 2x-3, y(4) = -3*2x-4, y(5) = 4!x-5, we get the general formulay(n)=(-1)n-1(n – 1)! x-n for n = 1, 2, 3 ..
Example: Expand Ln(x) at x = 1.
Ln(1) = 0, y(1)(1) = 1, y(2)(1) = -1, y(3)(1) = 2, y(4)(1) = -3*2, y(5)(1) = 4!,
We can't expand Ln(x) at x = 0 but we can expand it at x = 1. Taylor Expansions
At x = 1, we get
y(1) = x-1, y(2) = -x-2, y(3) = 2x-3, y(4) = -3*2x-4, y(5) = 4!x-5, we get the general formulay(n)=(-1)n-1(n – 1)! x-n for n = 1, 2, 3 ..
Example: Expand Ln(x) at x = 1.
Ln(1) = 0, y(1)(1) = 1, y(2)(1) = -1, y(3)(1) = 2, y(4)(1) = -3*2, y(5)(1) = 4!, … and the general formula is y(n)=(-1)n-1(n – 1)! for n = 1, 2, 3 ..
We can't expand Ln(x) at x = 0 but we can expand it at x = 1. Taylor Expansions
At x = 1, we get
y(1) = x-1, y(2) = -x-2, y(3) = 2x-3, y(4) = -3*2x-4, y(5) = 4!x-5, we get the general formulay(n)=(-1)n-1(n – 1)! x-n for n = 1, 2, 3 ..
Example: Expand Ln(x) at x = 1.
Ln(1) = 0, y(1)(1) = 1, y(2)(1) = -1, y(3)(1) = 2, y(4)(1) = -3*2, y(5)(1) = 4!, … and the general formula is y(n)=(-1)n-1(n – 1)! for n = 1, 2, 3 ..
P(x) = Σn=0
(x – a)n
n! f(n)(a)∞
We can't expand Ln(x) at x = 0 but we can expand it at x = 1. Taylor Expansions
At x = 1, we get
y(1) = x-1, y(2) = -x-2, y(3) = 2x-3, y(4) = -3*2x-4, y(5) = 4!x-5, we get the general formulay(n)=(-1)n-1(n – 1)! x-n for n = 1, 2, 3 ..
Example: Expand Ln(x) at x = 1.
Ln(1) = 0, y(1)(1) = 1, y(2)(1) = -1, y(3)(1) = 2, y(4)(1) = -3*2, y(5)(1) = 4!, … and the general formula is y(n)=(-1)n-1(n – 1)! for n = 1, 2, 3 ..
P(x) = Σn=0
(x – a)n
n! f(n)(a)∞
= Σn=0
(x – 1)n
n! y(n)(1)∞
We can't expand Ln(x) at x = 0 but we can expand it at x = 1. Taylor Expansions
At x = 1, we get
y(1) = x-1, y(2) = -x-2, y(3) = 2x-3, y(4) = -3*2x-4, y(5) = 4!x-5, we get the general formulay(n)=(-1)n-1(n – 1)! x-n for n = 1, 2, 3 ..
Example: Expand Ln(x) at x = 1.
Ln(1) = 0, y(1)(1) = 1, y(2)(1) = -1, y(3)(1) = 2, y(4)(1) = -3*2, y(5)(1) = 4!, … and the general formula is y(n)=(-1)n-1(n – 1)! for n = 1, 2, 3 ..
P(x) = Σn=0
(x – a)n
n! f(n)(a)∞
= Σn=0
(x – 1)n
n! y(n)(1)∞
= Σn=1
(x – 1)n
n! (-1)n-1(n – 1)!∞
=
We can't expand Ln(x) at x = 0 but we can expand it at x = 1. Taylor Expansions
At x = 1, we get
y(1) = x-1, y(2) = -x-2, y(3) = 2x-3, y(4) = -3*2x-4, y(5) = 4!x-5, we get the general formulay(n)=(-1)n-1(n – 1)! x-n for n = 1, 2, 3 ..
Example: Expand Ln(x) at x = 1.
Ln(1) = 0, y(1)(1) = 1, y(2)(1) = -1, y(3)(1) = 2, y(4)(1) = -3*2, y(5)(1) = 4!, … and the general formula is y(n)=(-1)n-1(n – 1)! for n = 1, 2, 3 ..
P(x) = Σn=0
(x – a)n
n! f(n)(a)∞
= Σn=0
(x – 1)n
n! y(n)(1)∞
= Σn=1
(x – 1)n
n! (-1)n-1(n – 1)!∞
= Σn=1
(x – 1)n
n (-1)n-1∞
We can't expand Ln(x) at x = 0 but we can expand it at x = 1. Taylor Expansions
At x = 1, we get
y(1) = x-1, y(2) = -x-2, y(3) = 2x-3, y(4) = -3*2x-4, y(5) = 4!x-5, we get the general formulay(n)=(-1)n-1(n – 1)! x-n for n = 1, 2, 3 ..
Example: Expand Ln(x) at x = 1.
Ln(1) = 0, y(1)(1) = 1, y(2)(1) = -1, y(3)(1) = 2, y(4)(1) = -3*2, y(5)(1) = 4!, … and the general formula is y(n)=(-1)n-1(n – 1)! for n = 1, 2, 3 ..
P(x) = Σn=0
(x – a)n
n! f(n)(a)∞
= Σn=0
(x – 1)n
n! y(n)(1)∞
= Σn=1
(x – 1)n
n! (-1)n-1(n – 1)!∞
= Σn=1
(x – 1)n
n (-1)n-1∞
(x – 1) – (x – 1)2
2+ (x – 1)3
3– (x – 1)4
4+ (x – 1)5
5 …P(x) =