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) =