34 taylor's remainder theorem ii

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Transcript of 34 taylor's remainder theorem ii

Taylor's Remainder Theorem II

Taylor's Remainder Theorem II

We state the general form of the Taylor's remainder formula.

Taylor's Remainder Theorem II

We state the general form of the Taylor's remainder formula.

Taylor's Remainder Theorem (General Form): Let f(x) be an infinitely differentiable function over some open interval that contains [a, b]

Taylor's Remainder Theorem II

We state the general form of the Taylor's remainder formula.

Taylor's Remainder Theorem (General Form): Let f(x) be an infinitely differentiable function over some open interval that contains [a, b]

a( )[ ]

b

f(x) is infinitely differentiable in here

Taylor's Remainder Theorem II

We state the general form of the Taylor's remainder formula.

Taylor's Remainder Theorem (General Form): Let f(x) be an infinitely differentiable function over some open interval that contains [a, b] and

pn(x) =

be the n'th Taylor-poly expanded around at a.

a( )[ ]

b

f(x) is infinitely differentiable in here

Σk=0

n (x – a)k

k! f(k)(a)

Taylor's Remainder Theorem II

We state the general form of the Taylor's remainder formula.

Taylor's Remainder Theorem (General Form): Let f(x) be an infinitely differentiable function over some open interval that contains [a, b] and

pn(x) =

be the n'th Taylor-poly expanded around at a.

Then there exists a "c" between a and b such that

f(b) = pn(b) + (b – a)n+1(n + 1)! f(n+1)(c)

a( )[ ]

bc

f(x) is infinitely differentiable in here

Σk=0

n (x – a)k

k! f(k)(a)

or in full detail,

f '(a)(b – a)f(2)(a)

+ 2!= f(a) + (b – a)2f(b) ..f(n)(a)

n! (b – a)n+

Taylor's Remainder Theorem II

+ Rn(b)

or in full detail,

where Rn(b) = (b – a)n+1

(n + 1)! f(n+1)(c)

is the Lagrange form of the Taylor-remainder.

f '(a)(b – a)f(2)(a)

+ 2!= f(a) + (b – a)2f(b) ..f(n)(a)

n! (b – a)n+

Taylor's Remainder Theorem II

+ Rn(b)

or in full detail,

where Rn(b) = (b – a)n+1

(n + 1)! f(n+1)(c)

is the Lagrange form of the Taylor-remainder.

f '(a)(b – a)f(2)(a)

+ 2!= f(a) + (b – a)2f(b) ..f(n)(a)

n! (b – a)n+

Again, we note the following

Taylor's Remainder Theorem II

+ Rn(b)

or in full detail,

where Rn(b) = (b – a)n+1

(n + 1)! f(n+1)(c)

is the Lagrange form of the Taylor-remainder.

f '(a)(b – a)f(2)(a)

+ 2!= f(a) + (b – a)2f(b) ..f(n)(a)

n! (b – a)n+

Again, we note the following

* the theorem also works for the interval [b, a]

Taylor's Remainder Theorem II

+ Rn(b)

or in full detail,

where Rn(b) = (b – a)n+1

(n + 1)! f(n+1)(c)

is the Lagrange form of the Taylor-remainder.

f '(a)(b – a)f(2)(a)

+ 2!= f(a) + (b – a)2f(b) ..f(n)(a)

n! (b – a)n+

Again, we note the following

* the theorem also works for the interval [b, a]

* the value c changes if the value of b or n changes

Taylor's Remainder Theorem II

+ Rn(b)

or in full detail,

where Rn(b) = (b – a)n+1

(n + 1)! f(n+1)(c)

is the Lagrange form of the Taylor-remainder.

f '(a)(b – a)f(2)(a)

+ 2!= f(a) + (b – a)2f(b) ..f(n)(a)

n! (b – a)n+

Again, we note the following

* the theorem also works for the interval [b, a]

* the value c can't be easily determined, we just know there is at least one c that fits the description

* the value c changes if the value of b or n changes

Taylor's Remainder Theorem II

+ Rn(b)

Example: A. Find the Taylor-series of f(x) = sin(x) at x = .π

2

Taylor's Remainder Theorem II

Example: A. Find the Taylor-series of f(x) = sin(x) at x = .π

2sin(x)

cos(x)

-sin(x)

-cos(x)

Taylor's Remainder Theorem II

Example: A. Find the Taylor-series of f(x) = sin(x) at x = .π

2sin(x)

cos(x)

-sin(x)

-cos(x)

At x = π2 , we get the sequence of coefficients

1, 0, -1, 0, 1, 0, -1, …

Taylor's Remainder Theorem II

Example: A. Find the Taylor-series of f(x) = sin(x) at x = .π

2sin(x)

cos(x)

-sin(x)

-cos(x)

At x = π2 , we get the sequence of coefficients

1, 0, -1, 0, 1, 0, -1, … So the Taylor expansions is

= 1P(x)

Taylor's Remainder Theorem II

Example: A. Find the Taylor-series of f(x) = sin(x) at x = .π

2sin(x)

cos(x)

-sin(x)

-cos(x)

At x = π2 , we get the sequence of coefficients

1, 0, -1, 0, 1, 0, -1, … So the Taylor expansions is

– = 1P(x)(x – π/2)2

2!

Taylor's Remainder Theorem II

Example: A. Find the Taylor-series of f(x) = sin(x) at x = .π

2sin(x)

cos(x)

-sin(x)

-cos(x)

At x = π2 , we get the sequence of coefficients

1, 0, -1, 0, 1, 0, -1, … So the Taylor expansions is

– = 1P(x)(x – π/2)2

2!

1(x – π/2)4

+ 4!

Taylor's Remainder Theorem II

Example: A. Find the Taylor-series of f(x) = sin(x) at x = .π

2sin(x)

cos(x)

-sin(x)

-cos(x)

At x = π2 , we get the sequence of coefficients

1, 0, -1, 0, 1, 0, -1, … So the Taylor expansions is

– = 1P(x)(x – π/2)2

2!

1(x – π/2)4

+ 4!–

1(x – π/2)6

6!..1(x – π/2)8

+ 8!

Taylor's Remainder Theorem II

Example: A. Find the Taylor-series of f(x) = sin(x) at x = .π

2sin(x)

cos(x)

-sin(x)

-cos(x)

At x = π2 , we get the sequence of coefficients

1, 0, -1, 0, 1, 0, -1, … So the Taylor expansions is

– = 1P(x)(x – π/2)2

2!

1(x – π/2)4

+ 4!–

1(x – π/2)6

6!..

or P(x) = Σ(-1)n(x – π/2)2n

(2n)!n=0

n =∞

1(x – π/2)8

+ 8!

Taylor's Remainder Theorem II

Example:

B. Describe the Taylor remainder Rn(b) and show that f(b) = P(b) for all values b. (a = ) π/2

Taylor's Remainder Theorem II

Example:

B. Describe the Taylor remainder Rn(b) and show that f(b) = P(b) for all values b. (a = )

The remainder Rn(b) = (b – a)n+1

(n + 1)! f(n+1)(c)

π/2

Taylor's Remainder Theorem II

Example:

B. Describe the Taylor remainder Rn(b) and show that f(b) = P(b) for all values b. (a = )

The remainder Rn(b) = (b – a)n+1

(n + 1)! f(n+1)(c)

=

π/2

(b – )n+1

(n + 1)! f(n+1)(c) π

2

Taylor's Remainder Theorem II

Example:

B. Describe the Taylor remainder Rn(b) and show that f(b) = P(b) for all values b. (a = )

The remainder Rn(b) = (b – a)n+1

(n + 1)! f(n+1)(c)

=

π/2

(b – )n+1

(n + 1)! f(n+1)(c) π

2

where f(n+1) may be ±sin(x) or ±cos(x) and for some c that is between a and b.

Taylor's Remainder Theorem II

Example:

B. Describe the Taylor remainder Rn(b) and show that f(b) = P(b) for all values b. (a = )

The remainder Rn(b) = (b – a)n+1

(n + 1)! f(n+1)(c)

=

π/2

(b – )n+1

(n + 1)! f(n+1)(c) π

2

where f(n+1) may be ±sin(x) or ±cos(x) and for some c that is between a and b.

Since |f(n+1)(c)| < 1,

Taylor's Remainder Theorem II

Example:

B. Describe the Taylor remainder Rn(b) and show that f(b) = P(b) for all values b. (a = )

The remainder Rn(b) = (b – a)n+1

(n + 1)! f(n+1)(c)

=

π/2

(b – )n+1

(n + 1)! f(n+1)(c) π

2

where f(n+1) may be ±sin(x) or ±cos(x) and for some c that is between a and b.

Since |f(n+1)(c)| < 1, we've (b – )n+1

(n + 1)! f(n+1)(c) π

2 <(n + 1)!

(b – )n+1π2

Taylor's Remainder Theorem II

Example:

B. Describe the Taylor remainder Rn(b) and show that f(b) = P(b) for all values b. (a = )

The remainder Rn(b) = (b – a)n+1

(n + 1)! f(n+1)(c)

=

π/2

(b – )n+1

(n + 1)! f(n+1)(c) π

2

where f(n+1) may be ±sin(x) or ±cos(x) and for some c that is between a and b.

Since |f(n+1)(c)| < 1, we've (b – )n+1

(n + 1)! f(n+1)(c) π

2 <(n + 1)!

(b – )n+1π2

Again, as n we've (n + 1)!

(b – )n+1π2 0∞,

Taylor's Remainder Theorem II

Example:

B. Describe the Taylor remainder Rn(b) and show that f(b) = P(b) for all values b. (a = )

The remainder Rn(b) = (b – a)n+1

(n + 1)! f(n+1)(c)

=

π/2

(b – )n+1

(n + 1)! f(n+1)(c) π

2

where f(n+1) may be ±sin(x) or ±cos(x) and for some c that is between a and b.

Since |f(n+1)(c)| < 1, we've (b – )n+1

(n + 1)! f(n+1)(c) π

2 <(n + 1)!

(b – )n+1π2

Again, as n we've (n + 1)!

(b – )n+1π2 0∞,

Hence Rn(b) 0 and that f(b) = P(b) for all b.

Taylor's Remainder Theorem II

Differentiation and Integration of Power Series

The Taylor series of f(x) is the only power series that could be the same as f(x).

Σk=0

Differentiation and Integration of Power Series

Let f(x) = ck(x – a)k for all x in an open interval

(a – R, a + R) for some R, then the series ck(x – a)k

is the Taylor series P(x) of f(x).

Σk=0

Theorem (Uniqueness theorem for Taylor series) :

The Taylor series of f(x) is the only power series that could be the same as f(x).