Taylor's Remainder Theorem I

Taylor's Remainder Theorem IIn would be great if that the Taylor power series of a function, if it exists, would be the same as the function itself.

Taylor's Remainder Theorem IIn would be great if that the Taylor power series of a function, if it exists, would be the same as the function itself. Such is not the case by the following example.

Taylor's Remainder Theorem I

Example:

In would be great if that the Taylor power series of a function, if it exists, would be the same as the function itself. Such is not the case by the following example.

Let f(x) = { e-1/x 2 if x = 00 if x = 0

Taylor's Remainder Theorem I

Example:

In would be great if that the Taylor power series of a function, if it exists, would be the same as the function itself. Such is not the case by the following example.

Let f(x) = { e-1/x 2 if x = 00 if x = 0

We leave it as an exercise that it's derivatives of all the orders are 0, i.e. f(n)(0) = 0 for all n.

Taylor's Remainder Theorem I

Example:

In would be great if that the Taylor power series of a function, if it exists, would be the same as the function itself. Such is not the case by the following example.

Let f(x) = { e-1/x 2 if x = 00 if x = 0

We leave it as an exercise that it's derivatives of all the orders are 0, i.e. f(n)(0) = 0 for all n. Therefore it's Maclaurin series P(x) = 0.

Taylor's Remainder Theorem I

Example:

In would be great if that the Taylor power series of a function, if it exists, would be the same as the function itself. Such is not the case by the following example.

Let f(x) = { e-1/x 2 if x = 00 if x = 0

We leave it as an exercise that it's derivatives of all the orders are 0, i.e. f(n)(0) = 0 for all n. Therefore it's Maclaurin series P(x) = 0. Clearly P(x) is not the same as f(x).

Taylor's Remainder Theorem I

Example:

In the following discussion, we will assume all the functions are infinitely differentiable over an open interval unless otherwise stated.

In would be great if that the Taylor power series of a function, if it exists, would be the same as the function itself. Such is not the case by the following example.

Let f(x) = { e-1/x 2 if x = 00 if x = 0

We leave it as an exercise that it's derivatives of all the orders are 0, i.e. f(n)(0) = 0 for all n. Therefore it's Maclaurin series P(x) = 0. Clearly P(x) is not the same as f(x).

Taylor's Remainder Theorem ITaylor's Remainder Theorem gives a formula for the difference between function value f(b) and the pn(b) where pn is the n'th Taylor polynomial

(expanded about some point a).

Taylor's Remainder Theorem ITaylor's Remainder Theorem gives a formula for the difference between function value f(b) and the pn(b) where pn is the n'th Taylor polynomial

(expanded about some point a). We will give the theorem where pn is the Maclaurin polynomial first (i.e. expanded about 0).

Taylor's Remainder Theorem I

Taylor's Remainder Theorem: Let f(x) be an infinitely differentiable function over some open interval that contains [0, b] and pn(x) be the n'th Mac-poly .

Taylor's Remainder Theorem gives a formula for the difference between function value f(b) and the pn(b) where pn is the n'th Taylor polynomial

(expanded about some point a). We will give the theorem where pn is the Maclaurin polynomial first (i.e. expanded about 0).

Taylor's Remainder Theorem I

Taylor's Remainder Theorem: Let f(x) be an infinitely differentiable function over some open interval that contains [0, b] and pn(x) be the n'th Mac-poly .

Taylor's Remainder Theorem gives a formula for the difference between function value f(b) and the pn(b) where pn is the n'th Taylor polynomial

(expanded about some point a). We will give the theorem where pn is the Maclaurin polynomial first (i.e. expanded about 0).

0( )[ ]

b

f(x) is infinitely differentiable in here

Taylor's Remainder Theorem I

Taylor's Remainder Theorem: Let f(x) be an infinitely differentiable function over some open interval that contains [0, b] and pn(x) be the n'th Mac-poly .

Taylor's Remainder Theorem gives a formula for the difference between function value f(b) and the pn(b) where pn is the n'th Taylor polynomial

(expanded about some point a).

Then there exists a "c" between b and 0 such that

f(b) = pn(b) +

We will give the theorem where pn is the Maclaurin polynomial first (i.e. expanded about 0).

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

0( )[ ]

bc

f(x) is infinitely differentiable in here

Taylor's Remainder Theorem I

or in full detail,

f '(0)b f(2)(0)+ 2!= f(0) + b2f(b) +.. f(n)(0)n! bn+

bn+1

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

Taylor's Remainder Theorem I

or in full detail,

The difference term denoted as Rn(b) bn+1

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

is the Lagrange form of the Taylor-remainder (or error) term (there are other forms).

f '(0)b f(2)(0)+ 2!= f(0) + b2f(b) +.. f(n)(0)n! bn+

bn+1

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

Taylor's Remainder Theorem I

or in full detail,

The difference term denoted as Rn(b) bn+1

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

is the Lagrange form of the Taylor-remainder (or error) term (there are other forms).

f '(0)b f(2)(0)+ 2!= f(0) + b2f(b) +.. f(n)(0)n! bn+

bn+1

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

Remarks:

Taylor's Remainder Theorem I

or in full detail,

The difference term denoted as Rn(b) bn+1

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

is the Lagrange form of the Taylor-remainder (or error) term (there are other forms).

f '(0)b f(2)(0)+ 2!= f(0) + b2f(b) +.. f(n)(0)n! bn+

bn+1

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

Remarks:* the theorem also works for the interval [b, 0]

Taylor's Remainder Theorem I

or in full detail,

The difference term denoted as Rn(b) bn+1

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

is the Lagrange form of the Taylor-remainder (or error) term (there are other forms).

f '(0)b f(2)(0)+ 2!= f(0) + b2f(b) +.. f(n)(0)n! bn+

bn+1

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

Remarks:* the theorem also works for the interval [b, 0]

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

Taylor's Remainder Theorem I

or in full detail,

The difference term denoted as Rn(b) bn+1

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

is the Lagrange form of the Taylor-remainder (or error) term (there are other forms).

f '(0)b f(2)(0)+ 2!= f(0) + b2f(b) +.. f(n)(0)n! bn+

bn+1

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

Remarks:* the theorem also works for the interval [b, 0]

* 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

Example: f(x) = ex

Taylor's Remainder Theorem I is infinitely differentiable everywhere.

Example: f(x) = ex

Mac-poly of ex is

Pn(x) = Σk=0

xk

k! n

= 1 + x + x2

2! + .. + xn

n!

Taylor's Remainder Theorem I is infinitely differentiable everywhere.

Example: f(x) = ex

The Mac-poly of ex is

Pn(x) = Σk=0

xk

k! n

= 1 + x + x2

2! + .. + xn

n! At x = b, by the theorem we have

f(b) = eb = 1 + b + b2

2! + .. + bn

n! + f

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

for some c that is between 0 and b.

bn+1

Taylor's Remainder Theorem I is infinitely differentiable everywhere.

Example: f(x) = ex

Mac-poly of ex is

Pn(x) = Σk=0

xk

k! n

= 1 + x + x2

2! + .. + xn

n! At x = b, by the theorem we have

f(b) = eb = 1 + b + b2

2! + .. + bn

n! + f

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

for some c that is between 0 and b.

bn+1

Since f(n+1)(x) = ex so f(n+1)(c) = ec.

Taylor's Remainder Theorem I is infinitely differentiable everywhere.

Example: f(x) = ex

Mac-poly of ex is

Pn(x) = Σk=0

xk

k! n

= 1 + x + x2

2! + .. + xn

n! At x = b, by the theorem we have

f(b) = eb = 1 + b + b2

2! + .. + bn

n! + f

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

for some c that is between 0 and b.

bn+1

Since f(n+1)(x) = ex so f(n+1)(c) = ec. Hence

f(b) = eb = 1 + b + b2

2! + .. + bn

n! + ec

(n+1)! bn+1

for some c that is between 0 and b.

Taylor's Remainder Theorem I is infinitely differentiable everywhere.

Hence

Example: f(x) = ex

Mac-poly of ex is

Pn(x) = Σk=0

xk

k! n

= 1 + x + x2

2! + .. + xn

n! At x = b, by the theorem we have

f(b) = eb = 1 + b + b2

2! + .. + bn

n! + f

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

for some c that is between 0 and b.

bn+1

Since f(n+1)(x) = ex so f(n+1)(c) = ec. Hence

f(b) = eb = 1 + b + b2

2! + .. + bn

n! + ec

(n+1)! bn+1

for some c that is between 0 and b.

Taylor's Remainder Theorem I is infinitely differentiable everywhere.

Hence

The point here is not to find c but to use the formula to calculate the behavior of the error as n .∞

Example: Show that in the above example, the remainder term goes to 0 as n ∞.

Taylor's Remainder Theorem I

Example: Show that in the above example, the remainder term goes to 0 as n Hence f(b) = P(b) for all values bwhere P(x) is the Mac-series.

∞.

Taylor's Remainder Theorem I

Example: Show that in the above example, the remainder term goes to 0 as n Hence f(b) = P(b) for all values bwhere P(x) is the Mac-series.

f(b) = eb = 1 + b + b2

2! + .. + bn

n! + ec

(n+1)! bn+1

where b is a fixed value and c is between 0 and b.

∞.

Taylor's Remainder Theorem I

We have that

Example: Show that in the above example, the remainder term goes to 0 as n Hence f(b) = P(b) for all values bwhere P(x) is the Mac-series.

f(b) = eb = 1 + b + b2

2! + .. + bn

n! + ec

(n+1)! bn+1

where b is a fixed value and c is between 0 and b.

∞.

The maximum possible value of ec is eb.

Taylor's Remainder Theorem I

We have that

Example: Show that in the above example, the remainder term goes to 0 as n Hence f(b) = P(b) for all values bwhere P(x) is the Mac-series.

f(b) = eb = 1 + b + b2

2! + .. + bn

n! + ec

(n+1)! bn+1

where b is a fixed value and c is between 0 and b.

∞.

The maximum possible value of ec is eb. Write eb as K.

Taylor's Remainder Theorem I

We have that

Example: Show that in the above example, the remainder term goes to 0 as n Hence f(b) = P(b) for all values bwhere P(x) is the Mac-series.

f(b) = eb = 1 + b + b2

2! + .. + bn

n! + ec

(n+1)! bn+1

where b is a fixed value and c is between 0 and b.

∞.

The maximum possible value of ec is eb. Write eb as K.

ec

(n+1)! bn+1 < bn+1

(n+1)! KHence

Taylor's Remainder Theorem I

We have that

Example: Show that in the above example, the remainder term goes to 0 as n Hence f(b) = P(b) for all values bwhere P(x) is the Mac-series.

f(b) = eb = 1 + b + b2

2! + .. + bn

n! + ec

(n+1)! bn+1

where b is a fixed value and c is between 0 and b.

As n we've that K ∞,

∞.

The maximum possible value of ec is eb. Write eb as K.

ec

(n+1)! bn+1 < bn+1

(n+1)! K

bn+1

(n+1)! 0

Hence

Taylor's Remainder Theorem I

We have that

Example: Show that in the above example, the remainder term goes to 0 as n Hence f(b) = P(b) for all values bwhere P(x) is the Mac-series.

f(b) = eb = 1 + b + b2

2! + .. + bn

n! + ec

(n+1)! bn+1

where b is a fixed value and c is between 0 and b.

As n we've that K ∞,

∞.

The maximum possible value of ec is eb. Write eb as K.

ec

(n+1)! bn+1 < bn+1

(n+1)! K

bn+1

(n+1)! 0 (e.g. use ratio test).

Hence

Taylor's Remainder Theorem I

We have that

Example: Show that in the above example, the remainder term goes to 0 as n Hence f(b) = P(b) for all values bwhere P(x) is the Mac-series.

f(b) = eb = 1 + b + b2

2! + .. + bn

n! + ec

(n+1)! bn+1

where b is a fixed value and c is between 0 and b.

As n we've that K ∞,

∞.

The maximum possible value of ec is eb. Write eb as K.

ec

(n+1)! bn+1 < bn+1

(n+1)! K

bn+1

(n+1)! 0 (e.g. use ratio test).

Hence the error term ec

(n+1)! bn+1 0 as n

Hence

∞.

Taylor's Remainder Theorem I

We have that

Example: Show that in the above example, the remainder term goes to 0 as n Hence f(b) = P(b) for all values bwhere P(x) is the Mac-series.

f(b) = eb = 1 + b + b2

2! + .. + bn

n! + ec

(n+1)! bn+1

where b is a fixed value and c is between 0 and b.

As n we've that K ∞,

∞.

The maximum possible value of ec is eb. Write eb as K.

ec

(n+1)! bn+1 < bn+1

(n+1)! K

bn+1

(n+1)! 0 (e.g. use ratio test).

Hence the error term ec

(n+1)! bn+1 0 as n

Hence

∞.

This means the f(b) = Σn=0

bn

n! ∞

= P(b).

Taylor's Remainder Theorem I

We have that

Taylor's Remainder Theorem I We state the following theorem.

Taylor's Remainder Theorem I We state the following theorem. Theorem: Given f(x), and [0, b] as in the Taylor's Remainder theorem. Let P(x) be the Mac-series of f(x), then f(b) = P(b) if and only if the error term

Rn(b) = bn+1(n + 1)! f(n+1)(c) 0 as n ∞.

Taylor's Remainder Theorem I We state the following theorem.

(reminder: c is not fixed, it changes as n changes.)

Theorem: Given f(x), and [0, b] as in the Taylor's Remainder theorem. Let P(x) be the Mac-series of f(x), then f(b) = P(b) if and only if the error term

Rn(b) = bn+1(n + 1)! f(n+1)(c) 0 as n ∞.

Taylor's Remainder Theorem I We state the following theorem.

(reminder: c is not fixed, it changes as n changes.)

Theorem: Given f(x), and [0, b] as in the Taylor's Remainder theorem. Let P(x) be the Mac-series of f(x), then f(b) = P(b) if and only if the error term

Rn(b) = bn+1(n + 1)! f(n+1)(c) 0 as n ∞.

A function that is equal to its Mac-series over an open interval around 0 is said to be analytic at 0.

Taylor's Remainder Theorem I We state the following theorem.

(reminder: c is not fixed, it changes as n changes.)

Theorem: Given f(x), and [0, b] as in the Taylor's Remainder theorem. Let P(x) be the Mac-series of f(x), then f(b) = P(b) if and only if the error term

Rn(b) = bn+1(n + 1)! f(n+1)(c) 0 as n ∞.

A function that is equal to its Mac-series over an open interval around 0 is said to be analytic at 0. The function f(x) = ex is analytic at 0.

Taylor's Remainder Theorem I We state the following theorem.

(reminder: c is not fixed, it changes as n changes.)

Theorem: Given f(x), and [0, b] as in the Taylor's Remainder theorem. Let P(x) be the Mac-series of f(x), then f(b) = P(b) if and only if the error term

Rn(b) = bn+1(n + 1)! f(n+1)(c) 0 as n ∞.

A function that is equal to its Mac-series over an open interval around 0 is said to be analytic at 0. The function f(x) = ex is analytic at 0. The function

f(x) = { e-1/x 2 if x = 00 if x = 0

is infinitely differentiable but not analytic at 0.

Example: Let f(x) = cos(x). Show that the Lagrange form of the error Rn(b) of it's Mac-poly goes to 0 as n and conclude from that f(x) is the same as it's Mac-series.

Taylor's Remainder Theorem I

∞

Example: Let f(x) = cos(x). Show that the Lagrange form of the error Rn(b) of it's Mac-poly goes to 0 as n and conclude from that f(x) is the same as it's Mac-series.

P(x) =

Taylor's Remainder Theorem I

+ 4!x4

6!x6

8!x8

+ 1 – – 2!x2

..

The Mac-series is

∞

Example: Let f(x) = cos(x). Show that the Lagrange form of the error Rn(b) of it's Mac-poly goes to 0 as n and conclude from that f(x) is the same as it's Mac-series.

P(x) =

At x = b, the remainder is Rn(b) = f(n+1)(c)(n+1)!

for some c that is between 0 and b.

bn+1

Taylor's Remainder Theorem I

+ 4!x4

6!x6

8!x8

+ 1 – – 2!x2

..

The Mac-series is

∞

Example: Let f(x) = cos(x). Show that the Lagrange form of the error Rn(b) of it's Mac-poly goes to 0 as n and conclude from that f(x) is the same as it's Mac-series.

P(x) =

At x = b, the remainder is Rn(b) = f(n+1)(c)(n+1)!

for some c that is between 0 and b.

bn+1

Since the higher derivatives of cos(x) are ±sin(x), or ±cos(x),

Taylor's Remainder Theorem I

+ 4!x4

6!x6

8!x8

+ 1 – – 2!x2

..

The Mac-series is

∞

Example: Let f(x) = cos(x). Show that the Lagrange form of the error Rn(b) of it's Mac-poly goes to 0 as n and conclude from that f(x) is the same as it's Mac-series.

P(x) =

At x = b, the remainder is Rn(b) = f(n+1)(c)(n+1)!

for some c that is between 0 and b.

bn+1

Since the higher derivatives of cos(x) are ±sin(x), or ±cos(x),we may assume that |f(n+1)(c)| < 1.

Taylor's Remainder Theorem I

+ 4!x4

6!x6

8!x8

+ 1 – – 2!x2

..

The Mac-series is

∞

Example: Let f(x) = cos(x). Show that the Lagrange form of the error Rn(b) of it's Mac-poly goes to 0 as n and conclude from that f(x) is the same as it's Mac-series.

P(x) =

At x = b, the remainder is Rn(b) = f(n+1)(c)(n+1)!

for some c that is between 0 and b.

bn+1

Since the higher derivatives of cos(x) are ±sin(x), or ±cos(x),we may assume that |f(n+1)(c)| < 1.

Taylor's Remainder Theorem I

+ 4!x4

6!x6

8!x8

+ 1 – – 2!x2

..

The Mac-series is

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

(n+1)! bn+1 0 as n ∞.

∞

Hence,

Example: Let f(x) = cos(x). Show that the Lagrange form of the error Rn(b) of it's Mac-poly goes to 0 as n and conclude from that f(x) is the same as it's Mac-series.

P(x) =

At x = b, the remainder is Rn(b) = f(n+1)(c)(n+1)!

for some c that is between 0 and b.

bn+1

Since the higher derivatives of cos(x) are ±sin(x), or ±cos(x),we may assume that |f(n+1)(c)| < 1.

Taylor's Remainder Theorem I

+ 4!x4

6!x6

8!x8

+ 1 – – 2!x2

..

The Mac-series is

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

(n+1)! bn+1 0 as n ∞.

Therefore f(x) = P(x) = + 4!x4

6!x6

8!x8

+ 1 – – 2!x2

.. for all x.

∞

Hence,