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Taylor’s Polynomials & LaGrange Error Review

Maclaurin Series:

(generated by f at )0x =

( ) ( ) ( ) ( ) ( )2 30 00 0

2! 3!

f fP x f f x x x

′′ ′′′′= + + + + ⋅⋅⋅

If we want to center the series (and it’s graph) at some point other than zero, we get the Taylor Series:

Taylor Series:

(generated by f at )x a=

( ) ( ) ( )( ) ( )( ) ( )( )2 3

2! 3!

f a f aP x f a f a x a x a x a

′′ ′′′′= + − + − + − + ⋅⋅⋅

→

-1

0

1

-5 -4 -3 -2 -1 1 2 3 4 5

cosy x= ( )2 4 6 8 10

1 2! 4! 6! 8! 10!

x x x x xP x = − + − + − ⋅⋅⋅

The more terms we add, the better our approximation.

→

Hint: On the TI-89, the factorial symbol is: ÷

example: ( )cos 2y x=

Rather than start from scratch, we can use the function that we already know:

( ) ( ) ( ) ( ) ( ) ( )2 4 6 8 102 2 2 2 2

1 2! 4! 6! 8! 10!

x x x x xP x = − + − + − ⋅⋅⋅

→

When referring to Taylor polynomials, we can talk about number of terms, order or degree.

2 4

cos 12! 4!

x xx = − + This is a polynomial in 3 terms.

It is a 4th order Taylor polynomial, because it was found using the 4th derivative.

It is also a 4th degree polynomial, because x is raised to the 4th power.

The 3rd order polynomial for is , but it is degree 2.

cos x2

12!

x−

The x3 term drops out when using the third derivative.

This is also the 2nd order polynomial.

A recent AP exam required the student to know the difference between order and degree.

→

There are some Maclaurin series that occur often enough that they should be memorized. They are on your formula sheet, but today we are going to look at where they come from.

Maclaurin Series:

(generated by f at )0x =

( ) ( ) ( ) ( ) ( )2 30 00 0

2! 3!

f fP x f f x x x

′′ ′′′′= + + + + ⋅⋅⋅

→

( ) ( ) ( ) ( ) ( )2 30 00 0

2! 3!

f fP x f f x x x

′′ ′′′′= + + + + ⋅⋅⋅( ) 11

11

xx

−= −

−

( ) 11 x

−−

( ) 21 x

−−

( ) 32 1 x

−−

( ) 46 1 x

−−

( ) 524 1 x

−−

( ) ( )nf x

List the function and itsderivatives.

List the function and itsderivatives.

Evaluate column onefor x = 0.

( ) ( ) ( ) ( ) ( )2 30 00 0

2! 3!

f fP x f f x x x

′′ ′′′′= + + + + ⋅⋅⋅( ) 11

11

xx

−= −

−

( ) 11 x

−−

( ) 21 x

−−

( ) 32 1 x

−−

( ) 46 1 x

−−

( ) 524 1 x

−−

1

1

2

6 3!=

24 4!=

( ) ( )0nf( ) ( )nf x

2 3 42 31

1 2! 3!

! 4!1

!1

4x x x x

x= + + + + +⋅⋅⋅

−

2 3 411

1x x x x

x= + + + + +⋅⋅⋅

−

This is a geometric series witha = 1 and r = x.

( )1 1 x−1

1 x−x

x+

2x x−

2x+

2x2 3x x−

3x

3x+ +⋅⋅⋅

We could generate this same series for with polynomial long division:

1

1 x−

→

( ) ( ) ( ) ( ) ( )2 30 00 0

2! 3!

f fP x f f x x x

′′ ′′′′= + + + + ⋅⋅⋅( ) 11

11

xx

−= +

+

( ) 11 x

−+

( ) 21 x

−− +

( ) 32 1 x

−+

( ) 46 1 x

−− +

( ) 524 1 x

−+

1

1−

2

6 3!− =−

24 4!=

( ) ( )0nf( ) ( )nf x

2 3 421

1 2! 3! 4!

3! 4!1 1x x x x

x

−= + + + + +⋅⋅⋅

+−

2 3 411

1x x x x

x= − + − + +⋅⋅⋅

+

This is a geometric series witha = 1 and r = -x.

→

We wouldn’t expect to use the previous two series to evaluate the functions, since we can evaluate the functions directly.

We will find other uses for these series, as well.

They do help to explain where the formula for the sum of an infinite geometric comes from.

A more impressive use of Taylor series is to evaluate transcendental functions.

( ) ( ) ( ) ( ) ( )2 30 00 0

2! 3!

f fP x f f x x x

′′ ′′′′= + + + + ⋅⋅⋅( )cos x

( )cos x

( )sin x−

( )cos x−

( )sin x

( )cos x

1

0

1−

0

1

( ) ( )0nf( ) ( )nf x

( ) 2 3 4cos2! 3!

1 0

4!

11 0x x x x x= + + + +

−+ ⋅⋅⋅

( )2 4 6

cos 1 2! 4! 6!

x x xx = − + − ⋅⋅⋅

Both sides are even functions.

Cos (0) = 1 for both sides.→

( ) ( ) ( ) ( ) ( )2 30 00 0

2! 3!

f fP x f f x x x

′′ ′′′′= + + + + ⋅⋅⋅( )sin x

( )sin x

( )cos x

( )sin x−

( )cos x−

( )sin x

0

1

0

1−

0

( ) ( )0nf( ) ( )nf x

( ) 2 3 4sin2

0 1 00 1

! 3! 4!x x x x x

−= + + + + + ⋅⋅⋅

( )3 5 7

sin 3! 5! 7!

x x xx x= − + − ⋅⋅⋅

Both sides are odd functions.

Sin (0) = 0 for both sides.→

2 3 411

1x x x x

x= − + − + +⋅⋅⋅

+

2

1

1 x+

If we start with this function:

and substitute for , we get:2x x 2 4 6 8

2

11

1x x x x

x= − + − + +⋅⋅⋅

+

This is a geometric series with a = 1 and r = -x2.

If we integrate both sides:

2 4 6 82

1 1

1dx x x x x dx

x= − + − + +⋅⋅⋅

+∫ ∫

( )3 5 7

1tan 3 5 7

x x xx x− = − + − + ⋅⋅⋅

This looks the same as the series for sin (x), but without the factorials. →

( ) ( ) ( ) ( ) ( )2 30 00 0

2! 3!

f fP x f f x x x

′′ ′′′′= + + + + ⋅⋅⋅( )ln 1 x+

( )ln 1 x+

( ) 11 x

−+

( ) 21 x

−− +

( ) 32 1 x

−+

( ) 46 1 x

−− −

0

1

1−

2

6 3!− =−

( ) ( )0nf( ) ( )nf x

( ) 2 3 4ln 12

1

! 3! 4

20 1

!

3!x x x x x+ = + + +

−+

−+ ⋅⋅⋅

( )2 3 4

ln 12 3 4

x x xx x+ = − + − + ⋅⋅⋅

→

( ) ( ) ( ) ( ) ( )2 30 00 0

2! 3!

f fP x f f x x x

′′ ′′′′= + + + + ⋅⋅⋅xe

xe

xe

xe

xe

xe

1

1

1

1

1

( ) ( )0nf( ) ( )nf x

2 3 41 1 11 1

2! 3! 4!xe x x x x= + + + + +⋅⋅⋅

2 3 4

12! 3! 4!

x x x xe x= + + + + +⋅⋅⋅

Finding Truncation Error in a Taylor Polynomial

Graph the function y1 = ln (1 + x) and it’s corresponding Taylor Polynomial

y

2 = x -

x2

2 +

x3

3 -

x4

4 +

x5

5

Finding Truncation Error in a Taylor Polynomial

Graph the function y1 = ln (1 + x) and it’s corresponding Taylor Polynomial

2 3 4 5

2

x x x xy = x - + - -

2 3 4 5

To find the error between the functions, graph y3 = abs (y2 – y1).

Finding Truncation Error in a Taylor Polynomial

To find the error between the functions, graph y3 = abs (y2 – y1).

Where is the error the smallest?

Finding Truncation Error in a Taylor Polynomial

Use Table to see where truncation error is least on (-1,1)

Finding Error in a Taylor Polynomial

Graph the function y1 = sin x and it’s corresponding Taylor Polynomial and find the interval in which the

Taylor polynomial is accurate to the thousandths place.

3

2

xy = x -

6

Finding Error in a Taylor Polynomial

The third order Taylor Polynomial for y = sin x is accurate to the thousandths place on the interval (-.65, .65). Graph

y3 = abs (y1(x) – y2(x)), and y4 = .001.

Taylor series are used to estimate the value of functions (at least theoretically - now days we can usually use the calculator or computer to calculate directly.)

An estimate is only useful if we have an idea of how accurate the estimate is.

When we use part of a Taylor series to estimate the value of a function, the end of the series that we do not use is called the remainder. If we know the size of the remainder, then we know how close our estimate is.

→

For a geometric series, this is easy:

ex. 2: Use to approximate over .2 4 61 x x x+ + + 2

1

1 x−( )1,1−

Since the truncated part of the series is: ,8 10 12 x x x+ + + ⋅⋅⋅

the truncation error is , which is .8 10 12 x x x+ + + ⋅⋅⋅8

21

x

x−

When you “truncate” a number, you drop off the end.

Of course this is also trivial, because we have a formula that allows us to calculate the sum of a geometric series directly.

→

Taylor’s Theorem with Remainder

If f has derivatives of all orders in an open interval I containing a, then for each positive integer n and for each x

in I:

( ) ( ) ( )( ) ( ) ( )( ) ( ) ( ) ( )2

2! !

nn

n

f fa af f f Rx a a x a x a x a xn

′′′= + + + ⋅⋅⋅ +− − −

Lagrange Form of the Remainder

( )( ) ( )( ) ( )

11

1 !

nn

n

f cR x x a

n

++

= −+

Remainder after partial sum Sn

where c is between

a and x.

→

Lagrange Form of the Remainder

( )( ) ( )( ) ( )

11

1 !

nn

n

f cR x x a

n

++

= −+

Remainder after partial sum Sn

where c is between

a and x.

This is also called the remainder of order n or the error term.

Note that this looks just like the next term in the series, but

“a” has been replaced by the number “c” in .( ) ( )1nf c+

This seems kind of vague, since we don’t know the value of c,

but we can sometimes find a maximum value for .( ) ( )1nf c+

→

This is called Taylor’s Inequality.

Taylor’s InequalityNote that this is not the formula that is in our book. It is from another textbook.

Lagrange Form of the Remainder

( )( ) ( )( ) ( )

11

1 !

nn

n

f cR x x a

n

++

= −+

If M is the maximum value of on the interval

between a and x, then:

( ) ( )1nf x+

( ) ( )1

1 !n

n

MR x x a

n+≤ −

+

→

ex. 2: Prove that , which is the Taylor

series for sin x, converges for all real x.

( )( )

2 1

0 1!2 1

kk

k

x

k

+∞

= −+

∑

Since the maximum value of sin x or any of it’s

derivatives is 1, for all real x, M = 1.

( )( )

110

!1n

nR x xn

+∴ ≤ −+ ( )

1

!1

nx

n

+

=+

( )

1

lim 0!1

n

n

x

n

+

→∞=

+

so the series converges.

Taylor’s Inequality

( ) ( )1

1 !n

n

MR x x a

n+≤ −

+→

ex. 5: Find the Lagrange Error Bound when is used

to approximate and .

2

2

xx −

( )ln 1 x+ 0.1x ≤

( ) ( )ln 1f x x= +

( ) ( ) 11f x x

−′ = +

( ) ( ) 21f x x

−′′ = − +

( ) ( ) 32 1f x x−′′′ = +

( ) ( ) ( ) ( ) ( )22

0 00

1 2!

f ff f x x Rx x

′ ′′= + + +

Remainder after 2nd order term

( ) ( )2

202

xf x Rx x= + − +

On the interval , decreases, so

its maximum value occurs at the left end-point.

[ ].1,.1− ( )3

2

1 x+

( )3

2

1 .1M =

+ − ( )3

2

.9= 2.74348422497≈

→

ex. 5:2

2

xx −

( )ln 1 x+ 0.1x ≤

On the interval , decreases, so

its maximum value occurs at the left end-point.

[ ].1,.1− ( )3

2

1 x+

( )3

2

1 .1M =

+ − ( )3

2

.9= 2.74348422497≈

Taylor’s Inequality

( ) ( )1

1 !n

n

MR x x a

n+≤ −

+

( )32.7435 .1

3!nR x ≤

( ) 0.000457nR x ≤

Lagrange Error Bound

x ( )ln 1 x+

2

2

xx − error

.1 .0953102 .095 .000310

.1− .1053605− .105− .000361

Error is less than error bound. →

Find the Lagrange Error Bound when is used to

approximate and .

Example using Taylor’s Theorem with Remainder

For approximately what values of x can you replace sin x by with an error magnitude no greater than 1 x 10-3 ?

( ) ( )5

5

- 34

5

Since f c = cos c 1, then

xR (x) 1 x 10

5!

x .12

x .65

≤

≤ ≤

≤

≤

x - x3

3!

Taylor’s Theorem with Remainder

( )n

7n

100 100 100 100 100 100 100Ex : lim = ... ... ...

n+1 ! 1 2 3 100 101 10→∞g g g g g g

Taylor’s Theorem with Remainder works well with the functions y = sin x and y = cos x because |f (n+1)(c)| ≤ 1.

Note: Factorial growth in the denominator is larger than the exponential growth in the numerator.

Since | Rn(x)| 0 then the Taylor Series converges as n → ∞