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> 1. Taylor polynomials

Taylor polynomials

1 Taylor polynomials1 The Taylor polynomial2 Error in Taylor’s polynomial3 Polynomial evaluation

1. Taylor polynomials Math 1070

> 1. Taylor polynomials > 1.1 The Taylor polynomial

Let f(x) be a given function, for example

ex, sinx, log(x).

The Taylor polynomial mimics the behavior of f(x) near x = a:

T (x) ≈ f(x), for all x ”close” to a.

Example

Find a linear polynomial p1(x) for which{p1(a) = f(a),p′1(a) = f ′(a).

p1 is uniquely given by

p1(x) = f(a) + (x− a)f ′(a).

The graph of y = p1(x) is tangent to that of y = f(x) at x = a.1. Taylor polynomials Math 1070


Example

Let f(x) = ex, a = 0. Then

p1(x) = 1 + x.

-2.4 -2 -1.6 -1.2 -0.8 -0.4 0 0.4 0.8 1.2 1.6 2 2.4

-1.2

-0.8

-0.4

0.4

0.8

1.2

y=p_1(x)

Figure: Linear Taylor Approximation ex



Example

Find a quadratic polynomial p2(x) to approximate f(x) near x = a.

Sincep2(x) = b0 + b1x+ b2x

2

we impose three conditions on p2(x) to determine the coefficients. Tobetter mimic f(x) at x = a we require

p2(a) = f(a)p′2(a) = f ′(a)p′′2(a) = f ′′(a)

p2 is uniquely given by

p2(x) = f(a) + (x− a)f ′(a) +12(x− a)2f ′′(a).



Example

Let f(x) = ex, a = 0. Then

p2(x) = 1 + x+12x2.

-1 -0.5 0 0.5 1

-0.25

0.25

0.5

0.75

1

1.25

1.5

y=p_1(x)

y=p_2(x)

y=e^x

Figure: Linear and quadratic Taylor Approximation ex (see eval exp simple.m)



Let pn(x) be a polynomial of degree n that mimics the behavior off(x) at x = a. We require

p(j)n (a) = f (j)(a), j = 0, 1, . . . , n

where f (j) the jth derivative of f(x).The Taylor polynomial of degree n for the function f(x) at point a:

pn(x) = f(a) + (x− a)f ′(a) +(x− a)2

2!f ′′(a) + . . .+

(x− a)n

n!f (n)(a)

=n∑j=0

(x− a)j

j!f (j)(a). (3.1)

Recall the notations: f (0)(a) = f(a) and the ”factorial”:

j! ={

1, j = 0j · (j − 1) · · · 2 · 1, j = 1, 2, 3, 4, . . .



Example

Let f(x) = ex, a = 0.Since f (j)(x) = ex, f (j)(0) = 1, for all j ≥ 0, then

pn(x) = 1 + x+12!x2 + . . .+

1n!xn =

n∑j=0

xj

j!(3.2)

For a fixed x, the accuracy improves as the degree n increases.

For a fixed degree n, the accuracy improves as x gets close toa = 0.



Table. Taylor approximations to ex

x p1(x) p2(x) p3(x) ex

−1.0 0 0.500 0.33333 0.36788

−0.5 0.5 0.625 0.60417 0.60653

−0.1 0.9 0.905 0.90483 0.90484

0 1.0 1.000 1.00000 1.00000

0.1 1.1 1.105 1.10517 1.10517

0.5 1.5 1.625 1.64583 1.64872

1.0 2.0 2.500 2.66667 2.71828



Example

Let f(x) = ex, a arbitrary, not necessarily 0.Since f (j)(x) = ex, f (j)(a) = ea, for all j ≥ 0, then

pn(x; a) = ea(

1 + (x− a) +12!

(x− a)2 + . . .+1n!

(x− a)n)

= ean∑j=0

(x− a)j

j!



Example

Let f(x) = ln(x), a = 1.

Since f(1) = ln(1) = 0, and

f (j)(x) = (−1)j−1(j − 1)!1xj

f (j)(1) = (−1)j−1(j − 1)!

then the Taylor polynomial is

pn(x) = (x− 1)− 12(x− 1)2 +

13(x− 1)3 − . . .+ (−1)(n−1) 1

n(x− 1)n

=n∑

j=1

(−1)j−1

j(x− 1)j



-0.25 0 0.25 0.5 0.75 1 1.25 1.5 1.75 2

-1.5

-1

-0.5

0.5

1

y=ln(x)

y=p_2(x)

y=p_3y=p_1(x)

Figure: Taylor approximations of ln(x) about x = 1 (see plot log.m)


> 1. Taylor polynomials > 1.2 The error in Taylor’s polynomial

Theorem (Lagrange’s form)

Assume f ∈ Cn[α, β], x ∈ [α, β].The remainder Rn(x) ≡ f(x) − pn(x), or error in approximating

f(x) by pn(x) =n∑j=0

(x− a)j

j!f (j)(a) satisfies

Rn(x) =(x − a)n+1

(n + 1)!f (n+1)(cx), α ≤ x ≤ β (3.3)

where cx is an unknown point between a and x.

[Exercise.] Derive the formal Taylor series for f(x) = ln(1 + x) ata = 0, and determine the range of positive x for which the seriesrepresents the function.

Hint: f(k)(x) = (−1)k−1(k − 1)! 1(1+x)k , f(k)(0) = (−1)k−1(k − 1)!, 0 ≤ x

1+cx≤ 1, if x ∈ [0, 1].

ln(1 + x) =Pn

k=1(−1)k−1 xk

k+

(−1)n

n+1xn+1

(1+cx)n+1



Example

Let f(x) = ex and a = 0.

Recall that pn(x) =n∑j=0

xj

j!= 1 + x+

12!x2 + . . .+

1n!xn.

The approximation error is

ex − pn(x) =xn+1

(n+ 1)!ec, n ≥ 0 (3.4)

with c ∈ (0, x).[Exercise.] It can be proved that for each fixed x,

limn→∞

Rn(x) = 0, i.e., ex = limn→∞

n∑k=0

xk

k!=

∞∑k=0

xk

k!.

(See the case |x| ≤ 1).

For each fixed n, Rn becomes larger as x moves away from 0.



-1 -0.75 -0.5 -0.25 0 0.25 0.5 0.75 1

0.25

0.5

0.75

y=R_1(x)

y=R_2(x)

y=R_3(x)y=R_3(x)

y=R_1(x)

y=R_2(x)

y=R_4(x)

y=R_4(x)

Figure: Error in Taylor polynomial approx. to ex (see plot exp simple.m)



Example

Let x = 1 in (3.4), so from (3.2):

e ≈ pn(1) = 1 + 1 +12!

+13!

+ · · ·+ 1n!

and from (3.4)

Rn(1) = e− pn(1) =ec

(n+ 1)!, 0 < c < 1

Since e < 3 and e0 ≤ ec ≤ e1 we have

1(n+ 1)!

≤ Rn(1) ≤ e

(n+ 1)!<

3(n+ 1)!



Suppose we want to approximate e by pn(1) with

Rn(1) ≤ 10−9.

We have to take n ≥ 12 to guarantee

3(n+ 1)!

≤ 10−9,

i.e., to have p12 as a sufficiently good approximation to e.



Exercise

Expand√

1 + h in powers of h, then compute√

1.00001.

If f(x) = x12 , then f ′(x) = 1

2x− 1

2 , f ′′(x) = − 14x

− 32 , f ′′′(x) = 3

8x− 5

2 , . . .

√1 + h = 1 +

12h− 1

8h2 +

116h3ε−

52 , 1 < ε < 1 + h, if h > 0.

Let h = 10−5. Then

√1.00001 ≈ 1 + .5× 10−5 − 0.125× 10−10 = 1.00000 49999 87500

Since 1 < ε < 1 + h, the absolute error does not exceed

116h3ε−

52 <

116

10−52 = 0.00000 00000 00000 0625

and the numerical value is correct to all 15 decimal places shown.



Approximations and remainder formulae

ex = 1 + x+x2

2!+ · · ·+ xn

n!+

xn+1

(n+ 1)!ec (3.5)

sin(x) = x− x3

3!+x5

5!−· · ·+(−1)n−1 x2n−1

(2n−1)!+(−1)n x2n+1

(2n+1)!cos(c) (3.6)

cos(x) = 1− x2

2!+x4

4!−· · ·+(−1)n x2n

(2n)!+(−1)n+1 x2n+2

(2n+2)!cos(c) (3.7)

11− x

= 1 + x+ x2 + · · ·+ xn +xn+1

1− x, x 6= 1 (3.8)

(1 + x)α = 1 +(α1

)x+

(α2

)x2 + · · ·+

(αn

)xn (3.9)

+(

αn+ 1

)xn+1(1 + c)α−n−1, α ∈ R

Recall that c is between 0 and x, and the binomial coefficients are(ακ

)=α(α− 1) · · · (α− κ+ 1)

κ!



Example

Approximate cos(x) for |x| ≤ π4 with an error no greater than 10−5.

Since cos(c) ≤ 1 and

R2n+1(x) ≤x2n+2

(2n+ 2)!≤ 10−5

we must have (π4

)2n+2

(2n+ 2)!≤ 10−5

which is satisfied when n ≥ 3. Hence

cos(x) ≈ 1− x2

2!+x4

4!− x6

6!



Indirect construction of a Taylor polynomial approximation

Remark

From (3.5), by replacing x with −t2, we obtain

e−t2=1− t2 +

t4

2!− t6

3!+ · · ·+ (−1)nx2n

n!+

(−1)(n+1)x2n+2

(n+ 1)!ec,

(3.10)

− t2≤c≤0.

If we attempt to construct the Taylor approximation directly, thederivatives of e−t

2become quickly too complicated.



Indirect construction of a Taylor polynomial approximation

From (3.8) we can easily get:

−∫ t

0

dx

1− x= −

∫ t

0

(1 + x+ x2 + · · ·+ xn

)dx−

∫ t

0

xn+1

1− xdx,

ln(1− t) (3.12)= −

(t+

12t2 + · · ·+ 1

n+ 1tn+1

)− 1

1− c

∫ t

0xn+1dx

ln(1− t) = −(t+

12t2 + · · ·+ 1

n+ 1tn+1

)− 1

1− cttn+2

n+ 2, (3.11)

where ct is a number between 0 and t, −1 ≤ t ≤ 1.

Theorem (Integral Mean Value Theorem)

Let w(x) b a nonnegative integrable function on (a, b), andf ∈ C[a, b]. Then ∃ at least one point c ∈ [a, b] for which∫ b

af(x)w(x)dx = f(c)

∫ b

aw(x)dx (3.12)



Infinite Series

By rearranging the terms in (3.8) we obtain the sum of a infinitegeometric series

1 + x+ x2 + · · ·+ xn =1− xn+1

1− x, x 6= 1. (3.13)

For |x| < 1, letting n→∞ we obtain the infinite geometric series

11− x

= 1 + x+ x2 + x3 + · · · =∞∑j=0

xj , x 6= 1. (3.14)



Infinite Series

Definition

The infinite series∞∑

j=0

cj

is convergent if the partial sums

Sn =n∑

j=0

cj , n ≥ 0

form a convergent sequence, i.e.,

∃S = limn→∞

Sn

and we then write

S =∞∑

j=0

cj



Infinite Series

For the infinite series (3.14)with |x| 6= 1, the partial sums are given by (3.13):

Sn =1− xn+1

1− x

Sn|x|<1−→n→∞

11−x

Sn is divergent when |x| > 1

What happens when |x| = 1?



Infinite Series

Definition

Assume that f(x) has derivatives of any order atx = a. The infinite series

∞∑j=0

(x− a)j

j!f (j)(a)

is called the Taylor series expansion of the function f(x) about thepoint x = a.

The partial sumn∑j=0

(x− a)j

j!f (j)(a)

is simply the Taylor polynomial pn(x).



Infinite Series

If the sequence {pn(x)} has the limit f(x), i.e. the error tends to zeroas n→∞

limn→∞

(f(x)− pn(x)

)= 0

then we can write

f(x) =∞∑j=0

(x− a)j

j!f (j)(a)



Infinite series

Actually, it can be shown that the errors terms in (3.5)-(3.9) and(3.11) tend to 0 as n→∞ for suitable values of x. Hence the Taylorexpansions

ex =∞∑

j=0

xj

j!, −∞ < x <∞

sinx =∞∑

j=0

(−1)jx2j+1

(2j + 1)!, −∞ < x <∞

cosx =∞∑

j=0

(−1)jx2j

(2j)!, −∞ < x <∞ (3.15)

(1 + x)α =∞∑

j=0

(αj

)xj , −1 < x < 1

ln(1− t) = −∞∑

j=0

tj

j, −1 ≤ x < 1



Infinite seriesDefinition

Infinite series of the form ∞∑j=0

aj(x− a)j (3.16)

are called power series.

They can arise from Taylor’s formulae or some other ways. Their convergencecan be examined directly.

Theorem (Comparison criterion)

Assume the series (3.16) converges for some value x0. Then the series (3.16)converges for all x satisfying |x− a| ≤ |x0 − a|.

Theorem (Quotient criterion)

For the series (3.16), assume that the limit

R = limn→∞

∣∣∣∣an+1

an

∣∣∣∣exists. Then for x satisfying |x− a| < 1

R , the series (3.16) converges to alimit S(x). When R = 0, the series (3.16) converges for any x ∈ R.



Infinite series

Example

Consider the power series in (3.15). Letting t = x2, we obtain theseries

∞∑j=0

(−1)jtj

(2j)!(3.17)

Applying the quotient criterion with

aj =(−1)j

(2j)!

we find R = 0, so the series (3.17) converges for any value of t, hencethe series in the formula (3.15) converges for any value of x.


> 1. Taylor polynomials > 1.3 Polynomial evaluation

Consider the evaluation of the polynomial

p(x) = 3− 4x− 5x2 − 6x3 + 7x4 − 8x5

1 simplest method: compute each term independently, i.e.,

c ∗ xk or c ∗ x ∗ ∗k

yielding1 + 2 + 3 + 4 + 5 = 15 multiplications

2 a more efficient method: compute each power of x using the precedingone:

x3 = x(x2), x4 = x(x3), x5 = x(x4) (3.18)

Since each term takes two multiplications for k > 1, the result will be

1 + 2 + 2 + 2 + 2 = 9 multiplications

3 nested multiplication:

p(x) = 3 + x(−4 + x(−5 + x(−6 + x(7− 8x))))

with only 5 multiplications



Consider now the general polynomial of degree n:

p(x) = a0 + a1x+ · · ·+ anxn, an 6= 0

If we use the 2nd method, with the powers of x computed as in(3.18), the number of multiplications in evaluating p(x) is 2n− 1.

For the nested multiplication, we write and evaluate p(x) in theform

p(x) = a0 + x(a1 + x(a2 + · · ·+ x(an−1 + anx) · · · )) (3.19)

using only n multiplications, saving about 50% over the 2ndmethod.

All methods use n additions.



Example

Evaluate the Taylor polynomial p5(x) for ln(x) about a = 1.

A general formula is (3.11), with t replaced by −(x− 1), yielding

p5(x) = (x− 1︸︷︷︸w

)− 12(x− 1)2 +

13(x− 1)3 − 1

4(x− 1)4 +

15(x− 1)5

= w

(1 + w

(−1

2+ w

(13

+ w

(−1

4+

15w

)))).



A more formal algorithm than (3.19)

Suppose we want to evaluate p(x) at number z. Let define the sequence ofcoefficients bi as:

bn = an (3.20)

bn−1 = an−1 + zbn (3.21)

bn−2 = an−2 + zbn−1 (3.22)

... (3.23)

b0 = a0 + zb1 (3.24)

Now the nested multiplication is the Horner Method:p(z) = a0 + z(a1 + z(a2 + · · ·+ z(an−1 + an︸︷︷︸

bn

z

︸︷︷︸bn−1

) · · ·

︸︷︷︸b2

)

︸︷︷︸b1

)

︸︷︷︸b0



Hence

p(z) = b0

With the coefficients from (3.20), define the polynomial

q(x) = b1 + b2x+ b3x2 + · · ·+ bnx

n−1

It can be shown that

p(x) = b0 + (x− z)q(x),

i.e., q(x) is the quotient from dividing p(x) by x− z, and b0 is theremainder.

Remark

This property we will use it later for polynomial rootfinding method toreduce the degree of a polynomial when a root z has been found, sincethen b0 = 0 and p(x) = (x− z)q(x).


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