## Calculus Of One Real Variable – By Pheng Kim Ving Chapter 15: Representations Of Functions By Power Series – Section 15.5: Taylor Polynomials And Taylor Theorem

15.5
Taylor Polynomials And Taylor Theorem

 1. Taylor Polynomials

Consider the Taylor series T(x) of a function f(x) centered at c:

### Fig. 1.1

Taylor series of f(x) centered at c:

### Fig. 1.2

Taylor polynomial of f(x) centered at c:

Clearly Tn(x) is a polynomial in xc that involves derivatives of f(x) of orders up to and including n. So it's called the nth
Taylor polynomial of
f(x) centered at c. See Fig. 1.2. If f(x) has derivatives of orders up to and including n at c, regardless of
whether or not it has derivatives of orders higher than n at c, Eq. [1.2] still makes sense, that is, the nth Taylor polynomial of
f(x) centered at c still exists.

##### Definition 1.1 – Taylor Polynomials
 Suppose the function f(x) has derivatives of orders from 1 to n at a point c. Then the polynomial in x – c defined at any point x by:     is called the nth Taylor polynomial of f (x) centered at or about or at c. If c = 0, then the polynomial is also called the Maclaurin polynomial. Note that the nth Taylor polynomial is denoted Tn(x).

##### Remarks 1.1

1. Like the Taylor series, the Taylor polynomial of f(x) centered at c is defined at any x by an expression that involves the
derivatives f (k)(c) of f(x) at c and the signed distance xc from x to c. See Fig. 1.2.

2. It's possible that f (n)(c) = 0, as shown in Example 1.2 below, where c = 0 and n = 10.

3. The nth Taylor polynomial is so called because it involves derivatives of orders up to and including n, not because its degree
is n. Its degree may be less than n, since f (n)(c) may be 0. It has n + 1 terms, some of which may be 0.

##### Determining Taylor Polynomials

Consider the 5th Taylor polynomial T5(x) of f(x) = ex centered at 0. As the derivative of any order of ex is ex itself and as e0 = 1,
we have:

which is different from the first one. So which T5(x) of sin x is the correct one?  It's the first one with degree 5 that's correct,
because it's determined by directly using the definition of the 5th Taylor polynomial: the sum of terms from the first term to the
term with the 5th derivative only. Utilizing the sigma summation in this case leads to an incorrect 5th Taylor polynomial
because in this sigma summation, the index k isn't the order of the derivative, which is 2k + 1.

Now we see that for the case of ex, using the sigma summation also yields the correct Taylor polynomial because in its sigma
summation, the index k is the order of the derivative, as evidenced by the fact that the exponent on x is the same as the index.

In general, in determining a Taylor polynomial, if in the sigma summation the running index isn't the order of the derivative (as
evidenced by, for example, the fact that the exponent on xc isn't the same as the running index), then don't use the sigma
summation; instead use the definition of Taylor polynomials directly. Remember, the nth Taylor polynomial is the sum of terms
from the first term to the term with the nth derivative only.

##### Taylor Polynomials And Partial Sums

Since the nth Taylor polynomial Tn(x) and the (n + 1)st partial sum sn+1(x) are both the sum of the first n + 1 terms, we may
think that automatically they're the same quantity. Well, the situation discussed under the heading “Determining Taylor
Polynomials” above brings our attention to the possibility that this isn't always the case. Let's find out. For ex and c = 0 we
have:

##### Example 1.1

1. Find the 3rd Taylor polynomial T3(x) of the polynomial function f(x) = 4x3 + 2x2 + x + 4 centered at c = 2.
2. Express T3(x) as a polynomial in x.
3. Find the 4th Taylor polynomial T4(x) of f(x) centered at c = 2.
4. Find the 100th Taylor polynomial T100(x) of f(x) centered at c = 2.
5. Find the 3rd Taylor polynomial T3(x) of f(x) centered at c = –2.

Solution

3. As f '''(x) = 24 we have f (4)(x) = 0 and thus f (4)(2) = 0. So T4(x) = T3(x) = 46 + 57(x – 2) + 26(x – 2)2 + 4(x – 2)3.

4. As f '''(x) = 24 we have f (k)(x) = 0 and thus f (k)(2) = 0 for all k = 4, 5, 6, ... . So T100(x) = T3(x) = 46 + 57(x – 2) +
26(x – 2)2 + 4(x – 2)3.

EOS

##### Example 1.2

Determine the 10th Taylor polynomial T10(x) of the function sin x centered at c = 0. What's the degree of T10(x)?

Solution
Let f(x) = sin x. Then:

The degree of T10(x) is 9.
EOS

 2. Taylor Theorem

##### Approximations Of Functions By Taylor Polynomials

In Section 8.3 we approximate the value of f at a + h by the tangent-line approximation:

Approximation [2.2] approximates the value of f at x by using the values of f itself and of its derivative f ' itself at c and the

 Fig. 2.1

 Fig. 2.2

signed distance from x to c. This of course is natural, since it's the value of f at x that's to be approximated.

Among all polynomials of degree less than or equal to n, the nth Taylor polynomial Tn(x) best describes the behavior of f(x) at
points x near c, because, as:

That is, the values of Tn and of its derivatives of orders from 1 to n at c are equal to those of f and of its corresponding
derivatives of orders from 1 to n at c respectively.

Consider a function f(x) and its nth Taylor polynomial Tn(x) centered at c. At any point x, f(x) is the value of f and Tn(x) is the
value of Tn. For example, as seen in Fig. 2.3, f(x1) and Tn(x1) are the values of f and Tn at the point x1 respectively. Let D be
the difference between the value f(x) and the value Tn(x), so that D = f(x) – Tn(x). Of course f(x) depends on x. As for Tn(x),
as a matter of fact it depends on x too. For example, in Fig. 2.3, Tn(x2) is different from Tn(x1). Now Tn(x) also depends on n:
if n increases and the derivatives of orders of all those larger n's of f(x) at c exist, then the number of non-0 terms of Tn(x)

### Fig. 2.3

f(x1) = Tn(x1) + En(x1),
f(x2) = Tn(x2) + En(x2).

may increase also, and thus the value of Tn(x) may change. So D depends on both n and x. From D = f(x) – Tn(x) we have
f(x) = Tn(x) + D.

Since among all polynomials of degree less than or equal to n, the nth Taylor polynomial Tn(x) best describes the behavior of
f(x) at points x near c, it describes the behavior of f(x) at points x near c better than the degree-1 polynomial on the
right-hand side of approximation [2.2], which approximates f(x). Thus we can approximate f(x) by Tn(x). When we do so, the
error of the approximation of course is D. Since D is the error of the approximation and depends on n and x, let's denote it by
En(x). Consequently:

f(x) = Tn(x) + En(x).                                                                                                                                               [2.3]

Eq. [2.3] is the approximation of f(x) by Tn(x) with error En(x). In Fig. 2.3, at the point x1 we have f(x1) = Tn(x1) + En(x1), and
at the point x2 we have f(x2) = Tn(x2) + En(x2). As f(x1) > Tn(x1) and f(x2) < Tn(x2), we get En(x1) > 0 and En(x2) < 0. We
re-write Eq. [2.3] as two relations as follows:

Approximation [2.4] or [2.5] approximates the value of f at x by using the values of f itself and of its derivatives of orders from
1 to n themselves at c and the signed distance from x to c. This of course is natural, since it's the value of f at x that's to be
approximated. This approximation is an extension of and an improvement on approximation [2.2].

Definition 2.1 –Remainder And Taylor Formula

 The error in the approximation:     The error En(x) is also called the remainder of the approximation of f(x) by Tn(x), and Eq. [2.6] is called the Taylor formula.

##### Taylor Theorem

An approximation is of little or no use unless we know how accurate it is. That is, how much farthest away from the actual
value the approximate value can be, which requires that we know a bound on the size or magnitude (absolute value) of the
error.

Taylor Theorem provides two formulas for the error En(x) of Eq. [2.5]. The first one involves the (n + 1)st derivative of f(x)
evaluated at a point d different from c and x and between them. It is:

not f(t) or f (k)(t) for any k. This formula is called the integral form of the remainder. For Definition 2.1 and Theorem 2.1 below
refer to Fig. 2.4, where (ch, c + k) is an interval containing c and x, where h > 0 and k > 0.

### Fig. 2.4

f(x) = Tn(x) + En(x).

Theorem 2.1 – Taylor Theorem

 Eq. [2.8] is called the integral form of the remainder or integral remainder.

Proof
1. Langrange Form Of The Remainder.

For x = c, Eq. [2.7] yields En(c) = 0, which satisfies Eq. [2.6], because for x = c, Eq. [2.6] yields f(c) = f(c) = f(c) + 0 =
f(c) + En(c).

Consequently g(c) = g(x). Now of course g(t) is continuous on [c, x] if x > c or on [x, c] if x < c. Hence, by Rolle
Theorem, there exists d strictly between c and x such that g'(d) = 0. We have:

B = f (n+1)(d),

and the proof of the Lagrange form of the remainder is completed.

2. Integral Form Of The Remainder.

We use mathematical induction to prove Eq. [2.8]. Since f(t) is an antiderivative of f '(t), applying the Fundamental Theorem
Of Calculus to f '(t) we have:

Again apply integration by parts, this time to the integral of ( f (k+1)(t)/k!)(xt)k. Let u = f (k+1)(t)/k! and dv = (xt)k dt, so
that du = ( f (k+2)(t)/k!) dt and v = – (xt)k+1/(k + 1). Then:

EOP

##### Remark 2.1

As expected, Eqs. [2.7] and [2.8] demonstrate that the size of the error En(x) in the approximation of f(x) by Tn(x) is smaller,
and thus the approximation is more accurate, if x gets closer to c or n gets larger or both.

 3. Applications Of Taylor Theorem

The Lagrange and integral forms of the remainder in the Taylor theorem is used in the applications of this theorem. We'll use
the Lagrange form whenever appropriate, as it's less complicated than integration. However it's not always able to yield what
we want. In such cases we'll of course have to try the integral form.

##### Approximations And Error Bounds

An application of the Taylor Theorem is the approximations of functions by Taylor polynomials including the determinations of
bounds on the sizes of the errors.

##### Example 3.1

a. Approximate sin 12o using the 3rd Taylor polynomial T3(x) centered at 0 (3rd Maclaurin polynomial).
b. Find a bound on the size of the error. Hint: You can use the fact that |sin x| < x for all x > 0.
c. Now approximate sin 12o using the 4th Taylor polynomial T4(x) centered at 0, noting that for sin x, the 3rd and 4th Taylor
polynomials are the same
d. Find a bound on the size of the error when using T4(x).
e. Compare the bound from using T4(x) to the one from using T3(x).
f. Can the bound from using T4(x) be used as a bound when using T3(x)? Why or why not?

## Notes

1. The points 0 and 30o are where the values of sin x and of all of its derivatives are readily calculated, but 0 is closer to 12o
than is 30o. So the question asks us to use 0 as the centre.

2. Don't forget to convert degrees to radians, as the calculus of trigonometric functions is developed using the radian measure.

Solution

A bound on the size of the error is 0.0000034.

e. The bound from using T4(x), which is 0.0000034, is smaller and thus better than the one from using T3(x), which is
0.000017.

f. Yes, because T4(x) = T3(x) and thus | f(x) – T4(x)| = | f(x) – T3(x)| for any x.

EOS

##### Remarks 3.1

1. Because:

5. In this example, T4(x) = T3(x), but E4(x) < E3(x).

##### Tables, Calculators, And Computers

Example 3.1 indicates how tables of values of trigonometric functions were constructed. The values are approximate, with a
pre-defined accuracy, and were calculated by using Taylor polynomials. As for calculators, rather than using space in memory
to store tables of trigonometric functions, the manufacturer builds into the calculator a program that the machine utilizes to
estimate values of these functions as it needs them, by employing Taylor polynomials, as we approximated values in Example
3.1. As for computers, rather than using space on the hard disk to store tables of trigonometric functions in file(s) and then
loading them into memory, the manufacturer builds into the computer a program that the machine utilizes to estimate values
of these functions as it needs them, by employing Taylor polynomials, as we approximated values in Example 3.1.

##### Tangent-Line Approximations

In Part 2 we see the tangent-line approximation:

So the tangent-line approximation, as developed in Section 8.3, amounts to the approximation of f(x) by the first Taylor
polynomial T1(x).

 Fig. 3.1

In Section 8.3 Theorem 2.1, in the approximation of f(a + h) using f(a), the error is E(h):

f(a + h) = f(a) + h f '(a) + E(h),

for some d strictly between c and x. We see that E(xc) is exactly the Lagrange remainder E1(x). Thus the tangent-line approximation and the formula of its error are the approximation by the first Taylor polynomial and the Langrange form of its
remainder or error.

##### Example 3.3

b. Determine a bound on the error size.

Solution

A bound on the error size is 0.0005.

## EOS

We choose the perfect square 100 because its square root is known and among the perfect squares, it's closest to 102. In
general, in approximating f(x), we choose c such that f(c) is known or readily calculated and among the numbers b where
f(b) is known, c is the closest to x.

##### Determining Error Bounds

The preceding examples illustrate the general procedure for determining a bound on the error size. For the error:

Because (n + 1)! and |xc|n+1 are numerically determined, a bound on |En(x)| is obtained by finding a bound on | f (n+1)(d )|.
We attempt to find the maximum possible value that | f (n+1)(d )| can have for any d over the entire interval strictly between c
and x. If found, it'll be used as a bound. If that's hard or impossible, that is, if the exact maximum of | f (n+1)(d )| is hard or
impossible to find, then we determine a bound on | f (n+1)(d )| that can be computed, keeping in mind that either c < d < x or
x < d < c, from which an inequality is taken that's utilized to establish a bound.

## Notes

1. 45o are closer to 43o than are 0 and 60o.

3. In Example 3.1, we're given n and asked to find Tn(x) and a bound on |En(x)|. Here, we're given a bound on |En(x)| and
asked to find n to achieve that given bound; it's the reverse process.

Solution

EOS

##### Determining Series

Another application of the Taylor Theorem is the determination of Taylor series including Maclaurin series.

##### Example 3.4

Determine the Maclaurin series that represents ex by using Taylor Theorem. Write the series in the sigma notation and specify
the interval where the representation is valid.

## Note

In Section 15.3 Example 3.2, we use a different method to determine the Maclaurin series of ex.

Solution
Let f(x) = ex. Then:

EOS

##### Remark 3.3

By Taylor Theorem we have:

In general, to determine the Taylor or Maclaurin series of a function by utilizing Taylor Theorem, we have to show that the limit
of the remainder as n approaches infinity is 0.

As for the interval where the representation is valid, it's one of all x such that the limit of the remainder as n approaches infinity
is 0.

##### Example 3.5

Determine the Taylor series of ln x centered at 3 by employing Taylor Theorem. Express the series in the sigma notation and
state the interval where the representation applies.

Solution
Let f(x) = ln x. Then:

So:

EOS

##### Remark 3.4

We calculate derivatives of f(x) till the order that's high enough that the pattern of the derivatives is clear and we can express
the general nth-order derivative as a general expression. In the answer, notice the “extra” term “ln 3”. Although it's not part of
the series written in the sigma notation, it's a part of the “larger” series that represents the function.

# Problems & Solutions

1. State Taylor Theorem for the case where n = 0. Identify the familiar theorem that's obtained.

##### Solution

It's the Mean-Value Theore.

2. Let f(x) = x2 + 3x + 2. Find:
a. The 2nd Maclaurin polynomial of f(x).
b. The 100th Maclaurin polynomial of f(x).
c. The 2nd Taylor polynomial of f(x) centered at –3.
d. The 100th Taylor polynomial of f(x) centered at –3.

##### Solution

3. For a given positive integer n, what function f(x) that's equal to its nth Maclaurin polynomial?

##### Solution

The nth Maclaurin polynomial of f(x) is:

4. Find the 4th and 5th Maclaurin polynomials of sec x. What do you notice?

##### Solution

Let f(x) = sec x. Then:

We notice that T4(x) = T5(x).

##### Solution

6.

a. Approximate sec 0.2 using the 2nd Maclaurin polynomial.
b. Find a bound on the error size.

Solution

a. Let f(x) = sec x. Then:

a bound on the error size is 0.0063.

7.

b. Determine a bound on the error size.

##### Solution

A bound on the error size is 0.024.

8.

b. Obtain a bound on the error size of the approximation in part a.

##### Solution

As a consequence:

|error|< 0.0000078125 + 0.000064 = 0.0000718125 < 0.000072.

Hence a bound on the error size is 0.000072.

## Note

are each of degree 2.

9. Find the Maclaurin series that representis sin2 x by using the Taylor theorem. Write the series in the sigma notation and
specify the interval where the representation is valid.

##### Solution

Let f(x) = sin2 x. Then:

Solution

Let f(x) = cos x. Then:

So: