Calculus Of One Real Variable – By Pheng Kim Ving


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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 T_{n}(x) is a polynomial in x
– c 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.
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 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 
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 x – c
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.
Consider the 5th Taylor polynomial T_{5}(x) of f(x) = e^{x} centered at 0. As the
derivative of any order of e^{x}
is e^{x} itself and as e^{0} = 1,
we have:
which is different from the first one. So which T_{5}(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 e^{x},
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 x
– c 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.
Since the nth Taylor
polynomial T_{n}(x) and the (n + 1)st
partial sum s_{n}_{+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 e^{x}
and c = 0 we
have:
1. Find the 3rd Taylor polynomial T_{3}(x)
of the polynomial function f(x) = 4x^{3} + 2x^{2} + x + 4
centered at c = 2.
2. Express T_{3}(x) as a polynomial in x.
3. Find the 4th Taylor polynomial T_{4}(x) of f(x) centered at c
= 2.
4. Find the 100th Taylor polynomial T_{100}(x) of f(x) centered at c
= 2.
5. Find the 3rd Taylor polynomial T_{3}(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 T_{4}(x) = T_{3}(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 T_{100}(x) = T_{3}(x) = 46 + 57(x – 2) +
26(x – 2)^{2} + 4(x
– 2)^{3}.
EOS
Determine the 10th Taylor polynomial T_{10}(x) of the function sin x centered at c = 0. What's the degree of T_{10}(x)?
Solution
Let f(x) = sin
x. Then:
The degree of T_{10}(x) is 9.
EOS
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2. Taylor Theorem 
In Section 8.3 we approximate the value of f at a + h by the tangentline 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. 2.2 
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 T_{n}(x) best describes the behavior of f(x) at
points x near c,
because, as:
That is, the values of T_{n}
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 T_{n}(x) centered at c.
At any point x, f(x) is the value of f
and T_{n}(x) is the
value of T_{n}.
For example, as seen in Fig. 2.3, f(x_{1}) and T_{n}(x_{1}) are the values of f and T_{n} at
the point x_{1} respectively. Let D be
the difference between the value f(x) and the value T_{n}(x),
so that D = f(x) – T_{n}(x). Of course f(x) depends on x. As for T_{n}(x),
as a matter of fact it depends on x too. For
example, in Fig. 2.3, T_{n}(x_{2}) is different from T_{n}(x_{1}). Now T_{n}(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 non0 terms of T_{n}(x)

Fig. 2.3 f(x_{1}) = T_{n}(x_{1}) + E_{n}(x_{1}), 
may increase also, and thus the value of T_{n}(x) may change. So D depends on both n
and x. From D = f(x) – T_{n}(x) we have
f(x) = T_{n}(x) + D.
Since among all polynomials of degree less than or equal to n, the nth Taylor
polynomial T_{n}(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 degree1 polynomial on the
righthand side of approximation [2.2], which approximates f(x). Thus we can approximate f(x) by T_{n}(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
E_{n}(x). Consequently:
f(x) = T_{n}(x) + E_{n}(x). [2.3]
Eq. [2.3] is the approximation of f(x) by T_{n}(x) with error E_{n}(x). In Fig. 2.3, at the point x_{1} we have f(x_{1}) = T_{n}(x_{1}) + E_{n}(x_{1}), and
at the point x_{2} we have f(x_{2}) = T_{n}(x_{2}) + E_{n}(x_{2}). As f(x_{1}) > T_{n}(x_{1}) and f(x_{2}) < T_{n}(x_{2}), we get E_{n}(x_{1}) > 0 and E_{n}(x_{2}) < 0. We
rewrite 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 E_{n}(x) is also called the remainder of the
approximation of f(x) by T_{n}(x), and Eq. [2.6] is called the Taylor 
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 E_{n}(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 (c – h, c + k) is an interval containing c
and x, where h
> 0 and k > 0.

Fig. 2.4 f(x) = T_{n}(x) + E_{n}(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 E_{n}(c) = 0, which satisfies Eq. [2.6], because for x = c, Eq. [2.6]
yields f(c) = f(c) = f(c) + 0 =
f(c) + E_{n}(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!)(x – t)^{k}.
Let u = f
^{(}^{k}^{+1)}(t)/k! and dv = (x – t)^{k} dt,
so
that du
= ( f ^{(}^{k}^{+2)}(t)/k!) dt and v = – (x – t)^{k}^{+1}/(k + 1). Then:
EOP
As expected, Eqs. [2.7] and [2.8] demonstrate that the size
of the error E_{n}(x) in the approximation of f(x) by T_{n}(x) is smaller,
and thus the approximation is more accurate, if x
gets closer to c or n
gets larger or both.
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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.
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.
a. Approximate sin 12^{o} using the 3rd
Taylor polynomial T_{3}(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 12^{o} using the 4th Taylor polynomial T_{4}(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 T_{4}(x).
e. Compare the bound from using T_{4}(x) to the one from using T_{3}(x).
f. Can the bound from using T_{4}(x) be used as a bound when using T_{3}(x)?
Why or why not?
1. The points 0 and 30^{o} are where the values of sin
x and of all of its derivatives are readily
calculated, but 0 is closer to 12^{o}
than is 30^{o}. 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 T_{4}(x), which is
0.0000034, is smaller and thus better than the one from using T_{3}(x),
which is
0.000017.
f. Yes, because T_{4}(x) = T_{3}(x) and thus  f(x)
– T_{4}(x) =  f(x) – T_{3}(x) for any x.
EOS
1. Because:
5. In this example, T_{4}(x) = T_{3}(x), but E_{4}(x) < E_{3}(x).
Example 3.1 indicates how tables of values of trigonometric
functions were constructed. The values are approximate, with a
predefined 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.
In Part 2 we see the tangentline approximation:
So the tangentline approximation, as developed in Section
8.3, amounts to the approximation of f(x) by the first Taylor
polynomial T_{1}(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(x – c) is exactly
the Lagrange remainder E_{1}(x). Thus the tangentline approximation and the
formula of its error are the approximation by the first Taylor polynomial and the
Langrange form of its
remainder or error.
b. Determine a bound on the error size.
Solution
A bound on the error size is 0.0005.
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.
The preceding examples illustrate the general procedure for determining a bound on the error size. For the error:
Because (n + 1)! and x
– c^{n}^{+1} are
numerically determined, a bound on E_{n}(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.
1. 45^{o} are closer to 43^{o} than are 0 and 60^{o}.
3. In Example 3.1, we're given n
and asked to find T_{n}(x) and a bound on E_{n}(x). Here, we're given a bound on E_{n}(x) and
asked to find n
to achieve that given bound; it's the reverse process.
Solution
EOS
Another application of the Taylor Theorem is the determination of Taylor series including Maclaurin series.
Determine the Maclaurin series that represents e^{x} by using Taylor
Theorem. Write the series in the sigma notation and specify
the interval where the representation is valid.
In Section 15.3 Example 3.2, we use a different method to determine the Maclaurin series of e^{x}.
Solution
Let f(x) = e^{x}. Then:
EOS
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.
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
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 nthorder 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.
It's the MeanValue Theore.
2. Let f(x) = x^{2} + 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.
3. For a given positive integer n, what function f(x) that's equal to its nth Maclaurin polynomial?
The nth Maclaurin
polynomial of f(x)
is:
4. Find the 4th and 5th Maclaurin polynomials of sec x. What do you notice?
Let f(x) = sec x. Then:
We notice that T_{4}(x) = T_{5}(x).
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.
A bound on the error size is 0.024.
8.
b. Obtain a bound on the error size of the approximation in part a.
As a consequence:
error< 0.0000078125 + 0.000064 = 0.0000718125 < 0.000072.
Hence a bound on the error size is 0.000072.
are each of degree 2.
9. Find the Maclaurin series that representis sin^{2} x by using the Taylor theorem. Write the series
in the sigma notation and
specify
the interval where the representation is valid.
Let f(x) = sin^{2} x. Then:
Solution
Let f(x) = cos x. Then:
So:
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