Calculus Of One Real Variable – By Pheng Kim Ving
Chapter 15: Representations Of Functions By Power Series – Section 15.3: Taylor Series


15.3
Taylor Series

 

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1. Taylor Series

 

Suppose the function f(x) is represented by a power series on an interval centered at x = c as follows:

 

f(x) = a0 + a1(x – c) + a2(x – c)2 + a3(x – c)3 + ... .

 

Then the value of f(x) can be approximated by a partial sum of the series, eg the 10th partial sum:

 

 

{1.1} Section 8.3.

 

Fig. 1.1

 

 

Fig. 1.2

 

 

as seen in Fig. 1.2. Note that approximation [1.2] approximates the value of f at x by using the values of f itself and its
derivative f ' itself at c and the signed distance x – c from x to c.

 

Since f(c) + f '(c)(x – c) is a degree-1 polynomial in x – c where f(c) and f '(c) are constants, let a0 = f(c) and a1 = f '(c).
Then Approximation [1.2] approximates f(x) by the 2nd partial sum of this power series in x – c:

 

g(x) = a0 + a1(x – c) + a2(x – c)2 + a3(x – c)3 + ... ,

 

where the coefficients an's are such that the series g(x) represents the function f(x) on a neighborhood of c if f(x) can be so
represented. A neighborhood of c is an open interval centered at c.

 

As we want to have g(x) = f(x) for all x in a neighborhood Nc of c, it's necessary that 1) (on the part of f ) f is infinitely often
differentiable on Nc, because g, as a polynomial, is so, and 2) (on the part of g) the values of g and its derivatives of all orders
at c equal those of f and its corresponding derivatives at c respectively: g(c) = f(c), g'(c) = f '(c), g''(c) = f ''(c), etc. Now:

 

f(x) = g(x) = a0 + a1(x – c) + a2(x – c)2 + a3(x – c)3 + a4(x – c)4 + a5(x – c)5 + a6(x – c)6 + a7(x – c)7 + a8(x – c)8 + ... ,
f '(x) = g'(x) = a1 + 2a2(x – c) + 3a3(x – c)2 + 4a4(x – c)3 + 5a5(x – c)4 + 6a6(x – c)5 + 7a7(x – c)6 + 8a8(x – c)7 +... ,
f ''(x) = g''(x) = 2a2 + (3)(2)a3(x – c) + (4)(3)a4(x – c)2 + (5)(4)a5(x – c)3 + (6)(5)a6(x – c)4 + (7)(6)a7(x – c)5 +
            (8)(7)a8(x – c)6 + ...,
f '''(x) = g'''(x) = (3)(2)a3 + (4)(3)(2)a4(x – c) + (5)(4)(3)a5(x – c)2 + (6)(5)(4)a6(x – c)3 + (7)(6)(5)a7(x – c)4 +
             (8)(7)(6)a8(x – c)5 + ...,
f (4)(x) = g(4)(x) = (4)(3)(2)a4 + (5)(4)(3)(2)a5(x – c) + (6)(5)(4)(3)a6(x – c)2 + (7)(6)(5)(4)a7(x – c)3 + (8)(7)(6)(5)a8(x – c)4
              + ...,
f (5)(x) = g(5)(x) = (5)(4)(3)(2)a5 + (6)(5)(4)(3)(2)a6(x – c) + (7)(6)(5)(4)(3)a7(x – c)2 + (8)(7)(6)(5)(4)a8(x – c)3 + ...,
etc.

Thus:

 

f(c) = a0,
f '(c) = a1,
f ''(c) = 2a2,
f '''(c) = (3)(2)a3 = (3)(2)(1)a3 = 3!a3,
f (4)(c) = (4)(3)(2)a4 = (4)(3)(2)(1)a4 = 4!a4,
f (5)(c) = (5)(4)(3)(2)a5 = (5)(4)(3)(2)(1)a5 = 5!a5,
etc,

 

in general: f (n)(c) = n!an, for all integer n = 0, 1, 2, ... .

 

Consequently:

 

 

Definition 1.1 – Taylor Series

 

Suppose the function f(x) is infinitely often differentiable at c. Then the power series centered at c:

 

 

is called the Taylor series of f  centered at c or about c or in powers of x – c. For c = 0, the series is also called the
Maclaurin series. As each coefficient f (n)(c)/n! is unique, the Taylor series of a function centered at a point is unique.

 

 

Refer to Fig. 1.3. For some h > 0 let Nc = (c – h, c + h) be a neighborhood of c. The Taylor series of f centered at c is
defined at any x in Nc as the infinite sum of terms each of which is the product of an nth derivative of f at c, divided by n!, and

Fig. 1.3

 

h > 0, Nc = (c – h, c + h),
x in Nc,
Taylor series of f centered at c involves f (n)(c) and x – c.

 

the nth power of the signed distance x – c from x to c. Just like power series in general, we can think of a Taylor series of f as
a function on the interval where it converges, function that may or may not be f  itself, as we'll see below.

 

Remark that at any point x the Taylor series of f centered at c is an infinite polynomial that involves the values of f itself and its
derivatives of all orders themselves at c and the signed distance from x to c.

 

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2. Analytic Functions

 

Radius Of Convergence Of Taylor Series

 

The Taylor series of a function f(x) centered at a point c exists provided f(x) has derivatives of all orders at c, which means
that f(x) has derivatives of all orders on a neighborhood of c. Of course a Taylor series of f(x) centered at c always converges
at c: it converges to f(c). If it diverges everywhere else then its radius of convergence is 0.  If it converges at a point x1
different from c, then it converges at least on an interval centered at c with positive radius R = |x1 – c| > 0 (radius = half
length), since it's a power series. Does every Taylor series has a positive radius of convergence? Let's find out. We examine the
radius of convergence of series [1.3]. Let an = f (n)(c)/n!. The radius of convergence is the reciprocal of this limit:

 

 

This limit may be 0 or positive and finite or infinity. The radius of convergence may be either infinite or positive and finite or 0.
So the answer is no: not every Taylor series has a positive radius of convergence The radius of convergence of the Taylor
series of a function centered at a point depends on each individual function.

 

What Converging Taylor Series Converge To

 

As the Taylor series of a function f(x) centered at a point c always converges to f(c) at x = c, it always represents f at the
point c.

 

As seen in the discussion above, the conditions that lead to the expression of the Taylor series is necessary. That is, if a power
series represents a function f(x) on an interval I with positive radius, then it must be the Taylor series of f(x) centered at the
centre of I. The conditions may or may not be sufficient. That is, either it's always that the Taylor series of a function f(x)
centered at a point c represents f(x) on an interval centered at c with positive radius or it's not always so. Let's find out. Of
course if the Taylor series diverges everywhere except at c then it can't represent the function on any interval centered at c
with positive radius. Now let the function f(x) be defined by:

 

 

Fig. 2.1

 

T(x) doesn't represent f(x) on any interval centered at 0 with positive radius.

 

Analytic Functions

 

Now we see that the Taylor series T(x) of a function f(x) centered at a point c exists if f(x) is infinitely often differentiable at c,
that T(x) has a radius of convergence that may be 0 or positive and finite or infinite, that if it has a positive radius of
convergence (as opposed to 0, not as opposed to negative, because a radius of convergence can't be negative), it may converge
to f(x) or may converge to some other function.

 

Definition 2.1 – Analytic Functions

 

Suppose that the function f(x) is infinitely often differentiable at a point c. If its Taylor series centered at c converges to it on
an interval centered at c with positive radius, then we say that f(x) is analytic at c. We say that a function is analytic if it's
analytic at every point of its domain.

 

 

Not every infinitely often differentiable function is analytic. However most of the elementary functions are analytic wherever
they're infinitely often differentiable.

 

Converging Power Series Generating Analytic Functions

 

Earlier on in this section we realize that if a power series represents a function on an interval with positive radius, then it must
be the Taylor series of the function centered at the centre of the interval. Now we can add that the function is analytic at the
centre of the interval, because the series converges to the function on the interval. We now prove this property as a theorem.
By “converging” we mean the series converges on an interval with positive radius, not just at only one point.

 

Theorem 2.1 – Converging Power Series Generating Analytic Functions

 

Suppose the power series:

 

 

As the Taylor series of f(x) converges to f(x) on I, f(x) is analytic at c.

 

 

Proof
f(x) = a0 + a1(x – c) + a2(x – c)2 + a3(x – c)3 + a4(x – c)4 + a5(x – c)5 + a6(x – c)6 + ...,
f '(x) = a1 + 2a2(x – c) + 3a3(x – c)2 + 4a4(x – c)3 + 5a5(x – c)4 + 6a6(x – c)5 + ...,
f ''(x) = 2a2 + (3)(2)a3(x – c) + (4)(3)a4(x – c)2 + (5)(4)a5(x – c)3 + (6)(5)a6(x – c)4 + ...,
f '''(x) = (3)(2)a3 + (4)(3)(2)a4(x – c) + (5)(4)(3)a5(x – c)2 + (6)(5)(4)a6(x – c)3 + ...,



EOP

 

Analyticity Of Sum Functions

 

An immediate and straightforward consequence of Theorem 2.1 is that if a power series converges on an interval, then it's the
Taylor series of its sum function centered at the centre of the interval, and its sum function is analytic at the centre of the
interval, since it converges to its sum function on the interval. We state this consequence as a corollary.

Corollary 2.1 – Analyticity Of Sum Functions

 

If a power series converges on an interval I, then it's the Taylor series of its sum function f(x) centered at the centre c of I
and f(x) is analytic at c.

 

 

Uniqueness Of Power-Series Representation

 

Another immediate and straightforward consequence of Theorem 2.1 is that if a function has a power-series representation
centered at a point, then the representation is unique, in the sense that there's only one such power series. This is due to the
uniqueness of the Taylor series of a function centered at a point. We also state this consequence as a corollary.

 

Corollary 2.2 – Uniqueness Of Power-Series Representation

 

If a function has a power-series representation centered at a point, then the representation is unique. That is, if:

 

 

then an = bn for every integer n = 0, 1, 2, ... .

 

 

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3. Maclaurin Series Of Elementary Functions

 

We know the power-series representation of the function 1/(1 – x). In Section 15.2, we obtain those of the functions ln (1 + x)
and arctan x. All three are Maclaurin series of their respective functions. We'll gather them later on with some more series. We
now develop Maclaurin series of sin x, cos x, ex, e–x, sinh x, and cosh x.

 

Example 3.1

 

Show that:

 

 

Given Facts: 1) both series converge absolutely for all x, by the RaT (Ratio Test), and thus converge for all x, and 2) the
general solution of the differential equation y'' + y = 0 is y = A cos x + B sin x, where A and B are arbitrary constants.

 

Note

We may think of using Definition 1.1 to get the Taylor series at c = 0. Of course sin x and cos x are both infinitely often
differentiable at 0. However at this stage we don't yet know if sin x and cos x are analytic, and we don't yet have a theorem
that gives the sufficient condition(s) on a function or its Taylor series or both for the Taylor series to converge to (equal) the
function, which may be used for this purpose. So if we use Definition 1.1, we wouldn't know whether or not the series
converges to the function. Thus we have to utilize a different approach.

 

Solution

Let:

 

 

f(x) = 0(cos x) + B sin x = B sin x,
g(x) = f '(x) = B cos x,

 


EOS

 

Note that, since the establishments of the Maclaurin series of sin x and cos x involve using derivatives, the series require that
x be in radians.

 

Example 3.2

 

Show that:

 

 

Given Facts: 1) the first series converges for all x, by the RaT (Ratio Test), and 2) the general solution of the differential
equation y' – y = 0 is y = Cex, where C is an arbitrary constant.

 

Note

Similar to Note in Example 3.1.

 

Solution
Let:

 


EOS

 

Example 3.3

 

Show that:

 

 

Note

Similar to the Note in Example 3.1.

 

Solution

 


EOS

 

We gather below the Maclaurin series obtained from previous sections and in the above three examples.

 

Maclaurin Series Of Some Elementary Functions

 

 

 

By Corollary 2.1, all the nine above functions are analytic at 0.

 

 

For example, as shown in Fig. 3.1, the Maclaurin series 1 + x + x2 + x3 + x4 + x5 + ... of the function 1/(1 – x) converges to
1/(1 – x) on (–1, 1).

 

Fig. 3.1

 

–1 < x < 1,

 

 

Memorizing The Maclaurin Series

 

The nine Maclaurin series listed above are useful and thus you should commit them to memory. Here's a suggestion on a way
to help memorizing them:

 

For 1/(1 – x): not even not odd, so exponents all 0, 1, 2, ..., non-alternating, denominators all 1, –1 < x < 1.
 
For sin x: odd, so exponents odd 1, 3, 5, ..., alternating, denominators factorials of exponents, all x. Must be alternating,
     otherwise can be > 1. Differs from arctan x only in denominators.
For cos x: even, so exponents even 0, 2, 4, ..., alternating, denominators factorials of exponents, all x. Must be alternating,
     otherwise can be > 1.
For ex: not even not odd, so exponents all 0, 1, 2, ..., non-alternating, denominators factorials of exponents, all x.
For e–x: change x in that of ex to –x.
For sinh x: change all “–” signs in that of sin x to “+”, all x.
For cosh x: change all “–” signs in that of cos x to “+”, all x.

 

For ln (1 + x) and arctan x, denominators aren't factorials. For remaining, denominators are factorials (remember, for
1/(1 – x), all denominators are 1).

 

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4. Maclaurin Series Of Other Functions

 

We can employ known Maclaurin series of elementary functions to find Maclaurin series of other functions, as the following
three examples show.

 

Example 4.1

 

Find the Maclaurin-series representations of the following functions. Write each series in the sigma notation and specify where
the representation is valid.

 

a. (sin2 x)/x.     b. (sin (x2))/x.

 

Note

For part a, we can multiply the series of sin x by itself to get the series of sin2 x. However we'll employ the formula sin2 x =
(1 – cos 2x)/2 and the series of cos 2x, which would be less complicated. As:

 

 

Solution

 


EOS

 

Example 4.2

 

 

Solution

EOS

 

Example 4.3

 

Obtain the first three non-0 terms of the Maclaurin series of the functions:

 

a. tan x, using long division of sin x by cos x.     b. ln cos x.     c. tan x, using part b.

 

Solution

 

 


EOS

 

Remarks 4.2

 

1. In Example 4.3, we need only the first three terms of the requiqred series. In part a, we use long division, so in the long
    division we can utilize just the first three terms of the divident sin x and the first three terms of the divisor cos x. In part b,
    the first series on the third line shows that the first three terms go up to the term containing x6, so in the calculations we
    keep only enough terms to ensure that we can get all the terms up to x6.

 

2. As tanx) = – tan x, tan x is an odd function, thus its Maclaurin series has only odd powers of x. As ln cosx) =
    ln cos x, ln cos x is an even function, thus its Maclaurin series has only even powers of x. These properties serve as a
    check to avoid obtaining a series that can't be correct.

 

 

4. The general terms of the required series in this example can't be found easily. This is the reason why the problem
     statement doesn't require us to write any of the series in the sigma notation.

 

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5. Taylor Series Centered at Points Other Than 0

 

We can also employ known Maclaurin series of elementary functions to find their Taylor series centered at points other than 0,
as the following two examples demonstrate.

 

Example 5.1

 

 

Note

 

Solution

 

Since the representations of cos x and sin x by their Maclaurin series are valid for all real x and since the algebraic operations
on series here involve only mutiplication by a constant and series subtraction, the above representation of cos x by its Taylor
series is also valid for all real x.
EOS

 
Example 5.2

 

Determine the Talylor series of ex about any real point c. Determine where the representation is valid.

 

Solution

 

Since the representation of ex by its Maclaurin series is valid for all real x and since the algebraic operation on series here
involves only multiplication by a constant, the above representation of ex by its Taylor series is valid for all real x.
EOS

 

Remark 5.1

 

As Examples 5.1 and 5.2 show, the Maclaurin series of elementary functions can be utilized to determine their Taylor series
about points other than 0. Similar calculations would enable us to expand sin x and cos x as Taylor series about any real c. So
sin x, cos x, as well as ex are analytic functions (analytic at every point of their domain, which is the real line).

 

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6. Finding Sums Of Series

 

Finding Sums Of Series
 
Example 6.1

 

Find the sum of the series:

 

 

Note

The part of the given series from the third term –x3/3! on is a part of the Maclaurin series for sin x. So we'll algebraically
manipulate the given series to get the Maclaurin series.

 

Solution

EOS

 

Finding Functions Represented By Series

 

Example 6.2

 

Find the function represented by this series:

 

 

Note

Finding the function represented by a given power series is the same as finding the sum of the series.

 

Solution

EOS

 

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Problems & Solutions

 

1. Find the Maclaurin-series representation of the function ln (1 – x). Write the series in the sigma notation and specify where
    the representation is valid.

 

Solution

 

 

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2. Determine the Maclaurin series of e5x+2. Express the series in the sigma notation and state where the representation applies.

 

Solution

 

 

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3. Obtain the first three non-0 terms of the Maclaurin series of the function sec x tan x.

 

Solution

 

 

long division of series of sin x by series of cos2 x:

 

 

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4. Find the Taylor series of the function 1/x2 centered at the point –3. Write the series in the sigma notation and state the
    interval where the representation is valid.

 

Solution

 

 

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Solution

 

 

 

 

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6. Find the sum of the series:

 

    

 

Solution

 

 

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7. Determine the function represented by this series:

 

    

 

Solution

 

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8. A complex number is a number of the form x + iy where x and y are real numbers and i is an imaginary number such that
    i2 = –1.

 

a. Find Maclaurin series of eix and e–ix.

b. Use part a to verify Euler's formula eix = cos x + i sin x.
c. Use part a to verify e–ix = cos x – i sin x.
d. Use parts b and c to find formulas for sin x and cos x in terms of eix and e–ix.
e. Compare the formulas for sin x and cos x found in part d with the formulas for sinh x and cosh x in terms of ex and e–x.

 

Solution

 

 


    The formulas for cos x and sin x are similar to those of cosh x and sinh x respectively, except for the presence of i in those

    for cos x and sin x.

 

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