Calculus Of One Real Variable – By Pheng Kim Ving
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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.
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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
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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 |
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
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Fig. 1.3 h > 0, Nc =
(c – h,
c + h), |
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 |
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.
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:
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Fig. 2.1 T(x) doesn't represent f(x) on any interval centered at 0 with positive radius. |
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
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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 |
Not every infinitely often differentiable function is analytic.
However most of the elementary functions are analytic wherever
they're infinitely often differentiable.
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
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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
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
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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 |
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
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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.
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.
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.
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.
Similar to Note in Example 3.1.
Solution
Let:
EOS
Show that:
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
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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).
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Fig. 3.1 –1 < x < 1, |
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.
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.
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
Solution
EOS
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
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 tan (–x) = – tan
x, tan x
is an odd function, thus its Maclaurin series has only odd powers of x. As ln cos (–x) =
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.
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
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
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 |
Find the sum of the series:
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
Find the function represented by this series:
Finding the function represented by a given power series is the same as finding the sum of the series.
Solution
EOS
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.
2. Determine the Maclaurin series of e5x+2. Express the series in the sigma notation and state where the representation applies.
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:
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
6. Find the sum of the series:
7. Determine the function represented by this series:
Solution
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.
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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