Calculus Of One Real Variable By Pheng Kim Ving


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1. Derivatives Of Power Series 
Recall that if two functions are equal on an interval, their
derivatives must also be equal on that interval, except at one or both
endpoints of the interval when one or both derivatives don't exist there.
Consider the function 1/(1 x) and its
powerseries representation:
So its radius of convergence is 1/1 = 1, the same as that of
the original power series. Let's find its interval of convergence. At x
= 0 1 = 1 the derivative series is:
Now consider the function ln (1 + x), of course for x
> 1. As will be seen in Example 2.1 later on in
this section, it can be
represented by a power series, as follows:
which diverges, At x = 0 + 1 = 1 the derivative series is:
The interval of convergence of the derivative series is (1,
1), different from that of the original series, which is (1, 1]. The
derivative series diverges at an endpoint where the original series converges.
The derivative series loses an endpoint of the
interval of convergence of the original series. Now we see why for the
derivative series obtained after differentiation we put
at most on before the interval of convergence of the original series.
In general:
Properties Of Derivative Power Series
A power series with non0 (and thus positive) radius of
convergence is differentiated term by term. The derivative series also The centre and radius of convergence of the derivative
series are the same as those of the original series, and its interval of So to determine the interval of convergence of the
derivative series, we have to check any endpoint where the original series 
Determine a power series that represents the function 2/(1
x)^{3} on an interval centered at x
= 0. Write the series in the sigma
notation and determine the interval where the representation is valid.
Note
We can use algebraic manipulation on 1/(1 x),
whose powerseries representation is well known, as done in
Section
15.1 Example 5.1 and Example
5.2. However we note that 2/(1 x)^{3} can be
obtained by differentiating 1/(1 x)
twice
and that the powerseries representation of 1/(1 x)
is known. So we'll try differentiations on 1/(1 x)
and its series. Perhaps,
in this case, it'll be simpler than algebraic manipulation.
Solution
EOS
On Example 1.1.
1. Indeed, in this case, differentiation must be simpler than algebraic manipulation.
2. Because each original series diverges at both
endpoints of its interval of convergence, for the derivative series we don't
need
to put at most on before the
interval of convergence of the original series, as the derivative series must
diverge there also.
We don't worry about determining the
behavior of the derivative series at any endpoint. If the original series
converges at
one or both endpoints, we must check
the derivative series obtained there, as it may converge or may diverge there.
For
example:
The original
series converges at the endpoint x = b. The derivative series may converge or may
diverge there. So for the
derivative series we put at most
on before the interval of convergence of the original series. We would have to
check the
behavior of the derivative series
obtained at x = b
to determine the interval of convergence of the derivative power series.
3. To keep the pattern of the coefficients clear, we
don't multiply out their factors. This is very useful when we have to write
the series in the sigma notation.
4. In the sigma notation, the running index and the
exponent on x are the same. In general, in the
sigma notation we prefer
to have the exponent on x or (x c) be the same as the running index.
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2. Integrals Of Power Series 
Recall that if two functions are equal on an interval, their
integrals must also be equal on that interval up to a constant (differ
by a constant). Consider the function 1/(1 x)
and its powerseries representation:
In this case the integral series converges at an endpoint
where the original series diverges. The integral series gains an
endpoint. Now we see why for the integral series obtained after integration we
put at least on before the interval of
convergence of the original series.
thus:
In general:
Properties Of Integral Power Series
A power series with non0 (and thus positive) radius of
convergence is integrated term by term. The integral series also is a The centre and radius of convergence of the integral
series are the same as those of the original series, and its interval of So to determine the interval of convergence of the
integral series, we have to check any endpoint where the original series 
Since the derivative power series can lose one or both
endpoints of the interval of convergence of the original power series and
can't gain any, it must be that the integral power series can gain one or both
endpoints and can't lose any, because the original
is the derivative of the integral.
Fig. 2.1 provides a way to easily remember the properties of
derivative and integral power series, without confusion as to
which can only lose endpoints and which can only gain endpoints of the interval
of convergence of the original power series.
Think of the derivative as smaller than the original and thus it can only
lose and can't gain endpoints, and think of the integral
as larger than the original and thus it can only gain and can't lose
endpounts.

Fig. 2.1 Derivative Can Only Lose And 
1. Determine a powerseries representation of the
function ln (1 + x) on an
interval centered at x = 0. Write
the series in the
sigma notation and determine the
interval where the representation is valid.
2. Find the sum of the alternating harmonic series.
Note
We know that ln (1 + x) is an
integral of 1/(1 + x) = 1/(1
( x)) and we know the powerseries
representation of 1/(1 x).
So we'll try to integrate 1/(1 x) and its powerseries
representation.
Solution
The centre of convergence of the power series is 0. Let a_{n} = (1)^{n}^{+1}/n. Then:
EOS
In Example 2.1, We can multiply both sides of:
Recall that multiplying a series by a non0 constant doesn't
change the behavior of the series. So the interval of convergence of
of a non0 constant multiple of a power series is the same as that of the
original series. Note that at any point x in their
interval
of convergence, the non0 constant multiple may have a different value (sum)
than that of the original. The range of the non0
constant multiple may be different from that of the original. The interval of
convergence is the domain (of the sum function), not
the range.
1. Determine a powerseries representation of the
function arctan x on an
interval centered at x = 0. Write
the series in the
sigma notation and determine the
interval where the representation is valid.
2. Find the sum of the converging alternating series:
Note
We know that arctan x is the
integral of 1/(1 + x^{2}). We know the
powerseries representation of 1/(1 x) for x < 1. We see
that 1/(1 + x^{2}) can be obtained by replacing x in 1/(1 x) by x^{2}, and that x^{2} < 1 iff x < 1. So well
try integration.
Solution
The centre of convergence of this power series is 0. Let a_{n} = (1)^{n}/(2n + 1). Then:
EOS
Examples 1.1, 2.1,
and 2.2 show that to determine a powerseries
representation of a given function, we can try differentiation
or integration of a known seriesrepresentation of a function. Whether to try
differentiation or integration depends on the given
function.
We gather here the powerseries representations and the sums of series obtained in Examples 2.1 and 2.2.
Two Power Series And Sum Of Each Of Two Series

Series Developments Of Two Numbers

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3. Determining A Function Represented By A Given Series 
Consider the power series 1 + x
+ x^{2} + x^{3} + ... If we
didn't know the formula for its sum, we would have to settle with an
approximation of the sum, eg by a partial sum. Now it's good that the series
equals the function 1/(1 x) on the
interval (1,
1). Its exact sum at any point x in (1, 1)
is the value of 1/(1 x) at that
point x, which can easily be calculated. So
for a
given power series, it may also be useful if there's a function equal to it on
some interval.
The process here is the reverse to those in Examples 1.1, 2.1, and 2.2.
Consider the power series:
1 3x + 9x^{2} 27x^{3} + 81x^{4} 243x^{5} + ...
a. Write it in the sigma notation.
b. Determine its interval of convergence.
c. Determine a function represented by it. Specify the interval where the representation is valid.
which diverges. Thus the required interval of convergence is (1/3. 1/3).
EOS
In Example 3.1 part c, we don't use differentiation or integration, we simply use algebraic manipulation.
Consider the power series:
a. Express it in the sigma notation.
b. Determine its interval of convergence.
c. Find a function represented by it. Specify the interval where the
representation applies.
Solution
In Example 3.2 part c, we factor the common factor x, we write a series as the derivative of its
integral, for as many series as
needed till we reach a series that's recognized to represent some function.
Then we replace that series by the function. Then
we differentiate, as many functions as there are series written as the
derivatives of their integrals. And we obtain the function
that's represented by the given series.
When we replace the series 1 + x
+ x^{2} + x^{3} + x^{4} + x^{5} + ... by the
function 1/(1 x), the condition on x is 1 < x < 1. So
the
condition on the derivative of 1/(1 x) is also
1 < x < 1, for the derivative can't
gain endpoints. This is in accordance with
the finding in part b that the interval of convergence of the given series is
(1, 1), the series that's to represent the function.
Consider the series:
a. Write it in the sigma notation.
b. Find its interval of convergence.
c. Determine a function that it represents. State the interval where the representation is valid.
Solution
which diverges. Thus the interval of convergence is [1, 1).
So the function
is continuous there, and as the series converges and is rightcontinuous there,
the sum of the series and the
value of the function there are
equal. At x = 1 the series diverges. Thus:
EOS
In Example 3.3 part c, we multiply the given series by x^{2}, we write the resulting series as the integral of its
derivative, we get a
series that's recognized to represent some function. Then we replace that
series by the function. Next we integrate the function.
We obtain a function represented by the resulting series. And we divide that
function by x^{2} to obtain the function that's
represented by the given series.
When we replace the series t
+ t^{2} + t^{3} + t^{4} + t^{5} + t^{6} + ... by the function 1/(1 t)
1, the condition on t is 1 <
t < 1. So the
condition on the integral of 1/(1 t) 1 is at
least on 1 < t < 1, for the integral may
gain one or both endpoints where the
given series converges. Since the interval of convergence of the given series
as found in part b is [1, 1), we check the
endpoint 1 where it converges and we don't check the endpoint 1 where it
diverges, to determine the interval where the
representation of the function by it is valid.
The condition on x is the same
as that on t, because t
is just a dummy variable used to represent x to avoid
confusion with the
upper limit of integration x.
A power series involves a variable, say x.
So its sum involves x too. That
is, its sum is a function of x. Its sum is
a function
represented by it. Thus to determine the sum of a power series is to determine a
function that's represented by the power
series. For example, in Example 3.2, the line Determine the function
represented by the power series: can be rephrased as
Determine the sum of the power series:. The solution would be the same.
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4. Differentiability, Integrability, And Continuity Of Power
Series 
In the discussions above we've implicitly assumed that power
series with non0 (and thus positive) radiuses of convergence
are differentiable and integrable on (R,
R), where R > 0 is the radius of convergence. Rigorously
speaking, these properties
have to be proved, since power series are infinite polynomials (of infinite
degree) while polynomials that we dealt with
previously in the contexts of differentiablity and integrability are finite (of
finite degrees). Appropriately these properties are
investigated and proved as a theorem in a more advanced calculus course or a
mathematical analysis course. Here for
reference to the theorem in later sections we state it without proof.
We've also assumed that power series with non0 (and thus
positive) radiuses of convergence are continuous on their intervals
of convergence. If the interval of convergence of a series is (R , R) for some R
> 0, then the series is continuous at every
point in (R, R). If the interval of
convergence of a series is [R
, R), then
the series is continuous at every point in (R, R)
and
rightcontinuous at x = R. Etc. The continuity
property is usually stated and proved as a theorem accompanying the theorem
on the differentiability and integrability properties.
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5. Tasks Performed 
As pointed out in Section 15.1 Part 6, we performed in that section, among other things, the following tasks:
1. Given a function, for any point in its domain we
determine a power series that represents it on an interval centered at the
point, by using algebraic
manipulation.
2. Given a function, for any point in its domain we
determine a power series that represents it on an interval centered at the
point, by using algebraic
operations.
Now in this section we perform the following tasks:
3. In Part 1, given a
function, for any point in its domain we determine a power series that
represents it on an interval centered
at the point, by using
differentiation.
4. In Part 2, given a
function, for any point in its domain we determine a power series that
represents it on an interval centered
at the point, by using integration.
5. In Part 3, conversely,
given a power series, we determine the function represented by the series on an
interval derived from
the interval of convergence of the
series, by using algebraic manipilation or differentiation or integration.
Problems & Solutions 
1. Determine a power series centered at x = 0 that represents the function 1/(1 + 3x). Write the series in the sigma notation
and specify the interval where the
representation is valid.
With the powerseries representation 1/(1 x) = 1 + x + x^{2} + x^{3} + x^{4} + x^{5} + ..., we use
algebraic manipulation. Clearly using
differentiation would lead to nowhere. Now let's see what will happen if we use
integration:
replace x by 3x, 1 < 3x < 1 iff 1/3 > x > 1/3 or 1/3 < x < 1/3:
differentiate both sides:
Well, in this case, we have to also use differentiation in
addition to integration, and it's way more lengthy and more complicated
than using algebraic manipulation.
2. Find a power series in powers of x 3 that represents the function 1/x^{2}. Write the series in the
sigma notation and specify
the interval where the
representation is valid.
Recall that a power series in powers of x 3 is one that's centered at x = 3.
Solution
3. Establish a power series in powers of x that represents the function ln
(3 x). Put the series in the sigma
notation and
specify the interval where the
representation is valid.
Solution
At x = 3 the series is:
4. Determine a powerseriesinpowersofx representation of the function (2 x)/(1 + x). Use the
sigma notation for the
series and find the interval where
the representation applies.
Solution
5. Consider the power series:
1 5x + 25x^{2} 125x^{3} + 625x^{4} 3,125x^{5} + ...
a. Write it in the sigma notation.
b. Determine its interval of convergence.
c. Find a function represented by it. Specify the interval where the representation is valid.
6. Consider the series:
2 + 4x^{2} + 6x^{4} + 8x^{6} + 10x^{8} + 12x^{10} + ...
a. Express it in the sigma notation.
b. Find its interval of convergence.
c. Determine its sum. State the interval where the expression for the sum is valid.
which diverges. So the interval of convergence of the series is (1, 1).
7. Consider the series:
a. Write it in the sigma notation.
b. Determine its interval of convergence.
c. Find a function represented by it. Specify the interval where the representation applies.
Solution
which diverges. Thus the interval of convergence is [1, 1).
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