-
For
,
,
,
and
.
In general, for
,
we have that
.
For the Maclaurin series—which is a Taylor series about
—we
need to find the values for the derivatives when
.
Now
,
,
and, for
,
The Maclaurin series for
is given by 
-
We will do this one by manipulating the given to get an expression that is related to the geometric series.
-
Since
as
,
.
Therefore, by the nth Term Test, the given series diverges.
-
First, we check for absolute convergence. Let
.
Since
by the Ratio Test, the given series is absolutely convergent. From the Absolute Convergence Test, we conclude that the given series is convergent. -
To check for absolute convergence, consider
.
Appeal to the Ratio Test will be inconclusive. Either the Integral Test or one
of the Comparison Tests works. Let
and
.
Then
diverges as the Harmonic Series. Since
,
by the Limit Comparison test,
also diverges. Therefore, the given series is NOT absolutely convergent. Next,
we check for convergence of the series as given; the series
is alternating and
.
For
real, if
,
then
for
.
Consequently,
is decreasing for
.
Because
,
we conclude that
as
,
by the Alternating Series Test, the given series in convergent. Combining the
discussion, we have that the given series is conditionally
convergent.
-
Since
,
by the nth Term Test, the given series diverges.
-
To check the given series for absolute convergence, we check
.
For each
,
.
Since
is convergent as a p-series with
,
by the Comparison Test,
is convergent; thus,
is absolutely convergent. From the Absolute Convergence Test, the given series
is also convergent.
For
,
,
,
,
and
.
We want to show that, for every
there exists a
such that
implies that
.
Suppose that
is given and
where
is ours to define. Note that, if
,
then
and
.
Now
whenever
.
Hence, taking
works. Therefore,
Note that the solution just presented is in a format that we are allowed to
use to save time; the placement of the word "whenever" and making sure that
the "
"
is at the end of the string of inequalities that starts with
are essential to that form of presentation. The following is a more formal
presentation that follows scratch work that leads you to a
that works.
Another solution: For
,
let
.
If
,
then
and
.
Thus,
Since
was arbitrary, we conclude that, for every
there exists a
such that
whenever
;
i.e.,
.
For
,
and
Then, we
have
Suppose
and
has continuous partials. Let
and
.
Then we have that
while
and
are functions of
and
.
If you draw a branch diagram, this gives two branches emanating from
followed by two branches emanating from each of
and
.
This
yields
and
Now,
the branch diagram for
matches the one for
except that it starts with
instead of
;
it has two branches emanating from
followed by two branches emanating from each of
and
.
Also, because the partials are continuous,
.
In order to simplify the expression we derive, recall the following
alternative notation:
and
.
From the product rule and the chain rule (branch
diagram),
The
final answer given here is more simplified than was needed.
The lamina
is bounded by
for
and
.
We want
where
and
;
in this case, we take the density
to be constant, to get that
and
.
The area of
is 
Next we find
the numerators for
and
;
the first is found using either integration by parts or a table of integrals,
while the second uses the double angle formula or an integral
table.
and
Finally,
.
Taking the mass at one yields a density of
where
denotes the volume of the solid
that is described by
,
,
.
where
the last integral is done using trigonometric substitution or appeal to a
Table of Integrals. Taking the mass to be one yields a density at point
of
.
Finally,
and
First we want to obtain a description of the solid
that is bounded between the cone
and the sphere
in terms of spherical coordinates
.
(If you haven't done so already, draw in a sample radial unit from the
origin the spherical part of the boundary. Imagine yourself moving
the unit around to fill the solid.) We can translate the given by making use
of the following relationships between rectangular and spherical coordinates:
,
,
and
.
With the appropriate substitutions, the given cone in spherical coordinates
consists of the set of all
such that
which, for
yields the set of all
such that
or simply
.
The translate the equation of the sphere, note that
For
,
we are have that the whole sphere can be realized as the set of all
such that
 |
that is bounded below by
and above by
;
i.e.,
We are looking for a number of terms of the Maclaurin series expansion for
that we can use in order to insure finding
with an error not exceeding
.
A bound on the error can be found using Lagrange's Remainder Theorem. To find
the Maclaurin series of
about
,
we can use series manipulations from the geometric series or we can generate
the formula for the nth derivative in terms of n and use it in the formula for
the Maclaurin series. I will do the latter, because it will automatically give
me the formula that I can use for bounding the error. For
,
,
,
.
From
and
for
,
we have that, for
,
From Lagrange's
Theorem, if
is a fixed real number such that
,
then there exists some
between
and
for which
.
In the given problem, we are interested in
;
so we want to take
in the set-up that we have developed. Now we have that, for some
between
and
Because
is between
and
,
we have that
.
Therefore, 
Now
we just need to check some values of
until we get one that will give us a value for
that does not exceed
.
Using our trusty calculator, we see that for
in
we get
,
,
and
,
respectively. Since
,
we can take
to get within the specified range. NOTE that since
is decreasing in
,
any
will work. To check our work, taking
in
gives
.
Entering
into my TI-85 yielded
;
this value less
is approximately
.
Using the 6th order Maclaurin polynomial for
to obtain an estimate for
is a bit more on the `plug and chug' side of the things that we have done. For
,
.
To get the estimate requested, we use
to obtain that 
Since
is a power series in powers of
we know that we are going for an interval of convergence that is centered
about
;
we will have to check the endpoints to determine if the set of convergence is
an open interval, a closed interval, or a half-open (half-closed) interval.
For fixed
,
let
.
Since
whenever
,
by the (absolute) ratio test, the given series is convergent at least for
every fixed real
such that
and is divergent at least for each
such that
.
We only have the two points
and
to check directly. When
we have
which is
convergent as a constant times a series that is convergent by Limit Comparison
with
\a
convergent p-series with p=3. When
,
we have
which is
convergent by the Alternating Series Test because
as
.
Combining our results, we have that the given series is convergent for all
in the closed interval
and divergent for all
in
.
In your numberline picture you want to label (and color in) the points
and
which should be centered around
and the write the appropriate information above the three intervals and above
the two points.
We need a formula for the cost of making a box subject to the restrictions
that we are given. Denote the dimensions of the base of the box by
ft and
ft; suppose that the length of the height of the box is
ft. Then the cost is
which simplifies to
in dollars. Using the additional restriction that the volume of the box must
be 20
yields that
.
Since
,
we can take
which leads us to a cost function in two real
variables:
We
want to minimize
.
The critical points occur when
and
.
From the first, we get
which when substituted into the second yields
.
The last expression is equivalent to
or
;
since
,
we need only consider the critical point
.
To check that the point found yields a (relative) minimum we make use of the
discriminant. From
,
and
,
we have that
.
Since
,
by the Second Partial Derivative Test, we conclude that we have a relative
minimum when
and
.
Therefore, the box that would be the cheapest to make is
.
The base of the solid
in the
-plane
is the triangle that is bounded by
,
and
;
it is bounded about by a wedge from the right circular cylinder. The density
at the point
is
.
This gives
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I will do this two different ways. We are after a power series expansion in powers of
.
-
From
,
we have that
,
,
. . . ,
.
The Taylor series expansion about
for
is given by
.
To see that this is valid for all real
,
let
.
Since, for all
,
by the (absolute) ratio test, the given series is convergent everywhere; i.e., the radius of convergence is infinite. -
For this way, we make use of the fact that
for all
.
In the given problem, we know that we want a power series expansion for
that is in powers of
.
To make use of the series that we know, note that
.
Substitution in the known series expansion yields that, for each
real,
which agrees with what we obtained in (i).
-
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I will do this one two ways also. This time, I will go with the series manipulations first followed by using the formula for the Taylor series about
.
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We know that we want a series expansion that is in powers of
.
It turns out that doing the needed adding and subtracting of
leads us to trying to manipulate what we are given into something on which we
can apply the Geometric Series Theorem. This
gives
whenever
;
i.e.,
.
Therefore, the radius of convergence is
.
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For
,
the requested Taylor series expansion is given by
.
From
,
we have that
,
,
and
.
In general,
,
from which we conclude that
.
Therefore,
which agrees with what we obtained in (i). To find the radius of convergence, we will use the ratio test. Let
.
Since
whenever
,
we conclude that the radius of convergence is
.
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