Transcript Calculus for the Natural Sciences

Math 104 - Calculus I
August 9
(but first, a quick review…)
Series of positive terms
Convergence questions for series of positive terms
are easiest to understand conceptually.
Since all the terms an are assumed to be positive, the
sequence of partial sums {S n} must be an
increasing sequence.
So the least upper bound property discussed earlier
comes into play -- either the sequence of partial
sums has an upper bound or it doesn't.
If the sequence of partial sums is bounded above,
then it must converge and so will the series. If not,
then the series diverges. That's it.
Tests for convergence of series
of positive terms:
The upper bound observations give rise to
several "tests" for convergence of series of
positive terms. They all are based pretty much
on common sense ways to show that the partial
sums of the series being tested is bounded are
all less than those of a series that is known to
converge (or greater than those of a series that
is known to diverge). The names of the tests we
will discuss are...
Tests...
TODAY
1. The integral test
TODAY
2. The comparison test
3. The ratio test
4. The limit comparison test (sometimes
called the ratio comparison test)
5. The root test
The integral test
Since improper integrals of the form

1
f ( x )dx
provide us with many examples of telescoping
series whose convergence is readily determined,
we can use integrals to determine convergence
of series:

For example, consider the series

n 1
1
n2
From the following picture, it is evident that
the nth partial sum of this series is less than
n
1  1
1
x2
dx
Integral test cont.
The sum of the terms is equal to the sum of the areas
What
is the
of the shaded
rectangles,
and sum?
if we start
integrating at 1 instead of 0, the
improper integral


1
1 x2
dx converges
(question: what is the integral? so what bound to
you conclude for the series?).
Since the value of the improper integral (plus 1)
provides us with an upper bound for all of the
partial sums, the series must converge.
It is an interesting question as to exactly what the
sum is. We will answer it next week.
The integral test...
says that if the function f(x) is positive and

decreasing , the series
 f ( x ) and the improper
x 1
integral

1f ( x )dx either both converge or both
diverge (only convergenc e at infinity needs to
be checked for the integral). This gives us a new,
easier proof of the divergence of the harmonic
series - -because we already knew the divergence of
the integral


1 dx .
1 x
Discussion and Connect
Question…
-- for which exponents p does the series


n 1
1
np
converge?
(These are sometimes called p-series, for obvious
reasons -- these together with the geometric series
give us lots of useful examples of series whose
convergence or divergence we know).
Error estimates:
Using the picture that proves the integral test for
convergent series, we can get an estimate on how far
off we are from the limit of the series if we stop
adding after N terms for any finite valueof N.
If we approximate the convergent series
by the partial sum s N 
N
 f ( n)
 f (n)
n 1
n 1
then the error we commit is less than the value of the

integral
f ( x)dx

N
Take a closer look...
For example,t hesum 1  14  19  161  251 
5269
3600
, which
is approximately 1.46. T hisdiffers by less t han


1
2
5 x

dx  15 , or 0.2,from t heinfit iesum  n12 . (As
n 1
we shall see laer, t hisest imat eisn' t far off - - t he
act ualsum is a lit t lebigger t han 1.6
Question

Does theseries  1nn2 convergeor diverge?
n 1
A) Converge
B) Diverge
Question

Does theseries 
n 1
A) Converge
B) Diverge
arctan(n )
1 n
2
convergeor diverge?
Exercise
Connect
For this latter series, find a bound on the
error if we use the sum of the first 100
terms to approximate the limit. (answer:
it is less than about .015657444)
The comparison test
This convergence test is even more commonsensical than the integral test. It says that if

all the terms of the series
a
n 1
are less than
n

the corresponding terms of the series

and if
b
n 1
n
n 1

converges, then
converges also.
a
n 1
b
n
n
Reverse
This test can also be used in reversed -- if
the b series diverges and the a’s are bigger
than the corresponding b’s, then  a n
n 1
diverges also.
Examples:


n 1
1
2n  n
converges.


n 1
n
n  sin( n )
diverges.
Question

Does theseries 
k 5
A) Converge
B) Diverge
1
k 2
convergeor diverge?
Question

Does theseries  2n 1 n 2 convergeor diverge?
n 1
A) Converge
B) Diverge
Convergence Tests...
1. The integral test
2. The comparison test
3. The ratio test
4. The limit comparison test (sometimes
called the ratio comparison test)
5. The root test
The ratio test
The ratio test is a specific form of the
comparison test, where the comparison series
is a geometric series. We begin with the
observation that for geometric series, the ratio
of consecutive terms
an 1
an
is a constant (we called it r earlier).
Ratio test (cont.)
• For other series, even if the ratio of
consecutive terms is not constant, it might
have a limit as n goes to infinity. If this is
the case, and the limit is not equal to 1, then
the series converges or diverges according
to whether the geometric series with the
same ratio does. In other words:
The ratio test:

an 1
If lim
 r , then theseries  an
n  a
n 1
n
- -convergesif r  1
- -divergesif r  1
If r  1 (or if thelimit does not exist),
then no conclusioncan be drawn.
Example:

1
For  , we have
n 1 n!
an 1
n!
1
lim
 lim
 lim
 0,
n  a
n  ( n  1)!
n  n  1
n
so theseries converges.
Another example:

1
For  p , the ratio is 1 and the ratio
n 1 n
test is inconclusive.
Of course, the integral test applies to these
p-series.
Question

n!
Does theseries  n convergeor diverge?
n 1 5
A) Converge
B) Diverge
Question

ln k
Does theseries  k convergeor diverge?
k 2 e
A) Converge
B) Diverge
Root test
• The last test for series with positive terms that
we have to worry about is the root test. This is
another comparison with the geometric series.
It's like the ratio test, except that it begins with
the observation that for geometric series, the nth
root of the nth term approaches the ratio r as n
goes to infinity (because the nth term is arn and
so the nth root of the nth term is a1/nr-- which
approaches r since the nth root of any positive
number approaches 1 as n goes to infinity.
The root test says...
thatif lim an
 1n 
n 
 r , then

- - theseries  an convergesif r  1
n 1

- - theseries  an divergesif r  1
n 1
and no conclusioncan be drawn if r  1.
Example

n
 n 



n 1  2n  5 
convergesby theroot test.
Question


Does theseries  1  e
n 1
A) Converge
B) Diverge

n n
convergeor diverge?
Series whose terms are not all
positive
• Now that we have series of positive terms under
control, we turn to series whose terms can change
sign.
• Since subtraction tends to provide cancellation
which should "help" the series converge, we begin
with the following observation:
• A series with + and - signs will definitely converge
if the corresponding series obtained by replacing
all the - signs by + signs converges.
Absolutely convergent series
Another way tosay thisis to thinkof the termsan

of theseries  an as being positiveand negative,
n 1
and then theseries obtainedby changingall minus

signs to plus would be  an - the" series of absolute
n 1
values."
A series whose series of absolute values converges,
which is itself then convergent, is called an
absolutely convergent series.
Examples...


n 1
( 1) n
n
2
 1  14  19  161  251  ... is absolutleyconvergent.

n
(

1
)
 1  1  1  1  1  1  ... is divergent(since an doesn't

n 0
tend to zero)and (or course)its series of absolute valuesis divergent .


n 1
( 1) ( n1)
n
 1  12  13  14  15  16  ... is convergent(as we will see)
even though its series of absolute valuesis divergent .
Series that are convergent although their series of absolute
values diverge (convergent but not absolutely convergent) are
called conditionally convergent.
Alternating series
A special case of series whose terms are of
both signs that arises surprisingly often is
that of alternating series . These are series
whose terms alternate in sign. There is a
surprisingly simple convergence test that
works for many of these:
Alternating series test:

Let  (1) n an  a0  a1  a2  a3  ... be an alternating series.
n 0
If (theabsolute valuesof) the termsare decreasing, thatis
a0  a1  a2  ...
and if lim an  0, then theseries converges. Moreover,thedifference
n 
between the limit of theseries and thepartialsum sn has thesame sign
as and is less in absolute value that thefirst omittedterm(an 1 ).
Example:
1
1
1
1



The alternating harmonic series
2
3
4  ...
clearly satisfies the conditions of the test
and is therefore convergent. The error
estimate tells us that the sum 1  12  13  14  127
is less than the limit, and within 1/5. Just to
practice using the jargon, the alternating
harmonic series, being convergent but not
absolutely convergent, is an example of a
conditionally convergent series.
Classify each of the following...
(1)

n  2 n ln n

A) Absolutely convergent
B) Conditionally divergent
C) Divergent
n
Classify each of the following...

sin k

3
k 1 k
A) Absolutely convergent
B) Conditionally divergent
C) Divergent
Classify each of the following...
n cos n

2
n 1 n  5

A) Absolutely convergent
B) Conditionally divergent
C) Divergent
Power series
Last week's project was to try and sum series
using your calculator or computer. The
answers correct to ten decimal places are:
Sum((-1)^n/(2*n+1),n=0..infinity) = evalf(sum((-1)^n/(2*n+1),n=0..
infinity));


n 0
( 1) n
2 n 1
 .7853981635
Sum(1/factorial(n),n=0..infinity)=evalf(sum(1/factorial(n),n=0..infinity)
);


n 0
1
n!
 2.718281828
Power series (cont.)
Sum(1/n^2,n=1..infinity)=evalf(sum(1/n^2,n=1..infinity));


n 1
1
n2
 1.6644934068
Sum((-1)^(n+1)/n,n=1..infinity)=evalf(sum((-1)^
(n+1)/n,n=1..infinity));


n 1
( 1) ( n1)
n
 .6931471806

We can recognize these numbers as 4 , e,
2
6
and ln(2).
Two directions:
1. Given a number, come up with a series that
has the number as its sum, so we can use it
to get approximations.
2. Develop an extensive vocabulary of
"known" series, so we can recognize
"familiar" series more often.
Geometric series revisited
We begin with our old friend,
thegeometricseries :

 ar
n
 a  ar  ar  ar  ar  ... 
2
n 0
a
(providedr  1).
1- r
3
4
r as a variable
Changing our point of view for a minute (or a
week, or a lifetime), let's think of r as a
variable. We change its name to x to
emphasize the point:

a
f ( x)   ax  a  ax  ax  ax  ... 
(for x  1)
1 x
n 0
n
2
3
So the series defines a function (at least for
certain values of x).
Watch out...
We can identify the geometric series when we see it,
we can calculate the function it represents and go
back and forth between function values and
specific series.
We must be careful, though, to avoid substituting
values of x that are not allowed, lest we get
nonsensical statements like
1  2  2  2  2  ...  1  2  4  8  16  ...  1!!
2
3
4
Power series
If you look at the geometric series as a function, it
looks rather like a polynomial, but of infinite degree.
Polynomials are important in mathematics for many
reasons among which are:
1. Simplicity -- they are easy to express, to add,
subtract, multiply, and occasionally divide
2. Closure -- they stay polynomials when they are
added, subtracted and multiplied.
3. Calculus -- they stay polynomials when they are
differentiated or integrated
Infinite polynomials
So, we'll think of power series as "infinite
polynomials", and write

a x
n 0
n
n
 a0  a1 x  a2 x  a3 x  ...
2
3
Three (or 4) questions arise...
a
1. Given a function (other than f ( x)  1  x ), can it be
expressed as a power series? If so, how?
2. For what values of x is a power series representation valid?
(This is a two part question -- if we start with a function f(x)
and form "its" power series, then
(a) For which values of x does the series converge?
(b) For which values of x does the series converge to f(x) ?
[There's also the question of "how fast".]
continued
3. Given a series, can we tell what function it came from?
4. What is all this good for?
As it turns out, the questions in order of difficulty, are 1, 2(a),
2(b) and 3. So we start with question 1:
The power series of a
function of f(x)
Suppose the function f(x) has the power series
f ( x)  a0  a1x  a2 x  a3 x  a4 x  ...
2
3
4
Q. How can we calculate the coefficients ai from a
knowledge of f(x)?
A. One at a time -- differentiate and plug in x=0!
Take note...
First (or zerot h): f (0)  a0  a1 0  a2 0 2  a3 03  ...  a0
so we have t hata0  f(0)
Second (first?): It seems reasonablet o writ e
f ( x)  a1  2a2 x  3a3 x 2  4a4 x 3  ..., so we should have
a1  f (0)
Anot herderivative:
f ( x)  2a2  6a3 x  12a4 x 2  20a5 x 3  ...,
so we should havea2  f (0)/2
Continuing in this way...
a3  f (0)/6,a4  f (0)/24,etc....
In general:
an  nth derivativeof f evaluatedat zero,
divided by n!
Example
Suppose we know, for the function f, that f(0)=1 and
f ' = f.
Then f '' = f ', f ''' = f '' etc... So f '(0) = f ''(0) = f '''(0)
= ... = 1.
From the properties of f we know on the one hand
x
f
(
x
)

e
that
So we get that...
2
3
4
n

x
x
x
x
e x  1  x    ...  
2! 3! 4!
n  0 n!
(Set x  1 to get one of thesums we did previously!)
Good night…
See you Wednesday!