Calc Notes

7/29/2019 Calc Notes

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Supplemental Notes for Calculus II

Ellen Ziliak and Alexander Hulpke

Fall


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Alexander HulpkeDepartment of MathematicsColorado State University Campus Delivery

Fort Collins, CO,

by the authors. is work is licensed under the Creative Commons Attribution-Noncommercial-

Share Alike . United States License. To view a copy of this license, visit http://creativecommons.org/

licenses/by-nc-sa/3.0/us/ or send a letter to Creative Commons, Second Street, Suite , San

Francisco, California, , USA.


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-

Convergence of Improper Integrals

In this chapter we will look at whether or not improper integrals converge or diverge.

DEFINITION 1

An improper integral converges if when it is evaluated the result is a nite value. An improper integral di-verges if when it is evaluated the result is an innite or undened value.

We begin with an example. Suppose a scientist is observing a particle that is moving in a force eld. He has

been able to determine from several measurements he made that the particles acceleration is proportional to

its velocity, resulting in a dierential equation

dv

dt= v

where v is the velocity of the particle. He also measured the initial velocity to be ms. Solving this initialvalue problem we get v(t) = et.e scientists now asks whether the particle will move arbitrarily far. To answer this question we need to

determine whether or not

etdt converges. In this case the value of the integral is not dicult to

calculate using the methods previously described in this section:

limb

b

etdt = lim

betb

= limb

eb +

=

So the total distance traveled by the particle at the end of time is m, which is nite.

is is the equation for exponential growth:

dv

dt= v

dv

v=dt

Aer evaluating this integral we get ln v = t+ Cwhich we can then solve for v(t) to get v(t) = Aet.e initial condition forv() then gives v(t) = et.


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Now suppose instead of such a simple function for the velocity we got instead that the velocity of the

particle is given byv(t) = sin(t)t

. If we want to know, whether such a particle travels a nite distance, we

would have to determine whether

sin(t)t

dtconverges or diverges. We cannot just evaluate the integral,

so we must discover some other techniques to answer this question.

Direct Comparison

Our goal is to determine if

af(x)dxconverges without having to compute the value of the integral. One

way to do this is to compare with an integral we know something about. Let us rst consider an example

where the limits of integration are nite to develop intuition about how we compare two functions.

EXAMPLE 1

If f(x) = x + and g(x) = x, what can we say about

f(x)dxcompared to

g(x)dx?Solution First we look at the graphs of f(x) and g(x) below: We know geometrically

f(x)dx can

0. 5

1

1. 5

2

2. 5

3

-1 -0.8 -0.6 -0.4 -0. 2 0 .2 0. 4 0 .6 0. 8 1x

f(x)=x+2

f(x)=x

0

1

2

3

4y

-1 -0.8 -0.6 -0.4 -0. 2 0 .2 0. 4 0 .6 0. 8 1x

be interpreted as the area under the curve of f(x) on [, ] and

g(x)dx can be interpreted as the areaunder the curve ofg(x) on [, ]. And we notice on the interval from [, ] that f(x) > g(x) (in fact this isalways true but we only care about this interval) the area under gclearly must be smaller than the area under

f.

We have seen that if f(x) > g(x) then

f(x)dx>

g(x)dx.is result holds of course for any other interval:

THEOREM 1

If on the entire interval [a, b] we have f(x) > g(x) thenba f(x)dx>b

ag(x)dx.

Proof is given by picture on p

We now want to consider what happens in the limit case:


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EXAMPLE 2

Does

exdxconverge?

Solution Consider f(x) = ex and g(x) = x on the interval [, ) Let us rst notice that ex is smaller

g(x)=1 x

0

0. 2

0. 4

0. 6

0. 8

1

2 4 6 8 1 0 1 2 1 4 1 6 1 8 2 0x

f(x)=1 ex

than x on the entire interval from [, ). We also know that

xp dx converged (to

p ) when p > ,

so the area under x

from [, ) is equal to = .erefore, extrapolating from the nite case, we wouldassume that

exdx< which would mean that this improper integral converges.

To verify this, let us calculate this integral directly, so we can verify that this guess was correct.

exdx = lim

b

b

exdx= lim

bexb

= limb

eb + e =

e= . <

is comparison property holds for any other pair of functions. We get the following generalization of

eorem :

THEOREM 2Let f(x) and g(x) be continuous on [a,) with f(x) > g(x) on the whole interval.en ifa f(x)dxconverges, then

ag(x)dxconverges.

With this theorem we can answer the question from the start of the chapter:

EXAMPLE 3 The Particle Revisited

In the example we wanted to know whether or not

sin(t)t

dtconverges.

Solution Since sin(t) < always, if we compare sin(t)t < t . Since t dtconverges we can con-clude that

sin(t)t

dt converges. We therefore know that the distance traveled by the particle is bounded

by

tdt= m, however we cannot give a concrete value for this distance.

A similar argument can be used to show that an improper integral diverges.


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EXAMPLE 4

Does

x

dxconverge?

Solution Consider f(x) = x

and g(x) = x

on the interval [, ). We see in the graph that

2

4

6

8

10

12

2 4 6 8 1 0x

10

f(x)=1 xg(x)=1 x-1

x

>x

for x > . We also know from previous work that

x

dx diverges. since

x

dx is a

nite value (namely .) throwing this part out wont change the conclusion that the value of the

integral is innite, so

x

dxdiverges.

Since x

>x

on the entire interval [,), the area under x

on [, ) must be larger than thearea under

xon [, ), so x dxshould also diverge. Again we check the conclusion by actually

calculating the value of this integral:

x dx= lim

b

b

x dx

let u = x then we have dudx

= and substituting the limits of integration in we have u() = and u(b) = b,yielding:

limb

b

u du = lim

bu

b

= lim

b(b ) =

erefore

x

dxdiverges.

Again we state this observation as a general result:

THEOREM 3Let f(x) and g(x) be continuous on [a,) with f(x) < g(x) on the whole interval.en if

af(x)dx

diverges, then

ag(x)dxdiverges.

In all of the examples we have seen so far we have had f(x) g(x) on the entire interval we have con-sidered. In some situations this makes it unnecessaily hard to nd a function to compare with. We will see in

the next example that one can in fact weaken this condition:


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EXAMPLE 5

Does

si n(x)x

+

dxconverge?

Solution Let f(x) = sin(x)x + and g(x) = x on the interval [, ).en initially we have f(x) oscil-

f(x)=sin(x)/x+

g(x)=1 x

0. 2

0. 4

0. 6

0. 8

1

5 10 15 20 25x

lating over g(x) from [, .], then aer that point we have x < sin(x)x + from [., ). Since

.

sin(x)x

+

dx = . is nite and

.

xdx = . is also nite, we know that

the behavior on the interval from to . will not aect the convergence or divergence of the improper

integral.

We also know that

xdxdiverges; similarly by splitting up the interval we have

xdx

.

xdx=

.

xdx

letting us conclude that

.

xdx diverges as well.en since x then

af

(x

)dxand

ag

(x

)dxboth converge or both diverge.

b) If limx

f(x)g(x) = and

ag(x)dxconverges thena f(x)dxconverges.

c) If limx

f(x)g(x) = and

ag(x)dxdiverges thena f(x)dxdiverges.

One can show that all previous theorems in this chapter, including the direct comparison test, are special

cases of this theorem.is means that any problem which can be done by direct comparison can also be done

by limit comparison. (Still, oen direct comparison is easier, as no limit needs to be computed.)

Also noticeagain that if limx

f

(x

)g(x) = and

a f(x)dxconverges, we get no conclusion about

a g(x)dxand similarly if lim

x

f(x)g(x) = and

af(x)dxdiverges, we get no conclusion abouta g(x)dx.

Let us look at an example for comparison of same decay order:

EXAMPLE 7

Does

x + x +

x xdxconverge?

Solution Let f(x) = x + x + x x

and g(x) = x

on the interval [, ).Direct comparison is dicult, so we look at

limx

f(x)g(x) = limx

x+x+xx

x

= limx

x(x + x + )x x

= limx

x + x + x

x x=


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erefore we can conclude that these two functions decay at the same rate. Since

xdx diverges we

thus know that

x + x +

x xdxalso diverges.

In this example we could calculate the integral directly, using partial fractions, giving (of course) the

same result, albeit with far more eort.

Finding Comparison Candidates

e nal question we want to study in this chapter is how to choose good comparison functions. For this we

can consider growth order of the reciprocal functions. In the previous example, the reciprocal function would

be

f(x) = x x

x + x + . From chapter ., we know that this is of the same growth order as x = x, which

leads us to a comparison with g(x) = x

.

Consideration of the growth order will oen lead to integrals of the form

a

xpdx as comparison

candidates. For such integrals we have explicitly determined the criterion that they converge if and only if

We have

x + x +

x xdx = lim

b

b

x + x +

x xdx = lim

b

b

x + x +

x(x ) dx

= limb

b

x + x +

x(x )(x+ )dxe method for solving the integral is partial fractions, we get

x + x +

x x=

A

x+

B

x+

C

x +

D

x+

=

A

(x

)+ B

(x x

)+ C

(x + x

)+ D

(x x

)x(x )(x+ )=

Ax A+ Bx Bx+ Cx + Cx + Dx Dx

x(x )(x+ )Comparing coecients for powers ofxin the numerator, we get the following equations

B + C+ D =

A+ C D =

B =

A =

We can then solve for the variables and get A = , B = , C=

, D =

. Substituting in the original integral, we get

limbb

x +

(x ) +

(x+ )dx = limb x+

lnx +

lnx+ b

= limb

b+

ln b +

ln b +

ln

ln

= ++

ln() =

us the integral diverges, as we found already by the limit comparison test with far less eort.which is the main reason we bothered with stating a theorem for such integrals


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p > .is not only indicates comparison candidates, but also whether the integral will converge or diverge.

If we cannotnd a function of the same growth order asf(x), parts b)and c)ofeorem oen can beused to nd functions of strictly faster decay for which the improper integral diverges, respectively functions

of slower decay for which the integral converges.

Going back to the example starting this chapter, we nd for example that for f

(t

)= et the reciprocal

f(t) = et

is of larger growth than any power of t, thus we could compare to g(t) = t

to showconvergence of

f(t)dtwithout explicitly calculating the antiderivative.

EXAMPLE 8Determine, by nding suitable comparisons, whether the following improper integrals converge:

.

x+

xdx

.

ln x

x dx

.

x

xdx

.

(ln x) dx.

x +

x

xdx

.

sin

x dx

Solution

. If f(x) =

x+x, then f(x) = (x+x).is is of the same growth order as x, thus the integraldiverges by limit comparison with

xdx.

. If f(x) = ln xx , then f(x) = x

ln x

. We know that ln xgrows slower than any power of x, thus we see

that x for arbitrary small , for example x, is a function of smaller growth. We get that

limx

ln xx

x

= limx

ln x xx = limx

ln x

x =

As

x dxconverges, we conclude by limit comparison that

ln xx dxconverges.

. Iff(x) = xx

,thenf(x) = xx

.is is of growth larger than any power ofx, for example x.erefore

limx

x

x

x

= limx

x

x= , and the integral converges by limit comparison with

xdx.


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. If f(x) = (ln x) , then f(x) = (ln x). is grows slower than x, thus the integral diverges bylimit comparison with

xdx. (Of course we also have that (ln x) grows less than x, but limit

comparison with

xis a case in which we cannot draw any conclusion.)

. If f(x) = x + x

x,thenf(x) = xx

x + , which is of the same growth order as

xxx

= x.e integral

converges by limit comparison with

x dx.

. is one is tricky, as we dont know the growth order for sin(x).e xsuggests limit comparisonwith x. We get

limx

x

sin xLHopital

= limx

x

x cos x = limx

cos x =

As

xdx, diverges, so does

sin x dx. (Once we know power series well see another expla-

nation for this behavior: For small values ofz, sin(z) behaves like z.us sin xwill behave like x forlarge x.)

We close this chapter with the remark that we considered only criteria for convergence tests for the case

of

af(x)dx. Obviously similar tests are possible for the other types of improper integrals.

EXERCISES

Determine whether the following improper integralsconverge or diverge. You may use direct comparison

or limit comparison (or explicit computation of an

antiderivative)

.

+ cos(x)x

dx

.

ln(ln(x))dx.

x

x dx

.

x+ x

xxdx

.

(ln(x)) dx

.

x

xdx

.

x

xdx

.

x

xdx

.

ln(x)x.

dx

.

x+

xdx

.

x +

dx


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-

Geometric Series

In this chapter we will discuss one of the simplest and most useful types of innite series, the geometric series.

ey are derived from instances where geometric growth occurs, this is growth that is proportional to the

current value, i.e the growth rate is constant. (Here we are looking at the discrete situation, in a continuous

situation a dierential equation for constant growth rate produces exponential functions as solutions.)A simple example of geometric growth is interest accrued annually on a bank account. Each year we take

the amount of money in the account the value of the account and multiply it by the interest rate and add it to

the value to get a new value.erefore the constant of proportionality is one plus the interest rate.

EXAMPLE 1If you put into a savings account that earned % interest accrued annually, how much would you have

aer one year? two years? n years?

Solution Let pn be the value of our account at time n, so p = since we begin initially with .Next at the end of one year we earn % interest on this account so p = (.) p = (.) , hencep is proportional to p with the constant of proportionality being ..is process repeats so that p =

(.

) p =

(.

) . At this point we notice a pattern pn =

(.

)n .

As a second example of geometric growth we consider university scholarships which come from endow-ments. In this case, an endowment (a xed sum) is given to the university. A scholarship then is funded from

the interest of this money (without using up the money, the scholarship therefore exists forever).

EXAMPLE 2 EndowmentsDelighted by the calculus course he took at CSU, an alumnus wants to endow a scholarship for mathematics

students that will pay each year.e university guarantees an interest rate of % per year. How much

must the alumnus donate to guarantee this scholarship will be available forever?

Solution If we let x=the value of the endowment, then aer one year the endowment is worth . xand we want to give a scholarship, so we want .x= x+ .erefore the endowment will never

be touched and the scholarship is funded by the interest earned. Solving this equation we get . x =

which gives us that x= ,.erefore the alumni must give an endowment of, to guarantee the

scholarship is funded forever.

A geometric series now is a series whose terms show geometric growth, i.e. sn = r sn for a constant r

independent ofn.is type of series has applications in several elds including physics, biology, economics

and nance. We will look at a few examples of where geometric series occur in our everyday lives including:

repeating decimals and calculating dosage of medicine.


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DEFINITION 1Let rbe a ratio and a any nonzero constant. A geometric series is a series of the form

a + ar+ ar + . . . + arn + . . . =

n=

arn = a

n=

rn = a

n=

rn

As an example of a geometric series we look at the Race course paradox of Zeno . Suppose a runner

wanted to travel a given distance, say one kilometer.en he must rst travel the rst half kilometer, and the

next half kilometer remains. Next the runner must travel half of the next half kilometer a quarter kilometer

and the next quarter kilometer remains. It might seem (so claimed Zeno) as if the runner never reaches the

goal.

However if we look at the distance traveled aer k iterations, we get a geometric sum with ratio :

+

+

+

+ +

k

If the process continues innitely many times, we thus get get the following formula for the distance traveled:

+

+

+ . . . +

n+ . . . =

n=

n

We will see that this sums up to (i.e. the full distance is traveled, resolving the paradox).

Notice that there is a slight shi in this formula from the one given in the denition of a geometric seriesin that we start not with r = but with r = while in the denition this power is zero.is can be xed by

letting the constant a be ,is sort of index change arises oen enough that we will rst look at how we can

manipulate such innite sums.

Reindexing a Series

We begin this discussion with a brief reminder about the notation for sums. Givenk

n=

an we call n the

index of summation, an the nth term of the sum, and the upper and lower bounds of summation are k and

respectively.

Now we consider reindexing our innite series, notice in the following examples we are working with

geometric series, however this algorithm will be useful with other types of series that will be discussed later.

Zeno of Elea, Greek philosopher, ca. BC - ca. BC


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Most of the innite series we have seen so far have a lower bound of summation of one, however their are

times when we are given an innite series where the lower bound of summation will be a value other than

one. We consider rst a geometric series which does not have a lower bound of one:

+

+ . . . +

(

)n

+ . . . =

n=(

)n

is is still a geometric series, however our denition requires the series to have the form

n=

arn, so rein-

dexing is required. To reindex a series we follow the following basic algorithm:

Algorithm 1 Reindexing a series

Choose a new letter to be the index of summation (m)

Relate m to the original index of summation (n) such that when n= we have m=. So we have the

equation m=n-.

Solve for original index of summation (n) and substitute into the nth term of the sum:

n = m +

m=

()m+ =

m=

()m+

Notice if the upper bound of summation were nite we would have to decrease its value by as well,

however since we are summing to innity our upper bound of summation will remain innite.

Solve for an a such that the power on r is m-.

m=(

)m+ =

m=(

) (

)msince m-+=m+.

erefore a = ( ) = and r= so we have a geometric series of the form m=

arm. Notice that

in this example the ratio was negative, therefore a geometric series can have negative or positive r values. We

now look at another example of this reindexing algorithm before we develop an equation for the value of a

geometric series.

EXAMPLE 3

Consider the series

n=

n + , rewrite this series so that the lower bound of summation is n=.

Solution Let m = n , then n = m + so we have

m=

(m + ) + =

n=

m +

Using this algorithm we can always manipulate a given geometric series so that it can be written in the

form

n=

arn or equivalently

n=

arn next we consider how to calculate the value of this series.


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The Formula for the Geometric Series

In this section we will derive a general formula for the value ofk

n=

arn. We can apply this in the limit case

k = to get the value of an innite series, for example to show that the series

n=

n

from Zenos paradox

has value .

DEFINITION 2

e kth partial sum sk of a Geometric series

n=

arn is the (nite) sum of the rst k terms:

sk = a + ar+ ar+ . . . + ark =

k

n=

arn

Using this denition we can consider the sequence s, s, s, . . . , sk , . . . of partial sums, which converges

to

n=

arn, since limk

sk = limk

k

n=

arn =

n=

arn.erefore, once we know a formula for the value ofsk we

can compute the value of

n=

arn as the limit.

We begin by considering

sk = a + ar+ ar+ . . . + ark

and

rsk = ar+ ar+ ar + . . . + ark .

Notice that sk and rsk share several terms in common, in fact

sk rsk =

(a + ar+ ar + . . . + ark

)

(ar+ ar + ar + . . . + ark

)= a ark

Now if we factor both sides of this equation, we get

sk( r) = a( rk).As long as r we can divide both sides by( r) to get

sk =k

n=

arn = a rk

r.

(What happens in the case where r=? We get sk = a + a

(

)+ a

(

)+ . . . + a

(

)k

= ka .)

Before we go on to get a formula for the innite sum, we see how the formula for sk might be used on its

own:

EXAMPLE 4 The magnitude of repeated growthSuppose we are given a chessboard and place one grain of rice on the rst square, two grains of rice on the

second square, four grains of rice on the third square continuing on so that there are n grains of rice on the

nth square. How many grains of rice are on the chess board? (ere are = squares on a chess board.)


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Solution Since there are squares on the chess board, we want to compute s. Our ratio is and a isone, hence we are asked to nd

s = + + + . . . + =

( )

= .x

A similar calculation underlies the repayment of loans or mortgages:EXAMPLE 5 Paying back a loanSuppose we take out a loan of L dollars which is paid back periodically (typically monthly). e periodic

payment is a dollars, the xed interest rate per period is i. If bk is the loan sum outstanding aer k time

periods, we have that

bk+ = bk ( + i) a.Using b = L and setting r= + i, in the rst step we have

b = Lr a

we then continue on to the next step where we get

b = b(r) a = (Lr a)(r) a= Lr (a + ar)

We might be starting to notice a pattern, however it becomes obvious in the next step

b = b(r) a = (Lr (a + ar))(r) a= Lr (a + ar+ ar)

At this point we can solve this recursion to

bk= L rk

a + ar+ ar + + ark

= L rk

k

n=

arnSum formula

= L rk a rk

r.

Abanknowwouldset bm = (where m isthetimeaerwhichtheloanshouldbepaido,e.g. m = =

for a year mortgage) and solve for a to determine the necessary monthly repayment, given the loan sum

and interest rate.

For example, if we have a mortgage ofL = , , an annual interest rate of % (leading to a monthly

rate of i = . = ., i.e. r = .) and a monthly repayment sum of a = , , we nd that theoutstanding sum aer k months is

bk = , .k ,

.k

.= , .k , .k .

Aer years ( months) this leaves an outstanding amount of, ., aer years , .,roughly half, aer years . (i.e. the loan is paid o aer years less one month ).

Moralistic remark: Incidentally, initial interest amounts to per month at the start of the loan. An interest onlyloan thusdoes not save much and is a very bad deal!

In general, this means that a loan is paid o half aer roughly

of its planned life timee total cost of the loan then will have been , , more than double the loan amount. (ough ination means that the

actual value will be less.)


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If the interest rate instead was % annually (i = . = . monthly), we get with same repaymentsum a remaining loan amount of, . aer years, which is not even halfway paid o. A monthly

repayment sum of would be needed to have the loan paid o aer years.

Now since we saw earlier that the sequence s, s, s, . . . , sk , . . . of partial sums converges to

n=

arn, we

can use the formula just derived to compute a value for the limit of a geometric series.

THEOREM 1

e geometric series a + ar+ ar + . . . + arn + . . . =

n=

arn converges toa

rwhen r < , and diverges

otherwise.

Proof Case : If r < then rk < , and as k gets larger this will cause rk to get smaller and yieldlim

krk = .us, by the rules for limits,

limk

sk = limk

a( rk

) r = a( ) r = a r.Case : When r then rk .erefore when k tends to innity the numerator a( rk) tends to plusor minus innity, and the denominator ris a xed value. So lim

k

a( rk) r

diverges.

Now that we have a formula for the value of a geometric series let us verify the solution to the Zenos

paradox example. We are looking at

n=

n. First notice that this sum is not in the correct form, we must

factor (

) out of the nth

term. We get

n=(

)n=

n=(

)

(

)n

=

n=(

)

(

)n

, hence a=

and r=

), thus

the sequence of partial sums is bounded from above. On the other hand (we are summing up positive terms)

the sequence of partial sums is increasing, thus this sequence (which is the series) must converge.

Let us consider the same example once more, but as Riemann sum for le endpoints (green boxes), ap-

proximating k+

xdx:

0 2 4 6 8 100

0.5

1.0

x

y

e Riemann sum now is an upper approximation, thus we get that

k+

xdx

k

n=

n +

k

xdx

As

x diverges, we need to consider the lemost blue box separately


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We therefore see that we can also consider the series as upper approximation of the integral, by a similar

argument as in the previous example we get that convergence of the series n=

nimplies convergence of

the improper integral.

xdx.

is observation is not specic to x

, it holds for any decreasing function. We thus get the following

theorem:

THEOREM 1 Integral TestLet an be a sequence of positive terms. Suppose that there is a function f dened for all positive real numbers

such that

. an = f(n) for all n,. f is piecewise-continuousa,

. f is decreasing for x> N (where N is some integer, e.g. N= if f is decreasing over the positive real

axis)

en

f(x)dxconverges if and only if

n=

an converges.

aotherwise we cannot guarantee that integrals exist even in the case of nite limits

Proof Convergence of an innite sum or an improper integral is not aected by the behavior over a niteinterval.us we can consider without loss of generality the case N= .e argument then is essentially the

same as in the discussion for x

.

EXAMPLE 2

Does

n=

enconverge?

Solution e function f(x) = ex is continuous and decreasing for x> . We have seen in the chapter onimproper integrals that

exdx converges.us by theorem also

n=

enconverges.

EXAMPLE 3 p-series

For which values of p does

n=

npconverge?

Solution e function f

(x

)=

xp

is continuous and decreasing for x . We know that

xp con-

verges, if and only if p > .us, by theorem ,

n=

npconverges if and only if p > .

us (we have seen this previously)

n=

nconverges and

n=

ndiverges.

In the book this is theorem on p.


26/49

Direct Comparison Test for Series

In the chapter on improper integrals we learned about powerful comparison tests that can be used to deter-

mine convergence. If we wanted to test convergence for a series, we therefore could consider the correspond-

ing improper integral, and resolve its convergence via a comparison test. Doing so however it turns out to be

unnecessarily formal; because theorem has an if and only ifcondition, we can (by dening for a series ana function f(x) with f(x) = an ifn x < n + ) translate the comparison test for improper integrals to acomparison tests for series. (Compare the statement to theorem .)

THEOREM 2 Direct Comparison Test for SeriesLet an and bn be non-increasing sequences of non-negative numbers.

a) If there exists an integer value N such that an bn for all n N and

n=

an converges then

n=

bn

converges.

b) If there exists an integer value N such that an bn for all n N and

n=

an diverges then

n=

bn

diverges.

As with improper integrals, this theorem does not reach any conclusion if an cn and cn diverges or ifan dn anddn diverges.

EXAMPLE 4

Does

n=

sin(n)n

converge?

Solution We know that sin

(n)n n for all values ofn. We also know that

n=

n

dxconverges. By theo-

rem , part a), we can conclude that

n=

sin(n)n

dxconverges.

EXAMPLE 5

Determine if

n=

n + converges.

Solution Let us consider

n=

n,whichisa p-series for p = > and therefore converges. Since

n+

n

for all n the direct comparison test shows

n=

n + converges.

Limit Comparison Test for Series

With the same argument as used for the ordinary comparison test, the limit comparison test can be adapted

to series. (Again, compare to theorem .)


27/49

THEOREM 3 Limit Comparison Test for SeriesSuppose that an > and bn > for all n N where N is an integer

a) If limn

an

bn= L > , then an and bn both converge or both diverge.

b) If limn

an

bn= and bn converges then an converges.

c) If limn

an

bn= and bn diverges then an diverges.

Again notice that if limn

an

bn= and bn diverges or lim

n

an

bn= and bn converges the theorem

gives us no conclusion.

EXAMPLE 6

Consider

n=

n +

, does this series converge or diverge?

Solution We know that

n=

n

diverges (it is a p-series for p = < ).

We calculate the limit of the quotient:

limn

n+n

= limn

n

n + = lim

n

n(

n)

n + ( n)

= limn

+ n

=

So by the limit comparison test we conclude that

n=

n + diverges. (Note that direct comparison will notwork in this example, as

n +

.


28/49

We now use limit comparison:

limn

(ln(n))n

n

= limn

n (ln(n))

n= lim

n

(ln(n))n

To apply LHopitals rule, requiring derivatives, we rewrite this expression in the variable x instead ofn:

= limx

(ln(x))x

LHopital= lim

x

(ln(x))( x) x

= limx

(ln(x))x

LHopital= lim

x

( x) x

= limx

x

= .

Since

n=

n

converges, by the limit comparison test we can conclude that

n=

(ln(n))n

converges.

As in this example (and with improper integrals), comparison to a p-series is oen useful.

Another fruitful comparison candidate is given by geometric series, the method for nding a suitablegeometric factor is given by the ratio test, which we shall study next:

The Ratio Test

In this section we want to see how the geometric series can be used along with the comparison test to reach

conclusions about the convergence of series. Recall that a geometric series is a series of the form

n=

arn,

which we have shown will converge ifr < .Suppose we want to test a given series

nan (with nonnegative values an) for convergence by comparing

with a geometric seriesn arn

. For comparison to work, we need that the terms an of our series fall as quickly

as the terms arn of the geometric series.

We can ensure this by looking at the ratio of subsequent terms: Assume for the moment that

an+an

arn+

arn= r

and that r < , so we know thatn

arn converges.

Because of the inequality just assumed, we get setting b =aa

that

a

ar=

a

a

a

a

r r b

r= b

aar

=aa

a

ar

r r b

r= b

an+arn+

=an+

an

an

arn

r r b

r= b


29/49

and therefore

limn

an

arn b .

THEOREM 4 Ratio Test for Series with positive termsLet an be a series with positive terms and suppose that

limn

an+an

= r

then

(a) the series converges ifr<

(b) the series diverges ifr> or if r is innite

(c) the test is inconclusive ifr= .

To see why we really cannot deduce anything in the case of r = , consider two examples:

n=

nwhich

diverges by the p-series test and

n=

nwhich converges by the p-series test. Let us look at the limits of the

ratios in each of these cases:

limn

n+

n

= limn

n

n + = limn

n

(n

)(n + )( n) = limn

+ n=

Similarly

limn

n+

n

= limn

n

n + = lim

n

n( n)

(n + )( n) = limn +

n

=

erefore we see that in the case where r = the test is inconclusive.


30/49

EXAMPLE 8

Consider the series

n=

n

n. We want to determine if this series converges or diverges.

Solution Consider the ratio

an+an

=

(n+)n+

nn

= (n+

)

n

n+n

=(n + )n

nn=(n + )

n

Now consider the behavior as n gets large (in other words take the limit as n ), we get

limn

(n + )n

= limn

(n + )( n)

n( n) = limn

( + n)

=

< .

By the ratio test we therefore conclude that the series

n=

n

nconverges.

EXAMPLE 9

Determine whether

n=

n!

enconverges or diverges?

Solution We begin by considering the ratio

an+an

=

(n+)!en+

n!en

=(n + )n!en

enen!=(n + )

e

Now we look at the behavior as n tends toward innity limn

n +

e

= therefore we can conclude the series

n=

n!

endiverges.

Sometimes the limit in the ratio test requires LHopitals rule for evaluation:

EXAMPLE 10

Determine whether

n=

ln(n)nn

converges or diverges?


an+

an

=

ln(n+)n+(n+)

ln

(n

)n n=

ln

(n +

)nn

n

(n + ) ln(n)=

ln

(n +

)n

(n + ) ln(n)Next we take the limit of this function lim

n

ln(n + )n(n + ) ln(n) . It has the indeterminant form , therefore we

have to use LHopitals rule.

Writing the function in x (to indicate that we now consider it as a continuous, not a discrete function),

we get


31/49

limx

ln(x+ )x(x+ ) ln(x) = limx

xx+ + ln(x+ )

(() ln(x) + x+x )Again we have the indeterminant form

, so we apply LHopitals rule once more, and get.

= limx

(x+

)(

)x

(

)(x+) +

x+

( x + (x)()(x+)()x ) = limx

(x+) + (x+

)(x+)( x

x

x) = limx

x+

(x+)x

x

= limx

(x+ )(x)(x+ )(x ) = limx x+ x

x + x x =

lim

x

x + x

x + x x

At this point we have only polynomials le. We know that the behaviour of such a quotient is determined

by the leading terms (formally, one applies LHopitals rule multiple times), thus limx

x + x

x + x x = and

limx

an+an

=

< .

erefore we can conclude by the ratio test that the series

n=

ln

(n

)nnconverges.

EXAMPLE 11

Determine whether

n=

n!

nnconverges or diverges?


an+an

=(n + )!nn(n + )n+n! = (n + )n!n

n

(n + )n(n + )n! = nn

(n + )nWe note that lim

n( n

n + )n has indeterminant form . Using the fact that logarithm and exponential

function are continuous, we therefore use that

limx

( nn +

)n = exp limx

ln( nn +

)nWe now consider the inner limit

limn

ln( nn +

)n = limn

n ln( nn +

) = limn

ln( nn+)n

.

is has the indeterminant form which again we will resolve by applying LHopitals rule. (Again, we write

this as a function in x, to indicate that this is using a continuous function.) By LHopitals rule we get

= limx

ln

(x

x+

)x= lim

x

x

x+(x+)()x()

(x+)x

= limx

x+x

(x+)x

= limx

x(x+)x

= limx

x

x(x+ ) = limx x(x+ )= lim

x

x( x)(x+ )( x) = limx( + x) = .


32/49

We thus know that

limn

an+an

== exp limx

ln( nn +

)n = exp() = e <

and therefore we can conclude that

n=

n!

nnconverges.

So far all the terms of the series we considered were positive. Once we get to power series, however also

negative terms will occur. We therefore want to establish that by considering absolute values the ratio

test also applies to these series.

THEOREM 5 Ratio Test for SeriesLet an be a series with positive terms and suppose that

limn

an+an

= rthen

(a) the series converges ifr<

(b) the series diverges ifr> or if r is innite

(c) the test is inconclusive ifr= .

EXAMPLE 12

Consider the series

n=

()nn!

does it converge or diverge?

Solution As not all terms are positive, we cannot immediately apply the ratio test.erefore we begin bylooking at how this series compares to the corresponding series of positive terms

n=

()nn!

= n=

n

n!.

Each term in our original series is smaller than or equal to the terms of this new series, since ()n

n!

n

n!.

However, the comparison test as given only holds for positive series, so this is not enough to make a conclu-

sion.

erefore, we also consider the corresponding series of negative terms:

n=

()nn!

. In this series each

term is smaller than or equal to the terms of our original series, since n

n! ()n

n! .


33/49

n

n!n

n!n

n!

erefore, it seems plausible that if both

n=

n

n!and

n=

n

n!converge, we can conclude that

n=

()nn!

converges since the value of its terms are squeezed between the two other series terms and partial sums of the

original series therefore are bounded in terms of the partial sums of the positive and negative series, both of

which are converging.

Let us look at this in the context of the ratio test. By taking the absolute value of quotients we treat all

three series simultaneously: For

n=

()nn!

we thus get the ratio

an+

an =

()n+(n+)!()nn!

=

n+(n+)!nn!

=

n+(n+)!nn!

=n+

(n + )!n!

n=

n n!

(n + )n! n=

n + .

As limn

n + = < , we conclude by the ratio test that this series converges.

Since

n=

()nn!

= () n=

n

n!this is just a scalar multiple of

n=

n

n!and scalar multiples do not aect


34/49

convergence.erefore, we can conclude

n=

()nn!

converges. Hence by the argument above we conclude

n=

()nn!

must converge as well. erefore if we have a series with negative terms, we can determine the

convergence using the ratio test on the absolute value of the series. In fact the proof of this statement is the

same as the proof of the ratio test, but using an instead ofan.EXERCISES

Determine whether the following series converge or

diverge. You may use the integral test, direct compar-

ison, limit comparison, or the ratio test.

.

n=

n

.

n=

ln

(n

)n.

n=

n

n

.

n=

n

n

.

n=

n

n +

.

n=

n

n +

.

n=

sin n

.

n=

nn

n.

n=

n

nn

.

n=

n

n

.

n=

n

n!


35/49

-

Working with Power Series

In this chapter we nally see the reason why we were bothering with series. A power series is a series involving

powers ofx, thus as long as it converges we obtain a series value for everyx-value. In other words: the series

describes a function. Such a function might not look as nice as the functions you have seen so far, but in fact

has many advantages. By calculating partial sums we can approximate the value of the function. (is is infact how your calculator evaluates functions such as sine or logarithm. When you press sine there is no little

man in the calculator measuring the length among the circle; instead power series is evaluated.)

We have seen in the main textbook that Power series have an (possibly innite) interval, centered around

the center of the series (thats why its called center), in which they are converging (absolute value of ratio

< ).ey diverge outside the closed interval. (In this course we ignore the behavior at the end points.)

Power Series as Polynomials

Power series look a lot like polynomials. We therefore would like to treat them like polynomials in fact this

is one of the main reasons for using power series. Before we can do so, however we will have to establish that

this is a valid thing to do, for example we have to show that adding two power series gives the same result asadding the coecients o the power series:

n=

an +

n=

bn =

n=

(an + bn)If we only had nite sums this would be immediately clear, the law of commutativity tells us that the result is

the same.is is something you are probably using without actually thinking about it because you learned it

as soon as you learn the addition of numbers.

Surprisingly (though you will see more examples of such behavior in advanced calculus), this is far more

dicult for innite sums.e problem is basically that the law of commutativity only lets us move numbers in

addition by nitely many places, but we need to move them innitely many places.at can cause diculties,

because it potentially allows us to keep a particular summand for later addition in perpetuity, thus changingthe value of any nite approximation (and thus also the value of the sum). e following example illustrates

this in a particular case.

EXAMPLE 1What value should the innite sum

+ + + + +


36/49

have?

Solution Let us rst note that it is not clear whether this innite series actually converges. What we willbe doing now therefore is also an illustration of the need to prove convergence.

What we will be doing looks rather harmless at rst sight. We will simply change the order of addition,

by introducing pairs of parentheses in the sum.is changes the order of addition and is equivalent to rear-

ranging the terms of this series. In each of the cases we can then evaluate the sum easily:

+ + + = ( ) + ( ) + ( ) + = + + + + = = + ( + ) + ( + ) + ( + ) + = + + + + =

is cannot be right (or the world is insane, a possibility which we do not want to consider here). In fact one

could do even worse things. We could group each with the + following three positions laterand therefore

get the value of .

e root of the problem turns out to be the negative summands in between. If all were positive, we had

the sum +

+

+

which clearly diverges. In such a case we say, that this series is not absolutely convergent.e example shows, that rearrangement is not a valid operation for such a series.

Fortunately a theorem from advanced calculus tells us that this obstacle we observed here is the only one.

If the series converges absolutely, i.e. it would converge even if all minuses were replaced by pluses, we are

permitted to rearrange.

THEOREM 1 The Rearrangement Theorem for Absolutely Convergent Series

Suppose that

n=

an converges absolutely, i.e.

n=

an converges as well, and b, b, . . . , bn , . . . is any arrange-ment of the sequence {an}, then bn converges absolutely, and

n=

bn =

n=

an

is theorem is stated aseorem in the textbook on p. , a sketch of the proof is outlined on pg

exercise . We will be using this theorem in this lecture only to establisheorems - below.

For power series we are now in luck: because of the ratio test (which we also could do with absolute values)

a power series converges absolutely inside its interval of convergence. We therefore are permitted to rearrange

its terms. In particular, when adding to a power series (centered at the same point) we can collect their termstogether in the same way as we would do for nite sums. We get the following important theorem:

Modern Physics has a method called Renormalization , which is essentially about such reordering.


37/49

THEOREM 2 Term-by-Term Summation for Power Series

Suppose that f(x) = n=l

cn(x a)n and g(x) = n=

dn(x a)n are two power series, both centered at thesame point a, and that b is chosen inside the interval of convergence of both f and g. (us f(b) and g(b)are both absolutely convergent series.)en

f(b) + g(b) = n=

(cn + dn)(b a)nWe write that

f(x) + g(x) = n=

(cn + dn)(x a)nfor x inside the interval of convergence for both series.

EXAMPLE 2Let f(x) =

n

xn

nand g(x) =

n

xn

n!. Determine the common interval of convergence for both series and

write for x inside this interval f(x) + g(x) as a single sum.Solution

e same holds for products. Again, once we are permitted to rearrange terms, we can take the same

process as used for multiplying polynomials, and generalize it to the case of power series, as long as long as

we are inside the interval of convergence.e resulting formula is in fact the same as holds for the general

multiplication of polynomials. (In the textbook this is theorem on p. .)

THEOREM 3 Term-by-Term Multiplication for Power Series

Suppose that f(x) = n=l

cn(x a)n and g(x) = n=

dn(x a)n are two power series, both centered at thesame point a, and that b is chosen inside the interval of convergence of both f and g.en

f(b) g(b) = n=

n

k=

ck dnk(b a)n

We write that

f(x)

g(x)=

n=n

k= ck

dnk(x

a)n

for x inside the interval of convergence for both series.

Proof To simplify notation we assume that a = . (e general proof works the same way by replacing x


38/49

byx a, but this is more messy to write down.) We multiply out:

cx + cx + cx + dx + dx + dx += cx

dx + dx + dx + + cx dx + dx + dx ++cx

dx

+ dx

+ dx

+

+

= cdxx + cdx

x + cdxx + + cdx

x + cdxx +

cdxx + cdx

x +

= cdx+

+ cdx+

+ cdx+

+ + cdx+

+ cdx+

+

cdx+

+ cdx+

+

Now we collect the summands according to the power of x.is is where we need to rearrange the terms,

which is permitted because of the absolute convergence.

=

(cd)n=, k= x+

(cd+

cd)n=, k=, x+

(cd+

cd+

cd)n=, k=,, x+

+ (cdn + cdn + cdn + + cnd)k=,,,...,n

xn +

EXAMPLE 3

Let f

(x

)=

n

xn

n

and g

(x

)=

n

xn

n!

. For x inside the common interval of convergence, write f

(x

) g

(x

)as

a sum.

Solution

Since we are doing calculus, we dont only want to do arithmetic with power series, but also dierentiate

and integrate. Two theorems from advanced calculus state, that this also can be done, as one would do it for

polynomials:

THEOREM 4 The Term-by-Term Differentiation Theorem

Let f(x) = n=

cn(x a)n.en f(x) is dierentiable for x inside the interval of convergence, and we havethat

f(x) = ddx

f(x) = n=

ncn(x a)n,and this derived series converges (absolutely).

Notice that the series now begins with n = because the derivative of the constant term is zero.


39/49

THEOREM 5 The Term-by-Term Integration Theorem

Let f(x) = n=

cn(x a)n.en for anyx inside the interval of convergence of this series also the series

F(x) =

n=

cn (x

a)n+

n +

converges (absolutely), and we have thatd

dxF(x) = f(x).erefore f(x) has an antiderivative, and

f(x)dx= n=

cn(x a)n+

n + + C

is establishes that inside the interval of convergence, we can treat power series just like polynomials.

A power series as solution to a differential equationTo give you a real world example consider the Bessel functions.ese functions rst arose in solving Keplers

equations for planetary motion. Since that time these functions have been applied in many dierent physical

situations for example the temperature distribution in a circular plate.

EXAMPLE 4 The Bessel functionWe dene the Bessel function of order as

J(x) = n=

()nxnn(n!)

What is its interval of convergence?

Solution To nd the interval of convergence, set an = ()nxn(n(n!)).enan+

an =

()n+x(n+)(n+)((n + )!)

n(n!)()n xn =

x(n + ) <

us the series converges for all real x, the Bessel function is dened on the whole real axis.

If we want to see how the Bessel function looks we can start looking atthe approximations given by partial

sums:

s(x) = x

, s(x) = x+

x

, s(x) = x

+

x

x

, s(x) = x

+

x

x

+

x

as shown below.e second graph is the full Bessel function J(x).As we have absolute convergence within the interval of convergence (and thus may reorder terms!) we

can treat arithmetic with power series in the same way as arithmetic with polynomials (i.e. sum up term-

wise or form the product from partial sums), which anyhow looks like the natural thing to do. We also see

Friedrich Bessel, German astronomer, -


40/49

Degree=2

Degree=4

Deg=6

Deg=8

that we can calculate derivative and anti-derivative of a power series very easily while in general calculating

anti-derivatives is hard.

Later we will see that we can express solutions to dierential equations easily in terms of power series.

EXAMPLE 5 The Bessel function as solution for a differential equationIn continuation of the previous example, we want to see that the Bessel function J(x) is a solution to thedierential equation

xy(x) + x y(x) + xy(x) = Solution For this, we calculate the rst two derivatives (note that we can leave out the n = term in thederivatives):

J(x) = n=

()nxnn(n!) , J(x) =

n=

()n n xnn(n!) , J (x) =

n=

()nn(n )xnn(n!)

and multiply appropriately (doing an index-shi to have all three series start at n = ) to get

x

J(x) =

n= (

)nxn+

n(n!) =

n=

(

)(n+)x(n+)

((n+))((n + ) )!)m = n +

=

m=

(

)mxm

(m)((m )!) ,as well as:

x J(x) =

n=

()n n xnn(n!) , xJ (x) =

n=

()nn(n )xnn(n!)


41/49

and therefore (collecting according to powers ofxn):

xJ + x J

+ xJ =

n=

()n (n)((n )!) + nn(n!) + n(n )n(n!) xn

=

n=(

)nn

+ n + n

(n

)n(n!)xn

=

n=

()n n(n!) xn =

is veries that the Bessel function is a solution to the given dierential equation.

ere are many other so-called special functions, dened as power series, which occur in applications.

Expressing known functions as power series

To be able to work with power series and ordinary functions, the next big question then is how to translatebetween traditional functions, given by a formula, and power series.

We can do some cases using partial fractions and geometric series.

EXAMPLE 6 Power series for rational functions

Determine a power series for

x + x .

Solution In a partial fraction decomposition, we write

x + x =

(x )(x+ ) = x x+ (x+ )en we note that (by geometric series):

x = x x x

x+ =

x+

x

x + =

n=

()n n+

xn

For the power series for(x+ ) , remember that (x+ ) dx = x+ , we can thus get a series for

(x+ ) by term-by-term dierentiation of a series for x+ (which we know already):

(x+ ) = ddx x+ = ddx

n=

()n n+ xn =

n=( )n+nxn

= () + (

) x+ (

) x + (

) x +

=

x+

x

x +


42/49

Note that one could get this result also by multiplying the series for x+

with itself, using the formula for products:

(x+ ) = + x+ + + x+

x +

=

x+

x

x +

by geometric series and multiplication (remember reindexing!)

In summary we get that

x + x =

x

x +.

e general case will have to wait for the next chapters.

EXERCISES

. Determine a power series for (x+ )(x ) .

. Determine a power series for(x ) from a

power series for its antiderivative.

. Determine the rst four terms of a power series

for(x ) , by multiplication. Compare with the re-

sult of problem .

. We know thatd

dxln(+ x) =

+ x. Determine a

power series for

+ x, and use term-wise integration

to obtain a power series for ln( + x).


43/49

-

Taylor polynomials as functionapproximations

One of the principal uses of Taylor polynomials is to approximate functions. e key to this is the formula

for the error term. Let us recall the formula:

Let f be a function that is innitely oen dierentiable and M be a positive constant such thatf(n+)(t) M for all tbetween x and a, then the remainder term Rn(x) = f(x) = Tn(x) inTaylorseorem satises

Rn(x) Mx an+(n + )! .If f is well-behaved (essentially any function you will consider in calculus is), then for each nite interval

there will be one (possibly very large) value ofM, such that

f(n+)

(t

) Mfor every n.e reason for this is,

that otherwise higher derivatives off must have larger and larger values, and therefore oscillate rather wildly.

e remainder estimate however then is essentiallyx an(n + )! . From the study of series we remember that

n! grows faster than xn.us the error becomes smaller and smaller, the larger n gets.

Vice versa, for a xed n we can make the error arbitrary small, by choosing x to be close to the center a.

is principle is fundamental to applications in Physics and Engineering:

Approximation principle for well-behaved functionsIff is a well-behaved function, and Tn the Taylor polynomial for f of degree n centered at a, then the function

Tn(

x

)is an approximation for f

(x

).e approximation quality gets better, the larger n is and the closer xis

to a.

For values ofxvery close to a the linear approximation T(x) is oen suciently good.We ilustrate this principle in the case of deciding which function is largest:

Once n > xthe subsequent factors in n! are larger than the factors in xn


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EXAMPLE 1 Using Power Series to Approximate FunctionsSuppose we wish to compare the values of functions xx, sin(x), and ex for small (positive) values of x.Which function is largest?

Solution We begin expressing each of these functions as power series, centered at a = :

x

x=

x(

x)=

x

n= xn=

n= xn+

=

x+

x

+

x

+

sin(x) = n=

()nxn+(n + )! = x x

!+

x

!

ex =

n=

xn

n! =

n=

xn

n!= x+

x

!+

x

!+

We see that all three functions agree on the Taylor approximations T(x) and T(x), thus we need to considerat least T(x). We get the following approximations:

x

x x+ x

sin(x) xex x+

x

For positive values ofx, we then note that x+ x > x+x

> x, thus for small values ofxwe have that

x

x> ex > sin(x).

A look at the graph veries this result:e graph for x [...] shows that for even smaller values of x

x

x

sin x

ex

x

x

sin x

ex

x

x

sin x

ex

the graphs are almost identical, this is a consequence of the fact that the linear approximation of all three

functions is the same. Of course, for larger values ofxthe behaviour can be quite dierent, as the third graph

shows.


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EXERCISES

. By looking at their Taylor series, decide which of

the following functions is largest, and which smallest

for xnear :

ex, x, x. Consider the power series

W(x) = n=

(n)nn!

xn .

a) Determine the interval (open interval: you do not

need to decide on the behavior at the end points) of

convergence for W(x). You may use without proofthat limn

n

n + n+

=

eb) Calculate a power series for W(x).c) Let I= [,.]. You are given the information, thatfor the second derivative W()(x) for x I.Using the formula for the error term of a Taylor poly-

nomial, show that W(x) x x (and thus xx W(x) x+ x) for x I.


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-

Homework Assignments

EXERCISES

..A Evaluate the integrals

sin xcos xdx

sin xcos xdx

..A Let f(x) = x+ . Find the inverse f(x)and identify the domain and range off. As a check,

show that f(f(x)) = f(f(x)) = x...B Let f(x) = (x ). Show that f is one-to-one on R. Find a formula for d f

dx at x= using

theorem .

..A Given that = arccos(), nd sin , tan and sec .

..B Simplify the expression tan(arcsin(x))...C Evaluate the integrals

x x dx

r r dr..A Dierentiate:

d

dxln cos x

x + (x )

..B Evaluate the integrals

(ln x)x dx

(x ) x+ x

dx


et

et dt

t t

dt

..A Verify that the function ex(x ) is a so-lution to the initial value problem

(x )y + x y= , y() =

..B Solve the separable dierential equation

y

x = x y

..C When cane sugar is dissolved in water, it con-

verts to invert sugar over time. e amount f

(t

)of

unconverted cane sugar at time t decreases, follow-ing the dierential equation f = .f. Suppose we

start with a solution containing g of cane sugar.

How much cane sugar remains aer hours? aer

hours?

How long will it take, until only g of cane sugar re-

mains?


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x

xdx

x x+

dx


x ln xdx

arccos td t


x

( + x)dx

x( x) dx

..A Expand the quotient

x+ (x )(x + )by partial fractions.

..B Evaluate the integral

(x )(x )dx

..A Evaluate the improper integral

(x+ ) dx..A Find a formula for the terms of the following

sequences:

, , , , , . . .

,

,

,

, . . .

..B Give an N-proof, that the sequence an = n

+ nconverges to .

..A Determine the open (i.e. ignoring the behav-

ior at the endpoints) interval of convergence for

n=

xn+

n +

n=

(x )n..B Consider the power series

f(x) = x x+

x

x

+

i) determine the (open, ignoring endpoints) interval,

in which f(x) converges.ii) Determine its term-by-teem derivative f(x).iii) Determine the value of f

(x

)as a fraction (Hint:

Geometric series).iv) Give a formula for the value of f(x), using an an-tiderivative of the function found in iii)

..C Determine a power series for

( + x) ...D We know that

d

dxln( + x) =

+ x. Deter-

mine a power series for

+ x, and use term-wise in-

tegration to obtain a power series for ln( + x)...A Find the rst four terms of a Taylor series

about a = forsin(x) x

..B Determine the Taylor series for f(x) = xabout a = .

..A Determine a Power series for the following

functions about a = :

f(x) = x exf(x) = cos(x)

f(x) = sin(x) = cos(x)


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..B Identify the following functions from their

Taylor series about a =

+ x +x

!+

x

!+

x

!+

x+ x x + x x +

x x

+

x

x

..C We approximate sin(x) byx x on the in-terval [, ]. What estimate can you make for the er-ror?

Suppose we want to approximate sin(x) on the inter-val [, ] with an error of at most , what degreeTaylor polynomial do we need to choose?

..D By looking at their Taylor series, decidewhich of the following functions is largest, and which

smallest for xnear :

ex,

x,

x

..A Find a power series solving the dierential

equation

y y= x

under the initial condition y() = . Identify the so-lution as a function.

..B Find a power series solving the dierential

equation

y + y=

under the initial conditions y() = , y() = . Iden-tify the solution as a function.

..A Calculate the arc length of ln(cos(x)) from to .

..A Convert the equation of the curve

r= cos

into cartesian coordinates

..B Convert the equation of the curve

x y=

into polar coordinates

..A Find the points of intersection of the polarcurves

r= + sin , r= sin

..A Sketch the curves r = + sin and r =

+ cos.

Determine the area inside the curve r = + sin

but outside the curve r = + cos.

A..A Determine all complex h roots of .

A..B Calculate +i

A..C Determine all complex roots of the equation

x + x + =

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