Chapter 15



CHAPTER 14 SOLUTIONS TO EXERCISES FOR ANALYSIS FOR ENGINEERS - LIMITS, SEQUENCES, ITERATION, SERIES AND ALL THAT

Exercises on 14.1

1. ( is often approximated crudely by . Explain why this expression cannot be exactly equal to (.

Solution

is a rational number, and can be expressed as a recurring decimal. Thus to eleven decimal places it is 3. 14285714286. On the other hand it can be shown that ( is an irrational number - to eleven decimal places it is 3.14159265359, and it never recurs. The decimal part for ( continues indefinitely. So, since is rational and ( is not, they can never be equal.

2. Using a calculator or computer evaluate to as many decimal places as you can. Square your result – do you retrieve 2?

Solution

On my calculator = 1.41421356237. My calculator squares this to 1.9999999998, so it is clearly not the square root of 2.

Exercises on 14.2

1. Investigate the (very useful) limit of as x tends to infinity, ie gets infinitely large. Evaluate i) ii)

Solution

We have only to look at what happens to as we take larger and larger values of x: = 1, = 0.1, = 0.01, = 0.0001, ... etc. Clearly, as x gets larger in magnitude, gets smaller in magnitude. We say that tends to zero as x tends to infinity, and write

= 0

i) In we use a little common sense and note that as x becomes very large the subtraction of 1 is neither here nor there - we say that x – 1 (( as x (( . So

= 0

ii)

In this case both top and bottom tend to infinity with x and so it is not quite so clear what is happening here. However, a little algebra can help. Rewrite

= 2 +

Then we have

= = 2 + 8 = 2 + 8 (0) = 2

2. Consider the values of the following functions as x gets closer and closer (but not equal) to x = 2:

i) x – 2 ii) x + 2 iii) x2 – 4

iv) v)

In each case evaluate the limit as x ( 2 using the techniques of this section.

Solution

i) There is nothing very surprising about x – 2. Evaluating it for values of x that get closer and closer to 2 yields the obvious conclusion that as x approaches 2, x – 2 approaches zero, as we would expect. So

= 0

ii) Again, nothing problematical here, x + 2 tends to 4 as x tends to 2, or

= 4

iii) Since x2 – 4 = (x – 2)(x + 2) it is pretty obvious that as x tends to 2, x2 – 4 tends to zero:

= 0

iv) = does not exist at x = 2 and does not tend to any value as x tends to 2.

v) = = provided x ( 2. So as x gets closer and closer to 2 without actually becoming equal to 2, will get closer and closer to . So

=

3. Evaluate the limits

i) x2 ii) iii)

Solution

i) x2 = 02 = 0

ii) = = = – 1

iii) = = = – 1

Exercise on 14.3

Evaluate the following limits

i) ii) iii)

Solution

i) = 2 = 2

since 2x ( 0 as x ( 0

= 2 (1) = 2

ii) = = 3 ( 33 – 1 = 33 = 27

iii) = e– 3x = 0

Exercise on 14.4

Consider the following functions for 0 < x < (, and discuss whether or not they are continuous for these values of x.

i) x + 1 ii) iii)

iv) sin x v) ln x vi)

vii) viii)

ix) f(x) = x ( 2

= 4 x = 2

Solution

i) x + 1 exists and is finite for all values of x and for any value of x, x = a,

= a + 1

So x + 1 is continuous for all x

ii) Apart from the point x = 0, exists and is finite everywhere. But the range given, 0 < x < (, excludes x = 0 and so this function is continuous everywhere on the range.

iii) clearly has a discontinuity at x = 1 which is within the range give, 0 < x < (, so it is discontinuous on this range

iv) sin x is a continuous 'wavy' function stretching from – ( to + ( as you can confirm from its graph in Chapter 6 (UEM 181)

v) ln x is defined and continuous for positive values of x, and within the range 0 < x < ( it is continuous (see UEM 133)

vi) The function f(x) = is only discontinuous at x = – 1 and for any other value of x we can cancel the x + 1 and write:

f(x) = = = x – 1 for x ( – 1

So, on the range given f(x) is continuous

vii) f(x) = = has discontinuities at x = – 1 and x = 1, and so it is discontinuous on the given range

viii) = x – 2 provided x ( – 2, and so on the given range this function is continuous

ix) f(x) = x ( 2

= 4 x = 2

This is a tricky one! It may look like the function has a discontinuity at x = 2 because of the denominator x – 2. But the point x = 2 is specifically excluded in the expression containing this denominator. Instead, the value of the function at x = 2 is defined to be 4. So the overall function does have finite value at x = 2, namely 4. We also note that in fact

= 4

and so, thanks to this definition, f(x) satisfies the condition for continuity at x = 2:

= f(2)

It is clearly continuous for all other values of x as well and so this function is in fact continuous in the range given.

Exercise on 14.5

Evaluate the slope of the function f(x) = 3x3 using the limit definition.

Solution

We will use the definition

Slope =

for the slope at a, but replace a by the general point x - so the slope will be

Now if f(x) = 3x3 then f(x + h) = 3(x + h)3 = 3(x3 + 3x2h + 3xh2 + h3)

= 3x3 + 9x2h + 9xh2 + 3h3

using the binomial theorem. Then

f( x + h) – f(x) = 3x3 + 9x2h + 9xh2 + 3h3 – 3x3

= 9x2h + 9xh2 + 3h3

So

=

= = 9x2

Exercise on 14.6

Adapt the argument of this section to show that the series

1 + + + + … + + …

does not add up to a finite quantity, ie it diverges.

Solution

First note that in general

>

So we can say that

1 + + + + … + + … > 1 + + + + …

But the RHS is just the harmonic series and we have just shown that this adds up to 'infinity' - ie it diverges. Therefore

1 + + + + … + + … > (

So this series diverges also.

Exercise on 14.7

Investigate the convergence of the sequences

un = i) (–1)n ii) 2n iii) n iv) –

v) | r | > 1 vi) | r | < 1

Solution

i) For un = (–1)n we simply get the sequence – 1, 1, – 1, 1, – 1, ... and we don't need limits to tell us this is not going to settle down to anything definite - the sequence diverges. In fact we say it 'oscillates'

ii) For un = 2n we get 2, 4, 8, 16, ... and we can see pretty clearly that this sequence is divergent - the terms get bigger and bigger without limit.

iii) As n gets larger, un = n clearly gets smaller and smaller:

, , , , ...

In fact, in general, if |x| < 1 then as n tends to infinity, xn tends to zero. So in this case the sequence is convergent to zero.

iv) For un = – we get the sequence

– 1, – , – , – , – , ...

and again it is pretty obvious that as n tends to infinity, the terms tend to zero - this sequence is again convergent to zero.

v) As already noted in iii), if |x| < 1 then as n tends to infinity, xn tends to zero. If |r| > 1 then| |< 1 and so tends to zero as n tends to infinity. This sequence therefore converges to zero.

vi) If |r| < 1 then| |> 1 and the terms un = clearly get larger and larger in magnitude as n increases indefinitely. This sequence therefore diverges.

Exercise on 14.8

Rewriting the equation x3 – 5 x – 3 = 0 in the form

x =

and starting with the value x0 = – 1, use your calculator to generate a sequence of approximate solutions to the equation. Do you think this sequence converges to a solution ? What happens if you try x0 = – 2 ?

Now rewrite the equation in the form

x =

and again start with x0 = – 2. What happens ?

Solution

Depending on the accuracy you use and how you do the calculations you should find that if you start with x0 = – 1 the first rearrangement above leads in a few iterations to the approximation x = –0.657 to 3dp. However, if you start with x0 = – 2 then you will find after a few iterations that the resulting sequence is diverging noticeably, and the values obtained clearly will not settle down to any particular solution. If on the other hand we use the second rearrangement and start with x0 = – 2 then you should get pretty quickly to the approximate solution x = – 1.83. The point of all this is that whether or not an iterative process converges to a solution, and how quickly it does so, is often dependent on the form in which the iterating formula is arranged, as well as on the starting value used. There are methods for checking which sorts of arrangements will or won't work, and most standard software packages will do all this for you.

Exercise on 14.9

Find Sn for the series

1 + + 2 + … + r + …

and hence evaluate the sum to infinity.

Solution

With a = 1 and r = in

Sn = a + ar + ar2 + …… + arn–1

=

we obtain the sum to n terms as

Sn = 1 + + 2 + … + n – 1 = [1 – ( )n ]

Now n( 0 as n ( (, so

Sn =

Thus, the sum to infinity of the geometric series is

S =

Exercise on 14.10

Test the following series for convergence

i) + 3 + 5 + 7 + …+ 2r – 1 + …

ii) + 3 + 5 + 7 + … + 2r – 1 …

iii) 2 + 23 + 25 + … + 22n–1 + …

Solution

i) + 3 + 5 + 7 + …+ 2r – 1 + …

In this case, as with most series, there are a number of ways we can go, but let us simply use the ratio test and see where that gets us. We have

un = 2n – 1 and un + 1 = 2n + 1

So

= 2n + 1 (2n – 1) 22n – 1 = 2

and as we let n ( (, ( 1 and so

= 2 = < 1

So, by the ratio test this series converges. In fact, we can also see this by applying the comparison test, but this requires a bit more cunning. Clearly, from each term of the series we have

+ 3 + 5 + 7 + …+ 2r – 1 + …

< + 3 + 5 + 7 + …+ 2r – 1 + …

< + 2 + 3 + 4 + …+ r + …

because we have added a positive quantity to the series. But the RHS now is a GP with common ratio < 1 which therefore converges and so by the comparison test the original series must also converge.

ii) + 3 + 5 + 7 + … + 2r – 1 …

In this case all we have done from i) is change the sign of the in all but the first term. It may at first look like an alternating series, but in fact after the first term the signs do not alternate - they are always negative. In fact, apart from this it is exactly the same as the series in i) and for the same reasons it is convergent.

iii) 2 + 23 + 25 + … + 22n–1 + …

un = 22n–1 seems to get larger and larger as n increases and so our first concern will be whether the nth term goes to zero or not as n tends to infinity. In fact

22n–1 ∫

so far as convergence properties are concerned, and the RHS is in term similar to

which we know from Section 14.3 (UEM 417) is infinite. So the nth term does not tend to zero, and therefore this series cannot converge - ie it diverges.

Exercise on 14.11

Derive the series tan–1x = x – + – + … + (– 1)n+1 + … . and discuss its convergence.

Solution

We use the form

f(x) = f(0) + f´(0)x + x2 + … xr + …

=

and so need to calculate the first few derivatives. We will do this in a routine way, but as you become more practiced you will find various short cuts in different circumstances. Remember of course that we can only expand the series over a restricted range of x (the principal value) in order that tan–1x is a single valued function.

f(0) = tan–10 = 0

f´(0) = = 1

f´´(0) = – = 0

Now the differentiation becomes messy, but note that you don't have to do it in full detail because putting x = 0 will eliminate some of the terms as we go along, for example

f´´´(0) = – – 2x(...) (at x = 0) = – 2

on using the product rule (we don't need to know (...) because it drops out when we put x = 0). Continuing in this way will give us the derivatives we want, but you will soon find that the differentiations get more and more onerous. This brings us to a more convenient method of calculating such derivatives, which is often used in finding series and which uses implicit differentiation, studied in Section 8.2.5 (UEM 238). The key is to start with y = tan–1 x and rewrite the first derivative

y´ =

in the form of the equation

(1 + x2) y´ = 1

Putting x = 0 in this gives y´(0) = 1, as we already know.

Now differentiate this equation through with respect to x using implicit differentiation. We obtain

(1 + x2) y´´ + 2xy´ = 0

Notice that using this approach all we need to do is differentiate simple products. Again putting x = 0 in this gives immediately y´´(0) = 0, again as we already know. Now differentiate the last equation through again and we get (two products to differentiate this time)

(1 + x2) y´´´ + 2xy´´ + 2xy´´ + 2y´ = 0

or, tidying up

(1 + x2) y´´´ + 4xy´´ + 2y´ = 0

Putting x = 0 in this, using what we already know about y´ (= 1) gives

y´´´(0) = – 2

You should now be able to see the general approach and I will simply state the results for you to check. Differentiating again gives, after simplification

(1 + x2) y(iv) + 6xy´´´ + 6y´´ = 0

from which y(iv)(0) = 0 immediately.

Differentiating again gives

(1 + x2) y(v) + 8xy(iv) + 12y´´´ = 0

and putting x = 0 in this, using what we already know, gives

y(v)(0) = – 12y´´´(0) = – 12(– 2) = 24 = 4!

and so on:

(1 + x2) y(vi) + 10xy(v) + 20y(iv) = 0 giving y(vi)(0) = 0

(1 + x2) y(vii) + 12xy(vi) + 30y(v) = 0 giving y(vii)(0) = – 6!

Substituting these results in the series

f(x) = f(0) + f´(0)x + x2 + … xr + …

gives the required result

tan–1 x = x – + – + … + (– 1)n+1 + … .

Finally, we should mention perhaps the easiest way of all to obtain this series - treat it as the integral of the series for . The series for the latter function can be obtained from the binomial theorem:

= 1 – x2 + x4 – x6 + x8 – ...

then

tan–1 x + C = =

= x – + – + … + (– 1)n+1 + …

on integrating term by term. To determine the constant of integration note that tan–1 0 = 0 so C = 0, and we obtain the series we got previously. This approach also gives the simplest way to determine the convergence properties – the series for is convergent on |x| < 1. Also, for x = ± 1 the series is an alternating series with terms decreasing to zero and so it is convergent for these values also. So the series for tan–1 x converges on – 1 ( x ( 1.

This exercise has certainly not been easy, but it brings together a lot of topics that we have met before and provides useful revision.

REINFORCEMENT EXERCISES IN ANALYSIS

1. i) Find the values of the function (3x +1)/x when x has the values 10, 100, 1000, 1,000,000.

ii) What limit does the function approach as x becomes infinite ?

Solution

i) x (3x + 1)/x

10 3.1

100 3.01

1000 3.001

1,000,000 3.000001

ii) It should be pretty obvious that as x gets larger and larger, (3x + 1)/ tends to 3.

2. Find

Solution

= = = = 2

3. Find

Solution

When letting x ( ( it is best to express everything in terms of , because

= 0

So

= =

4. The distance fallen by a particle from rest is given by s = 16tm. Representing an increase in time by δt and the corresponding increase in s by δs, find an expression for δs in terms of δt and hence find δs/δt. Using this result find the average velocity for the following intervals :

i) 2 secs. to 2.2 secs. ii) 2 secs. to 2.1 secs.

iii) 2 secs. to 2.01 secs. iv) 2 secs. to 2.001 secs.

From these results infer the velocity at the end of 2 secs.

Solution

With s = 16twe have s + δs = 16(t + δt)= 16(t+ 2t δt + δt) = 16t+ 32t δt + 16δt. So

δs = 16t+ 32t δt + 16δt– 16t= 32t δt + 16δtm

Hence

= 32t + 16δtms– 1

i) δt = 0.2 s, t = 2 s gives an average velocity of

= 32(2) + 16(0.2)= 64 + 3.2 = 67.2 ms– 1

ii) δt = 0.1 s, t = 2 s gives an average velocity of

= 32(2) + 16(0.1)= 64 + 1.6 = 65.6 ms– 1

iii) δt = 0.01 s, t = 2 s gives an average velocity of

= 32(2) + 16(0.01)= 64 + 0.16 = 64.16 ms– 1

iv) δt = 0.001 s, t = 2 s gives an average velocity of

= 32(2) + 16(0.001)= 64 + 0.016 = 64.016ms– 1

These results suggest that the velocity at t = 2 is 64 ms– 1. We can confirm this by taking the limit:

v = = = 32t

So at t = 2

v = 32 (2) = 64 ms– 1

5. Which of the following functions exist at the stated values of x ? If the functions do not exist, find the limit of the function at the values concerned, if it exists.

[pic]

Solution

i) x2 clearly exists at x = 4 where its value is 42 = 16

ii) does not exist at x = 1 – it is indeterminate at x = 1 because of the (x – 1 ) factors on the top and bottom. However, the limit does exist there and

= = 1 + 2 = 3

Note again how it is that while the function does not exist at x = 1, the limit does. The reason is, as noted in the book, that when we evaluate the limit we do not actually consider the value at the 'end-point' of the limit. We simply consider what happens as x gets infinitesimally close to the end-point, without actually reaching it. So, for example, if we are approaching the limit at x = a then we never actually put x = a and so x – a is never zero - but it is as close to zero as we want it to be.

iii) is indeterminate at x = – 4, but

= 3 = 0

iv) does not exist at x = 0, and neither does the limit

v) is indeterminate at x = – 2, but

= = = – 4

vi) exists at x = – 1 and = = 1

vii) is indeterminate at x = 2, but the limit is

=

=

== 32

viii) does not exist at x = 1 and neither does the limit at x = 1.

6. Given = 1 find the following limits :-

i) ii) iii)

iv) v) vi)

vii)

Solution

In these exercises we are usually seeking to manipulate things so that we can use the important limit

= 1

i) = = = 1

Here we have used the fact that as x tends to zero, so does 2x, and then we have replaced 2x by u.

ii) = 3 = 3 = 3 (1) = 3

on using the same result as in i).

iii) Provided that α ( 0 and β ( 0 we have

=

= =

iv) = = (1)3 = 1

v) = =

= = =

vi) = =

= 1 ( 0 = 0

vii) = = 0

Here the message is - don't get stuck on the tramlines!

7. Which of the functions in Q5 are continuous for all x ? Give any points of discontinuity and where possible give a function which is continuous everywhere and is equal to the given function away from the points of discontinuity.

Solution

i) f(x) = x2 is continuous for all x

ii) f(x) = is continuous for all x except for x = 1. We can form a continuous function by replacing the function f(x) at the point x = 1 by the value of its limit at that point, which we found in Q5 to be 3. So the following function g(x) is continuous everywhere, for all values of x:

g(x) = f(x) = for x ( 1

= 3 for x = 1

It is of course equivalent to

g(x) = x + 2 for all x

iii) f(x) = is discontinuous at x = – 4, but since the limit exists there we can define a continuous function by

g(x) = for x ( – 4

= 0 for x = – 4

This is equivalent to

g(x) = (x + 4)3 for all x

iv) f(x) = is discontinuous at x = 0 and since the limit at x = 0 does not exist we cannot define a continuous function incorporating f(x) over all real values in this case.

v) f(x) = is discontinuous at x = – 2, but we can define the continuous function g(x) by

g(x) = f(x) = for x ( – 2

= – 4 for x = – 2

or

g(x) = x – 2 for all x

vi) f(x) = is actually continuous for all values of x, because the x4 + 1 in the denominator can never be zero.

vii) f(x) = is discontinuous at x = 2, but the function

g(x) = f(x) = x ( 2

= 32 x = 2

is continuous and is equivalent to

g(x) = (x + 2)(x2 + 4) for all x

viii) is not continuous at x = 1 and since the limit does not exist there, it cannot be patched up into a continuous function.

8. Find the limit of as h ( 0. Hence find the slope of the curve y = xat the point x = 1.

Solution

Using the binomial theorem:

=

= =

= 3x

So the slope of the curve y = xat x = 1 is 3 ( 12 = 3

9. Write down the first four terms of the sequence whose nterm is :-

i) n ii) 5n iii) cosnπ iv) n

v) 1/ vi) vii) ar

viii)

Investigate the limits of these sequences.

Hints vii) limit depends on r viii) limit depends on x

Solution

This question is to get you used to writing and interpreting the nth terms of sequences or series, and obtaining the limit of such sequences.

i) u= n has the first four terms 1, 2, 3, 4.

u= n = (

ii) u= 5n = = and so:

u= , u= 1, u= , u=

and

u= = (

iii) u= cosnπ

u= 0, u= – 1, u= 0, u= 1

In this case the limit as n tends to infinity is not defined - the sequence oscillates.

iv) u= n

u= 0, u= 1, u= 0, u=

While the top always remains 'bounded' - it never gets larger than 2 - the bottom grows larger and larger and so in this case as n tends to infinity the sequence tends to zero:

u= 0

v) u= 1/

u= 1, u= , u= , u= =

and

u= 0

vi) u=

u= , u= , u= , u=

For the limit we write

u=

= +

= + 0 =

Here we note that in the second limit the top is always bounded as the denominator increases to infinity, and in the first limit 1/n tends to zero as n tends to infinity.

vii) u= ar

u= a, u= ar, u= ar, u= ar

If |r| > 1, then the sequence diverges. If r = 1 it converges to a. If |r| < 1 it converges to zero. If r = – 1 it oscillates.

viii) u= =

u= – , u= , u= – , u=

If |x| ( 1 then this sequence converges to zero, since the numerator will always be less than or equal to 1 in magnitude while the denominator increases indefinitely (the answer in the book, giving |x| < 1, is an error). On the other hand if |x| > 1 then the power in the numerator grows much faster than the linear term in the denominator and so the sequence diverges.

10. Write down the nterms of the sequence

i) , , , , … ii) 0, , , , , … iii) , , , …

iv) , , , ,… v) –1, , – , , – , …

vi) , , ,… vii) – , , – , … viii) , , , , …

Investigate the limits of these sequences.

Hints vi), vii), viii) depend on x.

Solution

i) , , , , … is clearly , , , , … and so the nth term is

u=

and in the limit

u= = 0

ii) 0, , , , , … is , , , , , … and the nth term is

u=

and in the limit

u= = = 1

iii) , , , … is , , , … and the nth term is

u=

and in the limit

u= = 0

iv) , , , ,… requires a little more thought but you should soon see that

u=

and

u= = = 1

v) –1, , – , , – , …

u=

and

u= = 0

vi) , , ,… involves x but this really doesn't matter much in writing the nth term:

u=

However, it does make a difference to the convergence properties of the sequence

u= = 0 if |x| ( 1

but if |x| > 1 then, because an exponential always outpaces a polynomial, the limit will grow infinitely large in magnitude as n tends to infinity, and the sequence diverges.

vii) – , , – , …

Again, the x determines the convergence properties:

u= n = 0 if |x| ( 1

whereas the sequence diverges otherwise.

viii) , , , , …

u=

and

u= = x = x

11. Evaluate the following limits :-

i) xe ii) iii)

iv) v) vi)

vii)

Solution

Many of these are (deliberately) either done for you already, or fairly straightforward

i) xe= = 0

from a slight modification of the result on UEM 417.

ii) = 0

from UEM 425.

iii) = 0

pretty obviously from ii).

iv) = = 1

v) = 0 provided x is finite.

vi) = |x| = |x|

vii) = = if | r | < 1, divergent otherwise.

(See UEM 428)

12. If l = find l in the following cases

i) u= ii) u= iii) u=

Solution

i) If u= then u= and so

l = = l = = 1

ii) If u= then u= and so

l = = = = 0

iii) If u= then u= and

=

and

l = = =

13. Plot or sketch  the  function  f(x) = 3x – 4x + 5  and verify that it has a root  near x = –1.5. Use the Newton–Raphson method to obtain this root to 2 decimal places.

Solution

We'll take for granted that you have made a quick sketch of the graph and confirmed that it does indeed cross the x-axis near – 1.5, which is therefore an initial approximation to a root of the equation f(x) = 3x– 4x + 5 = 0. So we will take as our starting value x1 = – 1.5. Note that in Exercise on 14.8 (UEM 427) we denoted our starting value by x0. This was an oversight, but in any case is not a serious issue – the only time we might wish to avoid taking n = 0 as a starting point is if xn is not actually defined for n = 0.

With f(x) = 3x– 4x + 5 we have f′(x) = 9x– 4 and so

xn+1 = xn –

becomes

xn+1 = xn – =

In practice one would normally use a computer package for the neccesary calculations, or write a bit of simple code. However, our intention here is to illustrate the idea of iterating on such a result and so we will do the calculations 'long-hand'. The simplest calculator is all that is needed for this.

Using the above iterative relation we substitute in x1 = – 1.5 to get

x2 = = – 1.554 to 3dp

We now recycle with x2 = – 1.55 (there is no point rushing too far ahead with the dps) to obtain

x3 = = – 1.552 to 3dp

and it is now clear that to two decimal places we are going to get the root as x = – 1.55.

14. Using the Newton–Raphson method find to 2 decimal places the value of

i) ii)

Hint i) Treat as the equation x– 291.7 = 0. Similarly for ii).

Solution

i) We treat x = as the equation f(x) = x– 291.7 = 0 and use Newton's method on this equation.

With f(x) = x– 291.7 we have f′(x) = 2x and so

xn+1 = xn –

becomes

xn+1 = xn – =

We need an initial guess for the solution. Noting that 172 = 289 it seems that x1 = 17 would be a good starting point. Then for x2 the above relation gives

x2 = = 17.08 to 2dp

We now recycle with x2 = 17.08 and find x3 = 17.079 and clearly to 2dp, x = 17.08

ii) We treat x = as the equation f(x) = x– 3.074 = 0

With f(x) = x– 3.074 we have f′(x) = 3xand so

xn+1 = xn –

becomes

xn+1 = xn – =

Noting that 1.43 = 2.744 and 1.53 = 3.375 suggests starting with x1 = 1.4

Then for x2 the above relation gives

x2 = = 1.456 to 3dp

Repeating the calculation with x2 = 1.46 gives x3 = 1.454 and to 2dp we find x = 1.45

15. Find the sequence of the npartial sums for the following series, i.e. the sequence :-

S, S, S, …

If possible, find a formula for Sand thus evaluate S

i) 1 + + + + +… ii) 1 – + – + …

iii) 1 + 3 + 5 + 7 + … iv) 1 + 1 + + + + ...

v) 1 + + + + ...

Solution

i) For 1 + + + + +… we have

S= 1

S= 1 +

S= 1 + +

...

S= 1 + + + + + ...

This is a geometric series with first term 1 and common ratio the sum of which is

S= = 2

and so the sum to infinity is

S = S= 2= 2

ii) For 1 – + – + … we have

S= 1

S= 1 –

S= 1 – +

...

S= 1 – + – + … (– 1)n – 1

Again this is a geometric series. The common ratio this time is – and the sum is

S= =

and so the sum to infinity is

S = S= =

iii) S = 1 + 3 + 5 + 7 + … is an arithmetic series (UEM 104) with first term a = 1 and common difference d = 2. It's sum to n terms is

S= n(2a + (n – 1)d) = n(2 + (n – 1)2) = n2

Clearly as n tends to infinity this also tends to infinity and so this series is divergent. We saw this in general on UEM 429.

iv) There is no simple formula for the sum to n terms, S= 1 + 1 + + + + ... + , in this case, but remembering the series for ex (UEM 124) we have

S = S= 1 + 1 + + + + ... = e

v) In this case there is again no simple formula for the sum to n terms

S= 1 + + + + ... +

but this is the harmonic series and we know from UEM 423 that this in fact diverges.

16. Obtain a formula for the npartial sum Sfor the following series and hence investigate their convergence (or otherwise).

i) 4 + + + + ≡ ii) –6 – 2 + 2 + 6 + 10 + 14 + ...

iii) 1 + 2 + 2+ 2+ ... iv) 1 + + + + ...

Solution

i) S= 4 + + + + ... = 4

= 4 = 6

So

S = S= 6

and the series converges. It is of course simply a multiple of a geometric series.

ii) –6 – 2 + 2 + 6 + 10 + 14 + ... is an arithmetic series with a = – 6 and d = 4 and sum

S= n(2a + (n – 1)d) = n(– 12 + (n – 1)4) = 2n(n – 4)

the limit of which, as we already know, is infinity, so this series diverges.

iii) S= 1 + 2 + 2+ 2+ ... 2= = 2– 1

So

S = S= (

so this series diverges.

iv) S= 1 + + + + ... + =

If |x| ( 2 this is divergent. If |x| < 2 then it converges to =

17. State, without proof, which of the following series are convergent/divergent. Write down the nterm of each series.

i) –1 + 1 – 1 + 1 – 1 + ... ii) 1.01 + (1.01)+ (1.01)+ ...

iii) (.99)+ (.99)+ (.99)+ ... iv) + + + + ...

v) 10+ + + + ... vi) – + – + ...

vii) + 2+ 3+ 4+ ...

viii) + + + + … ix) 1 + (.2) + (.2)+ (.2)+ …

Solution

While no rigorous proof is required in this exercise, you should of course understand the reasoning in each case.

i) –1 + 1 – 1 + 1 – 1 + ... is divergent since the nth term (– 1)n – 1 does not tend to zero.

ii) 1.01 + (1.01)+ (1.01)+ ... is a geometric series with common ratio greater than 1 and so it diverges. The nth term (1.01)does not tend to zero.

iii) (.99)+ (.99)+ (.99) +... is convergent being a geometric series with common ratio less than 1. The nth term is (.99)and tends to zero. as it should.

iv) + + + + ... is just the harmonic series with the first 49 terms removed, so it is divergent. The nth term is . This tends to zero, showing that even if the nth term tends to zero, the series may still be divergent. So the nth term tending to zero is only a necessary condition for convergence and is not sufficient.

v) 10+ + + + ... has the nth term . This tends to zero as n tends to infinity. Also, each term is clearly less than a geometric series with first term 10and common ratio . Since we know such a series to converge then the given series must also converge, by the comparison test.

vi) – + – + ... is an alternating series with the terms decreasing and tending to zero – it therefore converges. The nth term is .

vii) In + 2+ 3+ 4+ ... the nth term is n. It can be shown by the ratio test (check it!) that this is convergent.

viii) The nth term of + + + + … is which tends to 1 as n tends to infinity and so this series is divergent.

ix) 1 + (.2) + (.2)+ (.2)+ … is a geometric series with common ratio, 0.2, less than 1 and therefore converges. The nth term is (.2).

18. Find the nterm of the following series and test for convergence.

i) 1 + 1 + 1 + 1 + … ii) 1 – 2 + 3 – 4 + 5 + …

iii) 1 + 2(.9) + 3(.9)+ 4(.9)+ … iv) 1 – + – …

v) 1+ 2+ 3+ … vi) + + +…

vii) – + – + ... viii) 1 . + . + . + . +...

Solution

i) For 1 + 1 + 1 + 1 + … the nth term is u= 1. Since this does not tend to zero as n tends to infinity, this series diverges - in fact it is pretty obvious that the 'sum' after an infinite number of terms will be infinity.

ii) 1 – 2 + 3 – 4 + 5 + … Again the nth term, u= (– 1)n + 1n, does not tend to zero and so this series is also divergent.

iii) 1 + 2(.9) + 3(.9)+ 4(.9)+ …

In this case the nth term, u= n(.9)n – 1, does tend to zero as n tends to infinity, and so the series may converge. To investigate this, we use the ratio test. We have

u= n(.9)n – 1 and u= (n + 1)(.9)n

So

l = = = 0.9 = 0.9 < 1

and therefore this series converges.

iv) 1 – + – … is an alternating series with the nth term

u= (– 1)n + 1

tending to zero as n tends to infinity and so this series converges.

v) 1+ 2+ 3+ …

u= n

This tends to zero (the power is stronger than the multiplier). Then the same argument as in iii) can be used to show that this series is in fact convergent.

vi) For + + +… the nth term is u= and

u= = 1 ( 0

So the nth term does not tend to zero and therefore this series diverges.

vii) – + – + ... is, despite the signs, little different to vi) in that the nth term u= (– 1)n – 1 fails to tend to zero and so this series also diverges.

viii) 1 . + . + . + . + ...

The nth term is u= = and u= 0 so the series may or may not converge. Now

u=

and so

l = = = 1

So in this case the ratio test is inconclusive and we have to try something else. In fact, you might have noticed beforehand that

1 . + . + . + . + ... > + + + + ...

= = ( (Harmonic series)

So this series diverges.

19. Find the range of values for x for which the following series are convergent.

i) x – + – + ... ii) 2x + (2x)+ (2x)+ (2x)+ ...

iii) 1 + x + + + ... iv) 1+ 2x + 3x+ 4x+ 5x+ ...

v) 1+ 2x + 3x+ 4x+ …

Solution

In each case we use the ratio test, finding in general that the limit of the ratio depends on the variable x and will only be less than 1 for some range of x to be determined.

i) x – + – + ...

u= (– 1)n –1 and u= (– 1)n

So

l = = =

= |x| = |x|

So this series converges for |x| < 1. It definitely diverges for |x| > 1. For x = 1 we get the series

1 – + – + ...

which is convergent since it is an alternating series with terms decreasing to zero. For x = – 1 we get the series

– 1 – – – – ...

which is the negative of the harmonic series and is therefore divergent.

So this series converges for – 1 < x ( 1 and is divergent otherwise. (The answer in the book is incomplete).

ii) 2x + (2x)+ (2x)+ (2x)+ ... = 2x(1 + 2x + (2x)+ (2x)+ (2x)+ ... )

Since the series in brackets is a geometric series with common ratio 2x it is convergent only if |2x| < 1, ie convergent for |x| <

iii) 1 + x + + + ... is the series for ex, which we know to be convergent for all x (UEM 436)

iv) 1+ 2x + 3x+ 4x+ 5x+ ...

u= nxand u= (n + 1)x

So

l = = = |x|

Therefore this series is convergent for |x| < 1 and divergent for |x| ( 1 (in the cases of x = 1 or – 1 the nth term does not tend to zero)

v) 1+ 2x + 3x+ 4x+ … is essentially the same as iv) so far as convergence properties are concerned and so is convergent for |x| < 1 and divergent for |x| ( 1.

20. For the binomial series (1 + x)show that = | x | and deduce that the series converges if | x | < 1 and diverges if | x | > 1. (m not a positive integer).

Solution

For the binomial theorem (UEM 73)we have

(1 + x)m = 1 + mx + x2 + x3 + …

+ xr – 1 + ...

So

u= xn– 1

and

u= xn

Hence

=

= | x |

on simplification by various cancellations. So

= | x | = | x |

and by the ratio test this shows that the series converges if | x | < 1 and diverges if | x | > 1.

21. Expand (1 + 2x)as far as the term in x. How many terms of the series would be required to give (1.02)correct to three decimal places? For what range of values of x is the series valid ?

Solution

We can use the binomial theorem given in the last question, taking m = but replacing x by 2x:

(1 + x)m = 1 + mx + x2 + x3 + …

+ xr – 1 + ...

So

(1 + 2x)= 1 + + 2 + 3 + …

= 1 + 3x + x+ …

on tidying up the coefficients. This series is convergent for |2x| < 1, ie

|x| <

(1.02)= (1 + 2(0.01))and so to evaluate this using the above binomial theorem we would take x = 0.01. It is clear from the series that with this value of x the third decimal place won't be affected after four terms, which is all we need to take.

22. Obtain terms up to x4 in the Maclaurin's series for the functions

i) sin 2x ii) ex2 iii)

iv) ln(1 + 2x) v) tan x vi)

State the values of x for which the series are convergent.

Solution

In these exercises we will use Maclaurin series in some cases as illustration, but in others we will use appropriate short cuts - if you want more practise in use of Maclaurin series, simply use your expansions to check results obtained by other means. In general the task is to obtain, by repeated differentiation, the coefficients in

f(x) = f(0) + f´(0)x + x2 + … xr + …

=

i) For f(x) = sin 2x we have

f(0) = sin 0 = 0

f´(0) = 2 cos 0 = 2

f´´(0) = – 4 sin 0 = 0

f´´´(0) = – 8 cos 0 = – 8

f(iv)(0) = 16 sin 0 = 0

f(v)(0) = 32 cos 0 = 32

and so on, giving the series

sin 2x = 2x – x3 + x5 – ... = 2x – x3 + x5 – ...

which is convergent for all values of x.

ii) For ex2 we don't need to derive the series from scratch. Instead, we can simply substitute x2 for x in the series we already obtained for ex in Problem 14.15 (UEM 436):

ex = 1 + x + + + …

This gives

ex2 = 1 + x2 + + + …

= 1 + x2 + + + …

This, like the series for ex (UEM 436), is convergent for all x.

iii) For we could use Maclaurin's series, or use a binomial expansion of (1 – 3x)– 1. However, the 'mature' way to do this now is to simply recognise it as an example of the series

= 1 + x + x2 + x3 + x4 + ...

that we introduced on UEM 107. As stated there, this result is so important that you should be able to recall it easily. Another viewpoint is to treat it as the sum of an infinite geometric series with common ratio x. Using this series with x replaced by 3x gives directly

= 1 + 3x + (3x)2 + (3x)3 + (3x)4 + ...

= 1 + 3x + 9x2 + 27x3 + 81x4 + ...

This is convergent for |3x| < 1 or |x| <

iv) The series for ln(1 + 2x) can be obtained from that for ln (1 + x) that we gave in Problem 14.16 (UEM 437), by replacing x by 2x. However, since we didn't actually obtain the series for ln(1 + x) we will do this exercise from scratch by the Maclaurin series.

For f(x) = ln(1 + 2x) we have

f(0) = ln 1 = 0

f´(x) = so f´(0) = 2

f´´(x) = – so f´´(0) = – 4

f´´´(x) = so f´´´(0) = 16

f(iv)(x) = – so f(iv)(0) = – 96

and so on, giving the series

ln(1 + 2x) = 2x – x2 + x3 – x4 ... = 2x – 2 x2 + x3 – 4 x4 ...

This series is convergent for |2x| < 1 (same as for the corresponding binomial series), ie |x| <

v) For f(x) = tan x we have

f(0) = tan 0 = 0

f´(x) = sec2 x so f´(0) = 1

f´´(x) = 2sec x tanx so f´´(0) = 0

f´´´(x) = 4 sec2 x tan2 x + 2 sec4 x so f´´´(0) = 2

Clearly f(iv)(0) will be zero

So we obtain the series

tan x = x + x3 + ...

which is convergent for x2 <

vi) For we will simply use the binomial theorem again

= (1 + x)– 2 = 1 + (– 2)x + x2 + x3 + x4 + …

= 1 – 2x + 3x2 – 4x3 + 5x4 – ...

which is convergent for |x| < 1.

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