CHAPTER 14SOLUTIONS 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 – 2ii)x + 2iii)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)x2ii)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 + 1ii)iii)
iv)sin xv) ln xvi)
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 n0 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 seriestan–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
103.1
1003.01
10003.001
1,000,0003.000001
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 t and the corresponding increase in s by s, find an expression for s in terms of t and hence find s/t. 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 + s = 16(t + t)= 16(t+ 2t t + t) =16t+ 32t t + 16t. So
s = 16t+ 32t t + 16t– 16t= 32t t + 16tm
Hence
= 32t + 16tms– 1
i) t = 0.2 s, t = 2 s gives an average velocity of
= 32(2) + 16(0.2)= 64 + 3.2 = 67.2 ms– 1
ii) t = 0.1 s, t = 2 s gives an average velocity of
= 32(2) + 16(0.1)= 64 + 1.6 = 65.6 ms– 1
iii) t = 0.01 s, t = 2 s gives an average velocity of
= 32(2) + 16(0.01)= 64 + 0.16 = 64.16 ms– 1
iv) t = 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.
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 0 and 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 h0. 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) cosn iv) n
v) 1/vi) vii) ar
viii)
Investigate the limits of these sequences.
Hintsvii) limit depends on rviii) 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= cosn
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.
Hintsvi), 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= = 0if |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= 0if |x| 1
whereas the sequence diverges otherwise.
viii) , , , , …
u=
and
u= = x = x
11.Evaluate the following limits :-
i) xeii) 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.