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Problem 10: Β True or false (βœ“ βœ“) 1998 Paper I

Which of the following statements are true and which are false? Justify your answers.

aln b = bln a for all positive numbers a and b.
cos(sinπœƒ) = sin(cosπœƒ) for all real πœƒ.
There exists a polynomial P such that |P(x) βˆ’ cosx|≀ 10βˆ’6 for all (real) x.
x4 + 3 + xβˆ’4 β‰₯ 5 for all x > 0.


The four parts are (somewhat annoyingly) related only by the fact that you have to decide whether each statement is true of false.

If true, the justification has to be a proof. If false, you could prove that it is false in some general way, but it is nearly always better to find a simple counterexample β€” as simple as possible.

Part (iii) might look a bit odd. I suppose it relates to the standard approximation cosx β‰ˆ 1 βˆ’1 2x2 which holds when x is small. This result can be improved by using a polynomial of higher degree: the next term is + 1 4!x4. It can be proved that cosx can be approximated as accurately as you like for small x by a polynomial of the form

βˆ‘ n=0N(βˆ’1)n x2n (2n)!

(the truncated Maclaurin expansion). You have to use more terms of the approximation (i.e. a larger value of N) if you want either greater accuracy or larger x. In part (iii) of this question, you are being asked if there is a polynomial such that the approximation is good for all values of x.

I’ve awarded the question βœ“ βœ“for difficulty, just because quite a few good ideas are required.

Solution to problem 10

(i) True. The easiest way to see this is to log both sides. For the left hand side, we have

ln(aln b) = (lnb)(lna)

and for the right hand side we have

ln(bln a) = (lna)(lnb),

which agree.

Note that we have to be a bit careful with this sort of argument. The argument used is that A = B because lnA = lnB. This requires the property of the ln function that lnA = lnB ⇒ A = B. You can easily see that this property holds because ln is a strictly increasing function; if A > B, then lnA > lnB. The same would not hold for (say) sin (i.e. sinA = sinB⇏A = B).

(ii) False. πœƒ = 1 2Ο€ is an easy counterexample. Even though it is β€˜obvious’ we still need to show that cos1β‰ 0, which we could do by noting that 0 < 1 < 1 3Ο€ and sketching the graph of cosx.

(iii) False. Roughly speaking, any polynomial can be made as large as you like by taking x to be very large (provided it is of degree greater than zero), whereas |cosx|≀ 1; and there is obviously no polynomial of degree zero (i.e. no constant number) for which the statement holds.

But how can we write this out formally? First let us knock off the case when the polynomial is of degree zero, i.e. a constant, call it P. Then either P β‰₯ 1 2 or P < 1 2. In either case, P cannot be close to both cos0 and cos 1 2Ο€.

Now suppose

P(x) = aNxN + βˆ‘ n=0Nβˆ’1a nxn (βˆ—)

where N β‰₯ 1 and assume aN > 0. It is enough to show that P(x) > 2 for some value of x. We can find a number x so large that aNxN > N|an|xn + 2 for each integer n with 0 ≀ n ≀ N βˆ’ 1. The smallest possible value of P(x) for any given positive x would be achieved if all the coefficients in the sum were negative. Thus

P(x) β‰₯ aNxN βˆ’βˆ‘ n=0Nβˆ’1|a n|xn > a NxN + (2 βˆ’ a NxN) = 2.

and we are done.

(iv) True: x4 + 3 + xβˆ’4 = (x2 βˆ’ xβˆ’2)2 + 5 β‰₯ 5.


The important point here is that if you want to show a statement is true, you have to give a formal proof, whereas if you want to show that it is false, you only need give one counterexample. It does not have to be an elaborate counterexample β€” in fact, the simpler the better.

My proof for part (iii) is more elaborate than could have been expected of candidates in the examination. A sketch of a polynomial of degree N would have been enough, provided the special case of N = 0 was dealt with separately.