Administrative matters

Administrative matters

Three Lectures per week (virtual). One Review session per week (virtual).

One tutorial per fortnight (in person and recorded).

Five pieces of assessed coursework, each contributing 5% of the overall credit for the module.

The lecture notes are definitive. In particular, the final examination will be based on material covered in the notes.

The lectures will reinforce in the lecture notes, e.g., by providing examples, and going carefully through the trickier aspects of the module.

Position of the module within mathematics

MTH5104 and MTH5105 together form an introduction to analysis, the branch of mathematics devoted to a precise study of sequences, series, di↵erentiation and integration. These courses bring rigour to calculus.

Pre-calculus

Ancient Greek mathematicians anticipated the integral calculus, but were reluctant to use any kind of infinite process.

Zeno’s paradox (⇠ 460 BC),Achilles and the tortoise:

If Achilles starts at Aand the tortoise starts at B then Achilles can never catch the tortoise since by the time Achilles reaches B, the tortoise will be at some further pointC and by the time Achilles reaches C, the tortoise will be further ahead at D and so onad infinitum. So the tortoise will always be ahead!

Model this argument with some numbers. Assume the tortoise is at distance 1 from Achilles and moves with speed 1, while Achilles runs with speed 4. When will he catch the tortoise?

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-A B C D

The time needed for Achilles to reachB is 1/4.

During this time the tortoise moved a distance 1/4, i.e. C is 1/4 away from B.

The time for Achilles to reach C is 1/4 + 1/16.

The tortoise has now moved to D which is 1/16 away from C. So the time for Achilles to reach D is 1/4 + 1/16 + 1/64, etc.

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So Achilles runs for time 1 X k=1 ⇣₁ 4 ⌘k = 1 4 + 1 16 + 1 64 + 1 256+· · ·

behind the tortoise.

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Infinite sequence, finite sum?

Could it be that, although we keep on adding positive numbers to the sum, the value of the total sum is still finite (and actually less than or

equal to 1₃)?

Somewhat surprisingly, the answer is yes, in fact

1 X k=1 ⇣₁ 4 ⌘k = 1 4 + 1 16 + 1 64 + 1 256+· · ·= 1 3.

Infinite sequence, finite sum?

Let us now look at the series

1 X k=1 1 k = 1 + 1 2 + 1 3 + 1 4 + 1 5+· · ·

As before, we infinitely often add a positive number, and the numbers added become smaller and smaller.

Will the sum therefore also converge to a finite value, as in the previous example?

This time, the answer is no! We can see this as follows. 1 X k=1 1 k = 1 + 1 2 + 1 3+ 1 4 + 1 5 + 1 6 + 1 7+ 1 8+ 1 9 +· · ·

What is the di↵erence this example and the previous one? In this module,

we will develop general tools which allow us to investigate whether or not an infinite series converges.

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Of course, the first example is nothing else than a geometric series, and we know from calculus that

1 X k=1 ⇣₁ 4 ⌘k = 1 X k=0 ⇣₁ 4 ⌘k 1 = 1 1 1₄ 1 = 1 3, using the rule P1_k=0qk ₌ 1

1 q (which agrees with what we claimed

Another paradox

Using this rule, we have for example

x 1 x =x ⇣ ₁ 1 x ⌘ =x(1 +x+x2+. . .) =x+x2+x3+· · · (1) and also x x 1 = ⇣ ₁ 1 x 1 ⌘ = 1 +x 1+x 2+· · · (2)

Adding (1) and (2) thus yields

0 =· · ·+x 2+x 1+ 1 +x+x2+· · ·

Ifx >0, this means that adding up infinitely many positive numbers gives 0.

What is wrong with this argument?

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Another paradox (continued)

The answer is that we have to consider for which values of x each series converges. The series on the right hand side of (1), namely

x+x2+x3+x4+· · ·

only converges for|x|<1. The series on the right hand side of (2), namely 1 +x 1+x 2+x 3· · ·

only converges for |x|>1.

It becomes clear that we need to define convergencevery carefully and worry about the set of values for which a series converges.

Alternating harmonic series

This is the last series we consider. It is given by

1 X k=1 ( 1)k+11 k = 1 1 2 + 1 3 1 4+ 1 5 1 6 +. . .

As we will see later in this course, this series converges. Intuitively, this seems plausible if we draw the first few elements. Denoting

sn=Pn_k=1( 1)k+1 1_k, we have the following picture.

-0 s2 s1 = 1 s3 s4 s5 . . .s₁. . .

(One can prove that the value of the infinite series is ln 2, the natural logarithm of 2.)

Yet another paradox

Now re-order the sequence, taking always a positive element and then two negative ones, so that we obtain

S := 1 1 2 1 4 + 1 3 1 6 1 8+ 1 5 1 10 1 12 +. . .

Note that we still have all the elements from the original series, we just changed the order of summation! As addition is commutative, we might expect that the value should be the same and the order of summation should not matter. . .

Yet another paradox

So now, the value of the sum is halfof what it was before! How did this

happen? Did we do something wrong? One of the goals of this course is

to answer these questions.

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Two questions concerning sums of continuous functions

Is the sum of two continuous functions continuous? Yes!

Is the sum of infinitely many continuous functions continuous? No!

For example the function

f(x) = sin(x) 1 sin(2x) 2 + sin(3x) 3 sin(4x) 4 +· · ·, has the following graph:

-6 x f(x) 2⇡ 0 2⇡ s s s s

Questions concerning sequences of continuous functions

For any sequence of continuous functionsfn which converges to

f = limn!1fn, is f continuous?

No! At least not in general. The answer depends on the typeof convergence – we will define di↵erent notions of convergence for functions.

For a sequence of di↵erentiable functions converging to

f = limn!1fn, is the limit f is also di↵erentiable?

For a sequence of integrable functions fn, is it true that

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n!1fn= limn!1

Z fn?

The theory needed to understand this question and the previous one will be covered in the course MTH5105.

Conclusion

We will have to make precise definitions and carefully restudy topics from calculus such as limits, convergence, continuity, etc.

In order to do this precisely, we will have to begin in an elementary manner, namely with the properties of real numbers.

In preparation, we introduce a method which makes it easier to read, understand and prove statements like

8x2R8">09 >0 8y 2R,_|x y|< : |x2 y2|<". (3) As we will later see, this complicated looking expression simply states that f(x) =x2 is continuous.

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Demon Games

In this module we shall sometimes translate statements like

8n 2N9m2N : m>n (4) (read “for all natural numbersn there exists a natural number msuch that

m>n”) into a Demon Game, a thought experiment in which we play against an imaginary Demon. The rules for the game are as follows:

1 We read the statement from left to right.

2 Whenever there is a “for all” quantifier8, the Demon gets to choose.

If there are restrictions, he has to fulfil them.

3 Whenever there is a “there exists” quantifier 9, then we get to

choose. If there are restrictions, we have to fulfil them.

Example

Expression (4), namely

8n 2N9m2N : m>n

corresponds to the Demon Game First the Demon picksn2N.

Then we pickm2N.

Example

The expression (3), namely

8x2R8">09 >0 8y 2R,_|x y|< : |x2 y2|<",

corresponds to the Demon Game First the Demon picksx 2R.

Then the Demon picks "2Rwith ">0.

Then we pick 2R with >0.

Then the Demon picks y2Rwith |x y|< . We win if|x2 y2|<".

Notes

The order of quantifiers is important. Peaking into the future is not allowed.

We abbreviated the expression 8"2R,">0 by 8">0 and

9 2R, >0 by 9 >0. In this module," and are alwaysreal numbers, so these abbreviations should cause no confusion.

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Trial games

These are just examples of possible games that could have been played, including the statement of who would have won the game if these moves had been played.

Example For the game from our first example

8n 2N9m2N : m>n

we might have:

Trial Game 1 Trial Game 2

Demon picksn = We pick m= Who wins? 7 "

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Trial games (continued)

Example For the game from our second example

8x2R8">09 >0 8y 2R,_|x y|< : |x2 y2|<",

we might have

Trial Game 1 Trial Game 2

Demon picks x= Demon picks "= We pick = Demon picks y = Who wins? I l l l 0.5 0.5 1.4 1.5 - tiny 19 -

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Winning strategy

A winning strategy is one thatalways win the game, no matter how the opponent plays.

Example. For example in our first example, a winning strategy is the following.

The Demon picks n2N.

We pick m:=n+ 12N.

Then m>n, so we win.

Note that we could have also chosen m=n+ 37 to get another winning

strategy, or m=n3+ 12 to obtain yet another one. On the other hand, setting m= 3n₂+7 is nota winning strategy – it satisfiesm>n, but it is not an allowed move because in general this mis not a natural number.

Formal proofs

If we find a winning strategy for the Demon Game, then the original mathematical expression is true.

A winning strategy can then be translated into a formal proof, using the following simple rules.

1 Whenever the Demon picks x, we write “Givenx”.

2 Whenever we picky, we write “Lety be”, or “Choosey”. 3 At the end, replace the “we win” by an end of proof box, i.e., ⇤.

Example. The winning strategy for the first example translates to the following formal proof:

Negation

We prove that a statement is true by finding a winning strategy for us in the Demon Game.

We can prove a statement false by finding a winning strategy for the

Demon.

Take the statement

9m2N8n2N : m n (5) The Demon Game for this is:

First we pickm2N.

Then the Demon picks n2N.

Recall (5): 9m2N8n2N : m n.

To prove that the statement is false, we have to find a winning strategy for

the Demon for this game. So we swap places with the Demon and

consider the new game

First the Demon picksm2N. Then we pickn 2N.

The Demon wins ifm n (and hence we win ifm<n).

A winning strategy for us in the new game is a winning strategy for the Demon in the original game and hence a proof that the original statement is false. Such a winning strategy for the new game is for example

First the Demon picksm2N.

Then we pickn :=m+ 12N.

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By finding a winning strategy for the new game, we have proved the statement

8m2N9n2N : m<n

which is the negation of the original statement (5).

As the example illustrates, we can find the negation (¬E) of a

mathematical statementE in a mechanical way as follows: We change all

9 into 8and vice-versa and we negate the final expression after the colon.

Example

Consider the expression

9q 2N 8">0 8x 2R,x >0 9p2N: p

q x ".

Its negation is:

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