Lecture for September 13

7/29/2019 Lecture for September 13

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Introduction to Real Analysis MATH 2001

Juris Steprans

York University

September 13, 2012

Juris Steprans MATH 2001
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Archimedes and the method of exhaustion

Archimedes set himself the task of calculating the area between aparabola 1 x2 and the x axis. Using the methods of calculus thisis no more difficult than calculating the integral

11

1 x2dx

However, these methods were not available to Archimedes and heobtained his answer by considering successive families of triangles

contained in the region whose area he wanted to calculate.

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Archimedes and the method of exhaustion

At each stage n the number of new triangles added to theapproximation is 2n.

Each of those triangles has area 1/8n.

Hence the sum of the areas of the new triangles is 1/4n.

Therefore, at stage n the sum of the areas of all the trianglesis

1 +1

4

+1

42

+ . . . +1

4n

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Archimedesclever idea

Archimedes realized that:

1 = 43 1

3

1 + 14

= 43 1

43

1 + 14

+ 142

= 43

1423

1 +

1

4 +

1

42 + . . .

1

4n

=

4

3

1

4n

3

and this last equation can easily be proved by induction on n bynoting that adding 1

4n+1to both sides of the last equation yields:

1 +1

4+

1

42+ . . .

1

4n+

1

4n+1=

4

3

1

4n 3+

1

4n+1=

4n+2 4 + 3

4n+1 3

=4n+2 1

4n+1 3=

4

3

1

4n+1 3

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Archimedes reasoning

Archimedes was, justifiably, suspicious of infinite sums. Hence hereasoned by showing that

The area bounded by the parabola could not be less than 4/3.This is easy using the sum of of the areas of the trianglesinscribed in the area. (Exercise 2.1)

The area bounded by the parabola could not be greater than4/3. This is more subtle and uses that each new generation oftriangles reduces the remaining area by at least half.(Exercises 2.2 and 2.3)

Hence, there is no other possibility than that the area is equalto 4/3. In other words, all other possbilities have beenexhausted.

In this way Archimedes avoided resorting to infinite sums.

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Archimedes understanding of infinite sums

From the preceding argument it is possible to distill theArchimedean notion of an infinite sum it is not too far from themodern understanding.

Definition

An infinite sum has value T if and only if for all L < T andM> T there is a number n so large that any partial sum with atleast n terms falls between L and M.

What do we mean by a partial sum though? Can we take any n

terms or do they have be taken in the order given to us?Notice that the definition does not say that the more terms wetake the closer we get to T.

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Associativity fails for infinite sums

1 1 + 1 1 + 1 . . . = (1 1) + (1 1) + . . . = 0

On the other hand,

1 1 + 1 1 + 1 . . . = 1 (1 1) (1 1) + . . . = 1

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Associativity fails for infinite sums

Less trivially,

1 1

2+

1

3

1

4+

1

5

1

6+

1

7. . . 0.7

while rearranging yields

1 +1

3

1

2+

1

5+

1

7

1

4+

1

8+

1

9

1

6+ . . . 1.04

justifying Archimedes caution about using infinite series.

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Hausdorff-Banach-Tarski paradox

A further reason for Archimedes to have worried about usinginfinite sums is that, even if the arithmetic works out one should

wonder about adding up infinitely many areas.For example, it is possible to decompose a ball of radius one into30 pieces that can then be rearranged (no stretching or rescalingallowed) into two balls of radius one!

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Definition of convergence

A simple reformulation of the Archimedean understanding ofconvergence of series is the following:

Definition

A series converges if there is a target value T such that for anyL < T and any M> T, all of the partial sums from some point onare between L and M.

We will eventually arrive at a more precise definition than this for

convergence, but for the moment we will look at a key example,the geometric series.

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Even before a precise definition of the convergence of series wasformulated it was known that the geometric series1 + x + x2 + x3 + . . . converges and the value is given by

1 + x + x2 + + xm + . . . =1

1 x

With this formula it is easy to compute, for example

1 + 1/3 + 1/9 + + 1/3m + . . . =1

1 1/3= 3/2

or, with x = 1/2

1 1/2 + 1/4 1/8 + + (1/2)m + . . . = 11 + 1/2

= 2/3

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However, one has to be careful with this formula. For exampleusing x = 2 yields

1 + 2 + 4 + 8 + 2m + . . . =1

1

2

= 1

which certainly contradicts our intuition about infinite sums. So anargument is needed to justify this formula.

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An attractive argument is the following:

1 = 1 x + x x2 + x2 xm + xm . . .

= (1 x)1 + (1 x)x + (1 x)x2 . . . + (1 x)xm . . .

= (1 x)(1 + x + x2 . . . + xm . . .)

and hence

11 x

= 1 + x + x2 . . . + xm + . . .

However, we have already seen that the use of the associative lawwith infinite sums can lead to problems. Recall

1 1 + 1 1 + 1 . . . = (1 1) + (1 1) + . . . = 0

1 1 + 1 1 + 1 . . . = 1 (1 1) (1 1) + . . . = 1

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But this argument can be made precise using an approach similarto that of Archimedes. For a finite number m

1 = 1 x + x x2 + x2 xm + xm

and hence

1 = (1 x)1 + (1 x)x + (1 x)x2 . . . + (1 x)xm1 + xm . . .

Therefore

1

1 x

xm

1 x

=1 xm

1 x

= 1 + x + x2 + . . . xm1

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Now recall our provisional definition.

Definition

A series converges if there is a target value T such that for any

L < T and any M> T, all of the partial sums from some point onare between L and M.

It must be shown that T = 11x

satisfies this definition for thegeometric series. So suppose that L < T < M. All that needs tobe done is to find an m so large that if n m then x

m

1x< T L

and xm

1x< M T. Why does this suffice? For what values of x

can this be done?

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Formulas for

How is the following expression for obtained?

4

= 1 13

+ 15 1

7+ 1

9 1

11+ 1

13. . .

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From the geometric series equation derived in the last lecture weknow that

11 + x2

= 1 x2 + x4 x6 + x8 . . .

for values of x such that 1 < x2 < 1. Recall also also that

dx1 + x2 = arctan(x) + Cand hence

x

0

dt

1 + t2

= arctan(x) arctan(0) = arctan(x)

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x

0

dt

1 + t2

= x

0

1 t2 + t4 t6 + t8 . . . dt

=

x

0

dt

x

0

t2dt +

x

0

t4dt

x

0

t6dt +

x

0

t8dt. . .

= xx3

3

+x5

5

x7

7

+ . . .

and hence

arctan(x) = xx3

3+

x5

5

x7

7+ . . .

at least for x such that 1 < x< 1. Recalling that tan(/4) = 1yields

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/4 = arctan(1) = 1 13

+15

17

+ . . .

but at least two questions arise:

Why is the interchange of integration and infinite summation

justified?Why is it justified to replace x with 1?

It turns out that the convergence of this series for /4 is quiteslow and Newton was able to find a faster convergence.

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The area of a quarter unit circle is expressed in the following

equation:

/4 =

10

1 x2dx

In order to evaluate this Newton obtained the following infinite

series for (1 + x)a

:

(1 + x)a = 1 + ax +a(a 1)

2x2 +

a(a 1)(a 2)

2 3x3 . . .

and this yields the following:

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/4 = 10

1 x2dx = 10

(1 x2)1/2dx =

10

1

1

2x2 +

12

( 12 1)

2x4

12

( 12 1)( 1

2 2)

2 3x6 . . .

dx =

10

dx

10

x2

2dx+

10

12

( 12 1)

2x4dx+

10

12

( 12 1)( 1

2 2)

2 3x6dx. . . =

1 1

2 3+

1

4 2! 5

3

8 3! 7+

3 5

16 4! 9. . .

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There are again questions to be answered in justifying thisargument, but also the equation

(1 + x)a = 1 + ax +a(a 1)

2x2 +

a(a 1)(a 2)

2 3x3 . . .

needs to by established. This will be done as part of a moregeneral theory of Taylor series. But before looking at this we willlook at one more example, the harmonic series.

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Recall again the geometric series equation:

1

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

and recall also that

dx

1 x= ln(1 x)

and hence

ln(1 x) =

1 + x + x2 + x3 + x4

dx =

x +x2

2+

x3

3+

x4

4+

x5

5+ . . .

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Replacing x by x yields

ln(1 + x) = xx2

2

+x3

3

x4

4

+x5

5

. . .

Even assuming the justification of interchanging integration andinfinite summation, this equation would only be valid for x suchthat 1 < x< 1. Nevertheless it is also true for x = 1 yielding

ln(2) = 1 12

+ 13

14

+ 15 . . .

Certainly we should not expect a sensible result evaluating this atx = 1 which purely formally yields

ln(0) = 1 +1

2+

1

3+

1

4+

1

5+ . . .

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The right hand side is the harmonic series which diverges. In theArchimedean understanding this means that for any integer Nthere is some m such that

1 +1

2+

1

3+

1

4+

1

5+ . . . +

1

k> N

for any k m. Lets prove this. Begin by observing

1 +1

2+

1

3+

1

4+

1

5+ . . . +

1

2n=

1+1

2+

1

3+

1

4+

1

5+

1

6+

1

7+

1

8+

1

9. . .+

1

16+

1

17+ . . . +

1

32+

1

33+. . .

1

2n

1 +1

2+ 2

1

4+ 4

1

8+ 8

1

16+ 16

1

32+ 32

1

64+ . . .+2n1

1

2n

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1 +1

2

+1

2

+1

2

+1

2

+ . . . +1

2 n + 1 terms

= 1 + n/2

Hence, given N let m be so large that 1 + m/2 > N. Hence if

k 2m then

1 +1

2+

1

3+

1

4+

1

5+ . . . +

1

k

1 +1

2 +1

3 +1

4 +1

5 + . . . +1

2m 1 + m/2 > N

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