Schinazi R.B. From Calculus to Analysis

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130 4 Fifty Ways to Estimate the Number π

(a) Show there is a natural N such that if n ≥N , then

|< 1/2.

(b) Use Exercise 4(b) to show that for n ≥N ,wehave



ln(1 +a

) −a



∞



n=1



ln(1 +a

) −a



converges.

(d) Let



i=1

(1 +a

) and S



i=1

Use (c) to show that the sequence ln p

−S

converges.

(e) Conclude that



∞

n=1

(1 + a

) converges to a nonzero limit if and only if



∞

n=1

converges.

6. Apply the result in Exercise 5 to show that

∞



n=1



1 +

(−1)



converges.

7. In this exercise we describe another inﬁnite product that may be used to esti-

mate π . This method is due to Viète. Let α =0.

(a) Show that for every natural n,wehave

sin α

=cos

cos

···cos

sin(

)

(Use that sin(2x) =2 cos x sin x.)

(b) Show that

lim

n→∞

sin(

)

=cos0.

=cos

cos

···cos

Show that q

converges to

sin α

(d) Let a

=cos(π/2

n+1

). Show that

lim

n→∞



i=1

(Use (c) for a suitable value of α.)

4.4 The Number π Is Irrational 131

(e) Show that a

√

2/2 and that, for n ≥1,

n+1



(1 +a

(Prove and use that cos

x =

1+cos(2x)

(f) Use (d) and (e) to estimate π.

4.4 The Number π Is Irrational

All the methods we have seen so far to estimate π involve inﬁnity: inﬁnite products

and inﬁnite series. Is it possible to avoid that? If π were a rational, for instance, it

would be a fraction. No need for inﬁnite series or inﬁnite products! However, we

will show that π is not a rational. Worse than that, it is known that π is not even

an algebraic number. That is, there is no polynomial with integer coefﬁcients that

has π as a root. Recall that

√

2 is not a rational. However, it is algebraic: it is a

solution of x

− 2 = 0. This allows simple methods to estimate

√

2: 1.4

= 1.96

and 1.5

=2.25, and thus,

1.4 <

√

2 < 1.5.

Since (1.45)

> 2, we have

1.4 <

√

2 < 1.45,

and so on. We may get an arbitrarily precise estimate for

√

2 just using algebraic

methods (additions and multiplications). A number, such as π, which is not alge-

braic is called transcendental (the number e is another example). As Euler said “it

transcends the power of algebraic methods”. One cannot avoid inﬁnity to estimate

π or e.

To prove that π is transcendental is rather involved. We will prove the weaker

result that

The number π is irrational.

This already requires several steps. We follow the proof of Ivan Niven (1947)

(Bulletin of the American Mathematical Society, 509). Assume, by contradiction,

that π is a rational a/b where a and b are natural numbers. Pick a natural n such

that

n+1

< 1.

(Why does such an n exist?) Let f

be the polynomial deﬁned by

(x) =

(a −bx)

132 4 Fifty Ways to Estimate the Number π

Hence, n!f

(x) = x

(a − bx)

. Since a and b are integers, if we expand x

(a −

bx)

, all the coefﬁcients are going to be integers. Moreover, all the powers in the

expansion are going to be between n and 2n. More precisely, there are integers

n+1

,...,c

such that

n!f

(x) =



k=n

At this point we need the following:

Lemma 4.3 Assume that P is a polynomial of degree K. That is, there are K +1

real numbers c

such that

P(x)=



k=0

Let the j th derivative of P be P

(j)

. Then,

(j)

(0) =j!c

for 1 ≤j ≤K,

and

(j)

(x) =0 for j>K.

We prove Lemma 4.3.Wehave

P(x)=c

x +c

+...+c

+···+c

We differentiate to get



(x) =P

(1)

(x) =c

+2c

x +···+kc

k−1

+···+Kc

K−1

We differentiate again to get

(2)

(x) =2c

+···+k(k −1)c

k−2

+···+K(K −1)c

K−2

At the j th differentiation we get

(j)

(x) =j!c

+(j +1)!c

j+1

x +···+···+K(K −1) ···(K −j +1)c

K−j

for j ≤K. In particular, letting x =0 above yields

(j)

(0) =j!c

for 1 ≤j ≤K.

Since at each differentiation the polynomial loses one degree, at the Kth differenti-

ation P

(K)

has degree 0 (it is the constant K!c

). Since the derivative of a constant

is 0, we get

(j)

=0forj>K.

This completes the proof of Lemma 4.3.

We apply Lemma 4.3 to f

to get:

Lemma 4.4 We have that f

(j)

(0) and f

(j)

(π) are integers for all j ≥1.

4.4 The Number π Is Irrational 133

We now prove Lemma 4.4. Applying Lemma 4.3 to the polynomial n!f

, we get

n!f

(j)

(0) =j!c

for j =1,...,2n since the degree of f

is 2n. Since c

=0forj<n, we have that

(j)

(0) =0for1≤j<n.

Note that if j ≥n, then j!/n! is an integer, and so

(j)

(0) =

are all integers for j =n,...,2n.Forj outside this range, f

(j)

(0) is 0. In summary,

we have shown that f

(j)

(0) is an integer for every j ≥1.

Moreover,

(π −x) =f

(a/b −x) =

(a/b −x)

(a −b(a/b −x))

(a/b −x)

(bx)

(x).

That is, f

is symmetric with respect to π. By the chain rule we have, for all deriva-

tives,

(k)

(x) =(−1)

(k)

(π −x).

In particular, letting x =0, we get

(k)

(0) =(−1)

(k)

(π).

Therefore, f

(k)

(π) are also integers for any k ≥ 1. This completes the proof of

Lemma 4.4.

We deﬁne the function g

−f

(2)

(4)

−···+(−1)

(2n)

We have, by the product rule,





(x) sin x −g

(x) cos x







(x) sin x +g



(x) cos x −g



(x) cos x +g

(x) sin x



(x) sin x +g

(x) sin x.

Note that





−f

(2)

(4)

−···+(−1)

(2n)





(2)

−f

(4)

+···+(−1)

(2n)

+(−1)

(2n+2)



There is cancellation of all terms except for the ﬁrst and the last:



+(−1)

(2n+2)

But f

is a polynomial with degree 2n, and therefore, f

(2n+2)

=0 and



134 4 Fifty Ways to Estimate the Number π

This yields





(x) sin x −g

(x) cos x









(x) +g

(x)



sin x =f

(x) sin x.

By the fundamental theorem of Calculus,



(x) sin xdx=g



(x) sin x −g

(x) cos x



(π) +g

(0).

Since f

(k)

(0) and f

(k)

(π) are integers for all k, g

(π) and g

(0) are also integers,

and so is



(x) sin xdx. It is easy to check that f

and sin are positive on [0,π].

Moreover, x

≤π

and (a −bx)

≤a

for x in [0,π]. Thus,

(x) <

for x ∈[0,π].

Therefore,

0 ≤f

(x) sin x<

for x ∈[0,π].

We now integrate between 0 and π to get

0 ≤



sinxf

(x) dx ≤

n+1

< 1

by our initial choice of n. Thus,



sinxf

(x) dx is a positive integer strictly less

than 1. It must be 0. As we will now see, this is impossible and provides a contra-

diction. Since 0 ≤f

(x) sin x for x in [0,π],wehave



(x) sin xdx≥



3π/4

π/4

(x) sin xdx.

The minimum of sin on [π/4, 3π/4] is

√

2/2. It is easy to check that f

increases

on [0,

n+1

π]and decreases on [

n+1

π,π]. Thus, the minimum of f

on [π/4, 3π/4]

is at π/4 (since

n+1

π>π/4 for all naturals n). Therefore,



(x) sin xdx≥



3π/4

π/4

(x) sin xdx≥f

(π/4)

√

> 0,

where the last inequality comes from the fact that f

is 0 only at 0 and π. Thus,

we have our contradiction:



sinxf

(x) dx is 0 and is not 0. Our initial assumption

that π is a rational cannot be true. This completes the proof that π is irrational.

Exercises

1. Show that for any real x, there is a natural n such that

< 1.

2. (a) Show that the maximum of f

occurs at

n+1

π.

(b) Show that f

is 0 only for x =0 and x =π.

3. Show that a rational number is algebraic but that the converse is not true.

4.4 The Number π Is Irrational 135

4. Show that if x

is irrational, so is x. Is the converse true?

5. In this exercise we show that exp(r) is irrational for every rational r =0. We ﬁrst

show that exp(k) is irrational for every natural k.Let

(x) =

(1 −x)

Let k be a ﬁxed natural number and deﬁne

(x) =k

(x) −k

2n−1



(x) +k

2n−2

(2)

(x) −···+h

(2n)

(x).

(a) Show that



exp(kx)H

(x)



2n+1

exp(kx)h

(x).

(b) Integrate the equality above for x between 0 and 1 to get

exp(k)H

(1) −H

(0) =k

2n+1



exp(kx)h

(x) dx.

naturals. Show that

a exp(k)H

(1) −bH

(0) =bk

2n+1



exp(kx)h

(x) dx.

(d) Show that there is N such that for all n ≥N and all x in [0, 1],wehave

2n+1

exp(kx)h

(x) =bk

2n+1

exp(kx)

(1 −x)

< 1/2.

(e) Explain why a exp(k)H

(1) −bH

(0) is a positive integer.

(f) Use (d) and (e) to ﬁnd a contradiction. Thus, exp(k) is not a rational.

(g) Show that exp(r) is not a rational if r =0 is a rational.

6. Show that if r =1 is a rational, then ln r is irrational. Use Exercise 5.

Chapter 5

Continuity, Limits, and Differentiation

5.1 Continuity

We start with a deﬁnition.

Continuous functions

Assume that the function f is deﬁned on some set D.Leta be in D. f is said

to be continuous at a if for any sequence a

in D that converges to a,wehave

that f(a

) converges to f(a).

Note that for ANY a

converging to a,weMUSThavef(a

) converging to f(a)

in order for f to be continuous at a.

Example 5.1 The function f(x) = x

is continuous everywhere on the reals. To

prove that, take a a real number. Let a

be any sequence of reals converging to a.

Since the product of two converging sequences converges to the product of the lim-

its, we have that f(a

) =a

= a

converges to a

.Buta

is also f(a). Hence,

f is continuous at a, and we are done.

In order to show that a function is not continuous at a, it is enough to ﬁnd ONE

sequence a

in D with the following properties: a

converges to a,butf(a

) does

not converge to f(a). We do such an example next.

Example 5.2 Consider the function g deﬁned on D =[−1, 1] by

g(x) =|x|/x for x = 0 and g(0) =1.

Hence, g(x) =−1forx in [−1, 0) and g(x) = 1forx in [0, 1]. Intuitively, this

function has a jump at 0 and so cannot be continuous at 0. See the graph below. We

now prove this. Let a

=−1/n. This sequence is in D and converges to 0. Since a

is in [−1, 0] for all n ≥1, we have g(a

) =−1. Hence, g(a

) converges to −1. But

g(0) =1. Therefore, g is not continuous at 0. See the graph of g in Fig. 5.1.

R.B. Schinazi, From Calculus to Analysis,

137

138 5 Continuity, Limits, and Differentiation

Fig. 5.1 This is the graph of g(x) =|x|/x

Example 5.3 The absolute value function is continuous on the reals.

Let a be a real, and a

a sequence of reals converging to a. Recall that, as a

consequence of the triangle inequality, we have for all reals x and y:

0 ≤



|x|−|y|



≤|x −y|.

Hence,

0 ≤



|−|a|



≤|a

−a|.

By the squeezing principle, ||a

|−|a|| converges to 0. Hence, |a

|−|a| con-

verges to 0, and therefore |a

| converges to |a| (why?). Thus, the absolute value

function is continuous at every a.

We will now deﬁne the usual operations on functions. Given that f and g are

deﬁned at a, we have that the function f +g is deﬁned as

(f +g)(a) =f(a)+g(a),

we deﬁne fg as

(fg)(a) = f (a)g(a),

and if g(a) =0, we deﬁne f/g as

(f/g)(a) =f (a)/g(a).

The following properties allow us to construct many continuous functions from

basic ones.

5.1 Continuity 139

Operations on continuous functions

Let the functions f and g be deﬁned on D and be continuous at a in D. Then:

f +g is continuous at a;

fg is continuous at a;

if g(a) =0, then f/g is continuous at a.

The proofs are direct consequences of the operations on limits that we recall

below.

Operations on limits

Assume that the sequences a

and b

converge, respectively, to a and b.

(i) lim

n→∞

) =a +b.

(ii) lim

n→∞

=ab.

(iii) Assume that b

is never 0 and that b is not 0. Then

lim

n→∞

=a/b.

(iv) Assume that a

≤b

for all n. Then a ≤b.

We now prove that if f and g are continuous at a,soaref +g and fg.Takeany

sequence a

in I that converges to a. We know that

lim

n→∞

f(a

) =f(a) and lim

n→∞

g(a

) =g(a).

Hence, by the operations on limits,

lim

n→∞



f(a

) +g(a

)



=f(a)+g(a) and lim

n→∞



f(a

)g(a

)



=f (a)g(a).

For the ratio property, we ﬁrst need to show that g(a

) is never 0 for n large

enough since g(a

) converges to g(a) = 0. Assume that g(a) < 0. Then pick  =

−g(a)/2 > 0. There is N such that if n ≥N , then



g(a

) −g(a)



<=−g(a)/2.

Hence,

g(a

)<g(a)−g(a)/2 =g(a)/2 < 0 for all n ≥N.

That is, g(a

) remains negative for all n ≥N . Similarly, if we assume that g(a) > 0,

we can show that there is a natural N

such that g(a

)>0 for all n ≥N

. This allows

us to deﬁne the sequence f(a

)/g(a

) for n ≥N , and by (iii) we have

lim

n→∞

f(a

)/g(a

) =f (a)/g(a).