Schinazi R.B. From Calculus to Analysis
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90 3 Power Series and Special Functions
Note that before this definition we could not make sense of 3
√
2
.Nowwehave
defined this as exp(
√
2ln3).
We now turn to additional properties of ln.
Properties of the logarithmic function
(a) ln(xy) =ln(x) +ln(y) for all x>0 and y>0.
(b) ln(a
x
) =x ln(a) for all x ∈R and a>0.
Property (a) is a direct consequence of the multiplicative property of exp. Given
x>0 and y>0, let
s =ln x and lny =t.
Then,
exp(s +t) =exp(s) exp(t) =xy.
Taking ln on both sides yields
ln
exp(s +t)
=ln(xy).
Thus,
s +t =ln(x) +ln(y) =ln(xy).
Property (a) is proved.
We now turn to (b). By the definition of a
x
,wehave
a
x
=exp(x lna).
Hence,
ln(a
x
) =ln
exp(x lna)
=x lna.
This proves (b).
The number e
We define the number e as e =exp(1). That is,
e =
∞
n=0
1
n!
.
Application 3.2 Evaluate the number e.
We start by evaluating e
−1
.Wehave
e
−1
=
1
exp(1)
=exp(−1) =
∞
n=0
(−1)
n
n!
.
3.4 Exponential and Logarithmic Functions 91
The preceding series satisfies the conditions of the alternating series theorem (the
sequence a
n
=1/n! is positive, decreasing, and converges to 0). Taking the first 10
and 11 terms, we get
9
n=0
(−1)
n
n!
=
16687
45360
and
10
n=0
(−1)
n
n!
=
16481
44800
.
Since this is an alternating series, we have the following bounds:
16687
45360
<e
−1
<
16481
44800
.
Thus,
44800
16481
<e<
45360
16687
.
In particular, we get the first five decimals of e: 2.71828.
The importance of the number e comes from the following property.
A new expression for the exponential function
We have, for any x real, that
exp(x) =e
x
.
Recall that for any a>0 we defined
a
x
=exp(x lna).
We use this for a =e to get
e
x
=exp(x lne).
Since exp(1) =e, we have that lne =1. Thus,
e
x
=exp(x lne) =exp(x),
which proves the formula.
Application 3.3 The number e is irrational.
We do a proof by contradiction. Assume that e is a rational p/q. Since e is strictly
between 2 and 3, it is not a natural, and q>1. We have
q!e =q!
p
q
=(q −1)!p,
92 3 Power Series and Special Functions
and so q!e is a natural number. By definition
e =
∞
n=0
1
n!
=
q
n=0
1
n!
+
∞
n=q+1
1
n!
.
Observe that
q!
q
n=0
1
n!
=q!+q!+q!/2 +q!/3!+···+q!/q!.
Every q!/n! for n =0, 1,...,q is a natural (why?). Hence, q!
q
n=0
1
n!
is natural as
well. On the other hand,
q!
∞
n=q+1
1
n!
=q!/(q +1)!+q!/(q +2)!+q!/(q +3)!+···.
We have
q!/(q +1)!=1/(q +1)<1/2 and q!/(q +2)!=1/(q +1)(q +2)<1/2
2
.
More generally,
q!/(q +k)!< 1/2
k
for k =1, 2,....
Hence,
q!
∞
n=q+1
1
n!
<
∞
k=1
1
2
k
=1.
In summary, q!e is a natural, q!
q
n=0
1
n!
is also a natural, but
q!
∞
n=q+1
1
n!
=q!e −q!
q
n=0
1
n!
is strictly between 0 and 1. The difference of two natural numbers must be an integer
and cannot be in (0, 1). We have a contradiction. The number e cannot be a rational.
Exercises
1. (a) Show that for any x>0 and any natural n,wehave
exp
√
x
>
x
n+1
(2n +2)!
.
(b) Prove that
lim
x→+∞
exp(
√
x)
x
n
=+∞.
(c) Generalize (b).
2. Use Lemma 3.1 to prove (iii).
3.4 Exponential and Logarithmic Functions 93
3. (a) Let s>0 be a rational. Show that if t>1, then
t
−1
<t
−1+s
.
(b) Integrate both sides of (a) for t between 1 and x>1 to get
lnx ≤
x
s
−1
s
.
(c) Compute the limit in (vii).
4. Let a>0 and define f(x)=a
x
for every x. Compute the derivative of f .
5. Prove Lemma 3.2 for negative rationals.
6. Let f be a continuous function on R. Assume that
lim
x→+∞
f(x)=+∞ and lim
x→−∞
f(x)=0.
We are going to show that for a given y>0, the equation f(x)=y has at least
one solution.
(a) Show that there is C<0 such that if x<C, then f(x)<y/2.
(b) Show that there is A>0 such that if x>A, then f(x)>2y.
(c) Show that the equation f(x)=y has at least one solution.
7. Find the first 12 decimals of the number e.
8. Show that the function f(x)=x cos(π x/2) is unbounded but does not tend to
infinity as x tends to positive infinity.
9. Take a real x.
(a) Show that
ln
(1 +x/n)
n
=x
ln(1 +x/n) −ln 1
x/n
.
(b) Show that
lim
n→∞
ln
(1 +x/n)
n
=x.
(c) Prove that
lim
n→∞
(1 +x/n)
n
=e
x
.
(d) Use (c) to find an estimate for e.
10. In this exercise we are going to show that the multiplicative property charac-
terizes exponential functions. That is, the only continuous functions having the
multiplicative property are the functions exp(ax) for a real a and the zero func-
tion. Assume that f is continuous with f(1) =0 and is such that for all reals x
and y,
f(x+y) =f(x)f(y).
(a) Show that f(1)>0.
(b) For every natural n, f(n)=f(1)
n
.
(c) For every natural n, f(1/n) =f(1)
1/n
.
(d) For every rational r, f(r)= f(1)
r
.
94 3 Power Series and Special Functions
(e) Show that there is a real a such that for all real x,
f(x)=exp(ax).
11. Assume that f is continuous, with f(1) = 0 and is such that for all reals x
and y
f(x+y) =f(x)f(y).
Show that f is identically 0. (Follow the pattern of the preceding exercise.)
12. Does the series
∞
n=1
n
n
n!
converge?
Chapter 4
Fifty Ways to Estimate the Number π
4.1 Power Series Expansions
The number π has been defined in a rather abstract form so far: we have shown
that sin and cos are periodic functions and that their period is defined to be 2π.In
this chapter we will use several methods to estimate the number π. Estimating π
is important for applications (it appears in all kinds of mathematical formulas in
geometry, physics, probability, and so on) and also from a theoretical point of view.
What is this number? More generally, what is a number? What is its exact value?
Why do we need approximations?
The first methods we will see rely on power series expansions.
Power series expansion for
1
1−x
1
1 −x
=
∞
n=0
x
n
for all x ∈(−1, 1).
We start with the algebraic formula
(1 −x)
1 +x +x
2
+···+x
n
=1 −x
n+1
.
Thus, for x =1,
1 +x +x
2
+···+x
n
=
1 −x
n+1
1 −x
. (4.1)
If |x|< 1, then we know that
lim
n→∞
x
n+1
=0.
For fixed x in (−1, 1),weletn go infinity in (4.1) to get
1
1 −x
=
∞
n=0
x
n
for |x|< 1.
R.B. Schinazi, From Calculus to Analysis,
DOI 10.1007/978-0-8176-8289-7_4, © Springer Science+Business Media, LLC 2012
95
96 4 Fifty Ways to Estimate the Number π
This proves the formula above.
The algebraic formula (4.1) can be used to derive a number of other useful power
series expansions as we will now see. We will need the following facts from Calcu-
lus regarding integrals and inequalities.
Inequalities and integrals
Assume that f and g are continuous on [a,b], a<b.
(S1) If f(x)≤g(x) for all x in [a,b], then
b
a
f(x)dx ≤
b
a
g(x)dx,
(S2) If f(x)≥0 for all x in [a,c] and b<c, then
b
a
f(x)dx ≤
c
a
f(x)dx,
(S3)
b
a
f(x)dx
≤
b
a
f(x)
dx,
(S4)
b
a
f(x)dx =−
a
b
f(x)dx.
Note in particular that if f is continuous and positive on [a,b], then by S1we
have that
b
a
f(x)dx≥
b
a
0 dx =0.
We now turn to our second power series expansion. We make the substitution
x =−y
2
into (4.1) to get
1 −y
2
+y
4
−···+(−1)
n
y
2n
=
1 −(−y
2
)
n+1
1 +y
2
.
On both sides of the equality we have continuous functions that we now integrate
between 0 and x>0. This yields
x
0
1 −y
2
+y
4
−···+(−1)
n
y
2n
dy =
x
0
1 −(−y
2
)
n+1
1 +y
2
dy.
Thus,
x −
x
3
3
+
x
5
5
−···+(−1)
n
x
2n+1
2n +1
=
x
0
1
1 +y
2
dy +(−1)
n+2
x
0
y
2n+2
1 +y
2
dy.
Recall that
x
0
1
1 +y
2
dy =arctan x.
4.1 Power Series Expansions 97
Hence,
x −
x
3
3
+
x
5
5
−···+(−1)
n
x
2n+1
2n +1
−arctan x
=
(−1)
n+2
x
0
y
2n+2
1 +y
2
dy
.
We need to show that for fixed x in [−1, 1], the right-hand side goes to 0 as n goes
to infinity.
First, fix x in [0, 1]. Since y
2n+2
≥0, we have
(−1)
n+2
x
0
y
2n+2
1 +y
2
dy
=
x
0
y
2n+2
1 +y
2
dy.
Using that 1 +y
2
≥1 for all y,
1
1 +y
2
≤1,
and so
y
2n+2
1 +y
2
≤y
2n+2
.
Since 0 ≤x ≤1 and the functions on both sides of the inequality are continuous,
x
0
y
2n+2
1 +y
2
dy ≤
x
0
y
2n+2
dy ≤
1
0
y
2n+2
dy =
1
2n +3
,
where we use (S1) for the first inequality and (S2) for the second.
We now turn to x in [−1, 0]:
(−1)
n+2
x
0
y
2n+2
1 +y
2
dy
=
x
0
y
2n+2
1 +y
2
dy
=
0
x
y
2n+2
1 +y
2
dy,
where the last equality comes from (S4). By (S2) we get
0
x
y
2n+2
1 +y
2
dy ≤
0
−1
y
2n+2
1 +y
2
dy ≤
0
−1
y
2n+2
=
1
2n +3
.
Therefore, we have for any x in [−1, 1],
x −
x
3
3
+
x
5
5
−···+(−1)
n
x
2n+1
2n +1
−arctan x
≤
1
2n +3
.
For fixed x in [−1, 1],weletn go to infinity, and we get the following power series
expansion.
Power series expansion for arctan
arctanx =
∞
n=0
(−1)
n
x
2n+1
2n +1
for all x ∈[−1, 1].
Note that, unlike what happens for
1
1−x
, the power series expansion for arctan
converges at the end points.
98 4 Fifty Ways to Estimate the Number π
Application 4.1 We use the formula above to estimate π . Letting x =1 and using
that arctan(1) =π/4, we get
π/4 =
∞
n=0
(−1)
n
1
2n +1
.
We sum up to n =1000 to get the approximation 3.142591654 for π . Only the first
two decimals are right! This is due to the fact that the series converges rather slowly.
As seen above, the error in summing up to n is less than 1/(2n +3). In the present
case the error is less than 1/2003.
The same power series expansion converges much faster for an x<1 as we will
see next.
Application 4.2 Recall that
tan(π/6) =
1
√
3
.
Thus, arctan(
1
√
3
) = π/6. The power series expansion of arctan at x =
1
√
3
is going
to converge much faster than at 1 because
1
√
3
< 1. We have
arctan
1
√
3
=
∞
n=0
(−1)
n
(
1
√
3
)
2n+1
2n +1
.
Since
1
√
3
2n+1
=
1
√
3
1
3
n
,
we have
arctan
1
√
3
=
1
√
3
∞
n=0
(−1)
n
1
3
n
(2n +1)
.
Note that the alternating series theorem applies to
∞
n=0
(−1)
n
1
3
n
(2n+1)
(why?).
Thus,
9
n=0
(−1)
n
1
3
n
(2n +1)
≤
∞
n=0
(−1)
n
1
3
n
(2n +1)
≤
10
n=0
(−1)
n
1
3
n
(2n +1)
.
This yields the numerical bounds
0.90689906 ≤
∞
n=0
(−1)
n
1
3
n
(2n +1)
≤0.90689987.
We also have the following bounds for
1
√
3
:
0.57735026 ≤
1
√
3
≤0.57735027.
4.1 Power Series Expansions 99
Thus,
0.5235984081 ≤
1
√
3
∞
n=0
(−1)
n
1
3
n
(2n +1)
≤0.5235988848.
The infinite series above is π/6, and multiplying by 6 both bounds yields the fol-
lowing estimate for π :
3.141590449 ≤π ≤3.141592654.
That is, we get the first 5 decimals of π by summing only 11 terms!
We will now find a power series expansion for ln(1−x). Our starting point again
is (4.1):
1 +x +x
2
+···+x
n
=
1 −x
n+1
1 −x
for x =1.
We integrate between 0 and y to get
y
0
1 +x +x
2
+···+x
n
dx =
y
0
1
1 −x
dx −
y
0
x
n+1
1 −x
dx.
Therefore,
n
k=0
y
k+1
k +1
=−ln(1 −y) −
y
0
x
n+1
1 −x
dx.
We now bound the remainder. First assume that y is in [0,1). Then
1
1 −x
≤
1
1 −y
for x ∈[0,y].
Thus,
−
y
0
x
n+1
1 −x
dx
=
y
0
x
n+1
1 −x
dx ≤
1
1 −y
y
0
x
n+1
dx =
1
1 −y
y
n+2
n +2
.
Since for fixed y in [0, 1),
1
1−y
y
n+2
n+2
converges to 0 as n goes to infinity, we get the
following power series expansion for y in [0, 1):
ln(1 −y) =−
∞
k=0
y
k+1
k +1
.
We now take care of negative y. Assume that y is in [−1, 0]. By (S4) and (S3) we
have
−
y
0
x
n+1
1 −x
dx
=
0
y
x
n+1
1 −x
dx
≤
0
y
|x|
n+1
1 −x
dx.
Since x<0, we have |x|=−x and
0
y
|x|
n+1
1 −x
dx =
0
y
(−x)
n+1
1 −x
dx ≤(−1)
n+1
0
y
x
n+1
dx,