Download On the number e, its irrationality, and factorials

CALCULUS II
On the number e
February 2, 2015
This is our first application of integration by parts. By computing certain integals we will
• find a formula for e;
• prove that e is an irrational number;
• discover the factorial of numbers which are not integers.
1
Some integrals
R
Our first example of integrating by parts was xex dx. Now we will consider the integrals
Z 1
Z 1
Z 1
2 −x
−x
x3 e−x dx, . . . .
x e dx,
xe dx,
0
0
0
We will compute all of these integrals, and then in the next section we will compute their limit
as the power of x goes to ∞. Amazingly, this will give us a formula for e.
To compute all of these integrals at once, we take a general polynomial f = f (x), and compute
the indefinite integral
Z
f e−x dx
using integration by parts over and over again: Let u = f , so v 0 = e−x . Then u0 = f 0 and
v = −e−x . We get
Z
−x
fe
−x
dx = −f e
Z
+
f 0 e−x dx.
(1)
Formula (1) holds for any polynomial, so it also holds for f 0 . Replacing f by f 0 in (1), we get
Z
Z
0 −x
0 −x
f e dx = −f e + f 00 e−x dx.
R
Putting this expression for f 0 e−x dx back into (1), we get
Z
Z
−x
−x
0 −x
f e dx = −f e − f e + f 00 e−x dx.
Repeat: replacing f by f 00 in (1), we get
Z
Z
00 −x
00 −x
f e dx = −f e + f 000 e−x dx.
1
(2)
R
Putting this expression for f 00 e−x dx back into (1), we get
Z
Z
−x
−x
0 −x
00 −x
f e dx = −f e − f e − f e + f 000 e−x dx.
(3)
We keep going like this, getting higher and higher derivatives inside the integral. But remember
that f is a polynomial, say of degree n. So f (n+1) = 0. This means that when the derivative
inside the integral becomes f (n+1) , the integral vanishes, and we get
Z
f e−x dx = −f e−x − f 0 e−x − f 00 e−x − · · · − f (n) e−x .
Thus we have found our basic integration formula, for a polynomial f of degree n:
Z
f e−x dx = −(f + f 0 + f 00 + · · · + f (n) )e−x
(4)
Isn’t it remarkable that you can compute an antiderivative by differentiating?
2
A formula for e
We now use formula (4) with f (x) = xn . We have
f = xn ,
f 0 = nxn−1 ,
f 00 = n(n − 1)xn−2 ,
...
f n−1 = n(n − 1) · · · 2 · x,
f (n) = n!.
It will be convenient to factor n! out of each of these derivatives. Thus, we get
f = n!
xn
,
n!
f 0 = n!
xn−1
,
(n − 1)!
f 00 = n!
xn−2
,
(n − 2)!
...
x
f n−1 = n! ,
1!
f (n) = n!.
Putting these into our formula (4) and factoring out n!, we get
Z
n −x
x e
dx = −n! ·
xn
xn−1
xn−2
x
+
+
+ · · · + + 1 e−x
n! (n − 1)! (n − 2)!
1!
(5)
Now move n! over to the left side, and evaluate the integral from 0 to 1. This gives
Z
1 1 n −x
1
1
1
1
−1
x e dx = 1 − e
+
+
+ ··· + + 1 .
n! 0
n! (n − 1)! (n − 2)!
1!
Finally, multiplying through by e, we get our integral computation in the form we need:
e
n!
Z
1
n −x
x e
0
dx = e −
1
1
1
1
+
+
+ ··· + + 1
n! (n − 1)! (n − 2)!
1!
(6)
Thus, the integral is the difference between e and a certain rational number depending on n. We
will show this difference goes to zero as n goes to infinity.
2
For x ≥ 0 we have e−x ≤ 1, so 0 ≤ xn e−x ≤ xn . In general, if one function is less than another
function on [a, b], then the integral of the first is less than the integal of the second. Therefore,
Z 1
Z 1
1
n −x
xn dx =
x e dx ≤
0≤
.
(7)
n+1
0
0
The Squeeze Law implies that
Z
1
xn e−x dx → 0
as
n → ∞.
0
Here are the graphs of xn e−x for n = 1, 2, 3, 4.
y
n=4
n=3
n=2
n=1
x
1
lim
R1
n→∞ 0
Multiplying by
e
n!
xn e−x dx = 0
makes it go to zero even faster. From (6) it follows that
1
1
1
1
lim e −
+
+
+ · · · + + 1 = 0.
n→∞
n! (n − 1)! (n − 2)!
1!
Thus we get our formula for e:
1
1
1
e = lim 1 + + + · · · +
n→∞
1! 2!
n!
This means that e is approximated by the rational numbers
1
=2
1!
1
1
1 + + = 2.5
1! 2!
1
1
1
1 + + + = 2.6666 . . .
1! 2! 3!
..
.
1+
1+
1
1
1
1
1
1
1
1
1
1
+ + + + + + + + +
= 2.7182818011 . . .
1! 2! 3! 4! 5! 6! 7! 8! 9! 10!
In fact,
e = 2.7182818284 . . .
3
(8)
itself is the smallest real number that is greater than all of the rational numbers
1+
3
1
1
1
+ + ··· + .
1! 2!
n!
Irrationality of e
A real number r is called rational if r = a/b, where a and b are integers. Equivalently, a
number is rational if its decimal expansion eventually repeats, and conversely all numbers whose
decimal expansion eventually repeats are rational. Real numbers which are not rational are
called irrational.
√
√
For example, 2 is irrational. To see this, suppose 2 = a/b for some integers a and b. Then
2 = a2 /b2 , so
2b2 = a2 .
But this equation is absurd, for 2 divides 2b2 an odd number of times, and 2 divides a2 an odd
√
number of times. Therefore there can be no integers a and b for which 2 = a/b.
All proofs that a number r is irrational follow this pattern of logic, called proof by contradiction:
To prove that r is irrational we assume r = a/b for some integers a and b and then show
(somehow) that this assumption implies something absurd. So the assumption that r is rational
must have been false, and therefore r is irrational.
Theorem 1. e is irrational.
Proof. Suppose e = a/b for some integers a and b. We will use b to produce an integer N such
that 0 < N < 1. This is absurd, so we will have proved that e is irrational.
To find N , we will use our approximation for e. First choose any integer n > b. Then let N be
n! times the difference between e and its nth approximation:
1
1
1
N = n! e − 1 + + + · · · +
.
1! 2!
n!
Since e is bigger than all of its approximations, we have N > 0. For each k ≤ n the term
n! ·
1
k!
= n(n − 1) · · · (n − k + 1) is an integer, and n! · e = n! ·
4
a
b
is an integer because b < n.
Therefore N is an integer. Finally,
1
1
1
N = n! e − 1 + + + · · · +
1! 2!
n!
1
1
1
+
+
···
= n!
(n + 1)! (n + 2)! (n + 3)!
n!
n!
n!
=
+
+
···
(n + 1)! (n + 2)! (n + 3)!
1
1
1
+
+
+ ···
=
n + 1 (n + 1)(n + 2) (n + 1)(n + 2)(n + 3)
1
1
1
<
+
+
+ ···
n + 1 (n + 1)(n + 1) (n + 1)(n + 1)(n + 1)
1
1
1
=
+
+
+ ···
n + 1 (n + 1)2 (n + 1)3
=
1
n+1
1
1 − n+1
1
= < 1,
n
where in the next-to-last line we used the geometric expansion
r + r2 + r3 + · · · =
r
,
1−r
for r < 1.
To summarize: the assumption that e is rational implies the existence the above number N
which is an integer, yet satisfies the inequalities 0 < N < 1. This is absurd and therefore e is
irrational. QED.
4
Some Pictures
We have been working with integrals of the functions xn e−x on the interval [0, 1]. Let us pause
to look at the graphs of these functions on the whole real line. The first derivative
(xn e−x )0 = nxn−1 e−x − xn e−x = xn−1 (n − x)e−x
is zero when x = n, and this is the point where xn e−x reaches its maximum. This maximum
value
nn e−n =
is enormous for large n. However e−x
n n
e
eventually crushes xn , making xn e−x → 0 as x → ∞. Here
are some pictures of the functions xr e−x , including the case x1/2 e−x , which will be important
shortly. We replace n by r to indicate that the power of x need no longer be an integer.
5
y
r=4
r=3
r = 1/2
r=2
r=1
1
2
3
4
5
6
7
8
9
x
Graphs of xr e−x for r = 21 , 1, 2, 3, 4. The one shown for x4 e−x is wildly inaccurate beyond x = 8
because in that range the conflict between x4 (big) and e−x (small) confuses the computer.
5
The factorial function
In third grade we learn powers: 2n is 2 times itself n times. And 21/2 is the positive number
whose square is 2. So in third grade we more-or-less understand 2r , for r a rational number. But
√
one cannot understand 2
2
from this point of view. The general definition of powers is deeper:
2r = er log 2 ,
for any real number r.
That is, you compute the area under y = 1/x above [1, 2] and multiply by r to get r log 2. Then
2r is the number on the positive x-axis such that the area under y = 1/x from 1 to 2r is r log 2
(if r > 0) or −r log 2 (if r < 0). Most third graders would find this difficult to understand.
So it is with n!. Our third-grade definition is n! = 1 · 2 · · · n, the product of all the positive
integers up to n. Oh, and also 0! = 1, by Royal Decree. This last out-of-the-blue definition is a
clue that the true factorial lies deeper. Indeed, the factorial of any real number r ≥ 0 is defined
by an integral:
Z
∞
xr e−x dx
r! =
0
which is shorthand for
b
Z
xr e−x dx.
r! = lim
b→∞
0
Note that the advanced version of the power function xr appears in the integrand.
6
(9)
This advanced definition of r! agrees with the third-grade definition if r = n is a positive integer.
To see this, we use our integral formula (5), which I rewrite here:
n
Z
x
xn−1
xn−2
x
n −x
x e dx = −n! ·
+
+
+ · · · + + 1 e−x .
n! (n − 1)! (n − 2)!
1!
(Here we mean the third-grade n! = 1 · 2 · · · n.) Evaluating from 0 to a positive number b, we
have
Z
b
n −x
x e
0
b
xn
xn−1
xn−2
x
−x dx = −n! ·
+
+
+ ··· + + 1 e n! (n − 1)! (n − 2)!
1!
n
0
n−1
n−2
b
b
b
b
= n! − n!
+
+
+ · · · + + 1 e−b .
n! (n − 1)! (n − 2)!
1!
To get the integral from 0 to ∞ we take the limit as b → ∞. From Calculus I we recall that
exponential growth crushes polynomial growth:
lim bk e−b = 0
b→∞
for every integer k. So all the terms with b go to zero, and we get
Z ∞
xn e−x dx = n!
0
Thus, the advanced n! from (9) equals the third grade n!, so we haven’t contradicted anything
you learned before.
Here are pictures of the functions xn e−x for n = 1, 2, 3, 4 (continuations of the graphs shown in
section (1)
By this point it should be easy for you to check that
Z ∞
Z ∞
0 −x
x e dx =
x0 e−x dx = 1.
0
0
So the advanced definition (9) explains why 0! = 1, rather than just declaring it to be so.
More importantly, we can now take the the factorial of non-integers! For example,
Z ∞
√ −x
1
!=
xe dx.
2
0
Let’s try to compute this integral using the tools we have now. Making the substitution w =
leads to
Z
∞
√
−x
xe
Z
∞
2
w2 e−w dw,
dx = 2
0
0
2
and then integration by parts with u = w and v 0 = we−w leads to
Z ∞
Z ∞
2
2 −w2
2
w e
dw =
e−w dw.
0
0
7
√
x
Changing the dummy variable back to x, we see that
Z ∞
1
2
e−x dx,
!=
2
0
which is the infamous integral that we’re pretty sure we cannot compute. In fact we will be
able to compute this integral, using the Wallis formula for π. (So far there is no indication that
(1/2)! has anything to do with π.) But we have to learn a few other things first. For now we
just note that (1/2)! is the area under half of the bell-curve:
y
Area is
1
2
!
y = e−x
2
x
8