May 2, 2011
18.01 Problem Set 11 solutions
Part II: 15 points
1. (10 points) This problem is about the functions xm e−x , with m a non-negative integer.
1a) Calculate the average value A0 of e−x over the interval [0, 1].
By definition this is
1
Z
e−x dx = [−e−x ]10 = −e−1 + e0 = 1 − e−1 ≈ .632
0
b) Calculate the average value A1 of xe−x over the interval [0, 1].
R1
By definition this is 0 xe−x dx. This is a good candidate for integration by parts, with u = x
(since the derivative of x is simpler than x) and dv = e−x dx (since that’s easy to integrate). This gives
du = dx, v = −e−x , so
1
Z
−x
xe
dx =
− xe
−x
1
Z
+
0
1
e
−x
dx = −e
−1
Z
0
0
1
+
e−x dx.
0
The last integral we calculated in (a) was A0 = 1 − e−1 , so the average value is
A1 = −e−1 + A0 = 1 − 2e−1 ≈ .264.
c) Calculate the average value A2 of x2 e−x over the interval [0, 1].
R1
By definition this is 0 x2 e−x dx. Again we can integrate by parts, with u = x2 (since the derivative
of x2 is simpler than x2 ) and dv = e−x dx. We find du = 2xdx, v = −e−x , and
Z
1
2 −x
x e
2 −x
1
−x e
dx =
0
Z
1
+2
e−x dx = −e−1 + 2A1 .
0
0
Plugging in the value of A1 from (b) gives
A2 = 2 − 5e−1 ≈ .1606.
d) Prove a reduction formula of the form
Z
Z
m −x
m −x
x e dx = Cm x e + Dm xm−1 e−x dx.
Integrate by parts: use u = xm , dv = e−x dx, so that du = mxm−1 and v = −e−x . The formula is
Z
Z
m −x
m −x
x e dx = −x e + m xm−1 e−x dx.
e) Explain how to calculate the average value Am of xm e−x over [0, 1] from Am−1 .
Applying the reduction formula gives
Z
1
m −x
x e
0
m −x
1
dx = −x e
Z
+m
0
1
x
0
m−1 −x
e
dx = −e
−1
Z
+m
0
1
xm−1 e−x dx.
In terms of the average values we are interested in, this says
Am = mAm−1 − e−1 .
f ) Show that there are integers am and bm with the property that
bm
.
Am = am −
e
Explain how to calculate am and bm from am−1 and bm−1 .
Certainly A0 and A1 and A2 are shaped like this. We proceed by induction: suppose that Am−1 =
am−1 − bm−1
e , with am−1 and bm−1 (positive) integers. According to (e),
Am = mAm−1 −
1
mbm−1 + 1
= mam−1 −
.
e
e
This answer has the shape that we want, with
am = mam−1 ,
bm = mbm−1 + 1.
That’s all you needed to say. In fact you can easily calculate am and bm using these formulas and the
starting conditions (from (a))
a0 = 1,
b0 = 1.
The answers are
am = m!,
bm = m! + m(m − 1) · · · 2 + m(m − 1) · · · 3 + · · · m(m − 1) + m + 1.
1
(m+1)
g) Explain why Am is between
and
1
e(m+1) .
What makes this question hard is thinking that it has something to do with parts (a)–(f). On the
interval from 0 to 1, e−x decreases from 1 to e−1 . The function xm e−x is therefore always between xm
and xm /e. The average values therefore satisfy
(average value of xm ) ≥ Am ≥ (average value of xm /e).
The first average value is
Z
0
and the last is
1
xm+1
x dx =
m+1
m
1
=
0
1
,
m+1
1
e(m+1) .
1
1
1
1
e
1
+ 2!
+ 3!
+ · · · + m!
) is between (m+1)!
and (m+1)!
. (This
2. (5 points) Explain why e − (1 + 1!
means, for instance, that the error in the approximation
1
1
1
1
e ≈ 1 + + + + ··· +
1! 2! 3!
15!
is at most e/15! ≈ 2 × 10−12 .) Parts (f) and (g) of Problem 1 (together with the formulas written
there for am and bm ) say that
1
m! + m(m − 1) · · · 2 + m(m − 1) · · · 3 + · · · m(m − 1) + m + 1
1
≤ m! −
≤
.
e(m + 1)
e
m+1
Multiplying this by e/m! gives
1
1
1
1
1
e
≤ e − (1 + + + · · · +
+
) ≤
.
(m + 1)!
1! 2!
(m − 1)! m!
(m + 1)!
Actually the first formula is interesting if you just multiply it by e: it says
1
e
≤ e · m! − [m! + m(m − 1) · · · 2 + m(m − 1) · · · 3 + · · · m(m − 1) + m + 1] ≤
.
m+1
m+1
The expression in square brackets is an integer, so these inequalities (for m + 1 > e tell you that e · m!
is not an integer: the reason is that it differs from an integer by something positive but less than 1.
Since e · m! can never be an integer (for any big m), it follows that e must be an irrational number.