ECE/CS 313: Probability with Engineering Applications Fall 2002
Problem Set 1 Solutions
1. Geometric series
(a) Since x6= 1, the equation to be proved is equivalent to the equation we get by multiplying both sides by
(1 −x). So it is sufficient to prove that (1 −x)(1 + x+x2+...+xn−1) = 1 −xn. But
(1 −x)(1 + x+x2+...+xn−1) = (1 + x+x2+...+xn−1)−x(1 + x+x2+...+xn−1)
= (1 + x+x2+...+xn−1)−(x+x2+x3...+xn)
= 1 −xn,
which establishes the desired equation. (Another way to prove the equation is by induction on n.)
(b) Since limn→∞ xn= 0 under the assumption |x|<1, taking the limit as n→ ∞ in both sides of the
equation in part (a) yields that 1 + x+x2+...=1
1−x.
2. Inverse function
Sketching f, we see that fis monotonically decreasing, with limit 2 at −∞ and limit 0 at ∞. In particular,
an inverse function exists since fis one-to-one. The range of fis (0,2), so the inverse gneeds only to
be defined on that interval. We need f(g(u)) = ufor 0 <u<2. Solving this equation for g(u) yields
2
eg(u)+1 =u, or 2
u−1 = eg(u), or g(u) = ln( 2
u−1) for 0 <u<2.
3. Extrema of functions
(a) The function f(x) = x4(1.001)−x2can be written in the form xnexp(−ax2), where a= ln(1.001), n = 4.
The function xnexp(−ax2) has local extrema at the points x=±∞,0,±pn/2a. In this case n= 4 and
therefore the maxima occur at x=±pn/2a=±44.7325 (as can be verified by differentiating once again and
checking the sign of the result at the two points); the points x= 0,±∞ represent minima. The maximum
function value (at both maxima) is 5.4188 ×105.
(b) If f(x) = x4(1.001)x2, the function converges to infinity as |x| → ∞. Thus, f(x) is not a maximum value
of ffor any finite value of x. Incidentally, carrying out the differentiation gives us x=±p−2/ln(1.001) as
the other solutions, which is not possible since the quantity within the “√” sign is negative.
4. Some definite integrals
(a) Drawing a picture, you see that the integral from −1 to 1 is just the sum of the areas of two triangles,
each having height 1 and base length 1. Therefore, it is easy to see that the total area = 2 ×1/2 = 1.
Try and draw a little sketch before you tackle most problems. Those who used formal integration principles
should note that |x|=xif x≥0 and |x|=−xif x < 0. Therefore
Z1
−1|x|dx =Z0
−1−x dx +Z1
0
x dx =−x2
2
0
−1
+x2
2
1
0
= 1.
(b) Substitute y= (1 −x2). Then
Z1
0
x(1 −x2)11 dx =−1
2Z0
1
y11 dy =1
2Z1
0
y11 dy =1
24.
(c) Integrate by parts twice:
Z1
0
x2e−xdx =−x2e−x|1
0+Z1
0
2xe−xdx
=−e−1+Z1
0
2xe−xdx
=−e−1−2xe−x|1
0+Z1
0
2e−xdx
=−3e−1+Z1
0
2e−xdx
=−3e−1+ 2(1 −e−1) = 2 −5e−1≈0.1602
