University of Illinois Spring 2009
ECE 313: Problem Set 1: Solutions
Calculus Review
1. [Geometric, MacLaurin and Taylor series]
(a) Let n > 0 be an integer. Then,
(1 −x)1 + x+x2+· ·· +xn−1=1 + x+x2+· ·· +xn−1−x+x2+· ·· +xn= 1 −xn.
If x6= 1, divide both sides by (1 −x) to get 1 + x+x2+· · · +xn−1=1−xn
1−x.
(b) Since |x|<1, 1 + x+x2+·· · = limn→∞ 1 + x+x2+·· · +xn−1= limn→∞
1−xn
1−x=1
1−x.
(c) The first derivative is f(1)(x) = n(1 + x)n−1, the second is f(2) (x) = n(n−1)(1 + x)n−2, and
so on. The k-th derivative is f(k)(x) = n(n−1) ·· ·(n−k+ 1)(1 + x)n−k, and finally the n-th
derivative is f(n)(x) = n(n−1)· ·· 2·1 = n! which is a constant.
(d) Using the results from part (c) for the first n derivatives, the first n+ 1 terms of the Maclaurin
series for f(x) are
f(x) = (1 + x)n≈
n
X
k=0
f(k)(0)
k!xk=
n
X
k=0
n(n−1) ·· ·(n−k+ 1)
k!xk.
(e) Since f(n)(x) is a constant, f(k)(x) = 0 for all k > n. Thus, the above is the complete Maclaurin
series for f(x); there is no approximation.
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 because 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. Will will abbreviate
g(u) by the letter galone, but remember that gis a function of u. So we have f(g) = u. Solving this
equation for gyields 2
eg+1 =u, or 2
u−1 = eg, or g= ln( 2
u−1) for 0 < u < 2. That is, 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 differ-
entiating 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 because the quantity within
the “√” sign is negative.
4. [Some definite integrals]
(a) Z1
−1|x|dx =Z0
−1−xdx +Z1
0
xdx =−1
2x2
0
−1
+1
2x2
1
0
=1
2+1
2= 1
