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(MATH024)2009_s_MATH204_by_cs_gxx891.pdf
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Math 204 Midterm Exam

April 28, 2009

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Total


(1) (20 points) Let K1 and K2 be two compact sets in Rn . Show that there are .a K1 and .b K2 such that ..a ..b. = inf ..x . .y..
.yK2xK1,.

(2)
(20 points) Let A =(aij) be a symmetric n n matrix. Find the derivative of the function f(.x)= .xT A.x.

(3)
(20 points) Find the Taylor expansion of the function f(x, y) = sin(x + xy . 2) at (1, 1) to third order.

(4)
(20 points) Find the Jacobian matrix of the map



F : R2 R2 (x, y) . (z, w)
given implicitly by
x = f(x + y + z + w) y = g(x . y + z . w),

where f and g are di.erentiable.
(5) (20 points) Show that the function
f(x, y)= . sin x + ey(y . 1 . sin x) has in.nitely many local minima but no local maxima.
Answer to Math 204 Midterm, Spring 2009
(1) (20 points) Let K1 and K2 be two compact sets in Rn . Show that there are .a K1 and .b K2 such that ..a ..b. = inf ..x . .y..
.yK2xK1,.

Solution: For each n, there are two points xn K1 and yn K2 such that
inf ..x . .y..xn . yn. < inf ..x . .y. +1/n.
.xK1,.yK2 .xK1,.yK2
Since K1, there is a convergent subsequence .xnk . Since K2 is also compact, .ynk has a convergent subsequence .ynkp . Let .a = lim .xnk , .b = lim .ynkp . Then we have
inf ..x . .y...xnkp . .ynkp . < inf ..x . .y. +1/nkp .
.xK1,.yK2 .xK1,.yK2
Let p , we have

inf ..x . .y...a ..b. < inf ..x . .y..
.xK1,.yK2 .xK1,.yK2
Hence ..a ..b. = inf ..x . .y..
.xK1,.yK2
(2) (20 points) Let A =(aij) be a symmetric n n matrix. Find the derivative of the
function f(.x)= .xT A.x.
Solution: For the symmetric matrix A,

f(.x + .h) . f(.x)=(.x + .h)T A(.x + .h) . .x T A.x T A.
= .x h + .hT A.x + .hT A.h T A.
=2.x h + .hT A.h.

Since 2.xT A.h is linear in .h and
..hT A.h...h..A...h. = .A...h.2
we know that f.(.x)=2.x T A.
(3) (20 points) Find the Taylor expansion of the function f(x, y) = sin(x + xy . 2) at (1, 1) to third order. Solution: Since
x + xy . 2=(x . 1) + 1 + [(x . 1) + 1][(y . 1) + 1] . 2 = 2(x . 1) + (y . 1) + (x . 1)(y . 1)
using the Taylor expansion for the function sin x, we obtain
sin(x + xy . 2)
= sin[2(x . 1) + (y . 1) + (x . 1)(y . 1)]

= [2(x . 1) + (y . 1) + (x . 1)(y . 1)] . 1 [2(x . 1) + (y . 1) + (x . 1)(y . 1)]3 + O(5)
3!

= 2(x . 1) + (y . 1) + (x . 1)(y . 1)
. 4(x . 1)3 . 2(x . 1)2(y . 1) . (x . 1)(y . 1)2 . 1(y . 1)3 + O(4).
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Hence, the Taylor expansion of sin(x + xy . 2) at (1, 1) to third order is
2(x.1)+(y .1)+(x.1)(y .1). 4(x.1)3 .2(x.1)2(y .1).(x.1)(y .1)2 . 1(y .1)3 .

36
(4) (20 points) Find the Jacobian matrix of the map
F : R2 R2 (x, y) . (z, w)
given implicitly by

x = f(x + y + z + w)

y = g(x . y + z . w), where f and g are di.erentiable. Solution: Di.erentiating the implicit functions gives
. .
. .
.

1= f. (1 + zx + wx) 0= f.