MATH 1700 MIDTERM EXAM 2 SOLUTION
SPRING 2016 - MOON
(1) Consider a dynamical system xn+1 = −x3n .
(a) (5 pts) Find all 2-cycles.
f (x) = −x3 ⇒ f (f (x)) = −(−x3 )3 = x9
f (x) = x ⇒ −x3 = x ⇒ x3 + x = 0
⇒ x(x2 + 1) = 0 ⇒ x = 0
Therefore 0 is a fixed point.
f (f (x)) = x ⇒ x9 = x ⇒ x9 − x = 0
⇒ x(x8 − 1) = 0 ⇒ x(x4 − 1)(x4 + 1) = 0
⇒ x(x2 − 1)(x2 + 1)(x4 + 1) = 0 ⇒ x(x − 1)(x + 1)(x2 + 1)(x4 + 1) = 0
⇒ x = 0, 1, −1
3
f (1) = −1 = −1, f (−1) = −(−1)3 = 1
Therefore {−1, 1} is a 2-cycle.
• Finding a fixed point 0: +2 pts.
• Finding a 2-cycle {−1, 1}: +3 pts.
(b) (4 pts) Determine their stability.
f 0 (x) = −3x2
|f 0 (−1)f 0 (1)| = |(−3)(−3)| = 9 > 1
Therefore {−1, 1} is unstable.
• Finding f 0 (x) = −3x2 : 1 pt.
Date: April 18, 2016.
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MATH 1700 Midterm Exam 2
Spring 2016 - Moon
(2) Below is a plot of graphs of functions, y = f (x), y = f 2 (x), and y = x. Consider
the dynamical system given by xn+1 = f (xn ).
(a) (3 pts) Find the number of non-negative fixed points. Explain your answer.
The graph of y = f (x) meets y = x three times. So there are three fixed
points.
(b) (3 pts) Find the number of 2-cycles. Explain your answer.
Except the fixed points, there are two more intersections points of y =
f (f (x)) and y = x. These two points form a 2-cycle. So there is one 2-cycle.
(c) (3 pts) Is 0 a stable fixed point? Explain your answer.
From the graph, we can see that 0 ≤ f 0 (0) < 1. Therefore it is locally stable.
(d) (3 pts) Let p be the largest fixed point. Do solutions oscillate locally around
p? Explain your answer.
At p, f 0 (p) < 0. So solutions oscillate locally around p.
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MATH 1700 Midterm Exam 2
Spring 2016 - Moon
(3) (4 pts) State the Mean Value Theorem.
Let f be a continuous function on [a, b] and differentiable on (a, b). Then there
is c on the interval (a, b) such that
f (b) − f (a)
= f 0 (c).
b−a
• Stating that f is a differentiable function: +1 pt.
• Stating the existence of c on the interval (a, b): +1 pt.
f (b) − f (a)
• Giving the precise formula
= f 0 (c): +2 pts.
b−a
(4) Consider a one parameter family fr (x) = rx(3 − x2 ) of functions with r > 0.
(a) (4 pts) Find the interval of stability of the fixed point 0.
fr (x) = 3rx − rx3 ⇒ fr0 (x) = 3r − 3rx2 ⇒ fr0 (0) = 3r
|fr0 (0)| < 1 ⇔ 3r < 1 ⇔ r <
1
3
1
Therefore the interval of stability is (0, ).
3
• Finding the derivative fr0 (x) = 3r − 3rx2 : 1 pt.
1
• From the inequality |fr0 (0)| < 1 and r > 0, getting the interval (0, ): 4
3
pts.
(b) (5 pts) Find a positive fixed point p(r) and its interval of existence.
fr (x) = x ⇒ rx(3 − x2 ) = x ⇒ r(3 − x2 ) = 1
r
1
1
1
⇒ 3 − x2 = ⇒ x2 = 3 − ⇒ x = 3 −
r
r
r
r
1
So p(r) = 3 − .
r
1
Because 3 − has to be larger than 0 to obtain a positive fixed point,
r
1
1
< 3 ⇒ < r.
r
3
1
So the interval of existence of p(r) is ( , ∞).
3
• Setting up the equation rx(3 − x2 ) = x whose solutions are fixed
points: 2 pts. r
1
• Getting p(r) = 3 − : 3 pts.
r
1
• Finding the interval of existence ( , ∞): 5 pts.
3
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MATH 1700 Midterm Exam 2
Spring 2016 - Moon
(c) (4 pts) Find the interval of stability of p(r).
1
p(r)2 = 3 −
r
1
0
2
fr (p(r)) = 3r − 3rp(r) = 3r − 3r 3 −
= −6r + 3
r
|fr0 (p(r))| < 1 ⇔ | − 6r + 3| < 1 ⇔ −1 < −6r + 3 < 1
1
2
⇔ <r<
3
3
1 2
Therefore the interval of stability is ( , ).
3 3
• Evaluating fr0 (p(r)) = −6r + 3: 2 pts.
1 2
• Getting the interval ( , ): 4 pts.
3 3
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MATH 1700 Midterm Exam 2
Spring 2016 - Moon
3 −2
1 4
(5) Let A =
and B =
0 5
7 −2
(a) (3 pts) Find 3A − 5B.
3 −2
1 4
3A − 5B = 3
−5
=
0 5
7 −2
.
9 −6
0 15
−
5 20
35 −10
=
4 −26
−35 25
(b) (3 pts) Find AB.
3 −2
1 4
3 · 1 + (−2) · 7 3 · 4 + (−2) · (−2)
−11 16
AB =
=
=
0 5
7 −2
0·1+5·7
0 · 4 + 5 · (−2)
35 −10
(c) (3 pts) Write the definition of the inverse matrix A−1 .
For a square matrix A, the inverse matrix A−1 is a matrix such that AA−1 = I
and A−1 A = I.
1
d −b
is not the definition. 2 pts.
•
ad − bc −c a
(d) (3 pts) Find A−1 .
−1
A
1
=
3 · 5 − (−2) · 0
5 −(−2)
−0
3
5
1
=
15
5 2
0 3
=
1
3
0
2
15
1
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MATH 1700 Midterm Exam 2
Spring 2016 - Moon
(6) Consider the following prey-predator-migration model:
Pn+1 = Pn − Qn + 100
Qn+1 = 2Pn + Qn − 500.
Here Pn and Qn are populations of two species in n-th year and the unit is million.
(a) (3 pts) Which one is the population of the prey between Pn and Qn ? Explain
your answer.
Predators hunt prey. So the population of predators affects negatively to
the population of prey. Therefore Pn is the population of prey because Qn
contributes negatively to Pn .
(b) (3 pts) Suppose that in this year, P0 = 300 and Q0 = 25. Find the population
distribution in next year.
P1 = P0 − Q0 + 100 = 300 − 25 + 100 = 375
Q1 = 2P0 + Q0 − 500 = 600 + 25 − 500 = 125
Thus in next year, P1 = 375 million, and P2 = 125 million.
(c) (4 pts) Find the fixed point (P, Q).
P = P − Q + 100
Q = 2P + Q − 500
Q = 2P + Q − 500 ⇒ 2P − 500 = 0 ⇒ P = 250
P = P − Q + 100 ⇒ 0 = −Q + 100 ⇒ Q = 100
Therefore the fixed population is P = 250 million and Q = 100 million.
• Stating the system of linear equation for the fixed points: 2 pts.
• Getting the answer: 4 pts.
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