MATH 1700 Homework
Han-Bom Moon
Homework 6 Solution
Section 3.2 ∼ 3.3.
• For every problem, explain your answer. It is not acceptable to write the answer
only.
1. Find the closed formula for the exact solution of xn+1 = (xn − c)a + c, where c is
a constant, using the substitution un = xn − c.
xn+1 − c = (xn − c)a ⇒ un+1 = uan
n
⇒ un = ua0
n
n
⇒ xn − c = (x0 − c)a ⇒ xn = c + (x0 − c)a
2. Find all fixed points (if any) of f (x) = 4x5 algebraically.
2
4x5 = x ⇒ 4x5 − x = 0 ⇒ x(4x4 − 1) = 0 ⇒ x((2x2) − 1) = 0
⇒ x(2x2 − 1)(2x2 + 1) = 0 ⇒ x = 0, 2x2 = 1
1
⇒ x = 0, ± √
2
3. Find all fixed points (if any) of Pn+1 = 4Pn e−Pn algebraically.
f (x) = 4xe−x
f (x) = x ⇒ 4xe−x = x ⇒ 4xe−x − x = 0
⇒ x(4e−x − 1) = 0 ⇒ x = 0, 4e−x = 1
4e−x = 1 ⇒ ex = 4 ⇒ x = ln 4
x = 0, ln 4
4.
(a) Sketch the graph of f (x) = 1 − |2x − 1|.
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MATH 1700 Homework
Han-Bom Moon
(b) Find the number of fixed points.
There are two intersection points of y = f (x) and y = x. Therefore there are
two fixed points.
5. Use the derivative to determine the stability of and any oscillation around the
given fixed point p = 1 for
f (x) = (x2 − 3x + 5)/3.
2
f 0 (x) = (2x − 3)/3 = x − 1
3
1
f 0 (1) = −
3
0
Because |f (1)| = 1/3 < 1, 1 is a stable fixed point. Also f 0 (1) < 1, so there is an
oscillation around it.
6. Find all positive fixed points of
xn+1 = xn1/4
and use the derivative to determine their stability and any oscillation around
them.
f (x) = x1/4
f (x) = x ⇒ x1/4 = x ⇒ x4 = x
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Han-Bom Moon
⇒ x(x3 − 1) = 0 ⇒ x(x − 1)(x2 + x + 1) = 0 ⇒ x = 0, x = 1
But because x0 > 0, we should exclude x = 0.
1 3
f 0 (x) = x− 4
4
1
1
⇒ |f 0 (1)| = < 1
4
4
So 1 is a stable fixed point. Because the derivative is positive, there is no oscillation around them.
f 0 (1) =
7. Find all non-negative fixed points of Pn+1 = rPn e−Pn /1000 and use the derivative
to determine their stability.
(a) r = 0.5
f (x) = 0.5xe−x/1000
x = f (x) ⇒ x = 0.5xe−x/1000 ⇒ x(0.5e−x/1000 − 1) = 0
x = 0 or 0.5e−x/1000 = 1
In the second case,
e−x/1000 =
1
x
=2⇒−
= ln 2 ⇒ x = −1000 ln 2.
0.5
1000
Because −1000 ln 2 < 0, there is one non-negative fixed point, x = 0.
1
x 0
−x/1000
−x/1000
f (x) = 0.5e
+ 0.5xe
· −
= 0.5e−x/1000 1 −
1000
1000
|f 0 (0)| = |0.5 · e0 · 1| = 0.5 < 1
So 0 is locally stable.
(b) r = 5
f (x) = 5xe−x/1000
x = f (x) ⇒ x = 5xe−x/1000 ⇒ x(5e−x/1000 − 1) = 0
x = 0 or 5e−x/1000 = 1
In the second case,
e−x/1000 =
1
x
= 0.2 ⇒ −
= ln 0.2 ⇒ x = −1000 ln 0.2.
5
1000
From ln 0.2 < 0, we have −1000 ln 0.2 > 0. So there are two non-negative
fixed points, x = 0 and x = −1000 ln 0.2.
1
x f 0 (x) = 5e−x/1000 + 5xe−x/1000 · −
= 5e−x/1000 1 −
1000
1000
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MATH 1700 Homework
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|f 0 (0)| = |5 · e0 · 1| = 5 > 1
So 0 is unstable.
|f 0 (−1000 ln 0.2)| = |5eln 0.2 (1 + ln 0.2) | = 1 + ln 0.2 < 1
Therefore −1000 ln 2 is stable.
8. Let xn+1 = f (xn ) be a discrete dynamical system. Suppose thatf 0 is continuous
and let p be a fixed point. By using mean value theorem, show that if f 0 (p) < 0,
then solutions oscillate locally around p.
Because f is continuous, if f 0 (p) < 0, then there is δ > 0 such that if |x − p| < δ,
then f 0 (x) < 0. Suppose that |xn − p| < δ. By mean value theorem, there is c
between xn and p such that
f (xn ) − f (p)
= f 0 (c),
xn − p
in other words, f (xn ) − f (p) = f 0 (c)(xn − p). Because c is between xn and p,
|c − p| < δ, so f 0 (c) < 0. Now from
xn+1 − p = f (xn ) − f (p) = f 0 (c)(xn − p),
we can conclude that if xn < p, then xn+1 > p, and if xn > p, then xn+1 < p.
Therefore the solution oscillates around p.
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