Download Solution for Homework 7 1

Solution for Homework 7
1
[KA] Sec. 8.5, page 468
4. (a) The linearization of the equation at x = π is x0 = x − π.
(c) The general solution of the linearized equation is x = Aet + π. The initial
condition x(0) = 3 gives A = 3 − π. So the solution is x = (3 − π)et + π.
(d) The general solution of the linearized equation is Aet = csc x + cot x, i.e.,
x = 2 tan−1 (A−1 e−t ). The initial condition x(0) = 3 gives A = (tan(3/2))−1 .
So the solution is x = 2 tan−1 (e−t tan(3/2)).
(e) This question is impossible to answer precisely.
5. (a) The fixed points (equilibriums) are x = kπ for k integers; linearize
the equation at x = kπ: x0 = (−1)k (x − kπ); x = kπ is stable if k is odd and
unstable if k is even.
(b) The fixed points are y = kπ + π/2 for k integers; linearize the equation
at y = kπ + π/2: y 0 = (−1)k+1 (y − kπ − π/2); y = kπ + π/2 is stable if k is
even and unstable if k is odd.
(c) The fixed points are x = 1 and x = 2; linearize the equation at x = 1:
x0 = −(x − 1); linearize the equation at x = 2: x0 = x − 2; x = 1 is stable
and x = 2 is unstable.
(d) The fixed points are y = 1 and y = 2; linearize the equation at y = 1:
y 0 = y − 1; linearize the equation at y = 2: y 0 = −(y − 2); y = 1 is unstable
and the solution y = 2 is stable.
[KA] Sec. 3.1, page 137
1. (a) The solution is x = ( 12 − t)−1 . The interval of existence is (−∞, 12 ).
(b) The solution is x = (− 12 − t)−1 . The interval of existence is (− 12 , ∞).
(c) No such solution exists.
(d) The initial condition is x(0) = 10 and the corresponding solution is
1
http://www.math.ucsb.edu/˜xichen/math3c01s/hw7key.pdf
1
1
x = ( 10
− t)−1 .
−1
−1
(e) The solution is x = (x−1
0 − t) . The interval of existence is (−∞, x0 ).
−1
−1
(f) The solution is x = (−x−1
0 − t) . The interval of existence is (−x0 , ∞).
3. (b) Since x2 + t2 is increasing in both x and t when x > 0 and t > 0, the
Euler’s method underestimates the true value of x(t) for t > 0.
(c) The general solution is x = (C −t)−1 . By the initial condition x(0.9) = 14,
C = 34/35.
(d) Example 4 shows that the solution has a singularity greater than π/4
and (c) shows it has a singularity less than 34/35.
4. Solve x0 = x2 with x(0) = 2 and we obtain x = (1/2−t)−1 ; solve x0 = x2 +1
with x(0) = 2 and we obtain x = tan(t + tan−1 (2)). So the solution has an
singularity between π/2 − tan−1 (2) and 1/2.
p
10. The general solution is x = A |t2 − 1|.
√
(a) The solution is x = 1 − t2 . So the interval of existence is (−1, 1).
√
(b) The solution is x = − 1 − t2 . So the interval of existence is (−1, 1).
(c) The solution is x = 0. So the interval of existence is (−∞, ∞).
√
(d) The solution is x = 3−1/2 t2 − 1. So the interval of existence is (1, ∞).
√
(d) The solution is x = −3−1/2 t2 − 1. So the interval of existence is (1, ∞).
11. The general solution is x = (C − (p − 1)t)1/(1−p) .
(a) The solution is x = (1 − (p − 1)t)1/(1−p) . So the interval of existence is
(−∞, 1/(p − 1)).
(b) The solution is x = (x1−p
− (p − 1)t)1/(1−p) . So the interval of existence
0
1−p
is (−∞, x0 /(p − 1)).
[S] Chap. 2, page 4
2.2. The quadratic approximation of ln x at x = 1 is (x − 1) − 21 (x − 1)2 .
2
3.2. The Taylor series of cos 3x is cos 3x =
P∞
n=0
(−1)n (3x)2n
.
(2n)!
3.4. The 4th Taylor polynomial of M 4 − T 4 at T = M is
M 4 − T 4 = M 4 − ((T − M ) + M )4
= −4M 3 (T − M ) − 6M 2 (T − M )2 − 4M (T − M )3 − (T − M )4 .
4.1. The Taylor series of (1 + x)−1 at x = 0 is
∞
X
1
=
(−1)n xn .
1 + x n=0
The Taylor series of (1 − x2 )−1 at x = 0 is
∞
X
1
=
x2n .
1 − x2
n=0
The Taylor series of (1 + x2 )−1 at x = 0 is
∞
X
1
=
(−1)n x2n .
1 + x2
n=0
4.2. The Taylor series of cos(3x) at x = 0 is
cos(3x) =
∞
X
(−1)n
n=0
(3x)2n
.
(2n)!
Multiply it by x2 and we obtain
2
x cos(3x) =
∞
X
nx
(−1)
n=0
∞
2
(3x)2n X (−1)n 32n 2n+2
=
x
.
(2n)!
(2n)!
n=0
4.3. The Taylor series of (2 − x)−1 at x = 0 is
1
1
=
2−x
2
1
1 − (x/2)
∞
=
3
∞
1 X x n X x n
=
.
n+1
2 n=0 2
2
n=0
Replace x by x3 and we obtain the Taylor series of (2 − x3 )−1 at x = 0:
∞
X x3n
1
=
.
2 − x3
2n+1
n=0
4.4. The Taylor series of sinh(x) at x = 0 is
1 x
e + e−x
2
!
∞
∞
∞
∞
n
n
X
X
1 X xn X
x
1
x2n
n
n x
=
+
(−1)
=
(1 + (−1) )
=
.
2 n=0 n! n=0
n!
2
n!
(2n)!
n=0
n=0
sinh(x) =
The Taylor series of x/(x + 1) at x = 0 is
∞
∞
X
X
x
n n
=x
(−1) x =
(−1)n xn+1 .
1+x
n=0
n=0
The Taylor series of xe−x at x = 0 is
xe
−x
=x
∞
X
nx
(−1)
n=0
n
n!
=
∞
X
(−1)n
n=0
xn+1
.
n!
The Taylor series of x2 sin(x2 ) at x = 0 is
x2 sin(x2 ) = x2
∞
X
n=0
∞
(−1)n
X
(x2 )2n+1
x4n+4
=
(−1)n
.
(2n + 1)! n=0
(2n + 1)!
4.5. The Taylor series of sin(x)/x at x = 0 is
∞
∞
2n+1
X
X
sin(x)
x2n
−1
n x
=x
(−1)
=
(−1)n
.
x
(2n + 1)! n=0
(2n + 1)!
n=0
The Taylor series of (ex − 1)/x at x = 0 is
!
∞
∞
∞
n
X
X
ex − 1
x
xn X xn−1
= x−1
− 1 = x−1
=
x
n!
n!
n!
n=0
n=1
n=1
4
5.4. Since (arcsin(x))0 = (1 − x2 )−1/2 , we may first find the Taylor series of
(1 − x2 )−1/2 and then integrating it. By binomial theorem,
∞ ∞
X
X
−1/2
2 −1/2
2 n
n −1/2
(1 − x )
=
(−x ) =
(−1)
x2n .
n
n
n=0
n=0
So
arcsin(x) =
∞
X
n
(−1)
n=0
−1/2 x2n+1
.
n
2n + 1
5.5. Since
∞
et − 1 X tn−1
=
,
t
n!
n=1
Z
0
x
∞
X
et − 1
dt =
t
n=1
x n−1
Z
t
0
n!
dt =
∞
X
xn
.
n(n!)
n=1
5.6. Since
∞ n
∞ n−1
X
X
− ln(1 − t)
t
t
−1
=t
=
,
t
n
n
n=1
n=1
Z
0
x
∞
X
− ln(1 − t)
dt =
t
n=1
Z
0
x n−1
t
n
dt =
∞
X
xn
n=1
n2
.
7.3. The Taylor series of sinh(x) is
∞
X
x2n
sinh(x) =
.
(2n)!
n=0
Use ratio test:
2n+2
x
/(2n + 2)! x2
= 0 < 1.
lim = lim 2n
n→∞
n→∞ (2n + 1)(2n + 2) x /(2n)!
5
So the Taylor series always converges and hence the radius of convergence is
∞.
7.4. The Taylor series of x2 cos(3x) is
2
x cos(3x) =
∞
X
(−1)n 32n
n=0
(2n)!
x2n+2 .
Use ratio test:
(−1)n+1 32n+2 x2n+4 /(2n + 2)! 9x2
= 0 < 1.
= lim lim n
2n
2n+2
n→∞ (2n + 1)(2n + 2) n→∞
(−1) 3 x
/(2n)!
So the Taylor series always converges and hence the radius of convergence is
∞.
7.6. Use ratio test:
(n + 1)!xn+1 = lim |(n + 1)x|.
lim n→∞
n→∞ n!xn
Since limn→∞ |(n + 1)x| = ∞ > 1 for all x 6= 0, the power series diverges for
all x 6= 0. So the radius of convergence is 0.
6