Engineering Mathematics | CHEN30101
solutions to sheet 1
1. The equation 2x−1 = cos x has a root between x = 0 and x = 1. Compute
the root to an accuracy of two digits.
Running Newton–Raphson iteration with either starting point gives x∗ =
0.66. Note that f (x) = 2x−1 − cos x and f 0 (x) = 2x−1 ln 2 + sin x.
f 0 (x)
1.5346
1.2077
1.1616
1.1607
1.1607
(a)
iteration
0
1
2
3
4
x
1.0000
0.7005
0.6607
0.6600
0.6600
f (x)
0.4597
0.0479
0.0009
0.0000
0.0000
(b)
iteration
0
1
2
3
4
5
x
0.0000
1.4427
0.8060
0.6690
0.6600
0.6600
f (x)
-0.5000
1.2314
0.1817
0.0106
0.0000
0.0000
f 0 (x)
0.3466
1.9339
1.3274
1.1713
1.1607
1.1607
2. A tank is fed at a constant rate and also drains a second tank. In turn,
this tank drains into a third tank. The volume of fluid in the tank varies
like V = 10 − 9e−t − e−3t . Compute an accurate estimate (correct to 5
digits) of the time when the volume of fluid in the tank is equal to 8.
We need to solve the equation 10 − 9e−t − e−3t = 8 so we define f (t) =
9e−t + e−3t − 2 and f 0 (t) = −9e−t − 3e−3t and iterate using Newton
(because we want five-digit accuracy).
t
0
0.666666667
1.214950133
1.468378342
1.508629153
1.509490221
1.509490604
f (t)
8
2.756089355
0.696648614
0.08490229
0.001742279
7.73946×10−7
1.52767×10−13
f 0 (t)
-12
-5.026759921
-2.74889925
-2.109331204
-2.023392486
-2.021595126
-2.021594327
Thus x∗ = 1.50949 (to five decimal digits)
1
3. The coupled equations
2x + y + e−
5x − 2y −
1
10
xy
5
− 11 = 0
sin(x + y) − 14 = 0
have a solution near the point x = 4, y = 3. Construct a Newton iteration
to solve these two equations and use this strategy to compute the solution
to an accuracy of 5 digits.
Editing the function fdef.m and running the function newtonit gives the
following output:
>> newtonit([4;3],5)
Newton iteration in 2 dimensions ...
k
x_k
y_k
|f_k|
0
4.00000
3.00000
9.072e-02
1
3.98520
2.93323
1.607e-04
2
3.98514
2.93318
1.934e-10
3
3.98514
2.93318
0.000e+00
4
3.98514
2.93318
0.000e+00
done
|g_k|
6.570e-02
2.116e-04
3.325e-10
3.553e-15
3.553e-15
Hence the solution is x∗ = 3.98514, y ∗ = 2.93318.
4. The proportion of two kinds of impurities in a liquid vary with height and
temperature as follows:
T
A = e− 100 /100 + T × sin(πh)/11000 + h/10
T
B =e− 60 /80 + T × sin(πh)/5000 + (1 − h)/5.
Find a temperature near T = 600K and a position near h = 0.5 where
A = 0.1 and B = 0.2.∗ Record your final answer to an accuracy of 3 digits.
Editing the function fdef.m and running the function newtonit gives the
following output:
>> newtonit([600;0.5],5);
Newton iteration in 2 dimensions ...
k
x_k
y_k
|f_k|
0 600.00000
0.50000
4.570e-03
1 523.57708
0.52358
1.211e-04
2 524.96419
0.52354
5.935e-09
3 524.96415
0.52354
1.665e-16
4 524.96415
0.52354
0.000e+00
done
Hence the solution is T ∗ = 525, h∗ = 0.524.
2
|g_k|
2.000e-02
2.865e-04
2.381e-09
3.886e-16
0.000e+00
5. Consider the coupled equation system
x2 − 2xy + y 2 − 5 = 0
x2 + 4xy + 4y 2 − 7 = 0
Compute (on paper) the first Newton iteration for solving the system
starting from from x0 = 1, y0 = 1. What happens on a computer?
Repeat the above experiment, this time starting from the point x0 = 1,
y0 = 3. Does the iteration converge?
Defining the functions f = x2 − 2xy + y 2 − 5 and g = x2 + 4xy + 4y 2 − 7,
the Jacobian matrix associated with the nonlinear system is
fx fy
2x − 2y 2y − 2x
J=
=
.
gx gy
2x + 4y 4x + 8y
If the Jacobian matrix is evaluated at a point where x = y then the first
row is zero and so the matrix is singular. The Newton iteration breaks
down at this point. Indeed, running newtonit with the starting vector
[1;1] gives the error message:
Newton iteration in 2 dimensions ...
k
x_k
y_k
|f_k|
|g_k|
0
1.00000
1.00000
5.000e+00
2.000e+00
Warning: Matrix is singular to working precision.
Note that the iteration proceeds to completion (it does not stop at the
first step) by generating numbers Inf (∞), -Inf (−∞) and NaN (Not a
Number). Starting from a different point [1;3] converges to a root:
Newton iteration in 2 dimensions ...
k
x_k
y_k
|f_k|
0
1.00000
3.00000
1.000e+00
1 -0.16667
2.08333
6.250e-02
2 -0.53241
1.70370
1.929e-04
3 -0.60575
1.63032
1.860e-09
4 -0.60879
1.62728
1.776e-15
done
|g_k|
4.200e+01
9.000e+00
1.266e+00
4.845e-02
8.325e-05
6. Compute a solution to the system of equations
x2 + y 2 + z − 40 = 0
x2 − y + 3z 2 − 22 = 0
x + y 2 − 2z 2 − 20 = 0
that is accurate to two decimal places by constructing a Newton iteration
and starting from the point x = 5, y = 4, z = 0.
Editing the functions fdef.m and itntable and running the function
newtonit gives the following output:
3
>> newtonit([5;4;0],6);
Newton iteration in 3 dimensions ...
k
x_k
y_k
z_k
|f_k|
0
5.00000
4.00000
0.00000
1.000e+00
1
5.08642
3.86420 -0.77778
2.591e-02
2
4.99029
3.96682 -0.61883
1.977e-02
3
4.98483
3.97026 -0.61152
4.163e-05
4
4.98482
3.97027 -0.61151
2.544e-10
5
4.98482
3.97027 -0.61151
0.000e+00
done
Hence the solution is x∗ = 4.98, y ∗ = 3.97, z ∗ = −0.61.
4
|g_k|
1.000e+00
1.822e+00
8.503e-02
1.901e-04
7.597e-10
0.000e+00
|h_k|
1.000e+00
1.191e+00
4.000e-02
9.504e-05
3.082e-10
3.553e-15