Nonlinear Equations
Your nonlinearity confuses me
ax5  bx4  cx3  dx2  ex  f  0
tanh(x)  x
“The problem of not knowing what we
missed is that we believe we haven't
missed anything” – Stephen Chew on Multitasking
http://numericalmethods.eng.usf.edu
Numerical Methods for the STEM undergraduate
Example – General Engineering
You are working for ‘DOWN THE TOILET COMPANY’ that
makes floats for ABC commodes. The floating ball has a
specific gravity of 0.6 and has a radius of 5.5 cm. You
are asked to find the depth to which the ball is
submerged when floating in water.
x3  0.165x 2  3.993104  0
Figure Diagram of the floating ball
2
For the trunnion-hub problem discussed
on first day of class where we were
seeking contraction of 0.015”, did the
trunnion shrink enough when dipped in
dry-ice/alcohol mixture?
1. Yes
2. No
0%
1.
0%
2.
30
Example – Mechanical Engineering
Since the answer was a
resounding NO, a logical
question to ask would be:
If the temperature of
-108oF is not enough for
the contraction, what is?
4
Finding The Temperature of the Fluid
Tc
D  D   (T ) dT
7
Ta
6.5
6
80oF
5.5
Ta =
Tc = ???oF
D = 12.363"
∆D = -0.015"

5
4.5
4
3.5
3
2.5
2
-350
-300
-250
-200
-150
-100
-50
0
50
100
T
 T   6.033 0.009696T
 0.015 5.992108Tc  7.457105Tc  6.349103
2
f (Tc )  5.992108Tc  7.457105Tc  8.651103  0
2
5
Finding The Temperature of the Fluid
Tc
D  D   (T ) dT
Ta
Ta = 80oF
Tc = ???oF
D = 12.363"
∆D = -0.015"
  1.228105 T 2  6.195103 T  6.015
 0.015 5.059109 Tc  3.829106 Tc  7.435105Tc  6.166103
3
2
f (Tc )  5.059109 Tc  3.829106 Tc  7.435105Tc  8.834103  0
3
2
6
Nonlinear Equations
(Background)
http://numericalmethods.eng.usf.edu
Numerical Methods for the STEM undergraduate
How many roots can a nonlinear
equation have?
sin(x)=2 has no roots
3
2
1
0
-1
-2
-3
-20
-10
0
10
20
How many roots can a nonlinear
equation have?
sin(x)=0.75 has infinite roots
3
2
1
0
-1
-2
-3
-20
-10
0
10
20
How many roots can a nonlinear
equation have?
sin(x)=x has one root
3
2
1
0
-1
-2
-3
-20
-10
0
10
20
How many roots can a nonlinear
equation have?
sin(x)=x/2 has finite number of roots
3
2
1
0
-1
-2
-3
-20
-10
0
10
20
The value of x that satisfies f (x)=0 is
called the
1.
2.
3.
4.
root of equation f (x)=0
root of function f (x)
zero of equation f (x)=0
none of the above
0%
1.
0%
0%
2.
3.
0%
4.
A quadratic equation has ______ root(s)
1.
2.
3.
4.
one
two
three
cannot be determined
0%
1.
0%
0%
2.
3.
0%
4.
For a certain cubic equation, at least one of
the roots is known to be a complex root. The
total number of complex roots the cubic
equation has is
1.
2.
3.
4.
one
two
three
cannot be determined
0%
1.
0%
0%
2.
3.
0%
4.
Equation such as tan (x)=x has __ root(s)
1.
2.
3.
4.
zero
one
two
infinite
0%
1.
0%
0%
2.
3.
0%
4.
A polynomial of order n has
1.
2.
3.
4.
zeros
n -1
n
n +1
n +2
0%
1.
0%
0%
2.
3.
0%
4.
END
http://numericalmethods.eng.usf.edu
Numerical Methods for the STEM undergraduate
Bisection Method
http://numericalmethods.eng.usf.edu
Numerical Methods for the STEM undergraduate
Bisection method of finding roots of
nonlinear equations falls under the category
of a (an)
method.
1.
2.
3.
4.
open
bracketing
random
graphical
0%
1.
0%
0%
2.
3.
0%
4.
If for a real continuous function f(x),
f (a) f (b)<0, then in the range [a,b] for
f(x)=0, there is (are)
1.
2.
3.
4.
one root
undeterminable number of roots
no root
at least one root
0%
1.
0%
0%
2.
3.
0%
4.
The velocity of a body is given by
v (t)=5e-t+4, where t is in seconds and v is in m/s. We want
to find the time when the velocity of the body is 6 m/s. The
equation form needed for bisection and Newton-Raphson
methods is
1.
2.
3.
4.
f (t)= 5e-t+4=0
f (t)= 5e-t+4=6
f (t)= 5e-t=2
f (t)= 5e-t-2=0
25%
1.
25%
2.
25%
3.
25%
4.
To find the root of an equation f (x)=0, a student started
using the bisection method with a valid bracket of [20,40].
The smallest range for the absolute true error at the end of
the 2nd iteration is
1.
2.
3.
4.
0
0
0
0
≤
≤
≤
≤
|Et|≤2.5
|Et| ≤ 5
|Et| ≤ 10
|Et| ≤ 20
25%
1.
25%
25%
2.
3.
25%
4.
For an equation like x2=0, a root exists at
x=0. The bisection method cannot be
adopted to solve this equation in spite of the
root existing at x=0 because the function
f(x)=x 2
1.
2.
3.
4.
is a polynomial
has repeated zeros at x=0
is always non-negative
slope is zero at x=0
0%
1.
0%
0%
2.
3.
0%
4.
END
http://numericalmethods.eng.usf.edu
Numerical Methods for the STEM undergraduate
Newton Raphson Method
http://numericalmethods.eng.usf.edu
Numerical Methods for the STEM undergraduate
Newton-Raphson method of finding roots of
nonlinear equations falls under the category
of __________ method.
1.
2.
3.
4.
bracketing
open
random
graphical
0%
1.
0%
0%
2.
3.
0%
4.
The next iterative value of the root of the
equation x 2=4 using Newton-Raphson
method, if the initial guess is 3 is
1.
2.
3.
4.
1.500
2.066
2.166
3.000
0%
1.
0%
0%
2.
3.
0%
4.
The root of equation f (x)=0 is found by using Newton-Raphson
method.
The initial estimate of the root is xo=3, f (3)=5.
The angle the tangent to the function f (x) makes at x=3 is 57o.
The next estimate of the root, x1 most nearly is
1.
2.
3.
4.
-3.2470
-0.2470
3.2470
6.2470
0%
1.
0%
0%
2.
3.
0%
4.
The Newton-Raphson method formula for
finding the square root of a real number R
from the equation x2-R=0 is,
25%
25%
25%
2.
3.
25%
xi

2
1.
x i 1
2.
xi 1 
3.
1
R
xi 1   xi  
2
xi 
4.
xi 1 
3 xi
2
1
R
 3 xi  
2
xi 
1.
4.
END
http://numericalmethods.eng.usf.edu
Numerical Methods for the STEM undergraduate