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Solution to
Algebraic &Transcendental
Equations
A
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Algebraic functions
0011
1
fyfyfyfn
n
n
n
The general form of an Algebraic function:
fi = an i-th order polynomial.
Example : 073362 233 yxyxxf3 f2 f0
Polynomials are a simple class of algebraic function
011
1 axaxaxaxfn
n
n
nn
ais are constants.
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A transcendental function is non-algebraic.
May include trigonometric, exponential, logarithmic
functions
Examples:
Transcendental functions
12 xxf ln
).sin(. 50320 xexf x
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Equation Solving
Given an approximate location (initial value)
find a single real root
Root
Finding
Open
Methods
Brackting
Methods
Iterative
Newton-
Rapson
Secant
False-
position
Bisection
non-linear
Single variable
A
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Iterative method
A.1
-
6
Simple Fixed-point Iteration
... 2, 1,k ,given )(
)(0)(
1
okk xxgx
xxgxf
Now progressively estimate the value of x.
Rearrange the function so that x is on the
left side of the equation:
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Problem
Find the root of
f(x) = e-x x
There is no exact or
analytic solution
Numerical solution:
0
0)(
21
1
2
xfxfxf
xxf
exf
xexf
x
x
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Iterative Solution
1. Start with a guess say x1=1,
2. Generate
a) x2=e-x1 = e-1= 0.368
b) x3=e-x2= e-0.368 = 0.692
c) x4=e-x3= e-0.692=0.500
In general:
After a few more iteration we will get
nx
nex
156705670 .. e
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Iteration
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Convergence Examples
Convergent staircase pattern Convergent spiral pattern
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Divergence Example
Divergent staircase pattern Divergent spiral pattern
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Existence of Root
There exists one and only one root if
L is Lipschitz constant,
10
Liii
baxxxxLxgxgii
bxgabxai
.,,.
.
Lxg
xx
xgxg
xx
)(lim 1
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Convergence?
])( aagnn
nn
n
nn
aga
agagag
ag
xgx
and 0 [
221
1
If x=a is a solution then,
11
111
agif
ag
nn
anxnnn ] [
error reduces at each step
i.e. iteration will converge
If magnitude of 1st derivative
at x=a is less than 1
i
ni
i
axi
afafxf
!
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Problem
Find a root near x=1.0 and x=2.0
Solution:
Starting at x=1, x=0.292893 at 15th iteration
Starting at x=2, it will not converge
Why? Relate to g'(x)=x. for convergence g'(x) < 1
Starting at x=1, x=1.707 at iteration 19
Starting at x=2, x=1.707 at iteration 12
Why? Relate to
142 2 xxxf
412
21 xxgx
212 xxgx
21
212
xxg
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Aitkens Process
A.2
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kth Order Convergence
Pervious iterative method has linear (1st order)
convergence, since:
For kth order convergence we have:
Now consider a 2nd order method.
Aitkens 2 process
ag nn 1
A knn 1
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Aitkens process
If is a root of the equation i.e., =g() then,
Now if we use
10
2
1
1
n
n
nn
g
g
g
1
1
0
n
n A
Ag and
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Aitkens process
11
2
11
1
1
1
1
1
1
1
1
2
nnn
nnn
n
n
n
n
n
n
n
n
n
n
n
n
xxx
xxx
x
x
x
x
Ax
Ax
Ax
A
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Algorithm
guess_value;
while (! g()) {
}
11
2
11
1
1
1
2
nnn
nnn
nn
nn
n
xxx
xxx
xgx
xgx
x
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Why 2?
11
11
1
1
11
2
1
1
2
2
1
11
2
11
11
2
11
2
2
2
nnn
nnnn
nn
nn
nn
nnn
n
nn
nnn
nnn
nnn
nnn
xxx
xxxx
xx
xx
xx
xxx
where
x
xx
xxx
xxx
xxx
xxx
