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Solution of transcendental equations - Coggle Diagram
Solution of transcendental equations
Bracket Method
bisection Method
Find points a and b such that f(a)⋅f(b)<0
find x0 which is the average of and b
Conditions
If f(x0)=0 then x0 is an exact root
else if f(a)⋅f(x0)<0 then b=x0
else if f(x0)⋅f(b)<0 then a=x0
Termination criteria
We limit the iterations
We stop the calculation if the iterations increase the given limit
We specify a Tolerance
If the Error is less than the Tolerance, it will terminate and the current root will be the approximate root of the equation, Formula =
Regula Falsi
Find points x0 and x1 such that f(x0)⋅f(x1)<0.
we find x2
Condition
If f(x2)=0 then x2 is an exact root
else if f(x0)⋅f(x2)<0 then x1=x2
else if f(x2)⋅f(x1)<0 then x0=x2
termination Criteria
We limit the iterations
We stop the calculation if the iterations increase the given limit
We specify a Tolerance
If the Error is less than the Tolerance, it will terminate and the current root will be the approximate root of the equation
Open Method
Newton Raphson
Find points a and b such that f(a)⋅f(b)<0
we Find x0
Conditions
If f(x1)=0 then x1 is an exact root
else x0=x1
Termination Criteria
where 2 iterations repeat
Find f(x0) and f′(x0)
Simple Iteration
First write the equation x=ϕ(x)
Find points a and b such that a<b and f(a)⋅f(b)<0
Conditions
If f(a) is more closer to 0 then f(b)
then x0=a else x0=b
repeat until |f(xi)-f(xi-1)|~0
Newton + Horner's Method
we are given with x0
using x0 as root in synthetic division, we find P(x0) and P ́(x0)
using Newton Method we find x1
Conditions
else we find x2
termination criteria
|x1 - x0| < given tolerance , then we write the equation from synthetic division
approximating and factorizing to find the approximate roots
secant method
modified form of Newton-Raphson method
required two initial guesses
finding x0 and x1 such that x0<x1 and f(x0).f(x1)<0
if x2 is =0 the x2 is exact, else x0=x1 and x1=x2
find next value x2
