SOLUTION OF ALGEBRAIC AND TRANSCENDENTAL EQUATIONS(REGULA-FALSI METHOD )

SOLUTION OF ALGEBRAIC AND TRANSCENDENTAL EQUATIONS (REGULA-FALSI METHOD)

In this post, we discuss the Regula- Falsi method. Also, we solve a few questions and give a few numerical for practice. So that you are able to understand the Regula- Falsi method. As we discuss in our previous post about the definition of Algebraic and Transcendental equations. Also, we discuss the Bisection method. Now, we continue further.

REGULA-FALSI METHOD:
This method is the oldest method of finding the root of the equation `f(x)=0` and resembles to the bisection method. Here, also we find two initial points `x_0` and `x_1` such that `f(x_0)` and `f(x_1)` are of opposite signs. Thus it indicates that a root lies between `x_0` and `x_1` if `f(x_0) f(x_1) < 0.` ` x_2 = x_0 - \frac{x_1 -x_0}{f(x_1)-f(x_0)}f(x_0)`. This can be written as `x_2 = \frac{x_0 f(x_1) -x_1 f(x_0)}{f(x_1)-f(x_0)}`

If `f(x_0)` and `f(x_2)` are of opposite signs, then the root lies between `x_0` and `x_2`. So replacing `x_1` by `x_2` in equation (1), we obtain the next iteration `x_3`. The root could lie in between `x_1` and `x_2` and we would obtain `x_3` accordingly. This procedure is repeated till the root is found to desired accuracy. This method is also known as false position method. Now, we can solve few numerical based on Regula-Falsi method.

Example 1: Find a real root of the equation `x^3 -2x-5=0` by Regula-Falsi method correct to three decimal places.
Solution: Let `f(x)=x^3 -2x-5=0`.
Step1: First we find the initial points.
`f(0)= 0-0-5 `(-ive value)
`f(1)=1-2-5= -6 `(-ive value)
`f(2)=8-4-5=-1` (-ive value)
`f(3)=27-6-5 =16` (+ive value)
Now, the we get opposite sign. Thus we stop here. Therefore root lies between `2` and `3` and we proceed to second step.
Step 2: First iteration
Taking `x_0=2,  x_1=3,  f(x_0)= -1,  f(x_1)=16` and using the Regula-Falsi formula.

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`x_2=\frac{2(16)-3(-1)}{16-(-1)}=2.0588`.

Now, `f(2.0588) = -0.3908` (-ive value). Thus it replaces the previous negative value that is `2`. Therefore, root lies between `2.0588` and `3`.

Second Iteration
Taking `x_0=2.0588,  x_1=3,  f(x_0)= -0.3908,  f(x_1)=16` and in formula (1).

`x_3=\frac{2.0588(16)-3(-0.3908)}{16-(-0.3908)}=2.0813`

Now, `f(2.0813) = -0.14680`(-ive value). Thus it replaces the previous negative value that is `2.0588`. Therefore, root lies between `2.0813` and `3`.

Third Iteration

Taking `x_0=2.0813,  x_1=3,  f(x_0)= -0.14680,  f(x_1)=16` and in formula (1) .

`x_4=\frac{2.0813(16)-3(-0.14680)}{16-(-0.1468)}=2.08965`

Now, `f(2.08965) = -0.05452`(-ive value). Thus it replaces the previous negative value that is `2.0813`. Therefore, root lies between `2.08965` and `3`.
Fourth Iteration Taking `x_0=2.08965,  x_1=3,  f(x_0)= -0.05452,  f(x_1)=16` and in formula (1).
`x_5=\frac{2.08965(16)-3(-0.05452)}{16-(-0.05452)}=2.09274`

Now, `f(2.09274) = -0.02017` (-ive value). Thus it replaces the previous negative value that is `2.08965`. Therefore, root lies between `2.09274` and `3`.
Fifth Iteration
Taking `x_0=2.09274,  x_1=3,  f(x_0)= -0.02017,  f(x_1)=16` and in formula (1). `x_6=\frac{2.09274(16)-3(-0.02017)}{16-(-0.02017)}=2.093882`

Now, `f(2.093882) = -0.007465`(-ive value). Thus it replaces the previous negative value that is `2.09274`. Therefore, root lies between `2.093882` and `3`.
Sixth Iteration Taking `x_0=2.093882,  x_1=3,  f(x_0)= -0.007465,  f(x_1)=16` and in formula (1). `x_7=\frac{2.093882(16)-3(-0.007465)}{16-(-0.007465)}=2.094304`  Now, `f(2.094304) = -0.002754`(-ive value). Thus it replaces the previous negative value that is `2.093882`. Therefore, root lies between `2.094304` and `3`.
Seventh Iteration
Taking `x_0=2.094304,  x_1=3,  f(x_0)= -0.002754,  f(x_1)=16` and in formula (1). `x_8=\frac{2.094304(16)-3(-0.002754)}{16-(-0.002754)}=2.09445`
Therefore, root is `2.094` correct to three decimal places.

Example: Find the root of the equation `xe^x=cosx` in the interval `(0, 1)` using Regula-Falsi method correct to four decimal places.
Solution: Presume `f(x)= xe^x-cosx=0`.
Since here already given the intial roots that is `0` and `1`. To get more accuracy we can find as follow.
Step1:
`f(0.51)=-0.02344 ` (negative value)
`f(0.52)=0.00683`  (positive value). Note: In trigonometry we always calculate values in radian not in degree.
Here sign changes. Therefore root lies in `0.51` and `0.52`.
First Iteration
Taking `x_0=0.51, \; x_1=0.52, \; f(x_0)=-0.002344, \; f(x_1)=0.00683` and using the Regula-Falsi formula.
`x_2=\frac{0.51(0.00683)-0.52(-0.002344)}{0.00683-(-0.002344)}=0.517744`
Now, `f(0.517744) = -0.000041`(-ive value). Thus it replaces the previous negative value ie `0.51`. Therefore, root lies between `0.517744` and `0.52`.
Second Iteration
Taking `x_0=0.517744, \; x_1=0.52, \; f(x_0)=-0.000041, \; f(x_1)=0.00683` and using formula.
`x_3=\frac{0.517744(0.00683)-0.52(-0.000041)}{0.00683-(-0.000041)}=0.517757`
Therefore, root is `0.5177` correct to four decimal places.
EXERCISE
(1) Find the root of the equation `tanx+tanhx=0` which lies in the interval `(1.6, 3)` correct to four significant digits using Regula-Falsi method.
(2) Find the real root of the equation `x^4 -x-10=0` correct to four significant digits by using method of false position.
(3) Solve the equation `2x-log_{10}x=7` which lies between `3.5` and `4` correct to four significant digits using Regula-Falsi method.
(4) Solve the equation `x^3-3x+4=0` which lies between `-2` and `-3` correct to four significant digits using Regula-Falsi method.