Determination of any roots of any function by Newton Raphson method is known. One condition function must be continous and derivable.

According to the Newton Raphson method, root of a function can be calculated by the following method;

This equation gets closer to the root as a result of consecutive iterrations. However, the potential hardness is that taking derivatives becomes very though and time consuming for some functions.

This is the main idea of secant method. When we define calculations of derivatives, the formula to determine root becomes like the following form.

Last obtained equation is known as secant method. Secant method is an open method. It can be observed that how it is tending to the root.

Red line indicates the function we search for. Roots of blue lines are the ones determined after each iterrations.

**MatLab Script Code for Secant Method**

a=input('enter function:','s');

f=inline(a)

x(1)=input('enter first guess: ');

x(2)=input('enter second guess: ');

n=input('enter tolerance: ');

iteration=0;

for i=3:1000

x(i) = x(i-1) - (f(x(i-1)))*((x(i-1) - x(i-2))/(f(x(i-1)) - f(x(i-2))));

iteration=iteration+1;

if abs((x(i)-x(i-1))/x(i))*100<n

root=x(i)

iteration=iteration

break

end

end

** Output**

**Visual Basic Code for Secant Method**

Public Class Form1

Private Function f(ByVal X As Double) As Double

Dim a As Double

Dim b As Double

Dim c As Double

a = Val(TextBox1.Text)

b = Val(TextBox2.Text)

c = Val(TextBox3.Text)

Return a * X ^ 2 - b * X + c

End Function

Private Sub Button1_Click(ByVal sender As System.Object, ByVal e As System.EventArgs) Handles Button1.Click

Dim i As Integer

Dim u As Double

Dim d As Double

Dim k As Double

Dim tol As Double

u = Val(TextBox4.Text)

d = Val(TextBox5.Text)

tol = Val(TextBox6.Text)

ProgressBar1.Minimum = 0

ProgressBar1.Maximum = 100

For i = 1 To 10000

k = u - (f(u) * ((u - d) / (f(u) - f(d))))

u = k

ProgressBar1.Value = 100

If Math.Abs((k - u) / k) * 100 < tol Then

Exit For

End If

Next

MsgBox(k, MsgBoxStyle.OkCancel)

End Sub

End Class

**SECANT METHOD EXAMPLE MANUAL SOLUTION**

Using secant method, determine the roots of the function of

**SOLUTION**

iteration 1

iteration 2

iteration 3

Hi, may i know how to find the roots for;

Secant Method:

f(x)=x^3 – 2x^2 – 13x – 10

using the MATLAB ?

thank you…^_^

Roots of your function are -2,-1,5.

You can use this script code.

We suppose your first guess is 4 and second guess is 5. In this situation Secant Method is going to find 5(root).

f=inline(‘x^3-2*x^2-13*x-10′)
root=x(i)

x(1)=input(‘first guess: ‘);

x(2)=input(‘second guess: ‘);

n=input(‘tolerance: ‘);

iteration=0;

for i=3:1000

x(i) = x(i-1)-(f(x(i-1)))*((x(i-1)-x(i-2))/(f(x(i-1))-f(x(i-2))));

iteration=iteration+1;

if abs((x(i)-x(i-1))/x(i))*100

iteration=iteration

break

end

end

Hi

May i know how to find the roots for

f=(3*(p^2)*pi*u*d)+(.05*p*pi*(d^2)*(vo^2))+(.75*p*pi*(vo^2)*(d^2)*((1+(u^-.5)*(p^.5)*(vo^.5)*(d^.5))^-1))-(m*g);

using MATLAB

thanks

i forgot to add that

g=9.81;

u=1.8e-5;

p=1.29; %which are all constants

and i am trying to find vo as the root.

How do i write an m-file function using secant method, where there are two(2) inputs which are ‘m’ and ‘d’

Hi Jennifer

Also we sent e-mail to your adress.

m=input(‘enter m:’);

d=input(‘enter d:’);

g=9.81;

u=1.8e-5;

p=1.29;

f=@(x) ((3*(1.29^2)*1.29i*1.8e-5*d)+(.05*1.29*1.29i*(d^2)*(x^2))+(.75*1.29*1.29i*(x^2)*(d^2)*((1+(1.8e-5^-.5)*(1.29^.5)*(x^.5)*(d^.5))^-1))-(m*9.81))

x(1)=input(‘enter first guess: ‘);

x(2)=input(‘enter second guess: ‘);

n=input(‘enter tolerance: ‘);

for i=3:1000

x(i) = x(i-1) – (f(x(i-1)))*((x(i-1) – x(i-2))/(f(x(i-1)) – f(x(i-2))));
root=x(i)

if abs((x(i)-x(i-1))/x(i))*100

break

end

end

halo

i new in matlab

i want to know how to get the answer for fn f(x)=x^3-6x^2+11×-6.1 for x1=2.5, xU=3.5, ea<10^-3 by using secant method in matlab . i dont know how to start it in matlab. hope u can help me sir.

hello!

i wrote this code, but matlab doesn’t accept it.

It shows me

??? Error using ==> mupadmex

Error in MuPAD command: Index exceeds matrix

dimensions.

Error in ==> sym.sym>sym.subsref at 1381

B =

mupadmex(‘symobj::subsref’,A.s,inds{:});

What do i have to do??

my code:

>> f = inline(‘ tan(0.58*x+0.1)-x^2′,’x’);

>> x1 = input(‘first guess: -2′);

first guess: -2

>> x2 = input(‘second guess: 0′);

second guess: 0

>> n = input(‘tolerance: 0.0001′);

tolerance: 0.0001

>> iteration = 0;

>> for i = 3:1000

x(i) = x(i-1)-(f(x(i-1)))*((x(i-1)-x(i-2))/(f(x(i-1))-f(x(i-2))));

iteration = iteration+1;

if ((abs((x(i)-x(i-1))/x(i))*100 ) < n)

root = x(i)

iteration = iteration

break

end

end

f=inline(‘tan(0.58*x+0.1)-x^2′)
root=x(i)

x(1)=-2;

x(2)=0;

n=0.00001;

iteration=0;

for i=3:1000

x(i) = x(i-1) – (f(x(i-1)))*((x(i-1) – x(i-2))/(f(x(i-1)) – f(x(i-2))));

iteration=iteration+1;

if abs((x(i)-x(i-1))/x(i))*100

iteration=iteration

break

end

end

Can you help me to find how solve these exercises using secant iteration method(matlab script code for secant method)?

Find all zeros(five decimals correctly)

1. f(x)= 0.5(5x^3-3x)

2. f(x)=x cos(x)+sin(x)

3. f(x)= 35x^4-30x^2+3/

8

Can anybody help me please? Determine the highest real root of f(x) = x3 – 6×2 + 11x – 6.1

a) Graphically,

b) Using the Newton-Raphson method (three iterations, xi = 3.5)

c) Using the Secant method (three iterations, xi-1 = 2.5 and xi = 3.5).

i wrote this code, but matlab doesn’t accept it.

It shows me

??? Subscript indices must either be real positive integers or logicals.

Error in ==> newton at 3

f(C)=4.04+(3.94)*log(R*(sqrt(C)));

Error in ==> run at 57

evalin(‘caller’, [s ';']);

please help me

this is my code

C=0.0001;

R = 10^3:1000:10^5;

f(C)=4.04+(3.94)*log(R*(sqrt(C)));

df(C)=(1.97/C)+ (1/(2*C^1.5));

C0 = 0;

N = 10;

tol 10^(-14);

Iterations

for i = ({1 to N});

if df(C0)==0;

end

C=C0-(f(C0)/df(C0));

if (abs(C-C0)<tol)||(abs(f(C))<tol)

root=C;

break

end

end

if i==N;

end

C0 = C;

output

root;

plot(Cx)

how would you do this with modified secant method?

May i know how to find the roots for

f=x^3+3×-1=.0

using MATLAB

thanks

Hi Jerose,

You can use this;

how may get the root of the function using secant method (MATLAB), for three iterations with initial guess of X as (0 and 1). y(x)=cosX-Xexp^X

here is the question.

You are to write a short computer programme with MATLAB programming

language (without using existing Library functions) that will generally obtain

the bracket (xO1, xO2) containing the real root of this function within the

range (0, 4) at the interval of +1.0 and real root of this function using

SECANT and REGULA-FALSI methods. In developing the program, use the

bracket obtained as the initial guesses, maximum number of iteration equal

20 and error tolerance is to be less than 0.100E-03. Print in an output file:

the number of iterations, the initial guesses, and the real root and the value

of this function f(x) for this root.

Hello,

First of all: Thank you very much for your post! it has helped me A LOT!. However, I would like to know if you can help me by showing me how I can get to display the approximate root for each iteration. For instance, if it takes 9 iterations to get to the root I want to not only be able to see the root and the number of iterations, but also I want to see the approximate root for iteration 1, 2, 3, and so on.

Hope that you can help me!

Thanks!

You can use this script I did edit;

a=input(‘enter function:’,’s’);
root=x(i)

f=inline(a)

x(1)=input(‘enter first guess: ‘);

x(2)=input(‘enter second guess: ‘);

n=input(‘enter tolerance: ‘);

iteration=0;

for i=3:1000

x(i) = x(i-1) – (f(x(i-1)))*((x(i-1) – x(i-2))/(f(x(i-1)) – f(x(i-2))));

iteration=iteration+1

approximate=x(i)

if abs((x(i)-x(i-1))/x(i))*100

iteration=iteration

break

end

end