# Solving Equations Using Bisection, Newton, and Secant Methods

• MHB
• lunia
In summary: For the Newton method, try with several different initial guesses including x0 = 1.For the Secant method, use the same initial values as for the Bisection, as well as other appropriate values.For the bisection method, the number of iterations required by that method depends on the accuracy you're trying to achieve.For an accuracy of 10−6, it takes around 10 iterations to convergence.The behavior you observe in this problem seems to match theoretical predictions where they have been given.In summary, the bisection, Newton, and Secant methods all converge to a solution within 10−6 iterations for an accuracy of 10−6. They each have their own advantages and disadvantages, but
lunia
Use the Bisection, Newton, and Secant methods to solve (to at least 8
signicant figures) the equation
sin(x) = 0.98 cos(2x2)
over the interval [0, 2.5], in radiant units. For the Newton method, try with
several different initial guesses including x0 = 1. With the Secant method,
use the same initial values as for the Bisection, as well as other appropriate
values. Comment on the results by comparing the effectiveness of
each method for this problem. In particular, compare the number of iterations
required by each method to find the solution to a specied accuracy
(at least|f(xroot)| ≤ 10−6
), and see if the behavior you observe tallies with
theoretical predictions where they have been given, in particular the order of
convergence. Provide a careful explanation on your observations.
Hint
Draw a graph of the equation to understand the behaviors of the equation,
such as the number of solutions and the locations of the roots.

p/s: basicly i don't even know what this question want exactly, i do try my best to understand it but i cant. I am sorry

Well, the hint says to start with a plot, which is also what I would do. What do you get for that?

i don't know how to start plot the graph actually.

lunia said:
i don't know how to start plot the graph actually.

Hey lunia, welcome to MHB! ;)

For quick reference, you can copy and paste your formula into Wolfram|Alpha.
The graph looks like this:
View attachment 8600

Can you tell what the number of solutions is now?
And what the locations of the roots are approximately?

#### Attachments

• sin_vs_cosine_plot.png
5.6 KB · Views: 78
tbe number of solutions is 3.
locations of the roots are approximately: 0.7, 1.8 and 2.4.

did you just copy the exactly same like the formula or how did your write it? i didn't get it yet

lunia said:
tbe number of solutions is 3.
locations of the roots are approximately: 0.7, 1.8 and 2.4.

Couldn't there be 2 roots around 1.8 though?
lunia said:
did you just copy the exactly same like the formula or how did your write it? i didn't get it yet

I copied and pasted [M]sin(x) = 0.98 cos(2x2)[/M] into https://www.wolframalpha.com, which shows most of what you might want to know about that equation.

After that I refined it into [M]plot[sin(x) = 0.98 cos(2x2), {x, 0, 2.5}][/M] to get just the graph on the interval [0, 2.5].Anyway, to apply Bisection or any of the other methods up to 8 significant digits, you will need to use a computer or some other type of calculator.
Can you clarify how are you supposed to apply these methods?

i use excel to apply the method. but how to figure out the the root before as a and root after as b?

lunia said:
i use excel to apply the method. but how to figure out the the root before as a and root after as b?

It's not entirely trivial to do this with Excel.
Did you perhaps get something that was prepared for you?
You can google for the algorithm, but that really should have been provided to you.

Anyway, for bisection we start with the interval [0,2.5], and then make it smaller in each iteration (we 'bisect' it).
In the first iteration we consider [0,1.25] and [1.25,2.5].
If the function has opposite signs at each end of the sub interval, there must be a root in it.
We keep going with the interval that has a root until we reach the desired precision.

For Newton and the Secant method, your problem statement provides guidance how to start those methods.

## What is the purpose of solving equations using bisection, Newton, and secant methods?

The purpose of these methods is to find the roots or solutions of equations that cannot be solved algebraically. This is especially useful in scientific and engineering applications where equations can be very complex and have no exact solution.

## What is the difference between bisection, Newton, and secant methods?

Bisection method involves repeatedly dividing the interval in which the root lies and checking which subinterval contains the root. Newton method uses derivatives to approximate the root, while Secant method uses two points on the curve to create a secant line, and then iteratively approaches the root by finding the intersection of the secant line with the x-axis.

## What are the advantages of using these methods?

These methods are more efficient and accurate compared to other numerical methods for finding roots. They also have the advantage of being able to find both real and complex roots.

## What are the limitations of these methods?

These methods may not always converge to the true root, especially if the initial guess is far from the actual root. They also require an initial guess for the root, which may not always be readily available.

## How are these methods used in real-world applications?

These methods are used in various fields such as physics, engineering, economics, and finance to solve complex equations that cannot be solved algebraically. They are also used in computer programming and software development for solving equations and optimizing functions.

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