Bisection Method: Example

Learn via an example, the bisection method of finding roots of a nonlinear equation of the form f(x)=0. For more videos and resources on this topic, please visit nm.mathforcolle...

[@francismjenkins]

I love this guy, quick and crystal clear (if you don't get bisection after watching this vid, ur probably not made for math) :) But using Excel? Cmon people, write some code in a real programming language lol

[@numericalmethodsguy]

Thank you. To get even more help, subscribe to the numericalmethodsguy channel, and go to MathForCollege.com/nm and MathForCollege.com/ma for more resources and share the link with your friends through social media and email. Support the site by buying the textbooks at www.lulu.com/shop/search.ep?keyWords=autar+kaw&type= Follow my numerical methods blog at AutarKaw.org. You can also take a free online course at www.canvas.net/?query=numerical%20methods

[@statusworlds1726]

Yes you are right they are really not made for maths 🤣🤣🤣🤣🤣🤣🤣

[@bigears8296]

Frank J assholes like you are the reason few people go into maths. My guess is that you yourself are bad at math which is why you fee the need to put others down. :)

[@profautarkaw]

VBA for excel is a powerful language to learn.

[@citymoose]

u sound like a shitty person

[@abdulhalimaslanacier3847]

Shooting from the distance is the only thing that I criticise. Yet, your contents are among the best on youtube. WEll DONE!

[@prawgspinkz8015]

this really helped .. ,my lecturer didnt explain well in details like you did... God bless

[@chyndalefatejabutay56]

Thank you so much! It's understandable! My professor did this but I don't get it maybe because of new normal learning.

[@RSKingdom2012]

Hey, I just wanted to say thanks so much for the video! I had lots of trouble understanding it from my teacher and studied the lecture notes for hours... Then I came here and learnt and fully understand it in 10minutes. THANKS SO MUCH!

[@Pickle312]

Thanks, much more in depth walk through than my professor that really helped me understand the bisection method

[@thetick5321]

Thank you so much. I tanked this on an analysis quiz an was all screwed up. This is way simpler explanation than I had previous.

[@Jota92i]

Iam here after 14 years u are extremely amazing prof and i loved ur way😍🥰

[@numericalmethodsguy]

Thank you!

[@Kalanibrothers20]

sir , you are doing great job. you have cleared all concepts related to the topic.sir, your notes are brilliant.

[@vrushabh1816]

Simple and lucid explanation sir !! Thanks a lot !

[@numericalmethodsguy]

@dheerujma You can choose any real numbers such that f(xl) f(xu)

[@akashsatamkar]

In this example , first we find the midpoint(Xm) and then find the eqn of midpoint f(Xm) and f(Xl) and check for less than or greater than 0. Thats good but another method is only find eqn of f(Xm) and check for f(Xm)>0 or f(Xm)

[@3SmartyPants]

You are an awesome teacher!! I totally now understand the Bisection Method (with a few minor questions for clarification).

[@navnitlaxmiupadhyay1473]

Thanks Sir, I like you teaching way and its very clean for us.

[@Ultimatejellyfighter]

nice video... i understand it completely on my first watch... u taught me well.

[@Richard30006]

this is very clear and precise, none of that fancy jargon, thank you so much!

[@MaxVelez]

Great video needed to learn this concept for an early BME software course.

[@sharkbandre8278]

excellent explanation, Thank you!

[@FAHADkt22]

thank you sooooo much... I got an exam tomorrow before watching your videos i didn't understand now i can go to the exam with confidence. thanks again :)

[@michellefyyourelife]

Please make a video with an example of regula falsi (= false position method) P.S. you're video's are INCREDIBLY helpfull!

[@numericalmethodsguy]

Watch the videos made by my colleague: nm.mathforcollege.com/topics/false_position.html

[@omarshahid4447]

Explanation was amzing I can just say tht... Thnks alot

[@unwelcomedguest9741]

Excellent explenation, totally 10x better than our university Dr.

[@numericalmethodsguy]

For more videos and resources on numerical methods, please visit nm.mathforcollege.com

[@nishantsharma8776]

you are the best tutor over entire the world.

[@numericalmethodsguy]

Thank you. To get even more help, subscribe to the numericalmethodsguy channel, and go to MathForCollege.com/nm and MathForCollege.com/ma for more resources and share the link with your friends through social media and email. Follow my numerical methods blog at AutarKaw.org. You can also take a free online course at www.canvas.net/?query=numerical%20methods

[@TheSandy636]

Sir, Thanks a lot.. You made it so easy...

[@numericalmethodsguy]

@Luan Selimi if you do not have them, then use the physics of the problem to find the two guesses. If this is not possible, make sure that the two guesses bracket the root, that is, the function values at the two guesses are opposite in sign.

[@maryameqan]

I'm being from a total non-math background grasp everything he says really quickly

[@RafaelLima-ox9ul]

Thank you so much! This was very clarifying.

[@2112dim]

Very good,just started today studying such numerical methods.

[@RicardoTosin]

nice, bisection method is not so hard to understand and jus because you are nice to explain that, i got final exam and you will save me =D thanks !

[@Muck-qy2oo]

I could approximate logarithms like that! Might that work quicker if implemented in the newtons method?

[@numericalmethodsguy]

I do not understand the question. You can use Newton Raphson method for any equation that can be solved by bisection method until you can find f '(x), and it is finite.

[@numericalmethodsguy]

@TheManishchannel For example the value of f(1)=-19, f(4)=44. So if you would plot the function f(x) from x=1 to x=4, the function value is negative at x=1 and positive at x=4. So it changes sign as you plot it from x=1 to x=4. That means that the function has to cross the x-axis between x=1 and x=4. Click on the link at 0:20 of the video to follow the background. Go to numericalmethods(dot)eng(dot)usf(dot)edu and click on Keyword. Click on Bisection method. You will see more resources.

[@LilmasCheerPSP]

what's a good way to study for a numerical methods final?

[@rboro6969]

13. The equation x 3 + x 2 − 3x − 3 = 0 has a root on the interval (1,2), namely x = √ 3.

[@numericalmethodsguy]

Write the equation as f(x)=sin(x)-x=0 Check of f(xl) and f(xf) change sign. Then follow the same logic as the above example. Remember arguments of sin are in radians!

[@ankurrrrrrrrr]

Sir, you are awesome. Thankyou for help, so nice of you.

[@zahrahbahrah6814]

This guy is amazing

[@maheshmahee7092]

It`s depends upon comparing the relative approx error with pre specified tolerance. if |ea|

[@muffihottie]

When do we stop doing iteration if it is given to us that determine the root that is atleast accurate to within 10^-4?

[@autarkaw1826]

You can continue to iterate till absolute approximate error is

[@AshwinAshwinRamdas]

this information is RIGHT. thanks for the example problem

[@numericalmethodsguy]

The formula n=integer[ln(Xu-Xl)-ln(Ead)/ln2] is a little incorrect. It should be n=integer[(ln(Xu-Xl)-ln(Ead))/ln2]. Also, since Xu>Xl is not required, it should be n=integer[(ln|Xu-Xl|-ln(Ead))/ln2] =integer[ln(|Xu-Xl|/Ead)/ln2] - note the absolute value.

[@austinoligario2285]

What happens if we aren't given an initial bracket? f(x) = x^6 − x − 1 = 0. "Use bisection method to find the largest real root α of this equation with 4 decimal place accuracy." I am not looking for the solution, I just want to provide some context. How would I find the X_l and X_u?

[@numericalmethodsguy]

+Austin Oligario This equation could have 6, 4, 2 or 0 real roots. Unless, you plot the LHS of the equation, it would be hard to pinpoint largest real root, and hence a suitable XL and XU bracket.

[@CarlosDominguez-yr1ic]

I used a hand calculator, and I got the following roots for x^6-x-1=0 X1 = 1.134724138 X2 = 0.4510551586 + 1.002364572i X3 = 0.4510551586 - 1.002364572i X4 = -0.6293724285 - 0.735755953i X5 = -0.6293724285 + 0.735755953i X6 = -0.7780895987 So you can use the intervals [1, 1.5] and [-1,-0.5] to practice the bisection method. The complex roots could be found using Newton-Raphson method. Best regards from Venezuela. Carlos Vicente Dominguez

[@anooshiravanensafian7860]

sir, why your initial guesses are 1 and 4? according to which equations have you chosen these?? can we take 1 and 3 ? since 3^3 is greater than 20 again :-?

[@numericalmethodsguy]

Yes, you can take 1 and 3. I chose initial guesses only by observation.

[@divyapandey7276]

how can we guess the intial value of x1 and x2 so that we can finish in short iteration??

[@mohammedabbas5699]

awesome thank u did u have fix point method?

[@ManuelCollinsBarud]

Excellent video! Thanks from Mexico!

[@lapitburaytitibuday]

dude! do you have an example on fixed iteration method? I really needed it! I'll put five stars if you explained it well

[@sadahahmed5034]

man you are the BEST

[@lostdreamer3308]

Thank you very much sir.

[@9dubb9]

Thanks allot...I will ace this on my exam tomorrow/ 7.5hours

[@jalosaidu]

wow great, easy to understand

[@donaztek]

Hello sir , is there a mathematica sheet work for this method ? the one on your website is an old version thank you keep the good work

[@numericalmethodsguy]

Sorry, I am only keeping up with MATLAB.

[@donaztek]

Yes i understand MATLAB is more commonly used

[@ramchandrapaudel7890]

first of all thanks for all ur support. when desired accuracy exactly equals to functional value should i consider that the root or go for one more iteration????

[@numericalmethodsguy]

If it is equal, you are OK.

[@Jonny23Baller]

quick question, if f(a)f(b)>0 on the first try before getting into the iteration... would we conclude there aren't any roots on that interval? thank you

[@StrzelbaStian]

Most certainly

[@udayrallabhandi3345]

Or there are an even number of roots in that interval.

[@numericalmethodsguy]

@327372 The concept is based on the fact that the maximum true error in the root at the end of an iteration is width of the new bracket. Keep in mind that the specified error Ead is the error, not the relative error. Do a google search on "Bisection Error Analysis" - the #1 result shows the proof!

[@rutujachavan633]

Thank you so much sir 😊

[@kabronponcho]

thank you so much sir, ! thanks thanks thanks~

[@numericalmethodsguy]

@dheerujma Go to numericalmethods(.)eng(.)usf(.)edu and click on Keyword. Click on bisection method. You will see programs written in MTALAB, etc. Modify as needed!

[@foodtruckfactory1598]

You are amazing, thank you sir!

[@michaeljohnmagistrado1166]

is there a way of finding a root if the funcction doesnt pass through the x axis?

[@numericalmethodsguy]

miguel juan adjudicator The equation most probably has complex roots then. You can use Mullers' method for that. en.wikipedia.org/wiki/Muller%27s_method

[@TTabancaTT]

thank you very much for upload this video.

[@prajwolpaneru6975]

when do i know that i have to stop calculating if the no. of iteration is not given?

[@numericalmethodsguy]

Prajwol Paneru You can choose a pre-specified tolerance. When the absolute relative approximate error is less than or equal to the pre-specified tolerance, you can stop. To see how this works and its relationship to see how many significant digits are correct in your answer, see page 5 and 6 of this document: mathforcollege.com/nm/mws/gen/01aae/mws_gen_aae_spe_measuringerror.pdf

[@adityavajpayee6577]

u need to see previous parts of the bisection method series for your queries.. everything is as clear as it can be...

[@Silvyanutza]

Hi mister!I really apreciate your labour!It is really useful!What i wanna ask you is! After n iteration i should find the unique solution? sqrt of range 3 from 20? Thanks in advance for the answer!

[@sheikhejaz]

thank you sir very much very use ful for begginers

[@vaibhavjoshi8301]

i need to know till when do we perform the iterations?

[@fb-gu2er]

numericalmethodsguy can i suggest a topic: fixed point theorem?

[@kalidasrajendran1754]

sir if possible add fixed point theorem n banach fixed point theorem ....

[@mubarakmohammed3551]

do you have an illustration for regular false method please

[@numericalmethodsguy]

nm.mathforcollege.com/topics/false_position.html

[@MrKosova092]

what if we don't have Xl and Xu ?? do we have to put them as we wish .. for example if we have your exam but we don't have Xl and Xu , what to do ??

[@jorgemilhomem3274]

If f(a)*f(b)>0, it doesn't mean that between [a,b] there is no zero, because there can exist a x value, c (c is between a and b), which f(c)0. And I think you did that mistake in this video, am I wrong?

[@autarkaw1826]

Well, what is said in the video is that if f(a)*f(b)>0, there is no guarantee of a root between a and b. A root may or may not exist. Yes, if f(a)*f(b)>0, roots are possible but cannot be guaranteed! This is what is said in the video -"okay, hey, there is xl, and there's your xu, and says, okay, you have two points now, xl and xu right here, and what is happening is that the function is not changing sign, because here the value of the function is positive, and the value of the function here is positive, but you're getting two roots . . . you're getting two roots between xl and xu in spite of the function not changing sign. That is not a violation of the theorem which we just discussed. The reason why this is not a violation of the theorem which we just discussed is because the theorem only tells you if the function is changing sign. It's not . . . it does not tell you anything about if the function is not changing sign, whether there are going to be any roots between the two limits or not. So that's one thing which you've got to understand is that the theorem's only telling you if the function is changing sign, that you are going to get at least one root. If the function is not changing sign, as is the case here, then there may or may not be roots between those two points."

[@jorgemilhomem3274]

***** Correct! Ok, thank you!

[@harisanthoshkumar7449]

Sir, why do you xu = 4 but not 3

[@numericalmethodsguy]

You can use any lower and upper guess till the function changes sign. See the physical problems here under NONLINEAR EQUATIONS to see how the physics of the problem can help in deciding what the lower and upper guess should be. nm.mathforcollege.com/physical_problems_text.html

[@numericalmethodsguy]

There is a formula which can be used to determine the number of iterations. n=integer[ln(Xu-Xl)-ln(Ead)/ln­2] Ead= specified error for stopping criterion which should be given. The concept is based on that the maximum true error in the root at the end of an iteration is width of the new bracket. Do a google search on "Bisection Error Analysis" - the #1 result shows the proof! Go to nm(dot)mathforcollege(dot)com and click on Keyword. Click on Bisection method. You will see more resources.

[@topdecktunes]

Very helpful, thanks

[@kashishaggarwal9647]

sir if the initial values i.e x1 & x2 are not given that what should we do?

[@profautarkaw]

Well, you got to choose such that f(x1)f(x2)

[@kashishaggarwal9647]

@@profautarkaw thank you sir

[@numericalmethodsguy]

@Silvyanutza If you would rephrase your question, I would be able to answer your question!

[@namelastname7103]

Excellent

[@AnkitSaiyan]

why 1 and 4 why not 1 and 3?

[@ahmedmohamed-yx1ln]

great video

[@khirnafazlina3763]

hi sir, may i know what is pre-specified tolerence? and how do we know the value?

[@Aguvika]

Hello Sir, Since we know that F(XL)0. Why do we have to form a check around F(XL)*F(XU)??.. Why not just check the sign of F(XM) if it is negative it becomes XM becomes XL if positive it becomes XU.

[@numericalmethodsguy]

It is not guaranteed that f(xl)0. For example for f(x)= - x^2+4=0, it is f(xl= - 1)=3, f(xu=3)= - 5. Checking f(xl)*f(xu)

[@Aguvika]

numericalmethodsguy Ahh ok..thanks for the quick response!!!!!

[@ashutoshpatil26]

thank you sir

[@Endisupertramp]

This was incredibly helpful,hank you a lot!

[@adlanehichembriki1831]

Hello, Thank you for this For the fifth iteration, i found xm=2.59375, while u found it 2.78125, ?

[@numericalmethodsguy]

iter xl xm xu error(%) 1 1 2.5 4 2 2.5 3.25 4 23.076923 3 2.5 2.875 3.25 13.043478 4 2.5 2.6875 2.875 6.9767442 5 2.6875 2.78125 2.875 3.3707865

[@manjunathhipparagi8493]

How to obtain initial guesses if not given?

[@profautarkaw]

Physics of the problem can help. See how we can use this for example given here mathforcollege.com/nm/mws/gen/03nle/mws_gen_nle_txt_bisection.pdf

[@muhammadsyahiransuhaimi8488]

what if the question for sin(x) = x given xl = 0.5 and xu= 1 ?

[@VilleLaitila1]

# Bisection method coded in Python def function(x): return x*x*x-20 def iterate(bracket,accuracy): print bracket midpoint = ( bracket[0]+bracket[1] ) / 2.0 v1 = function( bracket[0] ) vm = function( midpoint ) if abs(v1-vm) < accuracy: return iterate( (bracket[0], midpoint) if v1*vm < 0 else (midpoint, bracket[1]), accuracy ) iterate( (1.0,4.0), 0.00001 )

[@ezmoney2008]

Thank you!

[@choudharynikunj]

thank you so much

[@AmitSealAmi]

thanks a lot :) a very good and understandable tutorial :)

[@AneesIsmail]

Iter 5 = 3.28125 NOT 3.78125 is that right

[@tristonveen9210]

How do you Calculate the solution error?

[@numericalmethodsguy]

You calculate the absolute relative approximate error. See videos or textbook chapter here: nm.mathforcollege.com/topics/measuring_errors.html

[@numericalmethodsguy]

What information is wrong. Please explain. Accusations without proof can discourage learners.

[@gracellove]

thank you very much!!!!

[@anumitasingharoy1819]

thank u so muuch its grt

[@anupadhikari4525]

plz tell where do we stop if it is not given where we stop...

[@numericalmethodsguy]

entertainment free You need to decide. You can say, for example, that I want at least 3 significant digits to be correct. Then you will continue iterations until absolute relative approximate error is less than 0.05%.

[@anupadhikari4525]

Wait I will provide you questions.

[@jasonlee4347]

thanks sir!!

[@wajidali-ew5pr]

very good

[@usmansaleh4511]

thanks for this video, 2maro is my paper of numerical, pry 40r me.

[@hexassasin6212]

Hey guys if you are solving a sinusoidal function then switch your calculator to RADIAN from degree. I made that mistake initially.

[@mahmoudsaad7469]

thank you my teacher