use Newton's Method, NOT WolframAlpha!

Рет қаралды 36,720

2 жыл бұрын

Here we will use Newton's method to approximate the solution to the quartic equation 3x^4-8x^3+2=0 on the interval [2, 3]. This is one of the most common ways to approximate the roots of an equation and the idea is based on the equation of tangent lines.
👉 How to get the newton's method formula: • Newton's method (intro...
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Пікірлер: 37

@bprpcalculusbasics 2 жыл бұрын

👉 How to get the newton's method formula: kzfaq.info/get/bejne/n7x_prhl2dTTpZ8.html

@maximeclere9902 2 жыл бұрын

4:58 XD the answer has 2.7 and 3.14 in it

@WerewolfLord 2 жыл бұрын

4:55 The answer is approximately 2.7.....[suspenseful music plays*] *it doesn't but it should

@owoLight 2 жыл бұрын

*audible gasp*

@tejarex 2 жыл бұрын

Newton's method works great when it works, but it fails when the initial guess is not good enough, which is pretty easy in general. Given a continuous function f and 2 points where f has opposite signs, finding an intermediate root by bisection, always keeping two points with opposite signs, is guaranteed to work. For this example, assuming that the 2.63 root is unique, one would check 2.5, 2.75, 2.625, 2.6875, 2.65625, etc.

@Fandikusnadi1979 2 жыл бұрын

12x^3 not 12x^2

@boldspider6923 6 күн бұрын

very straightforward, thanks Bro

@Tiwey 2 жыл бұрын

Isnt this the differentiating quotient of the normal?

@T-zm6ft 3 ай бұрын

Excellent video👍

@AviPars 2 жыл бұрын

Thanks

@robertveith6383 2 жыл бұрын

@ blackpenredpen -- I don't see your rationale for picking x_1 to be 3. I would have averaged the endpoint values and used 2.5 instead.

@NattyPi 2 жыл бұрын

my profesor didnt really explain it either and what sucks is we have an exam coming up and all the examples have been to 6 decimal places which means if you don't pick a really good starting point you will end up spending 30 minutes on this. In class we ended up on xsub11

@bradley9856 2 жыл бұрын

So the answer was closer to 3 anyway? That's interesting. Good video ty

@kiit8337 2 жыл бұрын

Where is the full lecture for newton method?? Help plz

@MyBroSux24 2 жыл бұрын

check on 3 blue 1 browns brilliant video where he not only shows how newtons method works but how its applied to resemble a fun pattern.

@kepler4192 2 жыл бұрын

now do it with euler's method now

@hanantoaditya4139 2 жыл бұрын

Is this numerical methods?

@bprpcalculusbasics 2 жыл бұрын

Yes

@woofy2000 2 жыл бұрын

π'th

@diniaadil6154 2 жыл бұрын

fancy way to just say you're third

@moskthinks9801 2 жыл бұрын

@@diniaadil6154 nah, according to the comments, π = 5

@riendmuskus4050 2 жыл бұрын

@@moskthinks9801 look at timestamps looks like he was after 3 but before 4 so it checks out Should've been 2.63rd though

@mathiasschembri5207 2 жыл бұрын

he means π in radians

@francocosta1 2 жыл бұрын

Wolfram rules!

@gatocomcirrose 2 жыл бұрын

i mean, if you have to do this by hand it's just better to use wolfram alpha

@TristanANain Жыл бұрын

Thank you sir for the lesson but the sad face tho😂

@lukopro3655 2 жыл бұрын

Fourth

@marmot1434 2 жыл бұрын

sixth

@robinson5923 2 жыл бұрын

Ωth

@priyanshrawat442 2 жыл бұрын

That pokemon ball mic is cool

@umarfayziev5235 2 жыл бұрын

Third

@dublindark7945 2 жыл бұрын

Second

@gamerpool9000 2 жыл бұрын

First

@henrybeenh7076 2 жыл бұрын

Nothing 's wrong with you

@robertveith6383 2 жыл бұрын

*@blackpenredpen* -- Here is a way without calculus: Set 3x^4 = 8x^3 - 2. Divide each term by 3x^3: x = 8/3 - 2/(3x^3). If you let x_1 = 3, then x_2 ~ 2.6419753, x_3 ~ 2.6305154, ... , x_6 ~ 2.6300203

@josevidal354 2 жыл бұрын

Though its correct that You approach the answer,, it doesnt always work. For example, lets Say You want to calculate another root of the same polynomial that is about 0.696889... Using the same method, lets start with a close one: x_0=0.6968, then x_1=0.6961, x_2=0.69044, x_3 = 0.6412, x_4 = 0.1378, x_5 = -251.83, x_6 = 2.66... if You continue You eventually aproach the other root 2.63002... but not the one we wanted. How can You know When is it ok to do this? The answer lies In calculus:: You have to make sure that the root You want to get has an absolute value of it's derivative at the function f(x) = 8/3-2/3x³ less than one.