Cube Root of a Non-Perfect Cube Number

[@note_07]

would be really helpful for entrance exam. hope you do more similar content!

[@nonsuspicious]

Approximating any n-th root: f(x) = xⁿ df/dx = d/dx (xⁿ) = n x ⁿ⁻¹ df = n x ⁿ⁻¹ dx df will be a change in y depending on an infinetely small change of x (dx) For approximating we can plug in our Δx instead of dx Δf ≈ n x ⁿ⁻¹ Δx => f(x + Δx ) ≈ f(x) + n x ⁿ⁻¹ Δx => Δx ≈ [ f(x + Δx) - f(x) ] / [ n x ⁿ⁻¹ ] x + Δx will be our answer f(x + Δx) will be our number inside the n-th root f(x) will be the the closest known perfect n-th power number x then is the n-th root of the known n-th power Final: ⁿ√a ≈ x + Δx = x + ( [ a - f(x) ] / [ n x ⁿ⁻¹ ]) Example: lets approximate 4th root of 17 plug in n = 4 known root -> ⁴√16 = 2 then x = 2 f(x) = 16 ⁴√17 ≈ 2 + ([ 17 - 16 ] / [ 4 2 ⁴⁻¹ ]) = 2 + (1 / [ 4 2³ ]) = 2 + (1 / [ 4 8 ]) = 2 + (1/32) = 2.03125 ⁴√17 ≈ 2.03125

[@sytherplayz]

Amazing! I liked your method coz its pure mathematical. 🎉❤

[@codenameconehead]

What the hell is this

[@nonsuspicious]

​​@@codenameconeheadapproximation by derivatives of any n-th root

[@sriramsriram2871]

I really appreciate your helping sense. Me - god please make more people like him to help students😅😅😅😅 like this. God - i can make more of him but i don't😝😝😝 Me - heartless god😒😒

[@riyansh4392]

How did you get 32 in your last step

[@vitowidjojo7038]

Is there a general formula of this for every value of root?

[@sugaku9455]

Linearization of the function ⁿ√x

[@vitowidjojo7038]

@@sugaku9455 how does that work?

[@nonsuspicious]

Approximate using derivative of the function

[@thehawk69]

use taylor series expansion upto 2nd term

[@nonsuspicious]

Let's approximate using derivatives: f(x) = xⁿ df/dx = d/dx (xⁿ) = n x ⁿ⁻¹ df = n x ⁿ⁻¹ dx df will be a change in y depending on an infinetely small change of x (dx) For approximating we can plug in our Δx instead of dx Δf ≈ n x ⁿ⁻¹ Δx => f(x + Δx ) ≈ f(x) + n x ⁿ⁻¹ Δx => Δx ≈ [ f(x + Δx) - f(x) ] / [ n x ⁿ⁻¹ ] x + Δx will be our answer f(x + Δx) will be our number inside the n-th root f(x) will be the the closest known perfect n-th power number x then is the n-th root of the known n-th power Final: ⁿ√a ≈ x + Δx = x + ( [ a - f(x) ] / [ n x ⁿ⁻¹ ]) Example: lets approximate 4th root of 17 plug in n = 4 known root -> ⁴√16 = 2 then x = 2 f(x) = 16 ⁴√17 ≈ 2 + ([ 17 - 16 ] / [ 4 2 ⁴⁻¹ ]) = 2 + (1 / [ 4 2³ ]) = 2 + (1 / [ 4 8 ]) = 2 + (1/32) = 2.03125 ⁴√17 ≈ 2.03125

[@NicholasOfAutrecourt]

I know that this is for students who haven't had calculus and therefore don't know where this approximation comes from, but it would be great to see a video with a generalized formula for how to take the nth root of any number.

[@rewolff2]

This is about derivatives. For any function, you can approximate it by taking the value at the closest known point and adding the difference from the wanted point times the derivative. Say I want sin(6.28) I know sin(2pi) ==0, so I take 0 + (6.28-2pi) * sin'(0) and get 0 + -0.003 * 1 = -0.00318 ... five digits because 6.28 is so close to 2pi. How much is 6.1 cubed? derivative of the cube function is 3 x^2, so 6 cubed = 216, add .1 * 3*36 = .1*108 , so you get 226.8 . Excecise: What's the fifth root of 1000? No calculator allowed.

[@NicholasOfAutrecourt]

@@rewolff2 Actually, it's not about derivatives at all. There's no mention of calculus at all in the video, and you don't have to know calculus to use the approximation. I realize that calculus is used in the derivation of the approximation. I was asking for the sake of students who might be watching. And thanks, but I don't need an "excecise" (sic). I teach this stuff for a living.

[@rewolff2]

@@NicholasOfAutrecourt OK. If you refuse to see the link, then you get to memorize a whole bunch of "calculation rules" for each function one. Fine by me. I've always been very bad at memorizing stuff. In primary school, I saw the structure before we hit the end of the times tables. I've never memorized the 8 and 9 tables. I was able to do the math quick enough so that the teacher didn't notice.

[@NicholasOfAutrecourt]

@@rewolff2 "Refuse to see" what link? I've already told you that I know the link between calculus and this formula. I can derive it myself. I TEACH IT FOR A LIVING, as I said in the previous message.

[@luisclementeortegasegovia8603]

Helpful and nice done professor 👍

[@christianmosquera9044]

Excellent video

[@RahulSharma_45]

The square root process is also similar to this. The difference is that we write 2 in the denominator multiplied by the closest square root we took.

[@Daniel31216]

It because they're both based off the Taylor expansion.

[@Calypso143]

Amazing! This is a very clever one

[@havenharrison8549]

Good teaching

[@caroleastmond9064]

So Smart!!!🎉🎉🎉💫

[@ealy3545]

Differentials are pretty neat

[@arepallivijayganeshgoud2120]

Excellent Sir

[@canadagooses]

this is actually using the second degree Taylor polynomial for y=x^(1/3) centered at x=64. I would recommend using a larger degree to be more precise. I usually do to 5 or 6. Factorial growth makes anything higher fairly minute.

[@anandvashishtha07]

Love from India

[@MathOrient]

It would be nice if you would explain to us, how you could come up with a formula for this approximation :)

[@reefu]

The particular one shown in this video is derived by using Taylor polynomials. It takes a bit of calculus, but if you are somewhat familiar with it from high school, we are essentially taking the first derivative of the cube root function, and using it to find a linear approximation of the cube root.

[@Daniel31216]

It's the first two terms of the Taylor expansion. He's taken the function f(x) = x^(1/3) and approximated it using it's derivatives. It looks something like this: f(x) ≈ a^(1/3) + (x-a)(1/3)a^(-2/3) where a is the closest perfect-cube to x. So in this case, x = 70, and 64 is the closest perfect-cube. therefore: f(70) ≈ 64^(1/3) + (70-64)(1/3)64^(-2/3) ≈ 4 + 6/(3*4^2) ≈ 4.125 For a more accurate answer, you'll need to add a few more terms to the Taylor expansion of f(x).

[@ThomasTheThermonuclearBomb]

Very cool, where does this come from?

[@mozamilosama6087]

Calculus, I think it's called the deffrential 😅

[@gamerpedia1535]

Essentially you're doing a tangent line approximation. y = cbrt(x) dy/dx=1/(3×cbrt(x)^2) x = n + dx dy = dx/(3×cbrt(n)^2) So 70 = 64 + 6 6 = dx dy = 6/(3×cbrt(64)^2) = 6/(3×16) = 2/16 = 1/8 cbrt(64) = 4 4 + 1/8 = 4.125 ≈ cbrt(70)

[@reefu]

This comes from the Taylor expansion of the cube root centred about the chosen number which you want to approximate the cube root.

[@surendranagdali6788]

​@@gamerpedia1535 please elaborate

[@Newtonapplepie]

Applications of derivative: errors and approximation

[@addinhowtobasic]

The 2 is always 2… me when i ask my parents why im not the eldest.

[@deeno1945]

Your hand writing is so good

[@mrhtutoring]

Thank you!

[@dcmhsotaeh]

Guess is good at table tennis Has strong supple wrist extensors

[@mrhtutoring]

I suck at table tennis. Pretty good at everything else though

[@dakoz]

Let f(x): D(f): (a; b) f(a) < 0 and f(b) > 0 (f(x) have root on (a;b)) f'(x) > 0 and f''(x) > 0 Let x0 be the root of the equation f(x) Number b > x0 The tangent equation (1) at point 'b' will look like this: y = f'(b)(x-b) + f(b) (1) = 0 => f'(b)(x1-b) + f(b) = 0 => x1 = b - f(b)/f'(b) Point x1 will be between x0 and b (from the drawing) We can repeat the process an infinite number of times, taking the point x1 instead of b (and in general, take x(n+1) instead of xn) x1 = b - f(b)/f'(b) x2 = x1 - f(x1)/f'(x1) x3 = x2 - f(x2)/f'(x2) ... x(n+1) = xn - f(xn)/f'(xn) x0 = lim(n=>inf)(xn - f(xn)/f'(xn)) For example, we wanted to find ³√70, eq x³ = 70 gives us this root x³ - 70 = f(x) x between 4 and 5 f'(x) = 3x² (always >= 0) f''(x) = 6x (> 0 on (4;5)) Choose number 5 and put it as b (1): x1 = 5 - f(5)/f'(5) = 5 - (125-70)/75 ≈ 4.267 ≈ ³√70 Choose number 4.267 x2 = 4.267 - f(4.267)/f'(4.267) ≈ 4.126 ≈ ³√70 ... Sorry for my English :)

[@gamerpedia1535]

Super cool! I don't know if this was just coincidence or not but I actually gave this exact answer to a commenter about a week ago here with an explanation and now it's up! Crazy how the world works

[@mercilessadmiral6314]

Thank you very much for this video, Sir😄

[@giuseppe3y3]

Great!

[@Sg190th]

I like this technique. Makes me wonder if the Newton method can work

[@snehaghosh4132]

Thank you 🙏

[@stephencindrich6787]

Very helpful

[@mrhtutoring]

Glad it helped

[@kojoakuffoaddo4699]

Best maths teacher

[@user-gx9tm6py9v]

After ages I found the easy and right way to find cube root❤❤

[@ahmad.martua10]

thanks prof

[@themadtitan728]

Yo can do this after studying chapter " Application of derivatives "

[@mohammadahmadi6772]

Always you’re just a genius wowww

[@kickazz2730]

My brain hurts

[@shivanshnigam4015]

we can use AOD or generalize using the infamaous Binomial Theorem

[@hasantao]

Tailor series ... good luck

[@grouchomarx6156]

Génial!

[@samarthpatil2053]

We could actually use methods like Differentials and Binomial Approximations to solve these kind of questions... But i found this method way faster than those...

[@shivanisingh-vs5ok]

Thanks sir

[@truptimohanty9197]

Sir please make another video of cube root

[@oliveiraphael]

Good one. Challenge us professor. Let us find out an answer, a hard answer not easy please.

[@thundercraft0496]

i just read about that in a calculus book

[@subindewan2457]

Its a bit lengthy than this, but there's also the good ol' newtonian way of approximating the root.

[@peterlohnes1]

You can also do intuition. cube root 64 is 4. cube root 125 is 5. 70 is 6 away from 64. The distance between 64 and 125 is 61. So 6/61 or roughly 0.1. 4.1. Sure its not accurate, but theres no reason you would need to approximate 2 decimal places of cube root 70 either.

[@TUB83709]

This is several years too late for me 😢 but great content❤

[@Unknown_1096_]

Is this just linear approximation?

[@pedroandre625]

man is just build different

[@igorartemchuk311]

(a+b)^3=a^3 + 3*a^2*b + 3*a*b^2 + b^3 ~ a^3 + 3*a^2*b. Where a >> b => (a^3 + 3*a^2*b)^(1/3) ~ (a + b)

[@DelhiRestaurant-kz3gz]

Was searching for someee time

[@aarontanchingco4387]

geometric series

[@omidsedighi-mornani1636]

taylor series left the chat

[@Daniel31216]

This is the Taylor expansion of the cubic function. Or the first two terms of it.

[@felixsebastian7208]

70^(1/3)= 4+((70-64)*4/64*3)=4.125

[@buelan.6525]

Gee, wish I understood math so I could understand what you just did. Some of us were meant to avoid math at all costs.

[@ThePigKnight]

Is that the special Japanese chalk?

[@SuperToughnut]

So this is a formula, but does it explain why/how we have this formula? I want to know the history of the discovery too. I want to solve problems organically. If we were taught how these models came about while we were in school, we could build these models ourselves when we see them in the real world in a real world form. It's especially challenging to learn math when we aren't getting real world problems (I don't mean word problems) and the history behind the problem. Teach the discovery. All we are doing now is solving puzzles. You are giving us the tools to solve abstract puzzles. I learned about binomials and trinomials. But I don't remember anything being taught about when I can put real life problems into a binomial or a trinomial form. Matrices are the same way. This is how you multiply matrices. Great, but why? How do I build working models in my head if I can't take a real world problem and put it in the correct form? If you can't recognize the puzzle in the organic world, you won't use the tools you were taught to solve it. You probably won't even see the puzzle. We don't have the language to see it. If we had the knowledge to recognize the puzzles in the real world, we can swim among the models in our mind. Create the connecting tissue between the models. We can solve the problems before they even happen because we have this language in our heads.

[@mirageshooter1474]

This one is can be derived using binomial theorem. Expansion of (a+b)^n for n

[@mryip06]

Taylor series

[@mryip06]

Taylor series f(x) = f(a) + f'(x)(x-a) +... When f(x) = ³√x , f'(x) = ⅓x^(⅓-1) = 1/(3x²) Choose a = 64, I believe you can do the rest

[@nahbro1321]

(TAKE copy and pen if u can't imagine things way clearly)👍 Differentiation... Just Simply think of it like this you have a number x^3. U want to go just a Little more forward (imagine graph y=x^3) ... That is nearly 0.... But not zero.... Then u need to find rate... That how it behaves or how it changes instantaneously or simply putting at the accurate time for example exact speed of car at 2.000345 seconds.. using laws of motion will not work .. as it will give u avg speed... So u will go just a little further... Now think what a rate is and how will u find it.... Lets say u go k distance forward in graph(y=x^3) ... Where k is nearly .. zero... So wot will the new distance.. indeed it is (x+k)^3 - x^3... ( I can't tell u here as we r not physically meeting but think while drawing graph the distance... Wot will it be... Simply final distance i.e (x+k)^3 - intital distance) Now to find the rate or the behaviour... U have to devide it by k (or the distance u moved ) just like distance/ time and many other formula out there... Now open up (x+k)^3 - x^3 ... It is 3x^2k + 3xk^2 + k^3 whole devided by k.... Now k will get cancelled (take k common on above) So u r left with 3x^2 + 3xk + k^2.. But as we know k is infinitely times close to zero ... So value ... 3xk + k^2... Will be meaningless or say will be zero only... So we r left with 3x^2.. and that is our rate.. Now can u figure out cube root of 70... Of course try for more .. x^2 or maybe x^n .... And try to think it through .. don't except anyone will give u spoon feeding.... DO READ ABOUT RAMANUJAN... IT WILL INSPIRE U ... AND IF U R TRUE MATHS FAN .. IT MAY SPARK A LIGHT IN YOU... THANK YOU..!!

[@SuperToughnut]

@@nahbro1321 Thank you for taking the time to write this out. 👍

[@Machtyn]

And 4.125 is just 4 1/8.

[@CarmenLC]

how are the digits just backwards 14 21 28 35…

[@Stadttaube3]

(70-64)/(3*4) = 0,5 unequal 0,121...

[@rohitraghunathan]

You're not relly expaining why. Admittedly, you can't fully explain a first order taylor approximation in under 1 minute, but atleast link to an explanation

[@mrhtutoring]

"Don't look a gift horse in the mouth"

[@patricklau8845]

Try cube root of 120. Used the same method and didn’t work.

[@mrhtutoring]

5-5/(3×25)=4.9333

[@eitanhaim3972]

We can't see the left of the board because the screen is covered by all the youtube signs

[@MiniMessi2012]

How did he get 4.125 I thought it was 4 .8??? Reply pls

[@ashenrashmika6279]

How you get 4.8

[@user-yt5rj3ly3p]

linear approximation in ma class 12

[@pierrettebalazut9407]

Vidéo à revoir. Ce n'est pas évident

[@MathTidbits]

here is my approximation of the cube root of 70 knowing cube root of 64 is 4. 70^(1/3) ~ 4 * (2*70 +64)/(2*64 +70) ~4*(140+64)/128+70) ~4*(204/198) ~816/198 ~4.121212..... true value=4.1212853..

[@jumpjump-oz2pr]

How it come from I heard some people said it from calfulqs

[@sarthaksingh9645]

he used calculus its a result, you can do ant root even 10th with calculus

[@ajeetsachan]

& My teacher had reduced a significant percentage of numbers due to this approximation in one esam☹️

[@rishabbansal7662]

SHUT UP GUYS!!! The taylor series has entered the chat now.

[@gwensimmons_gigi1629]

Does not work with 93???

[@muhammademran2464]

64^1/3 = 4 4 + 6/{3 × 4^(3×2/3)} = 4.125

[@IbrahimKhan-jw1xm]

Sir this method is not working for cube root of 25. Why?

[@СОЯЕИ]

³√27-2 =3-2/3×3² =3-2/27 =79/27 =2.92(ans) Take the nearest cube 25 is nearest to 27 i.e 3³ ,so 25 is smaller then just substract it.

[@popcria]

It works: ³√25 = ? The nearest cube number is 27. Thus, ³√27 = 3 Now put 3 in this equation: 3 - 2/27 Why 3 - 2/27? Numerator: 25 - 27 = (-2) Denominator: 3² • 3 = 9 • 3 = 27 The numerator is a negative so follow the rules of fractions of rational numbers: - When the numerator is a negative number meaning you have to subtract it not add it: Thus 3 - 2/27 = 79/27 = 2.925925926 And in calculator: 2.9240177

[@ommudliyar5429]

im getting 4.3125 for 79 (which is wrong). can someone show me the correct equation

[@bananaforscale1283]

4,125^3 = 70.1894531

[@Daniel31216]

This approximation uses the first two terms of the Taylor series. If you want a closer approximation, or want to try it on larger numbers, you'll need to add more terms.

[@Huh12758]

i did this with 33 it give me 3.125 the right answer was 3.207

[@popcria]

It's approximate so it's nearest answer

[@alejandro-hc6ms]

but 4.1212853.... aprox is 4.121..................

[@coloradomountain]

This seems to work only on the cubic root of 70.

[@mrhtutoring]

Give me another number and I'll show you that it works.

[@omarmohammed3969]

29 😅

[@mrhtutoring]

3+2/(3x3²)=3.074 on the calculator cube root of 29 is 3.072316 What do you think?

[@omarmohammed3969]

@@mrhtutoring That you’re the best math teacher

[@coloradomountain]

@@mrhtutoring 100

[@RohitKulan]

It's about 4

[@abdrahmanakjija9037]

Its rlly cool but useless at the same time

[@mrhtutoring]

Being cool is good~

[@user-mj5qx1ke9o]

5776？

[@Samy---963]

Didnt understand

[@susansuchenharris3727]

Very very confused. Need better explanation.

[@mrhtutoring]

Which part?

[@kookiecha7203]

r u chínα.

[@mrhtutoring]

No~

[@felixsebastian7208]

This is application of bimomial series.. I can do faster than you.. (1+x)^(1/n)=1+x/n, x

[@AliHassan-hb1bn]

Too long process

[@Syndicalism]

It's a simple plug and chug for (x)^(1/3). He explained the formula within a minute, how is that too long?

[@themathematicstutor4092]

You're just lazy

[@h.m.5724]

Clown 🤡

[@drgilbertourroz]

You still need the calculator to calculate that fraction. You may as well just use the calculator to get the cubic root.

[@Daniel31216]

You don't need a calculator for the fraction, but it will much easier to just use one.