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

After a long day of doing calculus, it's fun to go back to approximating irrationals in a new way. Thanks Steve. Stay being a Gigachad.

Fun fact: if you try to approximate √75 with the tangent/differential method Steve showed in one of his videos, and then do the same thing for the square root of the result (because the 4th root is basically the square root of the square root), you will get exactly what he got here the first time he applied the Newton method formula.

Thanks for this, Steve! I have learnt something new! 😀😀😀

Very clear method thank you, helped a lot!!

Loved it!

Thank you so much!!!!!

Learned something great bro. May you live healthy, wealthy brainy and long.

3:45 reading 2.9444444 aloud, reminded me about that olympiad problem you also did (you were also saying number 4 a lot in that video). Calculate the sum of digits of the sum of the digits of the sum of the digits of the number 4444⁴⁴⁴⁴(i think).

Thanks!

I have another method. The closest perfect power is 81 which is 3^4. Now we know that 81>75 so 3>fourth root(75). Now that means obviously, 3-fourth root(75)>0 . now raise this to the power 4. This becomes, 156-108 cube root(75)+270 root3-60 fourth root (675)>0 now upon some rearrangement we get that, cuberoot75>156+270 root3- fourth root(675) all divided by 108. This is tedious but it gives the very close approximation of fourth root of 75. Please correct me if there are errors

Isn't the algorithm used in the fast inverse square root of quake 3 ?

That's cool

Life. Saver.

I was working on the derivative on natural log in calculus and attempted to find the derivative of y=ln([x(x^(2)+1)^2)/sqrt(2x^(2)-1)] and I would like to see your approach. (sorry if it is hard to read based on how I typed it.

hey bprp why dont u post videos on ur blackpenredpen channel? as alwaz gr8 video tho

I got lost halfway through but then it all made sense by the end👌🏻 this is a cool method that I have never heard of before and I have a physics degree.

How u taken X1 value?

just when i thought this would be good for olympiad... also i just realised olympiad questions would probably have all the √s cancel out or be a perfect square

My man put microphone into pokemon ball💀💀

Hi Can you calculation ; (9797979797)^1/50 =X By Casio fx - 3600P calculater ; By the Newton Methode ?

fourth root of 75 is same as 75^1/4. this means to mupltiple 75 by 1/4 of itself, so 17.5/4 = 18.75

It looks similar to the approximation method used in calculus

well if i want to approx it in my head i will still stick to try and error i guess.

Does he forget to cut some parts in the video haha

Confused me a little bit with that looped intro lmao

I think this method is geometric beauty.

r^5

Imagine using a calculator to use newtons method but not being able to calculate sqrt(75) 😄

using differentials with linear approximation is far easier..... just an oπnion

Now do this in your head entirely lmao.

👍👍👍

Counting the seven fours as if 3-1/18 would not be fours all the way. :D

Why did you repeat first sentence twice

I would calculate square root twice with paper and pencil method In paper and pencil method I need to calculate twice as much digits for the first square root as I want in final result With paper and pencil method i calculated up to 4 digits after decimal point

I need a proof for the formula

lol he forgot to cut out his re-take at the start of the video, little joke :)

who else tryna get that web assign answer ?

why use newtons method with a calculator instead of calculating the irrational number with a calc itself lol

Just binary search tho?

These are useless, since it is too cumbersome to really compute in their mind or by hand each of these steps. If you used a calculator then that's ok, but then if I had a calculator I would just type in "75 (x^y) 0.25" and get the answer in one shot. This is OK to show that the method works in principle. You should choose the example of a function that is complicated enough that there is not a known way to compute the root easily even with a calculator. As for this function here is how I can do it in my mind with even fewer steps and without a calculator, that is pretty much analogous to Newton's method: 75^0.25 = sqrt(sqrt(75)) = sqrt(sqrt(25*3)) = sqrt(5 * sqrt(3)) = sqrt(10 * 1.732 / 2) = sqrt(10 * 0.866) = sqrt(8.66) Let: x = sqrt(8.66) x^2 = 8.66 Let x = k - y, where y is much smaller than k x^2 = (k - y)^2 = 8.66 k^2 - 2 k y + y^2 = 8.66 2 k y = k^2 - 8.66 + y^2 y = (k^2 - 8.66) / (2 k) + y^2 / (2 k) Since y is small, y^2 will be tiny and can be ignored as an approximation, giving, y = (k^2 - 8.66) / (2 k) We know that x is just slightly lower than 3, so lets start with approximation of k = 3 y = (3^2 - 8.66) / (2 * 3) y = (9 - 8.66) / 6 y = 0.34 / 6 = 34 / 600 = 17 / 300 = 5.666666 / 100 = 0.0566666.... y = 0.0566666.... x = 3 - y x = 3 - 0.0566666.... x = 2.94333333.... This is already very close to the correct answer of: 2.9428309563827118453573116740982 | ERROR | = 0.0005023769506214879760216592351 This is correct to 3 decimal places already, which is very good, but we could approximate much better with putting back the ignored term. Lets go back to equation before approximation: y = (k^2 - 8.66) / (2 k) + y^2 / (2 k) y = (current value of y) + (current value of y)^2 / 6 y = 0.0566666.... + (0.0566666....)^2 / 6 y = ~ 0.0566666.... + (0.06)^2 / 6 y = ~ 0.0566666.... + 0.0036 / 6 y = ~ 0.0566666.... + 0.0006 y = ~ 0.0572666.... x = 3 - y x = 3 - 0.0572666.... x = 2.9427333... | ERROR | = 0.0000976230493785120239783407649 This is correct to 4 decimal places If you want one more iteration, then choose k value equal to x, or at least much closer to x than before e.g. choose k2 = 2.95 based on above x approximation and similarly iterate on: y2 = (k2^2 - 8.66) / (2 k2) and x2 = k2 - y2 Next iteration after that would be: k3 = x2 y3 = (k3^2 - 8.66) / (2 k3) and x3 = k3 - y3 ...

What's the point of the method if you have to use a calculator anyway

Im the 999 like😂 26.6.22 22:56

:)

We may use a general formula to find the nth root of x given that x^n.=N. General formula ------------ x=[(n-1)x+{N/x^(n-1)}]/n m+1 m m Here n=4, N=75. Taking m =1 & x=3 we get 1 x=[(4-1)3+{75/3^3}]/4 =2.94444444 2 x=2.9428322282 3 x=2.942830956 4 x=2.942830956 5 Hence x=2.942830956.