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.
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
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).
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.
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.
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
13 This process’s load deflection curve is sawtooth like in your video Mechanical properties related to a unique variation of Euler’s Contain Column studies. It shows how materials (representing fields) naturally respond to induced stresses in a “quantized“ manor. This process, unlike harmonic oscillators can lead to formation of stable structures. The quantized responses closely models the behaviors known as the Quantum Wave Function as described in modern physics. The effect has been used to make light weight structures and shock mitigating/recoiled reduction systems. The model shows the known requirement of exponential load increase and the here-to-for unknown collapse of resistance during transition, leading to the very fast jump to the next energy levels. This is shown by the saw-tooth graph’s bifurcation during the quantum jump. In materials the process continues till the load passes the ultimate tensile strength. Fields are not bounded by these conditions. ruclips.net/video/wrBsqiE0vG4/видео.htmlsi=waT8lY2iX-wJdjO3
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.
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
Thank you! I was going to say the same thing. It is far quicker and more accurate to calculate 2 square roots by hand, than to do all these multiplications and divisions. Newton's method is not for 🖋️ and 📄.
This method sometimes works way better than binary search tho. I tested both binary search and Newton's method to calculate sqrt(2) and I found out that Newton method converges way faster than binary search (e.g the Newton's method double the correct digit every round of calculation, while binary search got 1 more correct digit every 3 round of calculation)
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 ...
![Noob To Max With DRAGON REWORK In Blox Fruits [FULL MOVIE]](http://i.ytimg.com/vi/LnBMOinoOvA/mqdefault.jpg)
👉 How to get the newton's method formula: ruclips.net/video/iVOsU4tnouk/видео.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.
It's because it is the same principle but newtons method just iterates it
@@Zeusbeer yes, I know it's the same principle
He used calculator 4times
Why not calculate it directly
@@Need4Speeeeed where did you see he used a calculator?
@@volodymyrgandzhuk361 hhhhh.how he calculated 8 fours
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
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).
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.
just split it up using logarithmic properties and take the derivative from there
@@jon2422 I always forget about log properties, I didn't think about that. Thanks
Thanks for this, Steve! I have learnt something new! 😀😀😀
Learned something great bro. May you live healthy, wealthy brainy and long.
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.
Really? This was taught first year in my physics degree.
@@danny89620 Well clearly not every university teaches the same material.
Probably you have never done numerical methods.
@@etgaming6063 you never had a course on numerical methods?
Very clear method thank you, helped a lot!!
Isn't the algorithm used in the fast inverse square root of quake 3 ?
absolute legend
Loved it!
Thank you so much!!!!!
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
No what you just did is simple mutiplication. 18.75^4 =/= 75
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
Thanks!
thank u
13
This process’s load deflection curve is sawtooth like in your video
Mechanical properties related to a unique variation of Euler’s Contain Column studies.
It shows how materials (representing fields) naturally respond to induced stresses in a “quantized“ manor.
This process, unlike harmonic oscillators can lead to formation of stable structures.
The quantized responses closely models the behaviors known as the Quantum Wave Function as described in modern physics.
The effect has been used to make light weight structures and shock mitigating/recoiled reduction systems.
The model shows the known requirement of exponential load increase and the here-to-for unknown collapse of resistance during transition, leading to the very fast jump to the next energy levels.
This is shown by the saw-tooth graph’s bifurcation during the quantum jump.
In materials the process continues till the load passes the ultimate tensile strength. Fields are not bounded by these conditions.
ruclips.net/video/wrBsqiE0vG4/видео.htmlsi=waT8lY2iX-wJdjO3
Life. Saver.
Watching 20 mins before exam, thanks.
How u taken X1 value?
just use a number that's close to the answer
hey bprp why dont u post videos on ur blackpenredpen channel?
as alwaz gr8 video tho
well if i want to approx it in my head i will still stick to try and error i guess.
My man put microphone into pokemon ball💀💀
Hi
Can you calculation ; (9797979797)^1/50 =X
By Casio fx - 3600P calculater ;
By the Newton Methode ?
Imagine using a calculator to use newtons method but not being able to calculate sqrt(75) 😄
Imagine saying the fourth root of 75 =sqrt(75)😂😂
Newtons method doesn't need more than the four basic mathematical operations: -+*/
@@Muck-qy2oo and because of it it is used by computers to calculate roots of any degree to very high accuracy in many iterations
r^5
It looks similar to the approximation method used in calculus
using differentials with linear approximation is far easier..... just an oπnion
I think this method is geometric beauty.
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.
It is based on Newton's formula & easy to use.
That's cool
Does he forget to cut some parts in the video haha
Confused me a little bit with that looped intro lmao
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
Thank you! I was going to say the same thing. It is far quicker and more accurate to calculate 2 square roots by hand, than to do all these multiplications and divisions. Newton's method is not for 🖋️ and 📄.
@@gkotsetube Also QR method for eigenvalues is not so great for finding numerical roots of polynomial equation with paper and pencil
I need a proof for the formula
Just search it up
Counting the seven fours as if 3-1/18 would not be fours all the way. :D
Why did you repeat first sentence twice
Now do this in your head entirely lmao.
Just binary search tho?
This method sometimes works way better than binary search tho. I tested both binary search and Newton's method to calculate sqrt(2) and I found out that Newton method converges way faster than binary search (e.g the Newton's method double the correct digit every round of calculation, while binary search got 1 more correct digit every 3 round of calculation)
What's the point of the method if you have to use a calculator anyway
Fun
lol he forgot to cut out his re-take at the start of the video, little joke :)
why use newtons method with a calculator instead of calculating the irrational number with a calc itself lol
who else tryna get that web assign answer ?
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
...
Ok
Some of us are not allowed to use a calculator...so no this is not useless
Im the 999 like😂 26.6.22 22:56
👍👍👍
:)