Square Root of 2, Newton's method vs Euler's method 

Thanks a lot, blackpenredpen. I can’t even begin to explain how much your help meant to me.

I use Newton's method as he had much better hair.

Engineering Students: sqrt(2) = 1.414 = 1

Why do you have a thermal detonator in your hand

You remind me how much fun Math has been for me over my years. I was never a whiz, reached my level of incompetence at the advanced undergraduate level, but it's fun and interesting to see your, Mathologer, and a few others' discussions. I was born in 1948, 71 as of this comment.

Lmao just draw an unit square and measure the diagonal with a ruler

IIRC newton's method basically doubles the precise number of digits at each step. It also has the advantage (in this situation) of being able to pick up where you left off - you can always add more accuracy by iterating for longer, but with Euler's you need to start over.

Little strange and unfair to compare two methods which are designed for two completely different things. Newton’s method - finds roots numerically Euler’s method - finds numerical solution to differential equations. Would have made more sense to compare Newton’s method to the secant method or some other numerical root solver :)

Try using both methods for a system of ordinary differential equations, and try using RK45 instead of vanilla Euler. That's where Euler methods are most effective. Newton's method and other fixed point iterative methods are pretty much the best for most solvers in numerical analysis.

Much prefer the Casio Method.

Another great method I found, with iteration: Say you wanted the square root of A. x = sqrt(A) x^2 = A x = A/x 2x = A/x + x x = (A/x + x)/2 So the formula we'd use is x_n = (A/x_{n-1} + x_{n-1})/2 Then iterate, begin maybe with x_0 = 1 or something you're sure is close, easy enough. Using a calculator, you just put in your first guess, then '=' to set your initial ANS value, then type the formula, using 'ANS' instead of 'x'. Then just spam that '=' key until the result in your calculator stops changing. Done. Extra, play around with a 'b' variable where we can say x = A/x (b+1)x = A/x + bx x = (A/x + bx)/(b + 1) x_n = (A/x_{n-1} + bx_{n-1})/(b + 1) 'b' can be anything

Start with 1.4 squared = 1.96. Then 2 = (1.4 + epsilon)^2 = 1.96 + 2.8 epsilon + epsilon^2, throw out last term. Then epsilon = 0.04/2.8 = 0.0143 So then 1.4 + epsilon = 1.4143 after one iteration. Keep going . . . This method actually does simplify to the Newton method but is more intuitive, in my opinion.

For some reason, I find it really cool when you say "well well"... it feels like you're a detective and you just figured out a big part of a case xD really cool!

Euler's Method isn't used for this purpose though; you got an inaccurate estimate for sqrt(2), but you also got estimates for the sqrt of 1.2, 1.4, 1.6 and 1.8. Newton's Method got you an accurate estimate of sqrt(2), but you learned nothing about the other values. Euler's Method and Newton's Method are used for different problems, and it's not fair to apply Euler's Method to a roots problem and then be disappointed when it doesn't produce as accurate a result as you would have hoped with that step size.

You are doing too much effort for youtube...3-4 videos in 24hours... Hats off to you... Hopr u reach a million soon

For "Well Behaved" functions ( ie. polynomials, trig, etc ) Newton's Method is excellent at approximating the root very quickly, especially if you do a bit of work in advance and smartly come up with a good initial value. For example: solve ( cos(x) = x ) , when you draw a graph of cos(x) and y= x together, you can see that they cross each other to the left of x = 1. Thus an initial guess of x = .8 will zero in on the root of .739 085 very quickly. Love Newton's Method for getting roots fast.

Recommended to my dog. He said you are cool ;) Still not at his level tho.

Very nice demonstration of the two methods. I solve a lot of cubic equations (equations of state), which are solved even faster using a third order Newton's method. This method is attributed to Hailey of Hailey's comet fame, and uses the first and second derivatives. The generalized method is known as Householder's method.

It seems that Euler's formula needs to be integrated because h needs to be 1/ infinity to give an very tiny step. that looks exactly like how we take a limit and figured out to perform integrations. While Newton's formula is an iterative process in which you can stop at any atribary point. The interesting part is that as the iterations tend toward infinity that the rate of change in the approximation tends towards zero. There seems to be some connection between the two. Is there a hybrid formula that uses both concepts?

Wow we just learned Newton's method in class today.

Newton will win for sure, got good vibes

Your videos are great and informative. Keep up the good work!

Sir Isaac Newton delivers the knockout punch!

Though I've never done a numerical analysis course, I knew Newton's approximation would win. I was first exposed to it as an adult through learning about the Fast inverse square root in the source code for the Quake computer game.

It was at about this level of math when I decided to be a history major.

As you were posing this contest, all I could do was smile and think, "This is gonna be a blowout!!" Fred

The 1.414 value is also used in sinusoidal electrical power calculations. RMS (root mean square) effective power vs Peak power. Certain numbers are magical.

Also worth looking at Goldschmidt's method. Let Y be your initial approximation to sqrt(n). Start with: x0 = n/Y, y0 = 1/2Y Then iterate: ri = 0.5 - xi yi x_{i+1} = xi + xi * ri y_{i+1} = yi + yi * ri Then the x's will converge to sqrt(n) and the y's will converge to 1/2sqrt(n). This is interesting because n doesn't appear anywhere in the iteration loop!

I like the method of: sqrt(2) = 1 + a 2 = (1+a)^2 2 = 1 + 2a + a^2 Since a is small, ignore a^2 2 = 1 + 2a a = 1/2 sqrt(2) = 1.5 then repeat like sqrt(2) = 1.5 + b solve for b after ignoring b^2 term, continue doing this to get some very good rational approximations of sqrt2 that approach it very fast (4 iterations and you are already very close)

They both do!!! Thinking of the tools they had to work with ... what an incredible feat by both ... both exceptional geniuses... with the pendulum leaning towards Newton in this round.

How about using the partial fractions from the continued fraction for sqrt(2) that you demonstrated 3 weeks ago? 1, 3/2 (1.5), 7/5 (1.4), 17/12 (1.416666), 41/29 (1.41379), 99/70 (1.41429), 239/169, (1.4142), 577/408 (1.4142156), 1393/985, 3363/2378, 8119/5741, 19601/13860, 47321/33461, 114243/80782, 275807/195025, 665857/470832.... wait, Newton's Method starting with 1 hits a few of these but much much quicker. Newton for the win!

I hardly understood a word, but I found this utterly fascinating!!

Here is a way to calculate the square root on a calculator without using the square root. First enter 1 and press = (or EXE). then enter (ans+xx/ans)/2 where xx is the number you want to take the root of and then press = a few times. This requires a calculator with line entry. Lets say yu want to calculate sqrt 2, you get 1.5 1.416666 1.414215686 1.414213562 and the last doe snot change anymore. You can of course start with a better estimate. I thought this method would be good if you have a calculator without a square root bit I found simple squaring of estimates and adjusting them faster.

If you know ln(2), there is another way you can approximate sqrt(2) with euler's method, use the function y = 2^x which at x=0.5 equals sqrt(2). It's defining differential equation is y' = y*ln(2) and you know that y(0) = 2^0 = 1, so if you know ln(2), this is also a way to approximate sqrt(2). I programmed all 3 ways on python, euler's method with y=2^x is more accurate than with y=sqrt(x), but still newton's method wins.

Did Newton actually write it that way, or was this done somehow with his favorite geometric method instead?

Wow, today I have studied the Newton method in Numerical Analysis class , and by chance I see this video. Great video 👍

Regardless of the fact that both methods have different purposes, the choice for x0 and by extension the convergence area are relevant. Of course the initial guess must be somewhat accurate, though choosing x0 < 0 would result in convergence to -sqrt(2) instead, to give an example

Great video as always! .. I'd like to see Newton's method for finding imaginary roots of polynomials (say 3rd degree for instance) if you please.

Team Newton!

Basically, you have 5 approximations with the Newton method, but only 1 with the Euler method. The second approximation would be Euler with a smaller step size. What I am getting at is, I am a big fan of Aitken's delta^2 process. This requires a series of estimates that will converge to the target. The Euler steps here don't converge; they only converge as a series of estimates taken with progressively small step sizes. Aitken's method for accelerating convergence can produce astounding results. For example, if you apply it iteratively to estimates of pi/4, you can get to 10+ digits using only odd numbers through 31 or 33 (e.g. 1 - 1/3 + 1/5 - 1/7 ...) You can get to pi^2/6 with the Bessel function with only a few thousand base iterations, if you take the sequence of estimates as powers of 2; e.g. 1, 2, 4, 8, 16, 32 etc to 1024 or 2048 - and then apply Aitken iteratively to those estimates. (Normally, you get 1 digit per power of 10, so 6 digits would require a million or more terms of the base sequence.) Here, I would suggest taking Euler using h=1, 0.5. 1/3, 1/4, then 1/5. Then I would compare the results of the convergences when accelerated with Aitken. Maybe I will do that, but not now.

Unlisted? Damn, I clicked fast!

The expression for Newton's Method for sqrt can be simplified to x[n+1]=(x[n]+c/x[n])/2 where c is the number of which we want the square root. In this form the method is almost intuitive. Also for floating point computations, we have a quick method of .getting a great first approximation. Mearly cut the exponent in half. Also in the NR method the error is determined only by the last step

It would have been a nice thing , if you had included a third method! It is a balancing method. It simply states:- X(n+1)= 1/2{X(n)+(N/X(n)} In present case, N=2, put n=1 & X(1)=1 X(2)=1/2{1+2/1}=1.5 X(3)=1/2{1.5+2/1.5}=1.4166666666 X(4)=1.414216018454 X(5)=1.414213562375 X(6)=1.41421356237 Above is the value of √2. Please comment.

This is awesome! Sweet comparison!

Euler will be better. - Wrong. But am I right that it is the root-sign, which lets you think of a wing?

What happened to Raphson?

I have never seen anything this beautiful ❤

Awesome the connection between equations and where they come from should be made while we are learning to solve them

Are there any function where doing it one way is easy and the other way hard, and vice versa? Given that you find different derivatives?

I have a TI-30 from my high school days and I pressed 2 and the sqrt key, and that was much faster than either approach.

I think Euler will win, but I prefer the Newton method. Being able to start the function and just do iterations to get closer and closer just seems more natural to me.

That’s what I said to myself the minute I jumped out of bed this morning, “Newton or Euler?” Because I just had to know the square root of bleedin’ two before I had any coffee…..

I did expect Euler's to be nowhere near the first four steps but be pretty close on the fifth x=2.0 step. Did you have a video on approximating roots? It was very easy (for roots below 1000) with this nice trick. You just need to know the squares of 1-32. The formula is: y1 = y0 + ∆ / 2*y0 with y0^2 the closest squared number and ∆ = y^2 - y0^2 Take y=√200. y0 = 14 and ∆ = y^2 - y0^2 = 200 - 196 = 4 y1 = 14 + 4 / 2*14 = 14 1/7 You can derive this by writing y as y0+f and squaring this. y^2 = (y0+f)^2 = y0^2+f^2+2*y0*f f

Excellent video!

Thanks man. I missed both of these in IB HL because I was sick.

"Is Newton going to win?" - he already gave you twice as many digits as you wrote down, so yeah

Nice vid!! Keep it up!

There's also the binary search method. Consider the interval [a,b] where f(a) is a different sign than f(b). Find the midpoint of the interval, if the sign of the function at that x value is the same as the sign of f(a), then the new interval becomes [f((a+b)/2),f(b)], otherwise the new interval becomes [f(a),f((a+b)/2)]. The final answer is the midpoint of the final interval. Starting with a = 1 and b = 2: [1,2] -> [1,1.5] -> [1.25,1.5] -> [1.375,1.5] -> [1.375, 1.4375] -> [1.40625, 1.4375], so the final answer is 1.421875. It's not as accurate, but it requires no calculus and it gives you more control over how accurate you want the approximation to be. Since the size of the interval halves every iteration, you get 1 decimal digit of precision for about every 4 iterations (because log_2(10) is about 3.3).

who else thought the video wasnt of balckpenredpen until this man appeared😂🤣

You win this time Newton, Euler gang will be back soon!

Oldton: I pick starting point 0 BGM: To be continued

I always recommend you for mathematical background to people trying to understand quantum mechanics. You forgot these.

And the winner is... Newton. Very interesting explanation and comparison. Thanks

euler was the greatest mathematical mind of all time

Start with X=1, Y=1, and h=1. Euler gives X = 1.5. Square to give Y = 2.25. and h= -0.25. Using this variation, h gets smaller each time around, and you can stop when you have the accuracy you want.

You used Old Doraemon theme song tune in starting of video😂😂

Best method used by students of 2020-21 --> calculator

The complaints that comparing both methods to each other as they are designed for different applications is an invalid complaint. He used both methods fairly, within their respective realms of application. Yes, Euler’s method is designed to solve differential equations, but solving the differential equation that he wrote on the board is identical to finding the square root of 2 in this case. In fact, as he himself clarified in the video, the problem is not the usage of the method, but rather the step size that he chose for the method. You must understand both methods have different convergence rates. Perhaps, next time, before complaining, pay more attention to the video instead of saying nonsense and criticizing a perfectly fine video.

x^2 = 2 x = 2/x 2x = x + 2/x x = 0.5(x + 2/x) x[n+1] = 0.5(x[n] + 2/x[n]) ...Newtons method derived not using calculus.

you are so great . really love your maths

Now it all makes sense. Thanks a lot.

I think Newton's is the favorite

Expectation:- I love Newton, I'll use Newton's Method. Reality:- If I have a calculator, I will use neither 😂😂😂

Congratulations from France

Show how to approximate the Lambert W function using Netwon's method.

I used Euler's method and has the value of h as 0.01, thus needing to use the formula 100 times. After doing this, i found the value of sqrt(2) to be approximately 1.414827967. After using Newton's method 3-4 times, it was clearly more accurate than Euler's method. I'll have to go with Newton on this one 👀

Newton has quadratic convergence, so not knowing anything about Eulers method I guess Newton is going to win because that's hard to beat.

If you started with a closer guess for Euler, the increments would be smaller and the final answer would be better. Start with: X = 2.25, Y = 1.5 and do 5 steps of -0.05. Or: X = 1.96, Y = 1.4 and do 5 steps of 0.008 Also- Newton's method is the best known, but Halley's method gives more digits per iteration.

Yay! Three cheers for Newton

I like approximating it this way: Let's choose first term somewhat close to true value - let's say x=2. Then we just iterate f(x) = 1/x + x+2. I believe it's using Banachs fixed point theorem. The inverse might be hard to calculate without calculator tho

Good tutorial.. Very complete

Seeing how Newton's works, it's going to be better because his method gets more precise as you keep stepping, while Euler's method compounds the imprecision of the previous steps

poor Euler u_u

My crocodile is now subscribed

I think I prefer Euler's method because it requres less knowledge about the function. All we need is an ODE and an initial condition. Thanks for this video, this explains a bit better how it's possible to numerically solve ODEs!

newton's method seems like a pd (proportional-derivative) controller algorithm, there is something that i barely remember about approximating a computer simulation, it was like 1/6*X1+2/6*X2+2/6*X3+1/6*X4 = X2 in short summing the values that computed using what we got at that point with 1,2,2,1 coefficients and getting our next value... maybe it can approximate quicker than that

would be great to see a follow up with a larger step size for faster convergence on eulers method.

Excellent!

I think the reason why Newton's method won out is simple. With Euler's method, your bad approximations compound as you do more work, getting you further and further away from the original function. With Newton's method, your bad approximations turn into better and better approximations as you plug the new result back into the formula.

While watching this video , I got an ad about Newton!

They both have their place

Great lecture !

Nice. Thank you.

thanks my cat loves you!

10/7 ~~ 2^0.5 100/49 ~~ 2

I do that to find squareroot of numbers For example=Sqrt(2) (Sqrt(2) ×10^n)/10^n (n is how many digit you want after the point) For example n=2 (Sqrt2 × 10^2) / 10^2 Sqrt (20000)/100 (sqrt 20000=《141-145》) Take 141 141/100=1,41

the music is relaxing

Euler's main use is approximating curves, not finding solutions (and from that you can also solve diff. equations). I think Newton's method is the fastest method for this, right?

I love the "Well...well..."

Winer heron with heron's formula : √a=(n+a/n)/2