Steffensen's Method with Aitken's Δ²

[@nikolaossmyrnioudis397]

Damn, this is so well made it deserves more recognition

[@mamaw8147]

OMG IM CRYING TYSM I REALLY NEEDED THIS TO PASS MY SUBJECT

[@hasmukhdhaduk3768]

Danke für die Klar Erklärung :)

[@soak2094]

THANK YOU VERY MUCH

[@THAdeeHARMONY]

Do you have videos on interpolation? If not you should, because that’s the next section in my numerical analysis class and your videos have helped me understand what’s going on geometrically in the root finding methods.

[@OscarVeliz]

I have one on Lagrange kzbin.info/www/bejne/faWtfIh_fJ5-jqs and one on Cubic Splines kzbin.info/www/bejne/nGXMf5WEh7CNgbc which I hope you'll find helpful. If there is a specific interpolation topic you'd like for me to cover please be sure to let me know. Note that it does take me a while to make each video so I likely won't get to it right away. The next video I will upload is also going to be on root finding.

[@OscarVeliz]

Also my video on Brent's Method kzbin.info/www/bejne/Y5OvhIWfpNCafM0 covers Inverse Quadratic Interpolation.

[@farzanamitu5542]

1. Use the false position method to find a root of the function f(x) = x2 - x - 2 = 0 in the range 1

[@OscarVeliz]

See False Position Method kzbin.info/www/bejne/ppiUemt3fJpsf80

[@fireknuckles2678]

How will you know the fixed point n that the aitken accelerated you to

[@annagladka841]

actuailly a lot of work you did, some easy examples to show how to apply would be really useful.

[@AJ-et3vf]

This is very nice! Although I'm honestly very baffled with the composite function evaluatio f(x + f(x)). Wouldn't x and f(x) have different units/dimensions thus they can't really be added?

[@OscarVeliz]

The idea is that f(x) would just be a really small number as you approach the root. It also explains the terrible performance of the method when far away.

[@AJ-et3vf]

@@OscarVeliz ohhh okay. I see. That's why. Thanks for the explanation! Although it's weakness of being terrible when far from the root is shared by the other Newton-type methods e.g. secant, newton, halley, etc

[@AJ-et3vf]

@@OscarVeliz update: i actually tried it in excel and gave steffensen's method an initial value far from the root. It wouldn't converge towards it unlike Newton's or Secant method that can still converge even if far away. So that's that. Still, Steffensen's method is still very handy when you know a good initial guess for the root and you don't want to calculate the derivative function

[@jonathanmontrose6102]

On step four of "solve for r" how did rc get a sign flip?

[@OscarVeliz]

Thanks for catching this. I think you're refering to 1:30 and that is just a typo. It immediately gets fixed in the next step. If you expand [ac - (a+c)r] you get [ac - ar - rc].

[@jonathanmontrose6102]

@@OscarVeliz yep, I saw that. Thanks for the response!

[@alexandrevachon541]

Can Aitken's delta squared accelerator and method can not only be used for Steffensen's method, but for other methods, such as Newton (which is a special case of fixed point iteration)?

[@OscarVeliz]

Aitken's can be applied to any linearly convergent series (like the example FPI). It can't be used on Newton's method because that is quadratic.

[@alexandrevachon541]

@@OscarVeliz Linearly series that includes bisection and regula falsi?

[@OscarVeliz]

Technically yes, but consider that those two methods aren't a series of x's but rather a set of shrinking [a,b]'s. You could try Aitken's on the c's but your mileage may vary.

[@alexandrevachon541]

@@OscarVeliz Remember that Laguerre's method is linear with multiplicity? It could work with that too.

[@OscarVeliz]

In that case the method is still cubic but behaving linearly. I suspect there might be some improvement, however there are other ways of dealing with multiplicity.

[@仕欣李-l3b]

I love your video

[@alexandrevachon541]

6:55 Out of order?!

[@Squish888]

ooohhhhhhhhhh......

[@sharoondanish4387]

sir kindly solve this sinx=15x/-3

[@arielfuxman8868]

Why the heck there are Steffensens method and Steffensens 2.0? This is confusing as fuck