Nice singing at the start btw

@siyangli4260

6 жыл бұрын

What song is it?

@vasishtvasudevan4059

4 жыл бұрын

sounded so much like Matthew McConaughey

@jasminesun3059

4 жыл бұрын

YES

You and Khan are life savers. Thanks so much

Very similar to Taylor series.

@Romis008

6 жыл бұрын

That's exactly what this is :)

@SohamChakraborty42069

4 жыл бұрын

How is that?

@chadschaefer5084

4 жыл бұрын

@@SohamChakraborty42069 If you kept adding 3rd, 4th, 5th derivatives, etc and moving on to "cubics", "quartics", "quintics", ad-infinitum, you would get a perfect approximation of your function as a polynomial (with infinite terms). This is simply taking the concept of the Taylor series, and expanding it to 2 dimensions (the x and y), instead of just one. (and technically, you could go to higher dimensions too, but computing the functions get more "expensive" as you go in higher dimensions) In practice, you can usually stop at the "Quadratic" term

@samirkhan6195

7 ай бұрын

It infact is multivariable Taylor series

Man !! This is like a Taylor series for multiple input

@malekalshemali1125

2 жыл бұрын

it is! but with him it makes sense!

I have relief when you start to calculate , when you try to explain my mind runs like a piston in the engine.

Brilliant!

"line things up a little bit right here" lmaoooo

very helpful

Anyone has the article he's talking about?

I watched 80 of these in a row, and I aged 80 years.

@Otochiro1

5 ай бұрын

Wisdom flows in your veins.

Really curious, why when you move from Linear approximation to Quadratic extra terms are added without correction? If we go back to your original expression Q(x,y) = f(x₀,y₀) + fₓ(x₀,y₀)(x-x₀) + fy(x₀,y₀) + a(x-x₀)² + b(x-x₀)(y-y₀) + c(y-y₀)², It's easy to express a, b and c with second derivatives as first three terms become zero after differentiation. However, initially fₓ and fy were defined as values of first derivatives, but now if you compute ∂Q(x,y)/∂x it's not just fₓ(x₀,y₀). It's: fₓ(x₀,y₀) + 2a(x-x₀) + b(y-y₀) Why we ignore those extra two terms?

@VladAdamenko

4 жыл бұрын

Ok, I guess I've answered that question while asking. Those two terms are zero when evaluated at (x₀,y₀).

@fahrenheit2101

7 ай бұрын

@@VladAdamenko 3 years late, but in case anybody else sees this, I'd like to point out that it's precisely analogous to normal single variable Taylor series, wherein the quadratic and higher order terms evaluate to 0 when taking a 2nd derivative.

Funny beginning 🤣🤣🤣

where can i find your article about this topic?

@Nur-vr6vb

6 жыл бұрын

www.khanacademy.org/math/multivariable-calculus/applications-of-multivariable-derivatives/quadratic-approximations/a/quadratic-approximation

@bikashthapa7316

6 жыл бұрын

thanks a lot

So beautiful

Actually it's also taught in single variable calculus

@chaoticcubes4929

4 жыл бұрын

Shraddha Publication how? I am genuinely curious. Is there a different explanation or something?

@johnmccrae2932

4 жыл бұрын

@@chaoticcubes4929 I think Shraddha is talking about taylor/maclaurin approximations of functions

@chaoticcubes4929

4 жыл бұрын

@@johnmccrae2932 Thanks

@That_One_Guy...

4 жыл бұрын

I've also tried to find quadratic approximation once in 2D before watching this video