Taylor Polynomials of a Function of Two Variables

Developing the 1st and 2nd degree Taylor Polynomials of a function of two variables and visualizing them on CalcPlot3D.

[@zedftofficial]

Best video regarding this on TH-cam

[@harwardw3237]

Thank you for making this tutorial, really appreciate it!

[@aghasaadfraz9186]

Nice work Dude!helped a lot ,keep up the good work

[@Kamnuma]

It was a very goog video. Thank you Paul!

[@andreatoth9329]

Thank you so much!!! It was a huge help :)

[@taylan5387]

thanks, you made it so clear !

[@viktorkarlsson9989]

Very good video! Thank you❤

[@cacao2539]

thanks from ARGENTINE!!!!!!!!
VERY VERY WELL

[@sanketh15]

thanks !! it really helped me .. :)

[@snakefromhell]

Thank you!

[@schelias9544]

underrated video!

[@kyri7716]

5:45 - Why did you consider only the partial derivative f_xy rather than both f_xy and f_yx when writing the Taylor's expansion of the function f? Does the Taylor's approximation to a function f consider only unique partial derivatives? Is that why only one of those 2 equivalent partial derivatives were considered?

[@pseeburger]

Kyr, did you notice that the second partial with respect to x and the second partial with respect to y both were divided by 2 (really 2!)? But the fxy was not divided by 2. Really it was, but since there were two identical terms (one with fxy and the other with fyx), we can combine them and get a factor of 1 from the 1/2 + 1/2.

[@kareemalbukai1779]

I know am a bit late but the reason for f_xy and f_yx being identical is that the function f is continuously differentiable (the partial derivatives exist and are continuous) which implies that f_xy will be equal to f_yx (theorem) and hence he can just sum the two guys up and continue on. Hope it helps.

[@oliverbeck6839]

@@pseeburger I don't understand why did you didn't square both the terms though, even if you sum you still have to square them no?

[@israelakinsanya3421]

Bro you helped a lot thank you

[@matthewhyland2677]

Nice one lad!

[@chathurikawijesinghe2819]

Thank you very much....🙏🙏

[@sickman112]

thanks so much

[@dhruvsharma5756]

thanks Sir, It helped

[@lazargugleta]

Thanks.

[@cjzhang101]

Thanks！

[@saheedkasali9793]

God bless you

[@SirTristanPM]

finaly i understand it

[@sarathesea]

thanks man

[@user-uw7ms4tt2m]

Can someone explain me how to do taylor to e^(x+y+xy) ?

[@salmankhanma1959]

Hello sir can i know how to decide the degree of the function if they dint mention that in the qtn
in this qtn they have given the degree but if they dint give then??

[@pseeburger]

Your original function is likely NOT a polynomial (where we usually discuss degree), although this process can be used to determine linear and quadratic approximations of even a polynomial function (of higher degree). The questions you are given on this topic should ask you to determine the Taylor Polynomial of a specified degree(s) for a given function, or it may ask for the Taylor Series that represents the given function (at least on an interval about the center point). In this example I show how to do this for the 1st and 2nd degree. The general case is a bit more interesting, but it does follow the same patterns. See the Java version of my CalcPlot3D app and use the Taylor Polynomial tool to view some higher order Taylor Polys for a function like z = cos(x) sin(y). Then select the option to Use Factorials in Taylor Polynomials from the Tools menu.

[@salmankhanma1959]

thank u sir

[@israeben3089]

شكرا جزيلا🌼

[@methmipavithra9646]

Sir can you please explain how to get fxy(x,y)=e^y

[@macchan1]

anyone knows how to get the 6th degree?

[@alvinlepik5265]

Follows the same logic, but extremely annoying to write out.

[@MazwiKhoza]

isn't 2!?

[@JohnCharlesRome]

ASMR voice

[@aashsyed1277]

Tell me the 3rd taylor polynomial of this now.!!!!