Developing the 1st and 2nd degree Taylor Polynomials of a function of two variables and visualizing them on CalcPlot3D.
Жүктеу.....
Пікірлер: 37
@haricalderom669217 күн бұрын
thank you, Paul the wise I was pocking my head into the wall trying to understand my mistakes 6 mins in and now i know!
@zedftofficial5 жыл бұрын
Best video regarding this on KZread
@harwardw32373 жыл бұрын
Thank you for making this tutorial, really appreciate it!
@Kamnuma7 жыл бұрын
It was a very goog video. Thank you Paul!
@aghasaadfraz91867 жыл бұрын
Nice work Dude!helped a lot ,keep up the good work
@andreatoth93293 жыл бұрын
Thank you so much!!! It was a huge help :)
@taylan53872 жыл бұрын
thanks, you made it so clear !
@viktorkarlsson99896 жыл бұрын
Very good video! Thank you❤
@cacao253910 жыл бұрын
thanks from ARGENTINE!!!!!!!! VERY VERY WELL
@sanketh158 жыл бұрын
thanks !! it really helped me .. :)
@snakefromhell7 жыл бұрын
Thank you!
@matthewhyland26778 жыл бұрын
Nice one lad!
@chathurikawijesinghe28194 жыл бұрын
Thank you very much....🙏🙏
@schelias95444 жыл бұрын
underrated video!
@kyri77168 жыл бұрын
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
7 жыл бұрын
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
5 жыл бұрын
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
5 жыл бұрын
@@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 Жыл бұрын
Bro you helped a lot thank you
@dhruvsharma57565 ай бұрын
thanks Sir, It helped
@user-uw7ms4tt2m3 жыл бұрын
Can someone explain me how to do taylor to e^(x+y+xy) ?
@sickman11211 жыл бұрын
thanks so much
@cjzhang1016 жыл бұрын
Thanks!
@saheedkasali97933 жыл бұрын
God bless you
@sarathesea5 жыл бұрын
thanks man
@lazargugleta5 жыл бұрын
Thanks.
@SirTristanPM11 жыл бұрын
finaly i understand it
@methmipavithra96462 жыл бұрын
Sir can you please explain how to get fxy(x,y)=e^y
@salmankhanma19597 жыл бұрын
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
7 жыл бұрын
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
7 жыл бұрын
thank u sir
@israeben3089 Жыл бұрын
شكرا جزيلا🌼
@macchan17 жыл бұрын
anyone knows how to get the 6th degree?
@alvinlepik5265
7 жыл бұрын
Follows the same logic, but extremely annoying to write out.
@MazwiKhoza11 жыл бұрын
isn't 2!?
@JohnCharlesRome Жыл бұрын
ASMR voice
@aashsyed12773 жыл бұрын
Tell me the 3rd taylor polynomial of this now.!!!!
Пікірлер: 37
thank you, Paul the wise I was pocking my head into the wall trying to understand my mistakes 6 mins in and now i know!
Best video regarding this on KZread
Thank you for making this tutorial, really appreciate it!
It was a very goog video. Thank you Paul!
Nice work Dude!helped a lot ,keep up the good work
Thank you so much!!! It was a huge help :)
thanks, you made it so clear !
Very good video! Thank you❤
thanks from ARGENTINE!!!!!!!! VERY VERY WELL
thanks !! it really helped me .. :)
Thank you!
Nice one lad!
Thank you very much....🙏🙏
underrated video!
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
7 жыл бұрын
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
5 жыл бұрын
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
5 жыл бұрын
@@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?
Bro you helped a lot thank you
thanks Sir, It helped
Can someone explain me how to do taylor to e^(x+y+xy) ?
thanks so much
Thanks!
God bless you
thanks man
Thanks.
finaly i understand it
Sir can you please explain how to get fxy(x,y)=e^y
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
7 жыл бұрын
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
7 жыл бұрын
thank u sir
شكرا جزيلا🌼
anyone knows how to get the 6th degree?
@alvinlepik5265
7 жыл бұрын
Follows the same logic, but extremely annoying to write out.
isn't 2!?
ASMR voice
Tell me the 3rd taylor polynomial of this now.!!!!