Yahoo Answers is shutting down on May 4th, 2021 (Eastern Time) and beginning April 20th, 2021 (Eastern Time) the Yahoo Answers website will be in read-only mode. There will be no changes to other Yahoo properties or services, or your Yahoo account. You can find more information about the Yahoo Answers shutdown and how to download your data on this help page.
Trending News
Taylor series problem?
: Find the second-order Taylor polynomial for f(x,y)=5yx^2 + 4xy^3 at the point (1,2)
(I get 42 + 52(x-1) + 80(y-2) + 10(x-1)^2 + 58(x-1)(y-2) + 23(y-2)^2 but thats wrong...)
Thanks!
Thanks but it says the answers wrong :(
nevermind, got it, it was 42 + 52(x - 1) + 53(y - 2) + .5(20(x - 1)^2 + 2*58(x - 1)(y - 2) + 48(y - 2)^2)
1 Answer
- kbLv 71 decade agoFavorite Answer
f(x,y) = 5x^2 y + 4xy^3 ==> f(1, 2) = 42
f_x(x,y) = 10xy + 4y^3 ==> f_x(1, 2) = 52
f_y(x,y) = 5x^2 + 12xy^2 ==> f_y(1, 2) = 53
f_xx(x,y) = 10y ==> f_xx(1, 2) = 20
f_xy(x,y) = 10x + 12y^2 ==> f_xy(1, 2) = 58
f_yy(x,y) = 24xy ==> f_yy(1, 2) = 48.
Therefore,
f(x, y) ≈ 42 + 52(x - 1) + 53(y - 2) + (1/2!) [20(x - 1)^2 + 2 * 58(x - 1)(y - 2) + 24(y - 2)^2]
.........= 42 + 52(x - 1) + 53(y - 2) + 10(x - 1)^2 + 58(x - 1)(y - 2) + 24(y - 2)^2.
I hope this helps!
