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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!

Update:

Thanks but it says the answers wrong :(

Update 2:

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

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  • kb
    Lv 7
    1 decade ago
    Favorite 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!

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