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$$f(x, y) = e^{2x+xy+y^2}$$

Find the 2nd order taylor polynomial to the above function about (0,0)

The formula is:

$$P(x,y)=f(a,b)+f_x(a,b)(x-a)+f_y(a,b)(y-b)+\frac 12[f_{xx}(x-a)^2+2f_{xy}(x-a)(y-b)+f_{yy}(y-b)^2]$$

$$f_x=e^{2x+xy+y^2}(2+y)$$ $$f_y=e^{2x+xy+y^2}(x+2y)$$ $$f_{xx}=e^{2x+xy+y^2}(2+y)^2$$ $$f_{yy}=e^{2x+xy+y^2}(x+2y)^2+2e^{2x+xy+y^2}$$ $$f_{xy}=e^{2x+xy+y^2}(2+y)(x+2y)+e^{2x+xy+y^2}$$

But I still get the wrong answer. What I am doing wrong?

Nash
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2 Answers2

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We have:

$$f(x, y) = e^{2x+xy+y^2}$$

Finding partials and evaluating them at the point $(a, b) = (0,0)$, yields:

  • $f_x(x, y) = (y+2) e^{x y+2 x+y^2} \implies f_x(0,0) = 2$
  • $f_y(x, y) = e^{x y+2 x+y^2} (x+2 y) \implies f_y(0,0) = 0$
  • $f_{xx}(x, y) = (y+2)^2 e^{x y+2 x+y^2} \implies f_{xx}(0,0) = 4$
  • $f_{xy}(x, y) = (y+2) e^{x y+2 x+y^2} (x+2 y)+e^{x y+2 x+y^2} \implies f_{xy}(0,0) = 1$
  • $f_{yy}(x, y) = e^{x y+2 x+y^2} (x+2 y)^2+2 e^{x y+2 x+y^2} \implies f_{yy}(0,0) = 2 $

Next, we have:

$P(x,y)=f(a,b)+f_x(a,b)(x-a)+f_y(a,b)(y-b)+\dfrac 12\left[f_{xx}(a,b)(x-a)^2+2f_{xy}(a,b)(x-a)(y-b)+f_{yy}(a,b)(y-b)^2\right]$

So, we get:

$$P(x, y) = 1 + 2 x + 0 +\frac 12[4 x^2 + 2(1) x y + 2 y^2] = 1 + 2x + 2x^2 + xy + y^2$$

Notice that the book forgot to divide $\frac 12 (4)$ and that is the source of their error. Your answer is correct.

Amzoti
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Hint: $f_{yy}=e^{2x+xy+y^2}[(x+2y)^2\color{red}{+2}]$.

Eclipse Sun
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