Question

Find the second Taylor polynomial of the function , f given by
f(x,y)=e(x+2y) at (-1,1) .

Answer 1

Answer on Question #41405, Integral Calculus

Find the second Taylor polynomial of the function $f$ given by

$$f(x, y) = e^{(x + 2y)} \text{ at } (-1, 1).$$

Solution.

For a function that depends on two variables, $x$ and $y$, the Taylor series to second order about the point $(a, b)$ is

$$\begin{aligned} f(x, y) &= f(a, b) + (x - a)f_x(a, b) + (y - b)f_y(a, b) + \frac{1}{2!} \left((x - a)^2 f_{xx}(a, b) + 2(x - a)(y - b)f_{xy}(a, b) \right. \\ &\quad \left. + (y - b)^2 f_{yy}(a, b)\right) \end{aligned}$$

For our function.

$$f(a, b) = f(-1, 1) = e^{(-1 + 2)} = e$$ $$f_x(x, y) = e^{(x + 2y)}$$ $$f_y(x, y) = 2e^{(x + 2y)}$$ $$f_{xy}(x, y) = \left(f_x(x, y)\right)_y = \left(e^{(x + 2y)}\right)_y = 2e^{(x + 2y)}$$ $$f_{xx}(x, y) = \left(e^{(x + 2y)}\right)_x = e^{(x + 2y)}$$ $$f_{yy}(x, y) = \left(2e^{(x + 2y)}\right)_y = 4e^{(x + 2y)}$$ $$f_x(-1, 1) = e^{(-1 + 2)} = e$$ $$f_y(-1, 1) = 2e^{(-1 + 2)} = 2e$$ $$f_{xy}(-1, 1) = 2e^{(-1 + 2)} = 2e$$ $$f_{xx}(-1, 1) = e^{(-1 + 2)} = e$$ $$f_{yy}(-1, 1) = 4e^{(-1 + 2)} = 4e$$

Thus,

$$\begin{aligned} f(x, y) &= e + (x + 1)e + (y - 1)2e + \frac{1}{2} \left[(x + 1)^2 e + 2(x + 1)(y - 1)2e + (y - 1)^2 4e\right] = \\ &= e + e(x + 1) + 2e(y - 1) + \frac{e}{2}(x + 1)^2 + e(x^2 - 1) + 2e(y - 1)^2 \end{aligned}$$

Answer:

$$f(x, y) = e + e(x + 1) + 2e(y - 1) + \frac{e}{2}(x + 1)^2 + e(x^2 - 1) + 2e(y - 1)^2$$

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