Calculate the second-degree Taylor polynomial of
| f (x, y) = e-(x2+y2) |
Solution: The second-degree Taylor polynomial at the point (a, b) is
p2(x, y) = f (a, b) + Df (a, b) . ((x, y) - (a, b)) + ![]() |
First compute all the derivatives:
| = - 2xe-(x2+y2) | ||
| = - 2ye-(x2+y2), | ||
| = (- 2 + 4x2)e-(x2+y2) | ||
| = (- 2 + 4y2)e-(x2+y2) | ||
| = 4xye-(x2+y2) |
At the point (a, b) = (0, 0)
| f (0, 0) | = e0 = 1 | |
| = - 2, | ||
| = - 2 | ||
| = 0 |
The second-degree Taylor polynomial at the point (0,0) is
| p2(x, y) | = f (0, 0) + Df (0, 0) . ((x, y) - (0, 0)) | |
+ ![]() |
||
= 1 + (0, 0) . (x, y) + ![]() |
||
| = 1 - x2 - y2 |
At the point (a, b) = (1, 2), things get uglier. But it is good practice.
| f (1, 2) | = e-5 | |
| = - 2e-5 | ||
| = - 4e-5, | ||
| = 2e-5 | ||
| = 14e-5 | ||
| = 16e-5 |
The second-degree Taylor polynomial at the point (1,2) is
p2(x, y) = f (1, 2) + Df (1, 2) . ((x, y) - (1, 2)) + ![]() |
||
= e-5 + e-5(- 2, -4) . (x - 1, y - 2) + ![]() ![]() |
||
| = e-5 - e-5(2(x - 1) + 4(y - 2)) + e-5(x - 1)2 + e-516(x - 1)(y - 2) + e-57(y - 2)2 | ||
| = e-5(1 - 2(x - 1) - 4(y - 2) + (x - 1)2 +16(x - 1)(y - 2) + 7(y - 2)2) |