Talyor polynomial example
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)) | ||
+ ![]() Hf(a,b)![]() |
First compute all the derivatives:
(x,y) | = -2xe-(x2+y2) | ||
(x,y) | = -2ye-(x2+y2), | ||
(x,y) | = (-2 + 4x2)e-(x2+y2) | ||
(x,y) | = (-2 + 4y2)e-(x2+y2) | ||
(x,y) = (x,y) | = 4xye-(x2+y2) |
At the point (a,b) = (0, 0)
| f(0, 0) | = e0 = 1 | ||
(0, 0) | = -2, | ||
(0, 0) | = -2 | ||
(0, 0) | = 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)) | ||
+ ![]() Hf(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 | ||
(1, 2) | = -2e-5 | ||
(1, 2) | = -4e-5, | ||
(1, 2) | = 2e-5 | ||
(1, 2) | = 14e-5 | ||
(1, 2) = (1, 2) | = 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)) + ![]() Hf(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) |

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