Prev | Next |
exp_eps(x, epsilon)
with
x
is equal to .5
and
epsilon
is equal to .2.
For this case, the mathematical function for the operation sequence
corresponding to exp_eps
is
@[@
f ( x , \varepsilon ) = 1 + x + x^2 / 2
@]@
The corresponding partial derivative with respect to @(@
x
@)@,
and the value of the derivative, are
@[@
\partial_x f ( x , \varepsilon ) = 1 + x = 1.5
@]@
Index |
| Operation |
| Zero Order |
| Derivative |
| First Order |
1 |
| @(@ v_1 = x @)@ | 0.5 | @(@ v_1^{(1)} = x^{(1)} @)@ | @(@ v_1^{(1)} = 1 @)@ | |||
2 |
| @(@ v_2 = 1 * v_1 @)@ | 0.5 | @(@ v_2^{(1)} = 1 * v_1^{(1)} @)@ | @(@ v_2^{(1)} = 1 @)@ | |||
3 |
| @(@ v_3 = v_2 / 1 @)@ | 0.5 | @(@ v_3^{(1)} = v_2^{(1)} / 1 @)@ | @(@ v_3^{(1)} = 1 @)@ | |||
4 |
| @(@ v_4 = 1 + v_3 @)@ | 1.5 | @(@ v_4^{(1)} = v_3^{(1)} @)@ | @(@ v_4^{(1)} = 1 @)@ | |||
5 |
| @(@ v_5 = v_3 * v_1 @)@ | 0.25 | @(@ v_5^{(1)} = v_3^{(1)} * v_1^{(0)} + v_3^{(0)} * v_1^{(1)} @)@ | @(@ v_5^{(1)} = 1 @)@ | |||
6 |
| @(@ v_6 = v_5 / 2 @)@ | 0.125 | @(@ v_6^{(1)} = v_5^{(1)} / 2 @)@ | @(@ v_6^{(1)} = 0.5 @)@ | |||
7 |
| @(@ v_7 = v_4 + v_6 @)@ | 1.625 | @(@ v_7^{(1)} = v_4^{(1)} + v_6^{(1)} @)@ | @(@ v_7^{(1)} = 1.5 @)@ |
exp_eps(x, epsilon)