The oracle function f(x,y) is presented below. For each point (x,y) you enter the oracle will tell you the value f(x,y). Estimate the partial derivative of the function at (1.6,−0.0999999999999999) using the Newton quotient definition.
I just did that exercise on my webwork assignment, do the following:
1) Calculate in the oracle f in the point 1 they gave you (1.6,−0.0999999999999999). Let's call this result f(x1,y1).
2) Calculate in the oracle f in a point slightly superior in the x axis. For example (1.601,−0.0999999999999999). Let's call this result f(x2,y1).
3) Use the Newton's quotient definition:
Fx (x1,y1,) = ( f(x2,y1) - f(x1,y1) ) / (x2-x1)
Where x2 is 1.601 and x1 = 1.6.
Be happy :)
Suppose you want ∂f/∂y
We can approximate that by
(f(x,y+h)-f(x,y)) / h
So just plug and chug. For ease of typing, I'll use (1.6,-0.099)
You can add as many zeroes as you want.
(f(1.6,-.1+0.001)-f(1.6,-0.1))/0.001
You have not said what f(1.6,-0.1) is, so that's all I can say at the moment.
To estimate the partial derivative of the function at (1.6, -0.0999999999999999) using the Newton quotient definition, you will need to evaluate the function at two points that are very close to each other.
The Newton quotient definition for the partial derivative with respect to x is:
f_x(x, y) = (f(x + h, y) - f(x, y)) / h
where h is a small value. In this case, we want to estimate the partial derivative with respect to x at (1.6, -0.0999999999999999). Therefore, we need to choose two x-values that are very close to each other in order to calculate the Newton quotient.
Let's choose h to be a small number, such as h = 0.001. Now we can evaluate the function at (1.6, -0.0999999999999999) and at (1.6 + 0.001, -0.0999999999999999):
f(1.6, -0.0999999999999999) = [Enter Oracle Function to get the value]
f(1.6 + 0.001, -0.0999999999999999) = [Enter Oracle Function to get the value]
Now, we can calculate the Newton quotient:
f_x(1.6, -0.0999999999999999) = (f(1.6 + 0.001, -0.0999999999999999) - f(1.6, -0.0999999999999999)) / 0.001
Evaluate the above expression using the values you obtained from the oracle, and you will have an estimate for the partial derivative of the function at (1.6, -0.0999999999999999) using the Newton quotient definition.
you need a difference. if you are looking at the given point, try these points into the oracle x=1.59, and x=1.61 (h=.02), and you now have f(x+h), and f(x)
then it will give you f(x) at those points, then do this https://www.youtube.com/watch?v=1O5NEI8UuHM