Question

[Note that, in this example, the mesh sizes in x and y are identical (h); strictly speaking, this need not be true. In some applications, we may need more resolution along the x- or y-axis; we could then use separate mesh sizes h_x and h_y.]

By definition, the partial derivative of a function f ( x , y ) with respect to x is

∂ f ∂ x = L i m h ⟶ 0 f ( x i + h , y j ) − f ( x i , y j ) h

and the partial derivative with respect to y is

∂ f /∂ y = L i m h ⟶ 0 f ( x i , y j + h ) − f ( x i , y j )/ h

If we applied these formulas to our grid values, we would get the finite difference expressions

∂ f /d x ( x i , y j ) ≅ f ( x i + 1 , y j ) − f ( x i , y j )/ h

Note: To avoid round-off error, retain at least six decimal places in all of your calculations.

• Assume the function f is defined as f(x, y) = 3 tan x cos y
• Use differentiation rules to find the exact partial derivatives ∂ f /∂ x and ∂ f /∂ y , and evaluate those exact partial derivatives at (1.56, -2.1).  
• Use the finite difference formulas to estimate ∂ f /∂ x and ∂ f/ ∂ y at (1.56, -2.1) for three different values of the mesh size
  • h = 0.01
  • h = 0.001
  • h = 0.0001

• Use your calculated values to fill in this table:

Estimated partial derivatives using finite difference formulas:
h	finite difference approx. to ∂ f/ ∂ x	exact ∂ f/ ∂ x	finite difference approx. to ∂ f/ ∂ y	exact ∂ f/ ∂ y
0.01
0.001
0.0001

Answer the following questions:

• For which partial derivative is the finite difference approximation more accurate?
• Why is the finite difference approximation for the other partial derivative less accurate?  Under what real-world conditions might we see such poor approximations?

Answer 1