Solving for this slope function

Question

f'(a) = [f(a+h)-f(a)]/h

its f'(a) not just f(a)

steps would be much appreciated thanks!

Answer 1

Hi Mohammed,

The derivation:

The slope function f'(x) = [f(x+h)-f(x)]/h is called a forward difference scheme and is derived using the Taylor Series expansion of a function about x,

f(x+h) ≈ f(x) + f'(x)h + f''(x)h²/2 + f⁽³⁾(x)h³/3 + ...

assuming that h is small.

Solving for f'(x) gives us the forward difference scheme that you wanted.

f'(x)h = f(x+h) - f(x) - f''(x)h²/2 - f⁽³⁾(x)h³/3 + ...

f'(x) = [f(x+h) - f(x)]/h - f''(x)h/2 - f⁽³⁾(x)h²/3 + ...

This formula is a bit larger than the one you listed and includes extra higher-order terms that we don't want to include. These terms represent the error of the approximation. Since h > h² > h³ ... the term f''(x)h/2 will be the driving factor behind how well the error scales and will be rewritten as O(h) to represent that the error will scale with order h.

f'(x) = [f(x+h) - f(x)]/h + O(h)

Example:

Solve the derivative of sin(x) at x = pi/4 deg and calculate the error of the forward difference approximation for h = 0.1, h = 0.01, and h = 0.001.

Exact value:

f(x) = sin(x)

f'(x) = cos(x)

f'(pi/4) = cos(pi/4) = 0.707107

Forward difference approximation:

f'(x) = [f(x+h) - f(x)]/h + O(h)

f'(x) = [sin(x+h) - sin(x)]/h + O(h)

for h = 0.1

f'(x) = [sin(x+0.1) - sin(x)]/0.1 + O(h)

f'(0) = [sin(pi/4+0.1) - sin(pi/4)]/0.1 = 0.670603

Error = 0.707107 - 0.670603 = 0.036504

for h = 0.01

f'(x) = [sin(x+0.01) - sin(x)]/0.01 + O(h)

f'(0) = [sin(pi/4+0.01) - sin(pi/4)]/0.01 = 0.703559

Error = 0.707107 - 0.703559 = 0.003547

for h = 0.001

f'(x) = [sin(x+0.001) - sin(x)]/0.001 + O(h)

f'(0) = [sin(pi/4+0.001) - sin(pi/4)]/0.001 = 0.703559

Error = 0.707107 - 0.703559 = 0.000354

Notice that each time we decreased h by an order of 10 the error decreased by an order of 10. This is what it means to have a first order error.

If we were to solve a more complex forward difference formula with O(h³) each time we decrease h by an order of 10 the error would decrease by an order of 1000!

I hope that helps. If you have any questions let me know.

Bryce