Derive the following formulas using Taylor series and then establish the error terms for each.
f'(x) approx= (1/2*h)[4f(x+h)-3f(x)-f(x+2h)]
f'(x) approx= (1/4*h)[f(x+2h)-f(x-2h)]
To derive the formulas using Taylor series, let's start with the first formula:
f'(x) approx= (1/2*h)[4f(x+h)-3f(x)-f(x+2h)]
Step 1: Taylor expansion around x
We can approximate the function f(x) using a Taylor series expansion around the point x:
f(x+h) = f(x) + hf'(x) + (h^2/2)f''(x) + (h^3/6)f'''(x) + ... (1)
f(x+2h) = f(x) + 2hf'(x) + (4h^2/2)f''(x) + (8h^3/6)f'''(x) + ... (2)
Step 2: Manipulating the expansions
Now let's manipulate equations (1) and (2) to find the expression for f'(x).
Multiply equation (1) by 4:
4f(x+h) = 4f(x) + 4hf'(x) + 2(h^2)f''(x) + (4h^3/3)f'''(x) + ... (3)
Multiply equation (2) by 3:
3f(x+2h) = 3f(x) + 6hf'(x) + 6(h^2)f''(x) + (12h^3/3)f'''(x) + ... (4)
Step 3: Combining equations
Now we'll combine equations (3) and (4) in order to eliminate the higher-order terms and obtain an expression for f'(x).
Equation (3) - equation (4):
4f(x+h) - 3f(x+2h) = f(x) + (2h/3)f'(x) + O(h^2) (5)
where O(h^2) represents the higher-order terms that are neglected.
Step 4: Solving for f'(x)
Now we can solve equation (5) for f'(x):
4f(x+h) - 3f(x+2h) = f(x) + (2h/3)f'(x) + O(h^2)
Multiplying both sides by (1/2h):
f'(x) approx= (1/2*h)[4f(x+h) - 3f(x+2h) - f(x)] + O(h^2)
So, the derived formula for f'(x) using the Taylor series expansion is:
f'(x) approx= (1/2*h)[4f(x+h) - 3f(x+2h) - f(x)]
Now, let's establish the error term for this formula.
Step 5: Error term (Remainder term)
The error term in this formula is given by the omitted higher-order terms, which are represented by O(h^2) in equation (5).
To bound the error, we can use Taylor's remainder formula for the specific case of a second-order derivative:
R = |(h^2/3)f''(η)|
where R is the error term, h is the step size, and η is some value between x and x + 2h.
Therefore, the error term for the first formula is given by:
R = |(h^2/3)f''(η)|
Now, let's move on to the second formula:
f'(x) approx= (1/4*h)[f(x+2h) - f(x-2h)]
We will follow a similar approach to find the Taylor series expansion and error term.
Step 1: Taylor expansion around x
Using the same logic as before, we can expand the function f(x) around the point x:
f(x+2h) = f(x) + 2hf'(x) + (4h^2/2)f''(x) + (8h^3/6)f'''(x) + ... (6)
f(x-2h) = f(x) - 2hf'(x) + (4h^2/2)f''(x) - (8h^3/6)f'''(x) + ... (7)
Step 2: Manipulating the expansions
Now let's manipulate equations (6) and (7) to find the expression for f'(x).
Subtract equation (7) from equation (6):
f(x+2h) - f(x-2h) = 4hf'(x) + (16h^3/6)f'''(x) + ... (8)
Step 3: Solving for f'(x)
Now we can solve equation (8) for f'(x):
f'(x) approx= (1/4*h)[f(x+2h) - f(x-2h)] + O(h^2)
So, the derived formula for f'(x) using the Taylor series expansion is:
f'(x) approx= (1/4*h)[f(x+2h) - f(x-2h)]
Now, let's establish the error term for this formula.
Step 4: Error term (Remainder term)
The error term in this formula is also given by the omitted higher-order terms.
To bound the error, we can again use Taylor's remainder formula for the specific case of a second-order derivative:
R = |(4h^3/6)f'''(η)|
where R is the error term, h is the step size, and η is some value between x-2h and x+2h.
Therefore, the error term for the second formula is given by:
R = |(4h^3/6)f'''(η)|
So, we have derived the formulas using Taylor series expansions and established the error terms for each formula.