Question

For \(h>0\), and \(\alpha, \beta, \gamma \in \mathbb{R}\), let \[ D_h f(a) = \frac{\alpha f(a-h) + \beta f(a) + \gamma f(a+2h)}{6h} \] be a three-point formula to approximate \(f'(a)\) for any differentiable function \(f: \mathbb{R} \to \mathbb{R}\) and \(a \in \mathbb{R}\).
If \(D_h f(a) = f'(a)\) for every polynomial f of degree less than or equal to 2 and for all \(a \in \mathbb{R}\), then

• A. \(\alpha + 2\gamma = -2\)
• B. \(\alpha + 2\beta - 2\gamma = 0\)
• C. \(\alpha + 2\gamma = 2\)
• D. \(\alpha + 2\beta - 2\gamma = 1\)

Hint:

The Taylor series method for finding coefficients of numerical differentiation formulas is systematic and powerful. Equating coefficients of \(f(a), f'(a), f''(a), ......\) on both sides of the approximation provides a system of linear equations for the unknown formula weights.

Answer 1

The Correct Option is A, B

Step 1: Understanding the Concept:
We need to find the constants \(\alpha, \beta, \gamma\) that make a given numerical differentiation formula exact for all polynomials up to degree 2. The standard method is to enforce exactness for the basis polynomials \(1, x, x^2\). A more robust method uses Taylor series expansions.
Step 2: Key Formula or Approach:
Expand \(f(a-h)\) and \(f(a+2h)\) in Taylor series around \(a\). Substitute these into the formula for \(D_h f(a)\). Collect terms by derivatives of f at a (\(f(a), f'(a), f''(a)\)). To make the formula exact for polynomials of degree \(\le 2\), we match the coefficients with those of \(f'(a)\).
Step 3: Detailed Calculation:
Using Taylor series: \[ f(a-h) = f(a) - hf'(a) + \frac{h^2}{2}f''(a) - O(h^3) \] \[ f(a+2h) = f(a) + 2hf'(a) + \frac{(2h)^2}{2}f''(a) + O(h^3) = f(a) + 2hf'(a) + 2h^2f''(a) + O(h^3) \] Substitute these into the numerator of the formula: \[ \alpha f(a-h) + \beta f(a) + \gamma f(a+2h) = \alpha(f(a) - hf'(a) + \frac{h^2}{2}f''(a)) + \beta f(a) + \gamma(f(a) + 2hf'(a) + 2h^2f''(a)) + ...... \] Group terms by derivatives of \(f\): \[ = (\alpha + \beta + \gamma)f(a) + h(-\alpha + 2\gamma)f'(a) + h^2(\frac{\alpha}{2} + 2\gamma)f''(a) + ...... \] Now, divide by \(6h\) to get \(D_h f(a)\): \[ D_h f(a) = \frac{\alpha+\beta+\gamma}{6h}f(a) + \frac{-\alpha+2\gamma}{6}f'(a) + \frac{h(\alpha/2+2\gamma)}{6}f''(a) + ...... \] For this to equal \(f'(a)\) for all polynomials of degree \(\le 2\), the coefficients must match. For such polynomials, \(f'''(a)=0\) and higher, so we only need to consider terms up to \(f''(a)\).
1. Coeff of \(f(a)\): \(\frac{\alpha+\beta+\gamma}{6h} = 0 \implies \alpha+\beta+\gamma = 0\)
2. Coeff of \(f'(a)\): \(\frac{-\alpha+2\gamma}{6} = 1 \implies -\alpha+2\gamma = 6\)
3. Coeff of \(f''(a)\): \(\frac{h(\alpha/2+2\gamma)}{6} = 0 \implies \frac{\alpha}{2}+2\gamma = 0 \implies \alpha+4\gamma = 0\)
We solve this system of 3 linear equations. From (3), \(\alpha = -4\gamma\). Substitute into (2): \(-(-4\gamma) + 2\gamma = 6 \implies 6\gamma = 6 \implies \gamma = 1\).
Then \(\alpha = -4(1) = -4\).
Substitute into (1): \(-4 + \beta + 1 = 0 \implies \beta = 3\).
So, \(\alpha=-4, \beta=3, \gamma=1\).
Now check the given statements:
(A) \(\alpha + 2\gamma = -2\): \(-4 + 2(1) = -4 + 2 = -2\).
This is TRUE.
(B) \(\alpha + 2\beta - 2\gamma = 0\): \(-4 + 2(3) - 2(1) = -4 + 6 - 2 = 0\).
This is TRUE.
(C) \(\alpha + 2\gamma = 2\): \(-4 + 2(1) = -2 \neq 2\).
This is FALSE.
(D) \(\alpha + 2\beta - 2\gamma = 1\): \(-4 + 2(3) - 2(1) = 0 \neq 1\).
This is FALSE.
Step 4: Final Answer:
The correct statements are (A) and (B).