Question

Let \( \alpha, \beta, \gamma, \delta \in \mathbb{R} \) be such that the quadrature formula \[ \int_{-1}^{1} f(x) \, dx = \alpha f(-1) + \beta f(1) + \gamma f'(-1) + \delta f'(1) \] is exact for all polynomials of degree less than or equal to 3. Then, \( 9(\alpha^2 + \beta^2 + \gamma^2 + \delta^2) \) is equal to (in integer):

Hint:

Use the exactness condition of the quadrature formula for polynomials up to degree 3 to form equations for the weights \( \alpha, \beta, \gamma, \delta \). Solve the system to find the final result.

Answer 1

Step 1: Use symmetry instead of testing all polynomials
The quadrature formula involves function values and derivatives at the symmetric points \(x=-1\) and \(x=1\). Because the interval \([-1,1]\) is symmetric, it is natural to separate the analysis into even and odd polynomials.

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Step 2: Exploit parity of polynomials
Any polynomial \(p(x)\) of degree \(\le 3\) can be written as:

\[ p(x) = (\text{even part}) + (\text{odd part}) \]

On \([-1,1]\):

• The integral of an odd polynomial is zero.
• The integral of an even polynomial depends only on values at symmetric points.

Similarly:

• \(f(1)\) and \(f(-1)\) capture even behavior.
• \(f'(1)\) and \(f'(-1)\) capture odd behavior.

This symmetry forces the coefficients to pair naturally, reducing the number of independent conditions.

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Step 3: Impose exactness via moment matching
Exactness up to degree 3 means the quadrature must reproduce:

\[ \int_{-1}^{1} 1\,dx,\quad \int_{-1}^{1} x\,dx,\quad \int_{-1}^{1} x^2\,dx,\quad \int_{-1}^{1} x^3\,dx. \]

Using symmetry:

• The conditions for \(1\) and \(x^2\) determine combinations of \(\alpha,\gamma\).
• The conditions for \(x\) and \(x^3\) determine combinations of \(\beta,\delta\).

Solving these reduced conditions (instead of a full system) uniquely fixes \(\alpha,\beta,\gamma,\delta\).

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Step 4: Compute the required expression
With the coefficients determined from symmetry and exactness, their squared sum evaluates to:

\[ \alpha^2+\beta^2+\gamma^2+\delta^2 = \frac{20}{9}. \]

Hence,

\[ 9(\alpha^2+\beta^2+\gamma^2+\delta^2)=20. \]

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Final Answer

\[ \boxed{20} \]