Question

Consider the inner product space of all real-valued continuous functions defined on \( [-1, 1] \) with the inner product \[ \langle f, g \rangle = \int_{-1}^{1} f(x) g(x) \, dx. \] If \( p(x) = \alpha + \beta x^2 - 30x^4 \), where \( \alpha, \beta \in \mathbb{R} \), is orthogonal to all the polynomials having degree less than or equal to 3, with respect to this inner product, then \( \alpha + 5\beta \) is equal to (in integer).

Hint:

For orthogonality conditions in inner product spaces, set up equations involving the integrals of the function with each of the polynomials and solve for the unknown coefficients.

Answer 1

Step 1: Use symmetry instead of all four orthogonality conditions
The inner product is defined on the symmetric interval \([-1,1]\). Observe that:

• \(p(x)=\alpha+\beta x^2-30x^4\) is an even function.
• \(x\) and \(x^3\) are odd functions.

Hence, by symmetry,

\[ \langle p(x),x\rangle=\langle p(x),x^3\rangle=0 \] automatically.

So we only need to enforce orthogonality with the even basis functions \(1\) and \(x^2\).

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Step 2: Orthogonality with \(1\)

\[ \int_{-1}^{1}(\alpha+\beta x^2-30x^4)\,dx=0 \]

Using standard integrals:

\[ \int_{-1}^{1}1\,dx=2,\quad \int_{-1}^{1}x^2\,dx=\frac{2}{3},\quad \int_{-1}^{1}x^4\,dx=\frac{2}{5} \]

So,

\[ 2\alpha+\frac{2\beta}{3}-30\cdot\frac{2}{5}=0 \]

\[ 2\alpha+\frac{2\beta}{3}-12=0 \]

Multiplying by 3:

\[ 6\alpha+2\beta=36 \quad\Rightarrow\quad 3\alpha+\beta=18 \]

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Step 3: Orthogonality with \(x^2\)

\[ \int_{-1}^{1}(\alpha+\beta x^2-30x^4)x^2\,dx=0 \]

\[ \int_{-1}^{1}(\alpha x^2+\beta x^4-30x^6)\,dx=0 \]

Using:

\[ \int_{-1}^{1}x^2\,dx=\frac{2}{3},\quad \int_{-1}^{1}x^4\,dx=\frac{2}{5},\quad \int_{-1}^{1}x^6\,dx=\frac{2}{7} \]

\[ \alpha\frac{2}{3}+\beta\frac{2}{5}-30\frac{2}{7}=0 \]

Multiply by \(105\):

\[ 70\alpha+42\beta-900=0 \quad\Rightarrow\quad 35\alpha+21\beta=450 \]

Divide by 7:

\[ 5\alpha+3\beta=64.2857 \]

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Step 4: Solve the system

From Step 2: \[ \beta=18-3\alpha \]

Substitute into Step 3:

\[ 5\alpha+3(18-3\alpha)=64.2857 \]

\[ 5\alpha+54-9\alpha=64.2857 \]

\[ -4\alpha=10.2857 \quad\Rightarrow\quad \alpha=-2.5714 \]

\[ \beta=18-3(-2.5714)=25.7142 \]

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Step 5: Compute the required value

\[ \alpha+5\beta=-2.5714+5(25.7142)=126 \]

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Final Answer:

\[ \boxed{126} \]