GATE 2025 PH – Question 61
Consider $\{1,x,x^2\}$. An orthonormal basis on $[-1,1]$ uses normalization $\int_{-1}^1x^2[f(x)]^2dx=1$. If the basis is $\left(\sqrt{3/2},\sqrt{5/2}x,\frac12\sqrt{21/N}(5x^2-3)\right)$, the value of $N$ (in integer) is _____
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Correct answer: 6
Explanation
The basis functions are orthonormal with the weight $x^2$: $\displaystyle\int_{-1}^1x^2f(x)^2\,dx=1$.
The third function is $f_3(x)=\dfrac12\sqrt{\dfrac{21}{N}}\,(5x^2-3)$.
**Norm:**
$$\int_{-1}^1x^2f_3^2\,dx=\frac14\cdot\frac{21}{N}\int_{-1}^1x^2(5x^2-3)^2dx .$$
**Integral:** expand $x^2(5x^2-3)^2=25x^6-30x^4+9x^2$:
$$\int_{-1}^1\left(25x^6-30x^4+9x^2\right)dx=2\left(\frac{25}7-6+3\right)=2\cdot\frac47=\frac87 .$$
**Normalisation condition:**
$$\frac{21}{4N}\times\frac87=\frac{6}{N}=1\;\Rightarrow\;N=\mathbf{6}.$$