Question:
If \(\mu\) is an averaging operator and \(\delta^n\) is the \(n^{\text{th}}\) central difference operator, then what is the value of \(\mu^2\)?
यदि \(\mu\) एक औसत ऑपरेटर (averaging operator) है और \(\delta^n\) \(n^{\text{th}}\) केंद्रीय अंतर ऑपरेटर (central difference operator) है, तो \(\mu^2\) का मान क्या है?
Options:
- \(\mu^2 = \frac{1}{4}(\delta^2 + 4)\)
- \(\mu^2 = \frac{1}{2}(\delta^2 + 4)\)
- \(\mu^2 = \frac{1}{4}(\delta^2 + 6)\)
- \(\mu^2 = \frac{1}{4}(\delta^2 + 8)\)
- \(\mu^2 = \frac{1}{4}(\delta^2 + 4)\)
- \(\mu^2 = \frac{1}{2}(\delta^2 + 4)\)
- \(\mu^2 = \frac{1}{4}(\delta^2 + 6)\)
- \(\mu^2 = \frac{1}{4}(\delta^2 + 8)\)
Correct Answer: a)
Explanation:
In finite difference calculus, the operators are related to the shift operator \(E\) as follows:
– Averaging operator: \(\mu = \frac{E^{1/2} + E^{-1/2}}{2}\)
– Central difference operator: \(\delta = E^{1/2} – E^{-1/2}\)
Squaring both:
\[
\mu^2 = \frac{(E^{1/2} + E^{-1/2})^2}{4} = \frac{E + E^{-1} + 2}{4}
\]
\[
\delta^2 = (E^{1/2} – E^{-1/2})^2 = E + E^{-1} – 2 \implies E + E^{-1} = \delta^2 + 2
\]
Substituting \(E + E^{-1}\) into \(\mu^2\):
\[
\mu^2 = \frac{(\delta^2 + 2) + 2}{4} = \frac{1}{4}(\delta^2 + 4) = 1 + \frac{1}{4}\delta^2
\]
Thus, \(\mu^2 = \frac{1}{4}(\delta^2 + 4)\).
संख्यात्मक विश्लेषण में, शिफ्ट ऑपरेटर \(E\) के संदर्भ में:
– औसत ऑपरेटर: \(\mu = \frac{E^{1/2} + E^{-1/2}}{2}\)
– केंद्रीय अंतर ऑपरेटर: \(\delta = E^{1/2} – E^{-1/2}\)
वर्ग करने पर:
\[
\mu^2 = \frac{E + E^{-1} + 2}{4}
\]
\[
\delta^2 = E + E^{-1} – 2 \implies E + E^{-1} = \delta^2 + 2
\]
\(E + E^{-1}\) का मान \(\mu^2\) में रखने पर:
\[
\mu^2 = \frac{\delta^2 + 2 + 2}{4} = \frac{1}{4}(\delta^2 + 4)
\]
अतः विकल्प (1) सही है।