मान लीजिए f(z) = exp\(\rm\left(z+\frac{1}{z}\right)\), z ∈ ℂ\{0} है। तब f का z = 0 पर अवशेष _______ है।

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CSIR UGC (NET) Mathematical Science: Held On (7 June 2023)
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  1. \(\sum_{l=0}^{\infty} \frac{1}{(l+1) !}\)
  2. \(\sum_{l=0}^{\infty} \frac{1}{l !(l+1)}\)
  3. \(\sum_{l=0}^{\infty} \frac{1}{l !(l+1) !}\)
  4. \(\sum_{l=0}^{\infty} \frac{1}{\left(l^2+l\right) !}\)

Answer (Detailed Solution Below)

Option 3 : \(\sum_{l=0}^{\infty} \frac{1}{l !(l+1) !}\)
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अवधारणा:

f(z) का z = 0 पर अवशेष, f(z) के मैक्लॉरिन श्रेणी प्रसार में \(\frac1z\) के गुणांक के बराबर होता है।

व्याख्या:

f(z) = exp\(\rm\left(z+\frac{1}{z}\right)\)

= \(e^z.e^{\frac1z}\)

= \((1+z+\frac{z^2}{2!}+\frac{z^3}{3!}+...)\)\(.(1+\frac1zz+\frac{1}{z^22!}+\frac{1}{z^33!}+...)\)

इसलिए उपरोक्त व्यंजक में \(\frac1z\) का गुणांक

= \(\frac11+\frac1{2!.1!}+\frac1{3!.2!}+\frac1{4!.3!}+...\)

= \(\sum_{l=0}^{\infty} \frac{1}{l !(l+1) !}\)

इसलिए विकल्प (3) सही है।

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