Let fp - - cos-p2-isin-p2 - cos2-p2-isin2-p2 - - cos-p-1 -p2-isin-p-1 -p2 - cos-p-isin-p then limn- 1-fn - -

f _ p (β)=cosβ #160; p ^ 2 #160; +i sinβ #160; p ^ 2 #160; +...+( cosβ #160; p #160; +i sinβ #160; p #160; #160;) = e ^ iβ #16 ...

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Byju's Answer

Standard XII

Mathematics

Integration by Parts

Let fp β = ...

Question

Let $f_{p} (β) = (cos \frac{β}{p^{2}} + i sin \frac{β}{p^{2}}) + (cos \frac{2 β}{p^{2}} + i sin \frac{2 β}{p^{2}}) \dots \dots (cos \frac{β (p - 1)}{p^{2}} + i sin \frac{β (p - 1)}{p^{2}}) + (cos \frac{β}{p} + i sin \frac{β}{p})$
then $lim n \to \infty 1 / f_{n} (π) =$ .

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Solution

$f_{p} (β) = cos \frac{β}{p^{2}} + i sin \frac{β}{p^{2}} + . . . + (cos \frac{β}{p} + i sin \frac{β}{p})$
$= e^{i \frac{β}{p^{2}}} + e^{i \frac{2 β}{p^{2}}} + e^{i \frac{3 β}{p^{2}}} + . . . + e^{i \frac{p β}{p^{2}}}$
$= e^{i \frac{β}{p^{2}}} ⎛ ⎜ ⎜ ⎜ ⎝ 1 + e^{i \frac{β}{p^{2}}} + e^{i \frac{2 β}{p^{2}}} + . . . + e^{i \frac{(p - 1) β}{p^{2}}} ⎞ ⎟ ⎟ ⎟ ⎠$
$= e^{i \frac{β}{p^{2}}} ⎛ ⎜ ⎜ ⎜ ⎜ ⎜ ⎝ \frac{e^{\frac{p β}{p^{2}}} - 1}{e^{\frac{β}{p^{2}}} - 1} ⎞ ⎟ ⎟ ⎟ ⎟ ⎟ ⎠$
$= e^{i \frac{β}{p^{2}}} ⎛ ⎜ ⎜ ⎜ ⎜ ⎜ ⎝ \frac{e^{\frac{β}{p}} - 1}{e^{\frac{β}{p^{2}}} - 1} ⎞ ⎟ ⎟ ⎟ ⎟ ⎟ ⎠$
$lim n \to \infty f_{n} (π) = lim n \to \infty e^{i (\frac{π}{n^{2}})} . \frac{e^{\frac{π}{n}} - 1}{e^{\frac{π}{n^{2}}} - 1}$ $= lim n \to \infty e^{i \times 0} . \frac{\frac{- π}{n^{2}} e^{\frac{π}{n}}}{\frac{- 2 π}{n^{3}} e^{\frac{π}{n^{2}}}}$ $= lim n \to \infty n = \infty$

$lim n \to \infty \frac{1}{f_{n} (π)} = 0$

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Let fp(β)=(cosβp2+isinβp2)+(cos2βp2+isin2βp2)⋯⋯(cosβ(p−1)p2+isinβ(p−1)p2)+(cosβp+isinβp) then limn→∞1/fn(π)=.

Let $f_{p} (β) = (cos \frac{β}{p^{2}} + i sin \frac{β}{p^{2}}) + (cos \frac{2 β}{p^{2}} + i sin \frac{2 β}{p^{2}}) \dots \dots (cos \frac{β (p - 1)}{p^{2}} + i sin \frac{β (p - 1)}{p^{2}}) + (cos \frac{β}{p} + i sin \frac{β}{p})$
then $lim n \to \infty 1 / f_{n} (π) =$ .