Let fp - - cos-p2 - i sin-p2 cos2-p2 - i sin2-p2 - cos-p - i sin-p then n-lim fn - - -

In the last term write βp as pβp^2 ∴ f_p (β) = e^iβ/p^2 + 2iβ/p^2 + ⋯ + piβ/p^2 = e^(iβ/p^2){1 + 2 + 3 + ⋯ + p ...

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Byju's Answer

Standard XII

Mathematics

Logarithmic Differentiation

Let fp β =...

Question

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

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Solution

In the last term write $\frac{β}{p} a s \frac{p β}{p^{2}}$
$∴ f_{p} (β) = e^{i β / p^{2} + 2 i β / p^{2} + \dots + p i β / p^{2}}$
$= e^{(i β / p^{2}) {1 + 2 + 3 + \dots + p}} = e^{(i β / p^{2}) ⎧ ⎪ ⎨ ⎪ ⎩ \frac{p (p + 1)}{2} ⎫ ⎪ ⎬ ⎪ ⎭}$
$= e^{\frac{i β}{2} ⎛ ⎝ 1 + \frac{1}{p} ⎞ ⎠}$
$∴ f_{n} (π) = e^{\frac{i π}{2} ⎛ ⎝ 1 + \frac{1}{n} ⎞ ⎠}$
${lim}_{n \to \infty} f_{n} (π) = e^{i π / 2} = cos \frac{π}{2} + i sin \frac{π}{2} = i$

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Let fp(β)=(cosβp2+isinβp2)(cos2βp2+isin2βp2)....(cosβp+isinβp)then limn→∞fn(π)=.....

In the last term write βpaspβp2∴fp(β)=eiβ/p2+2iβ/p2+⋯+piβ/p2=e(iβ/p2){1+2+3+⋯+p}=e(iβ/p2)⎧⎪⎨⎪⎩p(p+1)2⎫⎪⎬⎪⎭=eiβ2⎛⎝1+1p⎞⎠∴fn(π)=eiπ2⎛⎝1+1n⎞⎠limn→∞fn(π)=eiπ/2=cosπ2+isinπ2=i

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