Let f p a - e i a - p 2 - e 2 i a - p 2 - e 3 i a - p 2 - e i a - p Where i - - - 1 and p - N then lim n - f n -

The correct option is C i We have, [ f_ p ( a ) = e^ ia / p^ 2 + 2ia / p^ 2 + 3ia / p^ 2 +........+ pia / p^ 2 #160;; ...

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Byju's Answer

Standard XII

Mathematics

Theorems for Continuity

Let f p a...

Question

Let $f_{p} (a) = e^{\frac{i a}{p^{2}}} \cdot e^{\frac{2 i a}{p^{2}}} \cdot e^{\frac{3 i a}{p^{2}}} \dots \cdot e^{\frac{i a}{p}}$ (Where $i = \sqrt{- 1} and p \in N) then {lim}_{n \to \infty} f_{n} (π)$

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Solution

The correct option is C $i$
We have,
$\begin{matrix} f_{p} (a) = e^{\frac{i a}{p^{2}} + \frac{2 i a}{p^{2}} + \frac{3 i a}{p^{2}} + . . . . . . . . + \frac{p i a}{p^{2}}} f_{p} (a) = e^{\frac{i a}{p^{2}} [1 + 2 + 3 + . . . . . p]} f_{p} (a) = e^{\frac{i a}{p^{2}} \frac{p (p + 1)}{2}} f_{p} (π) = e^{\frac{i π (x + 1)}{2 n}} {lim}_{n \to \infty} f n (π) = e {lim}_{n \to \infty} \frac{i π (n + 1)}{2 n} = e^{i \frac{π}{2}} = cos (\frac{π}{2}) + i sin (\frac{π}{2}) = i H e n c e, o p t i o n B i s c o r r e c t a n s w e r . \end{matrix}$

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Similar questions

Let $f_{p} (α) = e^{\frac{i α}{p^{2}}} . e^{\frac{2 i α}{p^{2}}} . e^{\frac{3 i α}{p^{2}}} . e^{\frac{4 i α}{p^{2}}} \dots e^{\frac{i α}{p}}$ , (where $i = \sqrt{- 1}$ and $p \in N$ ) then ${lim}_{n \to \infty} f_{n} (π)$ is

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) =

Q. Let p be a nonsingular matrix, and

I + p + p^{2} + . . . . . + p^{n} = 0

, then find

p^{- 1}

Q. Let

P

be a non-singular matrix and

I + P + P^{2} + \dots + P^{n} = O

. Then

P^{- 1}

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n \in Z^{+}

)

Q. If

a + i b = \frac{p + i}{2 q - i}

, prove that

a^{2} + b^{2} = \frac{(p^{2} + 1)}{4 q^{2} + 1}

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Let fp(a)=eiap2⋅e2iap2⋅e3iap2…⋅eiap (Where i=√−1 and p∈N) then limn→∞fn(π)

The correct option is C iWe have,fp(a)=eiap2+2iap2+3iap2+........+piap2fp(a)=eiap2[1+2+3+.....p]fp(a)=eiap2p(p+1)2fp(π)=eiπ(x+1)2nlimn→∞fn(π)=elimn→∞iπ(n+1)2n=eiπ2=cos(π2)+isin(π2)=iHence,optionBiscorrectanswer.

Let $f_{p} (a) = e^{\frac{i a}{p^{2}}} \cdot e^{\frac{2 i a}{p^{2}}} \cdot e^{\frac{3 i a}{p^{2}}} \dots \cdot e^{\frac{i a}{p}}$ (Where $i = \sqrt{- 1} and p \in N) then {lim}_{n \to \infty} f_{n} (π)$