Question

let :-
$\sf f_{p}(\beta) = \bigg(\cos\dfrac{\beta}{p^{2} } +i\sin\dfrac{\beta}{p^{2} } \bigg)+\bigg(\cos\dfrac{2\beta}{p^{2} } +i\sin\dfrac{2\beta}{p^{2} } \bigg)\: .\:.\:. \: \bigg(\cos\dfrac{\beta(p-1)}{p^{2} }+i \sin\dfrac{\beta(p-1)}{p^{2} } \bigg)+\bigg(\cos\dfrac{\beta}{p} +i \sin\dfrac{\beta}{p} \bigg)$

Then Evaluate the following:-
$\bullet\quad\displaystyle\sf \lim_{x \to \infty} \dfrac{1}{f_n(\pi)}$

Answer 1

Answer:

0

Step-by-step explanation:

Given Question :

Let : -

$\mathrm{f_p(\beta) = \bigg(\cos\dfrac{\beta}{p^2}+ i\sin \dfrac{\beta}{p^2}\bigg) + \bigg(\cos\dfrac{2\beta}{p^2}+ i\sin \dfrac{2\beta}{p^2}\bigg) +... + }\\ \mathrm{ \bigg(\cos\dfrac{(p-1)\beta}{p^2}+ i\sin \dfrac{(p-1)\beta}{p^2}\bigg) + \bigg(\cos\dfrac{\beta}{p}+ i\sin \dfrac{\beta}{p}\bigg)}$

To find :

(Corrected part - It should be 'n' in place of 'x')

$\mathrm{\lim \limits_{n \rightarrow \infty} \dfrac{1}{f_n(\pi)}}$

Solution :

Given function :

$\longmapsto \mathrm{f_p(\beta) = \bigg(\cos\dfrac{\beta}{p^2}+ i\sin \dfrac{\beta}{p^2}\bigg) + \bigg(\cos\dfrac{2\beta}{p^2}+ i\sin \dfrac{2\beta}{p^2}\bigg) +... + }\\ \mathrm{ \bigg(\cos\dfrac{(p-1)\beta}{p^2}+ i\sin \dfrac{(p-1)\beta}{p^2}\bigg) + \bigg(\cos\dfrac{\beta}{p}+ i\sin \dfrac{\beta}{p}\bigg)}$

We know that, from Euler's form,

$\boxed{\mathrm{cosx + isinx = e^{ix}}}$

Here, we have

$\mathrm{\dfrac{\beta}{p^2}, \dfrac{2\beta}{p^2}, ...,\dfrac{(p-1)\beta}{p^2}, \dfrac{\beta}{p}} \text{ in place of x}$

So, this can be written as

$\implies \large{\text{ \mathrm{f_p(\beta) = e^{\frac{i\beta}{p^2}}+ e^{\frac{2i\beta}{p^2}} + ... + e^{\frac{(p-1)i\beta}{p^2}}+ e^{\frac{i\beta}{p}}} }}$

this is of the form of sum of definite GP with common ratio $\large{\text{ \mathrm{e^{\frac{i\beta}{p^2}}} }}$

We know that, Sum of definite G.P. is,

$\boxed{\mathrm{Sum_{GP} = \dfrac{a(r^n-1)}{r-1}}}$

• a = first term
• r = common ratio
• n = number of terms in the sum

Here,

$\mathrm{a = \large{\text{ \mathrm{e^{\frac{i\beta}{p^2}}} }} ; r = \large{\text{ \mathrm{e^{\frac{i\beta}{p^2}}} }}; n = p}$

So, applying the sum formula we get

$\implies \mathrm{f_p(\beta)} = \large{\text{ \mathrm{e^{\frac{i\beta}{p^2}}} }} \mathrm{\times \dfrac{\bigg(\Big(\large{\text{ \mathrm{e^{\frac{i\beta}{p^2}}} }}\Big)^p -1\bigg)}{\bigg(\large{\text{ \mathrm{e^{\frac{i\beta}{p^2}}} }} - 1\bigg)}}$

$\implies \mathrm{f_p(\beta)} = \large{\text{ \mathrm{e^{\frac{i\beta}{p^2}}} }} \mathrm{\times \dfrac{\bigg(\large{\text{ \mathrm{e^{\frac{i\beta}{p}}} }} -1\bigg)}{\bigg(\large{\text{ \mathrm{e^{\frac{i\beta}{p^2}}} }} - 1\bigg)}}$

Now, Replacing n in place of p and π in place of β

$\implies \mathrm{f_n(\pi)} = \large{\text{ \mathrm{e^{\frac{i\pi}{n^2}}} }} \mathrm{\times \dfrac{\bigg(\large{\text{ \mathrm{e^{\frac{i\pi}{n}}} }} -1\bigg)}{\bigg(\large{\text{ \mathrm{e^{\frac{i\pi}{n^2}}} }} - 1\bigg)}}$

Asked equation is :

$\longmapsto \mathrm{\lim \limits_{n \rightarrow \infty} \dfrac{1}{f_n(\pi)}} = \mathrm{\lim \limits_{n \rightarrow \infty} \dfrac{1}{\Bigg( \large{\text{ \mathrm{e^{\frac{i\pi}{n^2}}} }} \mathrm{\times \dfrac{\bigg(\large{\text{ \mathrm{e^{\frac{i\pi}{n}}} }} -1\bigg)}{\bigg(\large{\text{ \mathrm{e^{\frac{i\pi}{n^2}}} }} - 1\bigg)}}\Bigg)}}$

$\implies \mathrm{\lim \limits_{n \rightarrow \infty} \dfrac{\bigg(\large{\text{ \mathrm{e^{\frac{i\pi}{n^2}}} }} -1\bigg)}{\large{\text{ \mathrm{e^{\frac{i\pi}{n^2}}} }} \times \bigg(\large{\text{ \mathrm{e^{\frac{i\pi}{n}}} }} - 1\bigg)}}$

Separating the denominators into limits, we get,

$\implies \mathrm{\lim \limits_{n \rightarrow \infty} \dfrac{1}{\large{\text{ \mathrm{e^{\frac{i\pi}{n^2}}} }}} \times \lim \limits_{n \rightarrow \infty} \dfrac{\bigg(\large{\text{ \mathrm{e^{\frac{i\pi}{n^2}}} }} -1\bigg)}{\bigg(\large{\text{ \mathrm{e^{\frac{i\pi}{n}}} }} - 1\bigg)}}$

Executing the limit in first term, we get,

$\implies \mathrm{\dfrac{1}{\mathrm{e^0}} \times \lim \limits_{n \rightarrow \infty} \dfrac{\bigg(\large{\text{ \mathrm{e^{\frac{i\pi}{n^2}}} }} -1\bigg)}{\bigg(\large{\text{ \mathrm{e^{\frac{i\pi}{n}}} }} - 1\bigg)}}$

We know that e⁰ = 1, so,

$\implies \mathrm{\lim \limits_{n \rightarrow \infty} \dfrac{\bigg(\large{\text{ \mathrm{e^{\frac{i\pi}{n^2}}} }} -1\bigg)}{\bigg(\large{\text{ \mathrm{e^{\frac{i\pi}{n}}} }} - 1\bigg)}}$

We know that,

$\boxed{\mathrm{\lim \limits_{x \rightarrow 0} \dfrac{e^x - 1}{x} = 1}}$

So, multiplying the numerator and denominator with iπ/n and iπ/n²

$\implies \mathrm{\lim \limits_{n \rightarrow \infty} \dfrac{\bigg(\large{\text{ \mathrm{e^{\frac{i\pi}{n^2}}} }} -1\bigg)}{\dfrac{i\pi}{n^2}}\times \dfrac{\dfrac{i\pi}{n}}{\bigg(\large{\text{ \mathrm{e^{\frac{i\pi}{n}}} }} - 1\bigg)}\times \dfrac{\bigg(\dfrac{i\pi}{n^2}\bigg)}{\bigg(\dfrac{i\pi}{n}\bigg)}}$

$\implies \mathrm{\lim \limits_{n \rightarrow \infty} 1 \times 1\times \dfrac{\bigg(\dfrac{i\pi}{n^{\not{2}}}\bigg)}{\bigg(\dfrac{i\pi}{\not{n}}\bigg)} }$

$\implies \mathrm{\lim \limits_{n \rightarrow \infty} \dfrac{1}{n}}$

$\implies \mathrm{\dfrac{1}{\infty}}$

$\implies \underline{\underline{\Large{\text{0}}}}$

$\therefore \underline{\boxed{\mathbf{\lim \limits_{n \rightarrow \infty} \dfrac{1}{f_n(\pi)}} = \large{\textbf{0}}}}$

Hope it helps!!

Answer 2

Given Expression :-

$\\ \quad \bullet \quad \sf \bigg(\dfrac{x}{y} \bigg)^{a-b} \times \bigg( \dfrac{y}{z} \bigg)^{b-c} \times \bigg( \dfrac{z}{x} \bigg)^{c-a}$

We know that :-

$\\ \quad \mapsto \quad \boxed{ \frak{ \bigg(\dfrac{a}{b} \bigg) ^{c - d} = \dfrac{ {a}^{c} }{ {b}^{d} } } } \bigstar$

Calculation :-

$\\ \quad \longrightarrow \quad \sf \bigg(\dfrac{x}{y} \bigg)^{a-b} \times \bigg( \dfrac{y}{z} \bigg)^{b-c} \times \bigg( \dfrac{z}{x} \bigg)^{c-a}$

$\\ \quad \longrightarrow \quad \sf \dfrac{ {x}^{a} }{ {y}^{b} } \times \dfrac{ {y}^{b} }{ {z}^{c} } \times \dfrac{ {z}^{c} }{ {x}^{a} }$

$\\ \quad \therefore \quad \boxed{ \frak{1}} \bigstar$

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