Question-115645

Question Number 115645 by ZiYangLee last updated on 27/Sep/20

Commented by Dwaipayan Shikari last updated on 27/Sep/20

\frac{(n + 1) p + (n - 1) q}{(n - 1) p + (n + 1) q} = C

\frac{2 np + 2 nq}{(n + 1) p - (n - 1) p + (n - 1) q - (n + 1) q} = \frac{C + 1}{C - 1}

2 n . \frac{p + q}{p - q} = \frac{C + 1}{C - 1}

2 n \frac{\frac{p}{q} + 1}{\frac{p}{q} - 1} = \frac{{(\frac{p}{q})}^{k} + 1}{{(\frac{p}{q})}^{k} - 1}

2 n \frac{a + 1}{a - 1} = \frac{a^{k} + 1}{a^{k} - 1}

2 n (a^{k + 1} + a^{k} - a - 1) = (a^{k + 1} + a - a^{k} - 1)

2 n = \frac{a^{k + 1} + a - a^{k} - 1}{a^{k + 1} + a^{k} - a - 1}

n > 1

\frac{a^{k + 1} + a - a^{k} - 1}{a^{k + 1} + a^{k} - a - 1} > 2

- a^{k + 1} - {2 a}^{k} + a + 1 > 0

a^{k} (a + 2) - 1 (a + 1) < 0

a^{k} < \frac{a + 1}{a + 2}

klog (a) < \log (\frac{a + 1}{a + 2})

k < \log (\frac{a + 1}{a + 2}) . \frac{1}{\log (a)}

k < \log (\frac{p + q}{p + 2 q}) . \frac{1}{\log (\frac{p}{q})}

Commented by PRITHWISH SEN 2 last updated on 01/Oct/20

(n + 1) p - (n - 1) p + (n - 1) q - (n + 1) q

= np + p - np + p + nq - q - np - q

= 2 (p - q)

∴ \frac{2 nq + 2 np}{(n + 1) p - (n - 1) p + (n - 1) q - (n + 1) q}

= \frac{2 n (p + q)}{2 (p - q)} = \frac{n (p + q)}{(p - q)}

please check

Answered by PRITHWISH SEN 2 last updated on 27/Sep/20

\frac{n (p + q) + (p - q)}{n (p + q) - (p - q)} = {(\frac{p}{q})}^{k}

∵ p \to q \Rightarrow (p - q) \to 0

\frac{n (p + q)}{n (p + q)} = {(\frac{p}{q})}^{k}

k = 0 please check

Commented by ZiYangLee last updated on 27/Sep/20

B u t p i s j u s t a p p r o x i m a t e l y e q u a l

t o q, i t h i n k w e c a n n o t a s s u m e t h a t

p i s e q u a l t o q \dots \dots h m m m

Commented by PRITHWISH SEN 2 last updated on 27/Sep/20

that is why I have written

p - q \to 0 not as p - q = 0

Commented by PRITHWISH SEN 2 last updated on 27/Sep/20

now if p - q = ϵ ϵ = a very small positive quantity

then n (p + q) + ϵ \to n (p + q) - ϵ

∴ \frac{n (p + q) + ϵ}{n (p + q) - ϵ} \to 1 I think