Question-210441

Question Number 210441 by universe last updated on 09/Aug/24

Answered by aleks041103 last updated on 09/Aug/24

(i)

f o r o d d n, t h e f u n c t i o n {s i n}^{n} (n x) i s s i m i l a r

t o s i n (x) w i t h t h e f a c t t h a t t h e p r i m i t i v e f u n c t i o n o s c i l a t e s

b e t w e e n 0 a n d t h e m a x i m a l p o s s i b l e v a l u e

\int_{0}^{π / n} {s i n}^{n} (n x) d x = \frac{1}{n} \int_{0}^{π} {s i n}^{n} (x) d x ⩽

⩽ \frac{1}{n} \int_{0}^{π} s i n (x) d x = \frac{2}{n}

s_{n} = F_{n} (1) - F_{n} (0)

f o r o d d n w e h a v e F_{n} (x) \in [0, 2 / n]

t h e n s_{n} ⩽ \frac{2}{n}, o d d n .

Answered by aleks041103 last updated on 09/Aug/24

(i i) f o r e v e n n :

\int_{0}^{1} {s i n}^{n} (n x) d x = \frac{1}{n} \int_{0}^{n} {s i n}^{n} (x) d x =

= \frac{1}{n} [\underset{k = 0}{\sum^{⌊ n / π ⌋ - 1}} \int_{π k}^{π (k + 1)} {s i n}^{n} (x) d x + \int_{π ⌊ n / π ⌋}^{n} {s i n}^{n} (x) d x] =

= \frac{1}{n} (⌊ \frac{n}{π} ⌋ + c) \int_{0}^{π} {s i n}^{n} (x) d x

w h e r e 0 < c < 1 .

n o w l e t

I_{n} = \int_{0}^{π} {s i n}^{2 n} (x) d x =

= \int_{0}^{π} {s i n}^{2 n - 2} (x) d x - \int_{0}^{π} c o s (x) d (\frac{{s i n}^{2 n - 1} (x)}{2 n - 1}) d x =

= I_{n - 1} - {[\frac{{s i n}^{2 n - 1} (x) c o s (x)}{2 n - 1}]}_{0}^{π} - \frac{1}{2 n - 1} \int_{0}^{π} {s i n}^{2 n} (x) d x =

= I_{n - 1} - \frac{1}{2 n - 1} I_{n}

\Rightarrow I_{n - 1} = \frac{2 n}{2 n - 1} I_{n}

\Rightarrow I_{n} = \frac{2 n - 1}{2 n} I_{n - 1}

\Rightarrow I_{n} = \frac{(2 n - 1) (2 n - 3) \dots 3.1}{(2 n) (2 n - 2) \dots 4.2} I_{0}

I_{0} = \int_{0}^{π} d x = π

\Rightarrow I_{n} = \frac{(2 n - 1)!!}{(2 n)!!} π

(2 n)!! = 2^{n} n!

(2 n - 1)!! = \frac{(2 n)!}{(2 n)!!} = \frac{(2 n)!}{2^{n} n!}

\Rightarrow I_{n} = \frac{(2 n)!}{4^{n} {(n!)}^{2}} π

\Rightarrow s_{2 n} = \frac{1}{2 n} (⌊ \frac{2 n}{π} ⌋ + c_{n}) I_{n} = \frac{π}{2 n} (⌊ \frac{2 n}{π} ⌋ + c_{n}) \frac{(2 n)!}{4^{n} {(n!)}^{2}}

S t i r l i n g :

x! \sim \sqrt{2 π x} {(\frac{x}{e})}^{x}

\Rightarrow (2 n)! \sim \sqrt{4 π n} {(\frac{2 n}{e})}^{2 n} = \sqrt{4 π n} 4^{n} {(\frac{n}{e})}^{2 n}

\Rightarrow {(n!)}^{2} \sim 2 π n {(\frac{n}{e})}^{2 n}

\Rightarrow s_{2 n} = \frac{π}{2 n} (⌊ \frac{2 n}{π} ⌋ + c_{n}) \frac{(2 n)!}{4^{n} {(n!)}^{2}} \sim \frac{π}{2 n} (⌊ \frac{2 n}{π} ⌋ + c_{n}) \frac{\sqrt{4 π n}}{2 π n}

\Rightarrow s_{2 n} \sim \frac{π}{2 n} (⌊ \frac{2 n}{π} ⌋ + c_{n}) \frac{1}{\sqrt{π n}}

i t i s e a s y t o s e e t h a t \frac{π}{2 n} ⌊ \frac{2 n}{π} ⌋ \sim 1

\Rightarrow s_{2 n} \sim \frac{1}{\sqrt{π n}}

\Rightarrow s_{2 n} \to 0

f r o m (i) w e h a v e 0 < s_{2 n + 1} < \frac{2}{2 n + 1}

\Rightarrow s_{2 n + 1} \to 0

\Rightarrow s_{n} \to 0

\Rightarrow t h e o n l y l i m i t i n g p o i n t i s 0 .

Commented by hardmath last updated on 11/Aug/24

Dear professor, your solutions are perfect