Calculus-lim-1-pi-0-2pi-k-1-n-sin-kx-2-k-2-dx-

Question Number 144053 by mnjuly1970 last updated on 21/Jun/21

\dots \dots . . C a l c u l u s \dots \dots . .

Ω := l i m \frac{1}{π} \int_{0}^{2 π} {(\underset{k = 1}{\sum^{n}} \frac{s i n (k x)}{\sqrt{2^{k}}})}^{2} d x = ?

Answered by mindispower last updated on 21/Jun/21

l e t n, m \in N

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

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

= π, n = m \dots \dots . (1)

Ω = \lim_{n \to \infty} . \frac{1}{π} \int_{0}^{2 π} {(\underset{k = 1}{\sum^{n}} \frac{s i n (k x)}{2^{\frac{k}{2}}})}^{2} d x

= \lim_{n \to \infty} . \frac{1}{π} \int_{0}^{2 π} \underset{k = 1}{\sum^{n}} . \underset{m = 1}{\sum^{n}} \frac{s i n (k x) s i n (m x)}{2^{\frac{k + m}{2}}} d x

= \lim_{n \to \infty} . \frac{1}{π} \underset{k = 1}{\sum^{n}} \underset{m = 1}{\sum^{n}} \frac{1}{2^{\frac{m + k}{2}}} \int_{0}^{2 π} s i n (k x) s i n (m x) d x

w i t h e (1)

Ω = \lim_{n \to \infty} \frac{1}{π_{}} \underset{k = 1}{\sum^{n}} \sum_{m = k} \frac{1}{2^{k}} . π

= \lim_{n \to \infty} . \frac{1}{π} \underset{k = 1}{\sum^{n}} \frac{π}{2^{k}} = \underset{k = 1}{\sum^{n}} \frac{1}{2^{k}} = 1

Ω = 1

Commented by mnjuly1970 last updated on 21/Jun/21

\dots \dots t h a n k s a l o t m r p o w e r \dots . .

Commented by mindispower last updated on 22/Jun/21

p l e a s u r