........ Calculus........ Ω:=lim(1/π)∫_0 ^( 2π) (Σ_(k=1) ^n ((sin(kx))/( (√2^k ))))^2 dx=?


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

$. . . . . . . . C a l c u l u s . . . . . . . .$ $Ω := l i m \frac{1}{π} \int_{0}^{2 π} {(\underset{k = 1}{\sum^{n}} \frac{s i n (k x)}{\sqrt{2^{k}}})}^{2} d x = ?$
	Answered by mindispower last updated on 21/Jun/21

	$l e t n, m \in N$ $\int_{0}^{2 π} s i n (n x) s i n (m x) d x = \frac{1}{2} \int_{0}^{2 π} (c o s (m - n) x - c o s (n + m) x) d x$ $= \frac{1}{2} \int_{0}^{2 π} c o s (m - n) x d x = 0, n \neq m$ $= π, n = m . . . . . . . (1)$ $Ω = \lim_{n \to \infty} . \frac{1}{π} \int_{0}^{2 π} {(\underset{k = 1}{\sum^{n}} \frac{s i n (k x)}{2^{\frac{k}{2}}})}^{2} d x$ $= \lim_{n \to \infty} . \frac{1}{π} \int_{0}^{2 π} \underset{k = 1}{\sum^{n}} . \underset{m = 1}{\sum^{n}} \frac{s i n (k x) s i n (m x)}{2^{\frac{k + m}{2}}} d x$ $= \lim_{n \to \infty} . \frac{1}{π} \underset{k = 1}{\sum^{n}} \underset{m = 1}{\sum^{n}} \frac{1}{2^{\frac{m + k}{2}}} \int_{0}^{2 π} s i n (k x) s i n (m x) d x$ $w i t h e (1)$ $Ω = \lim_{n \to \infty} \frac{1}{π_{}} \underset{k = 1}{\sum^{n}} \sum_{m = k} \frac{1}{2^{k}} . π$ $= \lim_{n \to \infty} . \frac{1}{π} \underset{k = 1}{\sum^{n}} \frac{π}{2^{k}} = \underset{k = 1}{\sum^{n}} \frac{1}{2^{k}} = 1$ $Ω = 1$
	Commented by mnjuly1970 last updated on 21/Jun/21

	$. . . . . . t h a n k s a l o t m r p o w e r . . . . .$
	Commented by mindispower last updated on 22/Jun/21

	$p l e a s u r$
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