Product[Cos[(2^k) x] from k=0 to n
=
(Sin[(2^(n + 1)) x])/((2^(n + 1)) (Sin[x]))
Think anyone could help me out? I've been working on it for a while and can't seem to get things to meld together.
Blogs > Dave[9] |
Dave[9]
United States2365 Posts
Product[Cos[(2^k) x] from k=0 to n = (Sin[(2^(n + 1)) x])/((2^(n + 1)) (Sin[x])) Think anyone could help me out? I've been working on it for a while and can't seem to get things to meld together. | ||
EmeraldSparks
United States1451 Posts
| ||
bakesale
United States187 Posts
+ Show Spoiler + Use trig identities. | ||
Dave[9]
United States2365 Posts
| ||
EmeraldSparks
United States1451 Posts
On March 15 2010 16:47 Dave[9] wrote: i find using a double angle formula for cos . . . leads to a big mess. then maybe you shouldn't use the cosine double angle formula | ||
bakesale
United States187 Posts
\sin(2^{n+1} x) = 2 \sin(2^n x) \cos(2^n x) Here's the step for n=1 (presumably already shown for n=0): Since true for n=0, the LHS is the old result times \cos(2^1 x). So LHS can be written \frac{\sin(2^1 x)}{2^1 \sin(x)} \cos(2^1 x) and we want this to be \frac{\sin(2^2 x)}{2^2 \sin(x)}. Do cancellation, double-angle identity, and it should follow directly. Use my first tip and generalize this n=1 case to the general inductive step. | ||
15vs1
64 Posts
| ||
| ||
Next event in 1h 49m
[ Submit Event ] |
StarCraft 2 StarCraft: Brood War League of Legends Counter-Strike Other Games Organizations StarCraft 2 StarCraft: Brood War StarCraft 2 StarCraft: Brood War
StarCraft 2 • Berry_CruncH180 StarCraft: Brood War• OhrlRock 26 • aXEnki • intothetv • Gussbus • Kozan • IndyKCrew • LaughNgamez Trovo • Laughngamez YouTube • Migwel • Poblha League of Legends |
ESL Open Cup
ESL Open Cup
ESL Open Cup
Kung Fu Cup
GSL Code S
Maru vs TY
Creator vs SHIN
ESL Pro Tour
ESL Pro Tour
ESL Pro Tour
ESL Pro Tour
Online Event
[ Show More ] ESL Pro Tour
Hatchery Cup
BSL
ESL Pro Tour
Sparkling Tuna Cup
ESL Pro Tour
BSL
ESL Pro Tour
|
|