EPTProduct[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.
Mathematical Induction
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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. | ||
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EmeraldSparks
United States1451 Posts
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bakesale
United States187 Posts
+ Show Spoiler + Use trig identities. | ||
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Dave[9]
United States2365 Posts
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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 | ||
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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. | ||
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15vs1
64 Posts
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