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Old 2008-04-28, 09:21   #1
jinydu
 
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Default Signs of Remainders of Cosine and Sine Series

I'm stuck on a homework problem:

"For real y, show that every remainder in the series for cos y and sin y has the same sign as the leading term..."

The hint was to use induction; use the result for cos y at the Nth step to get the result for sin y at the Nth step to get the result for cos y at the (N+1)th step, etc.

I have reduced the first half of the induction step to showing that

\sum_{n=N}^{\infty}\frac{(-1)^n}{(2n)!}y^{2n+1} has the same sign as \sum_{n=N}^{\infty}\frac{(-1)^n}{(2n+1)!}y^{2n+1}.

But I'm not sure how to do this because the extra 1/(2n+1) factor in each term affects the terms unevenly.

Could I have some advice please?

Thanks

Last fiddled with by jinydu on 2008-04-28 at 09:21
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Old 2008-04-28, 13:50   #2
wblipp
 
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Isn't there something about absolutely convergent alternating series that would help?
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Old 2008-04-29, 01:22   #3
jinydu
 
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Never mind, I got it.

The Alternating Series Theorem doesn't work though because the terms do not necessarily decay monotonically. For large y, the first few terms can grow.
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