Signs of Remainders of Cosine and Sine Series

jinydu · 2008-04-28, 09:21

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

wblipp · 2008-04-28, 13:50

Isn't there something about absolutely convergent alternating series that would help?

jinydu · 2008-04-29, 01:22

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.

2008-04-28, 09:21	#1
jinydu Dec 2003 Hopefully Near M48 2×3×293 Posts	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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2008-04-28, 13:50	#2
wblipp "William" May 2003 New Haven 100101000011₂ Posts	Isn't there something about absolutely convergent alternating series that would help?

2008-04-29, 01:22	#3
jinydu Dec 2003 Hopefully Near M48 2·3·293 Posts	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.