# Mean value theorem & Cauchy sequence

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• Jan 21st 2010, 05:10 PM
kingwinner
Mean value theorem & Cauchy sequence
http://sites.google.com/site/asdfasdf23135/ra10.JPG

[note: also under discussion in s.o.s. math board]
• Jan 21st 2010, 05:51 PM
Drexel28
Quote:

Originally Posted by kingwinner
http://sites.google.com/site/asdfasdf23135/ra10.JPG

[note: also under discussion in s.o.s. math board]

Show that the proper hypotheses are made. Then we have that on the interval $\displaystyle [a_{n},a_{n+1}]$ that there exists some $\displaystyle c$ such that $\displaystyle \frac{\cos(a_n)-\cos(a_{n+1})}{a_n-a_{n+1}}=\frac{a_{n+1}-a_{n+2}}{a_n-a_{n+1}}=-\sin(c)$. Thus, we have that $\displaystyle \left|a_{n+1}-a_{n+2}\right|=|\sin(c)||a_n-a_{n+1}|\leqslant|a_n-a_{n+1}|$.
• Jan 21st 2010, 08:09 PM
kingwinner
Quote:

Originally Posted by Drexel28
Show that the proper hypotheses are made. Then we have that on the interval $\displaystyle [a_{n},a_{n+1}]$ that there exists some $\displaystyle c$ such that $\displaystyle \frac{\cos(a_n)-\cos(a_{n+1})}{a_n-a_{n+1}}=\frac{a_{n+1}-a_{n+2}}{a_n-a_{n+1}}=-\sin(c)$. Thus, we have that $\displaystyle \left|a_{n+1}-a_{n+2}\right|=|\sin(c)||a_n-a_{n+1}|\leqslant|a_n-a_{n+1}|$.

But when you write the interval [a(n),a(n+1)] and apply the mean value theorem, it is implictly assumed that a(n)<a(n+1), which is not true for our sequence, e.g. a0=0, a1=1, a2=0.54, a3=0.86,...

How can we remove the assumption a(n)<a(n+1) while reaching the same conclusion that |a(n+2)-a(n+1)| ≤ 0.85|a(n+1)-a(n)|?

thanks.
• Jan 21st 2010, 08:13 PM
Drexel28
Quote:

Originally Posted by kingwinner
But when you write the interval [a(n),a(n+1)] and apply the mean value theorem, it is implictly assumed that a(n)<a(n+1), which is not true for our sequence, e.g. a0=0, a1=1, a2=0.54, a3=0.86,...

How can we remove the assumption a(n)<a(n+1) while reaching the same conclusion that |a(n+2)-a(n+1)| ≤ 0.85|a(n+1)-a(n)|?

thanks.

I do leave holes on purpose you know? Do you not think I didn't realize this. You are supposed to do some of this. This is to help you learn.

But, I'll give you a hint. Does it matter which is bigger than the other? Clearly it does if we want to define an interval in the conventional way, but does it matter if we have the itneval I gave you or the reverse one? Is the result any different?
• Jan 22nd 2010, 03:13 PM
kingwinner
OK, I got it now.
Actually I got confused before about the "for ALL n" part in "a(n+1)>a(n) for ALL n" . I thought we need montone sequences, but acutally we don't.

Given any two consecutive terms a(n) and a(n+1), either a(n)<=a(n_1) or a(n)>a(n+1), and in either case the MVT gives the result. Done.
• Jan 24th 2010, 01:51 AM
kingwinner