My Math Forum [Complex Numbers] Inquiry about relationship

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April 21st, 2012, 04:10 PM   #1
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Hi,

I was not sure where to post this but apparently it had some kind of extended link to Euler's Theorem (although I do not know anything about it...) Feel free to move the post

I'm trying to do this proof by induction about complex numbers and I'm stuck..
I do not want to tell you exactly what it is because I wish to finish it myself. There is just this bit that I have never learnt and that I would like to understand..

Could someone tell me how I can substitute one into another or if any relationship exists mathematically here (see attachment)

XmathY
Attached Images
 relationship between.png (18.2 KB, 381 views)

 April 21st, 2012, 05:15 PM #2 Newbie   Joined: Apr 2012 From: (-?,+?) Posts: 4 Thanks: 0 Re: [Complex Numbers] Inquiry about relationship *Woops sorry, there is not ment to be an "i" at the bottom of the two PIs. "i" is "r"min.
 April 22nd, 2012, 12:46 AM #3 Senior Member   Joined: Apr 2010 Posts: 215 Thanks: 0 Re: [Complex Numbers] Inquiry about relationship Let $f(i)=2sin(\frac{\pi k}{i})$. $A=\prod_{i=2}^{r} f(i)$ $B=\prod_{i=2}^{r+1} f(i+1) = \prod_{i=3}^{r+2} f(i) = A\frac{1}{2sin(\pi k/2)} 2sin(\pi k/(r+1)) 2sin(\pi k/(r+2))=2A \frac{sin(\pi k / (r+1)) sin(\pi k / (r+2))}{sin(\pi k/2)}$ Is that what you're looking for?
April 22nd, 2012, 04:26 AM   #4
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Re: [Complex Numbers] Inquiry about relationship

Quote:
 Originally Posted by brangelito Let $f(i)=2sin(\frac{\pi k}{i})$. $A=\prod_{i=2}^{r} f(i)$ $B=\prod_{i=2}^{r+1} f(i+1) = \prod_{i=3}^{r+2} f(i) = A\frac{1}{2sin(\pi k/2)} 2sin(\pi k/(r+1)) 2sin(\pi k/(r+2))=2A \frac{sin(\pi k / (r+1)) sin(\pi k / (r+2))}{sin(\pi k/2)}$ Is that what you're looking for?
Thanks a lot but how about for this:
Attached Images
 Screen Shot 2012-04-22 at 13.25.46.png (26.9 KB, 355 views)

 April 22nd, 2012, 04:35 AM #5 Newbie   Joined: Apr 2012 From: (-?,+?) Posts: 4 Thanks: 0 Re: [Complex Numbers] Inquiry about relationship * No sorry, the beginning is [color=#FF0000]wrong[/color], ignore that, it would be : $A= \prod_{2}^{r} 2sin(\frac{\pi(r-1)}{r}$ and $B= \prod_{2}^{r+1} 2sin(\frac{\pi(r)}{r+1}$ Are you saying that $B= 2A*\frac{sin(\frac{\pi(r-1)}{r+1})*sin(\frac{\pi(r-1)}{r+2})}{sin(\frac{\pi(r-1)}{2})}$ Thanks in advance :P XmathY
 April 22nd, 2012, 09:42 AM #6 Senior Member   Joined: Apr 2010 Posts: 215 Thanks: 0 Re: [Complex Numbers] Inquiry about relationship I'm not sure what you're looking for. Could you please explain what you mean by your notation, because all I see is some number multiplied by itself (r-2) times.

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