January 2nd, 2019, 01:01 PM  #21  
Global Moderator Joined: May 2007 Posts: 6,710 Thanks: 675  Quote:
 
January 6th, 2019, 09:27 PM  #22  
Member Joined: Aug 2018 From: România Posts: 31 Thanks: 2  Quote:
For example , it is correct that some say that $\displaystyle \overrightarrow{x}^n=x^{2m}$ for $\displaystyle n=2m$ and $\displaystyle \overrightarrow{x}^n=x^{2m}\cdot \overrightarrow{x}$ for $\displaystyle n=2m+1$ , where $\displaystyle m\in \mathbb N$?Thank you very much! All the best, Integrator Last edited by Integrator; January 6th, 2019 at 09:30 PM.  
January 6th, 2019, 11:59 PM  #23  
Senior Member Joined: Oct 2009 Posts: 753 Thanks: 261  Quote:
 
January 11th, 2019, 06:52 AM  #24 
Banned Camp Joined: Mar 2015 From: New Jersey Posts: 1,720 Thanks: 125 
Let $\displaystyle \vec{z}=(x_{1},x_{2},...,x_{m})$ Definition: $\displaystyle \vec{z}^{n}=(x_{1}^{n}+x_{2}^{n}+,...,+x_{m}^{n})$ 

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calculation, vectorial 
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