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January 2nd, 2019, 01:01 PM   #21
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Quote:
 Originally Posted by Integrator Hello, Why can not be defined for $\displaystyle n>2$? However, some claim that according to Newton's binomial $\displaystyle \overrightarrow{z}^n=\overrightarrow{x}^n+C_n^1 \cdot \overrightarrow{x}^{n-1}\cdot \overrightarrow{y}+\cdots +C_n^{n-1}\cdot \overrightarrow{x}\cdot \overrightarrow{y}^{n-1}+\overrightarrow{y}^n$. All the best, Integrator
Meanwhile you haven't defined the higher powers of $\overrightarrow{x}$ or $\overrightarrow{y}$

January 6th, 2019, 09:27 PM   #22
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Quote:
 Originally Posted by mathman Meanwhile you haven't defined the higher powers of $\overrightarrow{x}$ or $\overrightarrow{y}$
Hello,

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
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Quote:
 Originally Posted by Integrator Hello, 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
Don't you read the posts? It is not correct and we have explained multiple times why.

 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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