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January 2nd, 2019, 01:01 PM   #21
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Quote:
Originally Posted by Integrator View Post
Hello,

Why can not be defined for $\displaystyle n>2$?
However, some claim that according to Newton's binomial $\displaystyle \overrightarrow{z}^n=\overrightarrow{x}^n+C_n^1 \cdot \overrightarrow{x}^{n-1}\cdot \overrightarrow{y}+\cdots +C_n^{n-1}\cdot \overrightarrow{x}\cdot \overrightarrow{y}^{n-1}+\overrightarrow{y}^n$.

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Integrator
Meanwhile you haven't defined the higher powers of $\overrightarrow{x}$ or $\overrightarrow{y}$
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January 6th, 2019, 09:27 PM   #22
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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.
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January 6th, 2019, 11:59 PM   #23
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Quote:
Originally Posted by Integrator View Post
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.
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January 11th, 2019, 06:52 AM   #24
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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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