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 December 28th, 2018, 10:08 PM #1 Newbie   Joined: Aug 2018 From: România Posts: 22 Thanks: 2 A vectorial calculation Hello all, Knowing $\displaystyle \overrightarrow{z}=\overrightarrow{x}+ \overrightarrow{y}$ to calculate $\displaystyle \overrightarrow{z}^n$ where the three vectors are not collinear , $\displaystyle n\in \mathbb N$ and $\displaystyle n\geq 2$. All the best, Integrator Last edited by Integrator; December 28th, 2018 at 10:29 PM.
 December 28th, 2018, 10:45 PM #2 Senior Member     Joined: Sep 2015 From: USA Posts: 2,264 Thanks: 1198 what does $\large \vec{z}^n$ mean? Thanks from topsquark
 December 29th, 2018, 02:26 PM #3 Global Moderator   Joined: May 2007 Posts: 6,663 Thanks: 649 What are you given? Thanks from topsquark
 December 29th, 2018, 03:05 PM #4 Senior Member   Joined: Dec 2015 From: Earth Posts: 327 Thanks: 42 The title of the thread is ”vectorial calculation” Maybe n is an exponent
December 29th, 2018, 04:24 PM   #5
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 Originally Posted by idontknow The title of the thread is ”vectorial calculation” Maybe n is an exponent
Yeah, but the only way to do that is to use the cross product. It's going to get messy. I'm not sure that is what the OP is trying to do.

-Dan

December 29th, 2018, 07:11 PM   #6
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 Originally Posted by topsquark Yeah, but the only way to do that is to use the cross product. It's going to get messy. I'm not sure that is what the OP is trying to do. -Dan
not to mention that $\vec{v} \times \vec{v} = \vec{0}$

December 29th, 2018, 09:51 PM   #7
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 Originally Posted by romsek not to mention that $\vec{v} \times \vec{v} = \vec{0}$
Heloo,

This is a trivial case....The scalar product $\displaystyle \overrightarrow {v}^n=\overrightarrow{v} \cdot \overrightarrow{v} \cdots \overrightarrow{v}\cdot \overrightarrow{v}$ is what would be interesting to calculate.

All the best,

Integrator

December 29th, 2018, 11:00 PM   #8
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Quote:
 Originally Posted by Integrator Heloo, This is a trivial case....The scalar product $\displaystyle \overrightarrow {v}^n=\overrightarrow{v} \cdot \overrightarrow{v} \cdots \overrightarrow{v}\cdot \overrightarrow{v}$ is what would be interesting to calculate. All the best, Integrator
this post shows you have no idea what you're talking about.

$\vec{v}\cdot \vec{v}$ is a scalar and thus cannot be further dotted with a vector.

 December 30th, 2018, 03:58 AM #9 Senior Member   Joined: Oct 2009 Posts: 696 Thanks: 235 A relevant product would be the geometric algebra product, which for vectors equals $$\overrightarrow{v}\overrightarrow{w} = \overrightarrow{v}\cdot \overrightarrow{w} + \overrightarrow{v}\times \overrightarrow{w},$$ the sum of cross and dot product. But to compute further and take the $n$th power of a vector, you'd need the formula for the geometric algebra product for arbitrary blades, and that's pretty ugly.
December 30th, 2018, 10:47 PM   #10
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 Originally Posted by Micrm@ss A relevant product would be the geometric algebra product, which for vectors equals $$\overrightarrow{v}\overrightarrow{w} = \overrightarrow{v}\cdot \overrightarrow{w} + \overrightarrow{v}\times \overrightarrow{w},$$ the sum of cross and dot product. But to compute further and take the $n$th power of a vector, you'd need the formula for the geometric algebra product for arbitrary blades, and that's pretty ugly.
Hello,

The problem I posted is from another forum...It is known that if $\displaystyle \overrightarrow{z}=\overrightarrow{x}+ \overrightarrow{y}$ , then $\displaystyle \overrightarrow{z} ^2=\overrightarrow{x}^2+\overrightarrow{y}^2+2 \overrightarrow{x}\overrightarrow{y}$ and so what expressions do they have $\displaystyle \overrightarrow{z} ^3$ , $\displaystyle \overrightarrow{z}^4$ , $\displaystyle \overrightarrow{z}^5$ ,......., $\displaystyle \overrightarrow{z}^{n-1}$ , $\displaystyle \overrightarrow{z}^n$?
Thank you very much!

All the best!

Integrator

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