My Math Forum (http://mymathforum.com/math-forums.php)
-   Linear Algebra (http://mymathforum.com/linear-algebra/)
-   -   A vectorial calculation (http://mymathforum.com/linear-algebra/345542-vectorial-calculation.html)

 Integrator December 28th, 2018 09:08 PM

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

 romsek December 28th, 2018 09:45 PM

what does $\large \vec{z}^n$ mean?

 mathman December 29th, 2018 01:26 PM

What are you given?

 idontknow December 29th, 2018 02:05 PM

The title of the thread is ”vectorial calculation”
Maybe n is an exponent

 topsquark December 29th, 2018 03:24 PM

Quote:
 Originally Posted by idontknow (Post 603791) 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

 romsek December 29th, 2018 06:11 PM

Quote:
 Originally Posted by topsquark (Post 603792) 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 $\vec{v} \times \vec{v} = \vec{0}$

 Integrator December 29th, 2018 08:51 PM

Quote:
 Originally Posted by romsek (Post 603797) 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

 romsek December 29th, 2018 10:00 PM

Quote:
 Originally Posted by Integrator (Post 603808) 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.

Please stop wasting everyone's time.

 Micrm@ss December 30th, 2018 02:58 AM

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.

 Integrator December 30th, 2018 09:47 PM

Quote:
 Originally Posted by Micrm@ss (Post 603820) 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

All times are GMT -8. The time now is 12:42 AM.