Calculating Nth Root of Vector: Non-Integer Solutions

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Calculating the nth root of a vector, particularly for non-integer values, raises questions about the definitions of multiplication and exponentiation being used. The discussion highlights that traditional vector operations, such as the dot product and cross product, do not allow for a square root of a vector in the conventional sense. Matrix multiplication rules further complicate the situation, as an n by 1 matrix cannot be multiplied by itself to yield a valid operation. The conversation emphasizes the need for a clear definition of what is meant by the "square root of a vector" before proceeding with calculations. Ultimately, the concept of a vector's square root is not generally defined in standard mathematical terms.
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Could someone please explain how to calculate the nth root of a vector, where n is not an integer?
 
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Hey sue132 and welcome to the forums.

What do you mean? Let's say you calculate y = v^n. What is the definition of multiplication you are using (or the definition for exponentiation if you are using that)?
 
Thanks for the reply, Chiro. What I meant was : if I have a column vector, say v, of length n, then how do I go about calculating v^(1/2) or v(-1/2)?

I should be able to get a vector, say x=v^1/2 such that I get v back when I do x*x. Is this right? If v is a column vector of dimension nX1, then x should be of dimension 1*n. Is this right?
 
You need to answer the question chiro asked, first. What kind of "multiplication" are you using? You have to have a square before you can talk about a square root! There are two different ways to multiply a vector by a vector in three dimensions: the dot product of a vector with itself gives a number (the square of the length of the vector) not a vector so there is no square root in that sense. The cross product of a vector with itself is always the 0 vector so there is no square root of a vector in that sense.

You talk about writing the vector as a column matrix so perhaps you are talking about matrix multiplication. But you multiply an "m by n" matrix by an "n by p" matrix- the number of columns of the first matrix has to be the same as the same as the number of rows of the second matrix and that is not possible for an "n by 1" matrix multiplied by itself. Some texts define the square of a column matrix as the product of the matrix with its transpose- but that gives a number (strictly speaking a "one by one matrix"), so that is equivalent to the dot product, and there is, again, no "square root" of a vector.

That is why we do NOT in general define the "square root of a vector". If you have been given a problem that asks you to find the square root of a vector, you are going to have to tell us what definition of "square root of a vector" you are using.
 
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I tried to combine those 2 formulas but it didn't work. I tried using another case where there are 2 red balls and 2 blue balls only so when combining the formula I got ##\frac{(4-1)!}{2!2!}=\frac{3}{2}## which does not make sense. Is there any formula to calculate cyclic permutation of identical objects or I have to do it by listing all the possibilities? Thanks
Since ##px^9+q## is the factor, then ##x^9=\frac{-q}{p}## will be one of the roots. Let ##f(x)=27x^{18}+bx^9+70##, then: $$27\left(\frac{-q}{p}\right)^2+b\left(\frac{-q}{p}\right)+70=0$$ $$b=27 \frac{q}{p}+70 \frac{p}{q}$$ $$b=\frac{27q^2+70p^2}{pq}$$ From this expression, it looks like there is no greatest value of ##b## because increasing the value of ##p## and ##q## will also increase the value of ##b##. How to find the greatest value of ##b##? Thanks
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