Need to rewrite linear combination as vector expression.

devonho
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Homework Statement


Given vectors
<br /> {\bf r}=\left[r_1,r_2,r_3\ldots{}r_n\right]^T<br />
<br /> {\bf e}=\left[e_1,e_2,e_3\ldots{}e_n\right]^T<br />

I need to write the sum

<br /> \sum_{i=1}^{n}r_ie_i^2<br />

in terms of {\bf r} and {\bf e}

Homework Equations


Nil.

The Attempt at a Solution



Without r_i, I am able to write \sum_{i=1}^{n}e_i^2 as

<br /> \sum_{i=1}^{n}e_i^2={\bf e}^T{\bf e}<br />

If r_i can be independant of i I should be able to move it out of the summation. I am looking for some expansion/reexpression of r_i.
 
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hi devonho! :smile:
devonho said:
I need to write the sum

<br /> \sum_{i=1}^{n}r_ie_i^2<br />

in terms of {\bf r} and {\bf e}

why?? :confused:

(i don't think you can)
 
This seems to work:

<br /> {\bf r}=<br /> \left[<br /> \begin{array}{ccccc}<br /> r_1 &amp; 0 &amp; \ldots &amp; &amp; 0\\<br /> 0 &amp; r_2 &amp; &amp; &amp; \vdots \\<br /> \vdots &amp; &amp; r_3 &amp; &amp; \\<br /> &amp; &amp; &amp; \ddots &amp; \\<br /> 0 &amp; \ldots &amp; &amp; &amp; r_n \\<br /> \end{array}<br /> \right]<br />

<br /> {\bf e}=<br /> \left[<br /> \begin{array}{c}<br /> e_1 \\<br /> e_2 \\<br /> \vdots\\<br /> e_n \\<br /> \end{array}<br /> \right]<br />

<br /> {\bf e}^T{\bf re}=\left[ e_1, e_2 \ldots e_n\right]<br /> \left[<br /> \begin{array}{ccccc}<br /> r_1 &amp; 0 &amp; \ldots &amp; &amp; 0\\<br /> 0 &amp; r_2 &amp; &amp; &amp; \vdots \\<br /> \vdots &amp; &amp; r_3 &amp; &amp; \\<br /> &amp; &amp; &amp; \ddots &amp; \\<br /> 0 &amp; \ldots &amp; &amp; &amp; r_n \\<br /> \end{array}<br /> \right]<br /> \left[<br /> \begin{array}{c}<br /> e_1 \\<br /> e_2 \\<br /> \vdots\\<br /> e_n \\<br /> \end{array}<br /> \right]<br /> =r_1e_1^2+r_2e_2^2\ldots +r_ne_n^2<br /> =\sum_{i=1}^{n}r_ie_i^2<br />
 
but devonho, how do you form that middle matrix out of the vector r ? :confused:
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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