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Kevin_B
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#1
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#1
\displaystyle \sum_{\text{sym}} \frac{a^2}{b} = \frac{a^2}{b} + \frac{b^2}{a} + \frac{a^2}{c} + \frac{c^2}{a} + \frac{b^2}{c} + \frac{c^2}{b}
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generalebriety
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#2
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What's "sym"? Can you give us a context for this?

This looks like something I might have written in a "roots of equations" context...
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Kevin_B
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#3
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Symmetric sum for a,b,c

The chapter in the book is symmetric and cyclic expressions
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SimonM
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#4
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Looks good. See my post from ages back that you already know the location of
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Kevin_B
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#5
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I got a question about that.

How come you had  \displaystyle \sum_{sym} f(a,b,c) = 2(a^2+b^2+c^2), where did the two come from? Wouldn't it be the same as the cyclic sum which is also symmetric
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SimonM
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#6
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\displaystyle \sum_{\mathbf{sym}} f(a,b,c) = f(a,b,c)+f(a,c,b)+f(b,a,c)+f(b,c  ,a)+f(c,a,b)+f(c,b,a) (Definition)

f(a,b,c)=a^2b^0c^0

Therefore f(a,b,c)=f(a,c,b)=a^2
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Kevin_B
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#7
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Oh so it means you will get b^2 and c^2 twice as well

So eg.  f(a,b,c) = a^2b^2c^0

\left \displaystyle \sum_{sym} a^2b^2 = a^2b^2 + b^2a^2 + a^2c^2 + c^2a^2 + b^2c^2 + c^2b^2 \\= 2(a^2b^2 + a^2c^2 + b^2c^2) right?
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SimonM
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#8
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Exactly. This distinction is quite important and often forgotten.
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Kevin_B
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#9
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Alright thanks
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Kevin_B
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#10
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So does that mean that \displaystyle \sum_{sym} f(a,b) = f(a,b) + f(a,b) + f(b,a) + f(b,a) or is it just  \displaystyle \sum_{sym} f(a,b) = f(a,b)+ f(b,a)
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SimonM
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#11
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The second. It is the sum across all permutations of the set \{ a_1, a_2, \ldots a_n \}

It could be expressed as

\displaystyle \sum_{\sigma} f(a_{\sigma (1)}, a_{\sigma (2)}, \ldots, a_{\sigma (n)} ) (if you are happy with that notation)
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Kevin_B
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#12
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OK so I got a question that says prove  \displaystyle \sum_{sym} a^3 + \sum_{sym} a = \sum_{sym} a^4 + \sum_{sym} a^3b. I proved it on the assumption that the variables a and b are the only ones involved. Is that correct? And is LHS the same as  \displaystyle (a^3+b^3)(a+b) ?
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SimonM
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#13
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Perhaps you mean

\displaystyle \left ( \sum_{\mathbf{sym}} a^3 \right ) \left ( \sum_{\mathbf{sym}} a \right ) = \sum_{\mathbf{sym}} a^4 + \sum_{\mathbf{sym}} a^3 b

?
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Kevin_B
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#14
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Yeah sorry LHS is a product not a sum
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SimonM
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#15
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Ok

\displaystyle \left ( \sum_{\mathbf{sym}} a^3 \right ) \left ( \sum_{\mathbf{sym}} a \right ) = (a^3+b^3)(a+b)

Unless they give more information on the number of variables?
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Kevin_B
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#16
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#16
Thanks again
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