Symmetric function: Difference between revisions

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<ul>
<li>Consider the [[Real number|real]] function
<math display=block>f(x_1,x_2,x_3) = (x-x_1)(x-x_2)(x-x_3).</math>
By definition, a symmetric function with <math>n</math> variables has the property that
<math display=block>f(x_1,x_2,\ldots,x_n) = f(x_2,x_1,\ldots,x_n) = f(x_3,x_1,\ldots,x_n,x_{n-1}), \quad \text{ etc.}</math>
In general, the function remains the same for every [[permutation]] of its variables. This means that, in this case,
<math display=block>(x-x_1)(x-x_2)(x-x_3) = (x-x_2)(x-x_1)(x-x_3) = (x-x_3)(x-x_1)(x-x_2)</math>
and so on, for all permutations of <math>x_1, x_2, x_3.</math>
</li>
 
<li>Consider the function
<li>Consider the function<math display=block>f(x,y) = x^2 + y^2 - r^2</math> If <math>x</math> and <math>y</math> are interchanged the function becomes<math display=block>f(y,x) = y^2 + x^2 - r^2</math> which yields exactly the same results as the original <math>f(x, y).</math>
<math display=block>f(x,y) = x^2 + y^2 - r^2.</math>
If <math>x</math> and <math>y</math> are interchanged the function becomes
<math display=block>f(y,x) = y^2 + x^2 - r^2,</math>
which yields exactly the same results as the original <math>f(x, y).</math>
</li>
 
<li>Consider now the function
<li>Consider now the function<math display=block>f(x,y) = ax^2+by^2-r^2</math> If <math>x</math> and <math>y</math> are interchanged, the function becomes<math display=block>f(y,x) = ay^2 + bx^2 - r^2.</math> This function is obviously not the same as the original if <math>a \neq b,</math> which makes it non-symmetric.
<math display=block>f(x,y) = ax^2+by^2-r^2.</math>
If <math>x</math> and <math>y</math> are interchanged, the function becomes
<math display=block>f(y,x) = ay^2 + bx^2 - r^2.</math>
This function is not the same as the original if <math>a \neq b,</math> which makes it non-symmetric.
</li>
</ul>