This is a self-made problem. I have a function $A$ of variables $x, m \geq 0$. A satisfies the following symmetry:

$$A(x, m) = 1 - A(m, x)$$

Is there an equation that relates the partial derivative of $A$ wrt the first argument to the partial derivative of $A$ with respect to the second argument? It seems like there should be something straightforward but I'm not sure how to deduce it.

Thank you!

  • $\begingroup$ Thanks @youknowme for the edit! $\endgroup$ – John_Krampf Nov 13 '20 at 5:20
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    $\begingroup$ You should take a look at Mathjax. $\endgroup$ – Wolgwang Nov 13 '20 at 5:21

All you need to do is differentiate both sides of the symmetry equation-

$$(\partial_1 A)(x,m) = \frac d{dx}[A(x,m)] = \frac d{dx}[1-A(m,x)] = - (\partial_2 A)(m,x)$$

or did you want something more?

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    $\begingroup$ I think his problem comes from the confusion notations arising with this type of question where we exchange the place of the variable names while still working on one defined by its place rather than its name. Althought correct I am not sure this will help him in this regard, we will see, I might be wrong. $\endgroup$ – nicomezi Nov 13 '20 at 5:53
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    $\begingroup$ @nicomezi Without more context I wouldn't know either. I was careful to bracket pedantically to prevent the confusion I had when learning this, namely that $\partial_1$ can be shorthand for $\partial/\partial x_1$ which could be confusing in e.g. $\partial_1f(a,b,x_1)$. $\endgroup$ – Calvin Khor Nov 13 '20 at 6:00
  • $\begingroup$ I see, yes I think I was hoping for something more that won't happen. I was hoping I could get a relationship between $\partial_1 A(x,m)$ and $\partial_2 A(x,m)$ without the arguments switched $\endgroup$ – John_Krampf Nov 13 '20 at 6:08
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    $\begingroup$ @John_Krampf such is life :) I'll just explcitly note; you can try to use the symmetry condition again, this time with $d/dm$, but you will end up with a 1=1 type equation that says nothing. Also, sometimes, its true: when $x=m$ :) $\endgroup$ – Calvin Khor Nov 13 '20 at 6:19

I suggest using the definition of partial derivative to have a clearer point of view on the situation. The partial derivative of $A$ with respect to the first variable at the point $(x_0,m_0)$ is defined by :

$$\partial_1 A(x_0,m_0)=\lim_{h\to0} \frac{A(x_0+h,m_0)-A(x_0,m_0)}{h}.$$

Using the symmetry relation, we have :

$$\lim_{h\to0} \frac{(1-A(m_0,x_0+h))-(1-A(m_0,x_0))}{h}=\lim_{h\to0} \frac{-A(m_0,x_0+h)+A(m_0,x_0)}{h}.$$

The last term is equal, also by the definition partial derivative, to $- \partial_2 A(m_0,x_0)$.

Thus :

$$\partial_1 A(x_0,m_0)=- \partial_2 A(m_0,x_0)$$

  • $\begingroup$ Thank you! I had been hoping I could get a relationship between ∂1A(x,m) and ∂2A(x,m) without the arguments switched but now see that can't come out of this $\endgroup$ – John_Krampf Nov 13 '20 at 6:09

Another way to look at this is to note that the function $B(x,m):=A(x,m)-\tfrac12$ is antisymmetric; $$B(x,m)=A(x,m)-\tfrac12=(1-A(m,x))-\tfrac12=\tfrac12-A(m,x)=-B(m,x).$$ What does that tell you about the partial derivatives?

  • $\begingroup$ This seems useful, but can I get anything from this about the partial derivative of B with respect to the first variable and the partial derivative of B with respect to the second variable? $\endgroup$ – John_Krampf Nov 15 '20 at 17:06

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