In my thermodynamics class, I came across an equation:
$$\Delta S_2 - \Delta S_1 = C_v\ln\frac{T_2}{T_1}$$
And many equations involving changes of two changes. Now my question is do mathematicians have any neat notations to represent such changes?
In calculus, I do have learnt that the second derivative $\frac{d^2y}{dx^2}$ does represent the rate of change of rate of change (which is similar to the change of change?) for some continuous functions. But I can't stuff that notation here properly. I did try it out when went back to its derivation of the formula, and it goes something like:
$$\frac{dS}{dT} = \frac{C_v}{T}$$ $$\frac{d^2S}{dT^2} = C_v\ln T$$
Which makes sense for a continuous function. My next guess was to convert the second derivative to a discrete form was:
$$\Delta^2S = C_v\Delta\ln T$$
But I'm not convinced because I have just exterminated the $dT^2$ without any reason.
You can certainly use $ \Delta ^ 2 $ to indicate a change of a change. Other people may not know what it means, but it should be easy to explain.
As for the derivation, you don't just eliminate the $ \mathrm d T ^ 2 $, you also change $ \ln T $ to $ \Delta \ln T $, so something more is going on.
Changing a differential to a discrete difference is a feature of linear approximation, but since we're willing to use logarithms, we don't need any approximation here, as long as $ C _ v $ is constant. We have $ \mathrm d S / \mathrm d T = C _ v / T $, so $ \mathrm d S = C _ v \, \mathrm d ( \ln T ) $, and integrating that gives us $ \Delta S = C _ v \, \Delta ( \ln T ) $ exactly. If the temperature $ T $ changes from $ T _ i $ to $ T _ f $, then $ \Delta ( \ln T ) = \ln T _ f - \ln T _ i = \ln ( T _ f / T _ i ) $, so we have $$ \Delta S = C _ v \ln \biggl ( \frac { T _ f } { T _ i } \biggr ) \text . $$ (This is the equation in the second comment by @QuantumSuperfield only with different subscripts because I don't want to clash with the meaning of your subscripts.) Then if this happens on two occasions, we have both $ \Delta S _ 1 = C _ v \ln ( T _ { 1 f } / T _ { 1 i } ) $ and $ \Delta S _ 2 = C _ v \ln ( T _ { 2 f } / T _ { 2 i } ) $; subtracting these gives $$ \Delta S _ 2 - \Delta S _ 1 = C _ v \ln \biggl ( \frac { T _ { 2 f } T _ { 1 i } } { T _ { 2 i } T _ { 1 f } } \biggr ) \text . $$ At no point did I make any approximation besides assuming that $ C _ v $ is truly constant throughout all of this (and assuming that both processes are reversible at constant volume so that $ \mathrm d S / \mathrm d T = C _ v / T $ applies in the first place).
So I'm struggling to make this exact equation match your equation. If $ T _ 1 $ and $ T _ 2 $ are the temperatures at which the two processes occur, and this is a sensible thing to say, then we're saying that $ T _ { 1 f } \approx T _ { 1 i } $ and $ T _ { 2 f } \approx T _ { 2 i } $, but this doesn't give anything like your equation. Instead, it allows us to write $ \Delta ( \ln T ) \approx \Delta T / T $, giving $$ \Delta S _ 2 - \Delta S _ 1 \approx C _ v \biggl ( \frac { \Delta T _ 2 } { T _ 2 } - \frac { \Delta T _ 1 } { T _ 1 } \biggr ) = C _ v \frac { T _ 1 \Delta T _ 2 - T _ 2 \Delta T _ 1 } { T _ 1 T _ 2 } \text . $$ That's not what you want at all.
Another simplification would be if the two processes occur right after each other, so that $ T _ { 2 i } = T _ { 1 f } $. But that still leaves us with three temperatures: an initial one, an intermediate one, and a final one. Perhaps we get back to the original temperature after the second process, so that $ T _ { 1 i } = T _ { 2 f } $ as well. Then if I write $ T _ 1 $ for $ T _ { 1 f } $ and $ T _ 2 $ for $ T _ { 2 f } $ (which is backwards from what I'd normally write but fits your equation better), I get $$ \Delta S _ 2 - \Delta S _ 1 = C _ v \ln \biggl ( \frac { { T _ 2 } ^ { \! 2 } } { { T _ 1 } ^ { \! 2 } } \biggr ) = 2 C _ v \ln \biggl ( \frac { T _ 2 } { T _ 1 } \biggr ) \text . $$ That's tantalizingly close, but different! If this is a situation that even could apply to your equation, then (unless $ T _ 1 = T _ 2 $ as well) it can't be right.
tl;dr: The answer to your question about notation is that $ \Delta ^ 2 $ is fine. But I have no idea how to fix the derivation. Possibly it was supposed to be $ \Delta S = C _ v \ln ( T _ 2 / T _ 1 ) $ all along.