What is the meaning of delta squared?

Question

I know that $\Delta x = x_2 - x_1$. This is well known Therefore, it follows that $\frac{\Delta a}{\Delta b}$ = $\frac{a_2 - a_1}{b_2 - b_1}$ (Unless I missed an elementary math class!)

Now, I found an equation in chemistry, while reading, that turns this principle on its head. It proposes that when considering the second in a set of values: $\frac{\Delta^2 a}{\Delta b^2} = (a_3 - a_2) \times \frac{1}{\Delta b}$

My issue with this is that $a_3$ hasn't been measured yet, we have only $a_1$ and $a_2$ so how can this be?

Secondly, they propose this: $\frac{\Delta \left(\frac{\Delta a}{\Delta b}\right)}{\Delta b} = \frac{\Delta^2 a}{\Delta b^2}$

Can anyone please help with the logic behind the first equation and algebra behind the second one?

Answer 1

$\frac{\Delta \left(\frac{\Delta a}{\Delta b}\right)}{\Delta b}$ can only be calculated when there has been two stages of change.

First set of change is from (a$_{1}$, b$_{1}$) to (a$_{2}$, b$_{2}$) and second set of change occurs from (a$_{2}$, b$_{2}$) to (a$_{3}$, b$_{3}$).

Hence your first $\frac{\Delta a}{\Delta b}$$_{1}$ = $\frac{a_2 - a_1}{b_2 - b_1}$

and your second $\frac{\Delta a}{\Delta b}$$_{2}$ = $\frac{a_3 - a_2}{b_3 - b_2}$

So, your $\frac{\Delta \left(\frac{\Delta a}{\Delta b}\right)}{\Delta b}$ = $\frac{\frac{\Delta a}{\Delta b}_{2} - \frac{\Delta a}{\Delta b}_{1}}{b_3 - b_1}$

Now I believe that algebraic calculation and manipulation should lead you to the fact that RHS and LHS are same..

Answer 2

If you should ever come across this issue, know that this is really a matter of differentiation and not simply change. Admittedly, there is some overlap between the two but the $\Delta b^2$ in the denominator will throw you off if you do not know that it is simply a notation for "wrt b", and not a divisor

Answer 3

The expression $\frac{\Delta^2a}{\Delta b^2}$ is likely meant to be a finite difference approximation to the second derivative. The numerator is the second forward difference: $$\Delta^2 a_i=\Delta a_{i+1}-\Delta a_i=(a_{i+2}-a_{i+1})-(a_{i+1}-a_i)=a_{i+2}-2a_{i+1}+a_i.$$

Finite differences can be used to approximate derivatives. This is more usually done with a uniform step $h_x$ between sample points. Thus, a first-order approximation to a function’s second derivative using differences is $\frac{a_{i+2}-2a_{i+1}+a_i}{h_x^2}+O(h_x)$. Based on what you’ve written, the text appears to use $(b_{i+2}-b_{i+1})(b_{i+1}-b_i)=\Delta b_{i+1}\Delta b_i$ for the denominator instead in order to account for non-uniform sample spacing. The notation then appears to be simplified at the expense of making it more confusing by dropping the indices and combining the two differences in the denominator into $\Delta b^2$. Of course, if the $b$’s are uniformly spaced, then everything is peachy. Indeed, the equation $\frac{\Delta^2a}{\Delta b^2}=\frac{\Delta\left(\frac{\Delta a}{\Delta b}\right)}{\Delta b}$ only makes sense if they are.

The definition $\frac{\Delta^2 a}{\Delta b^2} = (a_3 - a_2) \times \frac{1}{\Delta b}$, on the other hand, doesn’t make sense to me. It seems like there’s something missing. The above expression for $\frac{\Delta^2a}{\Delta b^2}$ can be rewritten as $\frac{\Delta a_{i+1}}{\Delta b_{i+1}}\cdot\frac1{\Delta b_i}-\frac{\Delta a_i}{\Delta b_i}\cdot\frac1{\Delta b_{i+1}}$, which is suggestive, but doesn’t quite explain this definition.