Math in small doses
The product of a function and its own change
Summary. Using finite differences alone, without differential calculus, we show that \(f(t)\,\Delta f(t) \approx \tfrac{1}{2}[f^2(t_2) - f^2(t_1)]\) and that the sum of these products is about \(\tfrac{1}{2}f^2(t)\): a result we meet again in the work–energy theorem, and a forerunner of the integral \(\int f\,df\).
1 - The product f(t) Δf(t)
In the sciences we often have to explain concepts that rely on how quantities change when they are expressed as functions of another variable.When we don't want to, or can't, use differential calculus, relations and equations between finite differences (quantities preceded by the symbol \(\small \Delta\), which denotes their change) are a good alternative. They can even make things easier to understand, because these quantities, and the algebra that involves them, feel more acceptable or, so to speak, more concrete.
Finite differences must, however, be handled with some care if we want results that are correct (though approximate) and consistent.
Here, given a function \(\small f(t)\) of the variable \(\small t\), we derive a useful relation for the product of \(\small f(t)\) and its own change around \(\small t\), \(\small \Delta f(t)\), that is:
(1) \( \qquad \qquad \qquad f(t)\,\Delta\,f(t) \)
Products of this kind come up often in physics. Suppose, for example, that we want the elementary work done by a constant force \(\small F\) that moves a mass \(\small m\) through a distance \(\small \Delta\,x\):
(2) \( \qquad \qquad \qquad \Delta\,W = F\,\Delta\,x \)
A few quick steps give:
(3) \( \qquad \qquad \qquad F = m\,a \qquad a = \dfrac{\Delta\,v(t)}{\Delta\,t} \qquad F = m\,\dfrac{\Delta\,v(t)}{\Delta\,t} \)
By the first of equations (3), a constant force implies a constant acceleration (assuming the mass is constant too), which is why we have not written the acceleration as a function of time. The velocity, on the other hand, must depend on time, otherwise the acceleration would be zero. Although we have not written it explicitly, for simplicity, the displacement \(\small \Delta\,x\) may depend on time as well.
Now we find the work:
(4) \( \qquad \qquad \qquad \Delta\,W = F\,\Delta\,x = m\,\dfrac{\Delta\,v(t)}{\Delta\,t}\,\Delta\,x \)
but \(\small \Delta\,x = v(t)\,\Delta\,t\), so:
(5) \( \qquad \qquad \qquad \Delta\,W = F\,\Delta\,x = m\,\dfrac{\Delta\,v(t)}{\Delta\,t}\,v(t)\,\Delta\,t \)
and, canceling \(\small \Delta\,t\):
(6) \( \qquad \qquad \qquad \Delta\,W = F\,\Delta\,x = m\,v(t)\,\Delta\,v(t) \)
It is therefore worth asking whether we can get from (6) an expression in which the change \(\small \Delta\,v(t)\) no longer appears.
Setting \(\small v(t) = f(t)\) and \(\small \Delta\,v(t) = \Delta\,f(t)\), we are clearly in the same situation described at the beginning.
To fix ideas, suppose that \(\small f(t)\) describes how some physical quantity evolves over time, and that we plot part of it on a graph (see the board on the left).
Take two instants \(\small t_1\) and \(\small t_2 = t_1 + \Delta\,t\), with the corresponding values \(\small f(t_1)\) and \(\small f(t_2)\) of the function.
The change of the function over the interval \(\small \Delta\,t\) is:
(7) \( \qquad \qquad \qquad \Delta\,f(t) = f(t_2) - f(t_1) \)
but within the interval the function may vary in a complicated, nonlinear way.
What we assume here is that, within the interval \(\small \Delta\,t\), the function can be approximated by a straight segment \(\small \overline{A B}\) with slope \(\small \dfrac{\Delta\,f(t)}{\Delta\,t}\).
This is clearly an approximation. We can see it by taking the midpoint of the interval, which we assume is \(\small t\), and comparing \(\small f(t)\) with the value we would have if \(\small f(t)\) were linear over \(\small \Delta\,t\), that is, if it were represented over that interval by the segment \(\small \overline{A B}\) itself. In that case the function would equal the average of \(\small f(t_1)\) and \(\small f(t_2)\), which we have called \(\small f_m\):
(8) \( \qquad \qquad \qquad f_m = \dfrac{f(t_1) + f(t_2)}{2} \)
so there is an error:
(9) \( \qquad \qquad \qquad \varepsilon = f(t) - f_m \)
This error, however, can be made small by reducing \(\small \Delta t\). The segment then approximates the function better and better, and the error \(\small \varepsilon\) tends to zero (try moving points A and B closer together on the curve and watch \(\small f(t)\) and \(\small f_m\) come together).
So, approximately, we write:
(10) \( \qquad \qquad \qquad f(t)\,\Delta\,f(t) \approx f_m\,\Delta\,f(t)= \dfrac{f(t_1) + f(t_2)}{2}\, [f(t_2) - f(t_1)] \)
and, multiplying out (sum times difference):
(11) \( \qquad \qquad \qquad f(t)\,\Delta\,f(t) \approx \dfrac{1}{2}\, [f^2(t_2) - f^2(t_1)] \)
Now we can go back to the opening example:
(6) \( \qquad \qquad \qquad \Delta\,W = F\,\Delta\,x = m\,v(t)\,\Delta\,v(t) \)
Applying (11) we get:
(12) \( \qquad \qquad \qquad \Delta\,W = m\,v(t)\,\Delta\,v(t) \approx \dfrac{1}{2}\,m\,v^2(t_2) - \dfrac{1}{2}\,m\,v^2(t_1) \)
which is the work–energy theorem.
2 - The sum of the products: Σ f(t) Δf(t)
Once we have a result like (11), we often need to add up \(\small n\) terms of the form:(13) \( \qquad \qquad \qquad \sum_{t_0}^{t_N} {f(t)\,\Delta\,f(t)} \)
evaluated from an initial value \(\small t_0\) of \(\small t\) to a final value \(\small t_N\), which we can simply call \(\small t\), since \(\small t_N\) can be any instant in the interval considered.
Carrying out the sum:
(14) \(\quad \sum_{t_0}^{t_N} {f(t)\,\Delta\,f(t)} \approx \dfrac{1}{2}\,[f^2(t_1)-f^2(t_0)+f^2(t_2)-f^2(t_1)+ f^2(t_3)-f^2(t_2) + ...] \)
we see at once that all the intermediate terms cancel, since each appears twice with opposite signs. So we get (setting \(\small f(t_N)=f(t)\)):
(15) \( \qquad \qquad \qquad \sum_{t_0}^{t_N=t} {f(t)\,\Delta\,f(t)} \approx \dfrac{1}{2}\,[f^2(t)-f^2(t_0)] \)
\(\small f(t_0)\) is the initial value of the function, which must be known. Often \(\small f(t_0)= 0\), which gives the remarkable result:
(16) \( \qquad \qquad \qquad \sum {f(t)\,\Delta\,f(t)} \approx \dfrac{1}{2}\,f^2(t) \)
This result turns up very often in the mathematical description of physical phenomena, and it is the conceptual forerunner of the integral:
\[ \int f(x)\,d\,f(x) = \int f(x)\,f'(x)\,d\,x = \dfrac{1}{2}\,f^2(x) + C \]
And that is what we wanted to highlight with this short note.
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