Differential approximation: If you’re here for the quick answer, here it is.
Step 1.) Identify your function
Step 2.) Write down this formula: f(x) \approx f(a) + f'(a)\Delta x, \quad \Delta x = x - a.
Step 3.) Find a, the center. This is the point nearby the troubling point, x, which happens to be easier to plug in.
Step 4.) Calculate f(a), f'(a), and \Delta x
Step 5.) Plug in.
And you’re done.
For those wanting some more detail, let’s dive in.
Differential approximation is, as the name suggests, the act of using differentials, which are related to derivatives, to estimate the value of a function at a point.
Suppose that we have a function f(x) = \sqrt{x} which we’d like to evaluate at some value, say, 28. Without a calculator, we run into a problem. 28 is not a perfect square, and so its square root is… messy.
Let’s draw a graph. Here, we have x along the horizontal axis and the square root of x along the vertical. Where are all of the perfect squares? That is, where are all the values of x which make f(x) easy enough to do in your head? They’re gonna be here, at these dots. Notice that the dot at (25, 5) seems to be pretty close! So let’s just take 5 as our answer and… hold up a minute. We can do better.
We can see that the true value of f(28) is a little bit higher than our estimate of 5. How can we get a handle on this “error term”? This is where differentials come in. Remember from your lecture that a differential gives the estimated change in y given two parameters, x and a change in x, which we call dx, or sometimes \Delta x. This is exactly what we want, because we want to see about how much f changes going from x = 25 to x = 28. In general, the differential of a function f(x) is df = f'(x)dx, so in our square root example here, \Delta f \approx f'(x) \Delta x. We have to use the squiggly equals sign here, because this is just an approximation. Now our starting x is 25, and our \Delta x is 28 - 25 = 3, so our differential becomes \Delta f \approx f'(25)\cdot3, and this is easily computable. Now we add this to our starting value of f(25), and we get an estimate, via differentials, of f(28).
f(28)\approx f(25) + f'(25)\Delta x,\quad\text{where }\Delta x = 28-25.
Generalizing this a bit, away from the specific values of 28 and 25, we have that f(x)\approx f(a) + f'(a)\Delta x,\quad\text{where } \Delta x = x - a. a is called the “center of approximation”, or sometimes just the “center”, as it is the basis from which we are attempting to approximate other values. I recommend thinking of a as the “easy” value, the one you can quickly plug in, and x as the “hard” value, the one you’re trying to approximate at.
Let’s rearrange the approximation, and assume that it is exact: f(x) - f(a) = f'(a)(x-a). Notice, now, what we have here. This is the point-slope form of the line passing through the point (a,f(a)) with slope f'(a). That is, this is the tangent line to f(x) at a.
With this intuition, it should be clear that taking a close to x will, in general, give a better approximation than picking a far from x. So it’s important to pick a value close to the one we’re trying to approximate when picking our “easy” value.
Now, of course, getting technical, you can find examples of functions where approximating, say, 10, is better done by 20 than by 5, but generally, in the common cases you’ll see in class, just taking the nearest possible “easy” value should get you close enough.
To review, when it comes to differential approximation problems, the can be broken down into simple steps.
First, identify your function. If you’re not given something off the bat, and are instead asked to do something like “approximate \sqrt[3]{122}”, then you find the function by looking to see what number you can nudge to make the expression easy to calculate by hand or in your head, and replace that number with x. In this case, that gives us f(x) = \sqrt[3]{x}.
Next, find the center, a, the “easy” value. In this case, that’s 125, because \sqrt[3]{125} = 5. You might also think to try something like 64, but that’s farther away from 122 than 125, so 125 it is.
Now, calculate f(a) and f'(a). Once you have these, just plug everything into the approximation formula to get f(122) \approx 5 + \frac{1}{75}(122-125) = \frac{124}{25} = 4.96.