Last week we did the sinc function. Let’s do it again!
The function, to refresh our collective memory, is this:
Now I was thinking about jumping right into some contour integration, but on actually doing it again I see that it’s a little hardcore for one post so eventually when we do it we’ll have to stretch it out over probably three posts. Probably it’ll be a Friday-Saturday thing culminating on an official Sunday Function. But it ain’t gonna be this week. This week we’re going to do three ways of computing a limit. There’s more, a bunch more, but we’re going to just do three.
As we noted last week, we can’t just plug in x = 0 to our function. If we do we’re dividing by zero, which is not possible. For some reason this is a frequent point of dissent among, uh, certain varieties of amateur mathematicians. But if you’re working with the plain ol’ real numbers, satisfying the usual axioms, there’s no consistent way to define it. On the other hand there’s no reason we can’t just say f(x) equals the rule above, except at the point x = 0. Which we can define to 0 or 1 or a billion or pi or whatever the heck we want. We would like for our definition to make sense and be continuous, fitting in with the rest of the function. From the graph we expect we should probably define f(0) = 1, but we’d like to prove it. Is f(x) really getting closer and closer to 1 as x gets closer and closer to 0?
Method 1: Wildly unwarranted surmise.
This is where we plug in numbers close to 0 and see what happens. Shall we?
f(1) = 0.8414709848
f(0.1) = 0.9983341665
f(0.01) = 0.9999833334
f(0.001) = 0.9999998333
f(0.0001) = 0.9999999983
Well that’s suggestive if nothing else. Though we won’t do it, it’s true that you’ll get the same numbers if you do the calculations with the negative numbers -0.01, etc so the limit appears to be the same on both sides. Rigorous it’s not. So let’s move on.
Method 2: Series expansion.
Like all functions*, the sine function can be expressed as an infinite sum of polynomials. For now we won’t pause to derive this, but this expression is true:
We’re dividing this by x, and fortunately that’s just the same as multiplying by 1/x which is easy. This will give us this nice expression:
No, no, don’t let your eyes glaze over! Instead, just look at that rightmost sum after the equals. There’s the number 1, and then a bunch of terms in x. But we’re looking to see what happens as x goes to 0. And clearly that means all those x terms themselves go to zero, leaving behind that lonely number 1. Ladies and gentlemen, that is an honest to goodness proof.
Method 3: l’Hopital’s Mathematical Hospital.
Early in the first calculus class, you’ll learn that it’s possible to figure out something about these kinds of problems using derivatives. Roughly speaking given two functions g and h that are both 0 at x = 0 (as ours are), then:
In our case g(x) = sin(x) and h(x) = x. Doing the differentiation,
Exactly the same as with methods 1 and 2.
Whew! We’re done for the moment. We could come up with a few more methods if we wanted, but there comes a point when the lily has been guilded to death. With that, we’ll call it a day.
*This statement has not been evaluated by the AMS and is not intended to prove, demonstrate, or verify any theorem.