Question

Question 1A-8 in Problem Set 1:

Suppose f(x) is odd and periodic. Show that the graph of f(x) crosses the x-axis infinitely often.

Solution: f(x) odd =⇒ f(0) = −f(0) =⇒ f(0) = 0.
So f(c) = f(2c) = · · · = 0, also (by periodicity, where c is the period).

My questions are:

1) Are they plugging in x=0 because we want to see that it crosses the x-axis infinitely often? I'm not understanding the solution at all and it's quite frustrating.

2) What does it mean if the graph is periodic?

Answer 1

Let's start by observing that the exercise is phrased incorrectly. The intention was to ask you to show that the graph crosses the x-axis infinitely many times, not infinitely often.

A function is periodic if it repeats itself after a fixed interval (called the period of the function) repeatedly without limit. The best known periodic functions are the trig functions sine, cosine and tangent.

Any value that occurs anywhere on a periodic function will appear infinitely many times because it will appear at the same location in every other period of the function. If we can show that f(x)=0 for any x in a periodic function, we'll know f(x)=0 for infinitely many x's, and this means the function crosses the x-axis an infinite number of times.

We can solve the problem, then, if we establish that f(x)=0 for any x, and this turns out to be easy because every odd function crosses the origin. This may be apparent from the fact that odd functions are symmetrical about the origin (they look the same if you rotate the graph 180 degrees), but we can also prove the fact with algebraic logic. By definition, in an odd function, -f(x)=f(-x), and this means -f(0)=f(-0)=f(0), and the only way -f(0) can be equal to f(0) is if f(0)=0.

The sine function is an example of one that meets the conditions of this example. The cosine function also crosses the x-axis infinitely many times, but it's an even function, not an odd one.

Answer 2

I've been trying to wrap my head around this but I can't seem to picture it.

So we're told that f(x) is odd and periodic. I get what that means now. I also understand the point you make where if we can show that f(x) = 0 for any x in a periodic function, it will cross the x-axis infinitely many times.

The confusion is here; " We can solve the problem, then, if we establish that f(x)=0 for any x, and this turns out to be easy because every odd function crosses the origin. This may be apparent from the fact that odd functions are symmetrical about the origin "

Doesn't that only show that it passes through the x-axis once? i.e. the Origin? Which is at point (0,0)? How does it prove that it crosses the x-axis infinitely many times if it only goes through the origin once?

The sine function only goes through the origin once as well?

Sorry for being dense!!!

Answer 3

You say, "I also understand the point you make where if we can show that f(x) = 0 for any x in a periodic function, it will cross the x-axis infinitely many times." If you understand that, then you're almost there. The connection you're missing is that when we say the graph of an equation crosses the x-axis, we're saying that f(x)=0 when x=0. We've shown one place where f(x)=0, and that means there will be another place where f(x) equals zero at the same location in each period of the graph.

As it happens, the sine function, which I've so beautifully drawn, crosses the x-axis twice in each period (with the period being 2pi). Finding that sin(x)=0 when x=0 assures us that sin(x) will also equal 0 at x=2pi and x=4pi and x=n*2pi for any integer n.