This actually came up yesterday in my linear algebra class, when we wondered if period functions form a vector space.
The answer is no. The sum of two periodic functions is periodic only if their periods are rationally related to each other, in which case the new period is the lcm of the two.
Thanks. I wanted to used this to help my 11th graders understand "irrationals" a little better, but I could not find any examples in the cute little high school text book I use and wanted to double check.
I love the idea of "irrationals never being in-synch" I hope it will help them more than it confuses them.
There are lots of fun ways to illustrate it. You can take a square, pick a point on the perimeter, and draw a line with a certain slope, looping back to the opposite side when you hit the edge (i.e. drawing along the torus). If the slope is rational, it closes into a fixed orbit; if it's irrational, it fills the square. Another one (the same thing, really), take a grasshopper hopping along the sidewalk with a certain jump distance and a certain "sidewalk panel" distance, and ask if he ever lands on the little dip 'between' the panels. If they're rationally related, he ends up in closed orbits, otherwise he hits every point over time, etc., i.e. same deal as before.
Ooo, oo, that reminds me, there are periodic functions that don't have a period. See if you can work that one out. (If not, leave a comment, I'll tell you.)
A neat way to explore "irrational numbers never being in sync" is to talk about graphs of r = sin (c*theta) in polar form...I don't know if your students are going to talk about polar graphs at all, but a type of curve that authors of books love to talk about are the roses, where c is an integer. Cool things happen if you play around with c and let theta range much further than 0 to 2 pi, which is the default on a lot of graphing calculators.
I know this is really off topic, but I thought you might appreciate it. :)
I wasted away my highschool years playing with such graphs, and finally settled on the graph of r=tan(theta/sqrt(5)). If one confines the viewing window apropriately and graphs with theta vraying from 0 to 18pi or so, it looks quite lovely.
Suppose that there is a positive number p such that f(x+p) = f(x) for all x. Let g(x) = f(x) + f"(x) and h(x) = 2f(x) + f"(x). Then g(x+p) = g(x) and h(x+p) = h(x) for all x.
But g(x) = -sin(sqrt(2)*x), which has fundamental period 2pi/sqrt(2), and h(x) = sin(x), which has fundamental period 2pi. Therefore, p is an integral multiple of both 2pi and 2pi/sqrt(2). But this is impossible, since the square root of 2 is irrational.
you should definitely tell your students about the discovery of irrational numbers by the greeks in the mystical pentagram. It goes something like this :
Ancient Greek mathematics (Pythagorean school) was chiefly based on geometry. They believed that all pairs of line segments are of commensurable length, that is, both segments are integer multiples of a common smaller segment. This is just saying that all numbers are rational.
Now take a pentagram inscribed into a (regular) pentagon and note the smaller pentagon inside the figure. For the Pythagoreans it was impossible to prove that the side lengths of the outer and the inner pentagon were commensurable. One can, in fact disprove this by contradiction.
Comments 26
The answer is no. The sum of two periodic functions is periodic only if their periods are rationally related to each other, in which case the new period is the lcm of the two.
Reply
You know what I'm sayin. Brain fart.
Reply
I love the idea of "irrationals never being in-synch" I hope it will help them more than it confuses them.
Reply
Reply
Reply
Reply
Reply
Reply
I know this is really off topic, but I thought you might appreciate it. :)
Reply
Reply
Very nice.
Reply
Let g(x) = f(x) + f"(x) and h(x) = 2f(x) + f"(x). Then g(x+p) = g(x) and h(x+p) = h(x) for all x.
But g(x) = -sin(sqrt(2)*x), which has fundamental period 2pi/sqrt(2),
and h(x) = sin(x), which has fundamental period 2pi.
Therefore, p is an integral multiple of both 2pi and 2pi/sqrt(2).
But this is impossible, since the square root of 2 is irrational.
Reply
Ancient Greek mathematics (Pythagorean school) was chiefly based on geometry. They believed that all pairs of line segments are of commensurable length, that is, both segments are integer multiples of a common smaller segment. This is just saying that all numbers are rational.
Now take a pentagram inscribed into a (regular) pentagon and note the smaller pentagon inside the figure. For the Pythagoreans it was impossible to prove that the side lengths of the outer and the inner pentagon were commensurable. One can, in fact disprove this by contradiction.
make a drawing like the following
( ... )
Reply
Reply
Leave a comment