I think you’re assuming that c must be related somehow to the periods of f and g, which is not the case. In the above, you claim that g(a) = f(a) and g(a+c) = f(a+c) (c ≠ 0) implies that there must be infinitely many solutions. Thanks! RT emergentmath On intersecting periodic functions wp.me/p1jLi5-in Keep looking and “getting messy” everyone! There’s still some hesitation here and on twitter as to whether you can have additional finite number of intersections. Check out this beauty from Dave ( see the functions used, click on the image. Sure enough, I’m incorrect and my intuition was wrong. ![]() Stay tuned, this post could probably get updated thanks to the collective brainpower of people that are much more clever than me. Or even better, make a more convincing argument (for or against) than I did. ![]() If someone finds a couple periodic functions that do intersect more than 1 but less than infinity times, let us know in the comments or share your graph. I’m not particularly moved by my own argument. G(a+c+P)=g(a+c)=f(a+c) <- third (and presumably, infinite more) solutions! However, according to the definition of a periodic function, this would mean that G(a+c)=f(a+c) where c is some real number. If we assume that there are two solutions to the periodic functions f(x)=g(x), then we assume that, I’m trying to figure out how you’d make two functions meet within one period twice, and then never again. In my head, a periodic function is one that satisfies the following:į(x)=f(x+P) for all values of x and some constant P. After all, you can’t add a certain number of √2 ‘s together and get an integer. Easy peasy.īut I’m still wary of the existence a pair of periodic functions with two and only two intersections. It makes sense that something rotating with an irrational period will never match up with something rotating with a rational period more than once, if at all. Seeing this through the lens of rotating motion certainly helped me. All you need is one of the periodic functions to have an irrational period (say, pi) and the other to have a rational period (say, 1) and both to be oppositely shifted vertically and the twain may meet just once.Īlso helpful, was this video from Joshua on irregular periods. Initially I was wary of even this, but Joshua has me convinced. So definitely you can have a pair of periodic functions with one intersection. Joshua Bowman February 18, emergentmath “Can 2 periodic functions intersect finitely many times?” Answer via desmos: /calculator/cwp… #mathchat #math Your sines, your cosines.Īfter that came a flurry of answers and conversations from all your favorite mathtwitterblogosphere emergentmath Or how about: f(x) = cos(x) – 1 and g(x) = 1 – cos(π*x). Today I asked the following question on twitter.Ĭan two periodic functions intersect a finite number of times? #mathchatĪnd, to clarify, I mean continuous, differentiable periodic functions. ![]() And moreover, suggests that we get messy together in asking ourselves challenging problems and developing clever arguments.Ĭonsider this an attempt at heeding Lockhart’s advice. ![]() He presents math as something that you just need to “get messy with”. I’m reading through Measurement by Paul Lockhart these days.
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