Is the sum of two periodic functions still periodic?

We know sin(x) is periodic and sin(πx) is also periodic. How about the sin(x)+sin(πx)?
Note: The "least common multiple" method helps to find "a" period of the sum but it does not guarantee that it is the shortest period.


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[panulli]

If you are familiar with a Fourier transform then the answer to your question should be pretty clear.

[Tosti_bakker]

Thank you so much! This really helped me with my signals and systems course!

[PerfectlyCrafted]

I love using your videos on algebra as a refresher, Keep making them!

[ProfRonconi]

I just recently subscribed to your channel, and I must congratulate you. Not only are you able to explain complex ides in simple terms, but also to relate simple ideas to complex concepts.

Before watching the whole video, I thought to myself: "No. If the periods are inconmensurable (so that one is a non-rational multiple of the other) it would take an infinite amount of time for them to coincide again." Which is what (I think) you managed to explain with an excellent example.

[andywright]

Before seeing video:
Sin(x) + sin(root2 x) never repeats so can not be periodic

After seeing video: Yay

[ChristopheRoux]

For me the simple counter example I was thinking of was:

f(x)=1 for all integers and f(x)=0 for non integers. f has a period of 1
g(x)=1 for all multiple of π and g(x)=0 for non multiple of π. g has a period of π
h(x)=f(x)+h(x) is obviously a non constant function and if we suppose h is periodic of period T.
then h(T)=h(0+T)=h(0)=2
then f(T)=1 and g(T)=1
this implies T=q with q integer and T=p*π with p integer
Contradiction as π is irrational

[Phantomuxas]

I don't understand why we need to multiply by integers, and why pi is unsuitable :/

[tiborpejic]

I actually tried to prove the statement and then realised it's impossible before coming up with the exact same counterexample. It is interesting to graph your function because it resembles a periodic one even though it obviously isn't.

[calyodelphi]

Paused at 0:16 to think about my answer to that question...

My answer is "no" if the ratio of the periods of the two functions is irrational. However, in order for this to be the case, the period of one of the two functions must first be irrational, but this does not necessarily guarantee that the ratio between the two functions' periods is irrational. But any combination of functions in which the ratios between their periods is rational, you will always have a periodic sum of the two functions that is the LCM of the functions' periods at the very least.

[unpauses]

Ha! I was right! :D

Sort of. XD

I'm pretty sure it'd still work out even if one of the two periodic functions has a period that includes a square root, but I don't have much time this morning before I have to go to work to work it out. I'd have to peek at it again after I get home.

[BeefGoulashJosh]

Thank you! This helped me with my Signals and System HW

[CloudSkywalker]

Basically: There won't be any common multiples if one period is rational and the other irrational. So there won't be a period of the sum.

[copperfield]

why you need to find a integer and not a real number for this to work?

[DonSolaris]

Excellent! I've learned two new things. BTW been playing with graphing calculator software for years (converting it into sound, since i'm a sound designer) and know this from a total different perspective, so it was cool seeing the same picture in equations. I never experimented with irrational coefficients of "t", so this gave me a new idea to try, although i am 99.999% sure the result will be just a white noise. 😎

[georgebakradze]

Studying math but still cant get those kind of valuable info. Thank you mr. Teacher

[SimchaWaldman]

03:48 All this and you forgot to use the coding \sin?

[HemmligtNavn]

so it might be worth noting that your counter example is indeed a counter example in general because no integer multiple of a irrational number will ever be rational or an integer, so therefore there is no integer to multiply to opposite side by that will make it irrational?

[aurithrabarua]

Would you please make videos on the solve of MIT Integration Bee???
I think none is more perfect than you to get the deal done😊😊😊

[otakurocklee]

Why does the period of the sum have to be an integer multiple of the period of the individuals? I see it's a sufficient condition, but I don't see why it's a necessary condition.

[MA-bmjz]

A better proof would be that the sum of the two functions has a limit at infinity since we know that periodic functions don't converge at x->infinity

[anmolchaurasia]

Thanks for your amazing explanation, love from INDIA 🇮🇳
🙏👍👍