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I found out after a few tries that the periods of both $$(\sin{\theta})^0, (\tan{\theta})^0~\text{are $\pi$}$$

Whereas i was told by my prof that period of $(\sin{\theta})^0 +(\tan{\theta})^0$ is $\frac{\pi}2$ as an exception to the LCM rule while finding the periods of two functions

Why is this anomaly happening? Logically speaking shouldn't the period of sum of two functions having periods $\pi$ be $\pi$ too?

Arjun
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    The period of a sum of two functions having periods $\pi$ is usually $\pi$, but can also be a divisor of $\pi$. – user3257842 Dec 21 '21 at 13:05
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    This is a very artificial question ! Are you conscious that you are dealing with functions that are equal to $1$ but for values where they are undefined ? Talking about periodicity in this context is very specious... – Jean Marie Dec 21 '21 at 13:22
  • How does your professor define powers with integer exponents? – Angel Dec 21 '21 at 15:26
  • @Angel the usual, base multiplied the number of times in exponent – Arjun Dec 21 '21 at 15:32
  • @Arjun So, to be completely clear, if you multiply $0$ copies of $x$ together, the product is equal to $1$? Is that what your professor told you? I am asking as a matter of clarification, so I can now how to help you. – Angel Dec 21 '21 at 15:36
  • @Angel there wasn't actually much discussion this exponent thing, or why 0 copies of x is equal to 1 – Arjun Dec 21 '21 at 15:59
  • @Arjun What does your profressor say regarding $0^0$? What does he say the domain of the function is? – Angel Dec 21 '21 at 16:00
  • he says $0^0$ is one of the 7 indeterminate forms, domain of function is all real no's - points where it isn't defined – Arjun Dec 21 '21 at 16:05
  • @Arjun Okay, so your teacher wants you to treat $0^0$ as undefined. Good to know. But in that case, it is impossible for the domain of the function to be all real numbers, because the expression is undefined whenever $\sin(x)$ or $\tan(x)$ is equal to $0,$ since $0^0$ is being treated as undefined, and it is also undefined whenever $\tan(x)$ itself is undefined. – Angel Dec 21 '21 at 18:35
  • @Arjun Anyway, in that case, the period is $\pi/2$ simply because the expression is undefined at every multiple of $\pi/2.$ – Angel Dec 21 '21 at 18:36

1 Answers1

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In this context, $0^0$ is undefined, so $(\tan\theta)^0$ is undefined at all multiples of $\pi$. And $\tan\theta$ is undefined at odd multiples of $\pi/2.$ So this function is constant except at all multiples of $\pi/2$, therefore its period is $\pi/2$ not $\pi$.

B. Goddard
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  • So that LCM 'rule' is not actually a rule, it was just a trick to be used in (most) questions – Arjun Dec 21 '21 at 13:02
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    I don't know this rule. I sort of get it that if two functions have periods $3$ and $5$ then their sum probably has period $15$. But $\sin\theta$ and $-\sin \theta$ both have period $\pi$, while their sum has period any positive real. – B. Goddard Dec 21 '21 at 13:06
  • i should have noticed that it doesn't work for all functions – Arjun Dec 21 '21 at 13:09
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    I am not convinced that any mathematician would say that $0^0$ is undefined in this context. If we are considering it as limit in analysis, then yes, a mathematician might consider it undefined by proxy, but in the context of arithmetic, almost every mathematician renders $0^0$ as $1.$ So this is just misleading. I think the question is not particularly meaningful, and I am not sure why the professor would ask it without being more specific. But really, the function in question is just a constant function on $\mathbb{R},$ if we go by arithmetic and functional conventions. – Angel Dec 21 '21 at 14:44
  • Besides, the main problem occurs with the $\tan$ function, which still has singularities, regardless of the exponent. The periodicity occur perhaps because of that. But again, it is unclear whether the singularities are being considered as removed (as you would if you, say, simplify $x/x=1$ informally), or not. – Angel Dec 21 '21 at 14:45
  • @Angel You say, "it is unclear". Know what makes it unclear? The fact that $0^0$ is not defined. We can only guess what the prof really said and meant. But I think my guess is pretty good. – B. Goddard Dec 21 '21 at 15:03
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    @B.Goddard It is not a "fact" that $0^0$ is undefined, at least not in the context of arithmetic and function theory. That is precisely the problem I have with your answer. You present it is a fact, but what I am saying is that doing so is misleading, because mathematicians would not agree, in this context. In fact, there is genuinely no reason to consider $0^0$ as undefined to start with. – Angel Dec 21 '21 at 15:22
  • @B.Goddard Also, I do not appreciate you telling me why you think it is unclear to me, when I already explained the reason why, regarding the removal of singularities. – Angel Dec 21 '21 at 15:57
  • @Angel Whether you appreciate it or not, it's still a contradiction. $0^0$ can be defined in some contexts. In this one it is not. What "is not"? Answer: $0^0$ is not. How do we know? Because the OP didn't give us a definition. And you're dead wrong. If we're going to vote about truth, then "mathematicians" would overwhelmingly side with me. Did the OP tell you what he was doing about singularities? No. He left it undefined. – B. Goddard Dec 21 '21 at 16:03
  • @Angel Maybe read the first answer of this: https://math.stackexchange.com/q/11150/362009 – B. Goddard Dec 21 '21 at 16:05
  • Let us continue this discussion in chat. – Angel Dec 21 '21 at 16:53