Periodic Function: Prove Smallest Positive Period

[Andrei1]

One period of the function \(\displaystyle f(x)=\operatorname{tg}\frac{11x}{34}+\operatorname{ctg}\frac{13x}{54}\) is \(\displaystyle 918\pi.\) Please help me to prove that this is the smallest positive period. I can not use the most of trigonometric identities.

 

[Opalg]

One period of the function \(\displaystyle f(x)=\operatorname{tg}\frac{11x}{34}+\operatorname{ctg}\frac{13x}{54}\) is \(\displaystyle 918\pi.\) Please help me to prove that this is the smallest positive period. I can not use the most of trigonometric identities.

The tangent and cotangent functions both have period $\pi$. So the function $\tan\frac{11x}{34}$ will repeat at intervals $\frac{34\pi}{11}$, and $\cot\frac{13x}{54}$ will repeat at intervals $\frac{54\pi}{13}$. You need to find the least common multiple of those two intervals.

 

[Andrei1]

You need to find the least common multiple of those two intervals.

\(\displaystyle T=918\pi\) is the least common multiple of those periods. I found this. But why it is the smallest positive period of \(\displaystyle f\)?
For example, I consider the functions \(\displaystyle f_1(x)=\sin x\) and \(\displaystyle f_2(x)=\operatorname{tg} x-\sin x\), which both have \(\displaystyle 2\pi\) as main period. But then \(\displaystyle \pi\) is the main period of \(\displaystyle f_1+f_2\), which is not the least common multiple of \(\displaystyle 2\pi.\)

 

[Opalg]

\(\displaystyle T=918\pi\) is the least common multiple of those periods. I found this. But why it is the smallest positive period of \(\displaystyle f\)?
For example, I consider the functions \(\displaystyle f_1(x)=\sin x\) and \(\displaystyle f_2(x)=\operatorname{tg} x-\sin x\), which both have \(\displaystyle 2\pi\) as main period. But then \(\displaystyle \pi\) is the main period of \(\displaystyle f_1+f_2\), which is not the least common multiple of \(\displaystyle 2\pi.\)

You could look at the set of points at which $f(x)$ becomes infinite (or is undefined). That will happen at the points $x = \bigl(\frac{34}{11}k + \frac12\bigr)\pi$ and $x = \frac{54}{13}k\pi$ (and nowhere else). I think you should find that that set of points does not repeat at intervals of less than $918\pi$.

 

[CellCoree]

To prove that 918π is the smallest positive period of the given function, we must show that the function repeats itself after every 918π units. This means that for any value of x, the function will have the same output as it did 918π units earlier.

To do this, we can use the definition of a periodic function. A periodic function is one that repeats itself after a certain interval, called the period. In this case, the period is 918π. So, to prove that this is the smallest positive period, we need to show that the function repeats itself after every 918π units.

To start, we can rewrite the given function as f(x) = tan(11x/34) + cot(13x/54). Now, let's consider two values of x, say x1 and x2, that are 918π units apart. This means that x2 = x1 + 918π. Now, let's plug these values into the function:

f(x1) = tan(11x1/34) + cot(13x1/54)
f(x2) = tan(11x2/34) + cot(13x2/54)
= tan(11(x1+918π)/34) + cot(13(x1+918π)/54)
= tan(11x1/34 + 918π/34) + cot(13x1/54 + 918π/54)
= tan(11x1/34) + cot(13x1/54)
= f(x1)

As we can see, the function has the same output for both x1 and x2. This means that the function has repeated itself after 918π units, proving that 918π is indeed the smallest positive period.

Another way to think about this is to consider the graph of the function. Since the function is periodic, its graph will repeat itself after every 918π units. So, if we were to graph the function over a range of 918π units, we would see the same pattern repeating itself. This is another way to show that 918π is the smallest positive period of the function.

In conclusion, we have shown that the given function has a period of 918π, and that this is indeed the smallest positive period. This can be verified by plugging in any two values of x that are 918π units apart and seeing