Can this function still be a constant function?

[tellmesomething]

Homework Statement

Let $f(x)=f(2x)$ for all $x\in R$where $f(x)$ is continuous function & $f(2024)=π/2$ if $L=lim_{x\to 0} \frac{(cos²(f(x))+1-sin²(f(x)))} {sin²x}$ then $4L$ may be?

Relevant Equations

None

So I know that since $x \in R$ that means $2x$ can achieve all possible values on the real number line meaning $f(x)$ is a constant function. And I know hwo to calculate the limit beyond that. However my teacher made a point which I dont necessarily agree with he said, if $f(x)$ wasn't continuous we could not have said Its a constant function.
He gave an example like
$f(x)=1$ for all $x \in Q$
$f(x)=0$ for all $x \not\in Q$
The above piecewise function also satisfies $f(x)=f(2x)$

But my doubt is since the original question already says that $x \in R$, isnt that enough information to conclude its a constant function.
What I mean is if the question was instead
Let $f(x)=f(2x)$ for all $x \in R$ & $f(2024)=π/2$, if $L=lim_{x\to 0} \frac{(cos²(f(x))+1-sin²(f(x)))} {sin²x}$ then $4L$ may be?
Could we still not have concluded that $f(x)$ is a constant function I.e on a graph it would be parallel to the $x$ axis for alll $x$ ?

Because for all I know, a continuous function means that the limiting value and the functional value are equal , and here since we already know that the function $f(x)=f(2x)$ for all $x\in R$ we know that the limiting value and the functional value are equal..

 

[fresh_42]

I do not understand why you come up with such a complicated example or what the roles of $4L$ or $\pi/2$ are. I don't even understand what $X$ is, or how the function $f(x)$ in the limit expression is defined.

But my doubt is since the original question already says that x€R , isnt that enough information to conclude its a constant function.

Obviously not. What's wrong with your teacher's example? It is not constant, not continuous, its domain is $x\in \mathbb{R},$ its codomain $f(x)\in \{0,1\},$ $f(2x)=1=f(x)$ if $x\in \mathbb{Q},$ and $f(2x)=0=f(x)$ if $x\not\in \mathbb{Q}.$ This example uses the fact that $x\in \mathbb{Q} \Longleftrightarrow 2x\in \mathbb{Q}.$

 

[tellmesomething]

I do not understand why you come up with such a complicated example or what the roles of $4L$ or $\pi/2$ are. I don't even understand what $X$ is,

Sorry I didnt realise some variables got capitalised. I have edited it now.

or how the function $f(x)$ in the limit expression is defined.

I dont understand this though, if you could clear it out.

Obviously not. What's wrong with your teacher's example? It is not constant, not continuous, its domain is $x\in \mathbb{R},$ its codomain $f(x)\in \{0,1\},$ $f(2x)=1=f(x)$ if $x\in \mathbb{Q},$ and $f(2x)=0=f(x)$ if $x\not\in \mathbb{Q}.$ This example uses the fact that $x\in \mathbb{Q} \Longleftrightarrow 2x\in \mathbb{Q}.$

I think I get it now. Please check .. Firstly if $x\in R$ then I believe $2x\in R$ as well I.e all real numbers are in the set of $x$ as well as in the set of $2x$
You obviously cannot get an irrational number when you multiply a rational number by 2 or vice versa. Domain of the function $f(x)$ is all real numbers.
$f(x)=f(2x)$ implies if $x\in Q$ we get a constant function and if $x\not\in Q$ we get a different constant function. And if we want these constant functions to be equal we need to declare that its a continuous function overall.

 

[fresh_42]

I dont understand this though, if you could clear it out.

I meant this monster here:
$$L=\lim_{x\to 0} \dfrac{\cos^2(f(x)) - \sin^2(f(x)) +1} {\sin^2 x}=2\lim_{x\to 0} \cot^2(f(x))\stackrel{\boldsymbol ?}{=}2\lim_{x\to 0} \cot^2(\pi/2)=0$$
You haven't said what $f(x)$ is. If it was meant to be $f(x)=\pi/2$ then why write $f(x)$?

I think I get it now. Please check .. Firstly if $x\in R$ then I believe $2x\in R$ as well I.e all real numbers are in the set of $x$ as well as in the set of $2x$
You obviously cannot get an irrational number when you multiply a rational number by 2 or vice versa. Domain of the function $f(x)$ is all real numbers.
$f(x)=f(2x)$ implies if $x\in Q$ we get a constant function and if $x\not\in Q$ we get a different constant function. And if we want these constant functions to be equal we need to declare that its a continuous function overall.

Yes. Continuity guarantees that $|f(y)-f(x)|<\varepsilon$ if $|y-x|<\delta ,$ i.e. the function values $f(y)$ and $f(x)$ cannot "jump" if $x$ and $y$ are close enough. We can always find a rational and an irrational number that are arbitrarily close, so their function values have to be, too.

The standard proof of such a statement would be to prove that $f(x)$ is constant on $\mathbb{Z},$ then that it is constant on $\mathbb{Q},$ say $f(q)=c,$ and continuity allows us to conclude:

Let $r\in \mathbb{R}$ and $(q_n)_\{n\in \mathbb{N}\} \subseteq \mathbb{Q}$ a sequence of rational numbers $q_n$ such that $\lim_{n \to \infty}q_n = r.$ Since $\mathbb{Q}\subseteq \mathbb{R}$ is dense, such a sequence exists. Continuity of $f(x)$ now says
$$
c=f(q_n)=\lim_{n \to \infty}f(q_n) =f(r).
$$

 

[tellmesomething]

I meant this monster here:
$$L=\lim_{x\to 0} \dfrac{\cos^2(f(x)) - \sin^2(f(x)) +1} {\sin^2 x}=2\lim_{x\to 0} \cot^2(f(x))\stackrel{\boldsymbol ?}{=}2\lim_{x\to 0} \cot^2(\pi/2)=0$$
You haven't said what $f(x)$ is. If it was meant to be $f(x)=\pi/2$ then why write $f(x)$?

This is how the question was framed. From what I gather if it said $\frac{π} {2}$ directly it would not be "tricky" as the whole point is to conclude from the question that $f(x)$ is a constant and its value will be same as $f(2024)$ which is $\frac{π}{2}$ . Beyond that its pretty simple to calculate the limit which would have been the same if x was approaching something like π.

This might seem like a very silly question understandably so its for high school calc.

Yes. Continuity guarantees that $|f(y)-f(x)|<\varepsilon$ if $|y-x|<\delta ,$ i.e. the function values $f(y)$ and $f(x)$ cannot "jump" if $x$ and $y$ are close enough. We can always find a rational and an irrational number that are arbitrarily close, so their function values have to be, too.

The standard proof of such a statement would be to prove that $f(x)$ is constant on $\mathbb{Z},$ then that it is constant on $\mathbb{Q},$ say $f(q)=c,$ and continuity allows us to conclude:

Let $r\in \mathbb{R}$ and $(q_n)_\{n\in \mathbb{N}\} \subseteq \mathbb{Q}$ a sequence of rational numbers $q_n$ such that $\lim_{n \to \infty}q_n = r.$ Since $\mathbb{Q}\subseteq \mathbb{R}$ is dense, such a sequence exists. Continuity of $f(x)$ now says
$$
c=f(q_n)=\lim_{n \to \infty}f(q_n) =f(r).
$$

Though I dont understand a bit of the notation, I think I get the gist of it. Many thanks!

 

[fresh_42]

This might seem like a very silly question understandably so its for high school calc.

Continuity and limits at the high school level? That's ambitious, especially if all this should be rigorous. I would like to see the proof for
$$
f(x)=f(2x) \;\forall \;x\in \mathbb{R} \wedge f(2024)=\pi/2 \Longrightarrow f(0)=\pi/2
$$

 

[tellmesomething]

Continuity and limits at the high school level? That's ambitious, especially if all this should be rigorous. I would like to see the proof for
$$
f(x)=f(2x) \;\forall \;x\in \mathbb{R} \wedge f(2024)=\pi/2 \Longrightarrow f(0)=\pi/2
$$

I am not at all familiar with proofs to do this any justice. But from the discussion we had,
Since $f(x)$ is defined for all $x\in R$
We know that $f(2x)$ is also defined for all $x\in R$ as
If we suppose x is any number y, we know that it can be written as $\frac{y}{2}$ and multiplying it with 2 would return us the number hence telling us that all numbers in the set of x are also in the set of 2x

Now its given that $f(x) = f(2x)$ which means the subset of all rational numbers' functional value is constant and the subset of all irrational numbers' functional value is constant. Since the function is continuous I.e has no breaks in the graph this would mean that very close to any number on the real number line the functional value has to be equal to the functional value at the number. And we know that there are infinite number of irrational as well as rational values near any number we can conclude that the function $f(x)$ is constant overall.

This would mean that $f(2024)$ = $f(x)$= $\frac{π}{2}$

I think I just reiterated everything I gained from your posts. I know you asked of me a formal proof but we dont do proof writing until college here so this is the only hand wavy thing I know..

 

[PeroK]

To show the function is constant, you could note that
$$\forall n: f(x) = f(\frac x {2^n})$$And use the continuity of $f$ to show that $f(x) = f(0)$.

 

[fresh_42]

Are you sure it isn't $f(2048)=\pi/2$? That would make sense.

Let's set $f(1)=c.$ Then $c=f(1)=f(2)=f(4)=f(8)=f(16)=\ldots =f(2048)=\pi /2.$ On the other end we get $\ldots =f(1/32)=f(1/16)=f(1/8)=f(1/4)=f(1/2)=f(1)=c.$ Since the sequence $2^{-n}$ get's closer and closer to $0$ we have $\lim_{n \to \infty}f(2^{-n})=c=\pi/2=f(0)$ by continuity of $f$ (no jumps). Now we can set $f(0)$ into the formula for $L.$

 

[tellmesomething]

Are you sure it isn't $f(2048)=\pi/2$? That would make sense.

Let's set $f(1)=c.$ Then $c=f(1)=f(2)=f(4)=f(8)=f(16)=\ldots =f(2048)=\pi /2.$ On the other end we get $\ldots =f(1/32)=f(1/16)=f(1/8)=f(1/4)=f(1/2)=f(1)=c.$ Since the sequence $2^{-n}$ get's closer and closer to $0$ we have $\lim_{n \to \infty}f(2^{-n})=c=\pi/2=f(0)$ by continuity of $f$ (no jumps). Now we can set $f(0)$ into the formula for $L.$

I think I didnt follow. I thought a lot about this but I dont know..
Isn't f a constant function? How does it matter if its 2024 or 2048.. The question says 2024 by the way..

 

[fresh_42]

I think I didnt follow. I thought a lot about this but I dont know..
Isn't f a constant function? How does it matter if its 2024 or 2048.. The question says 2024 by the way..

I still don't know whether the condition is sufficient to prove that $f$ is constant. I haven't seen a proof. If it is constant, then it doesn't matter since $f(2024)=f(2048).$ However, I do not know.

If $f$ is not constant, but $f(x)=f(2x)$ and $f(2048)=0$ then this would be sufficient to prove $L=0.$ We get from $f(x)=f(2x)$ that $f(2^n)=f(2^0)=f(1)$ for all integers $n\in \mathbb{Z}.$ This means that $f(1)=f(2^{11})=f(2048)=\pi/2$ in case we would have such a condition. Going on with that argument, we also get
$$
\lim_{n \to \infty}f(2^{-n})=\lim_{n \to \infty}f(1)=f(1)=\pi/2
$$
and by continuity of $f$ we get
$$
\pi/2= \lim_{n \to \infty}f(2^{-n})= f\left(\lim_{n \to \infty}2^{-n}\right)=f(0)
$$
and
$$
L=2\lim_{x \to 0} \cot^2(f(x))=2\lim_{n \to \infty}\cot^2(f(2^{-n}))=2\cot^2(f(0))=2\cot^2(\pi/2)=2\cdot 0 = 0
$$
Conclusion: I can prove that $L=0$ if $f(2^{11})=\pi/2$ without knowing that $f$ is constant. If I only have $f(2024)=\pi/2$ then I do not know that $L=0$ because I haven't seen a proof that $f$ is constant.

 

[PeroK]

I still don't know whether the condition is sufficient to prove that $f$ is constant. I haven't seen a proof.

Continuity at $0$ is sufficient:

To show the function is constant, you could note that
$$\forall n: f(x) = f(\frac x {2^n})$$And use the continuity of $f$ [edit: at 0] to show that $f(x) = f(0)$.