1. I am too lazy to make mine white (CaptainBlank do that for me). (Talk about a dictator who breaks his own rules)

|y|<=sin(sin(sin(sin(1))))
Anyone understand why?

I have a formal treatment of this if anyone is interested?
So how you like my puzzle, fun, easy, too hard?

2. The extension to this PoW is still unanswered.

I will repeat it in case people did not notice that someone had actually

What is the range of:

f(z) = sin(sin(sin(sin(sin(z)))))

when the domain is C the set of all complex numbers?

RonL

3. Originally Posted by ThePerfectHacker
I am too lazy to make mine white (CaptainBlank do that for me). (Talk about a dictator who breaks his own rules)

|y|<=sin(sin(sin(sin(1))))
Anyone understand why?

I have a formal treatment of this if anyone is interested?
So how you like my puzzle, fun, easy, too hard?
It was too easy I'm afraid. To the extent that I knew the answer straight away but was struggling to find a way of writing it exactly which apparently isn't possible.

4. Originally Posted by ThePerfectHacker
I am too lazy to make mine white (CaptainBlank do that for me). (Talk about a dictator who breaks his own rules)

|y|<=sin(sin(sin(sin(1))))
Anyone understand why?

I have a formal treatment of this if anyone is interested?
So how you like my puzzle, fun, easy, too hard?
That's it? I spent all those hours trying (and failing) to come up with some exact multiple of pi for the limits of the domain and you're leaving the answer like that?? (grumbles) Heck, I had THAT answer within 30 seconds of starting the problem! (I had thought that expressing the exact answer in another form was the meat of the problem. )

Me thinks that if the answer is going to be in that form the problem was a bit too easy.

-Dan

5. Find the maximum of $\displaystyle f(x) = sin(sin(sin(sin(x))))$ by setting the derivative equal to zero:

$\displaystyle cos(x) cos(sin(x)) cos(sin(sin(x))) cos(sin(sin(sin(x)))) = 0$

One obvious solution is $\displaystyle \frac{1}{2} \pi$ because $\displaystyle cos(\frac{1}{2} \pi) = 0$.

So either the maximum or minimum occur at $\displaystyle x = \frac{1}{2} \pi$

Since $\displaystyle f(x)$ is even, its range is $\displaystyle [ sin(sin(sin(sin(\frac{1}{2} \pi)))), -sin(sin(sin(sin(\frac{1}{2} \pi)))) ]$.

Which is the same as $\displaystyle [ sin(sin(sin(1))), -sin(sin(sin(1)))]$.

6. Originally Posted by CaptainBlack
The extension to this PoW is still unanswered.

I will repeat it in case people did not notice that someone had actually

What is the range of:

f(z) = sin(sin(sin(sin(sin(z)))))

when the domain is C the set of all complex numbers?

RonL
No one seems interested in this so I will post the answer:

sin(z) is entire, so the image of C under sin is C, so the range of f(z) is C

RonL

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