jsaddern

Have responded. As I said, I will work on characterizing all numbers that serve as counter-examples.

I'm curious to know an approach to this post too.

jsaddern

Thank you! I profoundly appreciate your response.

I understand your logic and see the four counter-examples you provide. I will continue to work on describing all numbers that work as counter examples. Thanks Melody,

jsaddern

Yes, it does include $0$ and $9$.

Also, yes! It looks like $11$ does disprove it. Can you describe the set of all numbers for which the given expression is a perfect square? (Maybe to get there, how did you come up with your example?)

Maybe it consists of the regions where the induced hemispheres overlap?

Does anybody have an idea?

I tried comparing this to choosing $n$ diameters for a circle. That clearly forms subsets of the circle (i.e. segments between the $2n$ points of intersection between the diameters and the circumference)...

Maybe that can help answer the original question?

Hmm... I'm getting $\frac{\pi\sqrt{2}}{2}$...

Is anyone getting another answer?

Thanks so much! That is indeed a great solution!

An alternative that I thought of was:

f'(0)^2 = f(0)f''(0) --> Since f(0) = 1, we conclude that f''(0) = f'(0)^2. Now, we can differentiate both sides to get f'(x)f''(x)=f(x)f'''(x). Now, we can differentiate this equation once again! And use the product rule! --> f'(x)f'''(x) + f''(x)^2 = f(x)f''''(x) + f'(x)f'''(x) --> f''(x)^2 = f(x)f''''(x).

Plugging in x=0 simply gives us the answer, that is, f''(0)^4 = f'''(0) = 0. Therefore, f''(0) = $\pm 3$ and further, since f'(0)^2 =f''(0), we get the end result, $f'(0) = \pm \sqrt{3}$. Yayyyy!

Doesn't integrating that simply give us $f(x) = c$?

In that case, "all possible values of $f'(0)$  are simply... $0$.

Thank you!

I'd love to interact!

I'm not quite sure what the answers are... How would we continue to integrate f(x) = Cf'(x) to achieve the possible values of f'(0)?

Thank you both and happy holidays!

The form of the answer is correct but it seems like the constant k is a bit off.

Thanks