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I need to find the continuous function $f(x)$ that satisfies $f(0)=0$ and:

$$\frac{f(\sin(\pi/6))^2}{\sin^4(\pi/6)}=135$$ $$\frac{f(\sin(\pi/4))^2}{\sin^4(\pi/4)}=63$$ $$\frac{f(\sin(\pi/3))^2}{\sin^4(\pi/3)}=39$$ $$\frac{f(\sin(\pi/2))^2}{\sin^4(\pi/2)}=27$$

The RHS integers above have an interesting interpretation in terms of Knoedel numbers. It is also known that $f(x)$ must satisfy:

$$\frac{f(\sin(\pi/6))^2}{\sin^6(\pi/6)}=540$$ $$\frac{f(\sin(\pi/4))^2}{\sin^6(\pi/4)}=126$$ $$\frac{f(\sin(\pi/3))^2}{\sin^6(\pi/3)}=52$$ $$\frac{f(\sin(\pi/2))^2}{\sin^6(\pi/2)}=27$$

And

$$\frac{f(\sin(\pi/6))^2}{\sin^8(\pi/6)}=2160$$ $$\frac{f(\sin(\pi/4))^2}{\sin^8(\pi/4)}=252$$ $$\frac{f(\sin(\pi/3))^2}{\sin^8(\pi/3)}=69.3333$$ $$\frac{f(\sin(\pi/2))^2}{\sin^8(\pi/2)}=27$$

I guess it's also kind of interesting that the digits in the RHS integers always add up to 9 except for the $f(\pi/3)$ case.

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  • $\begingroup$ There was a question with the same four numbers at the same points yesterday, but I can't find it - it may have been deleted, but there were some answers there. $\endgroup$ Commented Jun 29, 2013 at 17:58
  • $\begingroup$ @MarkBennet Yep. That was me. I deleted it because, as many commenters pointed out, I did not include enough info. This time I include more info. $\endgroup$
    – ben
    Commented Jun 29, 2013 at 18:04
  • $\begingroup$ You are just dividing the same four relationships you had last time by rational numbers e.g. $ \sin \left( 1/6\,\pi \right) ^{6}={\frac {1}{64 }}$ there is no new info here... $\endgroup$ Commented Jun 30, 2013 at 19:16

3 Answers 3

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This is a comment, but it is quite long, so I am writing it as an answer, hope you don't mind.

OK, you asked this question yesterday and you didn't specify what type of function you want. Again, there is not enough information in this question. So, what are all these people asking as more information, let me try to explain. For example, the factorial function $n!$. Similarly, I can ask a question as follows:

Find me a function defined on all positive reals such that $f(n+1) = n!$ for all positive integers $n$.

There are infinitely many such functions, and one cannot give a definitive answer. So, you have to add extra conditions. For example, let's say we want $f(x+1) = x f(x),\,x>0$ and $f(1)=1$. This is much more restrictive, but again, there are many such functions. If we add one more condition, i.e. if we require $f$ to be log-convex, then we get only one function, namely the Gamma function (This is known as the Bohr–Mollerup theorem). Similarly, you should specify what extra conditions you want. I am not familiar with those Knoedel numbers, but perhaps you have an intuition as to how to extend their definitions to reals.

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  • $\begingroup$ Thanks, this was a helpful comment/answer. $\endgroup$
    – ben
    Commented Jul 1, 2013 at 20:37
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$$f(b(t))=\sqrt{36b(t)^2-9b(t)^4} \:\:\:\:; \:\: b(t)=\sin{t}, \:\:\: 0 < t <\pi/2$$

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Since you are given $f()$ only at a finite number of points, what's wrong with a piecewise linear or polynomial interpolation?

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  • $\begingroup$ I want the "real" equation (as opposed to a cubic spline approximation fitted to the known data points etc.) because I think it will yield geometric insights. $\endgroup$
    – ben
    Commented Jun 29, 2013 at 18:03
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    $\begingroup$ @ben - the point is that there isn't "the" real equation, and there isn't enough information in the question to understand what kind of solution you need. I can find a cubic through four points which will give a continuous function (in fact differentiable as many times as you like). If that is not what you want, you will have to be clearer about what you do want. $\endgroup$ Commented Jun 29, 2013 at 18:09
  • $\begingroup$ @ben - the whole point is that there are infinitely many such functions, and non of them less real the others. $\endgroup$ Commented Jun 29, 2013 at 18:16
  • $\begingroup$ But there may be a simple solution, or a solution with a geometric meaning, extending to other values of the argument. Consider gamma function or Dobinski formula... $\endgroup$ Commented Jun 29, 2013 at 20:53
  • $\begingroup$ @arbautjc Yes, I figured it out. The answer is above. $\endgroup$
    – ben
    Commented Jul 1, 2013 at 20:42

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