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I've been studying functions $f:\mathbb R\to\mathbb R$ that satisfy $f(f(x))=e^x$ (or, half-iterates of the exponential function). I know that there's only one such analytic function, but it's really hard to study since it is almost certainly non-elementary and I only know how to find finitely many terms of its Maclaurin Series.

Instead, I'm studying all continuous and increasing functions $f$ satisfying $f(f(x))=e^x$, and I've alighted on the following problem (which I came up with out of curiosity). I propose this question to all interested residents of MSE:

Given that $f$ is continuous and increasing and $f(f(x))=e^x$, find some bounds for the integral $$\int_0^1 f(x)dx$$

I've managed to come up with some pretty sweet bounds (in fact, they are the best possible bounds), and I'll post them after this question gets some answers.

I'll accept whatever answer has the tightest bounds, with a proof.

NOTE: Most people probably wouldn't think of this as recreational "fun" math, but hey, I did it for fun, and I'm proposing it as a problem to be done just for fun. So please try to enjoy it, and please don't try to close it.

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  • $\begingroup$ Anyone care to explain the close vote? $\endgroup$ – Frpzzd May 15 '18 at 23:43
  • $\begingroup$ Well, puzzle problems like this aren't really what MSE is for, though I've seen such questions before and enjoyed them greatly. That being said, this seems like an interesting question, so I'm definitely not going to vote to close :) - Brevan, from Close Votes Review $\endgroup$ – Brevan Ellefsen May 16 '18 at 3:33
  • $\begingroup$ To me, the most interesting thing here isn't the question itself, but your proof technique for showing your bounds are tightest. Was it simply due to construction, or did you simply play around till you found really good bounds and then found a juicy proof? $\endgroup$ – Brevan Ellefsen May 16 '18 at 3:38
  • $\begingroup$ @BrevanEllefsen Oh, I found the juicy proof right off the bat using graphical techniques and functional equations. $\endgroup$ – Frpzzd May 16 '18 at 14:24
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Suppose $f(x)$ is an increasing, continuous function satisfying $f(f(x))=e^x$. Then there must be some $\alpha\in(0,1)$ such that $f(\alpha)=1$. We can compute $$ \begin{align*} \int_\alpha^1 f(x)\,dx&=\int_\alpha^1 e^{f^{-1}(x)}\,dx\\ &=\int_0^\alpha e^u\,df(u)\\ &=e^\alpha f(\alpha)-e^0f(0)-\int_0^\alpha f(u)e^u\,du\\ &=e^\alpha-\alpha-\int_0^\alpha f(x)e^x\,dx. \end{align*} $$ So, we have $$ \int_0^1 f(x)\,dx=\left(\int_0^\alpha+\int_\alpha^1\right) f(x)\,dx=e^\alpha-\alpha-\int_0^\alpha(e^x-1)f(x)\,dx. $$ For $0< x< \alpha$, we have $\alpha< f(x)< 1$, and thus $$ 1=e^\alpha-\alpha-\int_0^\alpha e^x\,dx<\int_0^1 f(x)\,dx<e^\alpha-\alpha-\int_0^\alpha \alpha e^x\,dx=(1-\alpha)e^\alpha+\alpha^2. $$ For $0< \alpha< 1$, the quantity $(1-\alpha)e^\alpha+\alpha^2$ takes a maximum value of $2+\log^2(2)-\log(4)$ at $\alpha=\log(2)$, so we have bounds $$ 1<\int_0^1 f(x)\,dx < 2+\log^2(2)-\log(4)\approx 1.0942. $$

These bounds are sharp: for $0\leq x\leq \log(2)$, we can take $f(x)$ to be an arbitrary continuous, strictly increasing function satisfying $f(0)=\log(2)$ and $f(\log(2))=1$, then extend $f$ to a continuous increasing function on the entire real line satisfying $f(f(x))=e^x$ by iteratively using the relation $f(x)=e^{f^{-1}(x)}$. To make $\int_0^1 f(x)\,dx$ arbitrarily close to the lower bound, we can choose $f(x)$ to be close to $1$ for an arbitrarily large portion of the interval $[0,\log(2)]$, and to make the integral arbitrarily close to the upper bound, we can choose $f(x)$ to be close to $\log(2)$ for an arbitrarily large portion of the interval $[0,\log(2)]$.

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  • $\begingroup$ The exact value seems to be 1.02525, so +1 :) $\endgroup$ – Oldboy May 16 '18 at 16:18
  • $\begingroup$ I suspect this is the same as OP’s answer, as he says the difference between bounds is 0.1. $\endgroup$ – Szeto May 16 '18 at 22:26
  • $\begingroup$ @JulianRosen Wowie zowie, this is exactly my answer (though I preferred $2\ln 2$ to $\ln 4$). This is so similar to my proof that you've essentially saved me the trouble of writing an answer (however, I do intend to demonstrate to everyone that these are the best possible bounds, unless you get around to it first). $\endgroup$ – Frpzzd May 18 '18 at 1:08
  • $\begingroup$ I have edited the answer to better explain the sharpness. $\endgroup$ – Julian Rosen May 18 '18 at 14:00
  • $\begingroup$ @JulianRosen Well done, and I so hope that you enjoyed the problem (as was the purpose of my posting it). $\endgroup$ – Frpzzd May 18 '18 at 22:54
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$\int_0^1f(x)dx=1.02525$, up to to five significant digits.

I tried to deal with this problem numerically by heavily using number crunching capabilities of Mathematica. I have simply supposed that:

$$f(x)\approx\sum\limits_{k=0}^nc_kx^k$$

Basically, you can pick $n$, calculate $f(f(x))$ and make a set of $n$ equations by matching the first $n$ coefficients of $x^k$ with the coefficients from Maclauirn's expansion of $e^x$. It's a highly non-lienar problem and I focused my efforts to find only one solution numerically. The trick is to use values $c_k$ obtained for $n-1$ as a starting point to calculate $c_k$ in the next iteration.

For $n=5$ I got something like:

$f(x)=0.503212 +0.884755 x+0.173988 x^2-0.0557584 x^3+0.17115 x^4+0.249233 x^5-0.159222 x^6-0.317672 x^7+0.172933 x^8+0.0335486 x^9$

...which is pretty bad:

enter image description here

Blue line represents $f(x)$, orange line represents $f(f(x))$ and green line represents $e^x$

But for $n=20$ I got something "almost" perfect:

$f(x)=-0.00110484 x^{19}+0.00423925 x^{18}-0.00285303 x^{17}-0.00653605 x^{16}+0.00430471 x^{15}+0.0114125 x^{14}-0.00188582 x^{13}-0.0167065 x^{12}-0.00616185 x^{11}+0.0170445 x^{10}+0.0165896 x^9-0.00860714 x^8-0.0207938 x^7-0.00356292 x^6+0.0140354 x^5+0.00731033 x^4+0.0207005 x^3+0.243846 x^2+0.876672 x+0.498799$

enter image description here

...with $\int_0^1f(x)dx=1.02526$

For $n=25$ I did not see much improvement.

I stopped at $n=30$. This is my final result:

$f(x)=3.19108553894\cdot 10^{-6} x^{29}-0.0000320371 x^{28}+0.000102301 x^{27}-0.0000852623 x^{26}-0.000155577 x^{25}+0.000202985 x^{24}+0.0003114 x^{23}-0.000304959 x^{22}-0.000659598 x^{21}+0.00024257 x^{20}+0.00120893 x^{19}+0.000250339 x^{18}-0.00172616 x^{17}-0.00138061 x^{16}+0.00165885 x^{15}+0.00293783 x^{14}-0.000399194 x^{13}-0.00403302 x^{12}-0.00203198 x^{11}+0.00346346 x^{10}+0.00438265 x^9-0.00094675 x^8-0.0047584 x^7-0.00170557 x^6+0.00287885 x^5+0.00131622 x^4+0.0240847 x^3+0.246667 x^2+0.876334 x+0.498618$

enter image description here

...with $\int_0^1f(x)dx=1.02525$

This value did not change even for $n=40$.

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  • $\begingroup$ I am not sure if the OP is looking fot this kind of answer. The OP looks for bounds, not a value, because the function needs not to be analytic. $\endgroup$ – Szeto May 16 '18 at 22:23
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    $\begingroup$ My bounds are $1.02525\pm0.00001$ :) $\endgroup$ – Oldboy May 17 '18 at 4:06
  • $\begingroup$ I really appreciated your tedious computation, but I think this function itself is paradoxical(see the important edit in my answer). Can you explain or resolve the paradox? $\endgroup$ – Szeto May 17 '18 at 15:04
  • $\begingroup$ I don't get your paradox. First you are stating that $f^{-1}(a)=b$ which basically means that $a=f(b)$ which is fine. But why is $\ln f(b)=f^{-1}(b)$? $\endgroup$ – Oldboy May 17 '18 at 20:17
  • $\begingroup$ because the fuction can be defined as $$f(x)=e^{f^{-1}(x)}$$. $\endgroup$ – Szeto May 17 '18 at 22:02
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IMPORTANT EDIT:

I find that $f(x)$ can create some paradoxical result.

Assume $0<a<1$(which is the region of interest) and $f^{-1}(a)=b$. Then $0<b<a$.

Then, $$a=f(b)\implies \ln a=\ln f(b)=f^{-1}(b)$$

Clearly, $L.H.S.\le0$. However, we have proved that $f^{-1}(b)=R.H.S.>0$.

Can anyone resolve it?

END EDIT

Not a tight bound, but too long for a comment.

Assume the $f(x)$ is:

  1. Continuous
  2. A real function on the interval $(-\infty,\infty)$
    1. Strictly increasing

Since $f(x)$ is strictly increasing, it has a well-defined inverse $f^{-1}(x)$.

Let’s redefine the function as $$f(x)=e^{f^{-1}(x)}$$

Firstly, since $e^x$ is positive for all real $x$, hence $f(x)$ is strictly positive.

Secondly, we are interested in the fixed point of $f(x)$. For fixed points $(t,f(t))$, $$t=f(t)= f^{-1}(t)$$

Thus, $$t=e^t$$ which has no real solution.

Therefore, we can see $y=f(x)$ has no intersection with $y=x$.

With these two observations, one can deduce $f(x)>x$ for all $x$.

$$f(x)>x\implies \int^1_0f(x)dx>\int^1_0xdx=\frac12$$

EDIT

From $$f(x)>x$$ we have $$f(f(x))>f(x)$$ $$e^x>f(x)$$ $$e-1>\int^1_0f(x)dx$$

EDIT 2:

I think that $f(0)$ cannot be defined(but maybe okay in the sense of limit).

Assume $f^{-1}(0)=C$.

It is easy to prove $f^{-1}(x)$ is strictly increasing.

Thus, from $f(x)>x$, we get $x>f^{-1}(x)$. So $C<0$.

We also observe that, due to $f(x)>0$, $f^{-1}(x)$ is undefined for $x<0$. Since $f(x)$ is defined wherever its inverse is defined, so $f(x)$ is undefined in $(0,-\infty)$.

$$f^{-1}(0)=C\implies f(C)=0$$

But $C<0$, $f(C)$ should be undefined! Also, I have proven that $f(C)>0$(above). That leads to a contradiction.

So a value of $C$ does not exist, and hence $f^{-1}(0)$ and $f(0)$ is undefined.

From the above observations, we can also see: $$0<f^{-1}(x)<x$$.

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  • $\begingroup$ I had this idea building on yours, to prove that $f(x) < e^x$. Suppose on the other hand that there was $x_0$ such that $f(x_0) \geq e^{x_0}$. Then $f(x_0) \geq e^{x_0} = f(f(x_0)) \geq f(e^{x_0}) \geq f(x_0) $ which means that $ f(f(x_0)) = f(e^{x_0}) = e^{x_0}$, a fixed point, which as you showed doesn't exist. Hence $f(x) < e^x$. There's a good chance I have made a mistake, but it provides a fairly good upper bound if it's right? It assumes that the function is strictly increasing again. $\endgroup$ – Cataline May 16 '18 at 11:17
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    $\begingroup$ @Cataline Yeah, I came up with this idea while bathing a few minutes ago(before seeing you comment). I’ve added a short, straight forward derivation of the upper bound. $\endgroup$ – Szeto May 16 '18 at 11:47
  • $\begingroup$ Indeed, your upper bounds are correct. But are they the best possible bounds? $\endgroup$ – Frpzzd May 16 '18 at 14:25
  • $\begingroup$ @Frpzzd I’ve never seen ways to prove bounds to be the best. I would certainly be amazed by your proof. $\endgroup$ – Szeto May 16 '18 at 14:36
  • $\begingroup$ @Szeto Well, I don't plan to post it until this gets at least a couple more answers. $\endgroup$ – Frpzzd May 16 '18 at 14:41

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