Half iteration of exponential function

Question

I'm working on the half iteration of the exponential function. No one has any idea what fractional iterations could mean but I think intuitively it should be a function $f(x)$ such that $f(f(x))=e^x$.

Here's how I'm finding $f(x)$ when $x\approx 0$:

If $x\approx 0$, then, we have, $$e^x\approx 1+x+\frac{x^2}{2}$$. ...(1)

Now, if we assume the required function $f(x)$ to be of the form $ax^2+bx+c$, then $$f(f(x))= a^3x^4+2a^2bx^3+(2a^2c+ab^2+ab)x^2+(2abc+b^2)x+ac^2+bc+c$$

But, since $x\approx 0$ therefore,

$$f(f(x))=e^x\approx ac^2+bc+c+(2abc+b^2)x+(2a^2c+ab^2+ab)x^2$$. ....(2)

Comparing coefficients of like powers of $x$ in equation (1) and (2), we get,

$$ac^2+bc+c=1 \tag {3.1}$$ $$2abc+b^2=1 \tag {3.2}$$ $$2a^2c+ab^2+ab=\frac{1}{2} \tag {3.3}$$

The problem is solving these equations. I've tried substitution but they get reduced to a polynomial of very high degree which I don't know how to solve. Is there some way to solve these to get $a$, $b$, and $c$ and hence get the required half iteration function of $e^x$ as $ax^2+bx+c$? Please tell me how to solve these three equations.

Answer 1

There is a method using Carlemanmatrices which gives increasingly well approximations.
Consider the family of polynomials $$g_t(x) = 1+ x + \frac{x^2}{2!} +...+ \frac{x^t}{t!} $$ with the attempt to find increasingly precise approximations with increasing $t$ $$ f_t(f_t(x)) \approx g_t(x) \approx e^x $$
For some chosen $t$ define the Carlemanmatrix $G$ for $g_t(x)$ (I use a version which is transposed against the wikipedia-entry), for instance $t=2$ $$ G_2 = \left[\small \begin{array} {} 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 \\ 0 & 1 & 2 & 3 & 4 & 5 & 6 & 7 \\ 0 & 1/2 & 2 & 9/2 & 8 & 25/2 & 18 & 49/2 \\ 0 & 0 & 1 & 4 & 10 & 20 & 35 & 56 \\ 0 & 0 & 1/4 & 9/4 & 17/2 & 45/2 & 195/4 & 371/4 \\ 0 & 0 & 0 & 3/4 & 5 & 37/2 & 51 & 469/4 \\ 0 & 0 & 0 & 1/8 & 2 & 45/4 & 41 & 931/8 \\ 0 & 0 & 0 & 0 & 1/2 & 5 & 51/2 & 92 \end{array}\right]$$ We see the coefficients of our function $g_2(x)$ in the second column, that of $g_2(x)^0 = 1$ in the fist columns, that of $g_2(x)^2 $ in the third , that of $g_2(x)^3 $ in the fourth column and so on. The key is here, that with a vector $V(x)=[1,x,x^2,x^3,...]$ up to the correct dimension we can do the dotproduct $$ V(x) \cdot G_2 = V(g_2(x))$$ Actually, when we using software then the columns of higher indexes are always truncated versions of the powers of $g_2(x)$ so empirically we must take the approximations with a grain of salt.
THe key is now, that because of the form of the Carlemanmatrices, the "output" has the same form as the "input", and we can repeat the application like $$ V(x) \cdot G_2 = V(g_2(x)) \\ V(g_2(x)) \cdot G_2 = V(g_2°^2(x)) \\ V(g_2°^2(x)) \cdot G_2 = V(g_2°^3(x)) \\ $$ or more concisely, because we have associativity in the matrix-products $$V(x) \cdot G_2^h = V(g_2°^h(x)) $$ We see, that the h'th power of $G_2$ gives the h'th iterate of $g_2(x)$ and we can assume, that inserting $h=1/2$ gives -at least- an approximation to $g_2°^{1/2}(x)=f(x)$
What we need is a matrix-function for finding the square-root; this can either be done by diagonalization (implemented in Pari/GP and larger CAS-systems) of by Newton-iteration.
What I find for $G_2^{0.5} $ is

  1.0000000      0.49649737      0.24650274       0.12207723   0.060459565  0.030432921  0.015814772  0.0088358869
          0      0.88272304      0.87964476       0.65278437    0.42951272   0.26484713   0.15906693   0.097969211
          0      0.29626378       1.0688901        1.3895246     1.2985405    1.0259175   0.73517218    0.50088082
          0    -0.073304928      0.44254563        1.4092345     2.1282729    2.3110280    2.0617264     1.6127751
          0     0.020386654   -0.0089884907       0.62013806     1.9423663    3.2482288    3.8802221     3.6776250
          0   -0.0023714159   -0.0087696684      0.062271773    0.87615938    2.7873325    5.0295289     6.2944776
          0  -0.00064038818    0.0039322053    -0.0081032492    0.14270358    1.2522409    4.3026809     8.3339822
          0   0.00019462828  -0.00056495459  -0.000028314112  0.0028871718   0.23073290    1.8777280     8.6538082

and we see, that the coefficients in the second column gives some approximation to Sheldon's halfe - function.
Also the first three coefficients $(c=) 0.49649737 , (b=) 0.88272304, (a=) 0.29626378 $ give some approximation to the values of $(c,b,a)$ which I gave in my earlier comment and which solve your system of equations in (3.1) to (3.2).

Now if we do a better approximation of $g_t(x) $ to the true $\exp(x)$ - function, by, say, $t=8$ we get better approximations to Sheldon's Kneser-solution. Let $G_8$ be defined with size 16x16 then we get its top left:

  1       1           1               1               1                  1
  0       1           2               3               4                  5
  0     1/2           2             9/2               8               25/2
  0     1/6         4/3             9/2            32/3              125/6
  0    1/24         2/3            27/8            32/3             625/24
  0   1/120        4/15           81/40          128/15             625/24
  0   1/720        4/45           81/80          256/45           3125/144
  0  1/5040       8/315         243/560        1024/315         15625/1008
  0       0   127/20160       1093/6720       5461/3360         19531/2016
  0       0    41/30240      3271/60480       5459/7560         32549/6048
  0       0    19/75600     3247/201600     21809/75600       162697/60480
  0       0     1/25200      871/201600      3953/37800        14779/12096
  0       0  19/3628800    2533/2419200   62843/1814400        24589/48384
  0       0   1/1814400       37/161280     2389/226800       47129/241920
  0       0  1/25401600   1537/33868800    7499/2540160    702839/10160640
  0       0           0  4099/508032000  72803/95256000  3481427/152409600

and the square-root $G_8^{0.5} $ is

  1.0000000            0.49857405            0.24856073           0.12386855         0.061729127      0.030790812
          0            0.87630311            0.87401746           0.65347137          0.43393924       0.27003773
          0            0.24751412             1.0147132            1.3342705           1.2683840        1.0263147
          0           0.024641942            0.45793460            1.3404901           2.0049681        2.2203057
          0        -0.00094891787            0.10348608           0.72416179           1.8842050        3.0275836
          0         0.00022657746           0.010924205           0.23298583           1.1130489        2.7397893
          0        0.000077390020         0.00063916208          0.046228369          0.43595415        1.7059468
          0       -0.000032210830         0.00014299601         0.0059529659          0.11748258       0.75561802
          0       -0.000014796522       -0.000021914892        0.00061375239         0.022696178       0.24536210
          0        0.000011471527       -0.000021337036       0.000018742074        0.0033435423      0.060458407
          0      -0.0000014276569       0.0000089535786      -0.000019453706       0.00035980043      0.011807188
          0     -0.00000088358673      0.00000019051670      0.0000042134686      0.000016547312     0.0018917104
          0      0.00000031550433     -0.00000099866936     0.00000031495830     0.0000035791002    0.00026146045
          0  -0.00000000068329542      0.00000026181266    -0.00000053463876   -0.00000019767557   0.000038804566
          0    -0.000000014747905    -0.000000019088871     0.00000011141548  -0.000000094870895  0.0000061666156
          0    0.0000000019555565  -0.00000000099414503  -0.0000000090606767   0.000000019074856  0.0000014943166

The coefficients in the second column are now better approximations to Sheldon's solutions and give a better solution for $f(x)$ to approximate $f(f(x))\approx \exp(x)$

You see the principle. Ideally the Carlemanmatrix is of infinite size and also the polynomial is of infinite order (or better: equals the exponential-series).

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By the logic of the Carleman-matrices the following method seems to be more inaccurate, but the approximation pattern towards the Kneser-solution seems to be even better.

Here is a list for the coefficients for $f_t(x)$ for $t=3..16$ where the Carlemanmatrices $G_t$ are also truncated to size $t \times t$ (and not $2t \times 2t$, $3t \times 3t$ or the like). I've written them horizontally for better visual comparability of approximation towards the Kneser-solution:

 t      at x^0      at x^1         x^2           x^3           x^4             x^5
 3      0.50000000  0.89442719  0.22360680            .               .               .
 4      0.49944144  0.88075164  0.23809540  0.022538920               .               .
 5      0.49907754  0.87768412  0.24309637  0.024235749   0.00082874617               .
 6      0.49887415  0.87676479  0.24517938  0.024772649   0.00013110906  0.000089433779
 7      0.49875947  0.87644601  0.24618449  0.024898334  -0.00030603063   0.00011382794
 8      0.49869216  0.87632858  0.24671872  0.024893887  -0.00055911402   0.00013292704
 9      0.49865090  0.87628661  0.24702315  0.024852593  -0.00070934134   0.00015154857
10      0.49862460  0.87627466  0.24720567  0.024805640  -0.00080017819   0.00016953631
11      0.49860726  0.87627481  0.24731954  0.024763154  -0.00085630807   0.00018510631
12      0.49859549  0.87627959  0.24739292  0.024727523  -0.00089167888   0.00019807499
13      0.49858729  0.87628581  0.24744148  0.024698565  -0.00091432971   0.00020860446
14      0.49858146  0.87629209  0.24747433  0.024675319  -0.00092902883   0.00021706224
15      0.49857724  0.87629790  0.24749699  0.024656725  -0.00093865651   0.00022382486
16      0.49857412  0.87630304  0.24751287  0.024641836  -0.00094499831   0.00022922925

Kneser

KN:     0.49856329  0.87633613  0.24755219  0.024571812  -0.00095213638   0.00025333982 ...

So I think, the Kneser-solution is the limit for this process when $t \to \infty$

Answer 2

Here is a short problem I proposed two years ago to my students (translated from french ...)

I can add later a detailed solution if required.

I originally found this material somewhere on the web but don't remember where.

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Let's make the assumption that there exists a continuous map $f:\mathbb{R}\rightarrow\mathbb{R}$ such that $f\circ f=\exp.$

We will prove, step by step, several properties of $f$.

1. Prove that $\forall x\in\mathbb{R},\thinspace e^{x}>x.$
2. Prove that the equation $f\left(x\right)=x$ doesn't have any solution.
3. Deduce that $\forall x\in\mathbb{R},\thinspace f\left(x\right)>x.$

4. Prove finally that $\forall x\in\mathbb{R},\thinspace f\left(x\right)

5. Prove that $\forall x\in\mathbb{R},\thinspace f\left(e^{x}\right)=e^{f\left(x\right)}.$

6. Compute ${\displaystyle \lim_{+\infty}f}.$

7. Prove that there exists $\lambda<0$ such that ${\displaystyle \lim_{-\infty}f=\lambda.}$
8. Prove that $f$ is strictly increasing.

Answer 3

I've written a few programs that calculate Kneser's sexp and slog functions. One of my recent efforts is fatou.gp a pari-gp program that calculates the slog or Abel function for $\exp(x)$, for a wide range of real and complex bases. Kneser's construction requires calculating a Riemann mapping which is very tricky to get accurate numerical results, so instead I do an equivalent iterated 1-cyclic $\theta(z)$ mapping from the $\alpha(z)$ or Abel functions at the two fixed points. Here is the Taylor series for the half iterate of $\exp(x)$. You can compare these results with your system of equations. $$f(f(x))=\exp(x);\;\;\;f(x)=\text{sexp}(\text{slog}(x)+0.5)$$

{halfe= 0.498563287941114
+x^ 1*  0.876336132224813
+x^ 2*  0.247552187310898
+x^ 3*  0.0245718116969028
+x^ 4* -0.000952136380204206
+x^ 5*  0.000253339819008525
+x^ 6*  7.09275516366956 E-5
+x^ 7* -4.81808433402200 E-5
+x^ 8*  2.63228465405932 E-6
+x^ 9*  5.96598826774286 E-6
+x^10* -1.30879479719986 E-6
+x^11* -7.47165552015529 E-7
+x^12*  2.68510892327235 E-7
+x^13*  1.12440534247329 E-7
+x^14* -4.80789869461353 E-8
+x^15* -2.20118629742874 E-8
+x^16*  8.17933994010676 E-9
+x^17*  5.30688749879415 E-9
+x^18* -1.23819700193839 E-9
+x^19* -1.41844961463076 E-9
+x^20*  1.05287927108075 E-10
+x^21*  3.89632939104118 E-10
+x^22*  3.51707444355649 E-11
+x^23* -1.04753098725701 E-10
+x^24* -2.89321996209624 E-11
+x^25*  2.62480364845324 E-11
+x^26*  1.38848625050719 E-11
+x^27* -5.58405052292307 E-12
+x^28* -5.57754465436342 E-12
+x^29*  6.94355445214551 E-13
+x^30*  2.00345639417985 E-12
+x^31*  1.89932160713883 E-13
+x^32* -6.47541759020915 E-13
+x^33* -2.08549856161407 E-13
+x^34*  1.82048262120683 E-13
+x^35*  1.15807785730247 E-13
+x^36* -3.92069387382815 E-14
+x^37* -5.14573309824226 E-14
+x^38*  2.35454506494555 E-15
+x^39*  1.97495294136643 E-14
+x^40*  3.87021261821085 E-15 }

Answer 4

With a second attempt I can now solve your problem to arbitrary precision, and can even generalize to polynomials up to order 7.
The problem you state is to find a polynomial $f(x)=a + bx + cx^2$ which when iterated $f(f(x))$ gives $$f(f(x)) = g(x)=1 + 1x + 1 x^2/2! + O(x^3) \approx \exp(x) \tag 1$$
Expanding $f(f(x))$ and collecting like powers of $x$ gives this set of equations: $$ \begin{array}{} 1 &= & a + ab + a^2c & (2.1)\\ 1 &= & 2abc + b^2 & (2.2) \\ 1/2! &= & b^2c + bc + 2ac^2 & (2.3)\\ \hline ?? &= & 2bc^2 & (2.4)\\ ?? &= & c^3 & (2.5)\\ \end{array}$$ Equations (2.4) and (2.5) occur by the iteration $(f(f(x))$ but are ignored here.
Using (2.2) one can express $b$ by $a$ and $c$; using (2.1) one can express then $c$ by $a$. Some terms in the set of rearrangements when solving this show, that $a$ must be smaller than $0.5$ and some trial and error show that a bisection solver using $a$ in the interval $0.45 \lt a \lt 0.5$ can find the value for $a$ such that equations (2.1) to (2.3) are satisfied.
I got the solving coefficients $$a=0.49789408...\\ b=0.87811291...\\ c=0.26179546...$$ Of course the approximations of $f(f(x))$ to $\exp(x)$ are only meaningful for $x$ in small intervals around zero, see the table of examples: $$\small \begin{array} {r|rr|r} x & f(x) & f(f(x)) & \exp(x) \\ \hline -1/2 & 0.12428649 & 0.61107564 & 0.60653066 \\ -1/4 & 0.29472807 & 0.77943937 & 0.77880078 \\ -1/8 & 0.39222052 & 0.88258179 & 0.88249690 \\ 0 & 0.49789408 & 1 & 1 \\ 1/8 & 0.61174875 & 1.1330520 & 1.1331485 \\ 1/4 & 0.73378452 & 1.2832008 & 1.2840254 \\ 1/2 & 1.0023994 & 1.6411672 & 1.6487213 \end{array}$$

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Generalization
I was not able to find similar analytical reductions for examples with polynomials with higher degree and had to switch to a process of iterations very similar to the Newton-iteration for finding the squareroot - but which I applied to truncated Carlemanmatrices.
Let's denote $F$ for the (truncated) Carlemanmatrix for $f(x)$ and $G$ for that for $g(x)$ where $g(x)=f(f(x))$ . Because we work with truncated Carlemanmatrices it is not exactly $F^2 = G$ - either $ F^2 = \hat G$ or $ \hat F^2 = G$ where either $\hat G$ or $\hat F$ is not of the Carleman type. In my papers in the tetration-forum I always used the version $ \hat F^2 = G$ which means, the matrix associated to $f(x)$ is not really Carleman.
Here however the polynomial descriptions in (2.1) to (2.3) or generalized to higher degrees force the solution $F^2 = \hat G$ where $\hat G$ is allowed to be non-Carleman, and only the second column contain the leading coefficients $(1,1,1/2!,1/3!,...)$ from the truncated exponential-series.

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Example
For polynomial degree 3, so $g_3(x)=1 + x + x^2/2 + x^3/6 + O(x^4)$ and $f_3(x)=a + bx + cx^2 + dx^3 $ and $$F = \small \left[ \begin{array} {rrrr} 1 & a & a^2 & a^3 \\ 0 & b & 2ab & 3ba^2 \\ 0 & c & 2ac+b^2 & 3a^2c+3ab^2 \\ 0 & d & 2bc+2ad & 6abc+3a^2d+b^3 \end{array} \right]$$ with the idea $F^2 = \hat G$ and actually using only the second column of $\hat G$ which reduces the notation to $$F \cdot F_{0..3,1} = \hat G_{0..3,1} = \left[1,1,\frac12,\frac16\right]$$ In a matrix-multiplication-scheme this looks like $$ \begin{array}{r|c} \begin{array}{lr} F \cdot F_{0..3,1} = \hat G_{0..3,1} \qquad \qquad & \\ & \times \end{array} & \small \begin{bmatrix} a \\b \\ c \\ d \end{bmatrix}\\ \hline \small \left[ \begin{array} {rrrr} 1 & a & a^2 & a^3 \\ 0 & b & 2ab & 3ba^2 \\ 0 & c & 2ac+b^2 & 3a^2c+3ab^2 \\ 0 & d & 2bc+2ad & 6abc+3a^2d+b^3 \end{array} \right] & \small \begin{bmatrix} 1 \\1 \\ 1/2 \\ 1/6 \end{bmatrix} \end{array}$$

The task is to find the matrix $F$ which solves that equation.
I used the Newton-algorithm for finding the square-root, which goes in general this way
1) Initialize $F$ , for instance with the identity matrix
2) iterate $F = ( G\cdot F^{-1} + F) / 2$ until convergence

But this shall not make sure that $F$ becomes a Carlemanmatrix, so we need to implement a matrix-function $\text{Carl}()$which makes a (truncated) Carlemanmatrix from a set of coefficients in a column-vector. Moreover, we have not the full matrix $G$ as target, but only one column in $\hat G$ so we must also rewrite the second step

Thus we go

1) Initialize $F = \text{Carl}([0.49,0.88,0.23,0.1])$ with some values from our solution for the earlier three-term solution $(a,b,c)$
Initialize $Z = \hat G_{0..3,1}$ for notational convenience

2) iterate
$\qquad \qquad$ $ \displaystyle (2.1) \qquad T=F_{0..3,1} \\ (2.2) \qquad T= (F^{-1} \cdot Z + T)/2 \\ (2.3) \qquad F = \text{Carl}(T) $
$\hspace{160px}$ until convergence.

For higher degree of the polynomials to avoid divergences in step (2.2) I introduced a weighting $w >1 $ such that
$\qquad \qquad$ $ \displaystyle (2.2) \qquad T= (w \cdot F^{-1} \cdot Z + T)/(w+1) $

For degree 3 it is already better to use $w=3$ and for degree 7 I needed about $w=30$

Here is a table of coefficents computed for polynomial degrees $t=3$ to $t=7$

deg at x^0      at x^1      at x^2      at x^3       at x^4          at x^5         at x^6          at x^7        
-----------------------------------------------------------------------------------------------------------------------
 2  0.49789408  0.87811291  0.26179546            .               .              .               .
 3  0.49857386  0.87638317  0.24739136  0.024116727               .              .               .
 4  0.49856631  0.87631901  0.24749709  0.024635189  -0.00068603672              .               .
 5  0.49856360  0.87633476  0.24754665  0.024575891  -0.00093201041  0.00026544139               .
 6  0.49856335  0.87633377  0.24755048  0.024583631  -0.00093843971  0.00023767204  0.000024982253
 7  0.49856319  0.87633587  0.24755370  0.024574086  -0.00095506183  0.00024635371  0.000070926074  -0.000035173120
======= Kneser ========================================================================================================
 K  0.49856329  0.87633613  0.24755219  0.024571812  -0.00095213638  0.00025333982  0.000070927552  -0.000048180843 ...

Here is a table of numerical evaluations of $f_7(x)$ for small $x$:

    x    f(x)         f(f(x))      exp(x)
  -----------------------------------------
    -1  -0.15588838  0.36787478  0.36787944
  -1/2   0.11914591  0.60653061  0.60653066
  -1/4   0.29456341  0.77880078  0.77880078
  -1/8   0.39284100  0.88249690  0.88249690
     0   0.49856319   1.0000000   1.0000000
   1/8   0.61202097   1.1331485   1.1331485
   1/4   0.73349976   1.2840254   1.2840254
   1/2    1.0016402   1.6487211   1.6487213
     1    1.6463539   2.7181939   2.7182818

The polynomial of degree 7 allows already 4 digits precision in the interval $-1 \le x \le 1$ and 7 digits in the interval $-0.5 \le x \le 0.5$

Additional remark : this is better than what I got using my to-date standard method of "polynomial interpolation" for tetration with 8 or 16 degree polynomials which emploays the ansatz $\hat F^2 = G$ .

Answer 5

Looking at the equations $$ac^2+bc+c=1 \tag 1$$ $$2abc+b^2=1 \tag 2$$ $$2a^2c+ab^2+ab=\frac{1}{2} \tag 3$$ we can eliminate $b$ from $(1)$ $$b=\frac{1-c-a c^2}{c}$$ Replacing in $(2)$ and solving for $a$ leads to $$a=\frac{\sqrt{1-2 c}}{c^2}$$ Replacing in $(3)$ leads to $$-c \left(c^3+12 c+6 \sqrt{1-2 c}-14\right)+4 \sqrt{1-2 c}=4$$ After squaring steps, this reduces to $$c^7+24c^5-28c^4+152c^3-264c^2+160c-32=0$$ which has only one real root close to $c=\frac 12$.

Using Newton method for finding the zero of the equation in $c^7$ leads to $$a=0.261795456735753$$ $$b=0.878112905194437$$ $$c=0.497894079064888$$ as already given in Gottfried Helms's answer. These numbers can be rationalized as $$a=\frac{37409}{142894}\qquad b=\frac{77821}{88623}\qquad c=\frac{18323}{36801}$$

Edit

Back to the problem eighteen months later, we could get expressions for the solution using $[1,n]$ Padé approximants for the septic equation (built around $c=\frac 12$). For different values of $n$, this would give $$a_1=\frac{146336 \sqrt{352121}}{331786225}\qquad b_1=\frac{18369-4 \sqrt{352121}}{18215}\qquad c_1=\frac{18215}{36584}$$ $$a_2=\frac{3257213 \sqrt{44685705147}}{2630063332009}\qquad b_2=\frac{1635466-\sqrt{44685705147}}{1621747}\qquad c_2=\frac{1621747}{3257213}$$