Developing methods for identifying the inflection point of a convex/concave curve

The Fisher-Pry sigmoid curve with total symmetry (not noisy)

Let’ s take the function: f(x) = 5 + 5 tanh(x − 5) after [3], which has p = 5, L = 10, x1 = 2.7024, x99 = 7.2976 and examine it at the interval [2, 8] in order to have data symmetry w.r.t. inflection point. The function is also symmetrical around inflection point, i.e. we have total symmetry.

From Corollary 1.1 of [1] we compute x_l = 5.970315941, x_r = 4.029684059, x_F1 = 3.850750196, x_F2 = 6.149249804, all inside [2, 8], thus all methods are theoretically applicable.

We first take n = 500 sub-intervals equal spaced without error just for checking our estimators. The results are presented at Table 1 of [1], while here we also present the BESE iterations done by ‘bese()’.

library(inflection)
data("table_01")
x=table_01$x
y=table_01$y
plot(x,y,cex=0.3,pch=19)
grid()
bb=ese(x,y,0);bb

##      j1  j2 chi
## ESE 170 332   5

pese=bb[,3];pese

## [1] 5

abline(v=pese)
cc=bese(x,y,0)
cc$iplast

## [1] 5

abline(v=cc$iplast,col='blue')

knitr::kable(cc$iters, caption = 'BESE')

We observe that χ_l = 5.9720, χ_r = 4.0280, χ_F1 = 3.8480, χ_F2 = 6.1520 are very close to the theoretically expected values, so we are on the results of Lemma 1.3 of [1]. The absolutely accuracy from the first apply of all methods confirms our theoretical analysis.

The Fisher-Pry sigmoid curve with total symmetry (noisy)

We next add the error term ϵ_i∼ U(−0.05, 0.05) via the process 14 of [1] and run our algorithms again.The results are presented at Table 2 of [1] and here we also present the BESE iterations done by ‘bese()’.

library(inflection)
data("table_02")
x=table_01$x
y=table_01$y
plot(x,y,cex=0.3,pch=19)
grid()
bb=ese(x,y,0);bb

##      j1  j2 chi
## ESE 170 332   5

pese=bb[,3];pese

## [1] 5

abline(v=pese)
cc=bese(x,y,0)
cc$iplast

## [1] 5

abline(v=cc$iplast,col='blue')

knitr::kable(cc$iters, caption = 'BESE')

The Fisher-Pry sigmoid curve with data left asymmetry (not noisy)

We continue with the same sigmoid function, but now we choose a proper [a, b] to show data asymmetry w.r.t. inflection point.

Let’ s take for example [4.2, 8]. If we do our theoretical computations we find x_l = 5.974322740, x_r = 4.029684059, x_F1 = 4.025677260, x_F2 = 5.974322740. We have that x_r < a, so χ_r has to estimate a = 4.2 and χ_S must be close to 4.703504993. Additionally, x_F1 < a, so χ_F1 must be also an estimation of a, thus χ_D must lie near the value 5.087161370. It’ s time to see if our theoretical predictions will be confirmed by experiment. We use for comparability the same Standard Partition as before and have the output presented at Table 3 and 4 of [1].

data("table_03_04")
x=table_03_04$x
y=table_03_04$y
tese=ese(x,y,0);tese

##     j1  j2    chi
## ESE  2 156 4.7928

pese=tese[,3]
tede=ede(x,y,0);tede

##     j1  j2    chi
## EDE  1 234 5.0854

pede=tede[,3]
cc=bese(x,y,0)
cc$iplast

## [1] 5.0018

dd=bede(x,y,0)
dd$iplast

## [1] 4.998

The Fisher-Pry sigmoid curve with data left asymmetry (noisy)

Let’ s add again an same error term ϵ_i∼ U(−0.05, 0.05) and run our algorithms. The results at Table 5 of [1] clearly are close enough to the theoretical expectations. Since ESE method did not estimate the inflection point with acceptable accuracy, after running BESE and BEDE iterative methods we find Table 6 of [1] which is a clear improvement of both estimations.

data("table_05_06")
x=table_05_06$x
y=table_05_06$y
tese=ese(x,y,0);tese

##     j1  j2  chi
## ESE  3 149 4.77

pese=tese[,3]
tede=ede(x,y,0);tede

##     j1  j2    chi
## EDE  3 231 5.0816

pede=tede[,3]
cc=bese(x,y,0)
cc$iplast

## [1] 5.036

dd=bede(x,y,0)
dd$iplast

## [1] 4.9828

plot(x,y,cex=0.3,pch=19)
grid()
abline(v=pese)
abline(v=cc$iplast,col='blue')
abline(v=dd$iplast,col='red')

knitr::kable(cc$iters, caption = 'BESE')

knitr::kable(dd$iters, caption = 'BEDE')

The Gompertz non symmetric sigmoid curve (not noisy)

Let’ s examine the function: f(x) = 10e^−e5e−x after [4], in the interval [3.5, 8]. It is easy to prove that f is (0.224, 1.0)-asymptotically symmetric around inflection point, so we can handle it similar to a symmetric sigmoid only for a distance of ±1 from p = 5.

We use, for comparison reasons, the same SP with 500 sub-intervals without error and obtain the Table 8 of [1] which is absolutely compatible with theoretical predictions. The ESE & EDE iterations are showed at Table 9 of [1] where we observe convergence to the real p for both two methods.

data("table_08_09")
x=table_08_09$x
y=table_08_09$y
tese=ese(x,y,0);tese

##     j1  j2   chi
## ESE 72 266 5.012

pese=tese[,3]
tede=ede(x,y,0);tede

##     j1  j2   chi
## EDE 67 311 5.192

pede=tede[,3]
cc=bese(x,y,0)
cc$iplast

## [1] 4.9985

dd=bede(x,y,0)
dd$iplast

## [1] 4.9985

The Gompertz non symmetric sigmoid curve (noisy)

We continue with our familiar SP by adding error uniformly distributed by U(−0.05, 0.05) and the results are given at Table 10 of [1] while ESE & EDE iterations are shown at Table 11 of [1].

data("table_10_11")
x=table_08_09$x
y=table_08_09$y
tese=ese(x,y,0);tese

##     j1  j2   chi
## ESE 72 266 5.012

pese=tese[,3]
tede=ede(x,y,0);tede

##     j1  j2   chi
## EDE 67 311 5.192

pede=tede[,3]
cc=bese(x,y,0)
cc$iplast

## [1] 4.9985

dd=bede(x,y,0)
dd$iplast

## [1] 4.9985

From these Tables we conclude that convergence to the true value of inflection point p = 5 occurs from the iterative application of ESE and EDE methods in one or two steps only.

A symmetric 3rd order polynomial with total symmetry

Let the polynomial function: $$f(x)=-\frac{1}{3}\,x^3+\frac{5}{2}\,x^2-4x+\frac{1}{2}$$ We study it at [-2, 7], it has inflection point at p = 2.5 and we have total symmetry. The SP with 500 sub-intervals without error gives Table 13 of [1] which is absolutely compatible with theoretical predictions. There is no need for any kind of iteration, because both methods agree with the true value.

data("table_13")
x=table_13$x
y=table_13$y
plot(x,y,cex=0.3,pch=19)
grid()
bb=ese(x,y,0);bb

##      j1  j2 chi
## ESE 126 376 2.5

pese=bb[,3];pese

## [1] 2.5

abline(v=pese)

The same SP with uniform error distributed by U(−2, 2) gives the results of Table 14 of [1] and two ESE iterations are presented at Table 15 of [1].

data("table_14_15")
x=table_14_15$x
y=table_14_15$y
plot(x,y,cex=0.3,pch=19)
grid()
bb=ese(x,y,0);bb

##      j1  j2   chi
## ESE 115 375 2.392

pese=bb[,3];pese

## [1] 2.392

abline(v=pese)
cc=bese(x,y,0)
cc$iplast

## [1] 2.473

abline(v=cc$iplast,col='blue')

knitr::kable(cc$iters, caption = 'BESE')

A symmetric 3rd order polynomial with data right asymmetry

For the same symmetric 3rd order polynomial as above we change the interval to [-2, 8], thus we have data right asymmetry now. The case of SP with 500 sub-intervals and no error gives Table 17 of [1], while ESE and EDE iterations are presented at Table 18 of [1]. First results are absolutely compatible with theoretical predictions for ESE method.

data("table_17_18")
x=table_17_18$x
y=table_17_18$y
bb=ese(x,y,0);bb

##     j1  j2  chi
## ESE 88 338 2.24

pese=bb[,3];pese

## [1] 2.24

plot(x,y,cex=0.3,pch=19)
grid()
cc=bese(x,y,0)
cc$iplast

## [1] 2.5

dd=bede(x,y,0)
dd$iplast

## [1] 2.5

abline(v=pese)
abline(v=cc$iplast,col='blue')
abline(v=dd$iplast,col='red')

knitr::kable(cc$iters, caption = 'BESE')

knitr::kable(dd$iters, caption = 'BEDE')

We add uniform error distributed by U(-2, 2) and we have the results of Table 19 of [1], while one ESE & one EDE iteration are given at Table 20 of [1].

data("table_19_20")
x=table_19_20$x
y=table_19_20$y
bb=ese(x,y,0);bb

##     j1  j2  chi
## ESE 88 338 2.24

pese=bb[,3];pese

## [1] 2.24

plot(x,y,cex=0.3,pch=19)
grid()
cc=bese(x,y,0)
cc$iplast

## [1] 2.65

dd=bede(x,y,0)
dd$iplast

## [1] 2.35

abline(v=pese)
abline(v=cc$iplast,col='blue')
abline(v=dd$iplast,col='red')

knitr::kable(cc$iters, caption = 'BESE')

knitr::kable(dd$iters, caption = 'BEDE')

There exist a problem here. Although we have a symmetric polyno- mial, the TESE is not equal to the true inflection point. A remedy for this problem for the class of 3rd order polynomials is given with Lemma 2.1 of [1]. Lets apply it here. We have that a = −2, b = 8 and from Table 19 of [1] is χ_r = −0.26, χ_l = 4.74, so we have that: $$\hat{p}=\frac{1}{3}\,\chi_{l} + \frac{1}{3}\,\chi_{r}+\frac{1}{6}\,a+\frac{1}{6}\,b=2.493333333$$ which is much closer to the true value of 2.5.

Please send your comments, suggestions or bugs found to [email protected]

References

[1] Demetris T. Christopoulos (2014), Developing methods for identifying the inflection point of a convex/concave curve. arXiv:1206.5478v2 [math.NA]. URL: https://doi.org/10.48550/arXiv.1206.5478

[2] Demetris T. Christopoulos (2016), On the Efficient Identification of an Inflection Point, International Journal of Mathematics and Scientific Computing , Volume 6 (1), June 2016, Pages 13-20, ISSN: 2231-5330. URL: https://veltech.edu.in/wp-content/uploads/2016/04/Paper-04-2016.pdf

[3] J.C. Fisher and R.H. Pry (1971), A Simple Substitution Model of Technological Change, Technological Forecasting and Social Change, 3, pp. 5–88. URL: https://doi.org/10.1016/S0040-1625(71)80005-7

[4] B. Gompertz (1825), On the Nature of the Function Expressive of the Law of Human Mortality, and on a New Mode of Determining the Value of Life Contingencies, Philosophical Transactions of the Royal Society of London, 115, pp. 513–585.