# Find inflection points: Mission Impossible!" In inflection: Finds the Inflection Point of a Curve

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r
knitr::opts_chunk$set(echo = TRUE) options(max.width = 1000) options(max.print = 100000)  ## Can we always find the inflection point? Let’ s take the function: $$f (x) = \frac{\frac{1}{8}x+\frac{1}{2}x^2}{1+\frac{1}{8}x+\frac{1}{2}x^2}$$ following . What are the inflection points? We must take the first derivative: $$f'(x)= \frac {8+64\,x}{ \left( 4\,{x}^{2}+x+8 \right) ^{2}}$$ and continue to second derivative: $$f''(x)= \frac {-768\,{x}^{2}-192\,x+496}{ \left( 4\,{x}^{2}+x+8 \right) ^{3}}$$ Now we must solve above expression for x and find two solutions: $$x_1=-1/8-1/24\,\sqrt {381}=-0.9383008876$$ $$x_2=-1/8+1/24\,\sqrt {381}=0.6883008876$$ If we study the function we'll see that has a minimum at$x_{min}=-\frac{1}{8}=-0.125$and it is concave/convex at interval$xx_{min}$. But what if we do not want to perform such an analytical approach? What if we just wanted the inflection points without more details? Then we must follow next steps: • define an interval that encloses an inflection point • run ESE or EDE to find a first approximation • choose an interval that encloses inflection with your desired accuracy • run BESE to find the closest available approximation Here are the steps for the positive inflection point: library(inflection) f=function(x){(1/8*x+1/2*x^2)/(1+1/8*x+1/2*x^2)} x=seq(0,5,0.1) y=f(x) # First approximation cc=check_curve(x,y);cc ese(x,y,cc$index)
# New interval with equal distances from first approximation
x=seq(0,1.3,0.001)
y=f(x)
# Second approximation
cc=check_curve(x,y);cc
ipbese=bese(x,y,cc$index) ipbese$iplast
plot(x,y,pch=19,cex=0.1)
abline(v=ipbese$iplast,col='blue')  knitr::kable(ipbese$iters, caption = 'BESE')


Since our interval was defined with equal size step 0.001 we expect our estimation to have the same accuracy and indeed it is truth.

If we wanted a better accuracy, for example with 4 digits, then we could define another interval with previous approximation inside it and with equal size stpe 0.0001:

# x=seq(0.68,0.69,0.0001)
# y=f(x)
# cc=check_curve(x,y);cc
# ipbese=bese(x,y,cc$index,doparallel = TRUE) # ipbese$iplast
# #  0.6883
# ipbese$iters # n a b ESE ## 1 101 0.6800 0.6900 0.68870 ## 2 25 0.6876 0.6887 0.68815 ## 3 12 0.6881 0.6886 0.68835 ## 4 6 0.6882 0.6884 0.68830  ## What if our data set is noisy? There exist no problem. Since ESE and EDE methods both give consistent estimators for the inflection point, we do not need to worry about noise. Lets perform the same task with error added: x=seq(0.0,3.,0.001) set.seed(20190628) y=f(x)+runif(length(x),-0.01,0.01) cc=check_curve(x,y) cc ipbese=bese(x,y,cc$index)
ipbese$iplast plot(x,y,pch=19,cex=0.1) abline(v=ipbese$iplast,col='blue')

knitr::kable(ipbese$iters, caption = 'BESE')  ## What about big data? If your data set gas million of cases, then EDE and BEDE are your best friends. Lets see why by using 1000001 cases from a function with inflection point equal to 500: f=function(x){500+500*tanh(x-500)} x=seq(0,1000,0.001) y=f(x) length(x) t1=Sys.time();ede(x,y,0);t2=Sys.time();t2-t1 ipbede=bede(x,y,cc$index)
ipbede$iplast  knitr::kable(ipbede$iters, caption = 'BEDE')


Situation will not change if we add error $U(-50,50)$:

f=function(x){500+500*tanh(x-500)}
x=seq(0,1000,0.001)
set.seed(20190628)
y=f(x)+runif(length(x),-50,50)
length(x)
t1=Sys.time();ede(x,y,0);t2=Sys.time();t2-t1
ipbede=bede(x,y,0)
ipbede$iplast plot(x[495000:505000],y[495000:505000],xlab="x",ylab="y",pch='.') abline(v=ipbede$iplast)

knitr::kable(ipbede$iters, caption = 'BEDE')  Even if our data are not symmetric, estimation will not change: f=function(x){500+500*tanh(x-500)} x=seq(0,700,0.001) set.seed(20190628) y=f(x)+runif(length(x),-50,50) length(x) t1=Sys.time();ede(x,y,0);t2=Sys.time();t2-t1 ipbede=bede(x,y,0) ipbede$iplast
plot(x[495000:505000],y[495000:505000],xlab="x",ylab="y",pch='.')
abline(v=ipbede$iplast)  knitr::kable(ipbede$iters, caption = 'BEDE')


Just try to solve above problems by using regression or other techniques that involve matrix inversion...

OK, we'll perform that task for you.

# library(nlme)
# x=seq(0,5,0.1)
# f=function(x){(1/8*x+1/2*x^2)/(1+1/8*x+1/2*x^2)}
# y=f(x)
# df=data.frame("x"=x,"y"=y)
# Asym=1;xmid=1;scal=1;
# fmla=as.formula("y~SSlogis(x,Asym,xmid,scal)");fmla
# try(nls(fmla,df))
# est=try(nls(fmla,df))
# coef(est)


When it works it gives the result xmid=1.2516 which is away from true inflcetion (0.6883008876), altough all results seem to be statistically 'super':

# y ~ SSlogis(x, Asym, xmid, scal)
# Nonlinear regression model
#   model: y ~ SSlogis(x, Asym, xmid, scal)
#    data: df
#   Asym   xmid   scal
# 0.8909 1.2516 0.5478
#  residual sum-of-squares: 0.05242
#
# Number of iterations to convergence: 0
# Achieved convergence tolerance: 1.401e-06
###
# est=try(nls(fmla,df))
# coef(est)
#      Asym      xmid      scal
# 0.8908913 1.2516410 0.5478263
# summary(est)
#
# Formula: y ~ SSlogis(x, Asym, xmid, scal)
#
# Parameters:
#      Estimate Std. Error t value Pr(>|t|)
# Asym 0.890891   0.007872  113.18   <2e-16 ***
# xmid 1.251641   0.025638   48.82   <2e-16 ***
# scal 0.547826   0.023718   23.10   <2e-16 ***
# ---
# Signif. codes:  0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1
#
# Residual standard error: 0.03305 on 48 degrees of freedom
#
# Number of iterations to convergence: 0
# Achieved convergence tolerance: 1.401e-06


# x=seq(0.0,3.,0.001)
# f=function(x){(1/8*x+1/2*x^2)/(1+1/8*x+1/2*x^2)}
# set.seed(20190628)
# y=f(x)+runif(length(x),-0.01,0.01)
# df=data.frame("x"=x,"y"=y)
# Asym=1;xmid=1;scal=1;
# fmla=as.formula("y~SSlogis(x,Asym,xmid,scal)");fmla
# try(nls(fmla,df))


When it works it gives next result xmid=1.0938 which is away from true inflcetion (0.6883008876) and again rsults seem to be 'excellent':

# Nonlinear regression model
# model: y ~ SSlogis(x, Asym, xmid, scal)
# data: df
# Asym   xmid   scal
# 0.8018 1.0938 0.4318
# residual sum-of-squares: 1.884
#
# Number of iterations to convergence: 0
# Achieved convergence tolerance: 1.768e-07
###
# est=try(nls(fmla,df))
# coef(est)
#      Asym      xmid      scal
# 0.8018405 1.0937758 0.4318175
# summary(est)
#
# Formula: y ~ SSlogis(x, Asym, xmid, scal)
#
# Parameters:
#      Estimate Std. Error t value Pr(>|t|)
# Asym 0.801840   0.001175   682.7   <2e-16 ***
# xmid 1.093776   0.002417   452.5   <2e-16 ***
# scal 0.431817   0.002063   209.3   <2e-16 ***
# ---
# Signif. codes:  0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1
#
# Residual standard error: 0.02507 on 2998 degrees of freedom
#
# Number of iterations to convergence: 0
# Achieved convergence tolerance: 1.768e-07


We set as initial parameter (xmid) the true solution (500), but it failed to proceed:

# f=function(x){500+500*tanh(x-500)}
# x=seq(0,1000,0.001)
# y=f(x)
# length(x)
# df=data.frame("x"=x,"y"=y)
# Asym=1000;xmid=500;scal=1;
# fmla=as.formula("y~SSlogis(x,Asym,xmid,scal)");fmla
#
# t1=Sys.time();
# try(nls(fmla,df))
# Error in nls(y ~ 1/(1 + exp((xmid - x)/scal)), data = xy, start = list(xmid = aux[[1L]],  :
#   step factor 0.000488281 reduced below 'minFactor' of 0.000976562
# t2=Sys.time();t2-t1
# Time difference of 19.29784 secs


So R package inflection can give you an answer for all extremely difficult cases.

## References

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

 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

 Bardsley, W. G. & Childs, R. E. Sigmoid curves, non-linear double-reciprocal plots and allosterism. Biochemical Journal, Portland Press Limited, 1975, 149, 313-328

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inflection documentation built on June 28, 2019, 5:03 p.m.