Finding the inflection point of a sigmoid function

Hi, here's my function:

% define params
p1 = 0.0389;
p2 = 1.9158;
p3 = 162.4272;
p4 = 0.012;

% simulate the function
syms x
f = p1+(p2-p1)/(1+10^((p3-x)*p4));
fplot(f,[1 245])

Now, I would like to find the inflection point (I manually drew where I roughly expect it to be).

Here's what I was trying to do:

double(solve(f,'MaxDegree',3))

but I get this:

ans = 2.1394e+01 - 1.1370e+02i

Which means that there is no real inflection point... Where's my mistake?

Thanks in advance

Iddo

Answer 1

[Star Strider]

Try this:

p1 = 0.0389;
p2 = 1.9158;
p3 = 162.4272;
p4 = 0.012;

% simulate the function
syms x
f(x) = p1+(p2-p1)/(1+10^((p3-x)*p4));
inflpt = vpasolve(diff(f,2) == 0);                          % Calculate Inflection Point

figure(1)
fplot(f,[1 245])
hold on
plot(inflpt, f(inflpt), 'pr', 'MarkerSize',10, 'MarkerFaceColor','g')
hold off

[Iddo Weiner]

Great, thanks

[Star Strider]

As always, my pleasure.

[Karan Gill]

Iddo, did you understand why that code works?

[Iddo Weiner]

Yes, setting the 2nd derivative to equal zero will give you the inflection point

Answer 2

[Roger Stafford]

The inflection point occurs at x = p3. You can show it by using the symbolic 'diff' to find the second derivative of f with respect to x and finding the x that makes it zero. That occurs when 10^((p3-x)*p4)) is equal to 1 which forces x to equal p3.

[Iddo Weiner]

Hey Roger, thanks for your help.

I understand what you say and it's right for this case, but I would like to have a piece of code that automatically calculates the inflection point.. Looking at the official documentation it looked like this line of code is supposed to do the work:

double(solve(f,'MaxDegree',3))

But, it didn't work.. So do you see a different way to automatically generate the inflection point? Or maybe I'm not using solve() properly?

Thanks, Iddo

Answer 3

[Lalitesh]

Hi,

I hope you have solved your inflection point problem. Could you please suggest me how to find inflection point in the given below case ?

y = [4941.325135
    5373.592952
    5341.265151
    5109.495841
    5164.985178
    4910.347906
    5036.098753
    5210.505118
    4897.930074
    5303.555152
    5059.674066
    5294.250906
    5045.080549
    5139.444791
    5184.016273
    4989.822273
    4879.351675
    5037.299623
    5193.317198
    5126.702477
    5114.934361
    5163.894699
    5250.845909
    5236.142352
    5425.261686
    5026.998422
    5181.941373
    5043.063249
    5226.955717
    5045.126978
    5285.330253
    5147.00929
    5089.199187
    5433.585096
    5259.022363
    5135.878393
    5075.663506
    5068.604444
    5240.425345
    5013.836134
    5235.932975
    5293.152198
    5362.597535
    5288.109111
    5034.519472
    5095.714828
    5267.142594
    5168.69091
    5050.240489
    4921.738751
    ];
x = [-0.013607785
    0.765345224
    1.10289259
    1.029090207
    -0.022931512
    -0.017396819
    -0.074107114
    0.769494196
    2.98E-06
    0.920557852
    0.738470623
    -0.025517765
    0.820330824
    0.561636238
    1.02437854
    0.748933493
    -0.024258105
    0.172846729
    -0.040833861
    1.015883315
    0.961682273
    0.974008018
    -2.782219696
    1.02138985
    0.556479342
    0.411214807
    1.056242842
    -3.368250299
    1.094017765
    1.128149932
    -0.040686495
    -0.013525985
    -0.028379761
    0.979391037
    0.662479735
    1.127613475
    0.488222296
    -0.067298217
    1.022925141
    -0.119633187
    1.04065864
    0.286375511
    1.195491684
    0.957996374
    1.103174243
    1.049128867
    1.080726806
    0.99141825
    1.198772497
    0.864385088
    ];
fn = @(b,x) b(5).*(b(1) + b(2)./(1 + exp(-b(3).*(x-b(4))))).^(b(6))  % Sigmoid Function (Objective Function)

fn = function_handle with value:
    @(b,x)b(5).*(b(1)+b(2)./(1+exp(-b(3).*(x-b(4))))).^(b(6))

SSECF = @(b) sum((y - fn(b,x)).^2);                 % Sum-Squared-Error Cost Function
B0 = [-0.1; -0.1; -5; 0.1; 0.8; 10];                          % Initial Parameter Estimates
[B,SSE] = fminsearch(SSECF, B0)                    % Estimate Parameters (B), Return Sum-Squared Error (SSE)

B = 6×1

-1.1849 -0.8181 51.3923 0.4650 6.0624 10.1272

SSE = 2.8669e+08 + 1.8180e+08i

xp = linspace(min(x), max(x))

xp = 1×100

-3.3683 -3.3221 -3.2760 -3.2299 -3.1837 -3.1376 -3.0915 -3.0453 -2.9992 -2.9531 -2.9069 -2.8608 -2.8147 -2.7685 -2.7224 -2.6763 -2.6301 -2.5840 -2.5379 -2.4918 -2.4456 -2.3995 -2.3534 -2.3072 -2.2611 -2.2150 -2.1688 -2.1227 -2.0766 -2.0304

figure(1)
plot(xp, fn(B,xp), '-r')

Warning: Imaginary parts of complex X and/or Y arguments ignored.

hold on
plot(x,y,'o')
hold off

[Star Strider]

@Lalitesh —

I could, however not here.

This should be posted as a new Question.

The function is definitely not a good fit to the data and produces a complex result (the imaginary part is approximately equal in magnitude to the real part), so you need to decide how to deal with those problems.

[Torsten]

@Lalitesh

Your data are just a point cloud.

How do you believe one could make a fit or find an inflection point ?

[Lalitesh]

Okay, I got your point. I need to get the correct the data to fit the trajectory of sigmoid then only I can find the inflection point.

[Lalitesh]

I will post this as a new question.

[Torsten]

Hopefully with better data.

Otherwise you can see the question as answered: impossible.