How to find local minimum of fit polynomial

I have some stock data (time and price).

I log the data, and then use fit function to fit a polynomial onto it

then i make a function f

This is shown in the three lines of code below

date_log10 = log10(datenum(T2.Time)); price_log10 = log10(T2.Close); 
[f_mdl,gof,output] = fit(date_log10,price_log10,'poly1','Normalize','on','Robust','Bisquare');
f(j) = 10.^f_mdl(log10(datenum(T2.Time(end))));

My question is, how to i now find the local minimum of this function f?

I was thinking i would use fminsearch, but just unsure how i would make use of that.

[dpb]

What local minimum? It's a linear fit, the minimum will be at either end of the range depending upon whether the slope is positve or negative.

Seems a bizarre thing to even contemplate doing to stock price data...is there some literature in the field that uses such a model?

Also, there's essentially nothing to be gained, anyway...for something to use I downloaded some past 5-6 years for GOOGLE and tried the model over it -- first, of course, visualizing the data as one should always do before just blindly fitting a model--here are the two figures -- real data and then the log10 version -- NB, they are identically the same shape; just changing the numerical values is all that has been done.

As is obvious, the minimum value of the fitted line is the first value since the slope is positive. If we were to arbitrarily flip the data to be reversed in time, then the slope would be the same numerically, but with negative sign and the minimum would be the last point evaluated.

No idea what must be trying to do here...

[Rizwan Khan]

Dear Sir,

Thanks for the quick response, while you're there, can i quickly ask, where are you downloading the data from, are you able to share the code?

Secondly, in relation to my query, although the model is linear 1, it actually then chagnes the log back and eventually is equivialent to poly3.

So even when i plot it, i get a polynomial, and my question is how do i find the local minimum and maximum.

So as an example, if you look at your data, and change your linear orange line to a polynomial of degree 3, then you will see that it turns at the top, now how can you find the local minimum of that, i believe this required fminsearch or fmin function.

The eventual goal is to draw a polynomial, and then find the most recent turning point, so that may not be the very extreme (global) min or max value of the chart, as the charts goes up and then down and so on, the these local peaks and trophs keep coming, and i want to find the most recent local min or local max

I think the answer is in using fmin or fminsearch function, but i'm unsure how to exactly do that.

Thanks again so much for your response

Answer 1

[Image Analyst]

I use polyfit() instead of fit(). Have you gotten a fitted y value vector yet? If so, just use min

miny = min(yFitted(index1 : index2))

[Rizwan Khan]

This just gives the minimum value overall the global min, i want the most recent min value in a polynomical

So imagine sinewave with altering magnitudes, going up and down and up and down, i want to find the most recent turning point.

[Image Analyst]

It gives the min between index1 and index2. I don't know how you were defining the region when you originally said "local minimum of this function f". You didn't say that you needed the last turning point originally like you did just now.

If you have your data in an array you can simply use islocalmin(y) and look at the last index

indexesOfMins = find(islocalmin(y)) % Of ALL mins in the array.
lastIndex = indexesOfMins(end) % Just the very last location ONLY.
lastValue = y(lastIndex)   % Value of y there.

Alternatively you can take the coefficients of the fit, take the derivative and set it to zero, and that will tell you the "flat spots" in the curve.

[Rizwan Khan]

Dear Sir,

Thanks you for your response, i hae a few questions

What is your y varirable, is that a polynomial?

When i use fit it returns me a fit object, and as you rightly say, that comes with coefficients.

So regarding your second approach, that may be quicker for me because i already have the object and therefore the co-efficients.

How would i take the derivative and set it to zero to get the "flat spots" in the curve?

[Walter Roberson]

roots(polyder(p)) like I posted earlier

Answer 2

[Walter Roberson]

poly1 is of the form a*x+b for some model parameters a and b. The minimum of a straight line is:

• at the lower bound of x if a is positive
• at the upper bound of x if a is negative

[Rizwan Khan]

Hi Walter,

Thanks for your response, yes i log it, but then i also reverse the log, so eventually it actully charts a polynomical, for the purpose of this excercise you can assume i'm getting a polynomial curve of degree 3.

But there can be a few turns, up and down, and i want to find the most recent turning point.

I think the answer is in using Fmin and Fminsearch, but unsure how to do that.

Its an optimisation question i think they call it.

[Walter Roberson]

If you take 10 to the power of a linear fit, then the result will still have its maximum at either the upper bound or the lower bound. It will not turn into a cubic!!

If you had the coefficients p for a cubic then use roots(polyder(p)) to find the locations of the critical points