Finding polynomials from a set of coordinates

[tdh79]

Hi.
I need to find polynomial functions from a given set of coordinates, does anyone know how to do this with the 50g?
Example:
Temperatures are measured 6 times during 24 hours:
time - temp (celsius)
0000 8
0400 5
0800 11
1200 16
1600 18
2000 14

How can i find the function and plot for this?

[Han]

http://support.hp.com/us-en/document/c01939749

EDIT: I don't have my HP50G in front of me and the screenshots in the link above do not list polynomial fit as a choice, so the link above my not apply directly to you if you are looking for a polynomial fit. So if there is no polynomial fit option, you would need to do it by hand using the Vandermonde matrix. https://en.wikipedia.org/wiki/Polynomial_regression

EDIT #2:

[[ 8 ] [ 5 ] [ 11 ] [ 16 ] [ 18 ] [ 14]]
[ 0 4 8 12 16 20 ]
VANDERMONDE
/

returns

[[ 8 ]
[ '-439/120' ]
[ '203/192' ]
[ '-151/1536' ]
[ '13/3072' ]
[ '-3/40960' ]]

This is a list of the cefficients \( a_0, a_1, \dotsm, a_5 \) of the polynomial
\[ a_0 + a_1 \cdot x + a_2 \cdot x^2 + a_3 \cdot x^3 + a_4 \cdot x^4 + a_5 \cdot x^5 \]
That is, \( a_0 = 8 \), \( a_1 = -\frac{439}{120} \), \( a_2 = \frac{203}{192} \), \( a_3 = -\frac{151}{1536} \), \(a_4 = \frac{13}{3072} \), and \(a_5 = -\frac{3}{40960} \).

[Juan14]

Hello, you have to use the Lagrange polynomial. First enter a matrix with the values of x in the first row and then the values of y in the second row, and then the LAGRANGE command, for example:
[[0 4 8 12 16 20] [8 5 11 16 18 14]] |ENTER|

|ALPHA| |ALPHA| LAGRANGE |ENTER|

[Han]

Come to think of it, there are so many polynomial fits that you may want to clarify which types of polynomials to use: Taylor Polynomials, Lagrange Polynomial, Cubic Spline Interpolation, or Least Squares Polynomials.

[tdh79]

(03-10-2017 09:09 PM)Han Wrote:  Come to think of it, there are so many polynomial fits that you may want to clarify which types of polynomials to use: Taylor Polynomials, Lagrange Polynomial, Cubic Spline Interpolation, or Least Squares Polynomials.

Honestly I don't know, i guess I'll just try the methods in your link and the LAGRANGE command and see if i get results that look like the ones in my math book

[Han]

(03-10-2017 09:13 PM)tdh79 Wrote:  

(03-10-2017 09:09 PM)Han Wrote:  Come to think of it, there are so many polynomial fits that you may want to clarify which types of polynomials to use: Taylor Polynomials, Lagrange Polynomial, Cubic Spline Interpolation, or Least Squares Polynomials.

Honestly I don't know, i guess I'll just try the methods in your link and the LAGRANGE command and see if i get results that look like the ones in my math book

Out of curiosity, what is the topic (in your book) from which you obtained the original problem? And what type of math book is it? (Stats, Linear Algebra, ... )

[tdh79]

(03-10-2017 09:18 PM)Han Wrote:  

(03-10-2017 09:13 PM)tdh79 Wrote:  Honestly I don't know, i guess I'll just try the methods in your link and the LAGRANGE command and see if i get results that look like the ones in my math book

Out of curiosity, what is the topic (in your book) from which you obtained the original problem? And what type of math book is it? (Stats, Linear Algebra, ... )

I don't know right now where excactly the problem is presented but it's somewhere in chapters covering polynomial functions, derivations and integrals. I'm studying electrical engineering, powerplants, communications etc..
I've been a sertified telco lineman for 18 years and I'm trying to get some more education, please excuse my poor english I'm Norwegian so my accent is almost as bad as Thor Heyerdahl's

[SlideRule]

(03-10-2017 07:39 PM)tdh79 Wrote:  I need to find polynomial functions from a given set of coordinates ... How can i find the function and plot for this?

Try this URL Online Curve Fitting for a free numerical / graphical (albeit PC based rather than HP-50g based) solution(s).

BEST!
SlideRule

ps: the answer is in decimal fraction versus integer fraction form.

[grsbanks]

(03-10-2017 07:39 PM)tdh79 Wrote:  I need to find polynomial functions from a given set of coordinates, does anyone know how to do this with the 50g?

Fitting a polynomial to a set of coordinates can be done by changing your system into a set of linear equations that you then resolve using the Gauss-Jourdain pivot method.

One thing you need to know is that you need n+1 points to fit a polynomial of degree n. Eg. you need 2 points to fit a line (a.x + b = 0), 3 points to fit a quadratic (a.x² + b.x + c = 0) etc.

Eg. You have the points (-3,1), (1,0) and (2,1).

For the quadratic to fit, all of these points must satisfy the equation, so substituting the coordinates' values of x and y in the equation you end up with something like this:

a.(-3)² + b.(-3) + c = 1
a.1² + b.1 + c = 0
a.2² + b.2 +c = 1

-- or --

9a -3b +c = 1
a + b + c = 0
4a + 2b + c = 1

That looks suspiciously like a system of 3 linear equations that you can solve directly on the HP 50g. You will then get your values of a, b and c that you can replace in your quadratic equation and hey presto, there's your polynomial that passes through your three points.

The solution to this system is a=1/4, b=1/4 and c=-1/2

If you plot f(x)=(x^2)/4 + x/4 - 1/2 you'll see that the curve passes through all three points.

[SlideRule]

(03-11-2017 10:25 AM)grsbanks Wrote:  ... One thing you need to know is that you need n+1 points to fit a polynomial of degree n. Eg. you need 2 points to fit a line (a.x + b = 0), 3 points to fit a quadratic (a.x² + b.x + c = 0) etc...

Clarification - I always use three or more cartesian coordinates for a curve fit because
a) two coordinates, by DEFINITION, are the end-points of a line segment {either rectilinear or curvilinear}
and
b) a single coordinate fails the basic definition of a line segment.

Alibis?

BEST!
SlideRule

[Csaba Tizedes]

(03-10-2017 07:39 PM)tdh79 Wrote:  I need to find polynomial functions

This is not really a good example.
The temperature during a day better to estimate with a periodic function with 24h period.

This is not your fault. This is the fault of your teacher.

Csaba

[Dieter]

(03-10-2017 07:39 PM)tdh79 Wrote:  I need to find polynomial functions from a given set of coordinates

You said it: there are infinitely many polynomial functions that may approximate the given data here. Question is: what do you want?

(03-10-2017 07:39 PM)tdh79 Wrote:  Temperatures are measured 6 times during 24 hours:
time - temp (celsius)
0000 8
0400 5
0800 11
1200 16
1600 18
2000 14

There are essentially two different approaches:

1. From n=6 data points you can determine an (n–1) = 5th order polynomial that passes exactly through these six points. That's quite trivial since this is nothing more than solving a 6th order linear equation system. A piece of cake for the 50g. The required equations are simple:

1*a + t*b + t^2*c + t^3 * d + t^4 * e +t^5* f = temperature

Do this six times for every time (t) and temperature combination, solve for a...f.
This is the method proposed by Han in the second post.

2. You can approximate (!) a set of n datapoints by a polynomial regression function of order m < n–1. In your case you can approximate the six points by a polynomial of order 4, 3 or 2. The higher m, the better the match. This is also done by solving a linear equation system, the required equations are found in the respective Wikipedia article already mentioned here.

Example: fit a cubic function (m=3) through these six points. The result is:
temperature ~ 7,7143 – 1,5714 t + 0,3281 t² – 0,0117 t³

Note: t is not given as 800 or 1200 (does not make sense) but as 8 or 12.

Compare this approximation with the original data:

Code:

time  temperature   estimate
 0         8          7,71
 4         5          5,93
 8        11         10,14
12        16         15,86
16        18         18,57
20        14         13,79

Using a 4th order polynomial, the results move closer to the original data. Here the result is:
temperature ~ 7,9643 – 2,4163 t + 0,6081 t² – 0,0345 t³ + 0,00057 t^4

Again compare this approximation with the original data...

Code:

time  temperature   estimate
 0         8          7,96
 4         5          5,18
 8        11         10,64
12        16         16,36
16        18         17,82
20        14         14,04

...and the estimate already matches the original data when rounded to whole degrees. Finally a 5th order polynomial would exacly match the data (cf. Han's post).

Regarding Csaba Tizedes' remark: I would simply add a 7th data point for t=24 and temperature = 8 (same as for t=0). A periodic function is not required, simply apply the polynomial on (time mod 24). In this case a 6th order polynomial would result in a perfect fit, and 5th, 4th, 3rd... order approximations would come more or less close.

Dieter

[tdh79]

(03-10-2017 08:44 PM)Juan14 Wrote:  Hello, you have to use the Lagrange polynomial. First enter a matrix with the values of x in the first row and then the values of y in the second row, and then the LAGRANGE command, for example:
[[0 4 8 12 16 20] [8 5 11 16 18 14]] |ENTER|

|ALPHA| |ALPHA| LAGRANGE |ENTER|

This seems to do what i need, thanks
Is there any way to go directly to a spesific X value when this has been plotted?
I can find it in TABLE but i was hoping there was a "GOTO" function in the plot view so i could easily read values at certain points without solving the function.

[Csaba Tizedes]

(03-12-2017 06:38 PM)Dieter Wrote:  A periodic function is not required (...) a 6th order polynomial would result in a perfect fit.

The real problem is that you teach this guy something which is MATHEMATICALLY right but not right if he want to understand the model behind his data. Do not teach him anything without you do not clarify that his model is not right for this case.

You can see later he will use this method for all of data what he must to analyse.

If he will work as CSI, he will use the polynomial fit for the body temperature cooling down - because he learned polynomial fit without any clarification.

Maybe he will design your medical stuffz when you will an old man - I hope, he will learn other models during thats years...

And two additional thing:

1.) Why do not teach him how he can decide which polynomial degree is sufficient to fit?
2.) What about your 6th degree model first derivative in t=0 and t=24? As I guess these values aren't same - so that is not a good solution.


Csaba

[Dieter]

(03-12-2017 08:27 PM)Csaba Tizedes Wrote:  1.) Why do not teach him how he can decide which polynomial degree is sufficient to fit?

Sure – why not? Go ahead! ;-)

(03-12-2017 08:27 PM)Csaba Tizedes Wrote:  2.) What about your 6th degree model first derivative in t=0 and t=24? As I guess these values aren't same - so that is not a good solution.

Yes, f'(0) and f'(24) being equal would be nicer.
Of course calculating a periodoc function is possible as well. Here is a possible result with a period of 24 hours:

temperature ~ 12 – 13/3 cos(r*t) + 1/3 cos(3r*t) – 8/sqrt(3) sin(r*t) – 1/sqrt(3) sin(2r*t)

where r = 2pi/24 = pi/12

This may be a better solution, although it's not what the OP asked for.

Dieter

[SlideRule]

(03-10-2017 07:39 PM)tdh79 Wrote:  ... I need to find polynomial functions from a given set of coordinates, ...find the function and plot ...

A starting point is polynomial interpolation. Do you have a background in Technical Mathematics sufficient to identify the following;
a) Local Minima / Maxima
b) Inflection Points
on a data plot of cartesian coordinates?
Comprehension of these topics is not essential but will greatly facilitate the formulation of a solution for your proposition, such as the shape of the plot or the degree of the polynomial constituting the same plot.
"If you need the curve to pass through all the points in the data set, the only fool-proof way is to use a polynomial of degree (n-1) where n is the number of data points. If not, regression can be used." This process can be overwhelming while an inexact or approximate function may suffice within the required / implied accuracy /precision of the data set.
As in every problem, an initial estimate or approximation of the desired answer is welcome prior to a more rigorous and exacting analysis.
With a better understanding of your background, the application of the methods and procedures the many members have proffered may facilitate a most positive and productive learning experience for us all.

BEST!
SlideRule

[Csaba Tizedes]

(03-12-2017 11:31 PM)Dieter Wrote:  

(03-12-2017 08:27 PM)Csaba Tizedes Wrote:  1.) Why do not teach him how he can decide which polynomial degree is sufficient to fit?

Sure – why not? Go ahead! ;-)

OK, let's see:

0.) I'm an engineer, my solution shows clearly how an engineer is thinking about this kind of problems
1.) This data has a minimal- and a maximal value
2.) If I want to select a polynom, at least 3rd order must to use - because the 3rd order has a min. and max. extremum
3.) We have 6 data points, therefore we can use max. 5th order of polynom
4.) Fit 1st, 2nd, 3rd, 4th and 5th order polynom and check where the correlation coefficient is changes dramatically - see the picture below

   

5.) It's looks the 1st and 2nd is worst and 3rd order and above is better solution, the question is that what magnitude of error in each points is acceptable
6.) If no any special requirement, the 3rd is the simplest to use


Csaba

[grsbanks]

(03-13-2017 03:00 PM)Csaba Tizedes Wrote:  4.) Fit 1st, 2nd, 3rd, 4th and 5th order polynom and check where the correlation coefficient is changes dramatically - see the picture below

The question is, is the OP asking for a polynomial that fits the provided data set or one that is close? That much is unclear.

[Csaba Tizedes]

(03-12-2017 11:31 PM)Dieter Wrote:  Of course calculating a periodoc function is possible as well. Here is a possible result with a period of 24 hours:

temperature ~ 12 – 13/3 cos(r*t) + 1/3 cos(3r*t) – 8/sqrt(3) sin(r*t) – 1/sqrt(3) sin(2r*t)

where r = 2pi/24 = pi/12

This may be a better solution

And the second part of my answer about the periodic function:

If I'm sitting in the middle of nowhere and I want to fit a periodic function, I will use the simplest solution:

1.) This is about a data of one day: the period is 24 hours
2.) The min and max values are exactly at t=4h and at t=16h
3.) Therefore the inflection is between them at t=10h
4.) The range of data between the max and min values of temperature, therefore we can calculate the amplitude of periodic function and the mean value for the shift of the function
5.) If I want to more exact value for the shift on the temperature axis, I can calculate the average of the measured temperatures - this requires little more keypress, no problem, I like it
6.) Just for fun, draw a little diagram and calculate the temperature differences and calculate RMS

Everything fits in this half letter paper:

   

And I used only one piece of paper, one of my favourite durable pencil and my most durable HP15C.

   

This IS the engineering problem solving, not the sooo looong equuuuaaatioooons with looots of coefficients.

Csaba

PS.: I have no problem with you, I hope you're feel it, I'm just this kind of men.

[SlideRule]

tdh79

I will assume we sufficiently answered your question and in future queries a common reference might be an edition of the Lineman and Cableman's Field Manual ISBN: 978-0-07-162122-9 (2nd edition) or related occupational manual?

BEST!
SlideRule