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Given four points how can I find an equation for any pattern? [on hold]

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0

$begingroup$

If for example I know that there's a function f(x) that equals 1 at x=8, 8 at x=32, 36 at x=64, 98 at x=128.
How can I find the expression for this pattern or any other pattern?

real-analysis

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asked May 6 at 2:33

Mohammad AlSaqqaMohammad AlSaqqa

101

New contributor

Mohammad AlSaqqa is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

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put on hold as off-topic by user21820, YuiTo Cheng, Cesareo, RRL, max_zorn 2 days ago

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – user21820, YuiTo Cheng, Cesareo, RRL, max_zorn

If this question can be reworded to fit the rules in the help center, please edit the question.

• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  $begingroup$
  You're not guaranteed one unique pattern/function. You can use polynomial interpolation to get a formula $f$ where $deg(f) = n-1$ if you have have $n$ pairs $(x,f(x))$. Said $f$ will even be unique. But higher degree polynomials could also fit it, infinitely many in fact. And this is all just polynomial functions mind you.
  $endgroup$
  – Eevee Trainer
  May 6 at 2:36
  
  

  
  
  
  

add a comment | 

0

$begingroup$

If for example I know that there's a function f(x) that equals 1 at x=8, 8 at x=32, 36 at x=64, 98 at x=128.
How can I find the expression for this pattern or any other pattern?

real-analysis

share|cite|improve this question

asked May 6 at 2:33

Mohammad AlSaqqaMohammad AlSaqqa

101

New contributor

Mohammad AlSaqqa is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

put on hold as off-topic by user21820, YuiTo Cheng, Cesareo, RRL, max_zorn 2 days ago

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – user21820, YuiTo Cheng, Cesareo, RRL, max_zorn

If this question can be reworded to fit the rules in the help center, please edit the question.

• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  $begingroup$
  You're not guaranteed one unique pattern/function. You can use polynomial interpolation to get a formula $f$ where $deg(f) = n-1$ if you have have $n$ pairs $(x,f(x))$. Said $f$ will even be unique. But higher degree polynomials could also fit it, infinitely many in fact. And this is all just polynomial functions mind you.
  $endgroup$
  – Eevee Trainer
  May 6 at 2:36
  
  

  
  
  
  

add a comment | 

$begingroup$

If for example I know that there's a function f(x) that equals 1 at x=8, 8 at x=32, 36 at x=64, 98 at x=128.
How can I find the expression for this pattern or any other pattern?

real-analysis

share|cite|improve this question

asked May 6 at 2:33

Mohammad AlSaqqaMohammad AlSaqqa

101

New contributor

Mohammad AlSaqqa is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

share|cite|improve this question

asked May 6 at 2:33

Mohammad AlSaqqaMohammad AlSaqqa

101

New contributor

Mohammad AlSaqqa is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

share|cite|improve this question

asked May 6 at 2:33

Mohammad AlSaqqaMohammad AlSaqqa

101

New contributor

Mohammad AlSaqqa is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

asked May 6 at 2:33

Mohammad AlSaqqaMohammad AlSaqqa

101

New contributor

Mohammad AlSaqqa is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

asked May 6 at 2:33

Mohammad AlSaqqaMohammad AlSaqqa

101

2 Answers
2

active

oldest

votes

4

$begingroup$

The process is called "interpolation". There are an indefinite number of ways to do this for a given set of points.

One way is Lagrange Interpolation which I discuss in the answer linked below :

Explanation of Lagrange Interpolating Polynomial

---

I have plotted the Lagrange Polynomial for your curve on Desmos. You can see that it matches the pattern you gave. You can interact with the plot by clicking on the link below.

https://www.desmos.com/calculator/enavzwsl09

share|cite|improve this answer

edited May 6 at 4:10

answered May 6 at 2:35

SpencerSpencer

9,27222357

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Thanks a bunch! this is very helpful
  $endgroup$
  – Mohammad AlSaqqa
  May 6 at 2:50
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I'm glad it was helpful :). I noticed that you were having issues implementing the Lagrange Interpolating polynomial provided by J. W. Tanner, so I implemented it my self in Desmos. Hopefully that can clear things up for you.
  $endgroup$
  – Spencer
  May 6 at 4:11
  

  
  
  
  

add a comment | 

3

$begingroup$

Here's one (not simplified):

$f(x)=dfrac(x-32)(x-64)(x-128)(8-32)(8-64)(8-128)1+dfrac(x-8)(x-64)(x-128)(32-8)(32-64)(32-128)8$

$;+dfrac(x-8)(x-32)(x-128)(64-8)(64-32)(64-128)36+dfrac(x-8)(x-32)(x-64)(128-8)(128-32)(128-64)98$

share|cite|improve this answer

edited May 6 at 3:47

answered May 6 at 2:37

J. W. TannerJ. W. Tanner

6,4641521

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  this equation gave back 68687616x^3-6932668416x^2+186621886464x-1080117166080 which did not fit the curve
  $endgroup$
  – Mohammad AlSaqqa
  May 6 at 3:40
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I had a typo. ($28$ instead of $128$), which I just corrected
  $endgroup$
  – J. W. Tanner
  May 6 at 3:48
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  It still does not fit the curve, is my pattern maybe can't be represented as a polynomial?
  $endgroup$
  – Mohammad AlSaqqa
  May 6 at 3:54
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  There are many polynomials that fit the four points you gave; I showed how to get one of them
  $endgroup$
  – J. W. Tanner
  2 days ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I see that the function I gave is now plotted in the other answer
  $endgroup$
  – J. W. Tanner
  2 days ago
  

  
  
  
  

add a comment | 

4

$begingroup$

The process is called "interpolation". There are an indefinite number of ways to do this for a given set of points.

One way is Lagrange Interpolation which I discuss in the answer linked below :

Explanation of Lagrange Interpolating Polynomial

---

I have plotted the Lagrange Polynomial for your curve on Desmos. You can see that it matches the pattern you gave. You can interact with the plot by clicking on the link below.

https://www.desmos.com/calculator/enavzwsl09

share|cite|improve this answer

edited May 6 at 4:10

answered May 6 at 2:35

SpencerSpencer

9,27222357

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Thanks a bunch! this is very helpful
  $endgroup$
  – Mohammad AlSaqqa
  May 6 at 2:50
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I'm glad it was helpful :). I noticed that you were having issues implementing the Lagrange Interpolating polynomial provided by J. W. Tanner, so I implemented it my self in Desmos. Hopefully that can clear things up for you.
  $endgroup$
  – Spencer
  May 6 at 4:11
  

  
  
  
  

add a comment | 

4

$begingroup$

The process is called "interpolation". There are an indefinite number of ways to do this for a given set of points.

One way is Lagrange Interpolation which I discuss in the answer linked below :

Explanation of Lagrange Interpolating Polynomial

---

I have plotted the Lagrange Polynomial for your curve on Desmos. You can see that it matches the pattern you gave. You can interact with the plot by clicking on the link below.

https://www.desmos.com/calculator/enavzwsl09

share|cite|improve this answer

edited May 6 at 4:10

answered May 6 at 2:35

SpencerSpencer

9,27222357

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Thanks a bunch! this is very helpful
  $endgroup$
  – Mohammad AlSaqqa
  May 6 at 2:50
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I'm glad it was helpful :). I noticed that you were having issues implementing the Lagrange Interpolating polynomial provided by J. W. Tanner, so I implemented it my self in Desmos. Hopefully that can clear things up for you.
  $endgroup$
  – Spencer
  May 6 at 4:11
  

  
  
  
  

add a comment | 

$begingroup$

The process is called "interpolation". There are an indefinite number of ways to do this for a given set of points.

One way is Lagrange Interpolation which I discuss in the answer linked below :

Explanation of Lagrange Interpolating Polynomial

---

I have plotted the Lagrange Polynomial for your curve on Desmos. You can see that it matches the pattern you gave. You can interact with the plot by clicking on the link below.

https://www.desmos.com/calculator/enavzwsl09

share|cite|improve this answer

edited May 6 at 4:10

answered May 6 at 2:35

SpencerSpencer

9,27222357

$endgroup$

share|cite|improve this answer

edited May 6 at 4:10

answered May 6 at 2:35

SpencerSpencer

9,27222357

answered May 6 at 2:35

SpencerSpencer

9,27222357

answered May 6 at 2:35

SpencerSpencer

9,27222357

3

$begingroup$

Here's one (not simplified):

$f(x)=dfrac(x-32)(x-64)(x-128)(8-32)(8-64)(8-128)1+dfrac(x-8)(x-64)(x-128)(32-8)(32-64)(32-128)8$

$;+dfrac(x-8)(x-32)(x-128)(64-8)(64-32)(64-128)36+dfrac(x-8)(x-32)(x-64)(128-8)(128-32)(128-64)98$

share|cite|improve this answer

edited May 6 at 3:47

answered May 6 at 2:37

J. W. TannerJ. W. Tanner

6,4641521

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  this equation gave back 68687616x^3-6932668416x^2+186621886464x-1080117166080 which did not fit the curve
  $endgroup$
  – Mohammad AlSaqqa
  May 6 at 3:40
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I had a typo. ($28$ instead of $128$), which I just corrected
  $endgroup$
  – J. W. Tanner
  May 6 at 3:48
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  It still does not fit the curve, is my pattern maybe can't be represented as a polynomial?
  $endgroup$
  – Mohammad AlSaqqa
  May 6 at 3:54
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  There are many polynomials that fit the four points you gave; I showed how to get one of them
  $endgroup$
  – J. W. Tanner
  2 days ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I see that the function I gave is now plotted in the other answer
  $endgroup$
  – J. W. Tanner
  2 days ago
  

  
  
  
  

add a comment | 

3

$begingroup$

Here's one (not simplified):

$f(x)=dfrac(x-32)(x-64)(x-128)(8-32)(8-64)(8-128)1+dfrac(x-8)(x-64)(x-128)(32-8)(32-64)(32-128)8$

$;+dfrac(x-8)(x-32)(x-128)(64-8)(64-32)(64-128)36+dfrac(x-8)(x-32)(x-64)(128-8)(128-32)(128-64)98$

share|cite|improve this answer

edited May 6 at 3:47

answered May 6 at 2:37

J. W. TannerJ. W. Tanner

6,4641521

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  this equation gave back 68687616x^3-6932668416x^2+186621886464x-1080117166080 which did not fit the curve
  $endgroup$
  – Mohammad AlSaqqa
  May 6 at 3:40
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I had a typo. ($28$ instead of $128$), which I just corrected
  $endgroup$
  – J. W. Tanner
  May 6 at 3:48
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  It still does not fit the curve, is my pattern maybe can't be represented as a polynomial?
  $endgroup$
  – Mohammad AlSaqqa
  May 6 at 3:54
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  There are many polynomials that fit the four points you gave; I showed how to get one of them
  $endgroup$
  – J. W. Tanner
  2 days ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I see that the function I gave is now plotted in the other answer
  $endgroup$
  – J. W. Tanner
  2 days ago
  

  
  
  
  

add a comment | 

$begingroup$

Here's one (not simplified):

$f(x)=dfrac(x-32)(x-64)(x-128)(8-32)(8-64)(8-128)1+dfrac(x-8)(x-64)(x-128)(32-8)(32-64)(32-128)8$

$;+dfrac(x-8)(x-32)(x-128)(64-8)(64-32)(64-128)36+dfrac(x-8)(x-32)(x-64)(128-8)(128-32)(128-64)98$

share|cite|improve this answer

edited May 6 at 3:47

answered May 6 at 2:37

J. W. TannerJ. W. Tanner

6,4641521

$endgroup$

share|cite|improve this answer

edited May 6 at 3:47

answered May 6 at 2:37

J. W. TannerJ. W. Tanner

6,4641521

answered May 6 at 2:37

J. W. TannerJ. W. Tanner

6,4641521

answered May 6 at 2:37

J. W. TannerJ. W. Tanner

6,4641521