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Smoothness of discrete data

1

$begingroup$

I'm having a hard time putting my question into words, so I made a few pictures.
Look at this plot:

Clearly, everyone will agree that these data points are following some nice smooth and continuous function. In the following plot, this is not the case.

What I'm looking for, is a word that describes this difference:

Data set A is much more ??? than data set B.

Is it smooth? Well-behaved maybe? Thanks in advance!

discrete-mathematics terminology

share|cite|improve this question

asked Feb 20 '16 at 20:17

murphymurphy

1313

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  I would classify these data according to their frequency content. Data set A has only low-frequency components whereas data set B has many high-frequency components.
  $endgroup$
  – Matt L.
  Feb 20 '16 at 20:22
  

  
  
  
  

add a comment | 

1

$begingroup$

I'm having a hard time putting my question into words, so I made a few pictures.
Look at this plot:

Clearly, everyone will agree that these data points are following some nice smooth and continuous function. In the following plot, this is not the case.

What I'm looking for, is a word that describes this difference:

Data set A is much more ??? than data set B.

Is it smooth? Well-behaved maybe? Thanks in advance!

discrete-mathematics terminology

share|cite|improve this question

asked Feb 20 '16 at 20:17

murphymurphy

1313

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  I would classify these data according to their frequency content. Data set A has only low-frequency components whereas data set B has many high-frequency components.
  $endgroup$
  – Matt L.
  Feb 20 '16 at 20:22
  

  
  
  
  

add a comment | 

$begingroup$

I'm having a hard time putting my question into words, so I made a few pictures.
Look at this plot:

Clearly, everyone will agree that these data points are following some nice smooth and continuous function. In the following plot, this is not the case.

What I'm looking for, is a word that describes this difference:

Data set A is much more ??? than data set B.

Is it smooth? Well-behaved maybe? Thanks in advance!

discrete-mathematics terminology

share|cite|improve this question

asked Feb 20 '16 at 20:17

murphymurphy

1313

$endgroup$

share|cite|improve this question

asked Feb 20 '16 at 20:17

murphymurphy

1313

share|cite|improve this question

asked Feb 20 '16 at 20:17

murphymurphy

1313

asked Feb 20 '16 at 20:17

murphymurphy

1313

asked Feb 20 '16 at 20:17

murphymurphy

1313

3 Answers
3

active

oldest

votes

0

$begingroup$

It looks like data set A is more smooth than data set B because you can easily see the trend that is there. But, the other thing to know that data set B is [most likely] more well-behaved than data set A because you can get from point to point in set B "freely" without "any traffic" by any points, as compared to set A.

Short answer: Data set A is much more smooth than data set B.

share|cite|improve this answer

answered Feb 20 '16 at 20:22

Obinna NwakwueObinna Nwakwue

801524

$endgroup$

add a comment | 

0

$begingroup$

The term you're looking for is "variance." When you look at the data, you fit a curve to it. The curve is smooth, but it's the same curve for both data points. The difference is how much the data varies form the curve, usually measured by $frac{1}{n}sum (f(x)-x)^2$ where $f(x)$ is the equation of the curve. The square root of this quantity is known as the "standard deviation"

You can read more here.

share|cite|improve this answer

edited Jan 23 '17 at 4:44

answered Feb 20 '16 at 20:28

Stella BidermanStella Biderman

26.7k63375

$endgroup$

add a comment | 

0

$begingroup$

You can use "smoother". And the reason will be following:

You can find the Total Variation (TV) of both the data sets. The variation of B will be much more than A.

The formula of total variation is given by,

$TV(u)=sum_i |u_{i+1}-u_i|$.

share|cite|improve this answer

answered Jun 5 '17 at 6:56

Biswarup BiswasBiswarup Biswas

1

$endgroup$

add a comment | 

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0

$begingroup$

It looks like data set A is more smooth than data set B because you can easily see the trend that is there. But, the other thing to know that data set B is [most likely] more well-behaved than data set A because you can get from point to point in set B "freely" without "any traffic" by any points, as compared to set A.

Short answer: Data set A is much more smooth than data set B.

share|cite|improve this answer

answered Feb 20 '16 at 20:22

Obinna NwakwueObinna Nwakwue

801524

$endgroup$

add a comment | 

0

$begingroup$

It looks like data set A is more smooth than data set B because you can easily see the trend that is there. But, the other thing to know that data set B is [most likely] more well-behaved than data set A because you can get from point to point in set B "freely" without "any traffic" by any points, as compared to set A.

Short answer: Data set A is much more smooth than data set B.

share|cite|improve this answer

answered Feb 20 '16 at 20:22

Obinna NwakwueObinna Nwakwue

801524

$endgroup$

add a comment | 

$begingroup$

It looks like data set A is more smooth than data set B because you can easily see the trend that is there. But, the other thing to know that data set B is [most likely] more well-behaved than data set A because you can get from point to point in set B "freely" without "any traffic" by any points, as compared to set A.

Short answer: Data set A is much more smooth than data set B.

share|cite|improve this answer

answered Feb 20 '16 at 20:22

Obinna NwakwueObinna Nwakwue

801524

$endgroup$

share|cite|improve this answer

answered Feb 20 '16 at 20:22

Obinna NwakwueObinna Nwakwue

801524

answered Feb 20 '16 at 20:22

Obinna NwakwueObinna Nwakwue

801524

answered Feb 20 '16 at 20:22

Obinna NwakwueObinna Nwakwue

801524

0

$begingroup$

The term you're looking for is "variance." When you look at the data, you fit a curve to it. The curve is smooth, but it's the same curve for both data points. The difference is how much the data varies form the curve, usually measured by $frac{1}{n}sum (f(x)-x)^2$ where $f(x)$ is the equation of the curve. The square root of this quantity is known as the "standard deviation"

You can read more here.

share|cite|improve this answer

edited Jan 23 '17 at 4:44

answered Feb 20 '16 at 20:28

Stella BidermanStella Biderman

26.7k63375

$endgroup$

add a comment | 

0

$begingroup$

The term you're looking for is "variance." When you look at the data, you fit a curve to it. The curve is smooth, but it's the same curve for both data points. The difference is how much the data varies form the curve, usually measured by $frac{1}{n}sum (f(x)-x)^2$ where $f(x)$ is the equation of the curve. The square root of this quantity is known as the "standard deviation"

You can read more here.

share|cite|improve this answer

edited Jan 23 '17 at 4:44

answered Feb 20 '16 at 20:28

Stella BidermanStella Biderman

26.7k63375

$endgroup$

add a comment | 

$begingroup$

The term you're looking for is "variance." When you look at the data, you fit a curve to it. The curve is smooth, but it's the same curve for both data points. The difference is how much the data varies form the curve, usually measured by $frac{1}{n}sum (f(x)-x)^2$ where $f(x)$ is the equation of the curve. The square root of this quantity is known as the "standard deviation"

You can read more here.

share|cite|improve this answer

edited Jan 23 '17 at 4:44

answered Feb 20 '16 at 20:28

Stella BidermanStella Biderman

26.7k63375

$endgroup$

share|cite|improve this answer

edited Jan 23 '17 at 4:44

answered Feb 20 '16 at 20:28

Stella BidermanStella Biderman

26.7k63375

answered Feb 20 '16 at 20:28

Stella BidermanStella Biderman

26.7k63375

answered Feb 20 '16 at 20:28

Stella BidermanStella Biderman

26.7k63375

0

$begingroup$

You can use "smoother". And the reason will be following:

You can find the Total Variation (TV) of both the data sets. The variation of B will be much more than A.

The formula of total variation is given by,

$TV(u)=sum_i |u_{i+1}-u_i|$.

share|cite|improve this answer

answered Jun 5 '17 at 6:56

Biswarup BiswasBiswarup Biswas

1

$endgroup$

add a comment | 

0

$begingroup$

You can use "smoother". And the reason will be following:

You can find the Total Variation (TV) of both the data sets. The variation of B will be much more than A.

The formula of total variation is given by,

$TV(u)=sum_i |u_{i+1}-u_i|$.

share|cite|improve this answer

answered Jun 5 '17 at 6:56

Biswarup BiswasBiswarup Biswas

1

$endgroup$

add a comment | 

$begingroup$

You can use "smoother". And the reason will be following:

You can find the Total Variation (TV) of both the data sets. The variation of B will be much more than A.

The formula of total variation is given by,

$TV(u)=sum_i |u_{i+1}-u_i|$.

share|cite|improve this answer

answered Jun 5 '17 at 6:56

Biswarup BiswasBiswarup Biswas

1

$endgroup$

share|cite|improve this answer

answered Jun 5 '17 at 6:56

Biswarup BiswasBiswarup Biswas

1

answered Jun 5 '17 at 6:56

Biswarup BiswasBiswarup Biswas

1

answered Jun 5 '17 at 6:56

Biswarup BiswasBiswarup Biswas

1