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How to find a formula relating three values?
Clash Royale CLAN TAG#URR8PPP
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I haven't studied maths since high-school (20 years ago) and would like to find a formula to relate these values:
x y z
-----|------|-------
1 | 0.25 | 0.25
2 | 0.25 | 0.625
3 | 0.25 | 0.75
4 | 0.25 | 0.8125
5 | 0.25 | 0.85
1.5 | 0.25 | 0.5
10 | 0.25 | 0.925
2 | 0.1 | 0.55
The formula would find z from known x and y values.
e.g. if x = 6 and y = 0.25 what is z?
Thanks for any help solving the problem or pointing me in the right direction. I can try to explain the context if that helps, but perhaps it's easier just to work with the numbers.
extrapolation
edited Jul 30 at 7:17
Gerry Myerson
142k7143292
asked Jul 30 at 7:08
Laurence Lord
64
 |Â
show 3 more comments
up vote
0
down vote
favorite
I haven't studied maths since high-school (20 years ago) and would like to find a formula to relate these values:
x y z
-----|------|-------
1 | 0.25 | 0.25
2 | 0.25 | 0.625
3 | 0.25 | 0.75
4 | 0.25 | 0.8125
5 | 0.25 | 0.85
1.5 | 0.25 | 0.5
10 | 0.25 | 0.925
2 | 0.1 | 0.55
The formula would find z from known x and y values.
e.g. if x = 6 and y = 0.25 what is z?
Thanks for any help solving the problem or pointing me in the right direction. I can try to explain the context if that helps, but perhaps it's easier just to work with the numbers.
extrapolation
edited Jul 30 at 7:17
Gerry Myerson
142k7143292
asked Jul 30 at 7:08
Laurence Lord
64
2
There are infinitely many formulas that fit the given numbers but give widely different answers for $x=6$, $y=.25$. You have to know more about what kind of function you are trying to fit, in order to give a useful answer.
– Gerry Myerson
Jul 30 at 7:20
@GerryMyerson thanks for the help. How might I better define the kind of function I am looking for? I can't even find one formula that fits. Visually I understand the problem and could try to describe it in those terms or I can also supply more results from the model I have. I'm looking for the simpliest solution that gives the same results as the model.
– Laurence Lord
Jul 30 at 7:42
OK, here's one way, but it probably doesn't give you what you want. Temporarily ignore that last data point. Then all your data points have the same $y$-value, so you can ignore $y$ and get a polynomial $p(x)$ that fits all the points by using Lagrange interpolation (look it up!). Then you can find a number $c$ such that $p(x)+c(y-.25)$ fits all your points.
– Gerry Myerson
Jul 30 at 9:36
Hi Gerry, thanks for trying me on that. To be honest I don't know what a polynomial is and having tried to read up on Lagrange Interpolation I'm struggling to find anything I can understand... I'll keep trying
– Laurence Lord
Jul 30 at 17:53
@GerryMyerson thanks again. I'm clearer on polynomials now. Still not sure how to go about interpolation. In your example you show c(y – .25) but y is .25 so that will be zero. I'm confused.
– Laurence Lord
Jul 30 at 19:17
 |Â
show 3 more comments
up vote
0
down vote
favorite
up vote
0
down vote
favorite
I haven't studied maths since high-school (20 years ago) and would like to find a formula to relate these values:
x y z
-----|------|-------
1 | 0.25 | 0.25
2 | 0.25 | 0.625
3 | 0.25 | 0.75
4 | 0.25 | 0.8125
5 | 0.25 | 0.85
1.5 | 0.25 | 0.5
10 | 0.25 | 0.925
2 | 0.1 | 0.55
The formula would find z from known x and y values.
e.g. if x = 6 and y = 0.25 what is z?
Thanks for any help solving the problem or pointing me in the right direction. I can try to explain the context if that helps, but perhaps it's easier just to work with the numbers.
extrapolation
edited Jul 30 at 7:17
Gerry Myerson
142k7143292
asked Jul 30 at 7:08
Laurence Lord
64
I haven't studied maths since high-school (20 years ago) and would like to find a formula to relate these values:
x y z
-----|------|-------
1 | 0.25 | 0.25
2 | 0.25 | 0.625
3 | 0.25 | 0.75
4 | 0.25 | 0.8125
5 | 0.25 | 0.85
1.5 | 0.25 | 0.5
10 | 0.25 | 0.925
2 | 0.1 | 0.55
The formula would find z from known x and y values.
e.g. if x = 6 and y = 0.25 what is z?
Thanks for any help solving the problem or pointing me in the right direction. I can try to explain the context if that helps, but perhaps it's easier just to work with the numbers.
extrapolation
edited Jul 30 at 7:17
Gerry Myerson
142k7143292
asked Jul 30 at 7:08
Laurence Lord
64
edited Jul 30 at 7:17
Gerry Myerson
142k7143292
edited Jul 30 at 7:17
Gerry Myerson
142k7143292
edited Jul 30 at 7:17
Gerry Myerson
142k7143292
142k7143292
asked Jul 30 at 7:08
Laurence Lord
64
asked Jul 30 at 7:08
Laurence Lord
64
asked Jul 30 at 7:08
Laurence Lord
64
64
2
There are infinitely many formulas that fit the given numbers but give widely different answers for $x=6$, $y=.25$. You have to know more about what kind of function you are trying to fit, in order to give a useful answer.
– Gerry Myerson
Jul 30 at 7:20
@GerryMyerson thanks for the help. How might I better define the kind of function I am looking for? I can't even find one formula that fits. Visually I understand the problem and could try to describe it in those terms or I can also supply more results from the model I have. I'm looking for the simpliest solution that gives the same results as the model.
– Laurence Lord
Jul 30 at 7:42
OK, here's one way, but it probably doesn't give you what you want. Temporarily ignore that last data point. Then all your data points have the same $y$-value, so you can ignore $y$ and get a polynomial $p(x)$ that fits all the points by using Lagrange interpolation (look it up!). Then you can find a number $c$ such that $p(x)+c(y-.25)$ fits all your points.
– Gerry Myerson
Jul 30 at 9:36
Hi Gerry, thanks for trying me on that. To be honest I don't know what a polynomial is and having tried to read up on Lagrange Interpolation I'm struggling to find anything I can understand... I'll keep trying
– Laurence Lord
Jul 30 at 17:53
@GerryMyerson thanks again. I'm clearer on polynomials now. Still not sure how to go about interpolation. In your example you show c(y – .25) but y is .25 so that will be zero. I'm confused.
– Laurence Lord
Jul 30 at 19:17
 |Â
show 3 more comments
2
There are infinitely many formulas that fit the given numbers but give widely different answers for $x=6$, $y=.25$. You have to know more about what kind of function you are trying to fit, in order to give a useful answer.
– Gerry Myerson
Jul 30 at 7:20
@GerryMyerson thanks for the help. How might I better define the kind of function I am looking for? I can't even find one formula that fits. Visually I understand the problem and could try to describe it in those terms or I can also supply more results from the model I have. I'm looking for the simpliest solution that gives the same results as the model.
– Laurence Lord
Jul 30 at 7:42
OK, here's one way, but it probably doesn't give you what you want. Temporarily ignore that last data point. Then all your data points have the same $y$-value, so you can ignore $y$ and get a polynomial $p(x)$ that fits all the points by using Lagrange interpolation (look it up!). Then you can find a number $c$ such that $p(x)+c(y-.25)$ fits all your points.
– Gerry Myerson
Jul 30 at 9:36
Hi Gerry, thanks for trying me on that. To be honest I don't know what a polynomial is and having tried to read up on Lagrange Interpolation I'm struggling to find anything I can understand... I'll keep trying
– Laurence Lord
Jul 30 at 17:53
@GerryMyerson thanks again. I'm clearer on polynomials now. Still not sure how to go about interpolation. In your example you show c(y – .25) but y is .25 so that will be zero. I'm confused.
– Laurence Lord
Jul 30 at 19:17
2
2
There are infinitely many formulas that fit the given numbers but give widely different answers for $x=6$, $y=.25$. You have to know more about what kind of function you are trying to fit, in order to give a useful answer.
– Gerry Myerson
Jul 30 at 7:20
There are infinitely many formulas that fit the given numbers but give widely different answers for $x=6$, $y=.25$. You have to know more about what kind of function you are trying to fit, in order to give a useful answer.
– Gerry Myerson
Jul 30 at 7:20
@GerryMyerson thanks for the help. How might I better define the kind of function I am looking for? I can't even find one formula that fits. Visually I understand the problem and could try to describe it in those terms or I can also supply more results from the model I have. I'm looking for the simpliest solution that gives the same results as the model.
– Laurence Lord
Jul 30 at 7:42
@GerryMyerson thanks for the help. How might I better define the kind of function I am looking for? I can't even find one formula that fits. Visually I understand the problem and could try to describe it in those terms or I can also supply more results from the model I have. I'm looking for the simpliest solution that gives the same results as the model.
– Laurence Lord
Jul 30 at 7:42
OK, here's one way, but it probably doesn't give you what you want. Temporarily ignore that last data point. Then all your data points have the same $y$-value, so you can ignore $y$ and get a polynomial $p(x)$ that fits all the points by using Lagrange interpolation (look it up!). Then you can find a number $c$ such that $p(x)+c(y-.25)$ fits all your points.
– Gerry Myerson
Jul 30 at 9:36
OK, here's one way, but it probably doesn't give you what you want. Temporarily ignore that last data point. Then all your data points have the same $y$-value, so you can ignore $y$ and get a polynomial $p(x)$ that fits all the points by using Lagrange interpolation (look it up!). Then you can find a number $c$ such that $p(x)+c(y-.25)$ fits all your points.
– Gerry Myerson
Jul 30 at 9:36
Hi Gerry, thanks for trying me on that. To be honest I don't know what a polynomial is and having tried to read up on Lagrange Interpolation I'm struggling to find anything I can understand... I'll keep trying
– Laurence Lord
Jul 30 at 17:53
Hi Gerry, thanks for trying me on that. To be honest I don't know what a polynomial is and having tried to read up on Lagrange Interpolation I'm struggling to find anything I can understand... I'll keep trying
– Laurence Lord
Jul 30 at 17:53
@GerryMyerson thanks again. I'm clearer on polynomials now. Still not sure how to go about interpolation. In your example you show c(y – .25) but y is .25 so that will be zero. I'm confused.
– Laurence Lord
Jul 30 at 19:17
@GerryMyerson thanks again. I'm clearer on polynomials now. Still not sure how to go about interpolation. In your example you show c(y – .25) but y is .25 so that will be zero. I'm confused.
– Laurence Lord
Jul 30 at 19:17
 |Â
show 3 more comments
1 Answer
1
active
oldest
votes
up vote
0
down vote
Welcome back to maths. I hope the the re-introduction will not be too fearful!
Visualizing 3-dimensional data is very hard.
Whenever possible, try to start with the simplest model possible to describe your situation and build upwards. Thus, as Gerry suggested, given that for all but one data point, the y-value is constant, consider the scenario if you temporarily ignore it and consider this a 2-dimensional question.
In this case, graphing $x$ versus $z$ on graph paper or via Excel should always be your first step.
In my experience as a mathematician and statistician, I always find it useful to mentally think what do I expect the answer to be, before I plug my data into any formulae or models.
Once you do this you will immediately realize how much your point $x=10, z=0.925$ is separated from all your other data points.
The second thing to notice is that if you ignore this point, too, then a line fits the rest of your data quite well. Using something like Excel, with its built-in capabilities of line-of-best-fit, you would get something like $z=0.15x+0.22$. This would give value of $z=1.12$ if $x=6$.
However, you will notice that the if you include that far-right point, the data does not really look like a straight line at all, but rather a curve (maybe a logarithmic graph). In this case you would probably get a $z$ vaue of $simeq 0.9$ if $x=6$.
This is where you have to balance up two questions:
Based on how I collected my data, would I expect to be a simple linear relationship or a more complex curve? If so, would I expect it to curve upwards or downwards.
Other questions like the following may also help: "Do i expect the line-of-best fit to go through the origin?", "Can we have negative $x$ values, and/or negative $z$ values?, "What might I expect for very large $x$?"How much confidence do I have that the point x=10, which lies much further away than the other points, is correct? Can we obtain more data in between $x=5$ and $x=10$ to strengthen our confidence?
Good luck.
answered Jul 30 at 14:38
Martin Roberts
1,204318
thank you for the help. Okay, keeping things two dimensional, I would expect the line to be a curve with z tending towards 1 as x increases. x can't be negative but it could be less than 1 which would lead to a negative value of z
– Laurence Lord
Jul 30 at 18:07
For a very large x I'd expect a value very close to 1.
– Laurence Lord
Jul 30 at 19:06
when x = 6, z = 0.875
– Laurence Lord
Jul 31 at 15:31
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
Welcome back to maths. I hope the the re-introduction will not be too fearful!
Visualizing 3-dimensional data is very hard.
Whenever possible, try to start with the simplest model possible to describe your situation and build upwards. Thus, as Gerry suggested, given that for all but one data point, the y-value is constant, consider the scenario if you temporarily ignore it and consider this a 2-dimensional question.
In this case, graphing $x$ versus $z$ on graph paper or via Excel should always be your first step.
In my experience as a mathematician and statistician, I always find it useful to mentally think what do I expect the answer to be, before I plug my data into any formulae or models.
Once you do this you will immediately realize how much your point $x=10, z=0.925$ is separated from all your other data points.
The second thing to notice is that if you ignore this point, too, then a line fits the rest of your data quite well. Using something like Excel, with its built-in capabilities of line-of-best-fit, you would get something like $z=0.15x+0.22$. This would give value of $z=1.12$ if $x=6$.
However, you will notice that the if you include that far-right point, the data does not really look like a straight line at all, but rather a curve (maybe a logarithmic graph). In this case you would probably get a $z$ vaue of $simeq 0.9$ if $x=6$.
This is where you have to balance up two questions:
Based on how I collected my data, would I expect to be a simple linear relationship or a more complex curve? If so, would I expect it to curve upwards or downwards.
Other questions like the following may also help: "Do i expect the line-of-best fit to go through the origin?", "Can we have negative $x$ values, and/or negative $z$ values?, "What might I expect for very large $x$?"How much confidence do I have that the point x=10, which lies much further away than the other points, is correct? Can we obtain more data in between $x=5$ and $x=10$ to strengthen our confidence?
Good luck.
answered Jul 30 at 14:38
Martin Roberts
1,204318
thank you for the help. Okay, keeping things two dimensional, I would expect the line to be a curve with z tending towards 1 as x increases. x can't be negative but it could be less than 1 which would lead to a negative value of z
– Laurence Lord
Jul 30 at 18:07
For a very large x I'd expect a value very close to 1.
– Laurence Lord
Jul 30 at 19:06
when x = 6, z = 0.875
– Laurence Lord
Jul 31 at 15:31
add a comment |Â
up vote
0
down vote
Welcome back to maths. I hope the the re-introduction will not be too fearful!
Visualizing 3-dimensional data is very hard.
Whenever possible, try to start with the simplest model possible to describe your situation and build upwards. Thus, as Gerry suggested, given that for all but one data point, the y-value is constant, consider the scenario if you temporarily ignore it and consider this a 2-dimensional question.
In this case, graphing $x$ versus $z$ on graph paper or via Excel should always be your first step.
In my experience as a mathematician and statistician, I always find it useful to mentally think what do I expect the answer to be, before I plug my data into any formulae or models.
Once you do this you will immediately realize how much your point $x=10, z=0.925$ is separated from all your other data points.
The second thing to notice is that if you ignore this point, too, then a line fits the rest of your data quite well. Using something like Excel, with its built-in capabilities of line-of-best-fit, you would get something like $z=0.15x+0.22$. This would give value of $z=1.12$ if $x=6$.
However, you will notice that the if you include that far-right point, the data does not really look like a straight line at all, but rather a curve (maybe a logarithmic graph). In this case you would probably get a $z$ vaue of $simeq 0.9$ if $x=6$.
This is where you have to balance up two questions:
Based on how I collected my data, would I expect to be a simple linear relationship or a more complex curve? If so, would I expect it to curve upwards or downwards.
Other questions like the following may also help: "Do i expect the line-of-best fit to go through the origin?", "Can we have negative $x$ values, and/or negative $z$ values?, "What might I expect for very large $x$?"How much confidence do I have that the point x=10, which lies much further away than the other points, is correct? Can we obtain more data in between $x=5$ and $x=10$ to strengthen our confidence?
Good luck.
answered Jul 30 at 14:38
Martin Roberts
1,204318
thank you for the help. Okay, keeping things two dimensional, I would expect the line to be a curve with z tending towards 1 as x increases. x can't be negative but it could be less than 1 which would lead to a negative value of z
– Laurence Lord
Jul 30 at 18:07
For a very large x I'd expect a value very close to 1.
– Laurence Lord
Jul 30 at 19:06
when x = 6, z = 0.875
– Laurence Lord
Jul 31 at 15:31
add a comment |Â
up vote
0
down vote
up vote
0
down vote
Welcome back to maths. I hope the the re-introduction will not be too fearful!
Visualizing 3-dimensional data is very hard.
Whenever possible, try to start with the simplest model possible to describe your situation and build upwards. Thus, as Gerry suggested, given that for all but one data point, the y-value is constant, consider the scenario if you temporarily ignore it and consider this a 2-dimensional question.
In this case, graphing $x$ versus $z$ on graph paper or via Excel should always be your first step.
In my experience as a mathematician and statistician, I always find it useful to mentally think what do I expect the answer to be, before I plug my data into any formulae or models.
Once you do this you will immediately realize how much your point $x=10, z=0.925$ is separated from all your other data points.
The second thing to notice is that if you ignore this point, too, then a line fits the rest of your data quite well. Using something like Excel, with its built-in capabilities of line-of-best-fit, you would get something like $z=0.15x+0.22$. This would give value of $z=1.12$ if $x=6$.
However, you will notice that the if you include that far-right point, the data does not really look like a straight line at all, but rather a curve (maybe a logarithmic graph). In this case you would probably get a $z$ vaue of $simeq 0.9$ if $x=6$.
This is where you have to balance up two questions:
Based on how I collected my data, would I expect to be a simple linear relationship or a more complex curve? If so, would I expect it to curve upwards or downwards.
Other questions like the following may also help: "Do i expect the line-of-best fit to go through the origin?", "Can we have negative $x$ values, and/or negative $z$ values?, "What might I expect for very large $x$?"How much confidence do I have that the point x=10, which lies much further away than the other points, is correct? Can we obtain more data in between $x=5$ and $x=10$ to strengthen our confidence?
Good luck.
answered Jul 30 at 14:38
Martin Roberts
1,204318
Welcome back to maths. I hope the the re-introduction will not be too fearful!
Visualizing 3-dimensional data is very hard.
Whenever possible, try to start with the simplest model possible to describe your situation and build upwards. Thus, as Gerry suggested, given that for all but one data point, the y-value is constant, consider the scenario if you temporarily ignore it and consider this a 2-dimensional question.
In this case, graphing $x$ versus $z$ on graph paper or via Excel should always be your first step.
In my experience as a mathematician and statistician, I always find it useful to mentally think what do I expect the answer to be, before I plug my data into any formulae or models.
Once you do this you will immediately realize how much your point $x=10, z=0.925$ is separated from all your other data points.
The second thing to notice is that if you ignore this point, too, then a line fits the rest of your data quite well. Using something like Excel, with its built-in capabilities of line-of-best-fit, you would get something like $z=0.15x+0.22$. This would give value of $z=1.12$ if $x=6$.
However, you will notice that the if you include that far-right point, the data does not really look like a straight line at all, but rather a curve (maybe a logarithmic graph). In this case you would probably get a $z$ vaue of $simeq 0.9$ if $x=6$.
This is where you have to balance up two questions:
Based on how I collected my data, would I expect to be a simple linear relationship or a more complex curve? If so, would I expect it to curve upwards or downwards.
Other questions like the following may also help: "Do i expect the line-of-best fit to go through the origin?", "Can we have negative $x$ values, and/or negative $z$ values?, "What might I expect for very large $x$?"How much confidence do I have that the point x=10, which lies much further away than the other points, is correct? Can we obtain more data in between $x=5$ and $x=10$ to strengthen our confidence?
Good luck.
answered Jul 30 at 14:38
Martin Roberts
1,204318
edited Jul 30 at 14:44
answered Jul 30 at 14:38
Martin Roberts
1,204318
answered Jul 30 at 14:38
Martin Roberts
1,204318
answered Jul 30 at 14:38
Martin Roberts
1,204318
1,204318
thank you for the help. Okay, keeping things two dimensional, I would expect the line to be a curve with z tending towards 1 as x increases. x can't be negative but it could be less than 1 which would lead to a negative value of z
– Laurence Lord
Jul 30 at 18:07
For a very large x I'd expect a value very close to 1.
– Laurence Lord
Jul 30 at 19:06
when x = 6, z = 0.875
– Laurence Lord
Jul 31 at 15:31
add a comment |Â
thank you for the help. Okay, keeping things two dimensional, I would expect the line to be a curve with z tending towards 1 as x increases. x can't be negative but it could be less than 1 which would lead to a negative value of z
– Laurence Lord
Jul 30 at 18:07
For a very large x I'd expect a value very close to 1.
– Laurence Lord
Jul 30 at 19:06
when x = 6, z = 0.875
– Laurence Lord
Jul 31 at 15:31
thank you for the help. Okay, keeping things two dimensional, I would expect the line to be a curve with z tending towards 1 as x increases. x can't be negative but it could be less than 1 which would lead to a negative value of z
– Laurence Lord
Jul 30 at 18:07
thank you for the help. Okay, keeping things two dimensional, I would expect the line to be a curve with z tending towards 1 as x increases. x can't be negative but it could be less than 1 which would lead to a negative value of z
– Laurence Lord
Jul 30 at 18:07
For a very large x I'd expect a value very close to 1.
– Laurence Lord
Jul 30 at 19:06
For a very large x I'd expect a value very close to 1.
– Laurence Lord
Jul 30 at 19:06
when x = 6, z = 0.875
– Laurence Lord
Jul 31 at 15:31
when x = 6, z = 0.875
– Laurence Lord
Jul 31 at 15:31
add a comment |Â
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2
There are infinitely many formulas that fit the given numbers but give widely different answers for $x=6$, $y=.25$. You have to know more about what kind of function you are trying to fit, in order to give a useful answer.
– Gerry Myerson
Jul 30 at 7:20
@GerryMyerson thanks for the help. How might I better define the kind of function I am looking for? I can't even find one formula that fits. Visually I understand the problem and could try to describe it in those terms or I can also supply more results from the model I have. I'm looking for the simpliest solution that gives the same results as the model.
– Laurence Lord
Jul 30 at 7:42
OK, here's one way, but it probably doesn't give you what you want. Temporarily ignore that last data point. Then all your data points have the same $y$-value, so you can ignore $y$ and get a polynomial $p(x)$ that fits all the points by using Lagrange interpolation (look it up!). Then you can find a number $c$ such that $p(x)+c(y-.25)$ fits all your points.
– Gerry Myerson
Jul 30 at 9:36
Hi Gerry, thanks for trying me on that. To be honest I don't know what a polynomial is and having tried to read up on Lagrange Interpolation I'm struggling to find anything I can understand... I'll keep trying
– Laurence Lord
Jul 30 at 17:53
@GerryMyerson thanks again. I'm clearer on polynomials now. Still not sure how to go about interpolation. In your example you show c(y – .25) but y is .25 so that will be zero. I'm confused.
– Laurence Lord
Jul 30 at 19:17