Finding exponential behaviour from log plots
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Suppose we have a plot of $y$ vs $x,$ where we expect to see an exponential behaviour and we want to extract the parameters. Let's give our expected behaviour the following expression: $$y = ae^bxtag1$$
Then upon taking the (natural) log of the (1) we obtain:
$$
lny = lna+bx tag2
$$
which tells us that if we simply take the $ln$ of our y-values and plot $ln(y)$ vs $x,$ we expect to see a linear behaviour if the data indeed exhibits an exponential behaviour, and then by taking the slope we'll be able to extract the key parameter $b$, let's call this method (i). Alternatively, we can simply plot $y$ vs $x$ on semilogy plot, where the values are the same except that the y-axis is now shown on a log scale, and again we can take the slope for $b,$ let's call this approach (ii). In addition, one can also plot $y$ vs $x$ on a loglog plot, as a third approach (iii).
- The question is, if we want to not only test that we have a behaviour such as in $(1)$ but to also easily extract the parameters, which of the three approaches (above) should one opt for? and is there any difference or all they really equivalent? (e.g., one difference could be that in one method one may have to perform further operation on the slope to extract $b$).
calculus statistics logarithms exponential-function
edited Aug 29 at 9:48
xbh
3,252320
asked Aug 29 at 9:26
user929304
515419
add a comment |Â
up vote
0
down vote
favorite
Suppose we have a plot of $y$ vs $x,$ where we expect to see an exponential behaviour and we want to extract the parameters. Let's give our expected behaviour the following expression: $$y = ae^bxtag1$$
Then upon taking the (natural) log of the (1) we obtain:
$$
lny = lna+bx tag2
$$
which tells us that if we simply take the $ln$ of our y-values and plot $ln(y)$ vs $x,$ we expect to see a linear behaviour if the data indeed exhibits an exponential behaviour, and then by taking the slope we'll be able to extract the key parameter $b$, let's call this method (i). Alternatively, we can simply plot $y$ vs $x$ on semilogy plot, where the values are the same except that the y-axis is now shown on a log scale, and again we can take the slope for $b,$ let's call this approach (ii). In addition, one can also plot $y$ vs $x$ on a loglog plot, as a third approach (iii).
- The question is, if we want to not only test that we have a behaviour such as in $(1)$ but to also easily extract the parameters, which of the three approaches (above) should one opt for? and is there any difference or all they really equivalent? (e.g., one difference could be that in one method one may have to perform further operation on the slope to extract $b$).
calculus statistics logarithms exponential-function
edited Aug 29 at 9:48
xbh
3,252320
asked Aug 29 at 9:26
user929304
515419
Are you thinking about regression problems ?
– Claude Leibovici
Aug 29 at 9:34
@ClaudeLeibovici Right, I have a dataset y-x, from observation guess an exponential behaviour, now I want to test and extract the corresponding parameters (if model (1) is adopted).
– user929304
Aug 29 at 9:40
What is the difference between (i) and (ii)?
– tst
Aug 29 at 9:54
This is what I was suspecting. In fact, since what is measured is $y$ and nothing else (such as $log(y)$), you will need to end with a nonlinear regression : this measn that you need starting values. So, the simplest is to use $(2)$, perform a linear rgeression $z=alpha+bx$ then $a=e^alpha$ and $b$ are your starting guesses. Never stop after any transformation of the data.
– Claude Leibovici
Aug 29 at 9:56
What @tst asked; and also: how does (iii) allow you to extract a parameter?
– joriki
Aug 29 at 11:31
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Suppose we have a plot of $y$ vs $x,$ where we expect to see an exponential behaviour and we want to extract the parameters. Let's give our expected behaviour the following expression: $$y = ae^bxtag1$$
Then upon taking the (natural) log of the (1) we obtain:
$$
lny = lna+bx tag2
$$
which tells us that if we simply take the $ln$ of our y-values and plot $ln(y)$ vs $x,$ we expect to see a linear behaviour if the data indeed exhibits an exponential behaviour, and then by taking the slope we'll be able to extract the key parameter $b$, let's call this method (i). Alternatively, we can simply plot $y$ vs $x$ on semilogy plot, where the values are the same except that the y-axis is now shown on a log scale, and again we can take the slope for $b,$ let's call this approach (ii). In addition, one can also plot $y$ vs $x$ on a loglog plot, as a third approach (iii).
- The question is, if we want to not only test that we have a behaviour such as in $(1)$ but to also easily extract the parameters, which of the three approaches (above) should one opt for? and is there any difference or all they really equivalent? (e.g., one difference could be that in one method one may have to perform further operation on the slope to extract $b$).
calculus statistics logarithms exponential-function
edited Aug 29 at 9:48
xbh
3,252320
asked Aug 29 at 9:26
user929304
515419
Suppose we have a plot of $y$ vs $x,$ where we expect to see an exponential behaviour and we want to extract the parameters. Let's give our expected behaviour the following expression: $$y = ae^bxtag1$$
Then upon taking the (natural) log of the (1) we obtain:
$$
lny = lna+bx tag2
$$
which tells us that if we simply take the $ln$ of our y-values and plot $ln(y)$ vs $x,$ we expect to see a linear behaviour if the data indeed exhibits an exponential behaviour, and then by taking the slope we'll be able to extract the key parameter $b$, let's call this method (i). Alternatively, we can simply plot $y$ vs $x$ on semilogy plot, where the values are the same except that the y-axis is now shown on a log scale, and again we can take the slope for $b,$ let's call this approach (ii). In addition, one can also plot $y$ vs $x$ on a loglog plot, as a third approach (iii).
- The question is, if we want to not only test that we have a behaviour such as in $(1)$ but to also easily extract the parameters, which of the three approaches (above) should one opt for? and is there any difference or all they really equivalent? (e.g., one difference could be that in one method one may have to perform further operation on the slope to extract $b$).
calculus statistics logarithms exponential-function
edited Aug 29 at 9:48
xbh
3,252320
asked Aug 29 at 9:26
user929304
515419
edited Aug 29 at 9:48
xbh
3,252320
edited Aug 29 at 9:48
xbh
3,252320
edited Aug 29 at 9:48
xbh
3,252320
3,252320
asked Aug 29 at 9:26
user929304
515419
asked Aug 29 at 9:26
user929304
515419
asked Aug 29 at 9:26
user929304
515419
515419
Are you thinking about regression problems ?
– Claude Leibovici
Aug 29 at 9:34
@ClaudeLeibovici Right, I have a dataset y-x, from observation guess an exponential behaviour, now I want to test and extract the corresponding parameters (if model (1) is adopted).
– user929304
Aug 29 at 9:40
What is the difference between (i) and (ii)?
– tst
Aug 29 at 9:54
This is what I was suspecting. In fact, since what is measured is $y$ and nothing else (such as $log(y)$), you will need to end with a nonlinear regression : this measn that you need starting values. So, the simplest is to use $(2)$, perform a linear rgeression $z=alpha+bx$ then $a=e^alpha$ and $b$ are your starting guesses. Never stop after any transformation of the data.
– Claude Leibovici
Aug 29 at 9:56
What @tst asked; and also: how does (iii) allow you to extract a parameter?
– joriki
Aug 29 at 11:31
add a comment |Â
Are you thinking about regression problems ?
– Claude Leibovici
Aug 29 at 9:34
@ClaudeLeibovici Right, I have a dataset y-x, from observation guess an exponential behaviour, now I want to test and extract the corresponding parameters (if model (1) is adopted).
– user929304
Aug 29 at 9:40
What is the difference between (i) and (ii)?
– tst
Aug 29 at 9:54
This is what I was suspecting. In fact, since what is measured is $y$ and nothing else (such as $log(y)$), you will need to end with a nonlinear regression : this measn that you need starting values. So, the simplest is to use $(2)$, perform a linear rgeression $z=alpha+bx$ then $a=e^alpha$ and $b$ are your starting guesses. Never stop after any transformation of the data.
– Claude Leibovici
Aug 29 at 9:56
What @tst asked; and also: how does (iii) allow you to extract a parameter?
– joriki
Aug 29 at 11:31
Are you thinking about regression problems ?
– Claude Leibovici
Aug 29 at 9:34
Are you thinking about regression problems ?
– Claude Leibovici
Aug 29 at 9:34
@ClaudeLeibovici Right, I have a dataset y-x, from observation guess an exponential behaviour, now I want to test and extract the corresponding parameters (if model (1) is adopted).
– user929304
Aug 29 at 9:40
@ClaudeLeibovici Right, I have a dataset y-x, from observation guess an exponential behaviour, now I want to test and extract the corresponding parameters (if model (1) is adopted).
– user929304
Aug 29 at 9:40
What is the difference between (i) and (ii)?
– tst
Aug 29 at 9:54
What is the difference between (i) and (ii)?
– tst
Aug 29 at 9:54
This is what I was suspecting. In fact, since what is measured is $y$ and nothing else (such as $log(y)$), you will need to end with a nonlinear regression : this measn that you need starting values. So, the simplest is to use $(2)$, perform a linear rgeression $z=alpha+bx$ then $a=e^alpha$ and $b$ are your starting guesses. Never stop after any transformation of the data.
– Claude Leibovici
Aug 29 at 9:56
This is what I was suspecting. In fact, since what is measured is $y$ and nothing else (such as $log(y)$), you will need to end with a nonlinear regression : this measn that you need starting values. So, the simplest is to use $(2)$, perform a linear rgeression $z=alpha+bx$ then $a=e^alpha$ and $b$ are your starting guesses. Never stop after any transformation of the data.
– Claude Leibovici
Aug 29 at 9:56
What @tst asked; and also: how does (iii) allow you to extract a parameter?
– joriki
Aug 29 at 11:31
What @tst asked; and also: how does (iii) allow you to extract a parameter?
– joriki
Aug 29 at 11:31
add a comment |Â
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Are you thinking about regression problems ?
– Claude Leibovici
Aug 29 at 9:34
@ClaudeLeibovici Right, I have a dataset y-x, from observation guess an exponential behaviour, now I want to test and extract the corresponding parameters (if model (1) is adopted).
– user929304
Aug 29 at 9:40
What is the difference between (i) and (ii)?
– tst
Aug 29 at 9:54
This is what I was suspecting. In fact, since what is measured is $y$ and nothing else (such as $log(y)$), you will need to end with a nonlinear regression : this measn that you need starting values. So, the simplest is to use $(2)$, perform a linear rgeression $z=alpha+bx$ then $a=e^alpha$ and $b$ are your starting guesses. Never stop after any transformation of the data.
– Claude Leibovici
Aug 29 at 9:56
What @tst asked; and also: how does (iii) allow you to extract a parameter?
– joriki
Aug 29 at 11:31