Proportionality and the ln function

[M. next]

Hi,

We know that if x=2, and y=4 for example, i.e that x is directly proportional to y.
What i am wondering about is when to say that x is proportional to lny? As in the case of Entropy in Statistical Physics where S proportional to lnW?

 

[Ryuzaki]

A variable x is said to be "directly proportional" (or simply proportional) to another variable y, if one is a linear multiple of the other, i.e. there exists some real number \lambda, such that x = \lambda y.

It's termed "inversely proportional" if x = \lambda/y.

So, if x is proportional to ln (W), there has to exist some real number \lambda such that x = \lambda ln(W).

 

[M. next]

Thanks for the reply, but how to know whether it is proportional to y or lny? Is there some rule other than plotting a graph perhaps?

 

[Ryuzaki]

Just compute the ratio x/y for different values of x and their corresponding y values. If this ratio remains a constant (the constant of proportionality \lambda ), then x is proportional to y.

Graphically, if x is directly proportional to y, i.e, if x = \lambda y then the graph of y as a function of x will be a straight line passing through the origin with the slope of the line equal to the constant of proportionality 1/\lambda.

 

[M. next]

Thanks again. But still I was concentrating on the lny and not the y. Anyway, I will depend on graphical illustrations.

 

[mfb]

It is straightforward to generalize Ryuzaki's reply to all other functions.

Just compute the ratio x/y for different values of x and their corresponding y values. If this ratio remains a constant (the constant of proportionality \lambda ), then x is proportional to y.

Graphically, if x is directly proportional to y, i.e, if x = \lambda y then the graph of y as a function of x will be a straight line passing through the origin with the slope of the line equal to the constant of proportionality 1/\lambda.

For $\ln(y)$:

Just compute the ratio x/\ln(y) for different values of x and their corresponding \ln(y) values. If this ratio remains a constant (the constant of proportionality \lambda ), then x is proportional to \ln(y).

Graphically, if x is directly proportional to ln(y), i.e, if x = \lambda \ln(y) then the graph of \ln(y) as a function of x will be a straight line passing through the origin with the slope of the line equal to the constant of proportionality 1/\lambda.

 

[Ryuzaki]

It is straightforward to generalize Ryuzaki's reply to all other functions.

For $\ln(y)$:

Just compute the ratio x/\ln(y) for different values of x and their corresponding \ln(y) values. If this ratio remains a constant (the constant of proportionality \lambda ), then x is proportional to \ln(y).

Graphically, if x is directly proportional to ln(y), i.e, if x = \lambda \ln(y) then the graph of \ln(y) as a function of x will be a straight line passing through the origin with the slope of the line equal to the constant of proportionality 1/\lambda.

Ah, you beat me to it.

 

[M. next]

Thanks guys. This is close to answering my question.