Do linear functions, in addition to power law functions, have the property of scale invariance?
I recently came across the concept of scale invariance in the context of power laws.
According to wiki:
Given a relation $f(x)=ax^{-k}$, scaling the argument $x$ by a constant factor $c$ causes only a proportionate scaling of the function itself.
Suppose we have a simple linear equation $f(x) = 3x$
If we scale the argument, $x$, by a constant $c$, we have $f(cx)$ = $3cx$.
In this case, the value of the function is scaled by the same factor, $c$, as the argument was scaled by. Does this mean that $f(x) = 3x$ exhibits scale invariance?
Is the idea that it's more interesting that a power law is scale invariant since it is scale invariant despite it dealing with exponents that are not equal to 1?
probability statistics
asked Nov 27 at 5:17
spacediver
33
add a comment |
I recently came across the concept of scale invariance in the context of power laws.
According to wiki:
Given a relation $f(x)=ax^{-k}$, scaling the argument $x$ by a constant factor $c$ causes only a proportionate scaling of the function itself.
Suppose we have a simple linear equation $f(x) = 3x$
If we scale the argument, $x$, by a constant $c$, we have $f(cx)$ = $3cx$.
In this case, the value of the function is scaled by the same factor, $c$, as the argument was scaled by. Does this mean that $f(x) = 3x$ exhibits scale invariance?
Is the idea that it's more interesting that a power law is scale invariant since it is scale invariant despite it dealing with exponents that are not equal to 1?
probability statistics
asked Nov 27 at 5:17
spacediver
33
The linear case is just $k = -1$.
– angryavian
Nov 27 at 5:21
Thanks, that's what I suspected, but wasn't sure if I was missing something. It just seems odd that a central definition of a power law is scale invariance. I suppose perhaps it's to help provide contrast to exponential functions, which don't exhibit scale invariance?
– spacediver
Nov 27 at 5:29
add a comment |
I recently came across the concept of scale invariance in the context of power laws.
According to wiki:
Given a relation $f(x)=ax^{-k}$, scaling the argument $x$ by a constant factor $c$ causes only a proportionate scaling of the function itself.
Suppose we have a simple linear equation $f(x) = 3x$
If we scale the argument, $x$, by a constant $c$, we have $f(cx)$ = $3cx$.
In this case, the value of the function is scaled by the same factor, $c$, as the argument was scaled by. Does this mean that $f(x) = 3x$ exhibits scale invariance?
Is the idea that it's more interesting that a power law is scale invariant since it is scale invariant despite it dealing with exponents that are not equal to 1?
probability statistics
asked Nov 27 at 5:17
spacediver
33
I recently came across the concept of scale invariance in the context of power laws.
According to wiki:
Given a relation $f(x)=ax^{-k}$, scaling the argument $x$ by a constant factor $c$ causes only a proportionate scaling of the function itself.
Suppose we have a simple linear equation $f(x) = 3x$
If we scale the argument, $x$, by a constant $c$, we have $f(cx)$ = $3cx$.
In this case, the value of the function is scaled by the same factor, $c$, as the argument was scaled by. Does this mean that $f(x) = 3x$ exhibits scale invariance?
Is the idea that it's more interesting that a power law is scale invariant since it is scale invariant despite it dealing with exponents that are not equal to 1?
probability statistics
probability statistics
asked Nov 27 at 5:17
spacediver
33
asked Nov 27 at 5:17
spacediver
33
asked Nov 27 at 5:17
spacediver
33
asked Nov 27 at 5:17
spacediver
33
asked Nov 27 at 5:17
spacediver
33
33
The linear case is just $k = -1$.
– angryavian
Nov 27 at 5:21
Thanks, that's what I suspected, but wasn't sure if I was missing something. It just seems odd that a central definition of a power law is scale invariance. I suppose perhaps it's to help provide contrast to exponential functions, which don't exhibit scale invariance?
– spacediver
Nov 27 at 5:29
add a comment |
The linear case is just $k = -1$.
– angryavian
Nov 27 at 5:21
Thanks, that's what I suspected, but wasn't sure if I was missing something. It just seems odd that a central definition of a power law is scale invariance. I suppose perhaps it's to help provide contrast to exponential functions, which don't exhibit scale invariance?
– spacediver
Nov 27 at 5:29
The linear case is just $k = -1$.
– angryavian
Nov 27 at 5:21
The linear case is just $k = -1$.
– angryavian
Nov 27 at 5:21
Thanks, that's what I suspected, but wasn't sure if I was missing something. It just seems odd that a central definition of a power law is scale invariance. I suppose perhaps it's to help provide contrast to exponential functions, which don't exhibit scale invariance?
– spacediver
Nov 27 at 5:29
Thanks, that's what I suspected, but wasn't sure if I was missing something. It just seems odd that a central definition of a power law is scale invariance. I suppose perhaps it's to help provide contrast to exponential functions, which don't exhibit scale invariance?
– spacediver
Nov 27 at 5:29
add a comment |
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The linear case is just $k = -1$.
– angryavian
Nov 27 at 5:21
Thanks, that's what I suspected, but wasn't sure if I was missing something. It just seems odd that a central definition of a power law is scale invariance. I suppose perhaps it's to help provide contrast to exponential functions, which don't exhibit scale invariance?
– spacediver
Nov 27 at 5:29