Can any expression be written in terms of the natural exponent?
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Let's start off with a simple function say $y = x$. Can it be written in terms of the natural logarithm? If so, are there any functions that cannot?
exponential-function
edited Nov 14 at 15:19
William Grannis
708418
asked Nov 14 at 15:10
Dale
1,1421334
add a comment |
up vote
0
down vote
favorite
Let's start off with a simple function say $y = x$. Can it be written in terms of the natural logarithm? If so, are there any functions that cannot?
exponential-function
edited Nov 14 at 15:19
William Grannis
708418
asked Nov 14 at 15:10
Dale
1,1421334
Does $ln y = ln x$ count?
– William Grannis
Nov 14 at 15:11
Could you elaborate the question?
– Atharva Kathale
Nov 14 at 15:12
Of course, this only works as long as both $x$ and $y$ are greater than $0$.
– William Grannis
Nov 14 at 15:12
@JohnNash That is not rigorous, as you assume x and y are both larger than 0 without expressing it explicitly.
– William Grannis
Nov 14 at 15:21
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Let's start off with a simple function say $y = x$. Can it be written in terms of the natural logarithm? If so, are there any functions that cannot?
exponential-function
edited Nov 14 at 15:19
William Grannis
708418
asked Nov 14 at 15:10
Dale
1,1421334
Let's start off with a simple function say $y = x$. Can it be written in terms of the natural logarithm? If so, are there any functions that cannot?
exponential-function
exponential-function
edited Nov 14 at 15:19
William Grannis
708418
asked Nov 14 at 15:10
Dale
1,1421334
edited Nov 14 at 15:19
William Grannis
708418
asked Nov 14 at 15:10
Dale
1,1421334
edited Nov 14 at 15:19
William Grannis
708418
edited Nov 14 at 15:19
William Grannis
708418
edited Nov 14 at 15:19
William Grannis
708418
708418
asked Nov 14 at 15:10
Dale
1,1421334
asked Nov 14 at 15:10
Dale
1,1421334
asked Nov 14 at 15:10
Dale
1,1421334
1,1421334
Does $ln y = ln x$ count?
– William Grannis
Nov 14 at 15:11
Could you elaborate the question?
– Atharva Kathale
Nov 14 at 15:12
Of course, this only works as long as both $x$ and $y$ are greater than $0$.
– William Grannis
Nov 14 at 15:12
@JohnNash That is not rigorous, as you assume x and y are both larger than 0 without expressing it explicitly.
– William Grannis
Nov 14 at 15:21
add a comment |
Does $ln y = ln x$ count?
– William Grannis
Nov 14 at 15:11
Could you elaborate the question?
– Atharva Kathale
Nov 14 at 15:12
Of course, this only works as long as both $x$ and $y$ are greater than $0$.
– William Grannis
Nov 14 at 15:12
@JohnNash That is not rigorous, as you assume x and y are both larger than 0 without expressing it explicitly.
– William Grannis
Nov 14 at 15:21
Does $ln y = ln x$ count?
– William Grannis
Nov 14 at 15:11
Does $ln y = ln x$ count?
– William Grannis
Nov 14 at 15:11
Could you elaborate the question?
– Atharva Kathale
Nov 14 at 15:12
Could you elaborate the question?
– Atharva Kathale
Nov 14 at 15:12
Of course, this only works as long as both $x$ and $y$ are greater than $0$.
– William Grannis
Nov 14 at 15:12
Of course, this only works as long as both $x$ and $y$ are greater than $0$.
– William Grannis
Nov 14 at 15:12
@JohnNash That is not rigorous, as you assume x and y are both larger than 0 without expressing it explicitly.
– William Grannis
Nov 14 at 15:21
@JohnNash That is not rigorous, as you assume x and y are both larger than 0 without expressing it explicitly.
– William Grannis
Nov 14 at 15:21
add a comment |
2 Answers
2
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0
down vote
For any $x$ we have that
$$y=x iff y=ln e^x$$
and more in general
$$y=f(x) iff y=ln e^{f(x)}$$
Otherwise if we are interested in a $log-log$ identity
$$y=f(x) implies ln y = ln (f(x))$$
is true only for $f(x)>0$.
answered Nov 14 at 15:12
gimusi
86.3k74392
1
Although this is cheating, as by log rules we get $y = f(x) ln e $
– William Grannis
Nov 14 at 15:14
And obviously $ln e$ is $1$.
– William Grannis
Nov 14 at 15:15
@WilliamGrannis Yes indeed it is an identity!
– gimusi
Nov 14 at 15:16
@WilliamGrannis I've added also the case for the log-log expression.
– gimusi
Nov 14 at 15:19
add a comment |
up vote
0
down vote
There are plenty of functions which cannot. For example, consider Ackermann's Function. It can be rigorously shown that $A(x)$ is not expressible in any way save recursively. You could add an $ln$ to the definition, but it would still not be expressible in closed form.
answered Nov 14 at 15:20
William Grannis
708418
I'm not sure what "it can be rigorously shown that $A(x)$ is not expressible in any way save recursively" means. It's true that the Ackermann function isn't primitive recursive, but I don't see why that implies what you've written.
– Noah Schweber
Nov 14 at 15:21
add a comment |
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
For any $x$ we have that
$$y=x iff y=ln e^x$$
and more in general
$$y=f(x) iff y=ln e^{f(x)}$$
Otherwise if we are interested in a $log-log$ identity
$$y=f(x) implies ln y = ln (f(x))$$
is true only for $f(x)>0$.
answered Nov 14 at 15:12
gimusi
86.3k74392
1
Although this is cheating, as by log rules we get $y = f(x) ln e $
– William Grannis
Nov 14 at 15:14
And obviously $ln e$ is $1$.
– William Grannis
Nov 14 at 15:15
@WilliamGrannis Yes indeed it is an identity!
– gimusi
Nov 14 at 15:16
@WilliamGrannis I've added also the case for the log-log expression.
– gimusi
Nov 14 at 15:19
add a comment |
up vote
0
down vote
For any $x$ we have that
$$y=x iff y=ln e^x$$
and more in general
$$y=f(x) iff y=ln e^{f(x)}$$
Otherwise if we are interested in a $log-log$ identity
$$y=f(x) implies ln y = ln (f(x))$$
is true only for $f(x)>0$.
answered Nov 14 at 15:12
gimusi
86.3k74392
1
Although this is cheating, as by log rules we get $y = f(x) ln e $
– William Grannis
Nov 14 at 15:14
And obviously $ln e$ is $1$.
– William Grannis
Nov 14 at 15:15
@WilliamGrannis Yes indeed it is an identity!
– gimusi
Nov 14 at 15:16
@WilliamGrannis I've added also the case for the log-log expression.
– gimusi
Nov 14 at 15:19
add a comment |
up vote
0
down vote
up vote
0
down vote
For any $x$ we have that
$$y=x iff y=ln e^x$$
and more in general
$$y=f(x) iff y=ln e^{f(x)}$$
Otherwise if we are interested in a $log-log$ identity
$$y=f(x) implies ln y = ln (f(x))$$
is true only for $f(x)>0$.
answered Nov 14 at 15:12
gimusi
86.3k74392
For any $x$ we have that
$$y=x iff y=ln e^x$$
and more in general
$$y=f(x) iff y=ln e^{f(x)}$$
Otherwise if we are interested in a $log-log$ identity
$$y=f(x) implies ln y = ln (f(x))$$
is true only for $f(x)>0$.
answered Nov 14 at 15:12
gimusi
86.3k74392
edited Nov 14 at 15:18
answered Nov 14 at 15:12
gimusi
86.3k74392
answered Nov 14 at 15:12
gimusi
86.3k74392
answered Nov 14 at 15:12
gimusi
86.3k74392
86.3k74392
1
Although this is cheating, as by log rules we get $y = f(x) ln e $
– William Grannis
Nov 14 at 15:14
And obviously $ln e$ is $1$.
– William Grannis
Nov 14 at 15:15
@WilliamGrannis Yes indeed it is an identity!
– gimusi
Nov 14 at 15:16
@WilliamGrannis I've added also the case for the log-log expression.
– gimusi
Nov 14 at 15:19
add a comment |
1
Although this is cheating, as by log rules we get $y = f(x) ln e $
– William Grannis
Nov 14 at 15:14
And obviously $ln e$ is $1$.
– William Grannis
Nov 14 at 15:15
@WilliamGrannis Yes indeed it is an identity!
– gimusi
Nov 14 at 15:16
@WilliamGrannis I've added also the case for the log-log expression.
– gimusi
Nov 14 at 15:19
1
1
Although this is cheating, as by log rules we get $y = f(x) ln e $
– William Grannis
Nov 14 at 15:14
Although this is cheating, as by log rules we get $y = f(x) ln e $
– William Grannis
Nov 14 at 15:14
And obviously $ln e$ is $1$.
– William Grannis
Nov 14 at 15:15
And obviously $ln e$ is $1$.
– William Grannis
Nov 14 at 15:15
@WilliamGrannis Yes indeed it is an identity!
– gimusi
Nov 14 at 15:16
@WilliamGrannis Yes indeed it is an identity!
– gimusi
Nov 14 at 15:16
@WilliamGrannis I've added also the case for the log-log expression.
– gimusi
Nov 14 at 15:19
@WilliamGrannis I've added also the case for the log-log expression.
– gimusi
Nov 14 at 15:19
add a comment |
up vote
0
down vote
There are plenty of functions which cannot. For example, consider Ackermann's Function. It can be rigorously shown that $A(x)$ is not expressible in any way save recursively. You could add an $ln$ to the definition, but it would still not be expressible in closed form.
answered Nov 14 at 15:20
William Grannis
708418
I'm not sure what "it can be rigorously shown that $A(x)$ is not expressible in any way save recursively" means. It's true that the Ackermann function isn't primitive recursive, but I don't see why that implies what you've written.
– Noah Schweber
Nov 14 at 15:21
add a comment |
up vote
0
down vote
There are plenty of functions which cannot. For example, consider Ackermann's Function. It can be rigorously shown that $A(x)$ is not expressible in any way save recursively. You could add an $ln$ to the definition, but it would still not be expressible in closed form.
answered Nov 14 at 15:20
William Grannis
708418
I'm not sure what "it can be rigorously shown that $A(x)$ is not expressible in any way save recursively" means. It's true that the Ackermann function isn't primitive recursive, but I don't see why that implies what you've written.
– Noah Schweber
Nov 14 at 15:21
add a comment |
up vote
0
down vote
up vote
0
down vote
There are plenty of functions which cannot. For example, consider Ackermann's Function. It can be rigorously shown that $A(x)$ is not expressible in any way save recursively. You could add an $ln$ to the definition, but it would still not be expressible in closed form.
answered Nov 14 at 15:20
William Grannis
708418
There are plenty of functions which cannot. For example, consider Ackermann's Function. It can be rigorously shown that $A(x)$ is not expressible in any way save recursively. You could add an $ln$ to the definition, but it would still not be expressible in closed form.
answered Nov 14 at 15:20
William Grannis
708418
answered Nov 14 at 15:20
William Grannis
708418
answered Nov 14 at 15:20
William Grannis
708418
answered Nov 14 at 15:20
William Grannis
708418
708418
I'm not sure what "it can be rigorously shown that $A(x)$ is not expressible in any way save recursively" means. It's true that the Ackermann function isn't primitive recursive, but I don't see why that implies what you've written.
– Noah Schweber
Nov 14 at 15:21
add a comment |
I'm not sure what "it can be rigorously shown that $A(x)$ is not expressible in any way save recursively" means. It's true that the Ackermann function isn't primitive recursive, but I don't see why that implies what you've written.
– Noah Schweber
Nov 14 at 15:21
I'm not sure what "it can be rigorously shown that $A(x)$ is not expressible in any way save recursively" means. It's true that the Ackermann function isn't primitive recursive, but I don't see why that implies what you've written.
– Noah Schweber
Nov 14 at 15:21
I'm not sure what "it can be rigorously shown that $A(x)$ is not expressible in any way save recursively" means. It's true that the Ackermann function isn't primitive recursive, but I don't see why that implies what you've written.
– Noah Schweber
Nov 14 at 15:21
add a comment |
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Does $ln y = ln x$ count?
– William Grannis
Nov 14 at 15:11
Could you elaborate the question?
– Atharva Kathale
Nov 14 at 15:12
Of course, this only works as long as both $x$ and $y$ are greater than $0$.
– William Grannis
Nov 14 at 15:12
@JohnNash That is not rigorous, as you assume x and y are both larger than 0 without expressing it explicitly.
– William Grannis
Nov 14 at 15:21