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Does this proof contain circular logic?
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I am trying to prove that $e^x$ is a solution to $f'(x)=f(x)$, and there is a point at which I am concerned that there might be circular logic.
Here's what I've got so far:
$$f'(x)=f(x)$$
$$fracf'(x)f(x)=1$$
$$intfracf'(x)f(x)dx=x+c_0$$
Letting $y=f(x)$ gives
$$intfracdyy=x+c_0$$
Which gives $$ln|y|=x+c_0$$
$$f(x)=c_1e^x$$
QED
The bit I'm concerned about is $intfracdyy=ln|y|$.
Is there any proof of $$intfracdxx=ln|x|$$
which doesn't use $fracddxe^x=e^x$?
Edit: If it wasn't clear, I am asking if there is a way that one can prove that $$intfracdxx=ln|x|$$
without relying on the fact that $fracddxe^x=e^x$?
proof-verification logic proof-writing
asked 1 hour ago
clathratus
1,005117
 |Â
show 3 more comments
up vote
1
down vote
favorite
I am trying to prove that $e^x$ is a solution to $f'(x)=f(x)$, and there is a point at which I am concerned that there might be circular logic.
Here's what I've got so far:
$$f'(x)=f(x)$$
$$fracf'(x)f(x)=1$$
$$intfracf'(x)f(x)dx=x+c_0$$
Letting $y=f(x)$ gives
$$intfracdyy=x+c_0$$
Which gives $$ln|y|=x+c_0$$
$$f(x)=c_1e^x$$
QED
The bit I'm concerned about is $intfracdyy=ln|y|$.
Is there any proof of $$intfracdxx=ln|x|$$
which doesn't use $fracddxe^x=e^x$?
Edit: If it wasn't clear, I am asking if there is a way that one can prove that $$intfracdxx=ln|x|$$
without relying on the fact that $fracddxe^x=e^x$?
proof-verification logic proof-writing
asked 1 hour ago
clathratus
1,005117
5
How do you define $e^x$? Some people define $ln(x) := int_1^x frac1t , dt, ; x>0$ and then define $e^x$ as its inverse. In that case, there is no circular reasoning.
– Sobi
1 hour ago
@Sobi which ever way makes my proof valid...?
– clathratus
1 hour ago
1
Assuming you have defined $e = lim_n to infty (1+1/n)^n$, you could differentiate $ln(x)$ directly and appeal to this limit. See math.stackexchange.com/questions/1341958/…
– welshman500
1 hour ago
@Sobi Using the definition of $exp(x)$ to be the inverse of $log(x)=int_1^x frac1t,dt$, one would need to still show that $exp(x)=(e)^x$, which is not directly evident!
– Mark Viola
1 hour ago
1
@MarkViola If one first define $log$ from the integral and then $exp$ as its inverse function, then one can define $a^x=exp(xlog a)$; in the particular case of $a=e=exp(1)$, we get $e^x=exp(x)$. No need to have predefined general exponential functions.
– egreg
1 hour ago
 |Â
show 3 more comments
up vote
1
down vote
favorite
up vote
1
down vote
favorite
I am trying to prove that $e^x$ is a solution to $f'(x)=f(x)$, and there is a point at which I am concerned that there might be circular logic.
Here's what I've got so far:
$$f'(x)=f(x)$$
$$fracf'(x)f(x)=1$$
$$intfracf'(x)f(x)dx=x+c_0$$
Letting $y=f(x)$ gives
$$intfracdyy=x+c_0$$
Which gives $$ln|y|=x+c_0$$
$$f(x)=c_1e^x$$
QED
The bit I'm concerned about is $intfracdyy=ln|y|$.
Is there any proof of $$intfracdxx=ln|x|$$
which doesn't use $fracddxe^x=e^x$?
Edit: If it wasn't clear, I am asking if there is a way that one can prove that $$intfracdxx=ln|x|$$
without relying on the fact that $fracddxe^x=e^x$?
proof-verification logic proof-writing
asked 1 hour ago
clathratus
1,005117
I am trying to prove that $e^x$ is a solution to $f'(x)=f(x)$, and there is a point at which I am concerned that there might be circular logic.
Here's what I've got so far:
$$f'(x)=f(x)$$
$$fracf'(x)f(x)=1$$
$$intfracf'(x)f(x)dx=x+c_0$$
Letting $y=f(x)$ gives
$$intfracdyy=x+c_0$$
Which gives $$ln|y|=x+c_0$$
$$f(x)=c_1e^x$$
QED
The bit I'm concerned about is $intfracdyy=ln|y|$.
Is there any proof of $$intfracdxx=ln|x|$$
which doesn't use $fracddxe^x=e^x$?
Edit: If it wasn't clear, I am asking if there is a way that one can prove that $$intfracdxx=ln|x|$$
without relying on the fact that $fracddxe^x=e^x$?
proof-verification logic proof-writing
proof-verification logic proof-writing
asked 1 hour ago
clathratus
1,005117
asked 1 hour ago
clathratus
1,005117
edited 57 mins ago
asked 1 hour ago
clathratus
1,005117
asked 1 hour ago
clathratus
1,005117
asked 1 hour ago
clathratus
1,005117
1,005117
5
How do you define $e^x$? Some people define $ln(x) := int_1^x frac1t , dt, ; x>0$ and then define $e^x$ as its inverse. In that case, there is no circular reasoning.
– Sobi
1 hour ago
@Sobi which ever way makes my proof valid...?
– clathratus
1 hour ago
1
Assuming you have defined $e = lim_n to infty (1+1/n)^n$, you could differentiate $ln(x)$ directly and appeal to this limit. See math.stackexchange.com/questions/1341958/…
– welshman500
1 hour ago
@Sobi Using the definition of $exp(x)$ to be the inverse of $log(x)=int_1^x frac1t,dt$, one would need to still show that $exp(x)=(e)^x$, which is not directly evident!
– Mark Viola
1 hour ago
1
@MarkViola If one first define $log$ from the integral and then $exp$ as its inverse function, then one can define $a^x=exp(xlog a)$; in the particular case of $a=e=exp(1)$, we get $e^x=exp(x)$. No need to have predefined general exponential functions.
– egreg
1 hour ago
 |Â
show 3 more comments
5
How do you define $e^x$? Some people define $ln(x) := int_1^x frac1t , dt, ; x>0$ and then define $e^x$ as its inverse. In that case, there is no circular reasoning.
– Sobi
1 hour ago
@Sobi which ever way makes my proof valid...?
– clathratus
1 hour ago
1
Assuming you have defined $e = lim_n to infty (1+1/n)^n$, you could differentiate $ln(x)$ directly and appeal to this limit. See math.stackexchange.com/questions/1341958/…
– welshman500
1 hour ago
@Sobi Using the definition of $exp(x)$ to be the inverse of $log(x)=int_1^x frac1t,dt$, one would need to still show that $exp(x)=(e)^x$, which is not directly evident!
– Mark Viola
1 hour ago
1
@MarkViola If one first define $log$ from the integral and then $exp$ as its inverse function, then one can define $a^x=exp(xlog a)$; in the particular case of $a=e=exp(1)$, we get $e^x=exp(x)$. No need to have predefined general exponential functions.
– egreg
1 hour ago
5
5
How do you define $e^x$? Some people define $ln(x) := int_1^x frac1t , dt, ; x>0$ and then define $e^x$ as its inverse. In that case, there is no circular reasoning.
– Sobi
1 hour ago
How do you define $e^x$? Some people define $ln(x) := int_1^x frac1t , dt, ; x>0$ and then define $e^x$ as its inverse. In that case, there is no circular reasoning.
– Sobi
1 hour ago
@Sobi which ever way makes my proof valid...?
– clathratus
1 hour ago
@Sobi which ever way makes my proof valid...?
– clathratus
1 hour ago
1
1
Assuming you have defined $e = lim_n to infty (1+1/n)^n$, you could differentiate $ln(x)$ directly and appeal to this limit. See math.stackexchange.com/questions/1341958/…
– welshman500
1 hour ago
Assuming you have defined $e = lim_n to infty (1+1/n)^n$, you could differentiate $ln(x)$ directly and appeal to this limit. See math.stackexchange.com/questions/1341958/…
– welshman500
1 hour ago
@Sobi Using the definition of $exp(x)$ to be the inverse of $log(x)=int_1^x frac1t,dt$, one would need to still show that $exp(x)=(e)^x$, which is not directly evident!
– Mark Viola
1 hour ago
@Sobi Using the definition of $exp(x)$ to be the inverse of $log(x)=int_1^x frac1t,dt$, one would need to still show that $exp(x)=(e)^x$, which is not directly evident!
– Mark Viola
1 hour ago
1
1
@MarkViola If one first define $log$ from the integral and then $exp$ as its inverse function, then one can define $a^x=exp(xlog a)$; in the particular case of $a=e=exp(1)$, we get $e^x=exp(x)$. No need to have predefined general exponential functions.
– egreg
1 hour ago
@MarkViola If one first define $log$ from the integral and then $exp$ as its inverse function, then one can define $a^x=exp(xlog a)$; in the particular case of $a=e=exp(1)$, we get $e^x=exp(x)$. No need to have predefined general exponential functions.
– egreg
1 hour ago
 |Â
show 3 more comments
4 Answers
4
active
oldest
votes
up vote
1
down vote
accepted
Here's a way to remove the circularity.
Define $g(x) = int_1^x dt/t$. Then
$$
g(xy) = int_1^xyfracdtt = int_1^xfracdtt + int_x^xyfracdtt = int_1^xfracdtt + int_1^yfracduu = g(x)+g(y)
$$
where $t = xu$ is used for the substitution. This property is unique to logarithm functions, and its base will be the number $e$ such that $g(e) = 1$. So $g(x) = log_e(x)$, and $g^-1(x) = e^x$.
answered 12 mins ago
eyeballfrog
5,336528
just for clarification: are you defining $e$ as the base of the logarithm?
– clathratus
11 mins ago
1
Yes. The base of any logarithm function is the value it maps to 1.
– eyeballfrog
9 mins ago
Okay I think I get it... We define $g(x)$ as the function that satisfies $$g(xy)=g(x)+g(y)$$ Then we note that its a logarithm, then we define $e$ as its base?
– clathratus
6 mins ago
add a comment |Â
up vote
2
down vote
This is one way to define the logarithmic functions.
$$ln x= int _1 ^x fracdtt, text for x > 0 $$
So with this convenience, your proof is OK.
answered 1 hour ago
Mohammad Riazi-Kermani
36.5k41956
does this definition ensure that $e^x$ and $ln x$ are inverses?
– clathratus
1 hour ago
@clathratus No, it does not. That is why they define $e^x$ to be the inverse of $ln x $
– Mohammad Riazi-Kermani
1 hour ago
2
Using the definition of $exp(x)$ to be the inverse of $log(x)=int_1^x frac1t,dt$, one would need to still show that $exp(x)=(e)^x$, which is not directly evident!
– Mark Viola
1 hour ago
@MarkViola how does one do so (prove that $exp(x)=e^x$)?
– clathratus
1 hour ago
1
See THIS and THIS, THIS, and THIS.
– Mark Viola
1 hour ago
 |Â
show 1 more comment
up vote
1
down vote
Assuming the question you are asking is indeed that which you are trying to solve, to show that $e^x$ is a solution to $f' = f$ all you have to do is show that $fracddxe^x = e^x$. Which can be fairly easily done by looking at the series definition of $e^x$ term by term and differentiating.
If you're interested in investigating whether or not $ke^x$ is the only solution to $f' = f$ (which it is), then other steps are necessary.
answered 1 hour ago
J. Arrillaga
463
I know how to prove that $De^x=e^x$ with series, but I want to know if solving the ODE is a valid proof
– clathratus
1 hour ago
Using the definition of $exp(x)=sum_n=0^infty fracx^nn!$, one would still need to show that $exp(x)=(e)^x$, which is not directly evident.
– Mark Viola
1 hour ago
add a comment |Â
up vote
1
down vote
The usual proof of this is by setting $g(x)=f(x)e^-x$ and prove $g'=0$.
See for instance: Proof that $Cexp(x)$ is the only set of functions for which $f(x) = f'(x)$
Nevertheless your reasoning is correct, but solving the ODE this way lacks of rigour (dividing by $f(x)$ without discussing the annulation). Instead it should be used as a hint to find that solutions looks like $Ce^x$ then by CL theorem you claim their maximality and uniqueness.
See for instance LinAlgMan's answer here: https://math.stackexchange.com/a/409974/399263
answered 26 mins ago
zwim
11.1k628
I like your approach here. Thank you
– clathratus
5 mins ago
add a comment |Â
4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
Here's a way to remove the circularity.
Define $g(x) = int_1^x dt/t$. Then
$$
g(xy) = int_1^xyfracdtt = int_1^xfracdtt + int_x^xyfracdtt = int_1^xfracdtt + int_1^yfracduu = g(x)+g(y)
$$
where $t = xu$ is used for the substitution. This property is unique to logarithm functions, and its base will be the number $e$ such that $g(e) = 1$. So $g(x) = log_e(x)$, and $g^-1(x) = e^x$.
answered 12 mins ago
eyeballfrog
5,336528
just for clarification: are you defining $e$ as the base of the logarithm?
– clathratus
11 mins ago
1
Yes. The base of any logarithm function is the value it maps to 1.
– eyeballfrog
9 mins ago
Okay I think I get it... We define $g(x)$ as the function that satisfies $$g(xy)=g(x)+g(y)$$ Then we note that its a logarithm, then we define $e$ as its base?
– clathratus
6 mins ago
add a comment |Â
up vote
1
down vote
accepted
Here's a way to remove the circularity.
Define $g(x) = int_1^x dt/t$. Then
$$
g(xy) = int_1^xyfracdtt = int_1^xfracdtt + int_x^xyfracdtt = int_1^xfracdtt + int_1^yfracduu = g(x)+g(y)
$$
where $t = xu$ is used for the substitution. This property is unique to logarithm functions, and its base will be the number $e$ such that $g(e) = 1$. So $g(x) = log_e(x)$, and $g^-1(x) = e^x$.
answered 12 mins ago
eyeballfrog
5,336528
just for clarification: are you defining $e$ as the base of the logarithm?
– clathratus
11 mins ago
1
Yes. The base of any logarithm function is the value it maps to 1.
– eyeballfrog
9 mins ago
Okay I think I get it... We define $g(x)$ as the function that satisfies $$g(xy)=g(x)+g(y)$$ Then we note that its a logarithm, then we define $e$ as its base?
– clathratus
6 mins ago
add a comment |Â
up vote
1
down vote
accepted
up vote
1
down vote
accepted
Here's a way to remove the circularity.
Define $g(x) = int_1^x dt/t$. Then
$$
g(xy) = int_1^xyfracdtt = int_1^xfracdtt + int_x^xyfracdtt = int_1^xfracdtt + int_1^yfracduu = g(x)+g(y)
$$
where $t = xu$ is used for the substitution. This property is unique to logarithm functions, and its base will be the number $e$ such that $g(e) = 1$. So $g(x) = log_e(x)$, and $g^-1(x) = e^x$.
answered 12 mins ago
eyeballfrog
5,336528
Here's a way to remove the circularity.
Define $g(x) = int_1^x dt/t$. Then
$$
g(xy) = int_1^xyfracdtt = int_1^xfracdtt + int_x^xyfracdtt = int_1^xfracdtt + int_1^yfracduu = g(x)+g(y)
$$
where $t = xu$ is used for the substitution. This property is unique to logarithm functions, and its base will be the number $e$ such that $g(e) = 1$. So $g(x) = log_e(x)$, and $g^-1(x) = e^x$.
answered 12 mins ago
eyeballfrog
5,336528
answered 12 mins ago
eyeballfrog
5,336528
answered 12 mins ago
eyeballfrog
5,336528
answered 12 mins ago
eyeballfrog
5,336528
5,336528
just for clarification: are you defining $e$ as the base of the logarithm?
– clathratus
11 mins ago
1
Yes. The base of any logarithm function is the value it maps to 1.
– eyeballfrog
9 mins ago
Okay I think I get it... We define $g(x)$ as the function that satisfies $$g(xy)=g(x)+g(y)$$ Then we note that its a logarithm, then we define $e$ as its base?
– clathratus
6 mins ago
add a comment |Â
just for clarification: are you defining $e$ as the base of the logarithm?
– clathratus
11 mins ago
1
Yes. The base of any logarithm function is the value it maps to 1.
– eyeballfrog
9 mins ago
Okay I think I get it... We define $g(x)$ as the function that satisfies $$g(xy)=g(x)+g(y)$$ Then we note that its a logarithm, then we define $e$ as its base?
– clathratus
6 mins ago
just for clarification: are you defining $e$ as the base of the logarithm?
– clathratus
11 mins ago
just for clarification: are you defining $e$ as the base of the logarithm?
– clathratus
11 mins ago
1
1
Yes. The base of any logarithm function is the value it maps to 1.
– eyeballfrog
9 mins ago
Yes. The base of any logarithm function is the value it maps to 1.
– eyeballfrog
9 mins ago
Okay I think I get it... We define $g(x)$ as the function that satisfies $$g(xy)=g(x)+g(y)$$ Then we note that its a logarithm, then we define $e$ as its base?
– clathratus
6 mins ago
Okay I think I get it... We define $g(x)$ as the function that satisfies $$g(xy)=g(x)+g(y)$$ Then we note that its a logarithm, then we define $e$ as its base?
– clathratus
6 mins ago
add a comment |Â
up vote
2
down vote
This is one way to define the logarithmic functions.
$$ln x= int _1 ^x fracdtt, text for x > 0 $$
So with this convenience, your proof is OK.
answered 1 hour ago
Mohammad Riazi-Kermani
36.5k41956
does this definition ensure that $e^x$ and $ln x$ are inverses?
– clathratus
1 hour ago
@clathratus No, it does not. That is why they define $e^x$ to be the inverse of $ln x $
– Mohammad Riazi-Kermani
1 hour ago
2
Using the definition of $exp(x)$ to be the inverse of $log(x)=int_1^x frac1t,dt$, one would need to still show that $exp(x)=(e)^x$, which is not directly evident!
– Mark Viola
1 hour ago
@MarkViola how does one do so (prove that $exp(x)=e^x$)?
– clathratus
1 hour ago
1
See THIS and THIS, THIS, and THIS.
– Mark Viola
1 hour ago
 |Â
show 1 more comment
up vote
2
down vote
This is one way to define the logarithmic functions.
$$ln x= int _1 ^x fracdtt, text for x > 0 $$
So with this convenience, your proof is OK.
answered 1 hour ago
Mohammad Riazi-Kermani
36.5k41956
does this definition ensure that $e^x$ and $ln x$ are inverses?
– clathratus
1 hour ago
@clathratus No, it does not. That is why they define $e^x$ to be the inverse of $ln x $
– Mohammad Riazi-Kermani
1 hour ago
2
Using the definition of $exp(x)$ to be the inverse of $log(x)=int_1^x frac1t,dt$, one would need to still show that $exp(x)=(e)^x$, which is not directly evident!
– Mark Viola
1 hour ago
@MarkViola how does one do so (prove that $exp(x)=e^x$)?
– clathratus
1 hour ago
1
See THIS and THIS, THIS, and THIS.
– Mark Viola
1 hour ago
 |Â
show 1 more comment
up vote
2
down vote
up vote
2
down vote
This is one way to define the logarithmic functions.
$$ln x= int _1 ^x fracdtt, text for x > 0 $$
So with this convenience, your proof is OK.
answered 1 hour ago
Mohammad Riazi-Kermani
36.5k41956
This is one way to define the logarithmic functions.
$$ln x= int _1 ^x fracdtt, text for x > 0 $$
So with this convenience, your proof is OK.
answered 1 hour ago
Mohammad Riazi-Kermani
36.5k41956
answered 1 hour ago
Mohammad Riazi-Kermani
36.5k41956
answered 1 hour ago
Mohammad Riazi-Kermani
36.5k41956
answered 1 hour ago
Mohammad Riazi-Kermani
36.5k41956
36.5k41956
does this definition ensure that $e^x$ and $ln x$ are inverses?
– clathratus
1 hour ago
@clathratus No, it does not. That is why they define $e^x$ to be the inverse of $ln x $
– Mohammad Riazi-Kermani
1 hour ago
2
Using the definition of $exp(x)$ to be the inverse of $log(x)=int_1^x frac1t,dt$, one would need to still show that $exp(x)=(e)^x$, which is not directly evident!
– Mark Viola
1 hour ago
@MarkViola how does one do so (prove that $exp(x)=e^x$)?
– clathratus
1 hour ago
1
See THIS and THIS, THIS, and THIS.
– Mark Viola
1 hour ago
 |Â
show 1 more comment
does this definition ensure that $e^x$ and $ln x$ are inverses?
– clathratus
1 hour ago
@clathratus No, it does not. That is why they define $e^x$ to be the inverse of $ln x $
– Mohammad Riazi-Kermani
1 hour ago
2
Using the definition of $exp(x)$ to be the inverse of $log(x)=int_1^x frac1t,dt$, one would need to still show that $exp(x)=(e)^x$, which is not directly evident!
– Mark Viola
1 hour ago
@MarkViola how does one do so (prove that $exp(x)=e^x$)?
– clathratus
1 hour ago
1
See THIS and THIS, THIS, and THIS.
– Mark Viola
1 hour ago
does this definition ensure that $e^x$ and $ln x$ are inverses?
– clathratus
1 hour ago
does this definition ensure that $e^x$ and $ln x$ are inverses?
– clathratus
1 hour ago
@clathratus No, it does not. That is why they define $e^x$ to be the inverse of $ln x $
– Mohammad Riazi-Kermani
1 hour ago
@clathratus No, it does not. That is why they define $e^x$ to be the inverse of $ln x $
– Mohammad Riazi-Kermani
1 hour ago
2
2
Using the definition of $exp(x)$ to be the inverse of $log(x)=int_1^x frac1t,dt$, one would need to still show that $exp(x)=(e)^x$, which is not directly evident!
– Mark Viola
1 hour ago
Using the definition of $exp(x)$ to be the inverse of $log(x)=int_1^x frac1t,dt$, one would need to still show that $exp(x)=(e)^x$, which is not directly evident!
– Mark Viola
1 hour ago
@MarkViola how does one do so (prove that $exp(x)=e^x$)?
– clathratus
1 hour ago
@MarkViola how does one do so (prove that $exp(x)=e^x$)?
– clathratus
1 hour ago
1
1
See THIS and THIS, THIS, and THIS.
– Mark Viola
1 hour ago
See THIS and THIS, THIS, and THIS.
– Mark Viola
1 hour ago
 |Â
show 1 more comment
up vote
1
down vote
Assuming the question you are asking is indeed that which you are trying to solve, to show that $e^x$ is a solution to $f' = f$ all you have to do is show that $fracddxe^x = e^x$. Which can be fairly easily done by looking at the series definition of $e^x$ term by term and differentiating.
If you're interested in investigating whether or not $ke^x$ is the only solution to $f' = f$ (which it is), then other steps are necessary.
answered 1 hour ago
J. Arrillaga
463
I know how to prove that $De^x=e^x$ with series, but I want to know if solving the ODE is a valid proof
– clathratus
1 hour ago
Using the definition of $exp(x)=sum_n=0^infty fracx^nn!$, one would still need to show that $exp(x)=(e)^x$, which is not directly evident.
– Mark Viola
1 hour ago
add a comment |Â
up vote
1
down vote
Assuming the question you are asking is indeed that which you are trying to solve, to show that $e^x$ is a solution to $f' = f$ all you have to do is show that $fracddxe^x = e^x$. Which can be fairly easily done by looking at the series definition of $e^x$ term by term and differentiating.
If you're interested in investigating whether or not $ke^x$ is the only solution to $f' = f$ (which it is), then other steps are necessary.
answered 1 hour ago
J. Arrillaga
463
I know how to prove that $De^x=e^x$ with series, but I want to know if solving the ODE is a valid proof
– clathratus
1 hour ago
Using the definition of $exp(x)=sum_n=0^infty fracx^nn!$, one would still need to show that $exp(x)=(e)^x$, which is not directly evident.
– Mark Viola
1 hour ago
add a comment |Â
up vote
1
down vote
up vote
1
down vote
Assuming the question you are asking is indeed that which you are trying to solve, to show that $e^x$ is a solution to $f' = f$ all you have to do is show that $fracddxe^x = e^x$. Which can be fairly easily done by looking at the series definition of $e^x$ term by term and differentiating.
If you're interested in investigating whether or not $ke^x$ is the only solution to $f' = f$ (which it is), then other steps are necessary.
answered 1 hour ago
J. Arrillaga
463
Assuming the question you are asking is indeed that which you are trying to solve, to show that $e^x$ is a solution to $f' = f$ all you have to do is show that $fracddxe^x = e^x$. Which can be fairly easily done by looking at the series definition of $e^x$ term by term and differentiating.
If you're interested in investigating whether or not $ke^x$ is the only solution to $f' = f$ (which it is), then other steps are necessary.
answered 1 hour ago
J. Arrillaga
463
edited 1 hour ago
answered 1 hour ago
J. Arrillaga
463
answered 1 hour ago
J. Arrillaga
463
answered 1 hour ago
J. Arrillaga
463
463
I know how to prove that $De^x=e^x$ with series, but I want to know if solving the ODE is a valid proof
– clathratus
1 hour ago
Using the definition of $exp(x)=sum_n=0^infty fracx^nn!$, one would still need to show that $exp(x)=(e)^x$, which is not directly evident.
– Mark Viola
1 hour ago
add a comment |Â
I know how to prove that $De^x=e^x$ with series, but I want to know if solving the ODE is a valid proof
– clathratus
1 hour ago
Using the definition of $exp(x)=sum_n=0^infty fracx^nn!$, one would still need to show that $exp(x)=(e)^x$, which is not directly evident.
– Mark Viola
1 hour ago
I know how to prove that $De^x=e^x$ with series, but I want to know if solving the ODE is a valid proof
– clathratus
1 hour ago
I know how to prove that $De^x=e^x$ with series, but I want to know if solving the ODE is a valid proof
– clathratus
1 hour ago
Using the definition of $exp(x)=sum_n=0^infty fracx^nn!$, one would still need to show that $exp(x)=(e)^x$, which is not directly evident.
– Mark Viola
1 hour ago
Using the definition of $exp(x)=sum_n=0^infty fracx^nn!$, one would still need to show that $exp(x)=(e)^x$, which is not directly evident.
– Mark Viola
1 hour ago
add a comment |Â
up vote
1
down vote
The usual proof of this is by setting $g(x)=f(x)e^-x$ and prove $g'=0$.
See for instance: Proof that $Cexp(x)$ is the only set of functions for which $f(x) = f'(x)$
Nevertheless your reasoning is correct, but solving the ODE this way lacks of rigour (dividing by $f(x)$ without discussing the annulation). Instead it should be used as a hint to find that solutions looks like $Ce^x$ then by CL theorem you claim their maximality and uniqueness.
See for instance LinAlgMan's answer here: https://math.stackexchange.com/a/409974/399263
answered 26 mins ago
zwim
11.1k628
I like your approach here. Thank you
– clathratus
5 mins ago
add a comment |Â
up vote
1
down vote
The usual proof of this is by setting $g(x)=f(x)e^-x$ and prove $g'=0$.
See for instance: Proof that $Cexp(x)$ is the only set of functions for which $f(x) = f'(x)$
Nevertheless your reasoning is correct, but solving the ODE this way lacks of rigour (dividing by $f(x)$ without discussing the annulation). Instead it should be used as a hint to find that solutions looks like $Ce^x$ then by CL theorem you claim their maximality and uniqueness.
See for instance LinAlgMan's answer here: https://math.stackexchange.com/a/409974/399263
answered 26 mins ago
zwim
11.1k628
I like your approach here. Thank you
– clathratus
5 mins ago
add a comment |Â
up vote
1
down vote
up vote
1
down vote
The usual proof of this is by setting $g(x)=f(x)e^-x$ and prove $g'=0$.
See for instance: Proof that $Cexp(x)$ is the only set of functions for which $f(x) = f'(x)$
Nevertheless your reasoning is correct, but solving the ODE this way lacks of rigour (dividing by $f(x)$ without discussing the annulation). Instead it should be used as a hint to find that solutions looks like $Ce^x$ then by CL theorem you claim their maximality and uniqueness.
See for instance LinAlgMan's answer here: https://math.stackexchange.com/a/409974/399263
answered 26 mins ago
zwim
11.1k628
The usual proof of this is by setting $g(x)=f(x)e^-x$ and prove $g'=0$.
See for instance: Proof that $Cexp(x)$ is the only set of functions for which $f(x) = f'(x)$
Nevertheless your reasoning is correct, but solving the ODE this way lacks of rigour (dividing by $f(x)$ without discussing the annulation). Instead it should be used as a hint to find that solutions looks like $Ce^x$ then by CL theorem you claim their maximality and uniqueness.
See for instance LinAlgMan's answer here: https://math.stackexchange.com/a/409974/399263
answered 26 mins ago
zwim
11.1k628
answered 26 mins ago
zwim
11.1k628
answered 26 mins ago
zwim
11.1k628
answered 26 mins ago
zwim
11.1k628
11.1k628
I like your approach here. Thank you
– clathratus
5 mins ago
add a comment |Â
I like your approach here. Thank you
– clathratus
5 mins ago
I like your approach here. Thank you
– clathratus
5 mins ago
I like your approach here. Thank you
– clathratus
5 mins ago
add a comment |Â
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5
How do you define $e^x$? Some people define $ln(x) := int_1^x frac1t , dt, ; x>0$ and then define $e^x$ as its inverse. In that case, there is no circular reasoning.
– Sobi
1 hour ago
@Sobi which ever way makes my proof valid...?
– clathratus
1 hour ago
1
Assuming you have defined $e = lim_n to infty (1+1/n)^n$, you could differentiate $ln(x)$ directly and appeal to this limit. See math.stackexchange.com/questions/1341958/…
– welshman500
1 hour ago
@Sobi Using the definition of $exp(x)$ to be the inverse of $log(x)=int_1^x frac1t,dt$, one would need to still show that $exp(x)=(e)^x$, which is not directly evident!
– Mark Viola
1 hour ago
1
@MarkViola If one first define $log$ from the integral and then $exp$ as its inverse function, then one can define $a^x=exp(xlog a)$; in the particular case of $a=e=exp(1)$, we get $e^x=exp(x)$. No need to have predefined general exponential functions.
– egreg
1 hour ago