Possible definitions of exponential function
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I was wondering how many definitions of exponential functions can we think of. The basic ones could be:
$$e^x:=sum_k=0^inftyfracx^kk!$$
also
$$e^x:=lim_ntoinftybigg(1+fracxnbigg)^n$$
or this one:
Define $e^x:mathbbRrightarrowmathbbR\$ as unique function satisfying:
beginalign
e^xgeq x+1\
forall x,yinmathbbR:e^x+y=e^xe^y
endalign
Can anyone come up with something unusual? (Possibly with some explanation or references).
real-analysis definition big-list
add a comment |Â
up vote
4
down vote
favorite
I was wondering how many definitions of exponential functions can we think of. The basic ones could be:
$$e^x:=sum_k=0^inftyfracx^kk!$$
also
$$e^x:=lim_ntoinftybigg(1+fracxnbigg)^n$$
or this one:
Define $e^x:mathbbRrightarrowmathbbR\$ as unique function satisfying:
beginalign
e^xgeq x+1\
forall x,yinmathbbR:e^x+y=e^xe^y
endalign
Can anyone come up with something unusual? (Possibly with some explanation or references).
real-analysis definition big-list
Your second definition is incorrect: the second $;1;$ must be $;x;$ , and the last one just doesn't make sense as it doesn't define anything at all...
– DonAntonio
32 mins ago
1
Related : math.stackexchange.com/questions/833962/…
– Arnaud D.
19 mins ago
See also math.stackexchange.com/questions/1558734/…
– Arnaud D.
13 mins ago
add a comment |Â
up vote
4
down vote
favorite
up vote
4
down vote
favorite
I was wondering how many definitions of exponential functions can we think of. The basic ones could be:
$$e^x:=sum_k=0^inftyfracx^kk!$$
also
$$e^x:=lim_ntoinftybigg(1+fracxnbigg)^n$$
or this one:
Define $e^x:mathbbRrightarrowmathbbR\$ as unique function satisfying:
beginalign
e^xgeq x+1\
forall x,yinmathbbR:e^x+y=e^xe^y
endalign
Can anyone come up with something unusual? (Possibly with some explanation or references).
real-analysis definition big-list
I was wondering how many definitions of exponential functions can we think of. The basic ones could be:
$$e^x:=sum_k=0^inftyfracx^kk!$$
also
$$e^x:=lim_ntoinftybigg(1+fracxnbigg)^n$$
or this one:
Define $e^x:mathbbRrightarrowmathbbR\$ as unique function satisfying:
beginalign
e^xgeq x+1\
forall x,yinmathbbR:e^x+y=e^xe^y
endalign
Can anyone come up with something unusual? (Possibly with some explanation or references).
real-analysis definition big-list
real-analysis definition big-list
edited 28 mins ago
asked 33 mins ago
Michal Dvořák
764113
764113
Your second definition is incorrect: the second $;1;$ must be $;x;$ , and the last one just doesn't make sense as it doesn't define anything at all...
– DonAntonio
32 mins ago
1
Related : math.stackexchange.com/questions/833962/…
– Arnaud D.
19 mins ago
See also math.stackexchange.com/questions/1558734/…
– Arnaud D.
13 mins ago
add a comment |Â
Your second definition is incorrect: the second $;1;$ must be $;x;$ , and the last one just doesn't make sense as it doesn't define anything at all...
– DonAntonio
32 mins ago
1
Related : math.stackexchange.com/questions/833962/…
– Arnaud D.
19 mins ago
See also math.stackexchange.com/questions/1558734/…
– Arnaud D.
13 mins ago
Your second definition is incorrect: the second $;1;$ must be $;x;$ , and the last one just doesn't make sense as it doesn't define anything at all...
– DonAntonio
32 mins ago
Your second definition is incorrect: the second $;1;$ must be $;x;$ , and the last one just doesn't make sense as it doesn't define anything at all...
– DonAntonio
32 mins ago
1
1
Related : math.stackexchange.com/questions/833962/…
– Arnaud D.
19 mins ago
Related : math.stackexchange.com/questions/833962/…
– Arnaud D.
19 mins ago
See also math.stackexchange.com/questions/1558734/…
– Arnaud D.
13 mins ago
See also math.stackexchange.com/questions/1558734/…
– Arnaud D.
13 mins ago
add a comment |Â
4 Answers
4
active
oldest
votes
up vote
7
down vote
The exponential function is the unique solution of the initial value problem
$y'(x)=y(x) , quad y(0)=1$.
add a comment |Â
up vote
5
down vote
We can also define $e^x$ as
- the inverse function of $ln x$ (defining $ln x$ independently)
- the solution to $f'(x)=f(x)$ with $f(0)=1$
add a comment |Â
up vote
0
down vote
The function which is invarient under differentiatin or integration and passing through $(0,1)$
That's the same thing as Fred's answer (also mentioned by gimusi).
– Arnaud D.
21 mins ago
add a comment |Â
up vote
0
down vote
Define the value at rationals via powers and roots and then show that there is a unique continuous function which agrees with these values.
First define it for the natural numbers:
Define $e^2 = e times e$, $e^3 = e times e times e $, etc.
Now define it for other integers:
$e^0 = 1$, $e^-n = frac1e^n$, etc.
Now for other rational numbers (getting a bit harder):
$e^fracpq = sqrt[q]e^p$
Finally for irrational numbers $x$, you will need to prove that this definition evaluated for any sequence of rational numbers which converge to $x$ has a limit and that it is the same for all sequences which converge to $x$.
This is hard, especially the last step, but I think that it fits a common naive idea of what exponentiation is. We usually learn it in this sequence.
Can you elaborate your answer with something concrete?
– Michal Dvořák
25 mins ago
I expanded my answer.
– badjohn
14 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
7
down vote
The exponential function is the unique solution of the initial value problem
$y'(x)=y(x) , quad y(0)=1$.
add a comment |Â
up vote
7
down vote
The exponential function is the unique solution of the initial value problem
$y'(x)=y(x) , quad y(0)=1$.
add a comment |Â
up vote
7
down vote
up vote
7
down vote
The exponential function is the unique solution of the initial value problem
$y'(x)=y(x) , quad y(0)=1$.
The exponential function is the unique solution of the initial value problem
$y'(x)=y(x) , quad y(0)=1$.
answered 29 mins ago


Fred
40.8k1540
40.8k1540
add a comment |Â
add a comment |Â
up vote
5
down vote
We can also define $e^x$ as
- the inverse function of $ln x$ (defining $ln x$ independently)
- the solution to $f'(x)=f(x)$ with $f(0)=1$
add a comment |Â
up vote
5
down vote
We can also define $e^x$ as
- the inverse function of $ln x$ (defining $ln x$ independently)
- the solution to $f'(x)=f(x)$ with $f(0)=1$
add a comment |Â
up vote
5
down vote
up vote
5
down vote
We can also define $e^x$ as
- the inverse function of $ln x$ (defining $ln x$ independently)
- the solution to $f'(x)=f(x)$ with $f(0)=1$
We can also define $e^x$ as
- the inverse function of $ln x$ (defining $ln x$ independently)
- the solution to $f'(x)=f(x)$ with $f(0)=1$
answered 29 mins ago
gimusi
82.3k74090
82.3k74090
add a comment |Â
add a comment |Â
up vote
0
down vote
The function which is invarient under differentiatin or integration and passing through $(0,1)$
That's the same thing as Fred's answer (also mentioned by gimusi).
– Arnaud D.
21 mins ago
add a comment |Â
up vote
0
down vote
The function which is invarient under differentiatin or integration and passing through $(0,1)$
That's the same thing as Fred's answer (also mentioned by gimusi).
– Arnaud D.
21 mins ago
add a comment |Â
up vote
0
down vote
up vote
0
down vote
The function which is invarient under differentiatin or integration and passing through $(0,1)$
The function which is invarient under differentiatin or integration and passing through $(0,1)$
answered 22 mins ago


Mohammad Riazi-Kermani
38.2k41957
38.2k41957
That's the same thing as Fred's answer (also mentioned by gimusi).
– Arnaud D.
21 mins ago
add a comment |Â
That's the same thing as Fred's answer (also mentioned by gimusi).
– Arnaud D.
21 mins ago
That's the same thing as Fred's answer (also mentioned by gimusi).
– Arnaud D.
21 mins ago
That's the same thing as Fred's answer (also mentioned by gimusi).
– Arnaud D.
21 mins ago
add a comment |Â
up vote
0
down vote
Define the value at rationals via powers and roots and then show that there is a unique continuous function which agrees with these values.
First define it for the natural numbers:
Define $e^2 = e times e$, $e^3 = e times e times e $, etc.
Now define it for other integers:
$e^0 = 1$, $e^-n = frac1e^n$, etc.
Now for other rational numbers (getting a bit harder):
$e^fracpq = sqrt[q]e^p$
Finally for irrational numbers $x$, you will need to prove that this definition evaluated for any sequence of rational numbers which converge to $x$ has a limit and that it is the same for all sequences which converge to $x$.
This is hard, especially the last step, but I think that it fits a common naive idea of what exponentiation is. We usually learn it in this sequence.
Can you elaborate your answer with something concrete?
– Michal Dvořák
25 mins ago
I expanded my answer.
– badjohn
14 mins ago
add a comment |Â
up vote
0
down vote
Define the value at rationals via powers and roots and then show that there is a unique continuous function which agrees with these values.
First define it for the natural numbers:
Define $e^2 = e times e$, $e^3 = e times e times e $, etc.
Now define it for other integers:
$e^0 = 1$, $e^-n = frac1e^n$, etc.
Now for other rational numbers (getting a bit harder):
$e^fracpq = sqrt[q]e^p$
Finally for irrational numbers $x$, you will need to prove that this definition evaluated for any sequence of rational numbers which converge to $x$ has a limit and that it is the same for all sequences which converge to $x$.
This is hard, especially the last step, but I think that it fits a common naive idea of what exponentiation is. We usually learn it in this sequence.
Can you elaborate your answer with something concrete?
– Michal Dvořák
25 mins ago
I expanded my answer.
– badjohn
14 mins ago
add a comment |Â
up vote
0
down vote
up vote
0
down vote
Define the value at rationals via powers and roots and then show that there is a unique continuous function which agrees with these values.
First define it for the natural numbers:
Define $e^2 = e times e$, $e^3 = e times e times e $, etc.
Now define it for other integers:
$e^0 = 1$, $e^-n = frac1e^n$, etc.
Now for other rational numbers (getting a bit harder):
$e^fracpq = sqrt[q]e^p$
Finally for irrational numbers $x$, you will need to prove that this definition evaluated for any sequence of rational numbers which converge to $x$ has a limit and that it is the same for all sequences which converge to $x$.
This is hard, especially the last step, but I think that it fits a common naive idea of what exponentiation is. We usually learn it in this sequence.
Define the value at rationals via powers and roots and then show that there is a unique continuous function which agrees with these values.
First define it for the natural numbers:
Define $e^2 = e times e$, $e^3 = e times e times e $, etc.
Now define it for other integers:
$e^0 = 1$, $e^-n = frac1e^n$, etc.
Now for other rational numbers (getting a bit harder):
$e^fracpq = sqrt[q]e^p$
Finally for irrational numbers $x$, you will need to prove that this definition evaluated for any sequence of rational numbers which converge to $x$ has a limit and that it is the same for all sequences which converge to $x$.
This is hard, especially the last step, but I think that it fits a common naive idea of what exponentiation is. We usually learn it in this sequence.
edited 14 mins ago
answered 26 mins ago
badjohn
3,6291618
3,6291618
Can you elaborate your answer with something concrete?
– Michal Dvořák
25 mins ago
I expanded my answer.
– badjohn
14 mins ago
add a comment |Â
Can you elaborate your answer with something concrete?
– Michal Dvořák
25 mins ago
I expanded my answer.
– badjohn
14 mins ago
Can you elaborate your answer with something concrete?
– Michal Dvořák
25 mins ago
Can you elaborate your answer with something concrete?
– Michal Dvořák
25 mins ago
I expanded my answer.
– badjohn
14 mins ago
I expanded my answer.
– badjohn
14 mins ago
add a comment |Â
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Your second definition is incorrect: the second $;1;$ must be $;x;$ , and the last one just doesn't make sense as it doesn't define anything at all...
– DonAntonio
32 mins ago
1
Related : math.stackexchange.com/questions/833962/…
– Arnaud D.
19 mins ago
See also math.stackexchange.com/questions/1558734/…
– Arnaud D.
13 mins ago