Show that exp(x) is of the form $e^x$?
Let me define the function $exp(x)$ as that which satisfies:
$$ frac{d}{dx}exp(x) = exp(x) $$
We can show that
$$ exp(x) = sum_{n=0}^{infty} frac{x^n}{n!}.$$
So far so good, we have just defined the exponential by the differential equation it satisfies.
How can we show that the exponential, as defined above, is of the form:
$$exp(x) = e^x,$$
where $e$ is a real number.
In other words, how can we show, for example that:
$$ e^a = prod_{m=1}^{a}e^1$$
$$sum_{n=0}^infty frac{a^n}{n!} = prod_{m=1}^a left[ sum_{l=0}^{infty} frac{1}{n!} right]$$
for any positive integer $a$.
algebra-precalculus differential-equations functions special-functions
asked Nov 22 at 21:23
Daniel Duque
267
|
show 2 more comments
Let me define the function $exp(x)$ as that which satisfies:
$$ frac{d}{dx}exp(x) = exp(x) $$
We can show that
$$ exp(x) = sum_{n=0}^{infty} frac{x^n}{n!}.$$
So far so good, we have just defined the exponential by the differential equation it satisfies.
How can we show that the exponential, as defined above, is of the form:
$$exp(x) = e^x,$$
where $e$ is a real number.
In other words, how can we show, for example that:
$$ e^a = prod_{m=1}^{a}e^1$$
$$sum_{n=0}^infty frac{a^n}{n!} = prod_{m=1}^a left[ sum_{l=0}^{infty} frac{1}{n!} right]$$
for any positive integer $a$.
algebra-precalculus differential-equations functions special-functions
asked Nov 22 at 21:23
Daniel Duque
267
An interesting question. This is so often considered trivial that it is often treated as a definition.
– The Count
Nov 22 at 21:25
2
"We can show that ..." You need an initial condition for the differential equation.
– user587192
Nov 22 at 21:32
The first identity after "how can we show, for example that:" does not make sense.
– user587192
Nov 22 at 21:34
Usually, the limit definition is used to prove this property
– Yuriy S
Nov 22 at 21:37
@user587192 Could you expand on why it doesn't make sense?
– Daniel Duque
Nov 22 at 21:41
|
show 2 more comments
Let me define the function $exp(x)$ as that which satisfies:
$$ frac{d}{dx}exp(x) = exp(x) $$
We can show that
$$ exp(x) = sum_{n=0}^{infty} frac{x^n}{n!}.$$
So far so good, we have just defined the exponential by the differential equation it satisfies.
How can we show that the exponential, as defined above, is of the form:
$$exp(x) = e^x,$$
where $e$ is a real number.
In other words, how can we show, for example that:
$$ e^a = prod_{m=1}^{a}e^1$$
$$sum_{n=0}^infty frac{a^n}{n!} = prod_{m=1}^a left[ sum_{l=0}^{infty} frac{1}{n!} right]$$
for any positive integer $a$.
algebra-precalculus differential-equations functions special-functions
asked Nov 22 at 21:23
Daniel Duque
267
Let me define the function $exp(x)$ as that which satisfies:
$$ frac{d}{dx}exp(x) = exp(x) $$
We can show that
$$ exp(x) = sum_{n=0}^{infty} frac{x^n}{n!}.$$
So far so good, we have just defined the exponential by the differential equation it satisfies.
How can we show that the exponential, as defined above, is of the form:
$$exp(x) = e^x,$$
where $e$ is a real number.
In other words, how can we show, for example that:
$$ e^a = prod_{m=1}^{a}e^1$$
$$sum_{n=0}^infty frac{a^n}{n!} = prod_{m=1}^a left[ sum_{l=0}^{infty} frac{1}{n!} right]$$
for any positive integer $a$.
algebra-precalculus differential-equations functions special-functions
algebra-precalculus differential-equations functions special-functions
asked Nov 22 at 21:23
Daniel Duque
267
asked Nov 22 at 21:23
Daniel Duque
267
edited Nov 22 at 21:52
asked Nov 22 at 21:23
Daniel Duque
267
asked Nov 22 at 21:23
Daniel Duque
267
asked Nov 22 at 21:23
Daniel Duque
267
267
An interesting question. This is so often considered trivial that it is often treated as a definition.
– The Count
Nov 22 at 21:25
2
"We can show that ..." You need an initial condition for the differential equation.
– user587192
Nov 22 at 21:32
The first identity after "how can we show, for example that:" does not make sense.
– user587192
Nov 22 at 21:34
Usually, the limit definition is used to prove this property
– Yuriy S
Nov 22 at 21:37
@user587192 Could you expand on why it doesn't make sense?
– Daniel Duque
Nov 22 at 21:41
|
show 2 more comments
An interesting question. This is so often considered trivial that it is often treated as a definition.
– The Count
Nov 22 at 21:25
2
"We can show that ..." You need an initial condition for the differential equation.
– user587192
Nov 22 at 21:32
The first identity after "how can we show, for example that:" does not make sense.
– user587192
Nov 22 at 21:34
Usually, the limit definition is used to prove this property
– Yuriy S
Nov 22 at 21:37
@user587192 Could you expand on why it doesn't make sense?
– Daniel Duque
Nov 22 at 21:41
An interesting question. This is so often considered trivial that it is often treated as a definition.
– The Count
Nov 22 at 21:25
An interesting question. This is so often considered trivial that it is often treated as a definition.
– The Count
Nov 22 at 21:25
2
2
"We can show that ..." You need an initial condition for the differential equation.
– user587192
Nov 22 at 21:32
"We can show that ..." You need an initial condition for the differential equation.
– user587192
Nov 22 at 21:32
The first identity after "how can we show, for example that:" does not make sense.
– user587192
Nov 22 at 21:34
The first identity after "how can we show, for example that:" does not make sense.
– user587192
Nov 22 at 21:34
Usually, the limit definition is used to prove this property
– Yuriy S
Nov 22 at 21:37
Usually, the limit definition is used to prove this property
– Yuriy S
Nov 22 at 21:37
@user587192 Could you expand on why it doesn't make sense?
– Daniel Duque
Nov 22 at 21:41
@user587192 Could you expand on why it doesn't make sense?
– Daniel Duque
Nov 22 at 21:41
|
show 2 more comments
3 Answers
3
active
oldest
votes
First note that apart from $exp'=exp$, you need to postulate $exp(0)=1$. Then it is the uniqueness of the solution to the differential equation that determines the desired properties:
The derivative of $f(x)= bcdot exp(a+x)+cexp(x)$ apparently is $$f'(x)=bexp'(a+x)+cexp'(x)=bexp(a+x)+cexp(x)=f(x).$$ So if we adjust $b,c$ to ensure $f(0)=1$, we conclude that $fequiv exp$.
If $exp(a)=0$, we take $b=c=1$ and infer $exp(x)=exp(a+x)+exp(x)$ for all $x$, i.e., $expequiv 0$, contradiction.
Hence we may assume $exp(a)ne 0$, can take $c=0$ and $b=frac 1{exp(a)}$, and find
$$ exp(x)=frac{exp(a+x)}{exp(a)}$$
for all $a,xinBbb R$. In other words,
$$ exp(a+b)=exp(a)exp(b)$$
for all $a,binBbb R$.
From this, we learn among others (by induction) that $$exp(n)=exp(1)^n$$ for all $ninBbb Z$.
answered Nov 22 at 21:40
Hagen von Eitzen
275k21268495
Great answer, I would only add that we can say that $f(x) = exp(x)$ given the uniqueness of the solution to the differential equation. Do you recall any text book which introduces the exponential in the way you have answered my question?
– Daniel Duque
Nov 22 at 22:02
add a comment |
It suffices to show that $$exp(x+y)=exp (x)exp(y)$$But this can be easily shown since $$exp(x)exp(y){=sum_{n=0}^{infty}{x^nover n!}sum_{m=0}^{infty}{y^mover m!}\=sum_{n,m=0}^{infty}{x^ny^mover n!m!}\=sum_{k=0}^infty{1over k!}sum_{m+n=k}{k!x^ny^{k-n}over n!(k-n)!}\=sum_{k=0}^infty {(x+y)^kover k!}\=exp(x+y)}$$
answered Nov 22 at 22:11
Mostafa Ayaz
13.7k3836
add a comment |
One way is through $$e^xequivlim_{nrightarrowinfty}left(1+frac{x}{n}right)^n
$$
You can show through the binomial expansion that this recipe gives the Taylor series for $exp x$ and thence that it solves the differential equation; or by differentiating the expression with 'lim' in it directly; and also showing that this expression is also the natural extension of exponentiation into the domain of reals from its definition for integer argument in terms of iterated multiplication. If you ply these various relations backward & forward in every conceivable direction, bringing in also the fundamental definition of differentiation in terms of limits, it becomes clear that it all constitutes a beautifully internally consistent system of relations. And you can also get
$$frac{dy}{dx}equiv afrac{y}{x}$$ &
$$x=1implies y=1$$
as a definition of $x^a$ for general real $a$.
answered Nov 23 at 3:24
AmbretteOrrisey
56710
add a comment |
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3 Answers
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active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
First note that apart from $exp'=exp$, you need to postulate $exp(0)=1$. Then it is the uniqueness of the solution to the differential equation that determines the desired properties:
The derivative of $f(x)= bcdot exp(a+x)+cexp(x)$ apparently is $$f'(x)=bexp'(a+x)+cexp'(x)=bexp(a+x)+cexp(x)=f(x).$$ So if we adjust $b,c$ to ensure $f(0)=1$, we conclude that $fequiv exp$.
If $exp(a)=0$, we take $b=c=1$ and infer $exp(x)=exp(a+x)+exp(x)$ for all $x$, i.e., $expequiv 0$, contradiction.
Hence we may assume $exp(a)ne 0$, can take $c=0$ and $b=frac 1{exp(a)}$, and find
$$ exp(x)=frac{exp(a+x)}{exp(a)}$$
for all $a,xinBbb R$. In other words,
$$ exp(a+b)=exp(a)exp(b)$$
for all $a,binBbb R$.
From this, we learn among others (by induction) that $$exp(n)=exp(1)^n$$ for all $ninBbb Z$.
answered Nov 22 at 21:40
Hagen von Eitzen
275k21268495
Great answer, I would only add that we can say that $f(x) = exp(x)$ given the uniqueness of the solution to the differential equation. Do you recall any text book which introduces the exponential in the way you have answered my question?
– Daniel Duque
Nov 22 at 22:02
add a comment |
First note that apart from $exp'=exp$, you need to postulate $exp(0)=1$. Then it is the uniqueness of the solution to the differential equation that determines the desired properties:
The derivative of $f(x)= bcdot exp(a+x)+cexp(x)$ apparently is $$f'(x)=bexp'(a+x)+cexp'(x)=bexp(a+x)+cexp(x)=f(x).$$ So if we adjust $b,c$ to ensure $f(0)=1$, we conclude that $fequiv exp$.
If $exp(a)=0$, we take $b=c=1$ and infer $exp(x)=exp(a+x)+exp(x)$ for all $x$, i.e., $expequiv 0$, contradiction.
Hence we may assume $exp(a)ne 0$, can take $c=0$ and $b=frac 1{exp(a)}$, and find
$$ exp(x)=frac{exp(a+x)}{exp(a)}$$
for all $a,xinBbb R$. In other words,
$$ exp(a+b)=exp(a)exp(b)$$
for all $a,binBbb R$.
From this, we learn among others (by induction) that $$exp(n)=exp(1)^n$$ for all $ninBbb Z$.
answered Nov 22 at 21:40
Hagen von Eitzen
275k21268495
Great answer, I would only add that we can say that $f(x) = exp(x)$ given the uniqueness of the solution to the differential equation. Do you recall any text book which introduces the exponential in the way you have answered my question?
– Daniel Duque
Nov 22 at 22:02
add a comment |
First note that apart from $exp'=exp$, you need to postulate $exp(0)=1$. Then it is the uniqueness of the solution to the differential equation that determines the desired properties:
The derivative of $f(x)= bcdot exp(a+x)+cexp(x)$ apparently is $$f'(x)=bexp'(a+x)+cexp'(x)=bexp(a+x)+cexp(x)=f(x).$$ So if we adjust $b,c$ to ensure $f(0)=1$, we conclude that $fequiv exp$.
If $exp(a)=0$, we take $b=c=1$ and infer $exp(x)=exp(a+x)+exp(x)$ for all $x$, i.e., $expequiv 0$, contradiction.
Hence we may assume $exp(a)ne 0$, can take $c=0$ and $b=frac 1{exp(a)}$, and find
$$ exp(x)=frac{exp(a+x)}{exp(a)}$$
for all $a,xinBbb R$. In other words,
$$ exp(a+b)=exp(a)exp(b)$$
for all $a,binBbb R$.
From this, we learn among others (by induction) that $$exp(n)=exp(1)^n$$ for all $ninBbb Z$.
answered Nov 22 at 21:40
Hagen von Eitzen
275k21268495
First note that apart from $exp'=exp$, you need to postulate $exp(0)=1$. Then it is the uniqueness of the solution to the differential equation that determines the desired properties:
The derivative of $f(x)= bcdot exp(a+x)+cexp(x)$ apparently is $$f'(x)=bexp'(a+x)+cexp'(x)=bexp(a+x)+cexp(x)=f(x).$$ So if we adjust $b,c$ to ensure $f(0)=1$, we conclude that $fequiv exp$.
If $exp(a)=0$, we take $b=c=1$ and infer $exp(x)=exp(a+x)+exp(x)$ for all $x$, i.e., $expequiv 0$, contradiction.
Hence we may assume $exp(a)ne 0$, can take $c=0$ and $b=frac 1{exp(a)}$, and find
$$ exp(x)=frac{exp(a+x)}{exp(a)}$$
for all $a,xinBbb R$. In other words,
$$ exp(a+b)=exp(a)exp(b)$$
for all $a,binBbb R$.
From this, we learn among others (by induction) that $$exp(n)=exp(1)^n$$ for all $ninBbb Z$.
answered Nov 22 at 21:40
Hagen von Eitzen
275k21268495
answered Nov 22 at 21:40
Hagen von Eitzen
275k21268495
answered Nov 22 at 21:40
Hagen von Eitzen
275k21268495
answered Nov 22 at 21:40
Hagen von Eitzen
275k21268495
275k21268495
Great answer, I would only add that we can say that $f(x) = exp(x)$ given the uniqueness of the solution to the differential equation. Do you recall any text book which introduces the exponential in the way you have answered my question?
– Daniel Duque
Nov 22 at 22:02
add a comment |
Great answer, I would only add that we can say that $f(x) = exp(x)$ given the uniqueness of the solution to the differential equation. Do you recall any text book which introduces the exponential in the way you have answered my question?
– Daniel Duque
Nov 22 at 22:02
Great answer, I would only add that we can say that $f(x) = exp(x)$ given the uniqueness of the solution to the differential equation. Do you recall any text book which introduces the exponential in the way you have answered my question?
– Daniel Duque
Nov 22 at 22:02
Great answer, I would only add that we can say that $f(x) = exp(x)$ given the uniqueness of the solution to the differential equation. Do you recall any text book which introduces the exponential in the way you have answered my question?
– Daniel Duque
Nov 22 at 22:02
add a comment |
It suffices to show that $$exp(x+y)=exp (x)exp(y)$$But this can be easily shown since $$exp(x)exp(y){=sum_{n=0}^{infty}{x^nover n!}sum_{m=0}^{infty}{y^mover m!}\=sum_{n,m=0}^{infty}{x^ny^mover n!m!}\=sum_{k=0}^infty{1over k!}sum_{m+n=k}{k!x^ny^{k-n}over n!(k-n)!}\=sum_{k=0}^infty {(x+y)^kover k!}\=exp(x+y)}$$
answered Nov 22 at 22:11
Mostafa Ayaz
13.7k3836
add a comment |
It suffices to show that $$exp(x+y)=exp (x)exp(y)$$But this can be easily shown since $$exp(x)exp(y){=sum_{n=0}^{infty}{x^nover n!}sum_{m=0}^{infty}{y^mover m!}\=sum_{n,m=0}^{infty}{x^ny^mover n!m!}\=sum_{k=0}^infty{1over k!}sum_{m+n=k}{k!x^ny^{k-n}over n!(k-n)!}\=sum_{k=0}^infty {(x+y)^kover k!}\=exp(x+y)}$$
answered Nov 22 at 22:11
Mostafa Ayaz
13.7k3836
add a comment |
It suffices to show that $$exp(x+y)=exp (x)exp(y)$$But this can be easily shown since $$exp(x)exp(y){=sum_{n=0}^{infty}{x^nover n!}sum_{m=0}^{infty}{y^mover m!}\=sum_{n,m=0}^{infty}{x^ny^mover n!m!}\=sum_{k=0}^infty{1over k!}sum_{m+n=k}{k!x^ny^{k-n}over n!(k-n)!}\=sum_{k=0}^infty {(x+y)^kover k!}\=exp(x+y)}$$
answered Nov 22 at 22:11
Mostafa Ayaz
13.7k3836
It suffices to show that $$exp(x+y)=exp (x)exp(y)$$But this can be easily shown since $$exp(x)exp(y){=sum_{n=0}^{infty}{x^nover n!}sum_{m=0}^{infty}{y^mover m!}\=sum_{n,m=0}^{infty}{x^ny^mover n!m!}\=sum_{k=0}^infty{1over k!}sum_{m+n=k}{k!x^ny^{k-n}over n!(k-n)!}\=sum_{k=0}^infty {(x+y)^kover k!}\=exp(x+y)}$$
answered Nov 22 at 22:11
Mostafa Ayaz
13.7k3836
answered Nov 22 at 22:11
Mostafa Ayaz
13.7k3836
answered Nov 22 at 22:11
Mostafa Ayaz
13.7k3836
answered Nov 22 at 22:11
Mostafa Ayaz
13.7k3836
13.7k3836
add a comment |
add a comment |
One way is through $$e^xequivlim_{nrightarrowinfty}left(1+frac{x}{n}right)^n
$$
You can show through the binomial expansion that this recipe gives the Taylor series for $exp x$ and thence that it solves the differential equation; or by differentiating the expression with 'lim' in it directly; and also showing that this expression is also the natural extension of exponentiation into the domain of reals from its definition for integer argument in terms of iterated multiplication. If you ply these various relations backward & forward in every conceivable direction, bringing in also the fundamental definition of differentiation in terms of limits, it becomes clear that it all constitutes a beautifully internally consistent system of relations. And you can also get
$$frac{dy}{dx}equiv afrac{y}{x}$$ &
$$x=1implies y=1$$
as a definition of $x^a$ for general real $a$.
answered Nov 23 at 3:24
AmbretteOrrisey
56710
add a comment |
One way is through $$e^xequivlim_{nrightarrowinfty}left(1+frac{x}{n}right)^n
$$
You can show through the binomial expansion that this recipe gives the Taylor series for $exp x$ and thence that it solves the differential equation; or by differentiating the expression with 'lim' in it directly; and also showing that this expression is also the natural extension of exponentiation into the domain of reals from its definition for integer argument in terms of iterated multiplication. If you ply these various relations backward & forward in every conceivable direction, bringing in also the fundamental definition of differentiation in terms of limits, it becomes clear that it all constitutes a beautifully internally consistent system of relations. And you can also get
$$frac{dy}{dx}equiv afrac{y}{x}$$ &
$$x=1implies y=1$$
as a definition of $x^a$ for general real $a$.
answered Nov 23 at 3:24
AmbretteOrrisey
56710
add a comment |
One way is through $$e^xequivlim_{nrightarrowinfty}left(1+frac{x}{n}right)^n
$$
You can show through the binomial expansion that this recipe gives the Taylor series for $exp x$ and thence that it solves the differential equation; or by differentiating the expression with 'lim' in it directly; and also showing that this expression is also the natural extension of exponentiation into the domain of reals from its definition for integer argument in terms of iterated multiplication. If you ply these various relations backward & forward in every conceivable direction, bringing in also the fundamental definition of differentiation in terms of limits, it becomes clear that it all constitutes a beautifully internally consistent system of relations. And you can also get
$$frac{dy}{dx}equiv afrac{y}{x}$$ &
$$x=1implies y=1$$
as a definition of $x^a$ for general real $a$.
answered Nov 23 at 3:24
AmbretteOrrisey
56710
One way is through $$e^xequivlim_{nrightarrowinfty}left(1+frac{x}{n}right)^n
$$
You can show through the binomial expansion that this recipe gives the Taylor series for $exp x$ and thence that it solves the differential equation; or by differentiating the expression with 'lim' in it directly; and also showing that this expression is also the natural extension of exponentiation into the domain of reals from its definition for integer argument in terms of iterated multiplication. If you ply these various relations backward & forward in every conceivable direction, bringing in also the fundamental definition of differentiation in terms of limits, it becomes clear that it all constitutes a beautifully internally consistent system of relations. And you can also get
$$frac{dy}{dx}equiv afrac{y}{x}$$ &
$$x=1implies y=1$$
as a definition of $x^a$ for general real $a$.
answered Nov 23 at 3:24
AmbretteOrrisey
56710
answered Nov 23 at 3:24
AmbretteOrrisey
56710
answered Nov 23 at 3:24
AmbretteOrrisey
56710
answered Nov 23 at 3:24
AmbretteOrrisey
56710
56710
add a comment |
add a comment |
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An interesting question. This is so often considered trivial that it is often treated as a definition.
– The Count
Nov 22 at 21:25
2
"We can show that ..." You need an initial condition for the differential equation.
– user587192
Nov 22 at 21:32
The first identity after "how can we show, for example that:" does not make sense.
– user587192
Nov 22 at 21:34
Usually, the limit definition is used to prove this property
– Yuriy S
Nov 22 at 21:37
@user587192 Could you expand on why it doesn't make sense?
– Daniel Duque
Nov 22 at 21:41