Is there a closed form, or cleaner way of writing a function satisfying $fracd^nf(x)dx^n|_x=0=f(n)$ for all $n$?Finding the closed form for a sequenceA closed form for $int_0^inftyfracln(x+4)sqrtx,(x+3),(x+4)dx$Prove that the composition of two “closed form functions” is itself a “closed form function”?No closed form for $sum_nin P frac1n^2$Closed form of planetary radial motion as time functionClosed form for a series IIClosed form for Logarithmic equationIs there a closed form for this “flowery” integral?Deriving the closed form of Gamma function using Euler-Chi functionClosed form for $1+frac12 + frac13 + dots + frac1n$?
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Is there a closed form, or cleaner way of writing a function satisfying $fracd^nf(x)dx^n|_x=0=f(n)$ for all $n$?
Finding the closed form for a sequenceA closed form for $int_0^inftyfracln(x+4)sqrtx,(x+3),(x+4)dx$Prove that the composition of two “closed form functions” is itself a “closed form function”?No closed form for $sum_nin P frac1n^2$Closed form of planetary radial motion as time functionClosed form for a series IIClosed form for Logarithmic equationIs there a closed form for this “flowery” integral?Deriving the closed form of Gamma function using Euler-Chi functionClosed form for $1+frac12 + frac13 + dots + frac1n$?
$begingroup$
Given the following, and assuming that $f(x)$ is infinitely differentiable:
$$fracd^nf(x)dx^nBigg|_x=0=f(n)$$
What functions $f$ could satisfy this equation? Do any functions of $f$ have a closed form, or if not does it have a form that is just a normal ODE form?
ordinary-differential-equations derivatives closed-form
edited May 5 at 8:00
Asaf Karagila♦
310k33444776
asked May 5 at 1:17
tox123tox123
574721
$endgroup$
add a comment |
$begingroup$
Given the following, and assuming that $f(x)$ is infinitely differentiable:
$$fracd^nf(x)dx^nBigg|_x=0=f(n)$$
What functions $f$ could satisfy this equation? Do any functions of $f$ have a closed form, or if not does it have a form that is just a normal ODE form?
ordinary-differential-equations derivatives closed-form
edited May 5 at 8:00
Asaf Karagila♦
310k33444776
asked May 5 at 1:17
tox123tox123
574721
$endgroup$
$begingroup$
From what you've written it looks like: $$f(x=n)=f^(n)(0)$$ I've written $x=n$ to show that we are primarily defining $f$ in terms of $x$ which applies of the right. However you could also say: $$f(n)=f^(n)(0)$$
$endgroup$
– Henry Lee
May 5 at 1:33
$begingroup$
@Henry Lee I used the notation I did only because I'm more familiar with it. Is the notation you suggest preferred, or the standard notation?
$endgroup$
– tox123
May 5 at 1:48
$begingroup$
They both mean the same thing and what you have showed is equal and shows what you mean
$endgroup$
– Henry Lee
May 5 at 1:51
add a comment |
$begingroup$
Given the following, and assuming that $f(x)$ is infinitely differentiable:
$$fracd^nf(x)dx^nBigg|_x=0=f(n)$$
What functions $f$ could satisfy this equation? Do any functions of $f$ have a closed form, or if not does it have a form that is just a normal ODE form?
ordinary-differential-equations derivatives closed-form
edited May 5 at 8:00
Asaf Karagila♦
310k33444776
asked May 5 at 1:17
tox123tox123
574721
$endgroup$
Given the following, and assuming that $f(x)$ is infinitely differentiable:
$$fracd^nf(x)dx^nBigg|_x=0=f(n)$$
What functions $f$ could satisfy this equation? Do any functions of $f$ have a closed form, or if not does it have a form that is just a normal ODE form?
ordinary-differential-equations derivatives closed-form
ordinary-differential-equations derivatives closed-form
edited May 5 at 8:00
Asaf Karagila♦
310k33444776
asked May 5 at 1:17
tox123tox123
574721
edited May 5 at 8:00
Asaf Karagila♦
310k33444776
asked May 5 at 1:17
tox123tox123
574721
edited May 5 at 8:00
Asaf Karagila♦
310k33444776
edited May 5 at 8:00
Asaf Karagila♦
310k33444776
edited May 5 at 8:00
Asaf Karagila♦
310k33444776
310k33444776
asked May 5 at 1:17
tox123tox123
574721
asked May 5 at 1:17
tox123tox123
574721
asked May 5 at 1:17
tox123tox123
574721
574721
$begingroup$
From what you've written it looks like: $$f(x=n)=f^(n)(0)$$ I've written $x=n$ to show that we are primarily defining $f$ in terms of $x$ which applies of the right. However you could also say: $$f(n)=f^(n)(0)$$
$endgroup$
– Henry Lee
May 5 at 1:33
$begingroup$
@Henry Lee I used the notation I did only because I'm more familiar with it. Is the notation you suggest preferred, or the standard notation?
$endgroup$
– tox123
May 5 at 1:48
$begingroup$
They both mean the same thing and what you have showed is equal and shows what you mean
$endgroup$
– Henry Lee
May 5 at 1:51
add a comment |
$begingroup$
From what you've written it looks like: $$f(x=n)=f^(n)(0)$$ I've written $x=n$ to show that we are primarily defining $f$ in terms of $x$ which applies of the right. However you could also say: $$f(n)=f^(n)(0)$$
$endgroup$
– Henry Lee
May 5 at 1:33
$begingroup$
@Henry Lee I used the notation I did only because I'm more familiar with it. Is the notation you suggest preferred, or the standard notation?
$endgroup$
– tox123
May 5 at 1:48
$begingroup$
They both mean the same thing and what you have showed is equal and shows what you mean
$endgroup$
– Henry Lee
May 5 at 1:51
$begingroup$
From what you've written it looks like: $$f(x=n)=f^(n)(0)$$ I've written $x=n$ to show that we are primarily defining $f$ in terms of $x$ which applies of the right. However you could also say: $$f(n)=f^(n)(0)$$
$endgroup$
– Henry Lee
May 5 at 1:33
$begingroup$
From what you've written it looks like: $$f(x=n)=f^(n)(0)$$ I've written $x=n$ to show that we are primarily defining $f$ in terms of $x$ which applies of the right. However you could also say: $$f(n)=f^(n)(0)$$
$endgroup$
– Henry Lee
May 5 at 1:33
$begingroup$
@Henry Lee I used the notation I did only because I'm more familiar with it. Is the notation you suggest preferred, or the standard notation?
$endgroup$
– tox123
May 5 at 1:48
$begingroup$
@Henry Lee I used the notation I did only because I'm more familiar with it. Is the notation you suggest preferred, or the standard notation?
$endgroup$
– tox123
May 5 at 1:48
$begingroup$
They both mean the same thing and what you have showed is equal and shows what you mean
$endgroup$
– Henry Lee
May 5 at 1:51
$begingroup$
They both mean the same thing and what you have showed is equal and shows what you mean
$endgroup$
– Henry Lee
May 5 at 1:51
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Let $f(x)=a^x+1$, where $a$ satisfies $ln(a)=a$. Then $f^(n)(0)=ln^n(a) a^1=a^n+1=f(n)$, as desired. Note that $a$ will be a complex number here, explicitly in terms of Lambert’s W: $a=e^-W(-1)approx 0.318+1.337i$.
answered May 5 at 1:43
Alex R.Alex R.
25.4k12454
$endgroup$
$begingroup$
Is this exclusively the only answer or are there other functions that satisfy my conditions?
$endgroup$
– tox123
May 5 at 1:55
$begingroup$
I can't see how $ln^n(a)a^1=a^n+1$...
$endgroup$
– Thehx
May 5 at 1:55
1
$begingroup$
@Thehx: look at the definition of $a$.
$endgroup$
– Alex R.
May 5 at 2:08
1
$begingroup$
oh god, this is beautiful.
$endgroup$
– Thehx
May 5 at 2:11
add a comment |
Your Answer
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Let $f(x)=a^x+1$, where $a$ satisfies $ln(a)=a$. Then $f^(n)(0)=ln^n(a) a^1=a^n+1=f(n)$, as desired. Note that $a$ will be a complex number here, explicitly in terms of Lambert’s W: $a=e^-W(-1)approx 0.318+1.337i$.
answered May 5 at 1:43
Alex R.Alex R.
25.4k12454
$endgroup$
$begingroup$
Is this exclusively the only answer or are there other functions that satisfy my conditions?
$endgroup$
– tox123
May 5 at 1:55
$begingroup$
I can't see how $ln^n(a)a^1=a^n+1$...
$endgroup$
– Thehx
May 5 at 1:55
1
$begingroup$
@Thehx: look at the definition of $a$.
$endgroup$
– Alex R.
May 5 at 2:08
1
$begingroup$
oh god, this is beautiful.
$endgroup$
– Thehx
May 5 at 2:11
add a comment |
$begingroup$
Let $f(x)=a^x+1$, where $a$ satisfies $ln(a)=a$. Then $f^(n)(0)=ln^n(a) a^1=a^n+1=f(n)$, as desired. Note that $a$ will be a complex number here, explicitly in terms of Lambert’s W: $a=e^-W(-1)approx 0.318+1.337i$.
answered May 5 at 1:43
Alex R.Alex R.
25.4k12454
$endgroup$
$begingroup$
Is this exclusively the only answer or are there other functions that satisfy my conditions?
$endgroup$
– tox123
May 5 at 1:55
$begingroup$
I can't see how $ln^n(a)a^1=a^n+1$...
$endgroup$
– Thehx
May 5 at 1:55
1
$begingroup$
@Thehx: look at the definition of $a$.
$endgroup$
– Alex R.
May 5 at 2:08
1
$begingroup$
oh god, this is beautiful.
$endgroup$
– Thehx
May 5 at 2:11
add a comment |
$begingroup$
Let $f(x)=a^x+1$, where $a$ satisfies $ln(a)=a$. Then $f^(n)(0)=ln^n(a) a^1=a^n+1=f(n)$, as desired. Note that $a$ will be a complex number here, explicitly in terms of Lambert’s W: $a=e^-W(-1)approx 0.318+1.337i$.
answered May 5 at 1:43
Alex R.Alex R.
25.4k12454
$endgroup$
Let $f(x)=a^x+1$, where $a$ satisfies $ln(a)=a$. Then $f^(n)(0)=ln^n(a) a^1=a^n+1=f(n)$, as desired. Note that $a$ will be a complex number here, explicitly in terms of Lambert’s W: $a=e^-W(-1)approx 0.318+1.337i$.
answered May 5 at 1:43
Alex R.Alex R.
25.4k12454
edited May 5 at 4:23
answered May 5 at 1:43
Alex R.Alex R.
25.4k12454
answered May 5 at 1:43
Alex R.Alex R.
25.4k12454
answered May 5 at 1:43
Alex R.Alex R.
25.4k12454
25.4k12454
$begingroup$
Is this exclusively the only answer or are there other functions that satisfy my conditions?
$endgroup$
– tox123
May 5 at 1:55
$begingroup$
I can't see how $ln^n(a)a^1=a^n+1$...
$endgroup$
– Thehx
May 5 at 1:55
1
$begingroup$
@Thehx: look at the definition of $a$.
$endgroup$
– Alex R.
May 5 at 2:08
1
$begingroup$
oh god, this is beautiful.
$endgroup$
– Thehx
May 5 at 2:11
add a comment |
$begingroup$
Is this exclusively the only answer or are there other functions that satisfy my conditions?
$endgroup$
– tox123
May 5 at 1:55
$begingroup$
I can't see how $ln^n(a)a^1=a^n+1$...
$endgroup$
– Thehx
May 5 at 1:55
1
$begingroup$
@Thehx: look at the definition of $a$.
$endgroup$
– Alex R.
May 5 at 2:08
1
$begingroup$
oh god, this is beautiful.
$endgroup$
– Thehx
May 5 at 2:11
$begingroup$
Is this exclusively the only answer or are there other functions that satisfy my conditions?
$endgroup$
– tox123
May 5 at 1:55
$begingroup$
Is this exclusively the only answer or are there other functions that satisfy my conditions?
$endgroup$
– tox123
May 5 at 1:55
$begingroup$
I can't see how $ln^n(a)a^1=a^n+1$...
$endgroup$
– Thehx
May 5 at 1:55
$begingroup$
I can't see how $ln^n(a)a^1=a^n+1$...
$endgroup$
– Thehx
May 5 at 1:55
1
1
$begingroup$
@Thehx: look at the definition of $a$.
$endgroup$
– Alex R.
May 5 at 2:08
$begingroup$
@Thehx: look at the definition of $a$.
$endgroup$
– Alex R.
May 5 at 2:08
1
1
$begingroup$
oh god, this is beautiful.
$endgroup$
– Thehx
May 5 at 2:11
$begingroup$
oh god, this is beautiful.
$endgroup$
– Thehx
May 5 at 2:11
add a comment |
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$begingroup$
From what you've written it looks like: $$f(x=n)=f^(n)(0)$$ I've written $x=n$ to show that we are primarily defining $f$ in terms of $x$ which applies of the right. However you could also say: $$f(n)=f^(n)(0)$$
$endgroup$
– Henry Lee
May 5 at 1:33
$begingroup$
@Henry Lee I used the notation I did only because I'm more familiar with it. Is the notation you suggest preferred, or the standard notation?
$endgroup$
– tox123
May 5 at 1:48
$begingroup$
They both mean the same thing and what you have showed is equal and shows what you mean
$endgroup$
– Henry Lee
May 5 at 1:51