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Is there a closed form, or cleaner way of writing this function?
Does a closed form solution to this nonlinear ODE exist?Is there a closed form solution to this equation?Prove that the composition of two “closed form functions” is itself a “closed form function”?Areas where closed form solutions are of particular interestClassifying functions whose inverse do not have a closed formClosed form of planetary radial motion as time functionNonlinear ODE, closed form solution?Closed form for a series IIIs there a closed form for this “flowery” integral?Deriving the closed form of Gamma function using Euler-Chi function
$begingroup$
Given the following, and assuming that $f(x)$ is infinitely differentiable:
$$fracd^nf(x)dx^n|_x=0=f(n)$$
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
asked 3 hours ago
tox123tox123
572721
$endgroup$
add a comment |
$begingroup$
Given the following, and assuming that $f(x)$ is infinitely differentiable:
$$fracd^nf(x)dx^n|_x=0=f(n)$$
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
asked 3 hours ago
tox123tox123
572721
$endgroup$
$begingroup$
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
3 hours ago
$begingroup$
@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
2 hours ago
$begingroup$
They both mean the same thing and what you have showed is equal and shows what you mean
$endgroup$
– Henry Lee
2 hours ago
add a comment |
$begingroup$
Given the following, and assuming that $f(x)$ is infinitely differentiable:
$$fracd^nf(x)dx^n|_x=0=f(n)$$
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
asked 3 hours ago
tox123tox123
572721
$endgroup$
Given the following, and assuming that $f(x)$ is infinitely differentiable:
$$fracd^nf(x)dx^n|_x=0=f(n)$$
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
asked 3 hours ago
tox123tox123
572721
asked 3 hours ago
tox123tox123
572721
asked 3 hours ago
tox123tox123
572721
asked 3 hours ago
tox123tox123
572721
asked 3 hours ago
tox123tox123
572721
572721
$begingroup$
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
3 hours ago
$begingroup$
@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
2 hours ago
$begingroup$
They both mean the same thing and what you have showed is equal and shows what you mean
$endgroup$
– Henry Lee
2 hours ago
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
3 hours ago
$begingroup$
@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
2 hours ago
$begingroup$
They both mean the same thing and what you have showed is equal and shows what you mean
$endgroup$
– Henry Lee
2 hours ago
$begingroup$
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
3 hours ago
$begingroup$
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
3 hours ago
$begingroup$
@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
2 hours ago
$begingroup$
@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
2 hours ago
$begingroup$
They both mean the same thing and what you have showed is equal and shows what you mean
$endgroup$
– Henry Lee
2 hours ago
$begingroup$
They both mean the same thing and what you have showed is equal and shows what you mean
$endgroup$
– Henry Lee
2 hours ago
add a comment |
2 Answers
2
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 2 hours ago
Alex R.Alex R.
25.3k12454
$endgroup$
$begingroup$
Is this exclusively the only answer or are there other functions that satisfy my conditions?
$endgroup$
– tox123
2 hours ago
$begingroup$
I can't see how $ln^n(a)a^1=a^n+1$...
$endgroup$
– Thehx
2 hours ago
1
$begingroup$
@Thehx: look at the definition of $a$.
$endgroup$
– Alex R.
2 hours ago
1
$begingroup$
oh god, this is beautiful.
$endgroup$
– Thehx
2 hours ago
add a comment |
$begingroup$
I only see the trivial f(x)=0 solution so far. Couldn't yet prove there are no other solutions tho.
UPD: alex-r provided an absolutely amazing non-trivial solution. Make sure to check it out.
answered 2 hours ago
ThehxThehx
687
$endgroup$
add a comment |
Your Answer
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2 Answers
2
active
oldest
votes
2 Answers
2
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 2 hours ago
Alex R.Alex R.
25.3k12454
$endgroup$
$begingroup$
Is this exclusively the only answer or are there other functions that satisfy my conditions?
$endgroup$
– tox123
2 hours ago
$begingroup$
I can't see how $ln^n(a)a^1=a^n+1$...
$endgroup$
– Thehx
2 hours ago
1
$begingroup$
@Thehx: look at the definition of $a$.
$endgroup$
– Alex R.
2 hours ago
1
$begingroup$
oh god, this is beautiful.
$endgroup$
– Thehx
2 hours ago
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 2 hours ago
Alex R.Alex R.
25.3k12454
$endgroup$
$begingroup$
Is this exclusively the only answer or are there other functions that satisfy my conditions?
$endgroup$
– tox123
2 hours ago
$begingroup$
I can't see how $ln^n(a)a^1=a^n+1$...
$endgroup$
– Thehx
2 hours ago
1
$begingroup$
@Thehx: look at the definition of $a$.
$endgroup$
– Alex R.
2 hours ago
1
$begingroup$
oh god, this is beautiful.
$endgroup$
– Thehx
2 hours ago
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 2 hours ago
Alex R.Alex R.
25.3k12454
$endgroup$
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 2 hours ago
Alex R.Alex R.
25.3k12454
edited 10 mins ago
answered 2 hours ago
Alex R.Alex R.
25.3k12454
answered 2 hours ago
Alex R.Alex R.
25.3k12454
answered 2 hours ago
Alex R.Alex R.
25.3k12454
25.3k12454
$begingroup$
Is this exclusively the only answer or are there other functions that satisfy my conditions?
$endgroup$
– tox123
2 hours ago
$begingroup$
I can't see how $ln^n(a)a^1=a^n+1$...
$endgroup$
– Thehx
2 hours ago
1
$begingroup$
@Thehx: look at the definition of $a$.
$endgroup$
– Alex R.
2 hours ago
1
$begingroup$
oh god, this is beautiful.
$endgroup$
– Thehx
2 hours ago
add a comment |
$begingroup$
Is this exclusively the only answer or are there other functions that satisfy my conditions?
$endgroup$
– tox123
2 hours ago
$begingroup$
I can't see how $ln^n(a)a^1=a^n+1$...
$endgroup$
– Thehx
2 hours ago
1
$begingroup$
@Thehx: look at the definition of $a$.
$endgroup$
– Alex R.
2 hours ago
1
$begingroup$
oh god, this is beautiful.
$endgroup$
– Thehx
2 hours ago
$begingroup$
Is this exclusively the only answer or are there other functions that satisfy my conditions?
$endgroup$
– tox123
2 hours ago
$begingroup$
Is this exclusively the only answer or are there other functions that satisfy my conditions?
$endgroup$
– tox123
2 hours ago
$begingroup$
I can't see how $ln^n(a)a^1=a^n+1$...
$endgroup$
– Thehx
2 hours ago
$begingroup$
I can't see how $ln^n(a)a^1=a^n+1$...
$endgroup$
– Thehx
2 hours ago
1
1
$begingroup$
@Thehx: look at the definition of $a$.
$endgroup$
– Alex R.
2 hours ago
$begingroup$
@Thehx: look at the definition of $a$.
$endgroup$
– Alex R.
2 hours ago
1
1
$begingroup$
oh god, this is beautiful.
$endgroup$
– Thehx
2 hours ago
$begingroup$
oh god, this is beautiful.
$endgroup$
– Thehx
2 hours ago
add a comment |
$begingroup$
I only see the trivial f(x)=0 solution so far. Couldn't yet prove there are no other solutions tho.
UPD: alex-r provided an absolutely amazing non-trivial solution. Make sure to check it out.
answered 2 hours ago
ThehxThehx
687
$endgroup$
add a comment |
$begingroup$
I only see the trivial f(x)=0 solution so far. Couldn't yet prove there are no other solutions tho.
UPD: alex-r provided an absolutely amazing non-trivial solution. Make sure to check it out.
answered 2 hours ago
ThehxThehx
687
$endgroup$
add a comment |
$begingroup$
I only see the trivial f(x)=0 solution so far. Couldn't yet prove there are no other solutions tho.
UPD: alex-r provided an absolutely amazing non-trivial solution. Make sure to check it out.
answered 2 hours ago
ThehxThehx
687
$endgroup$
I only see the trivial f(x)=0 solution so far. Couldn't yet prove there are no other solutions tho.
UPD: alex-r provided an absolutely amazing non-trivial solution. Make sure to check it out.
answered 2 hours ago
ThehxThehx
687
edited 2 hours ago
answered 2 hours ago
ThehxThehx
687
answered 2 hours ago
ThehxThehx
687
answered 2 hours ago
ThehxThehx
687
687
add a comment |
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
3 hours ago
$begingroup$
@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
2 hours ago
$begingroup$
They both mean the same thing and what you have showed is equal and shows what you mean
$endgroup$
– Henry Lee
2 hours ago