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Simplify recurrence $fracddx f_n-1(x)= f_n(x)- f_n-1(x) f_1(x)$

nth derivative of a finite amount of composite functionsA question on Taylor expansion/approximationGiven a set of sequences, compute a corresponding set of functionsa sequence of functions defined by inductionDifferentiation- proof by InductionIs it possible to find an explicit function for any given recursive sequence?$f_n$ converges uniformly on $[a,b]$ to some function $f$Proving smoothness for a sequence of functions.Is there any sequence of functions containing functions growing slower than any smooth function?Finding $f(x)$ when given a composite function?

1

2

$begingroup$

Suppose we have a sequence of infinitely differentiable functions $ f_k(x) $. Now suppose that these functions satisfy the following recursion:
beginalign
fracddx f_n-1(x)= f_n(x)- f_n-1(x) f_1(x)
endalign
where $f_0(x)=1$ for all $x$.

Can we re-write $f_n(x)$ only in terms of $f_1(x)$ and derivatives of $f_1(x)$? It is not difficult to see that this is possible. The difficulty is to create the exact formula.

calculus recursion

share|cite|improve this question

edited Mar 21 at 16:45

Lisa

asked Mar 20 at 17:20

LisaLisa

664314

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Is the recursion $fracddx f_n+1(x)= f_n(x)- f_n-1(x) f_1(x)$?
  $endgroup$
  – xpaul
  Mar 20 at 21:28
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @xpaul No, it with $n-1$ for the derivative. Why? Does that equation have a known form?
  $endgroup$
  – Lisa
  Mar 20 at 21:34
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  It comes from Faa di Bruno's formula and complete Bell polynomials. See the OEIS sequence A036040 and the very similar OEIS sequence A080575 for much more details.
  $endgroup$
  – Somos
  Mar 21 at 4:04
  
  

  
  
  
  

add a comment | 

1

2

$begingroup$

Suppose we have a sequence of infinitely differentiable functions $ f_k(x) $. Now suppose that these functions satisfy the following recursion:
beginalign
fracddx f_n-1(x)= f_n(x)- f_n-1(x) f_1(x)
endalign
where $f_0(x)=1$ for all $x$.

Can we re-write $f_n(x)$ only in terms of $f_1(x)$ and derivatives of $f_1(x)$? It is not difficult to see that this is possible. The difficulty is to create the exact formula.

calculus recursion

share|cite|improve this question

edited Mar 21 at 16:45

Lisa

asked Mar 20 at 17:20

LisaLisa

664314

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Is the recursion $fracddx f_n+1(x)= f_n(x)- f_n-1(x) f_1(x)$?
  $endgroup$
  – xpaul
  Mar 20 at 21:28
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @xpaul No, it with $n-1$ for the derivative. Why? Does that equation have a known form?
  $endgroup$
  – Lisa
  Mar 20 at 21:34
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  It comes from Faa di Bruno's formula and complete Bell polynomials. See the OEIS sequence A036040 and the very similar OEIS sequence A080575 for much more details.
  $endgroup$
  – Somos
  Mar 21 at 4:04
  
  

  
  
  
  

add a comment | 

$begingroup$

Suppose we have a sequence of infinitely differentiable functions $ f_k(x) $. Now suppose that these functions satisfy the following recursion:
beginalign
fracddx f_n-1(x)= f_n(x)- f_n-1(x) f_1(x)
endalign
where $f_0(x)=1$ for all $x$.

Can we re-write $f_n(x)$ only in terms of $f_1(x)$ and derivatives of $f_1(x)$? It is not difficult to see that this is possible. The difficulty is to create the exact formula.

calculus recursion

share|cite|improve this question

edited Mar 21 at 16:45

Lisa

asked Mar 20 at 17:20

LisaLisa

664314

$endgroup$

share|cite|improve this question

edited Mar 21 at 16:45

Lisa

asked Mar 20 at 17:20

LisaLisa

664314

share|cite|improve this question

edited Mar 21 at 16:45

Lisa

asked Mar 20 at 17:20

LisaLisa

664314

asked Mar 20 at 17:20

LisaLisa

664314

asked Mar 20 at 17:20

LisaLisa

664314

2 Answers
2

active

oldest

votes

2

$begingroup$

I don't know if this is what you want. It is easy to see
$$f_n(x) = f_n-1^prime(x) + f_n-1(x)f_1(x). tag1$$
Multiplying both sides of (1) by $e^int f_1(x)dx$ gives
$$f_n(x)e^int f_1(x)dx = bigg[f_n-1(x) e^int f_1(x)dxbigg]'. tag2$$
Let
$$ g_n(x)=f_n(x)e^int f_1(x)dx. $$
Then (2) gives
$$ g_n(x)=g_n-1'(x) $$
Using this recursion $(n-1)$-times, one has
$$ g_n(x)=g_1^(n-1)(x)=fracd^n-1dx^n-1bigg[f_1(x)e^int f_1(x)dxbigg] $$
and hence
$$ f_n(x)=e^-int f_1(x)dxfracd^n-1dx^n-1bigg[f_1(x)e^int f_1(x)dxbigg] $$

share|cite|improve this answer

answered Mar 21 at 15:11

xpaulxpaul

23.4k24655

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Can you clarify what are the limits of integration are?
  $endgroup$
  – Lisa
  Mar 21 at 16:31
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  It is simply the antiderivative, the indefinite integral.
  $endgroup$
  – IV_
  Mar 21 at 18:55
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @IV_ is it $int_-infty^x f(t) dt$?
  $endgroup$
  – Lisa
  Mar 21 at 19:13
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @Lisa, you can simply use $int_0^xf_1(t)dt$ instead of $int f_1(x)dx$.
  $endgroup$
  – xpaul
  Mar 21 at 19:26
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I have more question. From your proof, it seems like you have seen something similar before. Does this equation have a name?
  $endgroup$
  – Lisa
  Mar 21 at 22:43
  

  
  
  
  

 | 
show 1 more comment

1

$begingroup$

$$fracddxf_n-1(x)=f_n(x)-f_n-1(x)f_1(x), f_0(x)=1$$

$$f_n(x)=fracddxf_n-1(x)+f_n-1(x)f_1(x)$$

$$f_1(x)=fracddxf_0(x)+f_0(x)f_1(x)=f_1(x)$$

$$f_2(x)=fracddxf_1(x)+f_1(x)f_1(x)$$

$$f_3(x)=fracddxf_2(x)+f_2(x)f_1(x)$$

$$...$$

$f_n(x)$ is the following special complete exponential Bell polynomial:

$$f_n(x)=B_n(f_1^(0)(x),f_1^(1)(x),...f_1^(n-1)(x))=sum_sum_t=1^ntk_t=nfracn!prod_i=1^ni!^k_ik_i!prod_i=1^nf_1^(i-1)(x)^k_i$$

The vectors of running indices $(k_1,...,k_n)$ in the sum expression represent the integer partitions of $n$.

Applying Faà di Bruno's Formula (Higher chain rule), we get:

$$f_n(x)=e^-int f_1(x)dxfracd^ndx^ne^int f_1(x)dx=e^-int f_1(x)dxfracd^n-1dx^n-1left(f_1(x)e^int f_1(x)dxright)$$

share|cite|improve this answer

edited Mar 23 at 13:37

answered Mar 21 at 17:45

IV_IV_

1,556525

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thanks. Can you add more details about how you arrived at the exponential Bell polynomial answer? Specifically, at what point do you see that it an exponential Bell polynomial.
  $endgroup$
  – Lisa
  Mar 21 at 18:17
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I saw this on the coefficients. "Higher derivatives are set-partitions". But you can prove it by verifying the concluded integral formula and applying Faà di Bruno's formula to it.
  $endgroup$
  – IV_
  Mar 21 at 21:07
  

  
  
  
  

add a comment | 

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2

$begingroup$

I don't know if this is what you want. It is easy to see
$$f_n(x) = f_n-1^prime(x) + f_n-1(x)f_1(x). tag1$$
Multiplying both sides of (1) by $e^int f_1(x)dx$ gives
$$f_n(x)e^int f_1(x)dx = bigg[f_n-1(x) e^int f_1(x)dxbigg]'. tag2$$
Let
$$ g_n(x)=f_n(x)e^int f_1(x)dx. $$
Then (2) gives
$$ g_n(x)=g_n-1'(x) $$
Using this recursion $(n-1)$-times, one has
$$ g_n(x)=g_1^(n-1)(x)=fracd^n-1dx^n-1bigg[f_1(x)e^int f_1(x)dxbigg] $$
and hence
$$ f_n(x)=e^-int f_1(x)dxfracd^n-1dx^n-1bigg[f_1(x)e^int f_1(x)dxbigg] $$

share|cite|improve this answer

answered Mar 21 at 15:11

xpaulxpaul

23.4k24655

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Can you clarify what are the limits of integration are?
  $endgroup$
  – Lisa
  Mar 21 at 16:31
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  It is simply the antiderivative, the indefinite integral.
  $endgroup$
  – IV_
  Mar 21 at 18:55
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @IV_ is it $int_-infty^x f(t) dt$?
  $endgroup$
  – Lisa
  Mar 21 at 19:13
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @Lisa, you can simply use $int_0^xf_1(t)dt$ instead of $int f_1(x)dx$.
  $endgroup$
  – xpaul
  Mar 21 at 19:26
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I have more question. From your proof, it seems like you have seen something similar before. Does this equation have a name?
  $endgroup$
  – Lisa
  Mar 21 at 22:43
  

  
  
  
  

 | 
show 1 more comment

2

$begingroup$

I don't know if this is what you want. It is easy to see
$$f_n(x) = f_n-1^prime(x) + f_n-1(x)f_1(x). tag1$$
Multiplying both sides of (1) by $e^int f_1(x)dx$ gives
$$f_n(x)e^int f_1(x)dx = bigg[f_n-1(x) e^int f_1(x)dxbigg]'. tag2$$
Let
$$ g_n(x)=f_n(x)e^int f_1(x)dx. $$
Then (2) gives
$$ g_n(x)=g_n-1'(x) $$
Using this recursion $(n-1)$-times, one has
$$ g_n(x)=g_1^(n-1)(x)=fracd^n-1dx^n-1bigg[f_1(x)e^int f_1(x)dxbigg] $$
and hence
$$ f_n(x)=e^-int f_1(x)dxfracd^n-1dx^n-1bigg[f_1(x)e^int f_1(x)dxbigg] $$

share|cite|improve this answer

answered Mar 21 at 15:11

xpaulxpaul

23.4k24655

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Can you clarify what are the limits of integration are?
  $endgroup$
  – Lisa
  Mar 21 at 16:31
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  It is simply the antiderivative, the indefinite integral.
  $endgroup$
  – IV_
  Mar 21 at 18:55
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @IV_ is it $int_-infty^x f(t) dt$?
  $endgroup$
  – Lisa
  Mar 21 at 19:13
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @Lisa, you can simply use $int_0^xf_1(t)dt$ instead of $int f_1(x)dx$.
  $endgroup$
  – xpaul
  Mar 21 at 19:26
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I have more question. From your proof, it seems like you have seen something similar before. Does this equation have a name?
  $endgroup$
  – Lisa
  Mar 21 at 22:43
  

  
  
  
  

 | 
show 1 more comment

$begingroup$

I don't know if this is what you want. It is easy to see
$$f_n(x) = f_n-1^prime(x) + f_n-1(x)f_1(x). tag1$$
Multiplying both sides of (1) by $e^int f_1(x)dx$ gives
$$f_n(x)e^int f_1(x)dx = bigg[f_n-1(x) e^int f_1(x)dxbigg]'. tag2$$
Let
$$ g_n(x)=f_n(x)e^int f_1(x)dx. $$
Then (2) gives
$$ g_n(x)=g_n-1'(x) $$
Using this recursion $(n-1)$-times, one has
$$ g_n(x)=g_1^(n-1)(x)=fracd^n-1dx^n-1bigg[f_1(x)e^int f_1(x)dxbigg] $$
and hence
$$ f_n(x)=e^-int f_1(x)dxfracd^n-1dx^n-1bigg[f_1(x)e^int f_1(x)dxbigg] $$

share|cite|improve this answer

answered Mar 21 at 15:11

xpaulxpaul

23.4k24655

$endgroup$

share|cite|improve this answer

answered Mar 21 at 15:11

xpaulxpaul

23.4k24655

answered Mar 21 at 15:11

xpaulxpaul

23.4k24655

answered Mar 21 at 15:11

xpaulxpaul

23.4k24655

1

$begingroup$

$$fracddxf_n-1(x)=f_n(x)-f_n-1(x)f_1(x), f_0(x)=1$$

$$f_n(x)=fracddxf_n-1(x)+f_n-1(x)f_1(x)$$

$$f_1(x)=fracddxf_0(x)+f_0(x)f_1(x)=f_1(x)$$

$$f_2(x)=fracddxf_1(x)+f_1(x)f_1(x)$$

$$f_3(x)=fracddxf_2(x)+f_2(x)f_1(x)$$

$$...$$

$f_n(x)$ is the following special complete exponential Bell polynomial:

$$f_n(x)=B_n(f_1^(0)(x),f_1^(1)(x),...f_1^(n-1)(x))=sum_sum_t=1^ntk_t=nfracn!prod_i=1^ni!^k_ik_i!prod_i=1^nf_1^(i-1)(x)^k_i$$

The vectors of running indices $(k_1,...,k_n)$ in the sum expression represent the integer partitions of $n$.

Applying Faà di Bruno's Formula (Higher chain rule), we get:

$$f_n(x)=e^-int f_1(x)dxfracd^ndx^ne^int f_1(x)dx=e^-int f_1(x)dxfracd^n-1dx^n-1left(f_1(x)e^int f_1(x)dxright)$$

share|cite|improve this answer

edited Mar 23 at 13:37

answered Mar 21 at 17:45

IV_IV_

1,556525

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thanks. Can you add more details about how you arrived at the exponential Bell polynomial answer? Specifically, at what point do you see that it an exponential Bell polynomial.
  $endgroup$
  – Lisa
  Mar 21 at 18:17
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I saw this on the coefficients. "Higher derivatives are set-partitions". But you can prove it by verifying the concluded integral formula and applying Faà di Bruno's formula to it.
  $endgroup$
  – IV_
  Mar 21 at 21:07
  

  
  
  
  

add a comment | 

1

$begingroup$

$$fracddxf_n-1(x)=f_n(x)-f_n-1(x)f_1(x), f_0(x)=1$$

$$f_n(x)=fracddxf_n-1(x)+f_n-1(x)f_1(x)$$

$$f_1(x)=fracddxf_0(x)+f_0(x)f_1(x)=f_1(x)$$

$$f_2(x)=fracddxf_1(x)+f_1(x)f_1(x)$$

$$f_3(x)=fracddxf_2(x)+f_2(x)f_1(x)$$

$$...$$

$f_n(x)$ is the following special complete exponential Bell polynomial:

$$f_n(x)=B_n(f_1^(0)(x),f_1^(1)(x),...f_1^(n-1)(x))=sum_sum_t=1^ntk_t=nfracn!prod_i=1^ni!^k_ik_i!prod_i=1^nf_1^(i-1)(x)^k_i$$

The vectors of running indices $(k_1,...,k_n)$ in the sum expression represent the integer partitions of $n$.

Applying Faà di Bruno's Formula (Higher chain rule), we get:

$$f_n(x)=e^-int f_1(x)dxfracd^ndx^ne^int f_1(x)dx=e^-int f_1(x)dxfracd^n-1dx^n-1left(f_1(x)e^int f_1(x)dxright)$$

share|cite|improve this answer

edited Mar 23 at 13:37

answered Mar 21 at 17:45

IV_IV_

1,556525

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thanks. Can you add more details about how you arrived at the exponential Bell polynomial answer? Specifically, at what point do you see that it an exponential Bell polynomial.
  $endgroup$
  – Lisa
  Mar 21 at 18:17
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I saw this on the coefficients. "Higher derivatives are set-partitions". But you can prove it by verifying the concluded integral formula and applying Faà di Bruno's formula to it.
  $endgroup$
  – IV_
  Mar 21 at 21:07
  

  
  
  
  

add a comment | 

$begingroup$

$$fracddxf_n-1(x)=f_n(x)-f_n-1(x)f_1(x), f_0(x)=1$$

$$f_n(x)=fracddxf_n-1(x)+f_n-1(x)f_1(x)$$

$$f_1(x)=fracddxf_0(x)+f_0(x)f_1(x)=f_1(x)$$

$$f_2(x)=fracddxf_1(x)+f_1(x)f_1(x)$$

$$f_3(x)=fracddxf_2(x)+f_2(x)f_1(x)$$

$$...$$

$f_n(x)$ is the following special complete exponential Bell polynomial:

$$f_n(x)=B_n(f_1^(0)(x),f_1^(1)(x),...f_1^(n-1)(x))=sum_sum_t=1^ntk_t=nfracn!prod_i=1^ni!^k_ik_i!prod_i=1^nf_1^(i-1)(x)^k_i$$

The vectors of running indices $(k_1,...,k_n)$ in the sum expression represent the integer partitions of $n$.

Applying Faà di Bruno's Formula (Higher chain rule), we get:

$$f_n(x)=e^-int f_1(x)dxfracd^ndx^ne^int f_1(x)dx=e^-int f_1(x)dxfracd^n-1dx^n-1left(f_1(x)e^int f_1(x)dxright)$$

share|cite|improve this answer

edited Mar 23 at 13:37

answered Mar 21 at 17:45

IV_IV_

1,556525

$endgroup$

share|cite|improve this answer

edited Mar 23 at 13:37

answered Mar 21 at 17:45

IV_IV_

1,556525

answered Mar 21 at 17:45

IV_IV_

1,556525

answered Mar 21 at 17:45

IV_IV_

1,556525