3
$\begingroup$

What is the Mathematica command for the recursive formula:

   F[a_]:=Sum[(-1)^a Binomial[a,k] Log[2]^(a-k) F[k], {k,0,a}]

where

 F[0]=Log[2]

Thank you,

The latex formula is

$$\int_0^1\frac{\ln^{a}\left(\frac{x}{1-x}\right)}{1+x}dx=(-1)^a \int_0^1\frac{\ln^{a}\left(\frac{2x}{1-x}\right)}{1+x}dx$$ $$=\sum_{k=0}^\infty(-1)^a \binom{a}{k}\ln^{a-k}(2)\int_0^1\frac{\ln^{k}\left(\frac{x}{1-x}\right)}{1+x}dx$$

which, by the way, follows from subbing $x=\frac{1-y}{1+y}$ then using the binomial theorem.

$\endgroup$
9
  • 1
    $\begingroup$ You cannot determine F[a], where $a$ is even from these equations. For instance you get for $a=2$: F[2]==F[2] - 2 F[1] Log[2] + F[0] Log[2]^2. Your question has no solution. $\endgroup$
    – yarchik
    Jul 29, 2023 at 7:26
  • $\begingroup$ Can you give what the output should be for say F[2] or F[3]? Also it is Binomial not binomial $\endgroup$
    – Nasser
    Jul 29, 2023 at 7:27
  • $\begingroup$ Please include $\LaTeX$ for your formula to facilitate disambiguation. $\endgroup$
    – Syed
    Jul 29, 2023 at 7:28
  • $\begingroup$ @Syed sure I will $\endgroup$ Jul 29, 2023 at 7:33
  • $\begingroup$ @yarchik sorry it is $(-1)^a$ not $(-1)^k$. $\endgroup$ Jul 29, 2023 at 7:40

1 Answer 1

5
$\begingroup$
  • It is impossible to deduce all the terms from the original recursive formula.
r1 = Block[{a = 1}, 
  F[a] - Sum[(-1)^a Binomial[a, k] Log[2]^(a - k) F[k], {k, 0, a}] == 
    0 // FullSimplify]
r2 = Block[{a = 2}, 
  F[a] - Sum[(-1)^a Binomial[a, k] Log[2]^(a - k) F[k], {k, 0, a}] == 
    0 // FullSimplify]
r3 = Block[{a = 3}, 
  F[a] - Sum[(-1)^a Binomial[a, k] Log[2]^(a - k) F[k], {k, 0, a}] == 
    0 // FullSimplify]
r4 = Block[{a = 4}, 
  F[a] - Sum[(-1)^a Binomial[a, k] Log[2]^(a - k) F[k], {k, 0, a}] == 
    0 // FullSimplify]

enter image description here

Observe the first 4 recursive formulas. We can see that we can deduce F[1] from F[0] by r1 or r2( actually r1 is equivalent to r2).

And we can also deduce F[3] from F[0],F[1],F[2] by r3 or r4, but we still cann't deduce F[2]. (BTW, under the condition of r1, the r3 and r4 are equivalent)

  • The expression of f[2] as below indicate that f[2] it is unlikely can be deduce from f[0] and f[1].
f[a_] := Integrate[(Log[x/(1 - x)])^a/(1 + x), {x, 0, 1}];
{f[0], f[1], f[2]}
% // N
Table[NIntegrate[(Log[x/(1 - x)])^a/(1 + x), {x, 0, 1}], {a, {0, 1, 
   2}}]

enter image description here

$\endgroup$
4
  • $\begingroup$ Thank you and sorry its $(-1)^a$ not $(-1)^k$ in the OP. $\endgroup$ Jul 29, 2023 at 7:42
  • $\begingroup$ @AliShadhar How to deduce the recursive formula ? $\endgroup$
    – cvgmt
    Jul 29, 2023 at 8:24
  • $\begingroup$ I just added some steps in the OP showing how I came up with it. $\endgroup$ Jul 29, 2023 at 8:26
  • $\begingroup$ Very clear thank you. So basically this formula shows only relations among these integrals. $\endgroup$ Jul 29, 2023 at 15:29

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.