9

consider the function :$$f(x)=\frac{1}{1+e^{x}}$$

the nth derivative of the function is given by the following formula:

$$f^{(n)} (x)=\sum_{k=1}^{n+1}a_{n,k}\frac{1}{\left(1+e^{x}\right)^{k}}$$ where

$$a_{n,k}=\left(-1\right)^{n}\sum_{j=0}^{k-1}\left(-1\right)^{j}{{k-1}\choose{j}}\left(j+1\right)^{n}$$

my question is that:how the formula can be derived without using induction? I have no idea about that, so any hint or full proof would be highly appreciated.

Absurd
  • 369
  • 1
  • 14
  • 1
    Use induction! – Lucas Smits Oct 31 '19 at 07:20
  • I haven't done the computation, but I'd try induction – ajotatxe Oct 31 '19 at 07:37
  • 3
    @Lucas Smits, I know I can do it, but where does the formula come from? the first person who introduced the formula did not use induction to create it and had an idea for deriving that! – Absurd Oct 31 '19 at 07:38
  • @ajotatxe do you know any other way except induction? – Absurd Oct 31 '19 at 07:41
  • 6
    The person who came up with the formula probably wrote out the derivative for n = 1, 2, 3, ... and saw a pattern. With that pattern, you can "guess" a formula and then prove its correctness with induction. – slimmerikko Oct 31 '19 at 07:42
  • @slimmerikko the problem is that I cannot find the full pattern. – Absurd Oct 31 '19 at 07:46
  • 2
    This seems to be an application of the general Lebniz rule, which is basically a generalisation of the product rule ( Also known as Lebniz rule). – Hello Oct 31 '19 at 09:24
  • https://en.wikipedia.org/wiki/General_Leibniz_rule#Proof – Hello Oct 31 '19 at 09:25
  • Note that the proof by Wikipedia also uses induction – Hello Oct 31 '19 at 09:25
  • 1
    @ ONG SEE HAI HCI, I think so,but do we have both $f$ and $g$? – Absurd Oct 31 '19 at 09:33
  • 1
    Note that $$f'(x) = -f(x) + f(x)^2,$$ and $$[f(x)^n]' = nf'(x) f(x)^{n - 1} = n\Big(f(x)^{n + 1} - f(x)^n\Big).$$ You can now derive $f$ many times and see the pattern. (noticing $f^{(n)} $ is a polynomial in $f(x)$ with degree $n + 1$ and constant term zero might also help.) – Azlif Oct 31 '19 at 10:34
  • Writing $f^{(n)} = \sum_{j = 1}^{n + 1} a_{n,j} f(x)^j$ and using the identity above can atleast change the problem to the following recurrence relations $$a_{n+1,n+2} = a_{n,n+1} (n+1)$$ $$a_{n+1,j} = (j - 1)a_{n,j-1} - a_{n,j},\quad 1 \leq j \leq n+1.$$ But I don't know how to solve it. – Azlif Oct 31 '19 at 10:52
  • This might help: $\frac{d^n}{dx^n}y=\sum_{k=1}^\infty (-1)^{k-1}(-k)^ne^{-kx}$ – TurlocTheRed Nov 01 '19 at 14:32

1 Answers1

5

A calculation without recurrence relation and without induction.

Let’s use the following formula, both sides come from the mixed bivariate generating function $~\displaystyle e^{z(e^x-1)}~$ for the Stirling numbers of the second kind:

$$e^{-z}\sum\limits_{k=0}^\infty\frac{z^k}{k!}k^n = \sum\limits_{k=0}^n z^k S(n,k)$$

$S(n,k)~$ are the Stirling numbers of the second kind .

Setting $~z:=at~$, multiplicating by $~e^{-t}~$ and integrating from $~t=0~$ to $~\infty~$ we get:

$$\sum\limits_{k=0}^\infty k^n\left(\frac{a}{1+a}\right)^k = (a+1)\sum\limits_{k=0}^n a^k k!S(n,k)$$

Setting $~\displaystyle\frac{a}{1+a} = -e^{-x}~$, substracting $~S_{n,0}=0^n~$ and

using $~S(n+1,k+1)=(k+1)S(n,k+1)+S(n,k)~$ we get:

$$\sum\limits_{k=1}^\infty k^n\left(-e^{-x}\right)^k = \sum\limits_{k=0}^n \frac{(-1)^{k+1} k!S(n+1,k+1)}{(1+e^x)^{k+1}}$$

It follows:

$\hspace{1cm}~\displaystyle\frac{d^n}{dx^n}\frac{1}{1+e^x} = (-1)^{n+1}\sum\limits_{k=1}^\infty k^n\left(-e^{-x}\right)^k = \sum\limits_{k=0}^n \frac{a_{n,k+1}}{(1+e^x)^{k+1}}~$

with $~~~\displaystyle a_{n,k+1} = (-1)^{n+k} k!S(n+1,k+1) = \left(-1\right)^{n}\sum_{j=0}^{k}\left(-1\right)^{j}{{k}\choose{j}}\left(j+1\right)^{n}$

Notes:

$\displaystyle e^{z(e^x-1)}=e^{-z}e^{ze^x}=e^{-z}\sum\limits_{k=0}^\infty\frac{z^k}{k!}e^{xk}=e^{-z}\sum\limits_{k=0}^\infty\frac{z^k}{k!}\sum\limits_{n=0}^\infty\frac{x^n k^n}{n!}=\sum\limits_{n=0}^\infty\frac{x^n }{n!}\left(e^{-z}\sum\limits_{k=0}^\infty\frac{z^k k^n}{k!}\right)$

$\displaystyle e^{z(e^x-1)}=\sum\limits_{k=0}^\infty\frac{z^k}{k!}(e^x-1)^k=\sum\limits_{k=0}^\infty\frac{z^k}{k!}\sum\limits_{j=0}^k(-1)^{k-j}{\binom k j}e^{xj}=$

$\displaystyle\hspace{1.5cm}=\sum\limits_{k=0}^\infty\frac{z^k}{k!}\sum\limits_{j=0}^k(-1)^{k-j}{\binom k j}\sum\limits_{n=0}^\infty\frac{x^n}{n!}j^n=\sum\limits_{n=0}^\infty\frac{x^n}{n!}\left(\sum\limits_{k=0}^\infty\frac{z^k}{k!}\sum\limits_{j=0}^k(-1)^{k-j}{\binom k j}j^n\right)$

$\displaystyle\hspace{1.5cm}=\sum\limits_{n=0}^\infty\frac{x^n}{n!}\left(\sum\limits_{k=0}^\infty z^k S(n,k)\right)=\sum\limits_{n=0}^\infty\frac{x^n}{n!}\left(\sum\limits_{k=0}^n z^k S(n,k)\right)$

$\hspace{1.8cm}$ because of $~S(n,k)=0~$ for $~k>n$

Comparing the coefficients of $~x^n~$ we get the first formula.

And the second formula comes from integrating by $~\int\limits_0^\infty ... dt~$ ;

with $~c>0~$ it's $\displaystyle~\int\limits_0^\infty\frac{t^n}{e^{ct}}dt=\frac{1}{c^{n+1}}\int\limits_0^\infty\frac{(ct)^n}{e^{ct}}d(ct)=\frac{n!}{c^{n+1}}~$ :

$\displaystyle\sum\limits_{k=0}^\infty\frac{(at)^k}{e^{at}k!}k^n = \sum\limits_{k=0}^n (at)^k S(n,k) ~~~~ |\cdot e^{-t} ~~~~ |\int\limits_0^\infty ... dt$

$\displaystyle\sum\limits_{k=0}^\infty\frac{a^k k^n}{k!}\int\limits_0^\infty\frac{t^k}{e^{(a+1)t}}dt = \sum\limits_{k=0}^n a^k S(n,k)\int\limits_0^\infty\frac{t^k}{e^t}dt$

$\displaystyle\sum\limits_{k=0}^\infty \frac{a^k k^n}{(1+a)^{k+1}} = \sum\limits_{k=0}^n a^k k!S(n,k) ~~~~ |\cdot (1+a)$

From there to the third formula:

$\displaystyle\sum\limits_{k=0}^\infty k^n\left(\frac{a}{1+a}\right)^k = (a+1)\sum\limits_{k=0}^n a^k k!S(n,k)$

$\displaystyle = \sum\limits_{k=0}^n a^{k+1} k!S(n,k) + \sum\limits_{k=0}^n a^k k!S(n,k) = \sum\limits_{k=1}^{n+1} a^k (k -1)!S(n,k-1) + \sum\limits_{k=0}^n a^k k!S(n,k) $

$\displaystyle = a^{n+1} n!S(n,n) + \sum\limits_{k=1}^{n} a^k (k -1)!(S(n,k-1) + k S(n,k)) + a^0 0!S(n,0) $

$\displaystyle = a^{n+1} n!S(n,n) + \sum\limits_{k=1}^{n} a^k (k -1)!S(n+1,k) + S(n,0) $

$\displaystyle = \sum\limits_{k=1}^{n+1} a^k (k -1)!S(n+1,k) + S(n,0) = \sum\limits_{k=0}^{n} a^{k+1} k!S(n+1,k+1) + S(n,0) $

Substracting $~0^n=S(n,0)~$ leads to $\displaystyle \sum\limits_{k=1}^\infty k^n\left(\frac{a}{1+a}\right)^k = \sum\limits_{k=0}^{n} a^{k+1} k!S(n+1,k+1)~$ .

With setting $\displaystyle ~\frac{a}{1+a}=-e^{-x}~$ we get $\displaystyle ~a=-\frac{1}{1+e^x}~$ and therefore the last formula.

user90369
  • 11,518
  • I'm sure that your answer is great (I give +1),but I did not think that my question to be advanced that much, do you have any reference(s) which proves your used formulas? – Absurd Nov 01 '19 at 13:47
  • @Absurd: Thanks! E.g. the second formula you can see here, line $(29)+(30)$ . I've added the mixed bivariate generating function for the Stirling numbers of the second kind, which produces the first formula. So, you know how it's possible to start. ;) – user90369 Nov 01 '19 at 17:07
  • @Absurd : I everything understandable now or should I show how to get the first formula from $~e^{z(e^x-1)}~$ ? – user90369 Nov 03 '19 at 20:38
  • If you do that,I will appreciate it so much – Absurd Nov 03 '19 at 21:04
  • @Absurd : I've added some notes. ;) – user90369 Nov 04 '19 at 07:33
  • I got it,is it possible for you to explain the last yellow box too?thnks :) – Absurd Nov 04 '19 at 08:52
  • @Absurd : Yes, I've added. :) --- But: My answer depends on the Stirling numbers of the second kind and of course I don't know if this is o.k. for you; maybe you have specifications which exclude that. – user90369 Nov 04 '19 at 10:34
  • thnk you again,now I know where do the first and last parts come from,but still I have a problem in second part,could you please explain that with more details?(I'm sorry for that!) – Absurd Nov 04 '19 at 14:44
  • @Absurd: So, I hope, it's o.k. now. Or any question left ? :) – user90369 Nov 04 '19 at 16:14
  • now the only problem is when you say :as follows: and you take the derivative,I just have a problem there :) – Absurd Nov 04 '19 at 18:21
  • @Absurd: Sorry, I don't understand, what do you mean ? – user90369 Nov 04 '19 at 18:23
  • how do you take the nth derivative of $1/e^x +1$ using your formulas? – Absurd Nov 04 '19 at 18:25
  • @Absurd: The $~n^{th}~$ derivation of $~\frac{1}{1+e^x}~$ is equivalent to the $~n^{th}~$ derivation of it's series $~-\sum\limits_{k=1}^\infty(-e^{-x})^k~$ . You know $~\frac{x}{1-x}=\sum\limits_{k=1}^\infty x^k~$ ? – user90369 Nov 04 '19 at 18:31
  • yes,but how did you use that? – Absurd Nov 04 '19 at 19:40
  • $\displaystyle\frac{d^n}{dx^n}\frac{1}{1+e^x} = \frac{d^n}{dx^n}\left(-\sum\limits_{k=1}^\infty(-e^{-x})^k\right) = -\sum\limits_{k=1}^\infty(-1)^k\frac{d^n}{dx^n}e^{-xk} = (-1)^{n+1}\sum\limits_{k=1}^\infty k^n (-e^{-x})^k ~$ and for $~\displaystyle\sum\limits_{k=1}^\infty k^n (-e^{-x})^k ~$ see the $3^{rd}$ formula. – user90369 Nov 04 '19 at 20:28
  • should not it be $(-1)^{k+1}$? – Absurd Nov 04 '19 at 20:56
  • The missed $-1$ is together with $(-1)^n$ of the derivation. If you write $~(-1)^{n+1}\cdot(-1)^k~$ or $~(-1)^n\cdot(-1)^{k+1}~$ has no meaning. – user90369 Nov 04 '19 at 21:02
  • thnks, how did you use geometric series for the series of $1/1+e^x$? – Absurd Nov 04 '19 at 21:17
  • It's not clear for me what you like to know. It's $\displaystyle~\frac{1}{1+e^x}=-\frac{(-e^{-x})}{1-(-e^{-x})}=-\sum\limits_{k=1}^\infty(-e^{-x})^k~$ . ;) – user90369 Nov 05 '19 at 07:31
  • that's totally true,but don't we have any convergence problem? because I think the formula is not valid for all real $x$(even if we use geometric series which is $1/1+x$,we just can use this for $x$ which are greater than $-1$ and less than $1$. – Absurd Nov 05 '19 at 07:50
  • A good question. You know the analytic continuation? E.g. here. If one continues a function (here: the series) analytically in two different ways, for example through functions, then these two functions have the same value range within the intersection of their definition sets. In your case, the common definition range is $\mathbb{R}$. But for comparing the coefficients you don't need that, because they are independend of $x$ . You can also use $x>0$ or whatever is senseful. – user90369 Nov 05 '19 at 08:57
  • Any question left ? :) – user90369 Nov 05 '19 at 16:45
  • I guess no,the first time you gave me this answer of my question I was like "this is so hard,I never would be able to understand it",but you made it possible for me,Thank you for all of your answers . – Absurd Nov 05 '19 at 18:13
  • @Absurd: You are welcome, it was a pleasure. :) – user90369 Nov 05 '19 at 19:52