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I recently managed to prove these equalities:

$$E_n=\sum_{k=0}^{n}S(n,k)\frac{-k!\sqrt{2}}{\sqrt{2}^{k}}\cos(\frac{3\pi}{4}(k+1))$$

$$E_n=2\int_{0}^{\infty}e^{-t}\cos(t)T_n(-t)dt$$

where $E_n$ are the Euler number, $S(n,k)$ the Stirling numbers of second kind and $T_n(x)$ the Touchard Polynomials.

My question is if they are useful, or if they are already discovered, since I didn't find anything.

For example, in Wikipedia we can see: (but I think this it's too much complicated). $$E_{n}=2^{2n-1}\sum_{\ell=1}^{n}\frac{(-1)^{\ell}S(n,\ell)}{\ell+1}\left(3\left(\frac{1}{4}\right)^{(\ell)}-\left(\frac{3}{4}\right)^{(\ell)}\right)$$

tomascatuxo
  • 161

2 Answers2

2

For the first one OP proposes

$$E_n = -\sqrt{2} \;\sum_{k=0}^n {n\brace k} \frac{k!}{\sqrt{2}^k} \cos(3\pi (k+1)/4).$$

which is

$$E_n = -\sqrt{2} \; \Re \sum_{k=0}^n {n\brace k} \frac{k!}{\sqrt{2}^k} \exp(3\pi i (k+1)/4).$$

Using the Stirling set number EGF this becomes

$$- \Re \left[ n! [z^n] \; (-1+i) \sum_{k=0}^n (\exp(z)-1)^k \frac{1}{\sqrt{2}^k} \exp(3\pi i k/4) \right].$$

We may extend the inner sum to infinity due to the coefficient extractor and the fact that $\exp(z)-1 = z + \cdots$, getting

$$\Re \left[ n! [z^n] \; (1-i) \frac{1}{1-\exp(3\pi i/4)(\exp(z)-1)/\sqrt{2}} \right] = \Re \; n! [z^n] f(z).$$

We have that $f(z)$ simplifies to

$$\frac{2(1-i)}{1+i+\exp(z)(1-i)} = \frac{2}{i+\exp(z)}.$$

For the real part we need the EGF of the conjugates. Using Mittag-Leffler we start with (poles of $f(z)$ are simple with residue $2i$)

$$g(z) = \sum_k \frac{2i}{z-(-\pi i/2 + 2\pi i k)} = \sum_k \frac{i(2z+\pi i)}{(z+\pi i/2)^2 - (2\pi i k)^2}$$

(the latter is convergent) and we can evaluate the sum through the residues in $w$ using the function

$$h(z,w) = - \frac{1}{w+(z+\pi i/2)} \frac{1}{w-(z+\pi i/2)} i(2z+\pi i) \cot(-iw/2) \frac{1}{2i}.$$

We use the fact that the residues in $w$ of $h(z,w)$ sum to zero and the residue at infinity is zero so that the residue at the two simple poles $w=\pm(z+ \pi i/2)$ produces for $g(z)$ (flip sign due to residue sum)

$$g(z) = - \frac{1}{2} \cot(-i(-(z+\pi i/2))/2) + \frac{1}{2} \cot(-i(+(z+\pi i/2))/2) \\ = - \frac{1}{2} \cot(iz/2-\pi/4) + \frac{1}{2} \cot(-iz/2+\pi/4) \\ = \cot(-iz/2+\pi/4).$$

We get for $g(z)$

$$\frac{\cos(-iz/2+\pi/4)}{\sin(-iz/2+\pi/4)} \\ = i \frac{\exp(z/2)\exp(\pi i/4)+\exp(-z/2)\exp(-\pi i/4)} {\exp(z/2)\exp(\pi i/4)-\exp(-z/2)\exp(-\pi i/4)} \\ = i \frac{\exp(z)\exp(\pi i/4)+\exp(-\pi i/4)} {\exp(z)\exp(\pi i/4)-\exp(-\pi i/4)} = i\frac{\exp(z)i+1}{\exp(z)i-1} = \frac{\exp(z)i+1}{i+\exp(z)}.$$

The difference between $f(z)$ and $g(z)$ is exactly $-i$ and we finally have

$$f(z) = -i + \sum_k \frac{2i}{z-(-\pi i/2 + 2\pi i k)}.$$

We seek the generating function of the conjugates which can now be obtained by inspection and is seen to be (expand terms into a series about zero)

$$f_C(z) = i + \sum_k \frac{-2i}{z-(\pi i/2 - 2\pi i k)} = i - \sum_k \frac{2i}{z-(\pi i/2 - 2\pi i k)}.$$

Here we have applied conjugation to

$$\frac{C}{z-\rho} = -\frac{C}{\rho} \frac{1}{1-z/\rho} = -\frac{C}{\rho} \sum_{q\ge 0} \frac{z^q}{\rho^q}.$$

We claim this is $-f(z-\pi i)$ and check

$$- f(z-\pi i) = i - \sum_k \frac{2i}{z-\pi i -(-\pi i/2 + 2\pi i k)} = i - \sum_k \frac{2i}{z-(\pi i/2 + 2\pi i k)}.$$

We are iterating over $k$ in two possible directions. Returning to the original problem we have that our answer is given by

$$\Re \; n! [z^n] f(z) = \frac{1}{2} n! [z^n] (f(z)+f_C(z)).$$

Using the expression for $f_C(z)$ in terms of $f(z)$ this becomes

$$\frac{1}{i+\exp(z)} - \frac{1}{i+\exp(z-\pi i)} = \frac{1}{i+\exp(z)} - \frac{1}{i-\exp(z)} \\ = \frac{i-\exp(z)-i-\exp(z)}{(-1)-\exp(2z)} = \frac{2\exp(z)}{\exp(2z)+1} = \frac{1}{\cosh(z)}.$$

We have obtained the EGF of the Euler numbers and may conclude.

Remark. The computation of the residue at infinity being zero may be seen at the following MSE link.

Marko Riedel
  • 61,317
2

For the second one OP proposes in terms of Touchard polynomials

$$E_n = 2 \int_0^\infty \exp(-t) \cos(t) T_n(-t) \; dt.$$

This is

$$2 \sum_{k=0}^n {n\brace k} \; \Re \; \int_0^\infty \exp(-t(1-i)) (-1)^k t^k \; dt.$$

Now put $(1-i)t = u$ so that $\frac{1}{2} (1+i) u = t$ and we get

$$\sum_{k=0}^n {n\brace k} \frac{1}{2^k} (-1)^k \; \Re \; (1+i)^{k+1} \int_0^{\infty (1-i)} \exp(-u) u^k \; du.$$

Evaluating the integral with the Gamma function we get

$$\sum_{k=0}^n {n\brace k} \frac{1}{2^k} (-1)^k \; \Re \; (1+i)^{k+1} k!.$$

This has EGF

$$n! [z^n] \Re \left[ (1+i) \sum_{k\ge 0} (\exp(z)-1)^k \frac{1}{2^k} (-1)^k (1+i)^k \right]$$

where we have extended to infinity due to $\exp(z)-1=z+\cdots$. This is

$$n! [z^n] \;\Re\; \frac{1+i}{1+(\exp(z)-1)(1+i)/2} = n! [z^n] \;\Re\; f(z).$$

We have that $f(z)$ simplifies to

$$\frac{2(1+i)}{1-i+(1+i)\exp(z)} = \frac{2}{-i + \exp(z)} = - \frac{2}{i-\exp(z)}.$$

Note that we learned in the companion answer that the EGF of the conjugates of $f(z)$ is $\frac{2}{i+\exp(z)}.$ We thus obtain one more time that

$$\frac{1}{2} (f(z) + f_C(z)) = \frac{1}{\cosh(z)}$$

and may conclude.

Remark. For the Gamma function evaluation we use a pizza slice contour with an angle of $-\pi/4$ which contains no poles. So to apply the Gamma function we just need to show that the contribution from the arc $Q$ vanishes in the limit.

We get with $z=R\exp(i\theta)$ and $dz = iR\exp(i\theta) \; d\theta$

$$\int_Q \exp(-z) z^k \; dz = \int_{-\pi/4}^0 \exp(-R \exp(i\theta)) R^k \exp(ki\theta) iR\exp(i\theta) \; d\theta.$$

We have as an upper bound on the norm of this integral

$$\int_{-\pi/4}^0 \exp(-R\cos(\theta)) R^{k+1} \; d\theta \lt R^{k+1} \exp(-R/\sqrt{2}) \int_{-\pi/4}^0 1 \; d\theta \\ = \frac{\pi}{4} R^{k+1} \exp(-R/\sqrt{2}) \rightarrow 0 $$

as $R\to\infty.$

Marko Riedel
  • 61,317