User talk:StephanSp3

August 2020

Hello. Your recent edit to Machine learning appears to have added the name of a non-notable entity to a list that normally includes only notable entries. In general, a person, organization or product added to a list should have a pre-existing article before being added to most lists. If you wish to create such an article, please first confirm that the subject qualifies for a separate, stand-alone article according to Wikipedia's notability guideline. Please read also WP:COI, in case it applies. GermanJoe (talk) 14:58, 6 August 2020 (UTC)

N-th derivative of tanh

On the German page someone noted the n-th derivative of $tanh(x)$ as

$${\frac {d^{n}}{dx^{n}}}\tanh(x)={\frac {2^{n+1}\mathrm {e} ^{2x}}{(1+\mathrm {e} ^{2x})^{n+1}}}\sum _{k=0}^{n-1}(-1)^{k}A_{n,k}\,\mathrm {e} ^{2kx}$$ (1)

in which A_n,k represents the Eulerian number.

However, there is no reference for this formula. I found an article about derivatives and Eulerian numbers where an example for $tanh(x)$ is noted. I simplify this example to show that Equation (1) holds true.

We have $f(x)=tanh(x)=1-{\frac {2}{\mathrm {e} ^{2x}+1}}$ and for this function the identity

$f^{(n)}(x)=(-1)^{n}\sum _{k=0}^{n-1}A_{n,k}(f(x)-1)^{k+1}(f(x)+1)^{n-k}$ (2)

is described in the article. We find for

$(f(x)-1)={\frac {-2}{\mathrm {e} ^{2x}+1}}:=g(x)$

and

$(f(x)+1)={\frac {2\mathrm {e} ^{2x}}{\mathrm {e} ^{2x}+1}}=-\mathrm {e} ^{2x}g(x)$

and so we calculate

$(f(x)-1)^{k+1}(f(x)+1)^{n-k}=g(x)^{k+1}[-\mathrm {e} ^{2x}g(x)]^{n-k}=g(x)^{n+1}(-\mathrm {e} ^{2x})^{n-k}$

where

• $g(x)^{n+1}={\frac {2^{n+1}(-1)^{n+1}}{[\mathrm {e} ^{2x}+1]^{n+1}}}$ and
• $(-\mathrm {e} ^{2x})^{n-k}=\underbrace {(-1)~\mathrm {e} ^{2x}} _{\text{outside of sum}}~\underbrace {(-1)^{n-k-1}~\mathrm {e} ^{2x[n-k-1]}} _{\text{inside of sum}}$ .

Now, we write formula (2) in terms of $g(x)$ as

$f^{(n)}(x)=g(x)^{n+1}~\mathrm {e} ^{2x}(-1)^{n+1}\sum _{k=0}^{n-1}A_{n,k}(-1)^{n-k-1}~\mathrm {e} ^{2x[n-k-1]}$

For the part outside of the sum we find

$g(x)^{n+1}~\mathrm {e} ^{2x}(-1)^{n+1}={\frac {2^{n+1}\mathrm {e} ^{2x}}{[\mathrm {e} ^{2x}+1]^{n+1}}}(-1)^{2n+2}={\frac {2^{n+1}\mathrm {e} ^{2x}}{[\mathrm {e} ^{2x}+1]^{n+1}}}.$

The Eulerian numbers are symmetric as $A_{n,k}=A_{n,n-k-1}$ and this implies for the part inside of the sum

$(-1)^{n-k-1}A_{n,n-k-1}~\mathrm {e} ^{2x[n-k-1]}\equiv (-1)^{k}A_{n,k}~\mathrm {e} ^{2xk}.$

Finally, we find

$$f^{(n)}(x)={\frac {2^{n+1}\mathrm {e} ^{2x}}{(1+\mathrm {e} ^{2x})^{n+1}}}\sum _{k=0}^{n-1}(-1)^{k}A_{n,k}\,\mathrm {e} ^{2kx}$$

which is identical with Equation (1).

References

October 2023

Hello, I'm DVdm. I noticed that you added or changed content in an article, Hyperbolic functions, but you didn't provide a reliable source. It's been removed and archived in the page history for now, but if you'd like to include a citation and re-add it, please do so. You can have a look at referencing for beginners. If you think I made a mistake, you can leave me a message on my talk page. Thank you. - DVdm (talk) 14:05, 17 October 2023 (UTC)

Note, as I said in my revert, we don't copy from other wikis and we don't point to our own talk pages where possibly dubious derivations can be found with possibly dubious sources. Cheers - DVdm (talk) 14:08, 17 October 2023 (UTC)