Why do we use 'e' in the Sigmoid?

Why do we use the mathematical constant "e" in the sigmoid function?

[halibrahim]

Thank you for making us love math even more.

[pawanbhatt]

thankyou for bringing this intuitive video, I just had this thought yesterday.
Please keep uploading videos like this it makes my intuitive more strong and closer to statistics.

[haojiang]

Holy shit! I wish I could watch this video 6 years ago when I just got into machine learning. You did a great job! Thank you so much!

[ramirolopezvazquez]

Thanks! I really appreciate this bits of useful, subtle and insightful ideas about common objects in data science

[SuperMtheory]

Makes sense. dy/dx = y (1- y ) if k=e. Great video!

[uusserrrreesssuuu]

are operations with 'e' are more expensive then with 2 or 3?

[apoorvatiwari]

Nice explanation. Clarifies everything

[behrampatel]

Good one !
Video edit a picture in picture effect of showing this in desmos. That would help as an interactive visual aide.

[ChocolateMilkCultLeader]

The best in the game for this kind of conteht

[jasdeepsinghgrover]

Actually there is a better reasoning but I am still not sure about it... Sigmoid is derived through the linear regression on log odds of the two classes... So mx+c = ln(p/(1-p)) which gives p = 1/(1+e^-(mx+c))

[JeremiahLam-sd]

Huh, so I guess this is like a tradeoff of annoyances where using e upfront is just less annoying than discovering ln(k) much later.

[giorda]

Really good explanation. Keep it up :)

[masster_yoda]

This is a nice explanation, however one question is left open for me: We interpret the result of the sigmoid as probability. So sigmoid(x) results in some probability of something to be classified as some category. Let's assume the standard sigmoid(x) results in a value of 0.7. When I change sigmoid to use some other number k instead of e, this probability would change. Let's say it would now be 0.9 instead of 0.7. This appears to me as semantically completely different from 0.7. So I would conclude that with respect to the interpretation as probability, it is not arbitrary to choose e oder some other number k.

[nononnomonohjghdgdshrsrhsjgd]

sorry, you didn't explain anything.