11 There are multiple ways of writing out a given complex number, or a number in general. Usually we reduce things to the "simplest" terms for display -- saying $0$ is a lot cleaner than saying $1-1$ for example. The complex numbers are a field. This means that every non-$0$ element has a multiplicative inverse, and that inverse is unique.
In my AI textbook there is this paragraph, without any explanation. The sigmoid function is defined as follows $$\sigma (x) = \frac {1} {1+e^ {-x}}.$$ This function is easy to differentiate
Compute: $$\prod_ {n=1}^ {\infty}\left (1+\frac {1} {2^n}\right)$$ I and my friend came across this product. Is the product till infinity equal to $1$? If no, what is the answer?
Intending on marking as accepted, because I'm no mathematician and this response makes sense to a commoner. However, I'm still curious why there is 1 way to permute 0 things, instead of 0 ways.
The volume of this simplex is $\frac {1} {n!}$, but I am intuitively struggling to see why. I have seen proofs for this and am convinced, but I can't help but think there must be a slicker or more intuitive argument for why this is so than what I have already seen. Any help would be appreciated!
Thomson et al. provide a proof that $\lim_ {n\rightarrow \infty} \sqrt [n] {n}=1$ in this book (page 73). It has to do with using an inequality that relies on the binomial theorem: I have an alternat...
Prove that the sequence $\ {1, 11, 111, 1111, .\ldots\}$ will contain two numbers whose difference is a multiple of $2017$. I have been computing some of the immediate multiples of $2017$ to see how their congruence classes look like, but I am not really sure where that is taking me.
My only suggestion would be that you should adjust your terminology a bit. The Laplace transform of the integral isn't $\frac {1} {s}$. It'd be more accurate to say The Laplace transform of an integral is equal to the Laplace transform of the integrand multiplied by $\frac {1} {s}$.
