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- factorial - Why does 0! = 1? - Mathematics Stack Exchange
The theorem that $\binom {n} {k} = \frac {n!} {k! (n-k)!}$ already assumes $0!$ is defined to be $1$ Otherwise this would be restricted to $0 <k < n$ A reason that we do define $0!$ to be $1$ is so that we can cover those edge cases with the same formula, instead of having to treat them separately We treat binomial coefficients like $\binom {5} {6}$ separately already; the theorem assumes
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- Problem when integrating $e^x x$. - Mathematics Stack Exchange
This part looks right: $$\int {\frac {e^x} {x}} \, dx = \frac {e^x} {x} + \frac {e^x} {x^2} + \frac {2e^x} {x^3} + \frac {6 e^x} {x^4} + \frac {24 e^x} {x^5} + \cdots+ \frac {n!e^x} {x^ {n+1}}+ (n+1)!\int \frac {e^x} {x^ {n+1}}$$ When you say "repeating to infinity" you want to take the limit of that in order for your equality to hold, you need $$\lim_n (n+1)!\int \frac {e^x} {x^ {n+1}}=0
- What does it mean to have a determinant equal to zero?
Your answer is already solved, but I would like to add a trick If the rank of an nxn matrix is smaller than n, the determinant will be zero
- When 0 is multiplied with infinity, what is the result?
What I would say is that you can multiply any non-zero number by infinity and get either infinity or negative infinity as long as it isn't used in any mathematical proof Because multiplying by infinity is the equivalent of dividing by 0 When you allow things like that in proofs you end up with nonsense like 1 = 0 Multiplying 0 by infinity is the equivalent of 0 0 which is undefined
- What are the criteria for bad faith questions?
The main criteria is that it be asked in bad faith ;-) I'm not entirely insincere: The question is rather how can we tell that, and a big part of the answer is "context"; it's not mainly the question itself
- A notion of similarity in hyperbolic geometry
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