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- factorial - Why does 0! = 1? - Mathematics Stack Exchange
$\begingroup$ 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
- Mathematics Stack Exchange
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- Difference between ≈, ≃, and ≅ - Mathematics Stack Exchange
For a quadratic I would set f(x) to zero but would not define f(x) as zero I use ≡ for all cases, not only immediate ones Setting f(x) to zero creates the equivalency f(x) = 0 for the coordinate you are trying to solve but is not true for all coordinates that are solvable
- What does it mean to say a divides b - Mathematics Stack Exchange
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- Normal vector to plane - Mathematics Stack Exchange
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- Design a PDA for the language L= {0^n1^n | n gt;=1}
3)delta(q,1,0)=(p,epsilon), when we see a 1 in the input go to state p and pop one 0 from the stack 4)delta(p,1,0)=(p,epsilon), pop one 0 from stack, per 1 read from the input 5)`delta(p,epsilon,Z)=(final state, Z), now the whole input is consumed and we reach the final state and the stack is empty and we have hit the bottom of the stack
- Example of infinite field of characteristic $p\\neq 0$
Stack Exchange Network Stack Exchange network consists of 183 Q A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers
- Improper integral of sin(x) x from zero to infinity
I was having trouble with the following integral: $\int_{0}^\infty \frac{\sin(x)}{x}dx$ My question is, how does one go about evaluating this, since its existence seems fairly intuitive, while its solution, at least to me, does not seem particularly obvious
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