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Who first defined truth as adæquatio rei et intellectus? António Manuel Martins claims (@44:41 of his lecture quot;Fonseca on Signs quot;) that the origin of what is now called the correspondence theory of truth, Veritas est adæquatio rei et intellectus
Why is $\infty\times 0$ indeterminate? - Mathematics Stack Exchange "Infinity times zero" or "zero times infinity" is a "battle of two giants" Zero is so small that it makes everyone vanish, but infinite is so huge that it makes everyone infinite after multiplication In particular, infinity is the same thing as "1 over 0", so "zero times infinity" is the same thing as "zero over zero", which is an indeterminate form Your title says something else than
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
Programación Lineal (PL) - Mathematics Stack Exchange El resultado de correr el proceso 3 por una hora es 2 barriles de gasolina 3 Todas las semanas se podrían comprar 200 barriles de crudo 1 a 2 dólares el barril y 300 barriles de crudo 2 a 3 dólares el barril
Difference between PEMDAS and BODMAS. - Mathematics Stack Exchange Division is the inverse operation of multiplication, and subtraction is the inverse of addition Because of that, multiplication and division are actually one step done together from left to right; the same goes for addition and subtraction Therefore, PEMDAS and BODMAS are the same thing To see why the difference in the order of the letters in PEMDAS and BODMAS doesn't matter, consider the
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
Prove that $1^3 + 2^3 + . . . + n^3 = (1+ 2 + . . . + n)^2$ HINT: You want that last expression to turn out to be $\big (1+2+\ldots+k+ (k+1)\big)^2$, so you want $ (k+1)^3$ to be equal to the difference $$\big (1+2+\ldots+k+ (k+1)\big)^2- (1+2+\ldots+k)^2\; $$ That’s a difference of two squares, so you can factor it as $$ (k+1)\Big (2 (1+2+\ldots+k)+ (k+1)\Big)\; \tag {1}$$ To show that $ (1)$ is just a fancy way of writing $ (k+1)^3$, you need to
Pole-zero cancelation method for PI controller design I think it is ill-advised in practice to do pole-zero cancellation Unstable pole-zero cancellation is just plain bad (the closed loop will be unstable) but stable pole-zero cancellation is also not great for practical reasons The cause is due to not knowing the pole $-p$ exactly, but primarily it is the side-effects of a failed cancellation that is truly the problem Normally, like in your