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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
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
Difference between PEMDAS and BODMAS. - Mathematics Stack Exchange You shouldn't think of either rule as setting different priorities for multiplication and division, or for addition and subtraction You need to work left to right for these PEMDAS = Parentheses > Exponents > (Multiplication Division) > (Addition Subtraction) BODMAS = Brackets > Order > (Division Multiplication) > (Addition Subtraction)
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