Fueling Creators with Stunning

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019964d7bc201b22932fcc2063f4b0cc0cd3784f8b Youtube

019964d7bc201b22932fcc2063f4b0cc0cd3784f8b Youtube Why does 0! = 1 0! = 1? all i know of factorial is that x! x! is equal to the product of all the numbers that come before it. the product of 0 and anything is 0 0, and seems like it would be reasonable to assume that 0! = 0 0! = 0. i'm perplexed as to why i have to account for this condition in my factorial function (trying to learn haskell. Is there a consensus in the mathematical community, or some accepted authority, to determine whether zero should be classified as a natural number? it seems as though formerly $0$ was considered i.

0b2de2679aef43dbb5ce5817d15bf090 Youtube
0b2de2679aef43dbb5ce5817d15bf090 Youtube

0b2de2679aef43dbb5ce5817d15bf090 Youtube 0i = 0 0 i = 0 is a good choice, and maybe the only choice that makes concrete sense, since it follows the convention 0x = 0 0 x = 0. on the other hand, 0−1 = 0 0 1 = 0 is clearly false (well, almost —see the discussion on goblin's answer), and 00 = 0 0 0 = 0 is questionable, so this convention could be unwise when x x is not a positive real. 10 several years ago i was bored and so for amusement i wrote out a proof that 0 0 0 0 does not equal 1 1. i began by assuming that 0 0 0 0 does equal 1 1 and then was eventually able to deduce that, based upon my assumption (which as we know was false) 0 = 1 0 = 1. 0 0 = x 0 0 = x 0x = 0 0 x = 0 x x can be any value, therefore 0 0 0 0 can be any value, and is indeterminate. 1 0 = x 1 0 = x 0x = 1 0 x = 1 there is no such x x that satisfies the above, therefore 1 0 1 0 is undefined. is this a reasonable or naive thought process? it seems too simple to be true. Is a constant raised to the power of infinity indeterminate? i am just curious. say, for instance, is $0^\\infty$ indeterminate? or is it only 1 raised to the infinity that is?.

0 02 05 D3f7dd15a28b10599e6fb5d0aaf0e1ce72d6338ef6cd673dbbaa063bf974b671 7c9db4e99b033106 Youtube
0 02 05 D3f7dd15a28b10599e6fb5d0aaf0e1ce72d6338ef6cd673dbbaa063bf974b671 7c9db4e99b033106 Youtube

0 02 05 D3f7dd15a28b10599e6fb5d0aaf0e1ce72d6338ef6cd673dbbaa063bf974b671 7c9db4e99b033106 Youtube 0 0 = x 0 0 = x 0x = 0 0 x = 0 x x can be any value, therefore 0 0 0 0 can be any value, and is indeterminate. 1 0 = x 1 0 = x 0x = 1 0 x = 1 there is no such x x that satisfies the above, therefore 1 0 1 0 is undefined. is this a reasonable or naive thought process? it seems too simple to be true. Is a constant raised to the power of infinity indeterminate? i am just curious. say, for instance, is $0^\\infty$ indeterminate? or is it only 1 raised to the infinity that is?. How can i prove from first principles that $0!$ is equal to $1$?. The above picture is the full background to it. it does not invoke "indeterminate forms". it does not require you to write 0 0 0 0 and then ponder what that might mean. we don't divide by zero anywhere. it is just the case where limx→a g(x) = 0 lim x → a g (x) = 0 is out of scope of the above theorem. however, it is very common, in mathematical education, to talk about "indeterminate forms. In the set of real numbers, there is no negative zero. however, can you please verify if and why this is so? is zero inherently "neutral"?. 5 we have x0:= 1 x 0:= 1 for every complex number x x. (notice that this is the only convention which fits into the rules of arithmetic, and there is no need to exclude x = 0 x = 0. think about the binomial theorem, for instance.) by the way, your exercise ∑1000 n=0 in ∑ n = 0 1000 i n can be solved with the usual formula for geometric series.

0 02 05 C3de877ede0bd26b21154af8fe10316a64b425900c0b31ee2f8648c3352d3e34 1c6db23a6b1014 Youtube
0 02 05 C3de877ede0bd26b21154af8fe10316a64b425900c0b31ee2f8648c3352d3e34 1c6db23a6b1014 Youtube

0 02 05 C3de877ede0bd26b21154af8fe10316a64b425900c0b31ee2f8648c3352d3e34 1c6db23a6b1014 Youtube How can i prove from first principles that $0!$ is equal to $1$?. The above picture is the full background to it. it does not invoke "indeterminate forms". it does not require you to write 0 0 0 0 and then ponder what that might mean. we don't divide by zero anywhere. it is just the case where limx→a g(x) = 0 lim x → a g (x) = 0 is out of scope of the above theorem. however, it is very common, in mathematical education, to talk about "indeterminate forms. In the set of real numbers, there is no negative zero. however, can you please verify if and why this is so? is zero inherently "neutral"?. 5 we have x0:= 1 x 0:= 1 for every complex number x x. (notice that this is the only convention which fits into the rules of arithmetic, and there is no need to exclude x = 0 x = 0. think about the binomial theorem, for instance.) by the way, your exercise ∑1000 n=0 in ∑ n = 0 1000 i n can be solved with the usual formula for geometric series.

0 02 0b 3b24be75754496acf38fb8791be55aaf5d3aa0b2af0f4355b72114512d3f7bf2 Fe05d4dae8ccfa6d Youtube
0 02 0b 3b24be75754496acf38fb8791be55aaf5d3aa0b2af0f4355b72114512d3f7bf2 Fe05d4dae8ccfa6d Youtube

0 02 0b 3b24be75754496acf38fb8791be55aaf5d3aa0b2af0f4355b72114512d3f7bf2 Fe05d4dae8ccfa6d Youtube In the set of real numbers, there is no negative zero. however, can you please verify if and why this is so? is zero inherently "neutral"?. 5 we have x0:= 1 x 0:= 1 for every complex number x x. (notice that this is the only convention which fits into the rules of arithmetic, and there is no need to exclude x = 0 x = 0. think about the binomial theorem, for instance.) by the way, your exercise ∑1000 n=0 in ∑ n = 0 1000 i n can be solved with the usual formula for geometric series.

0 02 04 6d52c9497a1578271624653448400283c3d2414c46369487ba27a565093eb837 6fa0a133 Youtube
0 02 04 6d52c9497a1578271624653448400283c3d2414c46369487ba27a565093eb837 6fa0a133 Youtube

0 02 04 6d52c9497a1578271624653448400283c3d2414c46369487ba27a565093eb837 6fa0a133 Youtube

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