The reason this works by the way, is because it was the way (or analogous to the way) it was created by Euler and somewhat refined ( in my view) by Gauss.
The gamma function definition of 0!=0 has the exact same problem with (-1)!=1/0. Based on the gamma function definition of x!, we have (n+1)!=n!*(n+1) for all values where it is defined, and the limit as x-> -1 of x! (Gamma function definition) is 1/0 which is the complex infinity.
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Example; The product made up by (n+1)(n+1)/n(n+2) from n=1 to n=N is in numerical value [(N+1)!(N+1)!/N!(N+2)!] [2!•0!/1! •1!], for all N (as specified, N larger or equal to 1). Here, since ALL other quantities are known, 0! must be 1, and there is literally an infinite number of reasons for it. This is of course not an unique example. Rather, all cases of this type, is the same. The product (n+1)(n+2)/n(n+3), would give partial products [(N+1)!(N+2)!/N!(N+3)!] • 0!3!/1!2!, for example.
There’s a good reason to do so because 0! Shows up all the time combinatorially and as an expansion of Taylor series. If you’re going to define recursively it makes more sense to start at 0 then at 1. Of course the gamma function is a more general definition but this works for natural numbers and 0
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Obviously you want to show the connection between topics and even unite them. And broaden the area of investigation. I only wanted to show how this can be dealt with using only elementary techniques.
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I hope you saw my answer to that preceeding video on the topic, which avoids this problem if you look at what it says. It's totally different than the "similar looking" false arguments given in other comments.
The reason this works by the way, is because it was the way (or analogous to the way) it was created by Euler and somewhat refined ( in my view) by Gauss.
The gamma function definition of 0!=0 has the exact same problem with (-1)!=1/0. Based on the gamma function definition of x!, we have (n+1)!=n!*(n+1) for all values where it is defined, and the limit as x-> -1 of x! (Gamma function definition) is 1/0 which is the complex infinity.
Example; The product made up by (n+1)(n+1)/n(n+2) from n=1 to n=N is in numerical value [(N+1)!(N+1)!/N!(N+2)!] [2!•0!/1! •1!], for all N (as specified, N larger or equal to 1). Here, since ALL other quantities are known, 0! must be 1, and there is literally an infinite number of reasons for it. This is of course not an unique example. Rather, all cases of this type, is the same. The product (n+1)(n+2)/n(n+3), would give partial products [(N+1)!(N+2)!/N!(N+3)!] • 0!3!/1!2!, for example.
Why can’t we define 0!=1 as the base case and the recursion starting from there?
We can always define, but there is even better if we have some reason it should be so. See my other posts if you wish to do so.
There’s a good reason to do so because 0! Shows up all the time combinatorially and as an expansion of Taylor series. If you’re going to define recursively it makes more sense to start at 0 then at 1. Of course the gamma function is a more general definition but this works for natural numbers and 0
Obviously you want to show the connection between topics and even unite them. And broaden the area of investigation. I only wanted to show how this can be dealt with using only elementary techniques.
I hope you saw my answer to that preceeding video on the topic, which avoids this problem if you look at what it says. It's totally different than the "similar looking" false arguments given in other comments.