MWresearch Posted July 7, 2015 Share Posted July 7, 2015 (edited) Overall there doesn't seem to be a lot on multifactorials. Generalized formulas for multifactorials seem to only work for numbers evenly divisible by k, where k is the type of factorial "like double, triple, quadruple ect). But, my concearn is that the non-evenly-divided numbers are treated completely arbitrarily and not even conjectured, just arbitrary selected. For instance, 7!! (double factorial) would count down to 7*5*3*1, whereas 8!! would count as 8*6*4*2. See the difference? 8!! ends on 2, like its suppose to, and its evenly divisible by 2. 7!! however ends on 1, like its not suppose to because its not a mono-factorial. But, is it really "not suppose to" or is there some huge proof I'm missing that I've never see nor heard of before that proves those numbers are suppose to not end on k on the multiplication chain? If not, I would take it upon myself to redefine multifactorials at non-evenly-divisible values. Edited July 7, 2015 by MWresearch Link to comment Share on other sites More sharing options...

imatfaal Posted July 8, 2015 Share Posted July 8, 2015 You cannot really prove a definition; the definition is axiomatic/arbitrary. Multifactorials are defined as [latex]n(k.exclamation marks) = n(n-k)(n-2k)... [/latex] stop before the expression in brackets is equal to or less than zero. There is nothing magical or fundamental about that definition - so thus there can be nothing to prove. It is merely a shorthand way of writing a long expression - there is no added meaning Link to comment Share on other sites More sharing options...

Endy0816 Posted July 8, 2015 Share Posted July 8, 2015 Shouldn't it be the same difference to say they both stop at 1? Link to comment Share on other sites More sharing options...

imatfaal Posted July 8, 2015 Share Posted July 8, 2015 Shouldn't it be the same difference to say they both stop at 1? No because the last term might well not be one. As the OP mentions the double factorial of an even number has lowest term of 2 [latex]8!!=8*6*4*2[/latex] using n as the number being factorialised (?) and k as the number of exclamations - ie the level of multiplefactorisation the last term will be n mod k Link to comment Share on other sites More sharing options...

Endy0816 Posted July 8, 2015 Share Posted July 8, 2015 What I'm not seeing why you can't just pull a 1 out and multiply, as it wouldn't change your final result. Link to comment Share on other sites More sharing options...

MWresearch Posted July 9, 2015 Author Share Posted July 9, 2015 (edited) You cannot really prove a definition; the definition is axiomatic/arbitrary. Multifactorials are defined as [latex]n(k.exclamation marks) = n(n-k)(n-2k)... [/latex] stop before the expression in brackets is equal to or less than zero. There is nothing magical or fundamental about that definition - so thus there can be nothing to prove. It is merely a shorthand way of writing a long expression - there is no added meaning Well, perhaps you are not understanding the point I am trying to make then. I would say we know with certainty, that, if we follow the pattern of a regular factorial, that we at least can know that 9*6*3 is a correct representation of a triple factorial. However, we don't actually know all in any way shape or form that 8!!! is 8*5*2, because it doesn't actually "end" on a 3. The way I see it, a factorial isn't "stop before the expression gets to 0", its "stop on value of k." From what I can see, the later is quantifiable, where you "stop on what k value you're using" which has already been put into terms of a generalized equation and works for apparent numbers evenly divisible by "k." I don't however, see any mathematical interpolation for "stop wherever the hell it happens to get closest to but not less than or equation to 0" under a single generalized equation, with no special "mod k" condition. In short, if it is just "the definition," then that interpolation was an assumption, which, means it could be the completely wrong way that mutifactorials are actually suppose to be interpolated and non-divisible values with respect to the whole of mathematics. If you don't stop on the value of "k," then it should be the same as not stopping at 1 in a regular factorial, which, would be equivalent to either an incomplete gamma function or some weird fractional argument for the gamma function. Suppose there exists continuous extension for all multifactorials with no special rules like "mod k" and "x>1", just one single formula and this generalized formula works flawlessly for all multifactorials divisible by their respective k values and for non-divisible values it generates an output close to but not exactly the equal to the number generated by the previously assumed classical interpolation. Perhaps for instance, it could turn out that this formula yields 8!!!=72.4463983759257389579 instead of 80, and something like 9!!=947.34735345 instead of 945, but, this equation would still generate 9!!!=162 since 162 is evenly divisible by 3. If that were the case, how would I know which set of standards is actually correct? Since the previous definition of a multifactorial is nothing but an arbitrary assumption, would it be correct to disregard or alter that definition if such an equation existed as described? How would I know which one is right? I mean saying a mathematician is wrong to assume something is a bold statement so if I put the time into investigating this and find something out that challenges that definition, I want to make sure that first, its actually logical to challenge or alter that classical definition or that doing so is acceptable by the mathematical community , and second that its not going to be illogically thrown away just because I don't happen to be a famous, wealthy, dead European male. This area to me is especially grey because, even if I do find such an equation, how would I know that it cannot be secretly altered in order to generate the same output as the classical definition? You know what I really wish is if there was a more scientific method for the exploration and proof for mathematics, kind of like induction. A method where, if if an equation works perfectly for a certain set of known values, then it must work for all values. Like a parabola, if a certain parabola works perfectly to describe 3 points, then shouldn't it be correct to assume it works perfectly for all points on its curve, since only 1 parabola can perfectly fit those 3 points? With only 2 points, I could see a problem, because there is more than one parabola that can fit any given two points, but, not 3. Every time I think about that though, I think about that old issue with prime numbers. I forgot who it was, but someone investigating prime numbers generated some formula that was suppose to describe the frequency or the value of all prime numbers, and it worked perfectly up until like the 43rd prime and then the formula stopped working for what seems to me like no apparent reason. But perhaps that mathematician just made a specific mistake that can be avoided or that circumstance is special. Edited July 9, 2015 by MWresearch Link to comment Share on other sites More sharing options...

imatfaal Posted July 9, 2015 Share Posted July 9, 2015 What I'm not seeing why you can't just pull a 1 out and multiply, as it wouldn't change your final result. Why bother? AS you say it makes no difference once you are doing the multiplication - but the formulation I gave allows you to stop at the right point; ie the step before the term in brackets equals or is less than zero. The formulation I gave is also the accepted definition of the multifactorial. Adding an arbitrary "multiply by one" would destroy the pattern whilst not adding any information nor changing the answer Well, perhaps you are not understanding the point I am trying to make then. I would say we know with certainty, that, if we follow the pattern of a regular factorial, that we at least can know that 9*6*3 is a correct representation of a triple factorial. However, we don't actually know all in any way shape or form that 8!!! is 8*5*2, because it doesn't actually "end" on a 3. The way I see it, a factorial isn't "stop before the expression gets to 0", its "stop on value of k." its "stop on value of k." - well then you are wrong. That is not the definition. here is a link to the page at Mathworld - you will note that on the definition of Double Factorial the definition explicity shows that the double factorial can end on a multiple of 2 or of 1 Factorials stop at n mod k [latex]\prod_{i=1}^{\frac{1}{k}(n - n \: (mod\; k))} (n\: (mod\: k) + (i-1) \cdot k)[/latex] It is clunky and I have never seen anything written in such an ugly form - but that works for all positive n and k (I hope) Link to comment Share on other sites More sharing options...

MWresearch Posted July 9, 2015 Author Share Posted July 9, 2015 (edited) Ok, that function for the double factorial does output 5!! as 15, so it works! But there's still something I can't figure out: what is the analogue of "mod k"? For instance the analogue version of (-1)^n is e^(i*pi*x), since the imaginary part is 0 when x is an integer or something similar to x^cos(pi*x) Edited July 9, 2015 by MWresearch Link to comment Share on other sites More sharing options...

imatfaal Posted July 9, 2015 Share Posted July 9, 2015 Ok, that function for the double factorial does output 5!! as 15, so it works! But there's still something I can't figure out: what is the analogue of "mod k"? For instance the analogue version of (-1)^n is e^(i*pi*x), since the imaginary part is 0 when x is an integer or something similar to x^cos(pi*x) I have no idea what you mean in that second part n (mod k) is well defined - just search on modular arithmetic [latex]e^{\pi \cdot i \cdot x} = e^{\pi \cdot i} \cdot e^x = -e^x \neq \frac{1}{n} \neq \frac{1}{x}[/latex] Link to comment Share on other sites More sharing options...

MWresearch Posted July 10, 2015 Author Share Posted July 10, 2015 (edited) I don't know if I can really explain it on a more fundamental level. In your study of math, I'm sure you've dealt with both discrete mathematics and analogue mathematics, you have to know what I'm talking about on some level. Think of statistics, where originally you had a discrete distribution for the distribution of averages, forming a bell-like shape that had a bit of roughness to it. Then, you found that as the sample size went towards infinity, it became a continuous curve and turned perfectly into the Gaussian bell curve, that's the kind of thing I'm talking about. (-1)^(n) typically only occurs in a discrete series or summation or infinite product. However, if you wanted to describe (-1)^n over all real numbers including decimals and irrational numbers, not just natural integers, you would instead use e^(i*pi*x), where x replaces n. Or, a classic case, the integral. Again with the bell curve, you use to describe the probability resulting from the frequency distribution in a distribution of averages as a discrete summation. Then, when you let the sample size tend towards in infinity and the probability density function became a probability function via a discrete summation, you switched to a continuous, analogue form of an infinite sum when working with continuous data, an integral, called the "error function." It's a very very very very very very clear distinction between discrete and analogue math, I know for a fact you know what I'm trying to say but I don't know the proper terminology for distinguishing between systems of math like that. I want to take that same concept of transforming this discrete n mod k from a discrete system to an analogue system, which I suspect would involve trigonometric functions not limited to Euler's identity. Basically, how do you define n mod k for non-integer values? Edited July 10, 2015 by MWresearch Link to comment Share on other sites More sharing options...

## Recommended Posts

## Create an account or sign in to comment

You need to be a member in order to leave a comment

## Create an account

Sign up for a new account in our community. It's easy!

Register a new account## Sign in

Already have an account? Sign in here.

Sign In Now