Question for the math professors ...

[daba]

... I followed a link dealing with Permutations and Combinations, and learnt a (to me) startling fact !

0 * 0 = 0 agreed ?

0 * 0 * 0 = 0 agreed again ?


so how does 0! = 1 ?

This has got me confused, perplexed, bamboozled, and a few other cliches.....

How can 0 * anything else not be zero ??

 

[Mickey]

http://mathforum.org/library/drmath/view/57128.html

 

[daba]

http://mathforum.org/library/drmath/view/57128.html

reddit ! so it's an arbitrary definition that conflicts with the factorial math - hmm, sounds like a get-out clause to me

But, anything raised to the power of zero is 1, regardless of the sign of the number, but 0^0 does not exist ???

Perhaps we have discovered some dark secret of math ...

 

[Peter Nachtwey]

What is 1.5! There is a deeper and better definition than the one in the math forum.

 

[Mickey]

I looked it up. I'm not sure its deeper but it is a better definition.

Evaluate the expression 1.5!

The factorial is the product of all positive integers less than or equal to n.
The formula for a factorial is:
n! = n x (n - 1) x (n - 2) x ... x 2 x 1

Evaluate the expression 1.5!
Factorial Expansions are as follows:
1.5! = 1.5 x 0.5 = 0.75

Now multiply out our factorial expansions:
1.5! = 0.75

Edit: I still don't get the 0!=1 ,but I'm not a professor or a Doctor.

 

[Peter Nachtwey]

The factorial function is actually a continuous function. You can find the factor of real numbers too. Th function starts is not continuous for negative numbers.

n! = integrate(x^n*exp(-x),0,inf)
n is the integral of x^n*exp(-x) as x goes from 0 to infinity

1.5! = 1.329.
0.5! = 0.886
https://deltamotion.com/peter/Mathcad/Mathcad - factorial.pdf
Notice that is gets the answers you expect for all the integers including 0! = 1.

 

[drbitboy]

Not to provide an alternate reason to the math PeterN mentions, but I have a vague memory that when 0! is defined as one, then the binomial coefficients* can be calculated via the formula

i.e. neither denominator term will be zero when k=0 or k=n.



* see also Pascal's Triangle

 

[daba]

It's all too much for me !

Happy Christmas to everyone !

 

[Peter Nachtwey]

Do a search on YouTube for why 0!=1
https://www.youtube.com/watch?v=Mfk_L4Nx2ZI

 

[Mickey]

it's all too much for me !

Happy christmas to everyone !

+1.

 

[tarik1978]

The general formula of factorial can be written in fully expanded form as
n! = n·(n-1)·(n-2)·...·3·2·1
or in partially expanded form as
n! = n · (n-1)!
We know with absolute certainty that 1!=1, where n = 1. If we substitute that value of n into the second formula which is the partially expanded form of n!, we obtain the following:
n!=n×(n-1)!
1!=1×(1-1)!
1!=1×0!
For the equation to be true, we must force the value of zero factorial to equal 1, and no other. Otherwise, 1!≠1 which is a contradiction.

So yes, 0! = 1 is correct because mathematicians agreed to define it that way (nothing more and nothing less) in order to be consistent with the rest of mathematics.

 

[daba]

n!=n×(n-1)!
1!=1×(1-1)!
1!=1×0!
For the equation to be true, we must force the value of zero factorial to equal 1, and no other. Otherwise, 1!≠1 which is a contradiction.

So yes, 0! = 1 is correct because mathematicians agreed to define it that way (nothing more and nothing less) in order to be consistent with the rest of mathematics.

Yes, but the first term on the right when n=0 is "0 x {something}". Forcing 0! to be 1 makes the equation incorrect... So, as you said, it's an agreed fudge ....

 

[Peter Nachtwey]

Yes, but the first term on the right when n=0 is "0 x {something}". Forcing 0! to be 1 makes the equation incorrect... So, as you said, it's an agreed fudge ....

It isn't a fudge. I showed how n! can be calculated even it it is a real. Just look at the link I posted above.


n! is related to the gamma function.

If
5! = 120
then 4! = 5!/5 = 24
then 3! = 4!/4 = 6
then 2! = 3!/3 = 2
then 1! = 2!/2 = 1
then 0! = 1!/1 = 1
Simple, don't make it hard.