stathelpplease
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This is (I hope) a simple question.
What is -5!
Thank you
Frank Norman Stein
What is -5!
Thank you
Frank Norman Stein
factorial
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stathelpplease
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Mrspi said:
So -5! would equal -120, consistent with -5^2 equals -25
Is that correct? <<< Yes
However, to be ABSOLUTELY clear - I prefer to write those as -(5[sup:2iqc3txz]2[/sup:2iqc3txz]) and -(5!)
Frank Norman Stein
stathelpplease
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stathelpplease said:
Mrspi - (I assume it was you who added the qualifier about being ABSOLUTELY clear - how did you do that anyway :? ) why are () needed to be clear here?
This actually goes to the root of my question. () are not needed for 3+4x5 because we all accept the convention. Don't we all (I am being serious with this question) accept the convention that -5^2=-25? (As crazy as it looks). And if we do, mustn't we all accept that -5!=-120 (as crazy as that looks)?
Frank Norman Stein
Because there is a difference. \(\displaystyle (-5)^{2}=25\), but \(\displaystyle -5^{2}=-25\)
With the factorial, it's the same thing. \(\displaystyle -5!=-120\), but \(\displaystyle (-5)!=\text{undefined}\)
Because there are no negative factorials in the sense that we have \(\displaystyle (-n)!\)
If you plug (-5)! in your calculator, you will get undefined. If you plug in -5! you will get -120. Simple as that.
The Gamma function generalizes the idea of factorial for non-integer numbers, but also gives us undefined values for negative integer factorials.
With the factorial, it's the same thing. \(\displaystyle -5!=-120\), but \(\displaystyle (-5)!=\text{undefined}\)
Because there are no negative factorials in the sense that we have \(\displaystyle (-n)!\)
If you plug (-5)! in your calculator, you will get undefined. If you plug in -5! you will get -120. Simple as that.
The Gamma function generalizes the idea of factorial for non-integer numbers, but also gives us undefined values for negative integer factorials.
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stathelpplease said:
What is 5!?
-5! would be -1*5!
5! = 5*4*3*2*1
Do the multiplication first, then take the opposite of the result.[/quote]
So -5! would equal -120, consistent with -5^2 equals -25
Is that correct? <<< Yes
However, to be ABSOLUTELY clear - I prefer to write those as -(5[sup:179zgc99]2[/sup:179zgc99]) and -(5!)
Frank Norman Stein[/quote]
Mrspi - (I assume it was you who added the qualifier about being ABSOLUTELY clear - how did you do that anyway :? )
No - that was me (Subhotosh Khan) and I did it by mistake!
Pressed the wrong button and instead of "quote"-ing you - I managed to edit your post. Sorry about that.
Here is a small tutorial on the Gamma and its relation to factorials for those interested.
\(\displaystyle {\Gamma}(p)=(p-1)!, \;\ {\Gamma}(p+1)=p!\)
So, \(\displaystyle {\Gamma}(6)=5!=120\)
Now, \(\displaystyle {\Gamma}(-4)=\text{undefined}\), but \(\displaystyle -{\Gamma}(6)=-5!=-120\)
\(\displaystyle {\Gamma}(p+1)=p{\Gamma}(p)\)
Solve for \(\displaystyle {\Gamma}(p)\) and get:
Note that \(\displaystyle {\Gamma}(p)=\frac{1}{p}{\Gamma}(p+1)\)..........[1]
This defines \(\displaystyle {\Gamma}(p) \;\ \text{for} \;\ p<0\)
Example:
\(\displaystyle {\Gamma}(-\frac{1}{2})=\frac{1}{\frac{-1}{2}}{\Gamma}(\frac{1}{2})\)
\(\displaystyle {\Gamma}(\frac{-3}{2})=\frac{1}{\frac{-3}{2}}\cdot \frac{1}{\frac{-1}{2}}{\Gamma}(\frac{1}{2})\) and so on.
Since \(\displaystyle {\Gamma}(1)=1,\) we have:
\(\displaystyle {\Gamma}(p)=\frac{{\Gamma}(p+1)}{p}\rightarrow {\infty} \;\ \text{as} \;\ p\to\ 0\)
From looking at this and the successive use of [1], we can see that \(\displaystyle {\Gamma}(p)\) becomes infinite not only at 0 but also at all negative integers.
For positive p, \(\displaystyle {\Gamma}(p)\) is continuous passing through the points \(\displaystyle p=n, \;\ {\Gamma}(p)=(n-1)!\)
For negative p, \(\displaystyle {\Gamma}(p)\) is discontinuous at the negative integers. In the intervals in between the integers it alternates between positive and
negative. negative from 0 to -1, positive from -1 to -2, and so on.
Interesting identities:
\(\displaystyle {\Gamma}(p){\Gamma}(p-1)=\frac{\pi}{sin({\pi}p)}\)
And its relation to the Beta function:
\(\displaystyle B(p,q)=\frac{{\Gamma}(p){\Gamma}(q)}{{\Gamma}(p+q)}\)
This can be very handy in numerous application, as with solving tough integrals such as \(\displaystyle \int_{0}^{\infty}\frac{y}{(1+y^{3})^{2}}dy\)
Actually, if we want to mess around in the complex realm, we can get a solution for a negative factorial via gamma.
For instance, \(\displaystyle {\Gamma}(-5)=.052359890304i\) and \(\displaystyle {\Gamma}(-1)\) is very close to \(\displaystyle {\pi}i\) and we can see has \(\displaystyle p\to -\infty\), \(\displaystyle {\Gamma}(p)\to 0\). But that is another matter.
I know, I know, this is more than you needed to know. But I posted it for mine and others interests as well.
\(\displaystyle {\Gamma}(p)=(p-1)!, \;\ {\Gamma}(p+1)=p!\)
So, \(\displaystyle {\Gamma}(6)=5!=120\)
Now, \(\displaystyle {\Gamma}(-4)=\text{undefined}\), but \(\displaystyle -{\Gamma}(6)=-5!=-120\)
\(\displaystyle {\Gamma}(p+1)=p{\Gamma}(p)\)
Solve for \(\displaystyle {\Gamma}(p)\) and get:
Note that \(\displaystyle {\Gamma}(p)=\frac{1}{p}{\Gamma}(p+1)\)..........[1]
This defines \(\displaystyle {\Gamma}(p) \;\ \text{for} \;\ p<0\)
Example:
\(\displaystyle {\Gamma}(-\frac{1}{2})=\frac{1}{\frac{-1}{2}}{\Gamma}(\frac{1}{2})\)
\(\displaystyle {\Gamma}(\frac{-3}{2})=\frac{1}{\frac{-3}{2}}\cdot \frac{1}{\frac{-1}{2}}{\Gamma}(\frac{1}{2})\) and so on.
Since \(\displaystyle {\Gamma}(1)=1,\) we have:
\(\displaystyle {\Gamma}(p)=\frac{{\Gamma}(p+1)}{p}\rightarrow {\infty} \;\ \text{as} \;\ p\to\ 0\)
From looking at this and the successive use of [1], we can see that \(\displaystyle {\Gamma}(p)\) becomes infinite not only at 0 but also at all negative integers.
For positive p, \(\displaystyle {\Gamma}(p)\) is continuous passing through the points \(\displaystyle p=n, \;\ {\Gamma}(p)=(n-1)!\)
For negative p, \(\displaystyle {\Gamma}(p)\) is discontinuous at the negative integers. In the intervals in between the integers it alternates between positive and
negative. negative from 0 to -1, positive from -1 to -2, and so on.
Interesting identities:
\(\displaystyle {\Gamma}(p){\Gamma}(p-1)=\frac{\pi}{sin({\pi}p)}\)
And its relation to the Beta function:
\(\displaystyle B(p,q)=\frac{{\Gamma}(p){\Gamma}(q)}{{\Gamma}(p+q)}\)
This can be very handy in numerous application, as with solving tough integrals such as \(\displaystyle \int_{0}^{\infty}\frac{y}{(1+y^{3})^{2}}dy\)
Actually, if we want to mess around in the complex realm, we can get a solution for a negative factorial via gamma.
For instance, \(\displaystyle {\Gamma}(-5)=.052359890304i\) and \(\displaystyle {\Gamma}(-1)\) is very close to \(\displaystyle {\pi}i\) and we can see has \(\displaystyle p\to -\infty\), \(\displaystyle {\Gamma}(p)\to 0\). But that is another matter.
I know, I know, this is more than you needed to know. But I posted it for mine and others interests as well.