okay last time I showe guys the PI function and also the gamma function and they are just ways to
extend the concept of factorial and we also saw that zero factorial equals 1 and 1/2 factorial
it's equal to square root of pi over 2 and today we will do more and of course now the question
is can we actually have some negative factorial and the answer to that is yes and no so let me
show you this and yes first last time we did the positive 1/2 factorial right why don't we
try negative 1/2 factorial and let me tell you guys first in fact this is going to be nicer
that the past 1/2 factorial ok so just wait for it that's plugging negative 1/2 into this factorial
of course I can just plug in negative 1/2 into this X from the PI function right so that will
be the quicker way this is the integral from 0 to infinity T to the negative 1/2 power and then we
have e to the negative T DT and we just have to work this out now so to do that let me write this
down in the denominator so I will keep this as e to the negative T over this becomes squared of T
in the denominator and we have the DT right here of course and then we'll take a used substitution
that U is equal to square root of T and square both sides we get is equal to u square and then
differentiate this did he will be to you the you okay and then we will just take this integral from
the T word to the world so you know this is T going from 0 to T goes to infinity right when
you take that to the new world u goes from we're plugging into here screw it's also 0 so u goes
from 0 and we're plugging that into here school of infinity is still infinity so you also goes from 0
to infinity and then we have e negative this T is U square so we have negative u square power right
here over square root of T is the U and TT is the to you do you and now what well as we can see that
you cancelled each ow and if you were like we can put the two to the front so we have two and
that's the integral in the new world from zero to infinity and this is just e to the negative
you Square to you and fine now we should recognize this already huh this is just half of the Gaussian
integral right and we saw that last time as well so this is the tuna front and this is square root
of PI over two and in the end of course we can change all the tool and we get square root of pi
and as I told you this answer is nicer compared to the positive 1/2 factorial right so I can say
negative 1/2 factorial which and by the way is there any connection between the negative 1/2
factorial versus the parsley 1/2 factorial the answer to that is yes I will show you guys this
real quick and I will look at this right here and factorial okay so recall that I will just
put down recall this is just one of the properties of factorial now I will let equals to 1/2 and you
will see when I plug in 1 having two here I get 1/2 factorial and this is what we did last time
already okay and then this is equal to and it is 1/2 and here we have 1/2 minus 1 and then we
do the factorial and this right here is square root of pi over 2 right but let me not write
that don't get because I want to show you guys the next connection this is just 1/2 factorial
and this is equal to 1/2 times this is negative 1 over 2 factorial right and now if you want to
figure this out all we have to do is multiply both sides by 2 isn't it so now in the end we
see negative 1/2 factorial it's just 2 times that 2 times positive 1/2 factorial and of course we
know what this is from last time so that square root of pi over 2 and then of course they cancel
out so we see that this is equal to square of Pi right negative 1/2 factorial its equal to square
root of pi all right so this right here is pretty much it and if you do it with this approach you
can easily figure out 3 over 2 factorial 5 over 2 factorial 7 over 2 factorial and also the negative
versions namely negatives you have a 2 factorial negative 5 over 2 factorial negative 7 over 2
factorial exact track sector okay in fact any odd number over 2 factorial you can do it easily with
this kind of formula ok and maybe your circus that general formula next time because earlier I told
you sometimes it's possible for us to have a negative factorial sometimes not so let me
show us an example that we cannot have a negative factorial what if we have negative 1 factorial and
maybe you can just try to do the quality earlier right so that's first plug in negative 1 into
this X and see what we get so this is the integral going from 0 to infinity and we will have T raised
to the negative 1 power now and then e to the negative T DT and you can try if you may but let
me just tell you no matter what we do we cannot end out a nice for you for this integral and that
T is this right here it's actually a divergent ok and this is whatever to show you first of all let
me just read this integral as the integral from 0 to infinity and let me put both of them down
to the denominator so I will just write this down as 1 over 3 to the positive 1 power and e to the
positive 3 power and indeed he like this in fact this improper integral is both type 1 and type 2
situation this right here is a type 1 situation because we have infinite interval it goes from 0
to infinity that's an infinite interval right so that's why it's a type of improper situation for
the integral and then its type 2 because on this interval we actually have a vertical asymptote
because when T is equal to 0 1 over T times e to the T we don't disappear on 0 1 over 0 that's
the vertical asymptote that's why its type 2 as well but the theory is that yes unlike earlier
even though that was also improper integral with both type 1 type 2 situations but that Interpol
actually converges this one actually diverges and I don't deal with both type 1 type 2 together let
me just pull this into two parts and hopefully first part just has the type 1 and hopefully
the second part only has the type who was inside that all right so let me do this for you guys let
me split this into the integral from 0 to of our you and you can pick any snail you want and let me
just keep it simple I will just put on one you can put down 1725 whatever you want doesn't matter I
will still have that 1 over T e to the T DT and then we will have to add the second integral we
had to pick up where we left off because I put on 1 so I will have to put on one and then continue
to infinity and we will have the 1 over T ok and now the do is this integral here it's the
type 1 situation only because we are infinite interval from 1 to infinity ok and you can do a
quick comparison test with the integral from 1 to infinity with just 1 over e to the T DT and
you can show that this integral it's actually smaller than the integral from 1 to infinity 1
over e to the T DT because that one converges this right here also converges ok so this is good this
right here is a bad guy this right here is a type 2 situation because we still have the vertical
asset up when T is equal to 0 right and I'm just going to work this out for you guys unfortunately
though we cannot find antiderivative of 1 over T e to the T so we have to use some kind of computers
and test for this and this is how it works first of all because we cannot find an antiderivative
for this let me put on something that we actually know much better ok and here is the deal when we
integrate from 0 to 1 we know that if we have 1 over T like that we can handle that much
in fact that diverges and this one also tougher chests okay so I'm just going to write this down
for you guys we know that the integral from 0 to 1 1 over T DT and the reason that I didn't
use 1 over e 2t because the integral from 0 to 1 1 over e to the T DT that's actually converge it
I have to make sure that I use a divergent and I also show this is a divergent as well ok and
you can check out more of my videos about the compression test for improper integrals or in
a series but anyway this is so much easier to deal with isn't it and you can quote us the room if you
would like but this is not bad to integrate at all so I was just integrate 1 over T then we out
get Ln of absolute value of T and I don't need to have the absolute value but you know doesn't
really matter because 0 to 1 positive right put it down anyway now our plugging one to here this
is ll f to the power of 1 and the minus Ln of the value of 0 and technically we are saying 0 plus
okay because we're going from 0 to 1 and we just care about 0 plus we take the limit right here
technically anyway Ln 1 is 0 minus L and 0 plus is negative infinity and when you do this you get
positive infinity right but oh no we can conclude that this right here we know it diverges and now
I will use this divergence to show that this is also at I project and this is how we are going to
do it here is the check okay and when we are doing comparison test we have two kind one kind is that
we can just do compare the inequality so that's exactly why I did earlier I just didn't write
it down okay but if I want to use an inequality to work this out it's not going to work ok but
it's ok because we can also to calculus we can also check the limit okay and that's exactly who
I'm going to do I'm going to check the limit he goes to well goes to what in this case you have
to remember the zero is the number that caused the vertical isotope right so we will take the
limit as T goes to zero and technically it's a zero plus okay and just remember if you are
doing the limit comparison test and in fact this is the type 2 situation because once again we have
a vertical as the top situation on this interval you take the limit as T goes to the number that
caused the particular setup okay anyway we are going to compare the ratio of this and that and
it doesn't really matter which one you put it on what time would you want to put it on the bottom
let me just put this one on the top or actually let me for this one without 1 over T e to the
T and now put this one on the bottom seriously doesn't matter which one you put it on top one
about them because in the end you just want to hopefully come with a nice value by nice we mean
to have a finite value and the final value is just to know anything between zero and positive
infinity okay if we can come up with that then we can conclude that both of them have the same
convergence e either they both converge or they both diverge but anyway let's just compute this
real quick well this is not bad at all we can just take the limit as T goes to 0 plus and right here
we can just multiply no T and T on both sides top and bottom so this and that cancel ok so and we
will just have 1 over e to the T and when T goes to 0 plus this is just 1 over e to the 0 plus and
0 is this right here 0 plus it's just the same as 0 e to the 0 is just 1 1 over 1 still 1 this
is good and as you can see this is the finite value and the best part is that this is 1 ok
an indicator this is finite it's in between of 0 and infinity so what's happening is that when we
compare the value of the these two expressions when x is the poor in 200 plus we are saying
that they are pretty much other thing value the same size when you are going to work the vertical
isotope number right here and yeah especially the same size because we have one right so that
mean this is just as big as that vice-versa okay so we're saying that this is just as big as that
because the ratio is one right especially you take the limit and because this right here diverged and
technically you can also imagine that this is passed the Infinity this also has to be passed
deep infinity so we can come CUDA this it's also a type virtual okay and I will just write down
the conclusion right here for you guys and this is usually the three steps that I want my students to
show put on what you know and do the check either inequality or take the limit and in the end you
draw the conclusion so it can be legitimate here's the conclusion I'm going to state that
again the integral from 0 to 1 1 over T into that he did he this right here also diverges okay and
remember they have the same conclusion if this diverged you can also show that this diverged if
this works out perfectly ok so I would like to use the word also diverges and that's courtesy room by
the limit comparison test and in fact this is the type 2 situation for improper integral because we
have the vertical as tough situation ok so what does that mean this right here diverges and as I
told you guys earlier this actually converges but you can imagine a passing thing that he
plus a positive finite value in the end we still end up with past infinity right but we're trying
to calculate the value for negative 1 factorial that's the main question and we of course we're
trying to calculate a value right what they're not talking about the limit this right here negative
1 factorial and vote is defined okay this right here yes and defined once again we're trying to
calculate value of negative 1 factorial so to now say that it's equal to infinity whatsoever
okay but anyway hopefully this is clear and can we also to forward it by using the N factorial
relationship let's try so recall this is this tackling poverty earlier as well when we had
n factorial this is equal to n times n minus 1 factorial right I want to come up with negative
1 factorial well if I put 0 into n I will have the negative 1 factorial right here right so I
will just say that equal to 0 and now this right here we will have 0 factorial and this is equal to
0 times 0 minus 1 factorial do we know what this is yes zero factorial is 1 right and you see this
is going to be crazy because we're saying 1 it's equal to this 0 times this right here is negative
1 factorial is it possible for us to find value okay final value okay so therefore you multiply
this with 0 you end up with 1 it's not possible right and you can also imagine you get em to 2
pi plus PI 0 packing graduation you just guarantee find expression anyway so in the end we can also
see that negative 1 factorial is defined okay and you can try to do with negative 2 factorial naked
is 3 factorial and things like that and guess what will happen your get and defined answers produce
factorials as well and let me emphasize this once again some people might say negative 1 factorial
should be infinity by my opinion because we cannot find find for you for negative 1 factorial from
here and also we know there's no finite value for negative 1 factorial for here and we will
multiply this by 0 and we know it's what that's why I say negative 1 factorial should be undefined
and also if you look at the graph for the PI function what the gamma function you will see
that they will have vertical asset of 4 you have naked heat integer factorial okay it's a bunch of
vertical asymptotes so that's why I say negative 1 factorial should be undefined and likewise for
negative 2 factorial negative 3 factorial and so on anyway hopefully you guys liked this video and
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a lot of interesting things that you guessed me now have her before and in fact I've been coding
a lot to know more advanced mathematics and they are making two videos for you guys and as you can
see all of this right here we just need calc 2 to understand all this but unfortunately the negative
1 factorial will maybe the 1/2 factorial they were not usually mentioned in the text books right so
Brenda walk is a definitely a great place for everybody who loves math to go and just check out
the articles their pages their courses okay and hopefully you guys like this video and easier to
check out bilenda work slash black pen real pen thank you guys so much and as always that's it