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Kind: captions
Language: en
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Hello friends, so welcome to the 25th lecture
of this course . And in this lecture, I will
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introduce some more properties of beta and
gamma functions . So, in the past couple of
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lectures, I told about beta functions and
gamma function and then I have taken some
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identities on these two functions . So, let
us come across few more properties of these
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two functions .
So, before going to those properties, I would
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like to introduce Euler's constant. So, Euler's
constant is defined by gamma equals to limit
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p tending into infinity 1 plus 1 by 2 plus
up to 1 by p minus log p . And if we if we
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find out the value of this limiting function,
then we are getting this value, so 0.5772
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and so on . So, this particular constant is
quite important in the theory of special functions
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. We can relate gamma function also with this
particular constant.
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So, for any real number x except on the negative
integers that is a null integer 0 minus 1
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minus 2 and so on . We have the infinite products
x into e raise power gamma x, here gamma is
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the Euler's constant, and then a product of
the infinite series where p equals to 1 to
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infinity 1 plus x upon p. So, it will be like
1 plus x, 1 plus x by 2, 1 plus x by 3 and
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so on, e raise to power minus x upon p. So,
and this particular expression will be equals
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to 1 upon gamma x. So, this relation between
the gamma function and Euler's constant was
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given by v r starts.
So, from this product we see that Euler's
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constant is deeply related to the gamma function
and the poles are clearly the negative or
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null integers . As we are not going to define
it x 0 minus 1 minus 2 and so on, because
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these are the poles.
Based on this definition, we will see a very
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important property of gamma function that
is called the complement formula. So, there
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is an important identity connecting the gamma
function at the complementary value complementary
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values means x and 1 minus x; x, gamma x into
gamma 1 minus x equals to pi upon sin pi x
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. And this formula is called complement or
reflection formula and it is valid when x
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and 1 minus x are not negative or null integers.
So, please because we are not going we
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have not defined gamma function for these
negative integers and null integer . And it
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was discovered by Euler . So, let us try to
prove this relation from the relation between
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gamma function and Euler's constant .
So, we know that that 1 upon gamma x equals
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to x into e raise power gamma x and then product
from p equals to 1 to infinity 1 plus x upon
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p into e raise to power minus x upon p . Similarly,
if I define it for minus x so 1 upon gamma
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minus x will be minus x into e raise to power
minus gamma x then p from 1 to infinity infinite
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term product 1 plus x upon p . And here x
is minus x so I am writing minus x here e
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raise to power minus x is here, so minus,
minus will become plus x upon p .
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Now, if I multiply these two relations then
in left hand side I will be having gamma x
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into gamma minus x in the denominator . So,
here I am having e raise to power gamma x,
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here I am having e raise to power minus gamma
x. So, it will become e raise to power 0 and
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that is 1. So, I will be having x into minus
x for x form here minus x from here . And
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then p equals to 1 to infinity 1 minus x square
upon p square, because here you are having
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1 plus x upon p, here 1 minus x upon p. So,
1 plus a into 1 minus a will become 1 minus
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a square. And then e raise to power minus
x upon p into e raise to power x upon p, it
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will become e raise to power 0 that is 1.
So, after multiplying this we are having this
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relation .
Now, we know that that minus x plus 1 the
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gamma function value of this particular thing
that it is minus x plus 1 will be
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minus x into gamma minus x from the lesson
that gamma n plus 1 equals to n gamma n. So,
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the same thing here n is minus x. So, from
here I can write gamma minus x equals to gamma
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1 minus x that is the term; and here please
note that 1 minus x is positive we have assumed
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it in the beginning upon minus x .
So, substitute this value of gamma minus x
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in this relation . So, what I will be having
minus in the numerator upon gamma x gamma
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1 minus x equals to x into minus x . Now,
there is a famous relation between this particular
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infinite product (Refer Time: 07:36) and sine
function that is pi x p equals to 1 to infinity
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1 minus x square upon p square will be sin
pi x, and this we can prove .
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So, what I am having it means x into this
term will be sin pi x upon pi . Now, this
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I am having p equals to 1 to infinity 1 minus
x square upon p square. Now, this will be
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cancel out. Now, see this x p equals to 1
to infinity 1 minus x square p square. So,
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this value I can put from here.
So, it will become sin pi x upon pi. And from
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here I can write gamma x gamma 1 minus x equals
to pi upon sin pi x. And this formula is called
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reflection formula or complement formula . Also
in some references it will be written as pi
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cosec pi x which is oblivious 1 upon sin pi
x will become cosec pi x . So, this is the
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proof of complement formula for gamma function
. And here we are using the relation between
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gamma x and Euler constant which I have just
introduce you in the beginning of this particular
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lecture.
Now, take few values of x and see what we
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are getting using this formula. So, if I take
x equals to half, then by the complement formula
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it will be gamma x that is gamma half gamma
1 minus x, so 1 minus 1 by 2 again will become
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gamma 1 by 2 equals to pi upon sin pi into
1 by 2 because x is 1 by 2 . So, it means
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pi upon sin pi by 2 and sin pi by is 1. So,
it equals to pi . So, here gamma half square
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equals to pi. So, from here I am getting directly
gamma half equals to root pi, which is the
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result which we have dive in previous classes
also.
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Now if I take x equals to 1 by 3 , so x is
1 by 3 . So, let us see what will be the value
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or what lesson we are getting using the complement
formula. So, it will become gamma 1 by 3 that
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gamma x into gamma 1 by x, so 1 minus 1 by
3 equals to pi upon sin pi upon three . So,
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this becomes gamma 1 by 3 into gamma 2 by
3 and that is equals to sin pi by 3 is root
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3 by 2. So, it will become 2 pi over root
3.
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Similarly if I take x equals to 1 by 4 then
I can get the value of gamma 1 by 4 into gamma
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three by four because it will be gamma 1 by
4 1 minus 1 by 4 equals to pi into sin pi
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by 4 . So, from here I will be having gamma
1 by 4 gamma 3 by 4 and you know that sin
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pi by 4 each 1 by root 2 . So, it will become
root 2 into pi . So, in this where we can
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get the value of the gamma functions operating
on the complement pair, pair of two numbers
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that is x and 1 minus x; only thing we need
to take care that x as well as 1 minus x what
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should be positive ok.
So, our result is given by Legendre in 1809
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that is called duplication formula. So, it
is given as gamma x into gamma x plus 1 by
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2 and product of these two will be root pi
upon 2 raise to power 2 x minus 1 gamma 2
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x . So, let us try to prove this result also
.
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So, result is given as gamma x gamma x plus
half, and these I need to prove root pi over
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2 raise to power 2 x minus 1 into gamma 2
x as I told you this formula is called duplication
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formula and the proof will be something like
this. We know that gamma x or basically I
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will start with the definition of beta function
. So, I know that beta of x y equals to gamma
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x gamma y upon gamma x plus y . So, this is
the relation between beta and gamma function
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and you know that beat x y can be written
as two times 0 to pi by 2 sin theta raise
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to power 2 x minus 1 cos theta raise to power
2 y minus 1 d theta. So, this is the famous
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identity of beta function.
Now, if we put x equals to y then this formula
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I can write as gamma x into gamma x upon gamma
x plus x equals to 2 0 to pi by 2 sin theta
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raise to power 2 x minus 1 cos theta raise
to power 2 x minus 1 d theta; or this I can
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write as 2 over 2 raise to power 2 x minus
1 . So, what I did I have multiplied and divide
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this particular right hand side by 2 raise
to power 2 x minus 1 . So, it is coming in
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denominator, whatever multiplied in numerator
will be inside .
00:16:02.750 --> 00:16:10.750
So, it will become two sin theta into cos
theta raise to power 2 x minus 1 and you know
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that two sin theta cos theta equals to sin
2 theta . So, it will be 0 to pi by 2 sin
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2 theta raise to power 2 x minus 1 d theta
. Now, take 2 theta equals to 1 by 2 minus
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phi then this will give you 2 times d theta
equals to minus d phi. So, what I will be
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having gamma x whole square upon gamma two
x equals to put it here. So, 2 times d theta
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I will write at minus d phi. So, it will be
1 upon 2 into 2 raise to power x minus 1 sin
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sorry take it pi by 2 minus phi. So, sin pi
by 2 minus phi, and then minus d phi what
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will be the limit, when theta is 0, phi will
become pi by 2, when theta is pi by 2, so
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2 times pi by 2 will be pi. So, phi will become
minus pi by 2 .
00:18:08.660 --> 00:18:26.050
So, from here I can write gamma x square upon
gamma 2 x I am taking this term also in left
00:18:26.050 --> 00:18:34.030
hand side . So, it will become 2 raise to
power 2 x minus 1 into equals to so this is
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minus here. So, I can inter change the limit
ok. And then sin pi by 2 minus phi will become
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cos phi and cos is an even functions. So,
I can write limit from 0 to pi by 2 by taking
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2 outside. So, I can write it 2 times 0 to
pi by 2 cos phi d phi . So, this is my relation
00:19:07.830 --> 00:19:20.460
one .
Now, we will start from this relation only
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. Here take y equals to half because I need
some time like gamma x plus half . So, if
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I take y equals to half from the denominator,
I will get x plus half. So, it will become
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gamma x into gamma half upon gamma x plus
half equals to 2 times 0 to pi by 2. And here
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I am making I am writing because you know
beta is a symmetric function. So, beta x y
00:20:04.010 --> 00:20:11.299
with equals to beta y x, so I can inter change
the role of x and y here , it will not affect
00:20:11.299 --> 00:20:19.050
these particular relation .
So, from here I can write, so this I am taking
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y and this I am taking x x. So, when y is
half sin theta 1 minus 1 0, so this term will
00:20:32.929 --> 00:20:44.870
become 1 and this term will be as such . So,
cos theta raise to power 2 x minus 1 and then
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d theta or I can put theta equals to phi.
So, if I take theta equals to phi, then this
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will become cos phi d phi. So, from here I
will be having this I am writing relation
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second. So, now, you notice that the right
hand side of relation one and relation two
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are same . So, I can compare the left hand
side .
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So, when I comparing left hand side of one
and second, so that is
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on comparing one and two, we get that is gamma
x into gamma half upon gamma x plus half,
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this equals to left hand side of one that
is gamma x square into 2 raise to power 2
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x minus 1 upon gamma 2 x . Now, I need to
prove this relation. So, this is coming from
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here I will take gamma x plus half here. So,
gamma x will be cancel this. And then this
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gamma half will become root pi, I will be
having the gamma x into gamma x plus half
00:22:36.880 --> 00:22:44.610
because this I have taken here this I have
taken in the other side. So, it will come
00:22:44.610 --> 00:22:59.080
in denominator gamma half is here. So, this
will become root pi into gamma 2 x and that
00:22:59.080 --> 00:23:06.700
is the duplication formula which we need to
prove. So, this is the proof of duplication
00:23:06.700 --> 00:23:11.929
formula.
And here what we have done we had use this
00:23:11.929 --> 00:23:19.200
particular definition of beta function; also
we have used the relation between beta function
00:23:19.200 --> 00:23:26.020
and gamma function. The proof sketch means
I have done the complete proof , but if you
00:23:26.020 --> 00:23:33.650
want to remember, it is easy. In first step
in this definition put x equals to y; then
00:23:33.650 --> 00:23:40.980
you need to make one more substitution that
is 2 theta equals to pi by two minus phi that
00:23:40.980 --> 00:23:49.110
is quite important . Then you will get a relation.
Again you start from this formula take y equals
00:23:49.110 --> 00:23:57.070
to half and you got another formula . If you
compare these two formulas, you will get the
00:23:57.070 --> 00:24:02.580
duplication formula .
So, again from the duplication formula, if
00:24:02.580 --> 00:24:15.790
I put x equals to half in this formula what
I will get gamma half into gamma half plus
00:24:15.790 --> 00:24:29.820
half that is these two terms I have written
equals to 1 upon 2 raise to power 1 minus
00:24:29.820 --> 00:24:39.390
1 because x is half. So, 2 x will become 1
into gamma 1, and only thing I have left it
00:24:39.390 --> 00:24:58.520
out root pi ok. So, from here this will be
gamma half, gamma 1 will be 1 equals to again
00:24:58.520 --> 00:25:06.220
we got our famous relation that gamma half
equals to root pi and this valid that is the
00:25:06.220 --> 00:25:10.950
relation given by the duplication formula
ok.
00:25:10.950 --> 00:25:21.570
So, in this lecture, we started with the definition
of Euler's constant, then we learn a relation
00:25:21.570 --> 00:25:28.610
between Euler's constant and gamma function.
After that we have seen two important formulas
00:25:28.610 --> 00:25:36.549
related to gamma function, one is called complement
formula and the other one is called duplication
00:25:36.549 --> 00:25:44.530
formula. We have learned how to prove these
two formulas, and what properties we need
00:25:44.530 --> 00:25:53.820
to use to obtain the proof of these formulas.
And then we have seen in both of the cases
00:25:53.820 --> 00:26:02.880
that gamma half equals to root pi . In the
next lecture, we will take few more properties
00:26:02.880 --> 00:26:11.299
of beta and gamma functions, we will take
a few important examples also related to these
00:26:11.299 --> 00:26:14.669
two functions. So, I will end this lecture
here.
00:26:14.669 --> 00:26:16.179
Thank you very much.