so welcome to the lecture series on mathematical methods and the applications so now we will
discuss fourier half range series uh we have already discussed in last lecture what how
we can find out fourier series expansion of even and odd functions that we have already
seen in the last lecture now half range series fourier half range series let us see half
range series plays an important role in solving several engineering and physical applications
where it is required to get the fourier series expansion of a function in an interval say
zero to l so its a half range basically uh in fourier series basically in expansion of
fourier series we need interval minus l to l ok and function must be periodic that is
the condition for the fourier series of the expansion f x
now here the fourier series expansion of a function is in the interval zero to l where
l is an half of the period ok now how can we find out fourier series expansion of such
a function and these problems are important in several engineering and physical applications
ok now it is possible to extend the function f x to the other half that is minus l to zero
of minus l to plus l so that f x is either an even or an odd function basically what
we do we know the nature of the function from zero to l and that is given to us ok and depending
upon the depending upon the basically nature of the function we find out we either extend
the function uh considering function as an even function or we either extend the function
taking function as an odd function ok suppose suppose suppose function is given to you
from zero to l suppose this function f function is given to us from zero to l ok this function
is given to you from zero to l it is a half half period ok
now you can find out the even extension of this function you can extend this function
as an even function from minus l to plus l ok and then find out the fourier series expansion
of this function ok so you can take this function as an even function whatever function is given
to you from zero to l you can find out the even extension of this function uh taking
function as an even function on the other way you can also find the suppose this is
given from zero to l you can find an odd extension of the function ok you can you can extend
this function like this from zero to minus l minus l to plus l and in the same way over
here ok in the same way over here so basically we need a fourier series expansion of a function
which is given in the half range from zero to l so we can either to find the fourier
series expansion of such a function we can either extend the function taking function
as either as an odd function or taking function either as an even function ok
so in the first case it is called as even periodic extension of f x while in the second
case it is called an odd periodic extension of the function f x if we do an even periodic
extension of f x which is the first one which is the first case it is the even extension
of the function f x then f x is an even function in minus l to plus l of course it is an even
function then f x has fourier cosine series that we already know that if function is an
even function ok we extend the function as an even periodic function then f x will be
an even function and in even function we only have a cosine series cosine terms in the fourier
series expansion now if we do an odd periodic extension of
the f x then f x is an odd function in minus l to plus l and therefore f x has fourier
sine series ok suppose we extend the function as an odd periodic function so this f x will
be an odd function and we know that if function is an odd function then it contains only sine
terms the fourier series expansion of that function will contain only sine terms ok if
a function is defined on a half interval from zero to l then we can obtain a fourier cosine
or fourier sine series expansion by suitable periodic extensions depending on the problem
ok if you want a fourier cosine series expansion that means we take an even periodic function
or we or we take function as an even function and if we want fourier sine series that means
we will take as a odd extension odd periodic extension of the function ok that depends
on the problem so suppose you want fourier sine series ok
so we already know this result that if you want fourier sine series now function is given
in zero to l ok and you want an odd extension of the function odd extension means only sine
terms ok so what will be the fourier series of that that function the fourier series of
that function will be nothing but uh it will be given by summation n from one to infinity
b n sin n pi x by l and where b n is nothing but two upon l integral zero to l f x sin
n pi x by l into d x ok so this is an odd extension of the function ok uh now if we
want fourier cosine series expansion or the function defined on a half range from zero
to l so that will contain only cosine terms so that will be given by that will be given
by a naught by two plus summation a n cos n pi x by l n varying from one to infinity
and here a naught will be nothing but two upon l integral integral zero to l f x d x
and a n will be given by two upon l integral zero to l f x cos n pi x by l into d x so
this will be the even extension of the function function is given from zero to l and we defined
an even extension of a function as a cosine terms in the form of cosine terms
now let us try to solve this problem uh obtain cosine and sine series for f x equal to x
in the interval from zero to pi and hence show this result ok so first of all function
is given function is x f x equal to x so a very simple function f x equal to x interval
is zero to zero to pi now suppose you want cosine series expansion of this function cosine
series expansion means we want to extend this function as an even function we want to extend
this function as an even function so this function is like this it is from zero to pi
it is uh it is x ok f x equal to x from zero to pi and we want to extend the function as
an even periodic function so from minus pi also if it is an even function it must be
symmetrical about x axis so we will draw this function like this we extend this function
like this and similarly here also like this and similarly here also like this and similarly
here also this will be an even extension of this function or uh or now we can write this
function in the cosine series form how uh how f x will be nothing but a naught by two
plus summation a n cos n pi x by pi ok n varying from one to infinity and what will be a naught
now it is two upon pi integral zero to pi so here pi pi cancels out uh zero to pi f
x d x f x is nothing but uh f x is nothing but x d x so it is nothing but two upon pi
it is x square upon two from zero to pi so it is nothing but two upon pi pi square by
two minus zero so it is nothing but pi so a naught is nothing but pi ok and a n what
will be a n a n will be nothing but two upon pi integral zero to pi f x cos n x d x so
it is equal to two upon pi integral zero to pi now what is f x f x is x so x cos n x d
x so now we will apply integration by parts
to simplify this expression so this will be nothing but two upon pi so first as it is
integral of second sin n x upon n minus derivative of this and integration of this so this is
nothing but minus cos n x upon n square zero to pi ok so that is further equal to two upon
pi now the first term is zero when x is pi or x is zero ok now second term is one upon
n square times when x is pi it is minus one k to the power n and when x is zero it is
one so this would be this term ok so this a n is equal to now when n is odd when n is
odd so this is nothing but when n is odd it is minus one minus one minus two minus two
minus into two minus four upon pi into n square so this will be nothing but this term when
n is odd and when n is even it is nothing but zero because minus k to the power even
is one one minus one is zero ok so n is defined like this
so what will be the cosine series of this function f x so a naught is pi ok a naught
is pi so it will be nothing but pi by two plus summation n from one to infinity it is
two upon n square pi minus one k to the power n minus one into cos n x which is further
equal to pi by two plus two upon pi can come out and when n is odd it is minus four upon
it is i mean minus two will come out so when when n is odd it is minus four upon pi into
one by n square that is cos one upon one square plus cos three x upon three square plus cos
five x into five upon five square and so on and when n is even it is zero ok so this will
be the expansion of x f x is x ok so this is an even extension of the function x
so x is defined in only zero to pi we extend this function take assuming the function as
an even function ok and that will contain only cosine terms these are the cosine series
expansion of function f x equal to x now to reduce this expression one is one upon one
square one upon three square one upon five square and so on with all plus signs ok we
can simply substitute x equal to zero so put x equal to zero both sides so when you put
x equal to zero it is pi by two plus or minus minus four upon pi it is one upon one square
plus one upon three square plus one upon five square and so on so that implies one upon
one square plus one upon three square plus one upon five square is simply equal to pi
square by eight ok so that we have proved this result also
now to obtain fourier sine series of this function f x equal to x so how you will obtain
that the first part is over of this problem now we want to obtain fourier sine series
of this function that means we have to extend the function f x equal to x as an odd function
ok then only we can obtain fourier uh sine series that is the uh series containing sine
terms only ok so how can we do that we have the function f x equal to x here ok we have
the function f x equal to x here is y equal to x ok and we have to extend this function
as an odd function to obtain the fourier sine series of this function f x equal to x so
it must be like this and assuming function as periodic function we can extend this function
as an odd function ok so what will be the fourier sine series of this function it will
be summation n from one to infinity b n sin n x ok where b n is nothing but two upon pi
integral zero to pi f x sin n x d x ok so this is nothing but is equal to two upon pi
integral zero to pi f x is x sin n x d x so this is nothing but two upon pi integral of
this is x into minus cos n x upon n minus derivative of this and integration of this
will be minus sin n x upon n square from zero to pi
so after integration by parts we obtain this thing so this is nothing but it is two upon
pi when you put x equal to pi here or zero here this term is zero ok and here when you
put x equal to zero it is zero so only term is left here so negative will come here so
it is pi into minus one k power n upon n ok so pi pi cancel out it is nothing but minus
one k to the power n plus n plus one upon n so this will be two into ok two into minus
two n k to the power n plus one upon n so this will be b n so what will be f x f x will
be nothing but summation n from one to infinity b n b n is two into minus one k power n plus
one upon n into sin n x ok so this is nothing but two when n is one it is sin x upon one
minus sin two x upon two plus sin three x upon three and so on f x is x so this will
be the odd extension of the sine series expression of the function f x in terms of sine so this
will be the final solution of this problem ok
so now let us try to solve one more problem of this function which is we find as an this
function is also given from zero to one ok and that is in half range and we want function
to expand as an fourier sine series that means we want an odd extension of this function
ok so how we can expand this as an odd function so again for the odd expansion that is we
want the fourier sine series of this function eta f x will be given by this expression so
what will be b n b n will be nothing but two upon l now here l is one ok in this problem
l is one zero to one f x and here also it is sin pi x upon one because l is one ok f
x sin n pi x upon one into d x now we have to split the function from zero
to half it is one by four minus x from zero to half ok from zero to half ok sorry zero
to half it is it is one by four minus x into sin n pi x d x plus half to one it is x minus
three by four sin n pi x into d x and whole two times ok so we can simplify this to find
out the sin series expansion of this function ok so how we can simplify simplification is
simple we have to apply by parts only two times it is nothing but one by four minus
x into minus of cos of n pi x upon upon n pi ok and minus derivative of this which is
nothing but minus one and integration of this which is nothing but minus sin n pi x upon
n square pi square ok and that whole from zero to one by two ok i simply apply integration
by parts ok first as it is integration of second minus derivative of this and integration
of this which is nothing but this again for the second part plus two times again two is
here ok for second part it is x minus three by four that you will easily solve by your
own you can easily apply integration by parts and simplify to find the fourier series expansion
of this function ok so it is nothing but minus cos n pi x upon n pi and it is minus plus
one into minus sin n pi x upon n square pi square and it is varying from half to one
so so after simplification what we will get two times ok so when x is half one by four
minus half is minus one by four minus minus plus so it is one by four into cos n pi remains
same into cos n pi by two ok when x is half here it is nothing but minus of one upon n
square pi square and when n is half it is sin n pi by two ok when x is zero it is zero
when x is zero it it is one one upon one by four so it is minus minus plus one by four
into n pi ok now for this term ok when x is one it is one
minus three by four that is one by four ok negative is here so minus one by four and
pi into minus one k power n and when x is one here it is zero now when x is half it
is half minus three by four half minus three by four is minus one by four ok it is minus
one by four minus minus plus one by four ok and minus is again here so it is minus and
then it is cos n pi by two upon n pi and when n is half here it is nothing but minus minus
plus and minus again so it is sin n pi by two upon n square pi square so you can simplify
this entire expression ok to find out the uh value of b n ok and that b n you can substitute
here to find out the fourier series expansion fourier series expansion of this function
f x ok now on the in the same lines uh we can also
find out the fourier series expansion of x sin x as a cosine series cosine series means
you want an even extension of this function so that means for this function x x sin x
you will simply find a naught and a n and a naught will be given by a naught will be
given by so for this problem a naught will be given by uh two upon pi integral zero to
pi x sin x d x and a n will be given by two upon pi integral zero to pi it will be x sin
x into into cos n x d x ok so this is how we can find a naught and a n in this expression
and finally for some x we can also find out the series expansion of this function ok so
uh thank you uh for this lecture in the next lecture we will study an identity and some
problems based on that and also complex form of fourier series expansion ok
thank you