so welcome to the lecture series on the mathematical methods and its applications so we have discussed
in the last two lectures basically we have discussed fourier series and its convergence
that how we can expand a function f x periodic function f x in fact uh in the series form
and the series of trigonometric functions of sine and cosine ok that we have seen and
when a series when a x or function is continuous and having first and second derivatives are
continuous then uh function f x is convergent to that series of sine and cosine not even
continuity if we have piecewise continuous function piecewise periodic continuous function
in fact then also that f x will converge to the infinite series of sine and cosine uh
if left and right hand derivatives exist at each point that we have already seen
now we will see fourier series of even and odd functions ok we already know what even
function is even function means f of minus x equals to f x that function is called an
even function and if f of minus x is minus of f x that function is called as odd function
we already know that mod x x square e k to the power minus x square cos x etcetera are
all even function because if we replace x by minus x in these functions then the functions
remain as it is that is f of minus x is same as f x then so therefore these functions as
even functions however if you see x cube x sin x or minus cos x then these functions
odd functions because if we replace x by minus x then the function will be equal to minus
of f x that means f of minus x is minus of f x therefore these functions are odd functions
so we have functions which are neither even nor odd also like like x minus x square which
is neither even nor odd ok now we already know this result that if f is an even function
then integration minus l to plus l f x d x is nothing but two times zero to l f x d x
that we already know and if f is an odd function then minus l to l f x d x is simply zero this
is by the properties of definite integral we already know this now come to fourier series
representation ok now we already know that fourier series representation of any f x is
given by a naught upon two plus summation n from one to infinity a n cos n pi x by l
plus summation n from one to infinity b n sin n pi x by l
so this is when interval is minus l to plus l and the function is periodic with period
two l ok now here a naught is given by uh that is one by l integration minus l to plus
l f x d x a n is given by one by l integral minus l to plus l f x cos n pi x by l into
d x and b n is given by one by l integral minus l to plus l f x sin n pi x by l into
d x now suppose function which is given to us whose fourier expansion is to find out
is an even function ok so if it is even function then minus l to plus l f x d x will be nothing
but two times by the property of definite integral
and the cos cos is an even function it is cosine ok c o s it is cosine now cosine is
an even function and f x is even function so even into even is an even function ok so
minus l to plus l it will again gives two upon l integral zero to l f x into cos n pi
x by l into d x now if we see this function f x is an even
function and sine is an odd function even into odd is odd so this value will be zero
so hence if we have an even function so we have only cosine terms in the fourier series
expansion ok because b n is zero b n is zero means no sine terms only cosine terms will
be there ok so if function is an even function we have the fourier series representation
of that function will contain only cosine terms ok now suppose now suppose we want to
expand an odd periodic function ok if it is odd suppose it is odd ok now suppose function
is odd if it is odd minus l to plus l this will be zero and f x is odd cos is even odd
into even is again odd it will be zero and this is odd sine is also odd odd into odd
is even this is an even function so when it is even this will be nothing but two upon
l times zero to l f x sin n pi x by l that means that if we have an odd function then
there will be no cosine term in that series expansion fourier series expansion
we will be having only sine terms because there is only b n a n a naught and n all are
zero ok so that is how one can find even and odd extension odd i mean even and odd fourier
series expansion for even and odd functions ok now let us solve two examples on this suppose
this function is given to us ok now what is this function what is this function this function
is f x equal to minus x plus one when minus pi less than equals to x less than equal to
zero ok and x plus one when zero less than equals to x less than equals to pi of course
function is periodic ok with period two pi so we have to first check whether function
is even and odd ok so to check whether function is even and odd we will find f of minus x
if it is equal to f x so it will be an even function and if it is equal to minus of f
x then we say it is an odd function ok so you find f of minus x so you replace x
by minus x so it is x plus one when minus x less than equal to zero greater than equal
to minus pi and minus x plus one when minus x less than equal to pi greater than equal
to zero so this will be nothing but is equals to x plus one when x is less than greater
than equal to zero less than equals to pi and it is minus x plus one when x is greater
than equals to minus pi and less than equal to zero so from minus pi to zero it is minus
x plus one which is same as this function and from zero to pi it is x plus one which
is same which is same x plus one over here that means it is nothing but f of x so hence
it's an even function ok so one can also see graphically because if it is an even function
then it is symmetrical about it is symmetrical about y axis so you can just plot the function
and you can see the symmetry if it is symmetrical about y axis then you can simply say it is
an even function ok now you have to find the fourier series expansion
of this function now fourier series representation can be find out because its an even function
which we have already seen so it will contain only cosine terms ok no sine terms that is
b n will be zero that we have already seen so what will be a naught this we have already
seen it is two upon pi zero to pi f x d x so that will be nothing but two upon pi two
upon l that is zero to pi ok l is pi here so it is two upon pi x plus one d x and this
is nothing but two upon pi it is x square by two plus x from zero to pi and which is
nothing but zero to pi it is pi square by two plus pi which is equals to pi plus two
because two two cancels out it is pi and it is two yeah it is pi plus two ok
now if you compute a n a n is nothing but two upon pi integral zero to pi f x cos n
x d x it is again equal to two upon pi integral zero to pi what is f x f x is x plus one into
cos n x d x and when you integrate it it is two upon pi it is first as it is integral
of second is sin n x upon n we will apply integration by parts and derivative of this
is one integration of this is minus cos n x upon n square from zero to pi ok and this
is nothing but two upon pi now it is zero when x is pi or zero
now here it is minus minus plus one by n square it is minus one k power n minus one ok so
this will be a n and this will be a naught b n is of course zero because it is an even
function so what will be the fourier sine fourier representation of this function so
a naught is pi plus two so f x will be nothing but pi plus two a naught by two ok a naught
by two plus summation n varying from one to infinity a n a n is two upon pi n square minus
one k power n minus one into cos n x ok this term come here now so this will be fourier
representation of this function now we have determined the value of this series one upon
one square plus one upon three square plus one upon five square and so on so what the
series is basically it is pi by two plus one plus two upon pi
now when you open this summation when n is one it is minus two it is minus two cos x
upon one square now when n is two minus one k to the power two is one one minus one is
zero ok now when n is three it is again minus two minus two cos three x upon three square
again when n is four it is zero minus two cos five x upon five square and so on so it
only contain odd terms i mean one three five like this ok so and we have we have this series
ok now we want value of this series so we want all these terms to be one so substitute
x equal to zero so put x equal to zero so it is f zero will be equal to pi plus two
plus one minus four upon pi times one upon one square plus one upon three square plus
one upon five square and so on now what is f zero is the function continuous
at zero yes it is continuous at zero because from both the ends value is one we can easily
check so the value of this will be one only ok if it is not continuous then we will find
in the same way like half of f zero plus plus f zero minus ok if it is not continuous at
zero here it is continuous at zero so it will be one only ok it is pi by two plus one minus
four upon pi times this expression so what this value is one one cancel out this is minus
pi by two is equals to minus four by pi times this expression so hence this value is nothing
but pi square by eight this value is from here we can say that this value is nothing
but pi square by eight so hence we got this value ok because it's an even function calculation
become easy otherwise we have to split the integral from minus pi to zero and zero to
pi ok now see this problem now f x is minus k when
between x lying between minus pi to zero and k when it is lying between zero to pi first
let us check whether it is even function odd function or neither so we have to verify this
first so let us see it is minus k when x lying between minus pi to zero and it is k when
x lying between zero to pi and it is a periodic function with period two pi ok so first we
will find first we will see whether it is even function odd function so again we will
find f of minus x if it is equal to f x it is an even function if it is equal to f of
minus of f x it means it is an odd function so you replace x by minus x so it will be
minus k when minus x lying between zero and minus pi and it is k when minus x lying pi
and zero so it is nothing but minus k when x is greater than zero less than pi and it
is k when x is less than minus pi greater than minus pi and less than zero
so if we see this and this function from minus pi to zero it is k from minus pi to zero it
is minus k from zero to pi it is minus k and from zero to pi it is k that is this is nothing
but minus of f x so hence this function is an odd function now in odd function so it
will contain only sine terms that is a naught and a n are zero we already know this so this
will only contain fourier series expansion of this function it will only contain sine
terms so how you find b n so only b n will exist so b n will be nothing but two upon
pi integral zero to pi k sin x sin n x d x f x f x is k ok and it is nothing but two
upon pi k is outside integral of sine is minus cos n x upon n from zero to pi minus will
come outside two k upon pi and it is nothing but when you integrate this what you will
get find the apply the limits minus one power n minus one so which is which is equals to
this is equal to when n is odd when n is odd it is minus one minus one minus two minus
two into minus two is four that is four k upon pi ok so n is also there ok n is also
there it is n it will come here ok because n is also there now when n is odd
when n is even it is minus plus one minus one zero so when n is even it is zero ok so
in this way we can define b n now what will be the fourier series of this function then
fourier series of this function will be nothing but summation b n with b n is b n is this
term it is minus two k upon pi n minus minus one k power n minus one sin n x and when we
open this when n is odd it is and when n is even it is zero so you can substitute values
of n when n is when n is one so it is four k upon four k upon pi will come outside it
is sin one sin x upon one ok plus sin three x upon three plus sin five x upon five and
so on because when n is even it is zero we have the existence of b n only when n is odd
ok so we have this series expansion now we want
to compute this expression summation n from one to infinity minus one k power n plus one
two n minus one equal to pi by four this we have to show ok so because it is minus one
k power n plus one that means we have alternate negative positive sign in the series of this
so to have alternate sign of plus minus sign over here so put x equals to pi by two so
it will be f pi by two will be equals to four k upon pi sin pi by two is one one by one
sin three pi by two is minus one so minus one by three sin five pi by two is plus one
and so on and what is sin pi by two f pi by two f pi by two comes from here it is k so
this expression will be nothing but k is equals to four k upon pi one minus one by three plus
one by five and so on so hence this implies one minus one by three plus one by five minus
one by seven and so on will be nothing but pi by four so this you have to derive yeah
ok so it is pi by four it comes pi by four so hence we have proved the result ok
so this is simple problem fourier series expansion four minus x square is an even function one
can easily see and it has a period four so we have only cosine terms in the series so
how can we how can we find the fourier series expansion of this function you using the same
technique same expressions so f x is four minus x square x lying between minus two to
plus two and it is a periodic function with period four ok so again we know that it will
contain only cosine terms because it is an even function so what will be a naught a naught
will be two upon two because period is two zero to two f x d x so this will be nothing
but four x minus x cube by three from zero to two so this value we can compute it is
eight minus eight by three that is nothing but eight into two by three that is sixteen
by three ok ok now a n a n will be nothing but two by two
integral zero to two four minus x square cos n pi x by two into d x because l is two ok
so we can easily integrate this there is no problem in this it is four minus x square
integration of this will be sin n pi x by two upon n pi by two minus derivative of this
and integration of this will be minus cos n pi x by two upon n square pi square by four
plus derivative of this and integration of this will be nothing but minus sin n pi x
by two upon n cube pi cube by eight and whole multiplied by zero to two limit from zero
to two now when you take when you take limits two
here it is zero and when x is zero it is zero so it is zero now here also when x is two
it is n pi it is zero when it is zero it will be zero so only this term will there ok so
it is minus minus minus it is minus two comes outside when x is two so it four into it is
four upon n square pi square ok it will also come outside four upon this so it is this
will come outside two will also outside so it is two into cos n pi and at zero it is
zero so it is nothing but minus sixteen upon n square pi square into minus one k power
n ok it is minus sixteen upon this so so fourier series expansion of this function will be
nothing but sixteen upon three into two a naught by two plus summation n from one to
infinity sixteen upon n square pi square minus one k to the power n plus one cos n x so this
should be the fourier series this expression ok so
thank you