so welcome to the lecture series on mathematical
methods and their applications so in the last
lecture we have seen that what fourier series
are and how we can expand the function f x
in the ah form of cos and sine terms ok so
now we will see that ah when the fourier series
is convergent so let us see first convergence
of fourier series for continuous functions
if a periodic function f x with period two
l is continuous in interval minus l to plus
l and has continuous first and second derivatives
at each point in that interval then the fourier
series which we have already discussed a naught
by two plus summation a n cos n pi x by l
plus summation b n sin n pi x by l of f x
is convergent ok so whats the proof so proof
we can see
now in this series in this series a naught
by 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 in this
ah series a naught we already know is given
by one upon l minus one minus l to l integral
minus l to 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 so this this we already said
in the last lecture that these are the this
is given by eulers formula a naught a n and
b n now if we if we see a n ok a n is nothing
but what a n is given by one by l minus integral
minus l to plus l f x cos n pi x by l into
d x now you integrate it by parts ok you take
it first function this second function this
integrate by parts eta it is nothing but equals
to one by l now first as it is integration
of second it is nothing but sin n pi x by
l upon n pi by l from minus l to plus l minus
integral derivative of first and integral
of second now when x is l eta it is sin n
pi which is zero and when x is minus l which
is again sin minus n pi which is again zero
this first term is zero eta it is nothing
but minus it is l l cancels out it is minus
one upon n pi and it is integral minus l to
plus l f dash sin n pi x by l into d x
so this we ah obtain now again you apply by
parts so it is first and second so it is minus
one by n pi now first as it is integration
of second integral second is nothing but minus
of cos n pi x upon l divided by n pi by l
and it is from minus l to plus l minus integration
derivative of first integration of second
that is plus cos cos n pi x by l upon n pi
by l into d x now it is minus one upon n pi
ok now since f dash is continuous and f is
periodic so f dash l is same as f dash minus
l ok since f x is continuous and it is periodic
so f dash l will be same as f dash minus l
and at x equal to l and x equal to minus l
this value is same which is minus one k to
the power n so the first term will be zero
ok the plus it is l upon n pi integral minus
l to plus l f double dash x into cos n pi
x by l into d x ok this is this is zero because
because function is periodic that is f of
x plus two l is same as f of x and derivative
is continuous so f dash of x plus two l is
also equal to f dash x for all x and when
you substitute x as minus l so f dash l will
be same as f dash minus l ok so these two
term are same f dash l is same as f dash minus
l and upper and lower limit the value of this
is same this term is zero and we left with
this expression
now we have to show that anyhow that this
is convergent so to show that this is convergent
again since f double dash x is continuous
and in the interval minus l to l so we can
say that mod of f double dash x will be less
than m for some m because it is continuous
f double dash is continuous in the statement
as given in the statement itself that f double
dash is continuous so f double there will
exist some m such that f double dash will
be less than capital m so if we apply this
over here eta mod of a n and mod of cos of
n is always less than equal to one mod of
cos of n x upon l is always less than equals
to one so mod of a n will be less than strictly
less than one upon n pi integral minus l to
plus l we can take mod inside mod of f double
dash x into cos n pi x by l into d x which
is equal to one upon n pi ah it is less than
or ah it is less than mod of this it is less
than mod of this not the entire expression
now when you take this expression inside when
you take this inside and applying these inequalities
so this is nothing but less than minus l to
plus l m into one d x which is nothing but
ok
now this this into this is also there so it
is l upon sorry it is l upon n square pi square
l upon n square pi square so it is nothing
but two m l upon n square pi square ok so
a n will be less than this now similarly we
can obtain that mod of b n will be less than
two similarly it will be less than this this
we can similarly obtain so ah so what about
this series so what about this series a naught
by 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 this mod
will be less than mod a naught upon two plus
now this mod mod of a n in each for each n
mod of cos n x is less than equal to one mod
of b n which is less than this and mod of
sin n x by l which is less than equal to one
so what we will obtain finally this is ah
four m by l upon pi square will come outside
ah two from here two from here this is four
m l upon pi square and it is summation one
by n square that is one by one square plus
one by two square plus one by three square
and so on ok because when you take the mod
and when mod come inside in each in each term
then using these inequalities we will obtain
this expression because it is mod of a n a
naught by two plus mod of a n is less than
two m l by n square pi square and mod of this
is less than equal to one again this is less
than equal to one we know that summation one
by n to power p is convergent when p is greater
than one eta here p is two it is convergent
as this series is convergent and this series
is convergent means inside the series also
convergent ok
this series is convergent means because because
f x is always less than equal to mod of f
x ok and if f x is convergent then this f
x is also convergent ok so this eta this series
will be convergent so hence we can say that
if function is a periodic and first and second
order partial first and second order derivatives
are continuous then this series is convergent
this series of f x is convergent so here is
the this is the same proof which i have discussed
here ok
now piecewise continuous function we have
already discussed this thing in laplace transforms
also that piecewise continuous means that
you divide interval a comma b into sub into
finite number of subintervals and if in each
subinterval function is continuous we say
the function is piecewise continuous ok the
same same thing we will define here suppose
the function is said to be piecewise continuous
in minus l to plus l if function is defined
and continuous for all l except at finite
number of points ok if a function at a point
say x naught is not continuous then left hand
limit and right hand limit at that point exists
and finite ok that means discontinuity are
of jump type ok now at end point of interval
that is at minus l and plus l the if it is
minus l the right hand limit if it is plus
l then the left hand limit exist and they
are finite so then we say the function is
if these three properties hold then we say
that function is piecewise continuous in minus
l to plus l or in other words we can same
revise the same definition that to divide
interval into finite number of subintervals
and in each subinterval function is continuous
and if there is a discontinuity that must
be of jump type ok
now if a function is piecewise continuous
what is the convergence of condition for fourier
series a function f x can be expressed as
fourier series this in the interval minus
l to plus l where a naught a n b n are constants
provided function is periodic with period
two l single valued and finite function is
piecewise continuous in minus l to plus l
piecewise continuous we already discussed
what it is and f x has left hand derivative
and right hand derivative at each point in
that interval so this is if function is piecewise
continuous if these properties holds then
it is converges to ah this expression then
function f x will converge to this expression
ok can be expressed in this way ok now we
have a theorem that if f x and f dash x that
is the first derivative be piecewise continuous
function on the interval minus l to plus l
then the fourier series converges fourier
series f x converges to f x at the point of
continuity if there is if it is a continuous
point then f x will converge to suppose x
naught is the point of conti x naught is the
point in minus l to plus l where it is continuous
then f x will converge to f x naught ok but
if it is a point of discontinuity say x naught
belongs to minus l to plus l then fourier
series will converge to the average of left
hand limit and right hand limit at that point
will converge to the half of f x naught plus
plus f x naught minus
now at both point both the end points of the
interval minus l to plus l if you take minus
what is the value of the function at minus
l what is the value of function at l then
the fourier series will converge to half of
f of minus l plus that is the right hand deriv
right hand limit at x equal to minus l and
f l minus that is the left hand limit at x
equal to l the average of these two will be
the value of the function at the end points
of the interval minus l to plus l ok i am
not discussing the proofs here ok proof of
this theorem so the proof can be obtained
so these results we will used while solving
some problems
now let us try this problem find the fourier
series expansion of the function this and
deduce this expression ok so we will solve
the problem in the usual way as we did in
earlier to express to find out the fourier
series expansion of this function so function
is periodic ok we assume here function is
periodic of period two pi function is defined
from minus pi to plus pi but function is periodic
that is period is two pi ok so let us find
a naught first eta what will be a naught a
naught will be one by pi minus l to minus
pi to plus pi f x d x by the definition so
it is one by pi now from minus pi to zero
it is nothing but minus pi and from zero to
pi it is nothing but x so using this it is
nothing but one by pi and it is minus pi zero
minus minus plus pi and it is plus pi square
by two eta that will be nothing but minus
pi by two so that is a naught now if you find
a n a n is nothing but one by pi integral
minus pi to plus pi f x cos n x d x because
here l is pi here l is pi so instead of cos
n pi x by l if you take l equal to pi eta
it will be nothing but cos n x d x ok now
it can be expressed as written as one upon
pi integral minus pi to pi it is minus pi
to zero it is nothing but minus pi it is nothing
but minus pi cos n x d x plus zero to pi it
is nothing but x x cos n x d x
now it is one by pi minus pi will come out
it is sin n x upon n from minus pi to zero
plus it is x sin n x upon n minus one into
integral of this that is minus minus plus
cos n x upon n square from zero to pi so it
is nothing but one by pi it is minus pi now
it is zero when x is zero it is pi it is zero
again when x is minus pi this term is zero
plus again this term when x is pi and zero
it is zero eta it is zero now here it is one
by n square times it is minus one k to the
power n minus one this is nothing but one
upon pi square n minus one k to the power
n minus one so this is this will be a n ok
now what is b n b n will be nothing but again
one by pi i break it here minus pi to zero
it is minus pi sin n x d x plus zero to pi
it is nothing but x sin n x d x now when we
take it it is one by pi it is minus pi it
is minus cos n x upon n you integrate it ok
from minus pi to pi minus pi to zero plus
you apply by parts here eta we will obtain
it is x minus cos n x upon n minus one into
minus sin n x upon n square so it is zero
to pi so when you simplify this it is one
by pi minus pi minus minus plus cos zero is
one minus it is minus one k to the power n
one by n will come out ok
now here plus this term is zero when x is
pi and zero because of sin n x and this term
is zero when x is zero only one term remaining
that is minus of pi by n into minus one k
to the power n ok now this is minus this is
minus and now what we will obtain from here
it is it is nothing but it is minus and here
it is also minus it is nothing but ah it is
ok let us see it once again it is minus pi
it is minus cos n x upon n minus pi to zero
eta when zero it is one when minus pi it is
minus one k to the power n it is plus ok now
x is the integration of this is minus cos
n x upon n ok ah minus derivative of this
and integral of this so that is minus sin
n x upon n square from zero to pi now when
it is zero eta it is one it is this and when
it is pi it is minus pi upon n minus one k
to the power n ok so it is nothing but ah
something like one upon n will come out pi
will cancel and it is one minus it is one
it is minus of this and minus of this will
add up eta it is two times so it will be minus
two times minus one k to the power n ok
so this will come here so ah now we have a
naught a n and b n in our hand so ah what
will be the fourier series now a naught is
minus pi by two so fourier series of f x will
be nothing but a naught is minus pi by two
eta a naught by two plus summation n from
one to infinity ah it is one upon n square
pi minus one k to the power n minus one plus
into into cos n x plus summation n from one
to infinity one upon n into one minus two
into minus one k to the power n into sin n
x so this is the fourier series of f x which
we will obtain from here after simplifying
now this this will converge to this ok because
function is we have seen that the conditions
of ah convergence of fourier series is piecewise
continuous holds here ok now we want to deduce
this expression so it is one by one square
now n square is only here so that means this
term is here but we dont want this term so
we can put x equal to zero ok we put x equal
to zero so it is nothing but f zero is equals
to minus pi by four plus summation n from
one to infinity one upon n square pi and minus
one to power n minus one and cos zero is one
ok now what what will be f zero because it
is not continuous at x equal to zero left
hand limit is minus pi and right hand limit
is zero ok because what is f zero minus f
zero minus is nothing but minus pi and f zero
plus is nothing but zero by the definition
of function so what will be f zero is the
average of two one by two minus pi plus zero
which is nothing but minus pi by two
so this is minus pi by two is equals to minus
pi by four plus one upon pi come outside when
n is one eta it is minus one minus one minus
two minus two upon one square now when n is
two it is zero when n is three it is again
minus two upon three square minus two upon
five square and so on so when you simplify
this it is minus pi by two plus pi by four
is equals to minus two upon pi one upon one
square plus one upon three square plus one
upon five square and so on so it is minus
pi by two minus pi by four which is minus
two by pi into this expression 
so hence the value of this expression is nothing
but pi square by eight pi square by eight
so hence we obtain the required result so
in this way if the function is discontinuous
at a point what i want to illustrate in this
example i want to restate that if you want
to find out the value of function at a point
where it is discontinuous then to find out
the value of that point is nothing but the
average of the left hand limit at that point
and right hand limit at that point ok you
find out left hand limit and right hand limit
and the average of these two will be the value
of function at that point eta all other things
are as usual we did earlier but the important
point to note here is what how can you find
the value of function at the point of discontinuity
ok if function is discontinuous
now it is the same same type of example eta
we can easily solve this example also again
suppose you want to find out a naught so a
naught is nothing but now it is from minus
pi to plus pi it is one by pi integral minus
pi to plus pi f x d x so it is one by pi now
you can break it from minus pi to minus pi
by two it is it is minus one from minus pi
by two pi by two it is zero and from pi by
two to pi it is it is one so you can simplify
this it is a very simple thing you can very
easily simplify it is minus pi by two ah minus
plus ok negative will come outside actually
negative of minus pi by two minus minus plus
pi and it is zero it is pi minus pi by two
and which is nothing but when you simplify
eta it is zero ok now what will be a n again
a n you can compute a n will be one upon pi
again you will break the interval into three
ok from minus pi by from minus pi to minus
pi by two it is minus one into cos n x from
minus pi by two to pi by two it is zero no
need to calculate that this is zero plus integral
pi by two to pi it is one into cos n x d x
now it is nothing but one upon pi ah negative
of sin n x upon n from minus pi to minus pi
by two and plus sin n x upon n again from
pi by two to pi so it is nothing but one upon
pi minus of sin minus one by n comes out and
it is sin ok it is negative of negative of
sin n pi by two and sin pi is zero
again here sin n pi is zero and it is nothing
but minus of one by n sin n pi by two which
is nothing but again it is zero so it is zero
ok so that you can easily simplify similarly
when you find b n what will be b n it is one
upon pi again minus pi to plus pi minus pi
by two it is minus of sin n x d x and plus
pi by two to pi it is sin n x d x so when
you simplify it is one upon pi it is minus
minus plus cos n x upon n from minus pi to
minus pi by two and it is minus of cos n x
upon n from pi by two to pi so this value
is nothing but one upon pi it is cos n pi
by two upon n one by n one by n comes out
and minus minus one k to the power n minus
again minus one k to the power n and minus
cos n pi by two is it ok one upon n pi will
come out and it is sin cos n pi by two minus
of cos n pi minus one k to the power n negative
is here so it is nothing but minus one k to
the power minus minus one k to the power n
and it is minus of cos n pi by two ok eta
it is ok minus is outside so minus is outside
eta minus will be here something like this
ok minus minus is outside eta it is nothing
but one upon n pi so it is two times two will
come outside two times cos n pi by two minus
minus one k to the power n ok
so only only we have in this expression a
naught a n are zero so for this function f
x we have only the sine terms on the right
hand side because a naught and a n are zero
ok eta we have on the right hand side only
sine terms so what will be f x f x will be
nothing but summation n from one to infinity
ah two upon pi can come outside it is one
upon n and it is cos n pi by two minus minus
one k to the power n into sin n x so this
will be the expression of this function so
f at pi by two will be nothing but one by
two f of pi by two minus plus f of pi by two
plus so it is nothing but one by two f of
pi by two minus is zero and f of pi by two
plus is one eta it is nothing but one by two
so we substitute x equal pi by two both the
sides on the left hand side we have one by
two and the on the right hand side we have
two upon pi one minus one by three plus one
by five and so on so value of this is nothing
but one by two ok
so in this way if you want to find out fourier
series we first have to find out a naught
a n and b n and if we want a series expression
value of the series expression and we have
to substitute x as a point where function
is discontinuous then you first find f x naught
minus f x naught plus and the average of these
two will give the value of function at that
point so
thank you very much for this lecture
