okay last time we did the Fourier series and then we came up with a formula for this this and that
and they are all in terms of integrals right and if you guys haven't seen the video please you guys
go check that out well today we are going to make this complex and to do so look at we have cosine
and sine and we did talk about the complex definition of cosine and sine in a past and
if you guys haven't seen this video be sure you guys go check that out as well I will have the
links in the description for your convenience so what we are gonna do is we will just cost out an
X this is my fate huh I would just put that right here in here and now we'll replace this with the
cosine and we will do the simplest thing for sine and they will continue from there and in the yet
it's going to be really beautiful guess what love it and by the way if you guys haven't subscribed
be sure to guess - so thank you guys so much my goal is to hit 200,000 subscribers by the end
of this year and thank you guys for your support we can't do it together okay enough talking now
let's see let's just write down the change that we are not going to touch so this right here a cero
stage and this right here it's the sum as n goes from 1 to infinity and here we have a and then for
this once again we just put an X and X therefore we are going to have pregnancies here we have e
to the I times and X and we are going to add e to the negative I and an X and then divided by 2 so
this is pretty much the complex version of cos alpha and X now for the second part right here
yes look at this is also a summation from 1 to infinity you can put her getting 1 summation pi
it's actually better if you write it down again so we add the sum as n goes from 1 to infinity here
we have B add and to get the sine of an X well that do use this of course however this time I
don't like to be on the bottom so I'm going to just multiply the top and bottom by I to fix the
situation and notice I times I just negative one so here you have negative one so altogether you
are just saying you're multiplying by negative I with the top pretty much so right here I'm
going to write this down right here for you guys this is negative I multiplied with this you have
negative i e to the I theta which is an X so we put on the an X right here and then negative I
times negative becomes positive I and we will have the e to the negative I X and the denominator is
still just a 2 because we took up this on the top earlier already so we departed by 2 like that so
this is cool now what can we do next well here we have e to the iron X that was e to the iron X this
per has e to the negative ion X and this per has e to the negative ion X welcome to just collect the
similar terms right so that's C I'm going to once again write this down because we are now going to
touch this yet so a 0 and then here I will just say we add the sum as n goes from 1 to infinity
and perhaps let's collect all the terms that has e to the i NX first so that's my target I will put
it down right here e to the iron X well from this part we know when you multiply am with this am
will be right here right so we will have a N and yes it's over 2 and in fact right here it's also
over 2 so we can just put together and to care the iron into the iron X from this part it's actually
be n times negative I therefore you will have to have the minus I and then the BN like this and
this right here is like a coefficient for this e to the iron X so that's what we have now we do the
similar thing for the into the negative ion X and as I said we're going to write down the another
summation by still uncle's from 1 to infinity so let's see this time my target is to get the
e to the negative I and X and from this part is a n divided by two so we will have a n and then
yes just put it over to like that from this part we will have BN times I and then divided by 2 we
have 2 divided by 2 right here already so we'll just say we add I and then multiplied by B I like
that cool huh now what can we do next well as we can see here we have e to the i NX and this right
here is e to the negative ion X in fact I can make this into e to the iron XS well because here we
have a summation all we need to do is rename the index and to do that let me just write this down
right here for you guys I will just put all this right here so all we are going to do is we will
change all the ends right here to negative urn so just kind of change it and you'll see right
here for this part it's going to be e and imagine if you put a negative end right here for this add
negative times that negative will give you parsnips all together you get I and X here
which is very nice and you still have to do the same for the rest of the else right here so you
put a negative end here and here so right here you get a stop negative end and then you add I times B
sub negative N and then divide it by 2 well you also have to do the same right here originally
and goes from past the one to pass the infinity now you're just negating the ends in other word
you are going from negative 1 to negative infinity however we always write down a smaller number here
right so in other word we all say and goes from negative infinity to negative 1 and this right
here is so much better because this and that are the same now and in fact we can put this and that
into one summation and this is how first look at this right here which is just like a coefficient
for that right and we can rename this and we used a and B already so we'll just call this
to be C and from this we are talking about C 1 C 2 C 3 and so on because n goes from plastic want
to pass infinity well here we have this number H 0 why don't we just go ahead and call this to be
C 0 sure right especially if you imagine you have e to the I times 0 X yes you can adjust McAleese
to be C 0 and if you come here yes we can also call this right here to PC and not to be confused
that this DN and that's the end even though they look different the idea is that the end right here
goes from past they want to pass infinity but the end right here goes from negative infinity
to negative 1 so right here you're talking about C+ was he passed he 2 and so on this right here
you're talking about C negative 1 and C negative 2 and so on see positive one can be different and
C negative 1 right that's the idea you're just calling this to be the sequence in fact see
parsley 1 and C negative 1 they are the complex conjugate why because here you have the minus
here you have a plus so see past the N and C off the negative wire and you know there are complex
conjugate anyway now we can just write this down into one summation and here we go we will have the
sum as n going from negative infinity including 0 and then up to positive infinity so let me just
put this down right here and for the coefficients it's just all these so just put on stage n right
here and we multiply this by e to the i-x like that well this is it ladies and gentlemen this
is how the complex version of the Fourier series looks like so let me just write down F of X on the
demand side right here for you guys well we are not quite done yet because we shall still figure
formula for CN right otherwise this is just the look we cannot do any computations now this is
how we are going to figure out this yet it's very similar to the situations that we are trying to
figure out a NPN first let me just write this down what we'll do first is that we will multiply both
sides by e to the negative I M X and you might be wondering why do we do the negative is so that we
can almost end up with the complex conjugate and what we used M instead it's because similar reason
than the situation from the previous previous video to figure a n we multiply everything by
cosine M X and you will see we actually end up with a complex account you get rows from right
here right but okay we multiply e to the I M X on both sides and we are of course integrate M and we
will integrate this from negative PI to PI just I found the previous video and don't forget to put
on with the X right here otherwise people will complain well assuming this comfor just the sum
of the integrals is that integral to sums right so we can just say if you integrate this we can
just put the inter coincide and that would be just a constant multiple so we'll just do cat integral
right here negative and negative PI to pass the pie and you have that DX right here so we will
focus on this integral right so look at this part because this is to be done and you are trying to
figure out this right here this right here you have two possibilities I will write this down
this is equal to either you have zero if and and M are different all right and in fact I have another
video that works out the you know the computation for this if any mmm are different you end up with
zero and the only time they will end up with now serial number is two pi when n is equal to n and
of course when nm and my equal here we had I I and X and this is e to the negative I own X that's why
it's the complex conjugate so that's very cool and here's the deal from here you only get one term
then the two pi times C M because that's the time when n is equal to M right so you can just perhaps
write this down right here just put on this right here so all this yeah all this of this is equal
to 2 pi times C M right well here I use em yeah I used em but we are using n so we can just change
the mmm two ends and with e by 2 pi on both sides so let me just write this down for you guys C n
okay it's equal to put a 2 pi here you get 1 over 2 pi and you would integrate this from negative PI
to positive PI and you have the function of course f of X times e to the negative I and X because
once again you put the and back DX so yes you start to work on integral to get the coefficients
but this right here now you may like it a lot more than the sine cosine once you may like it
more than the real ones because I know that's what you guys like anyway so this right here
is said who took me a while to record so please give me a like and leave a comment down below and
let me know if you guys have any questions for any of this anyway as always that's it