Here's my favorite calculus infinite series, the sum of sin(n)/n as n goes
from 1 to infinity. We will be using Fourier Series to evaluate this sum.
okay let's do some fo fun here we're going to calculate a sum as n goes from 1 to infinity
of sin(n)/n again but this time we'll be using Fourier series for this however I didn't come with
the following solution I want to thank Boolean for this thank you so much as you can see we're happy
for Fourier series and Fourier coefficients and the board ready the first step is which was are
to figure out what the function to use right well we are all we are going to do is we are going to
the f of X equal to X that's all in fact we let f of X equal to X you can come up with a few really
interesting series and we're going to show you guys true of course that's one of them and your
circuits the alternative version of that anyway when f of X is equal to X we're going to put it
here and we got these things right when we have just X of course X is odd when we integrate this
over negative PI to PI of course we'll just end up with 0 likewise f of X right here is X
and that's odd this one has even at times even it's an odd function so again we get 0 here we
wish that this is also 0 PI unfortunately it's not this is r times r is even right an even function
and just how to integrate this and i think you guys can trust me that I can do my integrals I
believe you guys came to yours too so them just a guess the answer right here yes 2 over n times
negative 1 to the N minus 1 power so be sure just integrate this by parts and plugging embrace it
and do all that stuff right now take a look right here f of X is equal to all that but a 0 is 0 and
a n is also 0 so that makes this particle 0 that means we just had to declare f of X equal to this
part and of course f of X is just X so let's go ahead and write that down we have X being equal
to the series when n goes from 1 to infinity and we have the BN which is just that so let's just go
ahead and put that out 2 over N times negative 1 to the -1 power and then we multiply this by
sine of X so that's pretty much it well if you want to clean this part up first because you
do get this and that maybe we can divide both sides by two so why not let's go ahead and do
that in that case we get the sum as n goes from 1 to infinity divide both sides by 2 we get perhaps
that's put everything on top now negative 1 to the minus 1 times sine of an x over N this
right here it's nice to equal to x over 2 these are yes x over 2 and the interval from negative
PI to PI well well if you do X equal to 1 let me just put this down right here for you guys
just for fun if you let X equal to 1 which is of course on this interval put X here and put
X here and you have a really nice series and in fact you end up with the alternating version of
that guy so when we do that let me just write it down we will end up with the sum as n goes
from 1 to infinity this and a stays which is negative 1 to the minus 1 power to 1 here of
course we get sine of just over this series is nicely equal to 1 over 2 namely 1/2 ah very very
nice isn't it very very nice well how can we go from the alternating version to the original
don't take the absolute value doesn't work here is the deal we had to think about how to work with
the X right here so let's go ahead and continue let me tell you guys that we are going to let X
equal to PI minus y and you will see when we put this right here and right here things will work
out very nicely let's continue put this right here down first off and we get PI minus y over
2 and this is equal to of course this side which is just the sum as n goes from one to infinity and
we have negative 1 to the n minus ones power over N and we have this namely sine of X and the X is
this now so I'll just put up pi minus y right here and that's very nice even better of course we can
distribute this and we'll just get n PI minus and why and don't forget we have sine of this minus
that of course we have the angle subtraction formula so let's use that so we get the sum as
n goes from 1 to infinity and this is negative 1 to the n minus 1 power over n and this right
here it's going to be sine of the first which is n pi times cosine of the second which is and Y
and for the subtraction this is also subtraction and we do sine of the second which is and Y times
cosine of the first cosine of pipe like this well well as you can see this is very nice because when
we have sine of n PI keep in mind and this josa pass the whole number this is always going to be
0 so the first part is very nice because this is just gone and if you look at sine of n pi if n is
1 2 3 4 5 so our sin of PI is negative 1 sum of 2 pi is positive 1 so again negative 1 positive
1 negative 1 positive 1 so on this right here we can write yes negative 1 to the the power because
we have n goes from 1 2 3 4 5 so on so that's good and we do have another negative right here to help
myself check this out this is gone already that's continued to write down the following we have the
sum as n goes from 1 to infinity and here we have negative 1 to the N minus 1 power over N
own all we have the following this right here I'll just keep es sign of and Y but this times that of
course this is like negative 1 as well so we can just put them together and we'll read yes negative
1 to the n plus 1 power well well well the beauty is that oh my god you see this guy times this guy
all together they will give us just part of the wuhan right because when you add exponents up you
just get 2 and a negative 1 to the twins power is always positive 1 so oh no this right here right
it's just a sum as n goes from 1 to infinity and we just have sine or and y over like that
and let's see what we do next of course with that end over there PI we let y equals we let y equal
to 1 and instead the jet yes yes even though we shifted a little bit but if you check the Y well
why would be going from 0 to 2 pi you can just check that on your own so that's the deal from
the earth we're going to say as let y equal to 1 let me put on left hand side first name D we'll
just get the sum as n goes from 1 to infinity and we get sine of n times 1 so that's just that over
and and this is equal to put one right here so we get PI minus 1 over 2 oh my cup so cool isn't this
Wow yeah you hate figure so much watching and again thanks to coding for this - sure