
Fourier Series 9

Lecture1.1

Lecture1.2

Lecture1.3

Lecture1.4

Lecture1.5

Lecture1.6

Lecture1.7

Lecture1.8

Lecture1.9

Fourier Series Part 2
This is the second part of the Fourier series course.
For this tutorial, I would suggest switching your viewer to highdefinition. You [can] do so by clicking on this number There’s a number right in this area here [that] has a p next to it switch it [to] either 720 or to 1080 P because there will be some small writing on this tutorial that you will need the higher resolution.
In the previous tutorial we showed that this step wise function here this periodic Step wise function was able [to] be modeled mathematically modeled by a combination of cosine and sine functions We showed [that] the width and also added and added to it was a constant a naught we showed that a naught was equal to 0 at the tail end of the previous tutorial and We also show that the period was equal to [2PI] Because this function goes from negative [Pi] to 0 which is a change of PI this distance is Pi And then it goes Pi again And that’s one full period And so that [full] the distance of the period of one oscillation is [2] Pi for this example and so we set the period equal to 2 Pi For the for this for these generalized equations now there are many forms of the generalized equations that you’ll see out there in different books this is my preference [I’ve] Seen a lot of different ways of writing it down and I prefer this the best This is my personal [preference] if you find something that works better for you. That’s fine, but I like this the best ok so with this Form the way that you do it is you set the p you set p the period of the function equal to 2 L Then L is equal to [1/2] the period of the function whatever the period may be for that example for this example The period is equal to 2 pi so L is equal to [1/2] of that. Which is Pi Ok so now we can just plug into these respective Formulas to find these different coefficients, and then just plug [into] this in the end So a 0 we already found that to be 0 at the tail end of the previous tutorial So we’ve got this done right now. We’ll go to a sub n. So it’s a sub [n] and notice it goes from N Equals 1 to cap in where capital N. Could conceivably go out to infinity in theory But who has that amount of time right? So we’re going to do a couple we’re going to do I think we’re going to seven something like that we’ll see how the time goes okay, so let’s start with a1 so [a1] now we already discussed what all these age should come out to because look they involve cosines, so Cosines are not good at modelling Odd functions because the cosine [function] is an even function, and we don’t use even functions to the model. Odd functions We use odd functions to model odd functions so the sine function is going to be the most useful in modelling this function not the Cosine and so we expect all the cosine terms to go away, and let’s see if that happens, so let’s go through the motions here So 1 over L. So L is Pi so we have 1 over Pi time times the integral from negative Pi to Pi You see because it’s it’s 1 over L and then from negative L to L So negative Pi to Pi because L is equal to Pi times f of x And we’re not putting f of x in yet times the cosine of n, pi x over L Cosine n. Is 1 for this one. We’re doing the first one, so this is 1 times Pi Divided by L. But L is Pi so we see the positive canceling times x like so so we see these going away [and] So this is equal [to] Now we’re going to break this up Because we have to integrate this stepwise function. We have to integrate it Piecewise Because it has this break in at this sharp break in it right here So we’ll integrate from negative part is 0 and then we’ll add the integral from 0 to Pi so this will be 1 [over] Pi [open] Bracket integral from negative Pi to 0 f of x from negative Pi to [0] what is the function the functions negative see? So let’s put [that] in parentheses Negative C times the cosine of x because just 1 times x and the pi’s cancel and This is let me clean this up a little bit D x plus I’m going to come underneath 0 to [Pi] But from 0 to Pi the function is positive C So we put that in I always like to put the function value in [parenthesis] just to remind myself to keep my bookkeeping notation correct, and this is [x] the cosine [of] x Dx close bracket like so okay, so this is equal to 1 over Pi times Open Bracket now factor out this negative C and we have the integral of cosine of x dx from negative Pi to 0 and Then factor out this positive C And we have the integral of cosine x from 0 to pi so let’s take a look at the cosine function and see if we can do this quick by using symmetry from negative Pi To 0 is from here to here [I] drew these tail ends right here. Just because [it] looks more sinusoidal It looks funny if you don’t draw these in there It doesn’t look like a sinusoidal function, so I just drew [that] in there, so there’s more identifiable But the function goes from here to here is the [period] involved and so this part right here is negative and This part is positive so we see that on the interval from negative Pi to 0 we have this negative area in this positive area to Give 0 So this integral here whenever this negative C is factored out is going to give 0 so this whole term here Is 0 Now we’re going to integrate cosine x from 0 to pI well look we start here, and we integrate this way Right so we have a positive area in this negative area and they cancel to give 0 this term is zero So this is zero Plus zero like that this is zero so zero times. This is zero, so this is [equal] to zero so a 1 is Equal to zero [okay]? [I’m] gonna make sure that made it in and it did Okay, next a a [2] well. Here’s the thing Let’s let’s take a look at a 2 we can make it. We can make a generalised claim now Because we know all of a is are going to be equal to [zero] You can a 2 is going to involve the cosine of 2x Right because look this this number here This is the end this becomes a [2] the pi’s cancel again, so it’s going to vote the cosine of 2x Well that means you’re taking two wavelengths two periods if you will and shoving them shoving them in to the same interval right to this interval from negative Pi to Positive Pi and whenever you do that. You’re going to have the same business going on with the area’s canceling and those integrals are going to be equal to zero so the integral of cosine 2x from negative Pi to 0 is 0 and the integral of cosine 2x from 0 to 0 [2] Pi is 0 so and it’s true for [a] 2 it’s true for a 3 a 4 a 5 is true for all of them DotdotDot a sub N all of those are equal to 0 So we’re done with a sub n bang now B Sub in these will not all [be] equal to 0 [okay], [so] that’s why I’ve left as much [a] space as much space as possible So let’s move on to the [B’s] again, I’m going to step out of the way and let’s do b 1 B 1 will be equal to 1 over Pi negative Pi 2 Pi f of x we’re writing it writing it in general first times the sine of N [pi] n pi over L times x [n] is 1 [you] see this we’re dealing with the first one so this is 1 Times Pi divided by Pi because L is Pi Times x Dx so [this] is cancelling bangbang and We see that. This is [just] the sine of x in here So now we’re going to break the integral up and integrate Piecewise as before so [neck] 1 over Pi Open Bracket from negative Pi to 0 and Over that interval the function is equal to negative C and this is times the sine of x Dx like so plus the integral from 0 to Pi C times the sine of x Dx so like so close bracket [ok] so here we go from negative Pi to 0 take a look at the function for negative part is 0 this is the sign [of] this is just a sine of x so from here to here we have a negative a negative a negative area and That number comes out to be 2 so what we have is this we have 1 over Pi We have negative see this thing factored out in front right [here] negative C and that whole integral becomes out to be negative 2 Right because it’s below the [xAxis] Plus and then the integral this C factors out the integral of sine x from 0 to Pi is from 0 to 2 pi is 2 and not 0 to 2 pi it’s 0 going to by so this is 2 So you see these negatives cancelling. So now we have to [see] plus 2 C is for C. We have for C over the pot Okay, so that’s our first [b]. Let’s do b. Okay, so b. – now we’re doing the second one for this one over Pi Integral from and now we’re going to break it up right away. We’re not going to We’re just going to start breaking up right away because we know we’re going to have to anyway Negative L to L. So we’re going from negative Pi to 0 and then we’re adding 0 to pi so let’s go from negative Pi To 0 over that interval the function value is [negative] C times the sine of N pi [over] L. Well in and in for this one, this is the second B so this is 2 pi over pi the pi’s are cancelling pies are always cancelling, so That will that’s going to happen every time for this problem. So it’s going to be 2x Okay plus from 0 to Pi C because the function value is positive C over that interval [times] [the] sine of 2x Dx close bracket okay, this is equal to Okay, [now] if we evaluate this integral and this integral We will achieve and let me check this to make absolutely sure pressure [0] it is it will be 0 Now there’s a way to think about this. I’ll show you how to think about this so what we got to do now We have to break this up okay, so we’re going to draw the we’re going to draw the sine function on this axis, but [you] see that. This is sine 2x 2 sine of x had one The one wavelength in there, but now because the sine of 2x there’s [going] [to] be two wavelengths in there so watch this So it’s this is From right here to right here So now two wavelengths have to fit in there, so it’s going to go like [that] [and] like [that] right then when you integrate from negative part is [zero] this negative C factors out and look Now this is negative [Pi] right here, and this is zero you’ve got an area above and an area below zero, so this term is zero and Look when you integrate from zero to Pi You got a positive area and negative area that are equal magnitude, but opposite inside Zero okay, so I’m going to save some space here let’s equal to Zero b 3 equals 1 over [Pi] [zero] negative Pi to [Zero] Negative C Times the sine of 3x because we have a 3 now pi’s cancel always for this example sine of 3x Dx. [oh] no its notice. I forgot to put the Dx here Don’t make that mistake anywheres else now. I got it there. Yeah, we’re good. [okay] I just forgot [to] put the Dx right there, okay plus from 0 to Pi positive C sine of 3x Dx [close] Bracket like so Okay, so we’re all big boys and girls at this point Right so we can integrate the negative see factors out and then we integrate the sine of 3x now look now There’s not two [wavelets] being packed into that same space. There’s three wave less Okay, so let’s let’s take this thing. Let’s see if we can figure it out by looking so from here to here But this entire interval three wavelengths has to fit into it, so we have to break it up into three parts [so] [we] see this part this part and this part one way of what needs to fit into each of those So let’s go And this needs to come It needs to be turning quicker than that Okay, there it is Okay Here’s the thing though we integrate from and you might think oh look there’s one two three above one two three below zero, but no because your integral integrating Piecewise from negative Pi to zero and then from [zero] to PI so notice when you go from From negative part at zero you have below above below So this one cancel these this one cancels with this one, but this one remains right and that value is so we have 1 over [Pi] Sorry about the loop guys. It’s for a analysis. This is suspension she blows this is Tedious you just got to work out. You know if you got an exam I suggest [practicing] this because you want to be well practiced okay, so for B3 Let’s see here this number here. Well this will be negative C and This number will be negative [2/3] when you do when you carry out the integration, and then we have plus positive C Times positive 2/3 because look what happens to the other side you have [two] above when you go from zero to [Pi] You have this one both of these above and this one below So this one you can cancel with this one But you’re still left with this one above so and the value of that area above right here is 2/3 So [B3] is equal to We have to this– these negatives cancels, so we have 2/3 C plus 2/3 C Which is 2 plus 2 4 c 4 c 4 C over 3 PI, so this is 4 C? over 3 Pi [and] check that and that is correct okay [b] for now. I’m going to jump ahead a little bit You could carry out the integration as before the even beasts the b 2 b 4 [b] 6 [b] [d] 8 and 1 Are 0 okay, so b 4 is it going to be 6 you’re going [to] be 8 Dot dot [dot] you know And let’s just put let’s just put it out the 8 that’s fine 0 we get the iDea all the even B’s are 0 but the the Odd bees are not ok so b 5 And tell you what I’m going to do. Let’s see here. Yeah. I’m gonna have to do some racing Goodbye to these guys so we get the intuitive idea right we get how you know you can You’re squishing wavelengths into the same space So now we’re going to analytical [ok] so b 5 is equal to 1 over Pi and then this will be integral from negative Pi to 0 negative C times the sine of 5x Dx Plus integral from 0 to Pi of C sine of 5x Dx close Bracket This will be equal to [one] over Pi Okay, you see that. We have the negative C over this interval and positive C. The five is here because this is B Five right we did B2 B3. B3. You know we did B2 B3 before and all the evens and then these numbers are Be fought for B5. It will be negative C Times negative [2/5] Plus C times 2/5 like this and this number is forfeit for C over 5 Pi Okay, here. We go b6 0 b 7 [ok] one more b. 7 is equal to [one] over Pi and We’re going to run out of space so today. I want to do I’m not going to write down the integration I’m go write down the answer bummers although write down the numbers in between so and you know it’s good these calculators because you can just do all these integrations in here and you avoid errors this way, it’s a good idea to to have some symbolic Manipulating computer at your disposal, so [this] will be negative C Times so you courage you to work this problem, because you know it’s correct this problem did not get up online if I do not check it first and and I had other people check it, so This problem is definitely correct [to] sevens if it’s not correct we redo them for for C over seven Pi Starting to see [you] starting [to] see the pattern Look be three. Let’s see be [one] [for] C over PI. This is 1 times Pi isn’t it? then look b 3 is 4 C over 3 PI so look it’s basically this is equal to 1/3 times 4 C over Pi [and] Then this is equal to 1/5 times 4 C over [Pi] you see that