Hello, my name is Daniel Friedrich. This is part four of week two of the Fourier series topic of Engineering Mathematics 2A. In this part we're going to continue non-periodic extensions and we're looking at odd extensions for non-periodic functions. After this part, you will be able to calculate odd half range extensions for non-periodic functions. Before this, you should know how to extend not periodic functions into a full range series and to even extension. And you should know the properties of odd and even functions. A brief recap. In the previous two parts, we handled extensions for non-periodic functions. We started with the full range extension, which usually has no symmetry and requires us to calculate both the an and bn coefficients Then in the last part, we looked at even extensions and the sketch on this slide shows an even periodic extension of the square function. And the even extension has the benefit that the Fourier series only has cosine terms. So we only need to calculate the an coefficients. However, in some cases, it might be better to extend the function the odd way. And this is what we're going to handle in this part. The odd extension will only have sine terms in the Fourier series, and you only need to calculate the bn terms. The slides show an odd extension of the square function. The odd extension, and odd functions have symmetry at the origin. So we have the red solid line gives our square function, so t to the power of two, between 0 and 1. We extended in an odd way because we mirrored it at the origin. So we get this blue dashed line here. And this shows us an odd periodic extension. This extension doubles the period. So we're going from minus one to one instead of from 0 to one. And also, because it's odd, we know our Fourier series only  has sine terms. We will need to calculate the bn values. And let's do this in the next slides. Again, analyze the sketch. Our period is now 2 from minus one to one. Our angular frequency is pi, and the extension is odd. In the following, we denote the n times pi as xn. For less typing. Because this is an odd extension of the square function. We have a sine half series. Fourier series of the function capital F of t is only the, some of the sine terms. In the end, we are only interested in our original period. So as previously, we need to restrict the Fourier series to the original domain which was 0 to 1. In this case, we have to calculate our bn coefficients in the usual form. We set up our calculation. And here I already did all the simplifications we did last time. Originally we would have the interval from minus tau to tau. But because we have an odd times an odd function, so we have an even function inside. So we can restrict our integration to 0 to tau, but multiplying the other with two. So this already has all the simplifications which are explained in more detail in the previous part. For the even extension. If you are not sure how we get this, have a look at the previous video. So continue with this. Bn we can write it in this form, so its just this simplification, because our tau is one,  replace it with tau at 1. And this is the integral we need to solve. I'm not sure this is on this slides. If you're interested to solve these yourself or you can look at the example for the full range series, which has similar integral to be solved. And now we can plot this solution for our f of t once we have calculated the bn. This slide shows the sign half-range series truncated after 20 or 100 terms. We can see on the left for 20 terms, our convergence to original red line here is not very good. And even with a 100 terms, sort of oscillations here and with the over and under shots which are typical for discontinuous functions. The problem is, the odd extension has a jump discontinuity, which means our converges only one over n. So for the square function, the even extension is arguably better because even for the simple half range even extension, we get a convergence of one over n squared. Let us finish this part with some words about uniqueness. Previously, we said that the Fourier series is unique. If the Fourier series coefficients are the same, then the functions are the same. And if two functions are the same, then they have the same Fourier series coefficients. Now we have calculated at least three different Fourier series for the square function. Isn't this a contradiction to the uniqueness? That is not a contradiction. Because our periodic function which we're describing with the Fourier series is different for all three cases. The periodic function is the complete extension of our non-periodic function. While the non periodic part is always the same. The way how we extended this function is different for the full range, the sine half range, and the cosine half range. And for the different even extension which we had at the end of the last part. This means different extensions, different Fourier series coefficient. So this is not a contradiction to the Uniqueness Theorem which we handled previously. This concludes this part, as well as a small series of parts about the extension of non-periodic functions. If you have any questions, as always, head over to the discussion board. And in the next part, we're going to handle the Fourier Series of similar functions. Thank you.