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Periodic function of period is defined by2 < x < 0 0 < x < 2f(x) = {3Sketch the function and obtain its Fourier series.(10 marks)QUESTION ON NEXT PAGE6 ...

Question

Periodic function of period is defined by2 < x < 0 0 < x < 2f(x) = {3Sketch the function and obtain its Fourier series.(10 marks)QUESTION ON NEXT PAGE6 (a) Determine the Fourier series for the function f (x) = x2 in the range T<x < T The function has period ZT. marks)Bv lettingT; use your answer tO pant (a) to show that=T2/6(4 marks)

periodic function of period is defined by 2 < x < 0 0 < x < 2 f(x) = {3 Sketch the function and obtain its Fourier series. (10 marks) QUESTION ON NEXT PAGE 6 (a) Determine the Fourier series for the function f (x) = x2 in the range T<x < T The function has period ZT. marks) Bv letting T; use your answer tO pant (a) to show that =T2/6 (4 marks)



Answers

1. Find a power series representation for the function. (Give your power series representation centered at x = 0.)

f(x)= 2+x / 1-x

2. Determine the interval of convergence. (Enter your answer using interval notation.)

Problem, you're given the graph of a sine function and clearly it has some transformations happen. So the first thing we want to do is find the amplitude, the midline in the period. And I actually think finding the midline is going to be the easiest to find first. So remember when sine is equal to your typical sign curve would start the origin an increase from here and then it was, it would go through its cycle until it gets back to the origin. So we just need to determine what this uh cycle would look like. Well, if you know this, it looks like our midline is going to be at this line, Y equals negative two. Because if we were going to actually to go through and trace this starting at um are y intercept? We would go up, we go back through our midline and then we would come up and hit our midline again. So now we have determined are middling, it's wide equals negative two. Next, we're going to find our ample too well remember the amplitude represents the height of the curve, meaning it's the distance from our midline to the max or to the men. Well, if I find our max that happens when y is equal to two and our midline is down here at negative two. So this distance we represent our amplitude which is four units, so our amplitude is four. Alright, the next thing we need to do is we need to find the period. Remember the period represents how long it takes to go through one complete cycle. So if we were again to start the Y intercept, how long does it take to go through one cycle? Well, it goes along, goes down and comes back up and it goes from zero till five, which means our period would be equal to five because it takes five units to go through one full cycle. Okay well now do we know the midline to amplitude in the period? We can start to write an equation for this function. So remember our general form is why equals A. Times the sine of B times X minus C. And then re add t. Well remember A represents the amplitude. Which we already found two before now to get B. We actually used the periods of the graph. So remember the period is equal to two pi divided by B. Because your typical sine function goes through a period of two pi. Well remember we just found a period to be five. So we'll substitute that in for the period. And now we just need to solve for B. So I multiply both sides by B. So these peace cancel. So we get five B. Is equal to two pi. So the sulfur be. We just divide both sides of our equation by five. So now we've found that B. Is equal to two pi divided by five. All right, C. Represents the horizontal shift. We'll remember our typical graph starts at the origin. But all we did in this case was go down. We didn't move left or right. So R. C. Value is zero. Indeed represents the vertical shift. And as I just mentioned, we went down to units. So our devalue will be negative too. So now we just have to substitute these into our formula. So when we do that we'll have Y equals four times the sine of B. Which is two pi over five times X minus C. Which is zero. And then we have to add our devalue which is negative tip. So now we can clean this up a little bit. So we'll have Y. Equals four times to sign. Oh well if we distribute to pi over five, that just leaves us with two pi over five times X. And then we have plus negative two, which is really just minus two. So now we have found an equation to represent the ground.

All right so you asked for the 2nd 3 X. And you wanted on the T. 80 for now. Unfortunately I'm on a computer and I can't do that. But I pulled out my favorite calculator at the desk house and I can type in seeking of thanks. Now we actually wanted three apes and so I can go and put that in. So state the range of this function was we can see it goes comes down and then goes back up. If I zoom out it's going to keep going. So the range on this would be from negative infinity Up until -1. That's this negative one And from 1 to Infinity. So if I bring some stuff over here so we got the range head from negative infinity up to one And it's gonna hit -1. See how we got negative one right there and then we got the range from one to infinity. Um That's one way to do it now. Uh the other way that we could do it is we can take one over the sign of three X. Seeking. It's not the sign I forgot co sign. And So seconds one over coastline. You can definitely type this into your t. and then graph it and you would end up with the same thing. See how when I toggle one off and on through the same graph. Um and if we're just sketching one one cycle of it. Well I think you it's um no go from one assassin tote until you hit the next one. So we would need to be going from she was this assassin. Tote Hi six. I would say halfway in between there we go from Piper six. Um, that if I want to Mexico range six negative of it, that that's then their ascent owes like it's on the left side of this one. It's on the right side of that one. And we would want that until we got to well between that C46. Uh, actually course I have. That's invites. We got one side and then the other side Egypt sketch that um, you could get this red part in this red part. That's what I didn't tell you.

All right, So for the first function that were asked to uh consider here the to seek X. You just get this written down here. So for the first one we have to seek X. Now, seek X. C can't is the same thing as one over coast of X. So if we're looking at two ckx and that's the same thing as two, divided by CAs affects so to determine the period of this, the period is actually going to be the exact same as co sign. So the period it's going to be two pi ah and to get an idea of what it's going to look like, we can think about in the context of Cossacks, we'd be starting out at one. So we'd start out at one. If we're multiplying by two, that would bring it up to starting out at two, then we'd be going downwards, coming back up going downwards and coming back up when we're considering seek of X, then we're in a coast, we have these values of zero instead for seek. Um We would have something going off towards infinity. So what I'll do is draw seek in red here. Yeah. For to seek affects. We'd start out at two here because we have two divided by one. Then we'll go off ah not quite to scale there, but we'd go off towards infinity positively on either side of that initial spike. Then on the other side of that infinity. Since we're looking at now coast is taking on negative values. Actually I need to make an adjustment there. Since coast is taking on negative values, we'd be coming up from a negative infinity, they'd meet at that peak and then he'll shoot off towards negative infinity again and similar on the other side here. So the red the plot of Sikh affects next up. And what I'll do is um let's see, oh delete the plot that I had here and add a new grid on uh for the next one we had we are considering cosi can't of X. Specifically cosi cosi can't of X over four which is the same thing as one over sine of X over four. So sign typically has a period. So write this down here sine of X. It's period is two pi. But since we have this division by four, that is effectively um spreading out the values to get signed to have an argument of two pi rather than having XB two pi, we need to have XB eight pi. So pussy can't of X over four is going to have a period of eight pi because we have this division by four where the period is increased times for essentially um What and we can do we can consider a similar process for figuring out what cosi can't is going to look like If we think about sign, we start at zero. Uh there's no scaling. So what I'm actually going to do here is make that a little bit bigger. So sign in second year, you ever sign, going to start at zero, Go up to one, passed down through zero, come back up, hit zero again and do the same thing on the opposite side, something like that. So we'll have for cozy can't Yeah, it should be going off towards infinity towards the center, going off towards um So it's going to be going off towards negative infinity. Um On the left hand side of the origin, it's going to be going off toward positive infinity on the right hand side of the origin. Because we have over on this side we have positive values of X. Or positive values of sign effects rather. But they're getting smaller and smaller and smaller as the sine of X approaches zero. Cozy can is going to be approaching infinity. So our cozy can't is going to look like something like. So first of all should be one when sign of axes one we should be negative one when sign of access negative one, then it shoots off towards positive infinity and it shoots off towards positive infinity. And over here shooting off towards negative infinity. Shooting off towards negative infinity. On this side here, a similar deal And over here similar deal gives you an idea overall of what it's going to look like. Yeah, of course, I'm not specifying on this little sketch. Um the exact values of X just because um this is just to get an idea of the overall sort of shape, the qualitative sense of the function. Mhm. Uh Next up we have cozy can't add that to this text box here we have cozy can't now it was a little bit difficult to parse what was written but I'm assuming it's intended to be cozy can't of X divided by two plus pi divided by four. So one of the significant things here is that the only thing that's going to be affecting our period is what is multiplying or dividing our X. So the pi over four doesn't matter for the period. What matters is the fact that we have that division by two. So just like in part B. When we had cosi can't of X over four, that division by four multiplied the period by four. Here. Our period going to be the regular period of cozy can't. Which is the same, essentially the same as the period of sign. So it would be to pie than being multiplied by two because that division by two in the argument. So the period is going to be for pie. And note what that pi by four does. The pi by four is going to shift the plot. Mhm. Uh Specifically it's going to be essentially shifting sorry one second. Was that pi by four or pie by two? I'm just going to double check here. Yes sorry that should be high by two. Non pi by four. The high by two going to shift the plot specifically. It's essentially going to be shifting it slightly to, you can view it as shifting it to the left. Um That's the sort of proper way. But because we are looking at these periodic functions there's not actually a huge difference between shifting to the left or shifting to the right but sorry undo that. Uh huh. Okay. Apparently I'm not allowed to bring back that grid. Okay, so make it bigger. Um What we'll have is something that um where we would have um sign at X equals zero would be zero. Since we're since we have the inclusion of this plus pi by two, that shift that is going to completely change the phase of our sign. In fact one moment here, a sign of we can if we think about in black here, I'll draw sine of X than in red. Just for the explanation here, I'll draw sign of X plus pi by two. So for a sign of exit starts at zero, goes up to one, goes back down, it's negative one and vice versa on the opposite side. Yes, a little bit. Uh Too far there, whatever. Doesn't matter too much. Yeah. When we do that shift by pi by two. Yeah. Keep in mind Pie by two would be appearing here. Pie by two or X equals pi by two. Is when sign is going to reach its first it's going that's when it hits positive one for the first time and I lost my grid again. But I'll just work past that. So if we're shifting this to the left by pi by two we're going to get something that looks like this which should look familiar. If we shift sign by pi by two, ignore the fact that they end up lining up there. That is just because I'm not a great artist when trying to draw with a mouse. Uh When we shift signed by pi by two we end up getting something that is identical to co sign. So it actually have that cosi can't of why did that disappear? Okay. Um We should get that cozy can't of X over two plus high by two. It's actually the same thing as C pants of X over two. Because we would have the plus. Bye bye to shift makes the sign into the same thing as coast. So if we're looking at 1/1 over sine plus bye bye to that would be the same thing as looking at one over coast which is R. C. Can't. Of um X. Over two. So for our cozy can't of X over two plus pi by two. The period in this case, as I said, the period was going to be for pie. Uh And it's going to look overall. Something like this same sort of general shape as we saw for the first one. Just rather than having these peaks at two, those peaks are going to be that should be to down there that should be negative one. The peaks are going to be at positive one and negative one respectively. For the positive and negative peaks. So the last one for us to consider here and I'll write this down perhaps that is not what I meant to do. The last one that we want to consider here. It was Turkey just two plus two times C can't of two X. So here we're we've already looked at um you know, the behavior of C. Can't and so on that two out front is just going to be giving a vertical shift upwards. Uh The multiplying by two here is going to be basically just scaling everything making each number of larger. Um Then the other thing that we need to consider you for the question, we're asked, what is the period? So because we have this two X. Um Where in the previous question or in the previous sections parts of the question. Um We had division on the X. Which led to the period of being multiplied in this case. Since we have the multiplication on the X. The period is going to be divided. So if we have to X then our period, the regular two pi period is going to be reduced to high. Then let me just clear away everything here. He just needs to consider what a quick sketch of our function is going to look like. So we already know C can't has that shape of going up to positive and negative infinity or positive infinity on either side of the center. And then we have these explosions towards negative infinity on either side of it. Uh For this case, since we have that multiplication by or that addition of two out front, our our smallest value that we can hit is going to be or rather, let me rephrase this. So we would have um when the excuse me, when the co sign would take on a negative value, we would expect it to have a minimum value. You know if we just had this to see can't uh it's minimum we would expect to be negative too because we would have the co sign that we're dividing by is hitting a value of negative one, then we're multiplying by two. Then with this edition of two out front that's going to be increased up to zero. Similarly our largest part or are basically our minimum rather for the positive peak. Normally we would expect that to be one. If it was just a regular C. Can't. Then the fact that we're multiplying by two is going to make that into. We would expect it to be too and then we're shifting everything up by two so we'd have it go up to four. So you have for is there? So we have that is our sort of minimum point on the positive peak. It spreads outwards then for the negative peak. So our we have the points where we're going off towards um infinity for a negative sort of Kerala like shapes. You have this one over here peeking out at um y of zero and similar over here.

The given function is ffx is equal to two plus x divided by one minus x square. We can write is a stew plus X divided by one minus x into one plus X. We can force split into partial fractions. It's very easy to find the interval of convergence. So we split into partial fractions two plus x divided by one minus x into one. Plus X is equal to a by one minus X. Is B. Bye. One plus X. For NPR constants. So when you read the L c M, you get two plus X is equal to into one. Plus X has been to one minus x afghans in the denominators. So then you get two plus X is equal to x minus B X x into a minus B and plus eight plus B. Right now comparing corruptions in minus B should be one and a plus B should be too. So we can solve these two simultaneous equations. So you get a three by two bees, one by two. So that means you are function is two plus X divided by one minus X squared is equal to three by two divided by one minus X. There's one by two divided by one plus X. Because I replaced with three by two and gave it one by two. Now this is nothing but three by two into one minus X. Whole power negative one and this is half into one plus X. Whole power negative one. Now when does the series convergence? If you remember A plus A. R plus they are square plus A. R. Q. This is a famous geometric progression. The infinity is a by one minus R. And it's well known that this is value. Don't even models of rs less than one. What we do is we replace it with one or with X. In this formerly what you get is when he's one are actually at one place express extra square and so on. To infinity is one by one minus X. And that is nothing but one minus X. Hold power minus one. And this series converges only when memorials of our is less than one that is more years of excess, less than one implies minus one is less than X is less than one. And how about the others use one best example of ar minus one. Again in this formula replace is one, R is negative X. So then we get one minus X plus X squared minus execute plus on infinity is one by one plus X and it is one plus X, whole power minus one. And it converges when more minus X is less than one but more minus x. Sms marks. So that boils down to minus one, less than x, less than one. So that means in both cases the interval of convergence is this, so the intersection of both is also the same. So finally, the function our cities expansion of the function to plus X, divided by one minus X square, converges in the interval minus 1 to 1, so your ex should belong to open minus one to open one.


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