Advanced Engineering Mathematics Prof. Jitendra Kumar Department of Mathematics Indian Institute of Technology, Kharagpur

Save this PDF as:

Size: px
Start display at page:

Transcription

1 Advanced Engineering Mathematics Prof. Jitendra Kumar Department of Mathematics Indian Institute of Technology, Kharagpur Lecture No. # 28 Fourier Series (Contd.) Welcome back to the lecture on Fourier series. And in the last lecture, we have discussed Fourier series representation of a function of period 2 L in general; we were talking about the functions of period 2 pi as well. (Refer Slide Time: 00:35) So, if a function is periodic of period 2 L, then its Fourier series will be given by a 0 by 2 and it is a constant, which will depend on the functions. So here, 0 1 to infinity a n cos n pi x over L plus b n sin n pi x over L, and where these co-efficient. So, a n 1 over L minus L to L f (x) cos n pi x over L dx k or this n here, so n 0, 1, 2, and so on. And this b n 1 over L minus L to L f (x) sin n pi x over L dx, and this n was 1, 2 and all. And then we have seen that, if this function f (x) is piece wise continuous and one sided derivatives of f (x) is at each point in the domain. Then we have the series equal to, so we have the f (x) plus right limit of f at x and f (x) minus divided by 2, and this is equal to that series.

2 If the function is piece wise continuous and one sided derivatives exist, and they are finite; so then we have this series converges and that value of the series is equal to the average value of the function at that point. And we have also mentioned that at the point of continuity, since there we have f (x) plus is equal to f (x) and that is equal to f (x) minus 1. So at the point of continuity, this is simply f (x); so that series converges to the point to the functional value. (Refer Slide Time: 02:51) Now, we continue with more examples and the first example I consider here is, expand f (x) is equal to sin x in a Fourier series. So, now if we see this function, what we observe, so it is 0 to pi and so on. It is a modulus, so we do not have the negative part. So now, there are two possibilities to gap this Fourier series of this problem, because we can see here that this pi, so is the period of the function, because it repeats the same value after this pi. We can work out in this interval 0 to pi and taking this pi as period of this function or the second possibility is as this standard case we take. So, we take here minus pi to pi interval and consider this 2 pi as its period, because we can take a anyway this 2 pi also a period, because after the 2 pi also the function values, the function repeats this values. So, what we do? These two possibilities we will consider and see, definitely we will get the same series of course at the end. So that is just to make it more clear, let us just continue with the case 1, so that means we work with the with this treat f (x) as a function of period pi. And we work in the interval and take the interval 0 to pi.

3 And that is a good point to discuss that, we should not always it is not necessary to work with always with this symmetric interval minus pi to pi. We can also work with any interval, so we take here 0 to pi. And then this f (x), the general Fourier series a 0 by 2 and 1 to infinity a n cos n pi x over L plus b n sin n pi x over L. And this was the series for a function of period 2 L, so a n as 1 over L minus L to L in general case as, cos n pi x over L dx or we can take also 1 over L 0 to 2 pi periodic function. So we integrate in any range with length 2 pi or 2 L sorry 2 L, it will be the same. So f (x) cos n pi x over L and dx or this b n. Similarly, we have 1 over L 0 to 2 L f (x) sin n pi x over L dx. In this case, which we our considering now, our 2 L is the period taken in this form 2 L and that is pi, so basically, what we have L is pi by 2; now we calculate a 0 first. (Refer Slide Time: 06:44) This a 0 will be 1 over and L, L is pi by 2 so, pi by 2 and we have 0 to 2 L pi and f (x) d x; so we have, 2 over pi and 0 to pi f (x) sin x 0 to pi this mod sin x is positive. So we have sin x, dx on that value is minus cos x and 0 to pi. So, cos pi minus 1 and the is there already. So 1 and minus 1 will be again 2, so we have 4 over pi this value. Now, we calculate the a n so, a n as 2 over pi again. The same 1 over L factor is there and 0 to 2 pi will come with this function sin x and cos 2 n x dx. Then, we have 1 over pi and this 2 sin x cos 2 n x, we write in sin a plus b (( )) sin 2 n plus 1 x sin a plus b and then plus sin a minus b, but we will take b minus a. So, with sin get minus sin so, sin 2 n minus 1 x and dx we integrate now.

4 So we have, sin x cos with minus sign 2 n plus 1 x over 2 n plus 1 and the limit 0 to 5 and we have here to n minus 1 x over 2 n minus 1 and again 0 to pi. We have 1 over pi and then when we put this pi here, we get minus 1 over minus 1 power 2 n plus 1; so we have with minus and minus 2 n plus 1 over 2 n plus 1 and then minus minus plus cos 0 will be 1, so 2 n plus 1. Similarly here, we have minus 1 2 n minus 1 over 2 n minus 1 and we have minus n 1 over 2 n minus 1; this we can simplify and this after simplification, because this is 2 n plus 1, so always an odd number here. We will get minus, so we have here then plus and similarly here, this 2 n, so n is 1, 2 and so on. Here again, we have always the odd number so, get minus so minus 2 over 2 n minus 1 and here we get 2 over 2 n plus 1. And then the further simplification, we will get minus 4 over pi and 4 n square minus 1 for 4 n square minus 1. (Refer Slide Time: 1 0:07) Now we calculate b n, we have 2 over pi 0 to pi and we have sin x and sin 2 n x dx. Again, we take 1 over pi 0 to pi 2 sin x so, sin a sin b. We have cos a minus b cos b minus a so same, we write in this form, 2 n minus 1 x and minus cos a plus b so 2 n plus 1 x and then we have this dx. And now, this the integral will be the sin 2 n minus 1 x and here also sin and the sin function, when we take this have x is equal to pi or we take 0, it will be 0. The value we will get 0, then the Fourier series a naught by 2. So a naught was 4 by pi so, we have 2 over pi and we have n 1 to infinity minus 4 over pi and 4 n square minus 1 cos 2 n x. This is 2 over pi minus 4 over pi and n from 1 to infinity cos 2 n x over 4 n square minus 1 and this x was 0 to pi.

5 (Refer Slide Time: 11:45) Now we check this second approach; and in the second approach, we have so, if we would have taken, we have taken second possibility that is, the period this 2 L 2 pi. We take again minus pi to pi the standard interval and we will get here n is equal to pi. In this case again, this f (x) is a 0 by 2 1 to infinity, we have this standard in this interval. So, n x plus b n sin n x and this a n is 1 over pi minus pi to pi f (x) cos n x dx and b n is 1 over pi and minus pi to pi f (x) sin n x dx. Let us just see first this b n, because this f (x) is an odd even function and we have here the odd function so this was even function. We will come to this point in a minute more into the detail and this is the odd function, because sin minus x is minus sin and here sin over f minus x is f (x) so, its cos x. Then the product will be anyway in odd function so, the integrant is odd. And we are integrating over the symmetric interval minus pi to pi, so this value will be 0 without further calculation. Now this a naught, we can calculate 1 over pi and minus pi to pi and we have this f (x) dx and in this case also, let me just simplify this, a n will be a because it is a here even function of the product is even.

6 So minus pi to pi will be 2 times 0 to pi and we have this f (x) and cos n x dx. This is 2 over pi 0 to pi and this is sin x and dx. So, this cos x and then we have this value cos pi minus and with the minus sin, we get again the 2 out of these integrals. We will get 4 over pi and then a n 2 over pi 0 to pi sin x cos n x dx, which we can integrate some values and this values will be come for and 2, 3 and so on. So, will be 1 over pi times the 0, if n is odd and this will come 2 over n plus 1 minus 2 over n minus 1, when n is even. And this is true for 2, 3 and so on. Again for a 1, we need to calculate 2 over pi 0 to pi sin x cos x dx. This 2 times sin x cos sin 2 x, then cos 2 x over 2 so this will be again 0. What will be the Fourier series in this case? So what we have here finally, this is 2 n minus 2 and minus 2 n minus 2, so we have minus 4 over n square minus 1. (Refer Slide Time: 15:38) We write this Fourier series now, we have f (x) 2 over pi minus 4 over pi and we have cos 2 x over 3 plus cos 4 x over 15 plus cos 6 x over 35 and so on. This Fourier series and the Fourier series, we got by this function of period pi. They both are basically the same, just expand this minus 4 over pi will come and then we have n is equal to 1. So, cos 2 x over 3, then the second term, we have n is equal to 2 so 4 x so, 16 minus 1 15 and so on. If the same Fourier series would be get here. So here what we have seen, that we can work with any general integral not always the symmetric interval. Now, we go for the another important point here, which partially we have discussed in this example, that the Fourier series for even and odd functions for even and odd functions.

7 Again just a function is said to be even, if we have f minus x is equal to f (x) for all x and an odd function. If we have f (x) is equal to minus f (x) for all x. In this case, what will happen? Just let us consider Fourier series, a naught by 2 n 1 to infinity a n Cos n x plus b n sin n x with it is standard in period 2 pi and then we have this a n 1 over pi minus pi to pi f (x) cos n x dx. And now what will happen here, because we have actually we can say always that, when we have a even function and it is multiplied by even function, we will get even function only. And we have even function, odd function, we will get odd or odd into even or we have odd and odd so, again the minus minus will be positive, so we have the even function. With this consideration, what we see here that f (x) is if f is even and we have even function. So, we have the integral value 2 over pi and we can integrate simply 0 to pi f (x) cos n x dx, if f is even. And if f is odd, then this will be odd function and in that case this value will be 0, if f is odd. (Refer Slide Time: 19:20) Similarly, for the coefficient b n, so what we have for b n, we have 1 over pi minus pi to pi and f (x) sin n x d x; this will be 0, if f is even function, because even into odd will be odd and so if f is even and this will give us 2 over pi 0 to pi f (x) sin n x dx, if f is odd. In this case, what we get if f is even function, we are getting a simplified Fourier series. And if f is even, then the Fourier series so for the even case, we have this b n 0. So we will get only with the so a by 2 and 1 to infinity a n cos n x and this a n with the simplified formula, we can calculate 0 to pi f (x) cos n x dx and and so on. And the second case, if f is odd and this is also clear from here, this cos is even function.

8 If f is even function, we will get only the combination of even function on the right hand side. If f is odd, the a n will be 0 and we will get only the sin series. We have this summation and from 1 to infinity b n and sin n x and this b n will be given by 2 over pi 0 to pi f (x) and sin n x dx. (Refer Slide Time: 21:24) Now we consider example with this consideration, so obtain the Fourier series to represent f (x), which is given by x for 0 to pi and 2 pi minus x for pi to 2 pi and then the function is periodic, so f (x) plus 2 pi is f (x). The Fourier series will be given by a naught by 2 and we have a standard period, so n 1 to infinity a n cos n x plus b n sin n x and we have for x 0 to 2 pi. If you look at the graph of this function from 0 to pi, we have this x and then 2 pi minus x, so we have this. And again with this, we continue with this period 2 pi. So here 0 here pi and then 2 pi and so on, here minus pi. So this a naught, we can get with the 1 over pi and 0 to 2 pi f (x) dx 1 over pi 0 to pi we have the function x plus pi to 2 pi 2 pi minus x dx. And the simple integration will give us just pi, a n is 1 over pi and 0 to 2 pi f (x) cos n x dx. And again we can break into 2 intervals, 0 to pi x cos n x dx plus pi to 2 pi and 2 pi minus x cos n x dx. And again these simple calculations and we will get in this case it is 0, when n is even and we will get minus 4 over n square pi n square pi, when n is odd

9 (Refer Slide Time: 24:25) Here we have calculated, now we go for the b n. And we no need to calculate b n, because the function is the even function, so we have even function, f (x) is f minus x. And in this case, b n will be 0 without going for any calculation sin n x dx this is will, this will be 0 and therefore, we have this f (x) is equal to pi by 2 minus 4 over pi cos x plus cos 3 x 3 square, cos 5 x 5 square and so on. And our x is from 0 to 2 pi and not that we can write this equality, because our function is continuous. And the function is continuous then this values equal to f (x) so, this series converges to the function value f (x). (Refer Slide Time: 25:36)

10 Let us go for another example, and determine the Fourier series of f (x) is equal to x square, and then find the value of the series n 1 to infinity minus 1 n plus 1 one over n square and n 1 to infinity 1 over n square. Let us continue with this, first the so Fourier series of f (x) is equal to x square on minus pi to pi. In this interval, we want to have the Fourier series of this, f (x) is equal to x square. If you look at this functions parabola, so we have this from minus pi to pi. In this interval, we are interested to get this Fourier series expansion of x square. To get this, we have to extend we have to put a continuation of this function to the periodic. We need to continue this as a periodic function and so on. And then we find the Fourier series and that will be of course, it will in this interval. Since the function is an even function again, so clearly our b n will be 0 for all n. And now a n we can get so 1 over pi minus pi to pi we have x square and cos n x dx and that will be again 2 over pi and 0 to pi x square cos n x dx. This we can find out 2 over pi and we have here the x square and sin n x over n the integral of this limit minus. Again this n will come and we have 0 to pi this 2 x and sin n x dx. If it would pi 0, this is going to be 0 anyway. So, we have 2 over pi and this minus 1 over n; 2 over pi and we take this 2 n plus also outside so, we have minus 2 over n and then we integrate again. We have x and cos n x over n with minus sin, so this is 0 to pi and then we have with minus minus will be plus. So 0 to pi x is 1 and we have cos n x over n dx and should not be 0 and cannot be 0. So for n is equal to 0, we have to calculate separately, so we have minus 4 over n, we have minus pi and minus 1 over n so minus pi and minus 1 over n divided by n. And put 0 it will be 0, so we have n here, 1 over n square, because after integral we get 1 over n sin n x. So, sin n x will be 0 at 0 and also at pi. We end up with this 4 plus and this minus 1 over n over, we get n square.

11 (Refer Slide Time: 30:07) Now we calculate a 0. So, a 0 will be 1 over pi minus pi to pi x square dx. And this is just x cube over 3 and minus pi to pi and then we have this 2 pi square over 3, because 5 cube will come and that will be cancelled with this pi. So, 2 pi square by 3. Now, we can write in that form, so we have this a naught by 2. So pi square by 3 and plus. And the cosine terms n 1 to infinity, we have this 4 minus 1 over n our a n n square and we have cos n x for x minus pi to pi. And again the function is continuous, so we can have this equality of this series to x square. And now the question was that how to find the series? We have to get this minus 1 over n plus 1 one over n square. We are getting the similar series here, the only point is this cos n x. So if you put this x is equal to 0, we will get it of this cos here. Let us first put so this is the reason number 1; if we substitute if we substitute x is equal to 0. So left hand side we have 0,we have pi square over 3 plus n 1 to infinity and x is 0 the power 0 is 1 and we have 4 minus 1 n over n square and this will. So, we take to the other side, this summation and 1 minus we can accommodate here. We get n 1 to infinity and minus 1 n plus 1 over n square and this 4 will go to the right hand side. We have pi square over 12. We got this sum of this series. So, the second question was, if we substitute, x is equal to pi. In 1 again, what we get left hand side, we get pi square a pi square over 3 we have plus and 1 to infinity 4 over n square and minus over 2 n, because cos pi will give minus 1 power n and this is 1.

12 So we get n 1 to infinity and this 4 over n square or for we can also take to the right hand side. So 1 over n square, summation n to infinity 1 over n square will be just pi square minus pi square over 3. So, 2 pi square over 3 and then we have also this 4 there. So that will be pi square by 6. So this is a series of, so that is also 1 application of this Fourier series, we can get the sum of the series. (Refer Slide Time: 33:48) Now we go to another topic of this Fourier series that is the half range series. And this is just motivated by this sin over this even and odd function. But in this case, what we, so let us let me write this suppose, that this f (x) is a function defined on 0 to L and also we want to find the cosine or sine series of f (x). F (x) is defined in some interval and our aim is to have the cosine series or sine series of that function. So what we can do this, for this, this can be by extending f (x) to be an even or odd function. If you want to have the cosine series, we will extend the function as an even function or we want to have this sine series of that function, we will extend this as an odd function. So suppose, f (x) is equal to x is given in an interval 0 to pi, if you want to have the cosine series of this, so what we will do? So we have this is the given function 0 to pi f (x) is given is equal to x. Now, if you want to write this x in terms of the cosine. What we will do? We will extend this as an odd function, as an even function so this is and then make it periodic. We have minus pi to pi and we can make it then periodic, now this function is an even function. We can get the curve Fourier cosine series and that will give for this f (x) is equal to x in this interval.

13 (Refer Slide Time: 36:50) And if you want to have the sine series of the same function for example, if you want to have sine series. What we shall do? So, this was a function and 0 to pi. We can extend this as an odd function and then make this extension of this function. Now, we can write this function into sine series form and that will be valid for this f (x) is equal to x and 0 to pi. (Refer Slide Time: 37:41)

14 So let us just go for few examples, in for this so obtain the half range sine series for f (x) is equal to e x defined in 0 to 1. So you want to have sine series now, so we need to go for the odd extension of this function. If this function is given here 0 to 1, will get this odd extension of this function and then we continue with this period minus 1 to 1 and we can then extend this as the a periodic function and also in that side. So clearly in this case, since this is an odd extension of e power x in this 0 to x, that is the given function. That we have extended, if you want to have cosine series, you would have extended as an odd function. In this case, since this is the odd extension this a n will be 0 and b n we can get 2 over L 0 to L f (x) sin n pi x over L dx. So in this case now, our period 2 L is 2 from minus 1 to 1. That is the period, that means this L is 1, so this b n is 2 over 1 so, 0 to 1 e x and sin n pi x d x. (Refer Slide Time: 40:02) So this we integrate and we get b n 2 n pi 1 minus e minus 1 over n over 1 plus n square pi square. Now for this x between 0 and 1, this e power x, we can represent by that Fourier series. So 2 pi and n 1 to infinity and we have this p n so n and 1 minus e minus 1 over n over 1 plus n square pi square and we have sin n pi x.

15 (Refer Slide Time: 40:02) We now take an example, where expand f (x) is equal to x and x is given between 0 and 2 in a sine series and same function as a cosine series. We take the first case, as sine series that means we want to have the extension of this function 0 to 2. As an odd extension and then we continue with this period, as periodic function. So, in this case a n will be 0 and this b n be 2 over L 0 to L f (x) sin n pi x over L dx. So, 2 over 2 because our 2 L ahs 4 so L is 2 we have 2 over 2 0 to 2 f (x) is x and we have sin n pi x over 2 d x; so we integrate by part. So we have x and then this is cos n pi x over 2 and then we have the minus sin, because this sin will be minus cos minus 2 over n pi and 0 to 2. Then, we have minus 0 to 2 cos n pi x over L, the differentiation of x is 1 we have cos n pi x over L and again minus 2 over n pi d x; so this after simplification, what we get minus 4 over n pi n cos n pi.

16 (Refer Slide Time: 43:12) If we then we this Fourier series, 1 to infinity minus 4 over n pi cos n pi sin n pi x over 2 and x is between 0 to 2. When we write this x between 0 to 2, we can write this equality this is equal to x this is nothing else, but this is given as x. This is 4 over pi, if we expand this just see, what kind of form we are getting. So sin pi x over 2 minus this half and sin 2 pi x over 2 plus 1 over 3 and here minus 1, this is plus and then we have sin 3 pi x over 2 and minus and so on. Now we take the second case, we extend now function. This f (x) as an even function as an even function, so in this case as discussed above, so we will extend this as an even function and then this extension to be periodic functions where, minus 2 to 2 as an even function. In this case also our L is 2. Now b n will be 0 and a n we can get in a similar fashion, as we got their 2 over L 2 over 2 0 to 2 x cos and pi x over L to dx. And we integrate this to get, 4 over n square pi square and cos n pi minus 1 power n for a naught is equal to 0

17 (Refer Slide Time: 45:20) And if we put n is equal to take n is equal to 0; we will get 0 to 2 and x dx 2. In this case this f (x) will be or x is equal to 1 plus n 1 to infinity 4 over n square pi square cos n pi minus 1 and cos n pi x over 2 or we can expand this, we will get cos pi x over 2 plus 1 over 3 square cos 3 pi x over 2 and 5 square cos 5 pi x over 2 and so on. Now this is interesting to see, that for the same function so we have taken this x function, which was defined in 0 to 2 domains. And we get for the same function this cosine series and that was that is given here, that is the cosine series. This is also x is equal to this in this domain x from 0 to 2 and we also have this sine series for this function, which is also valid in this x 0 to 2. For the same function x, we have completely different expansion. One is in the cos another in the sin and both will equally, approximate this function in this domain. In fact, if we consider this series as a sum so this sum is equal to 2 x and this sum is also is equal to x for each x between 0 to 2.

18 (Refer Slide Time: 47:23) Now, we go to the last topic of this complex of this Fourier series and that is the complex Fourier series or complex form of Fourier series. So we consider, f (x) a by 2 we start with the standard form. We can also continue with this Fourier series of function period 2 L. So, a n cos n x plus b n and sin n x minus pi to pi. And with a n we have 1 over pi minus pi to pi and this f (x) cos n x dx and n is and we have this b n 1 over pi minus pi to pi f (x) and sin n x dx n is 1 to 3. So, what we know that this cos n x we can write with this formula, e i n x plus e minus i n x by 2 and this sin n x e i n x minus e minus i n x over 2 i. Now, we substitute this in this equation 1 for this cos n x and this sin n x. So, what we will get? That f (x) is a naught over 2 and 1 to infinity. We get a n for the cos, we have e i n x plus e minus i n x over 2 plus this b n e i n x minus e minus i n x over 2 i. Then, we have a naught by 2 and 1 to infinity and we take this for i n x from here and from there common so, we have half also a n and minus this i we can multiply. We get minus i b n so, if we take half a n minus i b n and we will get e i n x plus. Similarly, there a n plus i b n e minus i n x. We take it as C naught and this we denote it C n constant for this we denote it k n. Because n is equal to 0, we will see in a minute that we get C this a naught by 2 that is what here we name it C 0.

19 (Refer Slide Time: 50:59) So with this notation, what we have the f (x) is C naught plus n 1 to infinity C n e i n x plus k n e minus i n x this is 2. Now we also see, what is this C n again half a n minus i b n and we substitute this a n and b n 1 over pi coming there so 2 pi and minus pi to 2 pi f (x). Here you get cos n x, here you get sin n x, so n minus sin minus i sin n x dx and this again we can write minus pi to pi f (x) minus i n x dx. So, we can see that the 0 is just 1 over 2 pi minus pi to pi f (x) dx that is a naught by 2 that is what we have written. And also this k n take this k n that is a n plus i b n here plus will come and then we will have plus here. So it is simple C minus n. So, then this equation 2 becomes, that this f (x) Fourier series in this form we have C 0 and 1 to infinity and we have C n in x plus C minus n minus i n x or we can combine this an minus infinity to plus infinity and we take this C n e i n x. This will automatically cover this as well as this.

20 (Refer Slide Time: 53:07) And where our C n is the Fourier series in this complex form, for this f (x) is an minus infinity to plus infinity C n e i n x and this C n is 1 over 2 pi and minus pi to pi and we have f (x) i n x dx n is 0 plus minus 1 plus minus 2 and so on. So, this equation is called or this form of the courier series this is called Fourier series and this is called the Fourier of the complex form. If we have a period, for a function for a period 2 L, we have f (x) minus infinity to plus infinity C n i n pi x over L and where, C n will be 1 over 2 L minus L to L f (x) minus i n pi x over L dx. We quickly now go for one example of this Fourier series and the complex form (Refer Slide Time: 54:30)

21 If we take a function, f (x) e power x and this x is between minus pi and pi and then we have this periodicity. So, in this case we calculate the C n, that is 1 over 2 pi minus pi to pi function and e minus i n x dx so, we have 1 over 2 pi minus pi to pi e 1 minus i n x dx. And this we can integrate so, we have e power 1 minus i n x over 1 minus i n and we have the limits minus pi to pi, this just a bit simplification. So, we have 1 over 2 pi 1 minus i n and we will get e power pi minus 1 over n, because you will get e power n x this cos n x plus i sin n x. In that case, will give just minus 1 over 1 power n; we have minus e minus pi and again minus 1 power n. Since, this e power i n pi will be minus 1 over n or we get plus or minus in both the cases we will get cos n pi plus minus i sin n pi sin n pi will be 0 and in this cos n pi will give minus 1 power n. So in this case and this again, we can write this as so 1 over pi here 1 plus i n divided by 1 plus i n. So, 1 plus n square and this sin hyperbolic pi and we have minus 1 over n, This f (x), we have the sin hyperbolic pi over pi and n from minus infinity to plus infinity minus 1 over n 1 plus i n over 1 plus n square and e i n x. So, this is the complex form of the Fourier series of a function, which is just equivalent to what we have seen the earlier form of the Fourier series. Now, say we had enough insides into the Fourier series with the help of several different kinds of examples we have considered in this lecture. And in the next lecture, we shall extend this idea of a periodic function to non periodic function by extending the periods to infinity, and in that case this Fourier series will lead to the so called Fourier integral, and so more on that in the next lecture and thank you, good bye.

Regression Analysis Prof. Soumen Maity Department of Mathematics Indian Institute of Technology, Kharagpur. Lecture - 2 Simple Linear Regression

Regression Analysis Prof. Soumen Maity Department of Mathematics Indian Institute of Technology, Kharagpur Lecture - 2 Simple Linear Regression Hi, this is my second lecture in module one and on simple

Lecture VI. Review of even and odd functions Definition 1 A function f(x) is called an even function if. f( x) = f(x)

ecture VI Abstract Before learning to solve partial differential equations, it is necessary to know how to approximate arbitrary functions by infinite series, using special families of functions This process

Finite Element Analysis Prof. Dr. B. N. Rao Department of Civil Engineering Indian Institute of Technology, Madras. Lecture - 01

Finite Element Analysis Prof. Dr. B. N. Rao Department of Civil Engineering Indian Institute of Technology, Madras Lecture - 01 Welcome to the series of lectures, on finite element analysis. Before I start,

Probability and Statistics Prof. Dr. Somesh Kumar Department of Mathematics Indian Institute of Technology, Kharagpur

Probability and Statistics Prof. Dr. Somesh Kumar Department of Mathematics Indian Institute of Technology, Kharagpur Module No. #01 Lecture No. #15 Special Distributions-VI Today, I am going to introduce

Basic Quantum Mechanics Prof. Ajoy Ghatak Department of Physics. Indian Institute of Technology, Delhi

Basic Quantum Mechanics Prof. Ajoy Ghatak Department of Physics. Indian Institute of Technology, Delhi Module No. # 02 Simple Solutions of the 1 Dimensional Schrodinger Equation Lecture No. # 7. The Free

Finite Element Analysis Prof. Dr. B. N. Rao Department of Civil Engineering Indian Institute of Technology, Madras. Lecture - 28

Finite Element Analysis Prof. Dr. B. N. Rao Department of Civil Engineering Indian Institute of Technology, Madras Lecture - 28 In the earlier lectures, we have seen formulation for 3 node linear triangular

Power System Analysis Prof. A. K. Sinha Department of Electrical Engineering Indian Institute of Technology, Kharagpur. Lecture - 19 Power Flow IV

Power System Analysis Prof. A. K. Sinha Department of Electrical Engineering Indian Institute of Technology, Kharagpur Lecture - 19 Power Flow IV Welcome to lesson 19 on Power System Analysis. In this

Recap on Fourier series

Civil Engineering Mathematics Autumn 11 J. Mestel, M. Ottobre, A. Walton Recap on Fourier series A function f(x is called -periodic if f(x = f(x + for all x. A continuous periodic function can be represented

Fourier Series. 1. Full-range Fourier Series. ) + b n sin L. [ a n cos L )

Fourier Series These summary notes should be used in conjunction with, and should not be a replacement for, your lecture notes. You should be familiar with the following definitions. A function f is periodic

CHAPTER 2 FOURIER SERIES

CHAPTER 2 FOURIER SERIES PERIODIC FUNCTIONS A function is said to have a period T if for all x,, where T is a positive constant. The least value of T>0 is called the period of. EXAMPLES We know that =

Regression Analysis Prof. Soumen Maity Department of Mathematics Indian Institute of Technology, Kharagpur

Regression Analysis Prof. Soumen Maity Department of Mathematics Indian Institute of Technology, Kharagpur Lecture - 7 Multiple Linear Regression (Contd.) This is my second lecture on Multiple Linear Regression

Fourier Series Expansion

Fourier Series Expansion Deepesh K P There are many types of series expansions for functions. The Maclaurin series, Taylor series, Laurent series are some such expansions. But these expansions become valid

M344 - ADVANCED ENGINEERING MATHEMATICS Lecture 9: Orthogonal Functions and Trigonometric Fourier Series

M344 - ADVANCED ENGINEERING MATHEMATICS ecture 9: Orthogonal Functions and Trigonometric Fourier Series Before learning to solve partial differential equations, it is necessary to know how to approximate

A Basic Course in Real Analysis Prof. P. D. Srivastava Department of Mathematics Indian Institute of Technology, Kharagpur

A Basic Course in Real Analysis Prof. P. D. Srivastava Department of Mathematics Indian Institute of Technology, Kharagpur Lecture - 3 Continuum and Exercises So the last lecture we have discussed the

Advanced 3G and 4G Wireless Communication Prof. Aditya K. Jagannatham Department of Electrical Engineering Indian Institute of Technology, Kanpur

Advanced 3G and 4G Wireless Communication Prof. Aditya K. Jagannatham Department of Electrical Engineering Indian Institute of Technology, Kanpur Lecture - 3 Rayleigh Fading and BER of Wired Communication

Introduction to Fourier Series

Introduction to Fourier Series MA 16021 October 15, 2014 Even and odd functions Definition A function f(x) is said to be even if f( x) = f(x). The function f(x) is said to be odd if f( x) = f(x). Graphically,

Heat Transfer Prof. Dr. Aloke Kumar Ghosal Department of Chemical Engineering Indian Institute of Technology, Guwahati

Heat Transfer Prof. Dr. Aloke Kumar Ghosal Department of Chemical Engineering Indian Institute of Technology, Guwahati Module No. # 02 One Dimensional Steady State Heat Transfer Lecture No. # 05 Extended

Some Notes on Taylor Polynomials and Taylor Series

Some Notes on Taylor Polynomials and Taylor Series Mark MacLean October 3, 27 UBC s courses MATH /8 and MATH introduce students to the ideas of Taylor polynomials and Taylor series in a fairly limited

Engineering Mathematics II

PSUT Engineering Mathematics II Fourier Series and Transforms Dr. Mohammad Sababheh 4/14/2009 11.1 Fourier Series 2 Fourier Series and Transforms Contents 11.1 Fourier Series... 3 Periodic Functions...

(Refer Slide Time: 00:00:56 min)

Numerical Methods and Computation Prof. S.R.K. Iyengar Department of Mathematics Indian Institute of Technology, Delhi Lecture No # 3 Solution of Nonlinear Algebraic Equations (Continued) (Refer Slide

Fourier Series Chapter 3 of Coleman

Fourier Series Chapter 3 of Coleman Dr. Doreen De eon Math 18, Spring 14 1 Introduction Section 3.1 of Coleman The Fourier series takes its name from Joseph Fourier (1768-183), who made important contributions

Soil Dynamics Prof. Deepankar Choudhury Department of Civil Engineering Indian Institute of Technology, Bombay

Soil Dynamics Prof. Deepankar Choudhury Department of Civil Engineering Indian Institute of Technology, Bombay Module - 2 Vibration Theory Lecture - 8 Forced Vibrations, Dynamic Magnification Factor Let

Lecture 1 (Review of High School Math: Functions and Models) Introduction: Numbers and their properties

Lecture 1 (Review of High School Math: Functions and Models) Introduction: Numbers and their properties Addition: (1) (Associative law) If a, b, and c are any numbers, then ( ) ( ) (2) (Existence of an

(Refer Slide Time 1.50)

Discrete Mathematical Structures Dr. Kamala Krithivasan Department of Computer Science and Engineering Indian Institute of Technology, Madras Module -2 Lecture #11 Induction Today we shall consider proof

Digital System Design Prof. D. Roychoudhury Department of Computer Science and Engineering Indian Institute of Technology, Kharagpur

Digital System Design Prof. D. Roychoudhury Department of Computer Science and Engineering Indian Institute of Technology, Kharagpur Lecture - 03 Digital Logic - I Today, we will start discussing on Digital

TA: Sum as n goes from 0 to infinity, x minus 2 to the n over n. What's the center?

MAT 296 - Calculus II TA: Sum as n goes from 0 to infinity, x minus 2 to the n over n. What's the center? Student: Two. TA: Two. Why is the center 2? Student: Because of when it's equal to 0. TA: Yeah,

Geometry and Arithmetic

Geometry and Arithmetic Alex Tao 10 June 2008 1 Rational Points on Conics We begin by stating a few definitions: A rational number is a quotient of two integers and the whole set of rational numbers is

(Refer Slide Time: 2:03)

Control Engineering Prof. Madan Gopal Department of Electrical Engineering Indian Institute of Technology, Delhi Lecture - 11 Models of Industrial Control Devices and Systems (Contd.) Last time we were

No Solution Equations Let s look at the following equation: 2 +3=2 +7

5.4 Solving Equations with Infinite or No Solutions So far we have looked at equations where there is exactly one solution. It is possible to have more than solution in other types of equations that are

Maths 361 Fourier Series Notes 2

Today s topics: Even and odd functions Real trigonometric Fourier series Section 1. : Odd and even functions Consider a function f : [, ] R. Maths 361 Fourier Series Notes f is odd if f( x) = f(x) for

Managerial Economics Prof. Trupti Mishra S.J.M School of Management Indian Institute of Technology, Bombay. Lecture 9 Theory of Demand (Contd.

Managerial Economics Prof. Trupti Mishra S.J.M School of Management Indian Institute of Technology, Bombay Lecture 9 Theory of Demand (Contd.) Welcome to the second session of module two; module two talks

Tangent and normal lines to conics

4.B. Tangent and normal lines to conics Apollonius work on conics includes a study of tangent and normal lines to these curves. The purpose of this document is to relate his approaches to the modern viewpoints

Senior Secondary Australian Curriculum

Senior Secondary Australian Curriculum Mathematical Methods Glossary Unit 1 Functions and graphs Asymptote A line is an asymptote to a curve if the distance between the line and the curve approaches zero

Electrical Machines - I Prof. D. Kastha Department of Electrical Engineering Indian Institute of Technology, Kharagpur

Electrical Machines - I Prof. D. Kastha Department of Electrical Engineering Indian Institute of Technology, Kharagpur Lecture - 9 Harmonics and Switching Transients in Single Phase Transformers So far

Fourier Series, Integrals, and Transforms

Chap. Sec.. Fourier Series, Integrals, and Transforms Fourier Series Content: Fourier series (5) and their coefficients (6) Calculation of Fourier coefficients by integration (Example ) Reason hy (6) gives

Foundation Engineering Prof. Mahendra Singh Department of Civil Engineering Indian Institute of Technology, Roorkee

Foundation Engineering Prof. Mahendra Singh Department of Civil Engineering Indian Institute of Technology, Roorkee Module - 03 Lecture - 09 Stability of Slopes Welcome back to the classes of on this Stability

Taylor Polynomials and Taylor Series Math 126

Taylor Polynomials and Taylor Series Math 26 In many problems in science and engineering we have a function f(x) which is too complicated to answer the questions we d like to ask. In this chapter, we will

Fourier Series. Some Properties of Functions. Philippe B. Laval. Today KSU. Philippe B. Laval (KSU) Fourier Series Today 1 / 19

Fourier Series Some Properties of Functions Philippe B. Laval KSU Today Philippe B. Laval (KSU) Fourier Series Today 1 / 19 Introduction We review some results about functions which play an important role

Fourier Series. Chapter Some Properties of Functions Goal Preliminary Remarks

Chapter 3 Fourier Series 3.1 Some Properties of Functions 3.1.1 Goal We review some results about functions which play an important role in the development of the theory of Fourier series. These results

This is the 23 rd lecture on DSP and our topic for today and a few more lectures to come will be analog filter design. (Refer Slide Time: 01:13-01:14)

Digital Signal Processing Prof. S.C. Dutta Roy Department of Electrical Electronics Indian Institute of Technology, Delhi Lecture - 23 Analog Filter Design This is the 23 rd lecture on DSP and our topic

Introduction to Complex Fourier Series

Introduction to Complex Fourier Series Nathan Pflueger 1 December 2014 Fourier series come in two flavors. What we have studied so far are called real Fourier series: these decompose a given periodic function

Review of Key Concepts: 1.2 Characteristics of Polynomial Functions

Review of Key Concepts: 1.2 Characteristics of Polynomial Functions Polynomial functions of the same degree have similar characteristics The degree and leading coefficient of the equation of the polynomial

Sequences and Series

Contents 6 Sequences and Series 6. Sequences and Series 6. Infinite Series 3 6.3 The Binomial Series 6 6.4 Power Series 3 6.5 Maclaurin and Taylor Series 40 Learning outcomes In this Workbook you will

Processing of Semiconducting Materials Prof. Pallab Banerji Material Science Centre Indian Institute of Technology, Kharagpur

Processing of Semiconducting Materials Prof. Pallab Banerji Material Science Centre Indian Institute of Technology, Kharagpur Lecture - 27 Characterization II Let us define the following parameters for

Roots and Coefficients of a Quadratic Equation Summary

Roots and Coefficients of a Quadratic Equation Summary For a quadratic equation with roots α and β: Sum of roots = α + β = and Product of roots = αβ = Symmetrical functions of α and β include: x = and

Lecture No. # 02 Prologue-Part 2

Advanced Matrix Theory and Linear Algebra for Engineers Prof. R.Vittal Rao Center for Electronics Design and Technology Indian Institute of Science, Bangalore Lecture No. # 02 Prologue-Part 2 In the last

1. the function must be periodic; 3. it must have only a finite number of maxima and minima within one periodic;

Fourier Series 1 Dirichlet conditions The particular conditions that a function f(x must fulfil in order that it may be expanded as a Fourier series are known as the Dirichlet conditions, and may be summarized

Wave Hydro Dynamics Prof. V. Sundar Department of Ocean Engineering Indian Institute of Technology, Madras

Wave Hydro Dynamics Prof. V. Sundar Department of Ocean Engineering Indian Institute of Technology, Madras Module No. # 02 Wave Motion and Linear Wave Theory Lecture No. # 03 Wave Motion II We will today

Digital System Design Prof. D Roychoudhry Department of Computer Science and Engineering Indian Institute of Technology, Kharagpur

Digital System Design Prof. D Roychoudhry Department of Computer Science and Engineering Indian Institute of Technology, Kharagpur Lecture - 04 Digital Logic II May, I before starting the today s lecture

Data Communications Prof. Ajit Pal Department of Computer Science & Engineering Indian Institute of Technology, Kharagpur Lecture # 03 Data and Signal

(Refer Slide Time: 00:01:23) Data Communications Prof. Ajit Pal Department of Computer Science & Engineering Indian Institute of Technology, Kharagpur Lecture # 03 Data and Signal Hello viewers welcome

Question: What is the quadratic fmula? Question: What is the discriminant? Question: How do you determine if a quadratic equation has no real roots? The discriminant is negative ie Question: How do you

MATH 461: Fourier Series and Boundary Value Problems

MATH 461: Fourier Series and Boundary Value Problems Chapter III: Fourier Series Greg Fasshauer Department of Applied Mathematics Illinois Institute of Technology Fall 2015 fasshauer@iit.edu MATH 461 Chapter

Lecture5. Fourier Series

Lecture5. Fourier Series In 1807 the French mathematician Joseph Fourier (1768-1830) submitted a paper to the Academy of Sciences in Paris. In it he presented a mathematical treatment of problems involving

Sample Problems. Lecture Notes Equations with Parameters page 1

Lecture Notes Equations with Parameters page Sample Problems. In each of the parametric equations given, nd the value of the parameter m so that the equation has exactly one real solution. a) x + mx m

Chapter 4. Polynomial and Rational Functions. 4.1 Polynomial Functions and Their Graphs

Chapter 4. Polynomial and Rational Functions 4.1 Polynomial Functions and Their Graphs A polynomial function of degree n is a function of the form P = a n n + a n 1 n 1 + + a 2 2 + a 1 + a 0 Where a s

Arithmetic Series. or The sum of any finite, or limited, number of terms is called a partial sum. are shorthand ways of writing n 1

CONDENSED LESSON 9.1 Arithmetic Series In this lesson you will learn the terminology and notation associated with series discover a formula for the partial sum of an arithmetic series A series is the indicated

Section P.9 Notes Page 1 P.9 Linear Inequalities and Absolute Value Inequalities

Section P.9 Notes Page P.9 Linear Inequalities and Absolute Value Inequalities Sometimes the answer to certain math problems is not just a single answer. Sometimes a range of answers might be the answer.

Advanced Hydraulics Prof. Dr. Suresh. A. Kartha Department of Civil Engineering Indian Institute of Technology, Guwahati

Advanced Hydraulics Prof. Dr. Suresh. A. Kartha Department of Civil Engineering Indian Institute of Technology, Guwahati Module - 1 Open Channel Flow Lecture -6 Specific Energy and Critical Flow Very good

Problem Set 7 - Fall 2008 Due Tuesday, Oct. 28 at 1:00

18.781 Problem Set 7 - Fall 2008 Due Tuesday, Oct. 28 at 1:00 Throughout this assignment, f(x) always denotes a polynomial with integer coefficients. 1. (a) Show that e 32 (3) = 8, and write down a list

Introduction to Series and Sequences Math 121 Calculus II D Joyce, Spring 2013

Introduction to Series and Sequences Math Calculus II D Joyce, Spring 03 The goal. The main purpose of our study of series and sequences is to understand power series. A power series is like a polynomial

6.8 Taylor and Maclaurin s Series

6.8. TAYLOR AND MACLAURIN S SERIES 357 6.8 Taylor and Maclaurin s Series 6.8.1 Introduction The previous section showed us how to find the series representation of some functions by using the series representation

Applied Thermodynamics for Marine Systems Prof. P. K. Das Department of Mechanical Engineering Indian Institute of Technology, Kharagpur

Applied Thermodynamics for Marine Systems Prof. P. K. Das Department of Mechanical Engineering Indian Institute of Technology, Kharagpur Lecture - 12 Steam Turbine - Impulse Good afternoon. We were studying

Fourier Series. 1. Introduction. 2. Square Wave

531 Fourier Series Tools Used in Lab 31 Fourier Series: Square Wave Fourier Series: Triangle Wave Fourier Series: Gibbs Effect Fourier Series: Coefficients How does a Fourier series converge to a function?

Chapter 11 Fourier Analysis

Chapter 11 Fourier Analysis Advanced Engineering Mathematics Wei-Ta Chu National Chung Cheng University wtchu@cs.ccu.edu.tw 1 2 11.1 Fourier Series Fourier Series Fourier series are infinite series that

Diploma Plus in Certificate in Advanced Engineering

Diploma Plus in Certificate in Advanced Engineering Mathematics New Syllabus from April 2011 Ngee Ann Polytechnic / School of Interdisciplinary Studies 1 I. SYNOPSIS APPENDIX A This course of advanced

Manufacturing Processes II. Prof. A. B. Chattopadhyay Department of Mechanical Engineering Indian Institute of Technology, Kharagpur

Manufacturing Processes II Prof. A. B. Chattopadhyay Department of Mechanical Engineering Indian Institute of Technology, Kharagpur Lecture No. 4 Interrelations among the tool angles expressed in different

Graphs of Polar Equations

Graphs of Polar Equations In the last section, we learned how to graph a point with polar coordinates (r, θ). We will now look at graphing polar equations. Just as a quick review, the polar coordinate

Trigonometry (Chapter 6) Sample Test #1 First, a couple of things to help out:

First, a couple of things to help out: Page 1 of 20 More Formulas (memorize these): Law of Sines: sin sin sin Law of Cosines: 2 cos 2 cos 2 cos Area of a Triangle: 1 2 sin 1 2 sin 1 2 sin 1 2 Solve the

x 2 + y 2 = 1 y 1 = x 2 + 2x y = x 2 + 2x + 1

Implicit Functions Defining Implicit Functions Up until now in this course, we have only talked about functions, which assign to every real number x in their domain exactly one real number f(x). The graphs

2x + y = 3. Since the second equation is precisely the same as the first equation, it is enough to find x and y satisfying the system

1. Systems of linear equations We are interested in the solutions to systems of linear equations. A linear equation is of the form 3x 5y + 2z + w = 3. The key thing is that we don t multiply the variables

Advanced Hydraulics Prof. Dr. Suresh A Kartha Department of Civil Engineering Indian Institute of Technology, Guwahati

Advanced Hydraulics Prof. Dr. Suresh A Kartha Department of Civil Engineering Indian Institute of Technology, Guwahati Module - 1 Open Channel Flow Lecture - 9 Critical Flow Computations Part-2 (Refer

( ) = 3x 2, so this is not correct. If we multiply by 1/3 first to

Antiderivatives Today we begin a study of antiderivatives. Antiderivatives will be useful for evaluating integrals using the Fundamental Theorem, and for solving some initial value problems. Definition

Week 5 Vladimir Dobrushkin 5. Fourier Series In this section, we present a remarkable result that gives the matehmatical explanation for how cellular phones work: Fourier series. It was discovered in the

1. Periodic Fourier series. The Fourier expansion of a 2π-periodic function f is:

CONVERGENCE OF FOURIER SERIES 1. Periodic Fourier series. The Fourier expansion of a 2π-periodic function f is: with coefficients given by: a n = 1 π f(x) a 0 2 + a n cos(nx) + b n sin(nx), n 1 f(x) cos(nx)dx

Application of Fourier Transform to PDE (I) Fourier Sine Transform (application to PDEs defined on a semi-infinite domain)

Application of Fourier Transform to PDE (I) Fourier Sine Transform (application to PDEs defined on a semi-infinite domain) The Fourier Sine Transform pair are F. T. : U = 2/ u x sin x dx, denoted as U

BASIC ELECTRONICS PROF. T.S. NATARAJAN DEPT OF PHYSICS IIT MADRAS LECTURE-4 SOME USEFUL LAWS IN BASIC ELECTRONICS

BASIC ELECTRONICS PROF. T.S. NATARAJAN DEPT OF PHYSICS IIT MADRAS LECTURE-4 SOME USEFUL LAWS IN BASIC ELECTRONICS Hello everybody! In a series of lecture on basic electronics, learning by doing, we now

Prep for Calculus. Curriculum

Prep for Calculus This course covers the topics shown below. Students navigate learning paths based on their level of readiness. Institutional users may customize the scope and sequence to meet curricular

Inverse Trigonometric Functions

SECTION 4.7 Inverse Trigonometric Functions Copyright Cengage Learning. All rights reserved. Learning Objectives 1 2 3 4 Find the exact value of an inverse trigonometric function. Use a calculator to approximate

Classical Physics Prof. V. Balakrishnan Department of Physics Indian Institution of Technology, Madras. Lecture No. # 13

Classical Physics Prof. V. Balakrishnan Department of Physics Indian Institution of Technology, Madras Lecture No. # 13 Now, let me formalize the idea of symmetry, what I mean by symmetry, what we mean

Strength of Materials Prof: S.K.Bhattacharya Dept of Civil Engineering, IIT, Kharagpur Lecture no 23 Bending of Beams- II

Strength of Materials Prof: S.K.Bhattacharya Dept of Civil Engineering, IIT, Kharagpur Lecture no 23 Bending of Beams- II Welcome to the second lesson of the fifth module which is on Bending of Beams part

Math 115 Spring 2011 Written Homework 5 Solutions

. Evaluate each series. a) 4 7 0... 55 Math 5 Spring 0 Written Homework 5 Solutions Solution: We note that the associated sequence, 4, 7, 0,..., 55 appears to be an arithmetic sequence. If the sequence

Harmonic Formulas for Filtering Applications

Harmonic Formulas for Filtering Applications Trigonometric Series Harmonic Analysis The history of the trigonometric series, for all practical purposes, came of age in 1822 with Joseph De Fourier s book

Infinite series, improper integrals, and Taylor series

Chapter Infinite series, improper integrals, and Taylor series. Introduction This chapter has several important and challenging goals. The first of these is to understand how concepts that were discussed

Probability Foundations for Electrical Engineers Prof. Krishna Jagannathan Department of Electrical Engineering Indian Institute of Technology, Madras

Probability Foundations for Electrical Engineers Prof. Krishna Jagannathan Department of Electrical Engineering Indian Institute of Technology, Madras Lecture - 9 Borel Sets and Lebesgue Measure -1 (Refer

PURE MATHEMATICS AM 27

AM Syllabus (015): Pure Mathematics AM SYLLABUS (015) PURE MATHEMATICS AM 7 SYLLABUS 1 AM Syllabus (015): Pure Mathematics Pure Mathematics AM 7 Syllabus (Available in September) Paper I(3hrs)+Paper II(3hrs)

PURE MATHEMATICS AM 27

AM SYLLABUS (013) PURE MATHEMATICS AM 7 SYLLABUS 1 Pure Mathematics AM 7 Syllabus (Available in September) Paper I(3hrs)+Paper II(3hrs) 1. AIMS To prepare students for further studies in Mathematics and

1.5 ANALYZING GRAPHS OF FUNCTIONS Copyright Cengage Learning. All rights reserved. What You Should Learn Use the Vertical Line Test for functions. Find the zeros of functions. Determine intervals on which

5.1 Radical Notation and Rational Exponents

Section 5.1 Radical Notation and Rational Exponents 1 5.1 Radical Notation and Rational Exponents We now review how exponents can be used to describe not only powers (such as 5 2 and 2 3 ), but also roots

1. The algebra of exponents 1.1. Natural Number Powers. It is easy to say what is meant by a n a (raised to) to the (power) n if n N.

CHAPTER 3: EXPONENTS AND POWER FUNCTIONS 1. The algebra of exponents 1.1. Natural Number Powers. It is easy to say what is meant by a n a (raised to) to the (power) n if n N. For example: In general, if

Algebra. Exponents. Absolute Value. Simplify each of the following as much as possible. 2x y x + y y. xxx 3. x x x xx x. 1. Evaluate 5 and 123

Algebra Eponents Simplify each of the following as much as possible. 1 4 9 4 y + y y. 1 5. 1 5 4. y + y 4 5 6 5. + 1 4 9 10 1 7 9 0 Absolute Value Evaluate 5 and 1. Eliminate the absolute value bars from

Thinkwell s Homeschool Algebra 2 Course Lesson Plan: 34 weeks

Thinkwell s Homeschool Algebra 2 Course Lesson Plan: 34 weeks Welcome to Thinkwell s Homeschool Algebra 2! We re thrilled that you ve decided to make us part of your homeschool curriculum. This lesson

CHAPTER 3. Fourier Series

`A SERIES OF CLASS NOTES FOR 2005-2006 TO INTRODUCE LINEAR AND NONLINEAR PROBLEMS TO ENGINEERS, SCIENTISTS, AND APPLIED MATHEMATICIANS DE CLASS NOTES 4 A COLLECTION OF HANDOUTS ON PARTIAL DIFFERENTIAL

Lecture - 4 Diode Rectifier Circuits

Basic Electronics (Module 1 Semiconductor Diodes) Dr. Chitralekha Mahanta Department of Electronics and Communication Engineering Indian Institute of Technology, Guwahati Lecture - 4 Diode Rectifier Circuits

16 Convergence of Fourier Series

16 Convergence of Fourier Series 16.1 Pointwise convergence of Fourier series Definition: Piecewise smooth functions For f defined on interval [a, b], f is piecewise smooth on [a, b] if there is a partition

1.2. Mathematical Models: A Catalog of Essential Functions

1.2. Mathematical Models: A Catalog of Essential Functions Mathematical model A mathematical model is a mathematical description (often by means of a function or an equation) of a real-world phenomenon

COMP 250 Fall 2012 lecture 2 binary representations Sept. 11, 2012

Binary numbers The reason humans represent numbers using decimal (the ten digits from 0,1,... 9) is that we have ten fingers. There is no other reason than that. There is nothing special otherwise about

Faculty of Mathematics Waterloo, Ontario N2L 3G1 Centre for Education in Mathematics and Computing Grade 6 Math Circles November 4/5, 2014 Exponents Quick Warm-up Evaluate the following: 1. 4 + 4 + 4 +

Introduction to Fourier Series

Introduction to Fourier Series A function f(x) is called periodic with period T if f(x+t)=f(x) for all numbers x. The most familiar examples of periodic functions are the trigonometric functions sin and

Warm up Factoring Remember: a 2 b 2 = (a b)(a + b)

Warm up Factoring Remember: a 2 b 2 = (a b)(a + b) 1. x 2 16 2. x 2 + 10x + 25 3. 81y 2 4 4. 3x 3 15x 2 + 18x 5. 16h 4 81 6. x 2 + 2xh + h 2 7. x 2 +3x 4 8. x 2 3x + 4 9. 2x 2 11x + 5 10. x 4 + x 2 20

Chapter 1 Algebraic fractions Page 11 Chapter 2 Functions Page 22. Chapter 8 Differentiation Page 88. Chapter 3 Exponentials and Logarithms Page 33

Chapter 8 Differentiation Page 88 Chapter 1 Algebraic fractions Page 11 Chapter Functions Page Chapter 7 Further Trig Page 77 C3 Chapter 3 Exponentials and Logarithms Page 33 Chapter 6 Trigonometry Page