 Hello, Myself, MS Bhasargaon, Assistant Professor, Department of Humanities and Sciences, Walshan Institute of Technology, Solapur. Learning Outcome At the end of this session, students will be able to express the given function in four-year series in the interval minus pi to pi. Now, first of all, we will see the Euler's formulae. The four-year series expansion of the function f of x in the interval minus pi to pi is given by f of x is equal to a0 plus summation of n equal to 1 to infinity a n cos nx plus b n sin nx, where a0 is equal to 1 by 2 pi, integration from minus pi to pi f of x dx, and a n equal to 1 by pi, integration from minus pi to pi f of x cos nx dx, and b n equal to 1 by pi, integration from minus pi to pi f of x sin nx dx. Now, these Euler's formulae, we are using to calculate the three constants a0, a9, bn. Now, we will see the example, find the four-year series of f of x equal to x minus x square in the interval minus pi to pi. Let f of x equal to a0 plus summation of n equal to 1 to infinity a n cos nx plus b n sin nx, let us call this as equation number 1, where a0 is equal to 1 by 2 pi, integration from minus pi to pi f of x dx, which is equal to 1 by 2 pi, here f of x is x minus x square dx. Now, we can integrate with respect to x, which is equal to 1 by 2 pi, integration of x is x square by 2 minus, integration of x square is x cube by 3 with a limit minus pi to pi, which is equal to 1 by 2 pi, first we will put the upper limit, that is putting x equal to pi, we get pi square by 2 minus pi cube by 3 minus of lower limit, that is putting x equal to minus pi, we get minus pi whole square by 2 minus of minus pi cube by 3. By simplifying we get, which is equal to 1 by 2 pi, pi square by 2 minus pi cube by 3 minus pi square by 2 minus pi cube by 3, which is equal to we get minus pi square by 3. Now, we calculate a n, a n is equal to 1 by pi integration from minus pi to pi f of x cos n x which is equal to 1 by pi integration from minus pi to pi here f of x is x minus x square into cos n x dx. Now, here to integrate x minus x square into cos n x we can use generalized rule of integration by part that we already seen which is equal to 1 by pi first we will take u as a x minus x square as it is x minus x square and integration of cos n x is sin n x by n minus that is sin of the rule into bracket derivative of x minus x square that is 1 minus 2 x into into bracket integration of sin n x is minus cos n x by n square plus derivative of 1 minus 2 x is that is minus 2 into bracket integration of minus cos n x by n square is minus sin n x by n cube with a limit minus pi to pi. Now, first of all we will put the upper limit if you put x equal to pi here what we have sin of n pi is 0 therefore, the first term will be 0. Now, this minus into minus plus putting x equal to pi you get 1 minus 2 pi cos of n pi by n square this minus minus plus again we have sin of n pi is 0 minus of lower limit that is we have to put x equal to minus pi again here sin of minus n pi is 0 therefore, first term is 0 and in second term by putting x equal to minus pi you get plus 1 plus 2 pi cos of minus n pi by n square and sin of minus n pi is 0. Now, which is equal to 1 by pi 1 minus 2 pi and we know cos of n pi is minus 1 raise to n divided by n square minus 1 plus 2 pi cos of minus n pi is cos n pi therefore, you can write the value of this as a minus 1 raise to n upon n square which is equal to 1 by pi by simplifying these two terms we get minus 4 pi into minus 1 raise to n by n square which is equal to minus 4 into minus 1 raise to n upon n square this what a n. Now, we will calculate b n b n is equal to 1 by pi minus integration from minus pi to pi f of x sin n x dx which is equal to 1 by pi integration from minus pi to pi x minus x square that is f of x sin n x dx again using here the generalized rule of integration by part first we will keep x minus x square that is u as it is integration of sin n x is minus cos n x by n minus that is the sign of the rule now derivative of x minus x square is 1 minus 1 minus 2 x integration of minus cos n x by n is minus sin n x by n square plus derivative of 1 minus 2 x is minus 2 into integration of minus sin n x by n square is plus cos n x by n cube. Now, putting the upper limit first putting x equal to pi we get pi minus pi square into bracket minus cos n pi by n and here sin of n pi 0 therefore, second term will be 0 minus twice cos of n pi by n cube minus now we are putting the lower limit that is x equal to minus pi minus pi minus pi square into bracket minus cos of minus n pi by n and because of sin n pi you get here 0 minus 2 into bracket cos of minus n pi by n cube. Now, by simplifying we get which is equal to 1 by pi minus pi into minus 1 raise to n by n plus pi square into minus 1 raise to n by n minus twice minus 1 raise to n by n cube minus pi into minus 1 raise to n by n minus pi square into minus 1 raise to n by n plus twice minus 1 raise to n by n cube. Now, simplifying this we get which is equal to 1 by pi minus 2 pi into minus 1 raise to n by n which is equal to pi pi we get cancel minus 2 into minus 1 raise to n by n. Substituting all the values of a naught a n and b n in equation number 1 we get f of x is equal to a naught that is minus pi square by 3 minus 4 into summation of n equal to 1 to infinity minus 1 raise to n upon n square cos n x minus twice summation of n equal to 1 to infinity minus 1 raise to n by n sin n x. Now pause the video for a while and find the Fourier constant a naught for the function f of x equal to e raise to a x in the interval minus pi to pi. I hope you have completed here you can calculate a naught by using the formula 1 by 2 pi integration from minus pi to pi f of x dx is equal to 1 by 2 pi integration from minus pi to pi e raise to a x dx which is equal to 1 by 2 pi integration of e raise to a x is e raise to a x upon a with the limit minus pi to pi which is equal to 1 by 2 pi a putting upper limit we get e raise to a pi minus e raise to minus a pi and as you know e raise to a pi minus e raise to minus a pi divided by 2 as a sin hyperbolic a pi divided by pi a. Now we will see the second example find the Fourier series of f of x is equal to minus x minus pi and minus pi is less than x is less than 0 equal to x plus pi when 0 is less than x is less than pi. Now here let f of x is equal to a naught plus summation of n equal to 1 to infinity a n cos n x plus b n sin n x let us call this as a equation number 1. We have a naught is equal to 1 by 2 pi integration from minus pi to pi f of x dx which is equal to 1 by 2 pi. Now first we will take the limit minus pi to 0 in that f of x is minus x minus pi dx plus now we will take the limit 0 to pi in that f of x is x plus pi dx now which is equal to 1 by 2 pi integrating now with respect to x minus x square by 2 minus pi x with the limit of minus pi to 0 plus integrating x plus pi with respect to x we get x square by 2 plus pi x with the limit 0 to pi. Now here by putting upper limit we get 0 minus of lower limit that is putting x equal to minus pi we get minus of pi square by 2 plus pi square plus and now here putting x equal to pi upper limit that is pi square by 2 plus pi square and for lower limiting putting x equal to 0 we get 0 which is equal to by simplifying this we get pi by 2. Now an equal to 1 by pi integration from minus pi to pi f of x cos x dx which is equal to 1 by pi first we will take the limit minus pi to 0 in that f of x is minus x minus pi into cos x dx plus integration from 0 to pi x plus pi cos x dx which is equal to 1 by pi now integrating with respect to x keeping minus x minus pi as it is and integration of cos n x is sin n x by n. Now derivative of minus x minus pi is minus 1 and integration of sin n x by n is minus cos n x by n square with the limit minus pi to 0 plus now keeping x plus pi as it is and integration of cos n x is sin n x by n minus derivative of x pi x plus pi is 1 and integration of sin x by n is minus cos n x by n square with a limit 0 to pi, which is equal to 1 by pi. Now, we will put the upper limit as you know sin 0 is 0 therefore, first term is 0 then minus into minus into minus we get the sin minus and cos 0 is 1 that you get minus 1 by n square. Now minus again putting x equal to minus pi as a lower limit as you know sin of minus n pi is 0 then minus cos of n pi by n square plus for the second integral putting x equal to pi again sin of n pi is 0 minus minus plus cos of n pi by n square minus of lower limit that is putting x equal to 0 first term will be 0 because sin 0 is 0 then cos 0 is 1 that you get 0 plus 1 by n square which is equal to 1 by pi minus 1 by n square and minus 1 by n square minus 2 by n square and this minus minus plus 2 cos n pi by n square which is equal to 2 by pi n square and cos of n pi is minus 1 raise to n minus 1. Now, we will calculate b n b n equal to 1 by pi integration from minus pi to pi f of x sin n x dx which is equal to 1 by pi integration from minus pi to 0 minus x minus pi into sin n x dx plus integration from 0 to pi x plus pi sin n x dx which is equal to 1 by pi again integrating by part we get we will keep minus x minus pi as it is and integration of sin n x is minus cos n x by n minus derivative of minus x minus pi is minus 1 and integration of minus cos n x by n is minus sin n x by n square with the limit minus pi to 0 plus keeping x plus pi as it is integration of sin n x is minus cos n x by n minus derivative of x plus pi is 1 and integration of minus cos n x by n is minus sin x by n square with the limit 0 to pi which is equal to now putting the upper limit x equal to 0 here what you get pi by n minus sin 0 is 0 minus lower limit putting x equal to minus pi here we get minus pi minus pi plus pi we will get 0 minus of lower limit putting and next second term is also sin of minus sin pi that is 0 plus now putting x is equal to pi pi plus pi 2 pi that is minus 2 pi and cos of n pi is minus 1 is to n divided by n and sin of n pi is 0 now putting the lower limit 0 that is minus putting x equal to 0 here we get minus pi by n because we know cos 0 is 1 and plus sin 0 is 0 which is equal to 1 by pi into bracket by simplifying we get 2 by n minus 2 pi minus 1 is to n by n which is equal to 2 by n into bracket 1 minus of minus 1 is to n now substituting the values of a naught a n and b n in equation number 1 we get f of x is equal to pi by 2 plus 2 by pi summation of n equal to 1 to infinity minus 1 is to n minus 1 by n square cos n x plus twice summation of n equal to 1 to infinity 1 minus minus 1 is to n by n sin n x references higher engineering mathematics by Dr. B. S. Gravel. Thank you.