 Hello, Myself, M.S. Basurgaon, Assistant Professor, Department of Human Design Sciences, Walsh and Newstork Technologies, SolarPort. Now learning outcome, at the end of this session, students will be able to find Fourier series of the given function. Now we will solve the examples on Fourier series, find the Fourier series of f of x equal to e raised to Ax in the interval 0 to 2 pi. Now we know that 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 equation number 1. And now we can calculate a naught by the formula 1 by 2 pi integration from 0 to 2 pi f of x dx which is equal to 1 by 2 pi integration from 0 to 2 pi e raised to Ax dx. Now integration of e raised to Ax is 1 by 2 pi into bracket e raised to Ax upon a with a limit 0 to 2 pi which is equal to 1 by 2 a pi into e raised to Ax with a limit 0 to 2 pi which is equal to 1 by 2 a pi into bracket 2 a pi minus e raised to 0 as a 1. Now we calculate a n, a n equal to 1 by pi integration from 0 to 2 pi f of x cos n x dx which is equal to 1 by pi integration from 0 to 2 pi e raised to Ax cos n x dx which is equal to 1 by pi integration of e raised to Ax cos n x is e raised to Ax upon a square plus n square into bracket a cos n x plus n sin n x with a limit 0 to 2 pi since integration of e raised to Ax cos bx dx is equal to e raised to Ax upon a square plus b square into bracket a cos bx plus b sin bx here comparing with this formula we get this integration. Now we can take a square plus n square outside the integral that is 1 upon pi into bracket a square plus n square into bracket put the upper limit that is e raised to 2 a pi into bracket a cos of 2 n pi and we know sin of 2 n pi 0 therefore directly here we can add 0 minus of lower limit that is e raised to 0 into bracket cos 0 is 1 that we can write a plus sin 0 is 0 that we can write 0 taking a common a upon pi into bracket a square plus n square and using cos 2 n pi as a 1 we get e raised to 2 a pi minus 1. Now similarly bn equal to 1 by pi into integration from 0 to 2 pi f of x sin n x dx which is equal to 1 by pi into integration from 0 to 2 pi e raised to Ax sin n x dx which is equal to 1 by pi now integrating e raised to x sin n x e raised to Ax upon a square plus n square into bracket a sin n x minus n cos n x with a limit 0 to 2 pi since integration of e raised to Ax sin bx dx equal to e raised to Ax upon a square plus b square into bracket a sin bx minus b cos bx. Now putting the limit with 0 to 2 pi we get 1 upon pi into bracket a square plus n square and e raised to 2 a pi sin up to n pi that is 0 minus n into cos of 2 n pi minus e raised to 0 and sin 0 is 0 minus n cos 0 is 1 that is minus n. By simplifying this we get which is equal to taking minus n common from both minus n upon a square plus n square into bracket here 2 a pi and here minus 1 will remain. Putting the values of a naught a n and b n in equation number 1 by simplifying this we can write the simplified answer as e raised to Ax is equal to e raised to 2 a pi minus 1 by n into bracket 1 by 2 will remain plus summation of n equal to 1 to infinity a cos n x minus n sin n x by a square plus n square. Now find the Fourier series of f of x equal to x in the interval 0 to 2 pi. Pause the video for a while and complete the Fourier series of the given function. I hope you have completed. This is a very simple function 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 equation number 1 and you can calculate a n equal to 1 by 2 pi integration from 0 to 2 pi f of x dx that is equal to 1 by 2 pi integral of x is x square by 2 with the limit 0 to 2 pi by putting the limit we get pi. Similarly a n equal to 1 by pi integration from 0 to 2 pi f of x cos n x dx is equal to 1 by pi integration from 0 to 2 pi x cos n x dx again here by applying the rule of integration by part. First we write x integration of cos n x is sin n x by n minus derivative of x is 1 that is integration of sin n x by n is minus cos n x by n square with the limit 0 to 2 pi as we know sin of 2 n pi is 0 minus minus plus cos of 2 n pi we get 0 to 2 pi 1 by pi into bracket 1 by n square and for lower limit minus 1 by n square which is equal to 0. Now b n equal to 1 by pi into integration from 0 to 2 pi f of x sin n x dx which is equal to 1 by pi integration from 0 to 2 pi x sin n x dx again integrating by part we get which is equal to 1 by pi into a bracket x minus cos n x by n minus 1 into minus sin n x by n square with the limit 0 to 2 pi. Now putting upper limit we get minus 2 pi cos 2 n pi by n and plus the second term 0 minus of lower limit because of x here we get 0 and sin 0 is 0 which is equal to minus 2 by n by simplifying we get this as a Fourier series of the given function. Now we will see Fourier series for discontinuous functions. Let f of x be defined by f of x is equal to f 1 of x when 0 is less than x is less than x 0 and equal to f top of x when x 0 is less than x is less than 2 pi where x 0 is the point of finite discontinuity in the interval 0 to 2 pi. Then the value of a naught a n and b n are given by a naught is equal to 1 by 2 pi first of all you take the limit 0 to x 0 then in that f 1 of x plus x 0 to 2 pi f top of x dx. Similarly a n equal to 1 by pi integration from 0 to x 0 f 1 of x into cos n x dx plus integration from x 0 to 2 pi f top of x cos n x dx and b n equal to 1 by pi into integration from 0 to x 0 f 1 of x sin n x dx plus integration from x 0 to 2 pi f top of x sin n x dx. Now we will see the example based on discontinuous function find the Fourier series of f of x is equal to x when 0 is less than or equal to x is less than or equal to pi and equal to 2 pi minus x when pi is less than or equal to x is less than equal to 2 pi. Now 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 that is equation number 1. Now we will calculate a naught where a naught is equal to 1 by 2 pi integration from 0 to pi first you will take and in that function is x dx plus next integration from pi to 2 pi function is 2 pi minus x dx which is equal to 1 by 2 pi. Now integration of x is that is x square by 2 with only limit 0 to pi plus integration of 2 pi minus x is 2 pi x minus x square by 2 with a limit pi to 2 pi. Now putting the limit we get 1 by 2 pi into bracket that is pi square by 2 minus lower limit 0 plus for the second part 2 pi into 2 pi that is 4 pi square minus 4 pi square by 2 minus of lower limit is that you putting x equal to pi 2 pi square minus pi square by 2 and by simplifying we get y by 2. Now a n a n is equal to 1 by pi integration from 0 to 2 pi f of x cos n x dx which is equal to 1 by pi first we will take the limit integration from 0 to pi x cos n x dx plus integration from pi to 2 pi 2 pi minus x cos n x dx. Now here again we are applying integration by part keeping x as it is and integration of cos n x is sin n x by n minus derivative of x is 1 and integration of sin n x by n is minus cos n x by n square with a limit 0 to pi plus now integration of the second again 2 pi minus x as it is integration of cos n x is sin n x by n minus minus 1 into integration of sin n x by n is minus cos n x by n square with a limit pi 2 2 pi which is equal to 1 by pi. Now putting the limit for first part we know sin of n pi 0 therefore first part is 0 this minus minus plus cos n pi by n square and for lower limit because we know sin 0 is 0 and 0 get minus 0 and cos 0 is 1 that is minus minus plus 1 by n square plus putting the limit in the second part putting x equal to 2 pi first part will be 0 that is minus into minus into minus we get minus cos of 2 n pi by n square minus of lower limit as we know sin 0 is 0 minus cos of n pi by n square which is equal to 1 by pi into minus cos of n pi is minus 1 is to n by n square minus 1 by n square and cos of 2 n pi is 1 that is minus 1 by n square plus minus 1 is to n upon n square by simplifying it 2 by pi n square into bracket minus 1 is to n minus 1. Now similarly we will calculate b n b n is equal to 1 by pi integration from 0 to 2 pi f of x sin n x dx which is equal to 1 by pi integration from 0 to pi x sin n x dx plus integration from pi to 2 pi 2 pi minus x into sin n x dx which is equal to again using integration by part 1 by part keeping x as it is and integration of sin n x is minus cos n x by n minus derivative of x is 1 and integration of minus cos n x by n is minus sin n x by n square with the limit 0 to pi plus integral of the second that is keeping 2 pi minus x as it is and integration of sin n x is minus cos n x by n minus derivative of 2 pi minus x is minus 1 into integration of minus cos n x by n is minus sin n x by n square with the limit pi to 2 pi. Now by putting the limit we get b n equal to 1 by pi minus cos pi cos n pi divided by n plus 0 minus for lower limit we get 0 plus for upper limit we get both the terms are 0 minus into bracket minus cos n pi by n minus 0 which is equal to 1 by pi we know minus pi into cos of n pi is minus 1 raise to n divided by n plus pi into minus 1 raise to n by n and which is equal to what we get 0. Therefore f of x is equal to here n what is pi by 2 plus summation of n equal to 1 to infinity n is 2 by pi n square into bracket minus 1 raise to n minus 1 into cos n x plus b n that is 0 sin n x by simplifying we get f of x is equal to pi by 2 plus 2 by pi into summation of n equal to 1 to infinity minus 1 raise to n minus 1 upon n square cos n x.