 Hello, Myself, MS Bhastragam, Assistant Professor, Department of Humanities and Sciences, Walshan Institute of Technology, Solapur. Now, learning outcome, at the end of this session, students will be able to find four-year series of the given function. Now, before four-year series, we will see about the periodic function. A function f of x, which satisfies the relation f of x plus t equal to f of x for all x, and some fixed t is called a periodic function. The smallest positive number t for which this relation holds is called the period of f of x. If t is the period of f of x, then f of x is equal to f of x plus t equal to f of x plus 2t and so on, f of x plus n into t, where n is positive integer. For example, sin x cos x sec x cos x are periodic functions with period 2 pi, while tan x cot x are periodic functions with period pi. Now, we will see about the four-year series. The four-year series is an infinite series, which is represented in terms of the trigonometric sin and cosine functions of the form f of x is equal to a naught plus a 1 cos x plus a 2 cos 2 x plus a 3 cos 3 x and so on, plus a n cos n x and so on, plus b 1 sin x plus b 2 sin 2 x plus b 3 sin 3 x and so on, plus b n sin n x. Now, all these expression can be written in summation form as 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, where the constants a naught a n b n are called four-year coefficients. Now, we will see convergence of four-year series that that are called Dirichlet's conditions. The sufficient condition for the informed convergence of a four-year series are called Dirichlet's conditions. A function f of x, depending the interval c 1 to c 2, can be expressed as four-year series if it satisfies the following conditions. First condition is function f of x and its integrals are finite and single value, function f of x as finite number of discontinuities, function f of x as finite number of maxima and minima. These conditions are known as Dirichlet's conditions. Now, we will see one of the important rule, generalized rule of integration by parts. To integrate the product of two functions, one of which is a power of x, we apply generalized rule of integration by parts that is, integration of u v dx is equal to u into v suffix 1 minus u dash into v suffix 2 plus u double dash into v suffix 3 and so on, where dashes denotes the derivatives and suffix denotes the integration. Note for an integer n, sin n pi is 0, sin up to n pi is 0 and cos up n pi is equal to minus 1 this to n and cos up to n pi is equal to 1. Now, evaluate the integral integration from 0 to 2 pi x square sin 3 x dx. Pause the video for a while and evaluate this integral. I hope you have completed. Now, here we are applying the generalized rule of integration by part that integration from 0 to 2 pi x square sin 3 x dx is equal to that is keeping u as it is that x square and integration of sin 3 x is minus cos 3 x by 3 minus derivative of x square is 2x integration of minus cos 3 x by 3 is minus sin 3 x by 9 plus derivative of 2 x is 2 and integration of this minus sin 3 x by 9 is cos 3 x by 27 with the limit 0 to 2 pi. Pi simplifying this you get this and then putting the limit with 0 to 2 pi you get minus of 2 pi whole square into cos of 6 pi that is 3 into 2 pi divided by 3 plus 2 into x 2 pi sin 6 pi divided by 9 plus 2 cos 6 pi divided by 27 minus of lower limit as we know here x is there 0 sin 0 is also 0 and cos 0 is 1 that is 2 by 27 which is equal to by simplifying we get minus 4 pi square by 3 which is the value of the integral I hope you have completed this. Now, first of all we will see the Fourier series in the interval 0 to 2 pi. Now, for that we will see the Euler's formulae directly the Fourier series for the function f of x in the interval 0 to 2 pi is given by 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 where a naught is equal to 1 by 2 pi integration from 0 to 2 pi f of x dx a n is equal to 1 by pi integration from 0 to 2 pi f of x cos n x dx and b n equal to 1 by pi integration from 0 to 2 pi f of x sin n x dx. Now, we will see the example find the Fourier series of f of x is equal to 3 x square minus 6 x pi plus 2 pi square divided by 12 in the interval 0 to 2 pi. Now, Fourier series of f of x with period 2 pi is given by 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 where a naught is equal to now here we have to calculate a naught a n and b n a naught is equal to 1 by 2 pi integration from 0 to 2 pi f of x dx which is equal to 1 by 2 pi and here f of x is 3 x square minus 6 x pi plus 2 pi square by 12 dx. Now, we will take this total outside that is 1 by 24 pi now we will integrate this f of x that is 3 is constant as it is an integration of x square is x cube by 3 minus 6 pi into x square by 2 that is integration of x plus 2 pi square x integration of 2 pi square is 2 pi square x with the limit 0 to 2 pi which is equal to 1 by 24 pi into bracket by simplifying we get x cube minus 3 pi x square plus 2 pi square x with the limit 0 to 2 pi which is equal to 1 by 24 pi putting the value of x as a 2 pi here we get 8 pi cube minus 12 pi cube plus 4 pi cube and for lower limit what we get all the three terms are 0 that is minus of 0 minus 0 plus 0 8 pi cube plus 4 pi cube that is 12 pi cube and minus 12 pi cube is 0 therefore the constant a naught is 0. Now, we calculate a n and a n equal to be of the formula 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 f of x is 3 x square minus 6 x pi plus 2 pi square by 12 into cos n x dx now taking 12 outside the integral that is 1 by 12 pi integration from 0 to 2 pi 3 x square minus 6 x pi plus 2 pi square into cos n x dx now here we can apply the general law of integration by part by taking this complete first bracket as a u and v as a cos n x which is equal to 1 by 12 pi that is 3 x square minus 6 x pi plus 2 pi square as it is and integration of cos n x is sin n x by n minus that is sin of the rule now derivative of u that is 3 x square minus 6 x pi plus 2 pi square is 6 x minus 6 pi into the integration of sin n x by n is minus cos n x by n square plus that is sin of the rule and now the derivative of 6 x minus 6 pi is 6 and integration of minus cos n x by n square is minus sin n x by n cube with a limit 0 to 2 pi which is equal to 1 by 12 pi first we will simplify 3 x square minus 6 x pi plus 2 pi square into sin n x by n as it is this minus into minus plus 6 x minus 6 pi into cos of n x by n square minus 6 sin n x by n cube with a limit 0 to 2 pi now we put first here upper limit see if you put the upper limit as a 2 pi we know sin of 2 n pi is 0 that is here first term we can write here 0 plus second term will be 6 pi into cos of 2 n pi by n square and third term again it contains sin function therefore the value by putting x equal to 2 pi we get 0 minus of lower limit by putting x equal to 0 we get 0 minus 6 pi cos 0 divided by n square and minus 0 which is equal to 1 by 12 pi now 6 pi into cos of 2 n pi we know 1 that is 6 pi by n square and this minus minus plus 6 pi and cos 0 is 1 divided by n square that is 1 by 12 pi into 12 pi by n square which is equal to 1 by n square now we will calculate next constant b n that 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 2 pi 3 x square minus 6 x pi plus 2 pi square by 12 sin n x dx which is equal to 1 by 12 pi integration from 0 to 2 pi 3 x square minus 6 x pi plus 2 pi square into sin n x dx again here we can apply your v rule taking u as a first bracket and sin n x as a v now which is equal to 1 by 12 pi keeping u as it is that is 3 x square minus 6 x pi plus 2 pi square and integration of sin n x is minus cos n x by n minus again the derivative of this first bracket will be 6 x minus 6 pi into integration of minus cos n x by n is minus sin n x by n square plus derivative of 6 x minus 6 pi is 6 the integration of minus sin n x by n square is plus cos n x by n cube with the limit 0 to 2 pi now first of all we put the upper limit which is equal to 1 by 12 pi into bracket here 3 into x square means 2 pi whole square means 4 pi square minus 6 into x as a 2 pi into pi plus 2 pi square into minus cos of 2 n pi by n and now by putting x equal to 2 n pi sin of 2 n pi is 0 therefore middle term will be 0 plus 6 cos 2 n pi by n cube this about the upper limit now minus of lower limit that is by putting x equal to 0 first 2 term will be 0 2 pi square and cos 0 is 1 that is minus 1 by n plus middle term contains sin therefore sin 0 is 0 that is 0 and plus 6 into cos 0 is 1 that you get 6 by n cube which is equal to 1 by 12 pi into bracket 2 pi square minus 1 by n plus 6 by n cube plus 2 pi square by n minus 6 by n cube by simplifying this our relation and which is equal to what we get 2 pi square minus 1 by n and plus 2 pi square by n will get cancel 6 by n cube and minus 6 by n cube get cancer which is equal to 0 here b n is also 0 therefore by putting the values of a naught a n and b n we get squared 4 series for the given function