 Hello, Myself, M.S. Basragaon, Assistant Professor, Department of Humanities and Sciences, Valksandvistok Technology, Solapur. Learning outcome, at the end of this session, students will be able to express 4-year series of U1 and odd function in the interval minus pi to pi. Now we will see the examples. Find the 4-year series of f of x is equal to sin A x in the interval minus pi to pi. Now here f of x is equal to sin A x. Now first of all we will test the U1 and odd function. Now here f of minus x that is replacing x by minus x, we get f of minus x equal to sin of minus x which is equal to minus sin A x that is equal to minus f of x. Hence f of minus x is equal to minus f of x. Therefore f of x equal to sin A x is an odd function. And for odd function A naught is equal to 0 and A n equal to 0. The 4-year series of an odd function is given by f of x is equal to summation of n equal to 1 to infinity B n sin n x where B n is equal to 2 by pi integration from 0 to pi sin A x sin n x dx. Now before integrating we make use of the formula of sin A sin B. Now which is equal to 2 by pi integration from 0 to pi cos of A minus n x minus cos of A plus n x whole divided by 2 dx. Here 2, 2 will get cancelled. And here we know since sin A sin B is equal to cos of A minus B minus cos of A plus B divided by 2 which is equal to 1 by pi. Now we will integrate with respect to x that is integration of cos A minus n x is sin A minus n x divided by A minus n and integration of cos of A plus n x is that is minus sin of A plus n x by A plus n with a limit 0 to pi. Which is equal to now putting the upper limit 1 by pi sin of A minus n pi by A minus n minus sin of A plus n pi divided by A plus n and for lower limit as you know sin 0 is 0 therefore for lower limit you get 0 which is equal to 1 by pi. Now expanding sin of A minus n pi you get sin A pi cos n pi minus cos A pi sin n pi that is by using the formula of sin A minus B divided by A minus n minus. Now expanding sin of A plus n pi that is sin A pi cos n pi plus cos A pi sin n pi divided by A plus n which is equal to we can write 1 by pi you know cos of n pi is minus 1 raise to n and sin A pi as it is divided by A minus n and as you know sin of n pi is 0 therefore this second term will be 0 minus here again cos of n pi is minus 1 raise to n keeping sin A pi as it is and sin n pi is again 0 divided by A plus n since sin n pi is 0 and cos n pi is minus 1 raise to n. Now simplifying this you get minus 1 raise to n sin A pi by pi into bracket 1 by A minus n minus 1 by A plus n which is equal to minus 1 raise to n sin A pi divided by pi into bracket A plus n minus A plus n divided by A square minus n square that taking the cross multiplication and multiplying A minus n and A plus n which is equal to here A will get cancelled 2 n minus 1 raise to n sin A pi divided by pi into bracket A square minus n square hence f of x is equal to 2 sin A pi divided by pi summation of n equal to 1 to infinity minus 1 raise to n into n divided by A square minus n square sin n x. Now pause the video for a while and find the Fourier series of f of x equal to x in the interval minus pi to pi. I hope you have completed here f of x is equal to x and clearly here f of x equal to x is an odd function hence A naught is equal to 0 and A n equal to 0. The Fourier series of an odd function is given by f of x equal to summation of n equal to infinity b n sin n x where b n is equal to 2 by pi integration from 0 to pi f of x sin n x dx which is equal to 2 by pi integration from 0 to pi x sin n x dx. Now we can integrate this integral with respect to x by using generalized rule of integration by part which is equal to 2 by pi 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 which is equal to 2 by pi putting the upper limit that x equal to pi as pi minus cos n pi by n minus minus plus sin of n pi is 0 and if you put lower limit because of multiplication of x we get first term is 0 and sin of 0 is also 0 which is equal to we get the minus sin minus 2 into cos of n pi is minus 1 raise to n divided by n and here pi pi will get cancelled therefore f of x is equal to summation of n equal to 1 to infinity minus 2 into minus 1 raise to n divided by n sin n x. Now we will see one more example find the Fourier series of f of x is equal to 1 plus 2 x by pi and minus pi is minus pi is less than or equal to x is less than or equal to 0 and which is equal to 1 minus 2 x by pi when 0 is less than or equal to x is less than or equal to pi and hence deduce that pi square by 8 is equal to 1 by 1 square plus 1 by 3 square plus 1 by 5 square plus so on. Now here first of all you will test here the Fourier 1 and odd function for that you replace here x by minus x that is f of minus x is equal to 1 minus 2 x by pi that is by replacing x by minus x where minus pi is less than or equal to again here also replacing x by minus x that we get minus x less than or equal to 0 or we can write this inequality as 0 is less than or equal to x is less than or equal to pi which is equal to now we here replacing x by minus x we get 1 plus 2 x by pi where 0 is less than or equal to minus x is less than or equal to pi the same inequality can be written as minus pi is less than or equal to x is less than or equal to 0. Now you look at the given function f of x and f of minus x we are getting the same therefore here f of minus x is equal to f of x therefore f of x is an even function hence the value of b n is 0. Now the Fourier series of an even function 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 where a naught is equal to 1 by pi integration from 0 to pi f of x dx which is equal to 1 by pi integration from 0 to pi in the interval 0 to pi f of x is 1 minus 2 x by pi dx which is equal to 1 by pi and integration of 1 is x minus 2 by pi into x square by 2 that 2 2 will get cancelled you get minus x square by pi with the limit 0 to pi which is equal to 1 by pi putting x equal to pi you get pi minus pi square by pi and for lower limit both the terms are 0 that you get pi minus pi which is equal to 0 here a naught is also 0. Now we will calculate a n a n is equal to 2 by pi integration from 0 to pi f of x cos n x dx which is equal to 2 by pi integration from 0 to pi here f of x is 1 minus 2 x by pi cos n x dx now integrating by using generalized rule of integration by part taking u as a 1 minus 2 x by pi and v as a cos n x will integrate which is equal to 2 by pi keeping 1 minus 2 x by pi as it is an integration of cos n x is sin n x by n minus the derivative of 1 minus 2 x by pi is minus 2 by pi and integration of sin n x by n is minus cos n x by n square with the limit 0 to pi. Now putting the upper limit putting x equal to pi we know sin of n pi is 0 therefore the first term is 0 and here minus into minus into minus we get the minus sign twice cos of n pi by pi n square. Now minus of lower limit as we know sin 0 is 0 therefore this term is 0 then minus cos 0 is 1 that is we get minus 2 upon pi n square. Now here taking 2 by pi n square common we get which is equal to 4 by n pi square n square into bracket 1 minus minus 1 raised to n. Hence f of x is equal to 4 by pi square summation of n equal to 1 to infinity 1 minus minus 1 raised to n by n square cos n x. Now here putting the values of n we get 4 by pi square putting n equal to 1 we get the value of 1 minus minus 1 raised to n as 2 2 by n square means 1 square cos x plus now here for n equal to 2 we get the value of this as a 0 and for n equal to 3 we get 2 that is 2 by 3 square cos 3 x similarly putting n equal to 4 we get the value of this bracket is 0 and for n equal to 5 we get 2 that 2 by 5 square cos of 5 x plus so on which is equal to taking 2 common here 8 by pi square into bracket 1 by 1 square cos x plus 1 by 3 square cos 3 x plus 1 by 5 square cos 5 x and so on. Now to get the one more result here we have to put x equal to 0 in equation number 1 that we get f of 0 from the given function f of 0 is 1 which is equal to 8 by pi square as we know the values of cos 0's are 1 that is we get 1 by 1 square plus 1 by 3 square plus 1 by 5 square plus so on and by simplifying get therefore pi square by 8 is equal to 1 by 1 square plus 1 by 3 square plus 1 by 5 square and so on. Now reference is higher engineering method is by Dr. B. S. Graywald.