 Welcome to the session of Expansion of Functions by using Taylor Series Part 2. This is Swati Nikam, Assistant Professor, Department of Humanities and Sciences, Valchand Institute of Technology, Sulapur. At the end of this session, students can create the expansion of function about any point with the help of Taylor Series. So friends, we have seen different forms of Taylor Series. Let us solve few examples in this video with the help of some of the forms. For example, number one, by using Taylor Series arranged in powers of x, here given expression is 7 plus bracket x plus 2 plus 3 times bracket x plus 2 whole cube plus bracket x plus 2 whole raise to 4. Answer, we know that f of x plus h is the representation for Taylor Series as f of h plus x times f dash of h plus x square by 2 factorial f double dash of h plus x cube by 3 factorial f triple dash of h plus dash dash dash. Now by comparing this x plus h with the term x plus 2, we get h is equal to 2. So that f of x plus 2 in the expanded form becomes f of 2 plus x times f dash of 2 plus x square by 2 factorial f double dash of 2 plus x cube by 3 factorial f triple dash of 2 plus x raise to 4 upon 4 factorial into f fourth order derivative at 2. Now here f of x plus 2 is given as 7 plus x plus 2 plus 3 times x plus 2 cube plus x plus 2 raise to 4. Now from this we can get the value of f of x as removing plus 2 from each term we get 7 plus x plus 3 x cube plus x raise to 4. First order derivative f dash of x is derivative of 7 is 0, derivative of x is 1 plus 339 x square plus 4 x cube. Second order derivative f double dash of x is 18x plus 12x square. Third order derivative f triple dash of x is 18 plus 24x and fourth order derivative f fourth derivative at x is 24. Now if you look carefully at this representation we need to get value of f of 2 f dash of 2 f double dash of 2 and so on. And that is done by putting x is equal to 2 in all these representations. So here we get successively all the values at x is equal to 2. Now putting these values in equation number 1 we get so see here f of 2 is 49 f dash of 2 is 69 f double dash of 2 is 84 f triple dash of 2 is 66 f fourth derivative at 2 is 24. So let us put all these values in this representation so that f of x plus 2 becomes 49 plus x into 69 plus x square by 2 factorial into 84 plus x cube by 3 factorial into 66 plus x raise to 4 by 4 factorial into 24 which is equal to in a simplified form 49 plus 69x plus 42x square plus 66 upon 3 factorial is 6 11x cube plus 24 upon 4 factorial is again 24 so it is simply x raise to 4. Now friends before we solve one more example pause your video for a minute and write the Taylor series expansion in ascending powers of x minus a. I hope you have written your answer. So we have seen three types of forms of Taylor series among that the Taylor series expansion in ascending powers of x minus a is f of x is equal to f of a plus x minus a into f dash of a plus x minus a whole square upon 2 factorial into f double dash of a plus x minus a whole cube upon 3 factorial into f triple dash of a plus dash dash dash. Now let us solve example based upon these expression example number 2 expand f of x is equal to 5 plus 4 into bracket x minus 2 bracket square minus 3 times x minus 2 bracket cube plus bracket x minus 2 whole raised to 4 in ascending powers of x minus 1. Now here we will use the Taylor series expansion which we have seen in earlier slide that is f of x in ascending powers of x minus a. Let us call it as expression number 1. Now we want to expand f of x in terms of x minus 1 here according to our example. Therefore value of a is 1 here so that f of x is writing as a given expression and calculating few successive derivatives we have. Therefore first order derivative f dash of x is equal to derivative of 5 is 0 which is constant. So 4 to the 8 times x minus 2 minus 3 to the 9 x minus 2 whole square plus this 4 into bracket x minus 2 whole cube. Second order derivative f double dash of x is 8 minus 18 into bracket x minus 2 plus 12 into bracket x minus 2 whole square. Third order derivative f triple dash of x is minus 18 plus 24 in bracket x minus 2 and fourth order derivative f4 at x is 24 which is constant. So all other higher order derivative becomes 0. So we have to stop our expansion up to order 4. Now putting all these values in equation number 1 and a is equal to 1 we get f of x is equal to f of 1 plus x minus 1 into f dash of 1 plus x minus 1 whole square upon 2 factorial into f double dash of 1 plus x minus 1 whole cube upon 3 factorial into f triple dash of 1 plus x minus 1 whole raised to 4 upon 4 factorial into f fourth derivative at 1. So friends see here we need to find out the value of function and their derivatives at point x is equal to 1. So here we have calculated it f of x gives you f of 1 as 7. From f dash of x f dash of 1 is minus 21 f double dash of x gives f double dash of 1 as 48 f triple dash of 1 is obtained from f triple dash of x by replacing x with 1 as minus 42. And finally the fourth order derivative at x is 24 which remains same because we don't have the term of x here. So f fourth derivative at 1 is equal to 24. Now putting all these values in the expression of Taylor series whatever we have written just now is f of x is equal to 7 plus x minus 1 in bracket minus 21 plus x minus 1 whole square upon 2 factorial is 2 into f double dash at 1 is 48 plus x minus 1 whole cube upon 3 factorial is 6 into f triple dash at 1 is minus 42 plus x minus 1 whole raised to 4 upon 4 factorial is 24 into f fourth order derivative at 1 is 24. So finally after simplification f of x takes the form 7 minus 21 in bracket x minus 1 plus 24 in bracket x minus 1 whole square minus 7 times x minus 1 whole cube plus x minus 1 whole raised to 4. I have referred a text of applied mathematics by NP Bali for creation of this video. Thank you so much for watching this video and have a happy learning.