 Hello everyone! Welcome to the session of Expansion of Function by Taylor series part 1st. This is Swati Nikam, Assistant Professor, Department of Humanities and Sciences, Valchan 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. Now, let us see what are these different Taylor series. Assuming that f of x plus h can be expanded in ascending powers of h, it is expressed as f of x plus h is equal to f of x plus h times f dash of x plus h square by 2 factorial f double dash of x plus h cube by 3 factorial f triple dash of x plus dash dash dash plus and so on. In general, the nth term can be written as h raised to n upon n factorial into nth derivative of f at x plus and so on. Let us call this as representation number 1. This is known as Taylor series. Now, interchanging x and h, we can express f of x plus h in ascending powers of x as f of x plus h is equal to 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 and so on. Plus, the nth term is x raised to n upon n factorial nth derivative of f at h plus and so on. This is representation number 2. This is also known as Taylor series. Taylor series number 1 is the expression in ascending powers of h that is constant term while representation number 2 is the expression in ascending powers of x which is variable. Now, replacing x by a and h by x minus a so that a plus h is equal to a plus x minus a is equal to x. In equation number 1, we get f of x as a power series of x minus a as 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 f double dash of a plus x minus a whole cube upon 3 factorial f triple dash of a plus and so on. The nth term is x minus a whole raised to n upon n factorial f nth derivative at a plus and so on. Let us call this as representation number 3. Representation number 1, 2 and 3. All these 3 represents Taylor series. The first one represents Taylor series in ascending powers of h, second in ascending powers of x and third is in ascending powers of x minus a. We are going to use one of the three series according to the expectation of our example. So, let us see an example so that we will be clear about use of any one of the series. Before that, friends please pause your video for minute and write first two derivatives of function f of x is equal to tan inverse x. I hope you have written your answer. So, the first two derivatives of function f of x is equal to tan inverse x is, we know that tan inverse of x is a standard function and its derivative f dash of x is 1 upon 1 plus x square. Now, second order derivative f double dash of x is applying u by v rule. It is minus 2x upon 1 plus x square whole square. Let us see an example. Example number 1, expand tan inverse of x in powers of x minus 1 up to second power of x minus 1. Now friends see here how to choose the correct expansion of Taylor series for this example. Now here for tan inverse of x we need to expression in terms of x minus a. So, among the three forms we have to use form number 3. That is why in answer let us consider function f of x is equal to tan inverse x and comparing with x minus a form value of a is 1. Let us put this value of a in representation number 3. So, this is representation number 3 of Taylor series. Therefore, our Taylor series takes the form f of x is equal to f of 1 plus x minus 1 into f dash of 1 plus x minus 1 square upon 2 factorial f double dash of 1 plus x minus 1 whole cube upon 3 factorial f triple dash of 1 plus and so on. Let us call this as representation number 1. Now, f of x is equal to tan inverse of x. So, it is first order derivative. You have calculated already in previous slide f dash of x is equal to 1 upon 1 plus x square and second order derivative is minus 2x upon 1 plus x square whole square. Now, according to the need in this representation number 1 we want to calculate the value f at 1 f dash at 1 f double dash at 1 and f triple dash at 1. So, here let us replace each x by 1 and hence we get f of 1 is equal to tan inverse of 1 which is pi by 4. Now, from f dash of x f dash of 1 is 1 upon 1 plus 1 square that is 1 by 2. From f double dash of 1 is minus 2 into 1 that is minus 2 upon 1 plus 1 square is 2 square that is minus half. So, now putting all these values in equation number 1 we have f of x is equal to tan inverse of x is equal to pi by 4 plus x minus 1 into half plus x minus 1 square upon 2 factorial into minus 1 by 2 plus and so on. Therefore, tan inverse of x is equal to pi by 4 plus x minus 1 into 1 by 2 minus x minus 1 whole square divided by 4 plus and so on. So, this is the required expression of tan inverse of x in ascending powers of x minus 1. In example, it is as that we have to do the expansion up to power 2. Now, friends let us solve one more example for the practice of Taylor series example number 2. Expand 3x cube minus 2x square plus x minus 4 in powers of x plus 2. Now, again we have to choose the correct form of Taylor series among the three. So, in this example also we need expression in terms of x plus 2 that is x minus a form. We know that here Taylor series a is given by 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 f triple dash of a. Now, friends see here the given polynomial is a cubic polynomial. We know that the cubic polynomial has three non-zero derivatives all higher order derivative becomes 0. So, we will stop our expansion up to f triple dash of a only. Now, let us calculate all these required first, second and third order derivative from the given function f of x. So, f dash of x is 9x square minus 4x plus 1. Now, according to the requirement of our expression we need to find out f of minus 2 which is minus 38 f dash of minus 2 is 45. Second order derivative at x is equal to 18x minus 4 and third order derivative at x is equal to 18. This implies that f double dash at minus 2 is equal to minus 40 and f triple dash at x is constant 18. So, f triple dash at minus 2 is without any change 18. Now, putting all these values in the representation f of x is equal to f of minus 2 plus x plus 2 into f dash of minus 2 plus x plus 2 square upon 2 factorial into f double dash of minus 2 plus x plus 2 whole cube upon 3 factorial f triple dash of minus 2. Therefore, f of x is equal to minus 38. Value of f of minus 2 is minus 38. We have calculated here. Then x plus 2 into value of f dash of minus 2 which is 45. So, substituting 45 here plus x plus 2 whole square upon 2 factorial is 2 into f double dash of minus 2 which is minus 40. Substitute here plus x plus 2 whole cube upon 3 factorial is 6 into f triple dash at minus 2 is 18. Do the simple calculations so that f of x turns to be minus 38 plus x plus 2 into 45 plus x plus 2 whole square minus 40 upon 2 is minus 20 plus x plus 2 whole cube upon 6 and it is multiplied by 18 is 18 divided by 6 which is 3 simply. So, this is the representation of given cubic polynomial f of x in ascending powers of x plus 2 with the help of Taylor series. For creation of this video, I have applied a text of applied mathematics by NP Bali. Thank you so much for watching this video and have a happy learning.