 Welcome to the session of Expansion of Functions part 4. This is Swati Nikam, assistant professor, Department of Humanities and Sciences, Balchan Institute of Technology, Sulapur. At the end of this session, students can create the expansion of function about any point with the help of standard series which is obtained by using Metallorean series which we have seen in earlier video. Now, let us see some standard series which we are going to use in this video. First one is exponential series e raised to x which is equal to its expansion is 1 plus x plus x square by 2 factorial plus x cube by 3 factorial plus x raised to 4 by 4 factorial plus x raised to 5 by 5 factorial plus and so on. And second series is logarithmic series log of 1 plus x is equal to x minus x square by 2 plus x cube by 3 minus x raised to 4 by 4 plus x raised to 5 by 5 plus and so on. Now, let us see the use of these standard series to solve few examples. Example number one, expand function f of x is equal to log of 1 plus x whole raised to 2 in ascending powers of x up to x raised to 4. Answer, we know that series expansion of log of 1 plus x is, the expansion I have explained in earlier side for log of 1 plus x is written over here. Then log of 1 plus x whole square can be written as, so write down the bracket term which we have explained over here x minus x square by 2 plus x cube by 3 minus x raised to 4 by 4 and so on whole raised to 2. Now, in order to do this expansion, let me rewrite the given expression as first term x taking minus sign common and writing all other terms in one bracket as x minus x square by 2 minus x cube by 3 plus x raised to 4 by 4 minus x raised to 5 by 5 and so on. Whole raised to 2. We have really arranged the terms. Now, we have done this rearrangement to use the standard result a minus b whole square which is a square minus 2ab plus b square. So, here a is x and b is this bracket term. And therefore, log of 1 plus x whole square now becomes a square is x square minus 2 into a is x into b is bracket plus bracket square. Now, we will expand this first term x square as it is minus. So, this is twice x into bracket terms x cube by 2 minus and minus plus twice x raised to 4 by 3 next term plus and so on. Now, remember in our example it is as that we have to do the expansion up to x raised to 4 only. So, let us leave all other higher power terms and hence we will put dash dash for all other higher power terms. From this third expansion with plus sign we will rewrite this as now see in this expansion again we will use the same formula and rearrange the terms as a minus b whole square. But first term is x square itself. So, first term square is x square whole square is x raised to 4 by 4. So, no need to consider further expansion terms as they are greater than power 4. So, we will keep them as dash dash dash. Therefore, log of 1 plus x whole square is equal to x square minus x cube plus x raised to 4 into bracket since there are two terms. So, I have collected the coefficients 2 by 3 plus 1 by 4 plus all other terms dash dash dash dash and therefore finally the expression for log of 1 plus x whole square is equal to x square minus x cube plus x raised to 4 into bracket 11 by 12 plus and so on. Now friends before we proceed further by using laws of logarithm state the following first one log of a by b and second log of a into b. I hope you have written your answer. So, by using laws of logarithm we have log of a by b is equal to log a minus log b and log of a into b is equal to log a plus log b. Now, let us solve one more example prove that log of big bracket numerator x into e raised to x upon denominator e raised to x minus 1 big bracket complete is equal to x minus 2 minus x square by 24 plus x raised to 4 upon 2880 minus dash dash dash. Now in answer we will use the laws of logarithm which we have stated earlier log of a by b is equal to log a minus log b. So, log of numerator x e raised to x upon e to the power x minus 1 becomes log of a that is numerator x into e raised to x minus log of b that is denominator e raised to x minus 1. Once again this is log of a into b. So, which is log a that is log x plus log of e raised to x minus log of e to the power x minus 1. Here we have to use that is a expansion of e raised to x as 1 plus x plus x square by 2 factorial plus x cube by 3 factorial plus x raised to 4 by 4 factorial and so on and minus this one. Friends put the brackets properly. Therefore LHS becomes here it is log x log x as it is plus log of e to the power x now becomes x minus log of. So, see here if you open the bracket then first term is 1 and last term is minus 1 which it cancels and so all other terms are in ascending powers of x as x plus x square by 2 factorial plus x cube by 3 factorial plus x raised to 4 by 4 factorial and so on. Now which is equal to log x plus x log of since in this bracket all other terms contains x and hence let us have x common. So, writing all the terms in one bracket taking x common it is 1 plus x upon 2 factorial plus x square upon 3 factorial plus x cube upon 4 factorial and so on which is equal to log x plus x. Now see here it is of the form log of a into b where a is x and b is this bracket. So, log of a into b is log of a plus log of b but because of this minus sign we will rewrite it as minus log of x minus log of b which is bracket term. Now log x and minus log x cancels out and therefore it is equal to x minus log of in big bracket let me rearrange it as again first term 1 plus all other terms in one bracket as x by 2 factorial plus x square by 3 factorial plus x cube by 4 factorial and so on. So, we have done this arrangement in order to use the standard series log of 1 plus x. As I have said let us call the bracket term as capital X and so by using the result log of 1 plus capital X is equal to x minus x square by 2 plus x cube by 3 minus x raise to 4 by 4 plus and so on. And therefore log of x into e to the power x divided by e raise to x minus 1 becomes x minus log of 1 plus capital X remember capital X this big bracket now which is equal to x minus inside the big bracket capital X that is bracket minus capital X square bracket square divided by 2 plus capital X cube that is bracket cube divided by 3 minus capital X that is bracket minus 4 upon 4 plus and so on. Now since capital X is this bracket now in this expansion if you write down the terms directly in the expanded form the expression becomes very lengthy and confusing also. So, let me calculate the terms of powers of x separately as x square is equal to so for writing this bracket as x upon 2 factorial plus all the other terms in second bracket x square by 3 factorial plus x cube by 4 factorial and so on whole square. We have done this arrangement because we want to use the result a plus b whole square which is equal to a square that is first term square x by 2 factorial whole square plus 2 times a b that is 2 into x upon 2 factorial into bracket plus b square that is bracket square. So, it is a plus b whole square similarly to write down x cube we will again rewrite it as first term a x by 2 factorial plus second term b as bracket. So, a plus b whole cube is it is equal to inside the big bracket a cube that is x by 2 factorial whole cube plus 3 times a square that is x by 2 factorial square into b which is bracket plus 3 into a x by 2 factorial into b is bracket whole square plus b cube that is bracket cube. Similarly, x raised to 4 is a is x by 2 factorial plus b is bracket whole raised to 4 is equal to first term raised to 4 that is x by 2 factorial raised to 4 plus and so on. Now, all these expanded terms we will write down and get the expression as LHS log of x into e raised to x divided by e raised to x minus 1 is equal to x minus x by 2 factorial minus x square by 3 factorial minus x cube by 4 factorial minus x raised to 5 by 5 factorial plus dash dash dash plus x square upon 2 into 2 factorial square plus 2 into x cube upon 2 into 2 factorial into 3 factorial plus 2 into x raised to 4 upon 2 into 2 factorial into 4 factorial plus dash dash dash plus x raised to 4 upon 2 into 3 factorial square plus and so on minus x cube upon 3 into 2 factorial cube minus 3 into x raised to 4 upon 3 into 2 factorial square into 3 factorial dash dash dash plus x raised to 4 upon 4 into 2 factorial whole raised to 4 plus and so on. After calculating that is which is finally equal to x by 2 minus x square upon 24 here the term of x cube get cancelled plus x raised to 4 upon 2880 minus dash dash dash.