Find a MacLaurin Series of fx x sin 3x and determine the ra

Solution

LET\'S WRITE IT AGAIN AS (1/x) * sin(3x) ====> therefore, let\'s find the sin(3x)as the following: first by definition, the correct mclaurin series expansion for the function f(x) = sin(x) is 8 ? ( -1 )^n * [ x^(2n+1) / (2n+1)! ] ..............but how can we get this? as the following: n=0 first, i am going to define some terms such as n = the number of derivatives & number of factorials f^(n)(x) = is first original and the derived equations. f^(n)(0) = plugging 0 in the original equation and derived equations. an = is coefficient numbers of the variables. IT WILL BE AS TABLE: n ---- f^(n)(x) ------f^n(0)-------an 0------sin(3x)---------0-----------(0)… = 0/1 = 0 (0! is equal to one ) 1-----3cos(3x)-------3-----------3/1! = 3 2----(-9sin(3x)-------0-----------0/2! = 0 3---(-27cos(3x))---(-27)--------(-27/3… = -27/(1*2*3) = -27/6 = -9/2 4---(81sin(3x)) ------0------------ 0/4! = 0 5---(243cos(3x))-----243--------(243/5… = 243/(1*2*3*4*5) = 243/120 = 81/40 now sin(3x) = 0 + 3*x^1 + 0 - [ 27*x^3/3! ] + 0 + [ 243*x^5/5! ] then it will be since its alternating, and its increment for both: the exponent of the variable and to the denominator (with factorials) are odd number, it will be as 2n+1 but if its increment even numbers, it will be 2n so the series is 8 ? ( -1 )^n * [ 3^(2n+1) * x^(2n+1) / (2n+1)! ] = 3*x^1 - [ 27*x^3/3! ] + [ 243*x^5/5! ] - .... n = 0 then, we have (1/x), therefore, we are going to divide our series by x as the following: 8 ? (1/x) * ( -1 )^n * [ 3^(2n+1) * x^(2n+1) / (2n+1)! ] = 3*x^1 - [ 27*x^3/3! ] + [ 243*x^5/5! ] - .... n = 0 8 ? (1/x) * ( -1 )^n * [ 3^(2n+1) * x^(2n) * x^1) / (2n+1)! ] = 3*x^1 - [ 27*x^3/3! ] + [ 243*x^5/5! ] - .... n = 0 8 ? ( -1 )^n * [ 3^(2n+1) * x^(2n) / (2n+1)! ] = 3*x^1 - [ 27*x^3/3! ] + [ 243*x^5/5! ] - .... n = 0