- #1
NastyAccident
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Homework Statement
[tex]f(x) = \frac{ln(3x)}{6x}, a = \frac{1}{3}, n=3[/tex]
Find T3
Homework Equations
Taylor Series - f(n)(x)/n! * (x-a)^n
The Attempt at a Solution
So, I isolated ln(3x) from 1/6x.
I created the series based off of ln(3x).
f(0)(x)=ln(3x) ->f(0)(1/3)=ln(3(1/3)) =0
f(1)(x)=1/x ->f(1)(1/3)=1/(1/3) =3
f(2)(x)=-1/x2 ->f(2)(1/3)=-1/(1/3)2 =-32
f(3)(x)=2/x3->f(3)(1/3)=2/(1/3)3 =2*33
f(4)(x)=-2*3/x4->f(4)(1/3)=-2*3/(1/3)4 =-2*3*34
f(n)(x)=(-1)(n+1) * (n-1)!/xn ->f(n)(1/3)=(-1)(n+1) * (n-1)!/(1/3)n =(-1)(n+1) * (n-1)!*3n
Incorporating that back in:
[tex]\frac{1}{6x}*\sum^{3}_{n=1}(-1)^{n+1} * \frac{(n-1)! * 3^{n}}{n!}* (x-\frac{1}{3})^{n}[/tex]
Now, as you see I started the series at 1, since ln(1) [aka 0] does not fit with the general description of f(n)(x).
[tex]\frac{1}{6x}*\sum^{3}_{n=1}(-1)^{n+1} * \frac{3^{n}}{n}* (x-\frac{1}{3})^{n}[/tex]
So, I end up getting these first three terms:
[tex]\frac{3(x-\frac{1}{3})}{6x}-\frac{3^{2}(x-\frac{1}{3})^{2}}{6x*2}+\frac{3^{3}(x-\frac{1}{3})^{3}}{6x*3}[/tex]
Simplified:
[tex]T_{3}=\frac{x-\frac{1}{3}}{2x}-\frac{3(x-\frac{1}{3})^{2}}{4x}+\frac{3(x-\frac{1}{3})^{3}}{2x}[/tex]
However, that answer is being marked as incorrect for some odd reason. Where am I going wrong with this Taylor series? I retraced everything and it should work.
NastyAccident