# Finding the Taylor polynomial for the first three terms

1. Nov 9, 2009

### NastyAccident

1. The problem statement, all variables and given/known data
$$f(x) = \frac{ln(3x)}{6x}, a = \frac{1}{3}, n=3$$

Find T3

2. Relevant equations
Taylor Series - f(n)(x)/n! * (x-a)^n

3. 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:
$$\frac{1}{6x}*\sum^{3}_{n=1}(-1)^{n+1} * \frac{(n-1)! * 3^{n}}{n!}* (x-\frac{1}{3})^{n}$$

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).

$$\frac{1}{6x}*\sum^{3}_{n=1}(-1)^{n+1} * \frac{3^{n}}{n}* (x-\frac{1}{3})^{n}$$

So, I end up getting these first three terms:

$$\frac{3(x-\frac{1}{3})}{6x}-\frac{3^{2}(x-\frac{1}{3})^{2}}{6x*2}+\frac{3^{3}(x-\frac{1}{3})^{3}}{6x*3}$$

Simplified:

$$T_{3}=\frac{x-\frac{1}{3}}{2x}-\frac{3(x-\frac{1}{3})^{2}}{4x}+\frac{3(x-\frac{1}{3})^{3}}{2x}$$

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.

Sincerely,

NastyAccident

2. Nov 9, 2009

### Dick

It's not odd it was marked wrong because it is wrong. You can't separate log(3x) and 1/(6x). You need to find the first three derivatives of the function log(3x)/(6x) at x=(1/3). Use the quotient and chain rules. Your series should be an expansion in powers of (x-1/3), there shouldn't be x's floating around as well.

Last edited: Nov 9, 2009
3. Nov 9, 2009

### NastyAccident

So, because this is a Taylor series vs. a Maclaurin series, separating ln(3x) from 1/6x causes a rift when actually calculating the series since it isn't based at 0?

I was under the assumption that you could separate Taylor series exactly like Maclaurin series. Mm, back to the drawing board.

Thanks!

Sincerely,

NastyAccident

4. Nov 9, 2009

### Dick

Well, you just have to compute f^(n)(1/3) for a few values of n. Just do it directly. That shouldn't be totally back the drawing board. You're welcome!

5. Nov 10, 2009

### NastyAccident

Aye, just had to differentiate and simplify a lot... But, I ended up with:

T3=3/2(x-1/3)-27/4(x-1/3)^2+99/4(x-1/3)^3

Which is correct!

Thank you for the guidance Dick!

Sincerely,

NastyAccident