Finding the Taylor polynomial for the first three terms

[NastyAccident]

Homework Statement

f(x) = \frac{ln(3x)}{6x}, a = \frac{1}{3}, n=3

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:
\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.



NastyAccident

 

[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.

 

[NastyAccident]

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.

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!



NastyAccident

 

[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!

 

[NastyAccident]

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!

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!



NastyAccident