1. ## Taylor Series Question

Hi, this question regarding taylor series. i tried solve for 7 hours straight with no ans. differentiate the 1/(x^2 + x +1) up to 4 times without finding any fixed patterns. Experts please help.

2. ## Re: Taylor Series Question

Since at x = 0, there is a part where taking out x^n and leaving behind (f(x) / x) ^n, this caused a undefined value since when x = 0, there f(0) / 0 is undefined.

3. ## Re: Taylor Series Question

Originally Posted by hxhx
Hi, this question regarding taylor series. i tried solve for 7 hours straight with no ans. differentiate the 1/(x^2 + x +1) up to 4 times without finding any fixed patterns. Experts please help.
The result of this post are not correct for the reason explained in a succesive post...

Kind regards

$\chi$ $\sigma$

4. ## Re: Taylor Series Question

this is better
$1-x+x^3-x^4-\frac{x^6}{\left(1+2 (-1)^{2/3}\right) \left((-1)^{2/3}-x\right)}-\frac{x^6}{\left(-1+2 (-1)^{1/3}\right) \left((-1)^{1/3}+x\right)}$
chao

5. ## Re: Taylor Series Question

\displaystyle \begin{align*} \frac{1}{x^2 + x + 1} &= \frac{1}{1 - \left(-x^2 - x\right)}\textrm{ which is the closed form of a geometric series} \\ &= \sum_{n = 0}^{\infty}(-x^2 - x)^n\textrm{ provided }|-x^2 - x| < 1 \end{align*}

6. ## Re: Taylor Series Question

Just a little precisation: it was requested the Taylor expansion of $f(x)=\frac{1}{1+x+x^{2}}$ around x=0 [in that case we properly have a McLaurin extension...], so that the result must be in the form...

$\frac{1}{1+x+x^{2}} = \sum_{n=0}^{\infty} a_{n}\ x^{n}$ (1)

Kind regards

$\chi$ $\sigma$

7. ## Re: Taylor Series Question

$f(x) = \frac{1}{x^2+x+1} = \frac{1}{x^2+x+1}\cdot \frac{1-x}{1-x}$
$= \frac{1-x}{1-x^3} = (1-x)(1+x^3 +x^6+x^9+\dots)$
So if
$f(x) = \sum_{n=0}^{\infty} c_n x^n$
then
$c_{18}=1$
$c_{19}=-1$
$c_{20}=0$

8. ## Re: Taylor Series Question

Originally Posted by chisigma
Here we have the opportunity to use some properties of the Z-Tranform. The function...

$h(z)= \frac{1}{1+z+z^{2}}$ (1)

... can be written as Taylor series $h(z)= \sum_{n=0}^{\infty} a_{n}\ z^{n}$ where the $a_{n}$ satisfy the difference equation...

$a_{n} + a_{n-1} + a_{n-2}=0\ ;\ a_{0}=1\ ,\ a_{1}=0$ (2)

The solutions of the 'characteristic equation' $\alpha^{2}+\alpha+1=0$ are $e^{\pm i\ \frac{\pi}{3}$, so that the equation of (2) is...

$a_{n} = c_{1}\ \cos n\ \frac{\pi}{3} + c_{2}\ \sin n\ \frac{\pi}{3}$ (3)

Taking into account the 'initial conditions' we arrive to write...

$\frac{1}{1+z+z^{2}} = 1 -z^{2} - z^{3} + z^{5} + z^{6} - z^{8} - z^{9} +...$ (4)

Kind regards

$\chi$ $\sigma$
I started with the right way but soon I lost it!... In terms of Zeta-Tranform, difference equation corresponding to the function $h(z)=\frac{1}{1+z+z^{2}}$ is...

$a_{n}+a_{n-1}+a_{n-2}= \delta_{n}$ (1)

... where...

$\delta_{n} =\begin{cases}1 &\text{if }n=0\\ 0 &\text{if } n>0\end{cases}$ (2)

... so that the coefficients of the Taylor series are found symply expanding (1)...

$a_{0}= 1 - a_{-1}-a_{-2}= 1$

$a_{1}= 0 - a_{0}-a_{-1}= -1$

$a_{2}= 0 - a_{1}-a_{0}= 0$

$a_{3}= 0 - a_{2}-a_{1}= 1$

$a_{4}= 0 - a_{3}-a_{2}= -1$

... so that is...

$\frac{1}{1+z+z^{2}}= 1 - z + z^{3} - z^{4} + z^{6}-z^{7}+...$ (3)

... as awkward found...

Kind regards

$\chi$ $\sigma$

9. ## Re: Taylor Series Question

Originally Posted by awkward
$f(x) = \frac{1}{x^2+x+1} = \frac{1}{x^2+x+1}\cdot \frac{1-x}{1-x}$
$= \frac{1-x}{1-x^3} = (1-x)(1+x^3 +x^6+x^9+\dots)$
So if
$f(x) = \sum_{n=0}^{\infty} c_n x^n$
then
$c_{18}=1$
$c_{19}=-1$
$c_{20}=0$
I am puzzled about the part on how the C18 C19 C20 became 1, -1, 0 since Cn = (1-x)(x^2)^n, so since x = 0, arent C18 C19 and C20 all zero instead?

10. ## Re: Taylor Series Question

Originally Posted by hxhx
I am puzzled about the part on how the C18 C19 C20 became 1, -1, 0 since Cn = (1-x)(x^2)^n, so since x = 0, arent C18 C19 and C20 all zero instead?
Wrong.

What is the expansion of $(1-x) (\dots x^{15} + x^{18} + x^{21} \dots)$?
That is part of the series.

11. ## Re: Taylor Series Question

Originally Posted by awkward
Wrong.

What is the expansion of $(1-x) (\dots x^{15} + x^{18} + x^{21} \dots)$?
That is part of the series.
so the Cn stands for the coefficient of the x^n at nth position of the series? so Cn is not a general form of the coefficent of x^n?

12. ## Re: Taylor Series Question

$f(x) = \sum_{n=0}^{\infty} c_n x^n$

So $c_n$ is the coefficient of $x^n$ in the series for f(x).

But the problem did not ask for a general form of the coefficient, it simply asked for the value of
$c_{18} + c_{19} + c_{20}$,
which can be found without necessarily finding the general coefficient.

13. ## Re: Taylor Series Question

Originally Posted by awkward

$f(x) = \sum_{n=0}^{\infty} c_n x^n$

So $c_n$ is the coefficient of $x^n$ in the series for f(x).

But the problem did not ask for a general form of the coefficient, it simply asked for the value of
$c_{18} + c_{19} + c_{20}$,
which can be found without necessarily finding the general coefficient.
So what is the general form of the coefficient in this case?

14. ## Re: Taylor Series Question

Try multiplying out the first few terms of

$(1-x)(1+x^3 +x^6+x^9+\dots)$

and see if there is a pattern.