The prompt is to find the 8th derivative of the function f(x) defined as,

$$f(x) = \frac{3}{1+x-2x^2}$$ $$f^{(8)}(0) = ?$$

To find the maclaurin series, I proceeded by finding the derivatives of the function at 0 as follows,

$$f^{(1)}(x) = \frac{ -3 + 12x }{ (1 + x -2x^2)^2 } $$ $$f^{(2)}(x) = \frac{18(4x^2-2x+1)}{(-2x^2+x+1)^3} $$

such that, $$f(0) = 3$$ $$f^{(1)}(0) = -3$$ $$f^{(2)}(0) = 18$$ $$f^{(3)}(0) = -90$$

This makes the maclaurin series, $$f(x) = 3 - 3x + \frac{18x^2}{3!} - \frac{90x^3}{4!} + ...$$

I understand from the series, we have to have $(-1)^n$ since the negative sign is alternating, also in the denominator we have n!

$$\sum_{n=0}^{\infty} \frac{(-1)^n(x)^n}{n!}$$ but it is incomplete, since I don't see the pattern, I would appreciate I someone can help with completing the series hence finding the 8th derivative.

  • 2
    $\begingroup$ $$ \frac{3}{1+x-2x^2} = \frac{2}{2x+1} - \frac{1}{x-1} $$ and $\sum x^n = \frac{1}{1-x} ,\,\, x\in(-1,1)$ $\endgroup$ – Dando18 Jun 19 '17 at 0:50

Here's a hint.

Factor the denominator, and expand via "partial fractions". Then find the series for each term, and sum.


$$f (x)=\frac {1}{1-x}+\frac {2}{1+2x} $$

$$=\Bigl(1+x+x^2+...x^8\Bigr)+$$ $$2 \Bigl (1-(2x)+(2x)^2-....+(2x)^8\Bigr)+x^8\epsilon (x) $$


$$f^{(8)}(0)=8!(1+2^9)=8!.513 $$

put $t=-2x $ then $1/(1+2x)=1/(1-t)=1+t+t^2+... $.

  • $\begingroup$ I understand $\frac{1}{1-x} = \sum x^n$ but how did you expand $\frac{2}{1+2x}$? Would appreciate some more detail, thanks. $\endgroup$ – Archetype2142 Jun 19 '17 at 0:57
  • $\begingroup$ should be $\frac{-1}{1-x}$, I believe $\endgroup$ – Dando18 Jun 19 '17 at 0:58
  • 1
    $\begingroup$ @Ritwick Note that $$\frac2{1+2x}=\frac2{1-(-2x)}=2\sum_{n=0}^\infty(-2x)^n$$ $\endgroup$ – Simply Beautiful Art Jun 19 '17 at 0:59

Suppose we want to do this without expanding in partial fractions. Then we can just perform a long division of the fraction to obtain the coefficient of $x^8$. But a faster division algorithm is Newton-Raphson division, $n$ steps will yield the coefficients up to $x^{2^n -1}$ This method can be generalized such that it also works for arbitrary closed form functions, it is used in some computer algebra programs.

Newton-Raphson division for polynomials works in an analogous way as it does for real numbers. In the latter case this boils down to finding the reciprocal of some number $y$ by solving the equation $$\frac{1}{x} - y = 0$$ using the Newton-Raphson method, which yields a sequence of approximants $x_n$ satisfying the recurrence equation:

$$x_{n+1} = 2 x_{n} - y x_{n}^2\tag{1}$$

Then if $y$ is a polynomial and we take $x_0$ equal to the reciprocal of the constant term of that polynomial then iterating with Eq. (1) will yield successive approximations for the series expansion of the reciprocal of the polynomial, such that the number of correct terms doubles after each iteration.

If we put $p(x) = 1 + x - 2 x^2$ then the successive approximants $q_n(x)$ of the reciprocal of $p(x)$ satisfy the recurrence equation

$$q_{n+1}(x) = 2 q_n(x) - p(x) q_n(x)^2 \bmod x^{2^{n+1}}$$

where $q_0(x) = 1$. Here we compute modulo $x^{2^{n+1}}$ because only the coefficients of $x^r$ for $r<2^{n+1}$ will be correct. We then find:

$$ \begin{split} q_1(x) &= &1-x\\ q_2(x) &= -&5 x^3+3 x^2-x+1\\ q_3(x) &= -&85 x^7+43 x^6-21 x^5+11 x^4-5 x^3+3 x^2-x+1 \\ q_4(x) &= -&21845 x^{15}+10923 x^{14}-5461 x^{13}+2731 x^{12}-1365 x^{11}+\\ & &683 x^{10}-341 x^9+171 x^8-85 x^7+43 x^6-21 x^5+11 x^4-5 x^3+3 x^2-x+1 \end{split} $$

The 8th derivative of $\dfrac{3}{p(x)}$ is thus equal to $8!\times 3\times 171$


Just another way.

For the most general case of finding $f^{(n)}(0) = ?$, you received nice answers. Since I am lazy and since $n=8$ is quite small, I used long division to get $$\frac{3}{1+x-2x^2}=3-3 x+9 x^2-15 x^3+33 x^4-63 x^5+129 x^6-255 x^7+513 x^8+\cdots$$ Then $$f^{(8)}(0) =513\times 8!$$


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