Partial fraction of $\frac 1{x^6+1}$

Question

Can someone please help me find the partial fraction of $$1\over{x^6+1}$$ ?

I know the general method of how to find the partial fraction of functions but this seems a special case to me..

Answer 1

We have $$\begin{align}x^6+1&=(x^2)^3+1\\&=(x^2+1)(x^4-x^2+1)\\&=(x^2+1)(x^4+2x^2+1-3x^2)\\&=(x^2+1)((x^2+1)^2-(\sqrt 3\ x)^2)\\&=(x^2+1)(x^2-\sqrt 3\ x+1)(x^2+\sqrt 3\ x+1)\end{align}$$

Now setting $$\frac{1}{x^6+1}=\frac{a}{x^2+1}+\frac{bx^2+c}{x^4-x^2+1}$$ will give you $a=\frac 13,b=-\frac 13,c=\frac 23.$

Then, set $$\frac{-\frac 13x^2+\frac 23}{x^4-x^2+1}=\frac{dx+e}{x^2-\sqrt 3\ x+1}+\frac{fx+g}{x^2+\sqrt 3\ x+1}$$ to get $d,e,f,g$.

Answer 2

For starters, write $x^6+1 = (x^2+1)(x^4-x^2+1)$. If you are allowing complex roots you can rewrite $x^2+1 = (x-i)(x+i)$. Then you can continue to factor $x^4-x^2+1$.

The way I came up with this initial factorization is by realizing that $\pm i$ is a root of $x^6+1$, which means $(x-i)$ and $(x+i)$ divide $x^6+1$, which means so does $(x-i)(x+i) = x^2+1$. Then I used polynomial division to divide $x^6+1$ by $x^2+1$ which left me with the quantity $x^4-x^2+1$. Hence, $x^6+1 = (x^2+1)(x^4-x^2+1)$. You can find the roots of $x^4-x^2+1$ by treating it as a quadratic in terms of $x^2$ (that is, make the substitution $x^2 = y$ and solving the quadratic $y^2-y+1=0$)

Answer 3

\begin{align} \frac 1{x^6+1}&=\frac 1{(x^2)^3+1}\\ &=\frac 1{((x^2)1+)((x^2)^2-(x^2)+1)} \\ &= \frac 1{(x^2+1)(x^4-x^2+1)} \\ &= \frac {Ax+B}{x^2+1}+\frac{Cx+D}{x^4-x^2+1} \end{align} So $$1=(x^4-x^2+1)(Ax+B)+(x^2+1)(Cx+D)$$ At $x=0$, the equation becomes $$1=B+D$$ At this point, equating the coefficients is the only way to go.

Answer 4

One way to do it is is to note that the roots of this polynomial are the six complex sixth roots of $-1$, that is, $(-1)^{1/6}=\left (e^{\pi i + 2 k \pi i} \right )^{1/6}= e^{(2k+1)\pi i/6}$ for $k=0,1,\dots,5$. So it is

$$(x-e^{\pi i/6})(x-e^{3 \pi i/6})(x-e^{5 \pi i/6})(x-e^{7 \pi i/6})(x-e^{9 \pi i/6})(x-e^{11 \pi i/6}).$$

This is already a factorization over the complex numbers, which for certain purposes might be what you want. But if you want a factorization over the real numbers instead, notice that these roots come in conjugate pairs. Note that

$$(x-z)(x-\overline{z})=x^2-x(\overline{z}+z)+|z|^2=x^2-2Re(z)x+|z|^2.$$

So by grouping the conjugate pairs of roots, you have

$$(x^2-2\cos(\pi/6)x+1)(x^2-2\cos(3 \pi/6)x+1)(x^2-2\cos(5 \pi/6)x+1).$$

Now evaluate the trig functions. Then you partial fraction expand like usual; so you should get

$$\frac{1}{x^6+1}=\frac{Ax+B}{x^2-\sqrt{3}x+1}+\frac{Cx+D}{x^2+1}+\frac{Ex+F}{x^2+\sqrt{3}x+1}$$

for some real numbers $A,B,\dots,F$.

Answer 5

Let's grant for one moment that complex numbers are legal according to the rules of your game. Then you'd know that the zeros of $x^6+1$ are $\omega_k = e^{2\pi i k/12}$ for $k=1,3,5,7,9,11$. Since all the factors $x-\omega_k$ are simple, there is a general pattern $$ {1\over x^6+1} \;=\; {f'(\omega_1)\over x-\omega_1} + \ldots + {f'(\omega_{11})\over x-\omega_{11}} $$ The derivative $f'(x)=6x^5$ is easy to compute and evaluate. Then if your groundrules demand only real numbers, you combine pairs of terms...

Answer 6

It might be easier to use HET (Heaviside Expansion Theorem): $f(x)={N(x)\over D(x)}=\sum_k^n\frac{a_k}{x-x_k}$ for distinct root $x_k$, so $a_k=\sum_k^n{N(x)\over{D'(x_k)}}$. Let $y=x^2$, so $$ {1\over x^6+1}={1\over y^3+1}={{a}\over{y+1}}+{{a_1}\over{y-y_1}}+{{a_2}\over{y-y_2}}={(a_1+a_2)y+(-a_1y_2-a_2y_1)\over(y-y_1)(y-y_2)}={by+c\over(y-y_1)(y-y_2)}$$ where $y_k={1\pm\sqrt{3}\imath\over2}=e^{\pm{i\pi\over3}},\;k=1,2$. Then just apply HET for derivative of $(y^3+1)$, i.e., $a={1\over(y^3+1)'|_{y=-1}}={1\over3}$, and so on to get $a_k=-{1\over3}y_k$ or $b=a_1+a_2=-{1\over3}(y_1+y_2)=-{1\over3}$ and $c={2\over3}y_1y_2={2\over3}$. One might noticed that it is a perfect for Vieta's formulas for $y_1y_2=y_1+y_2=1$.