0
$\begingroup$

I am trying to solve it with "partial fractions"

$$\frac {x^2}{x^4+1}=\frac{x^2}{(x^2+x\sqrt{2}+1)(x^2-x\sqrt{2}+1)}=\frac{Ax+B}{(x^2+x\sqrt{2}+1)}+\frac{Cx+D}{(x^2-x\sqrt{2}+1)}$$

and I get the following system of equations:

$A+C=0$

$-\sqrt{2}A+B+\sqrt{2}C+D=1$

$A-\sqrt{2}B+C+\sqrt{2}D=0$

$B+D=0$

How can I find $A,B,C,D$?

$\endgroup$
5
  • 4
    $\begingroup$ It's a pretty standard system of linear equations, you could use, for example en.wikipedia.org/wiki/Gaussian_elimination $\endgroup$
    – Matti P.
    Apr 8, 2020 at 10:05
  • $\begingroup$ @MattiP. Is it possible to solve every system of equations with Gaussian elimination method? I am not familiar with solving system of equations $\endgroup$
    – Aligator
    Apr 8, 2020 at 10:08
  • 2
    $\begingroup$ How did you get $A+C=1$? Multiplying both sides by $x$ and then taking $x\to \infty$, you obtain $A+C=0$. $\endgroup$ Apr 8, 2020 at 10:10
  • $\begingroup$ @Soheil Every linear system of equations, yes. $\endgroup$ Apr 8, 2020 at 10:10
  • 1
    $\begingroup$ @WETutorialSchool thanks it is typo i fixed $\endgroup$
    – Aligator
    Apr 8, 2020 at 10:11

3 Answers 3

11
$\begingroup$

Observe that the integrand is an even fraction, so it is invariant when we change $x$ to $-x$: $$\frac{Ax+B}{(x^2+x\sqrt{2}+1)}+\frac{Cx+D}{(x^2-x\sqrt{2}+1)}=\frac{-Ax+B}{(x^2-x\sqrt{2}+1)}+\frac{-Cx+D}{(x^2+x\sqrt{2}+1)}.$$ This implies that $\;C=-A,\; D=B$, and you have only a linear system in two unknowns.

$\endgroup$
1
  • $\begingroup$ Nice trick :) ${}$ $\endgroup$
    – Botond
    Apr 8, 2020 at 10:30
2
$\begingroup$

From $A+C=0$ and $A-B\sqrt2+C+D\sqrt2=0$ we have $$-B\sqrt2+D\sqrt2=0$$ which with $B+D=0$, gives $B=D=0$.

Now from $A+C=0$ and $-A\sqrt2+B+C\sqrt2+D=1$, noting that $B=D=0$, we have $$A=-C\implies C\sqrt2+C\sqrt{2}=1\implies C=\frac{1}{2\sqrt2}$$ and $$A=-\frac{1}{2\sqrt2}$$

$\endgroup$
2
$\begingroup$

You can solve the (easy) $4\times4$ system of linear equations, but notice that

$$(x^2+\sqrt2 x+1)-(x^2-\sqrt2 x+1)=2\sqrt2x$$ so that

$$\frac x{2\sqrt2}\frac{(x^2+\sqrt2 x+1)-(x^2-\sqrt2 x+1)}{(x^2+\sqrt2 x+1)(x^2-\sqrt2 x+1)}$$ does the trick.


Notice that this will work for any positive power of $x$ at the numerator, hence any polynomial. Anyway, the constant term requires an extra twist:

$$(x^2+\sqrt2 x+1)+(x^2-\sqrt2 x+1)=2x^2+2,$$ from which you can cancel out $2x^2$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.