0
$\begingroup$

If I have a term in the following form:

$$ 2\frac{(ak + bk - ab)}{(a^2+b^2+k^2)} $$

is it possible to rearrange it into a term like this?

$$ 2*f(a, k, b) + f(a, k, b)^2 $$

f can by any type of function with a, k and b.

$\endgroup$
0
$\begingroup$

Yes it is possible.

Define $ c = 2 \frac{ ak+bk-ab }{a^2+b^2+k^2} $. You're looking for a function $f$ that satisfies the equation:

$f^2 + 2f = c$

Then: just solve the quadratic equation $f^2 + 2 f - c = 0$. There are two solutions to this quadratic equation:

$f=-\sqrt{c+1}-1 = -\sqrt{2 \frac{ ak+bk-ab }{a^2+b^2+k^2}+1}-1 $

and

$f=\sqrt{c+1}+1=\sqrt{2 \frac{ ak+bk-ab }{a^2+b^2+k^2}+1}+1$

So there are two ways in which you can do this. Note that sometimes this function will be complex valued (depending on your values of $a$, $b$ and $k$).

$\endgroup$

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.