Is it possible to reformulate this expression to only list $x$ and $y$ once using common math functions? The ranges of $x$ and $y$ are both $0$ to $1$ inclusive if that helps.
Common being those typically found on a scientific calculator.
$\begingroup$
$\endgroup$
1
-
6$\begingroup$ Use the reciprocal $\frac{x+y}{x}=1+\frac yx$ $\endgroup$xxxx036– xxxx0362021-10-13 23:29:40 +00:00Commented Oct 13, 2021 at 23:29
Add a comment
|
2 Answers
$\begingroup$
$\endgroup$
2
This is slightly more complicated, but $\frac{1}{1+\frac{y}{x}}$ is one possibility.
-
1$\begingroup$ Not to be a pedant, but I would argue that this is not technically a reformulation, as it doesn't include the possibility of $x = 0$ (which is explicitly included in the question). $\endgroup$Theo Bendit– Theo Bendit2021-10-13 23:38:01 +00:00Commented Oct 13, 2021 at 23:38
-
$\begingroup$ if x = 0 then there is no need for reformulaton. $\endgroup$Aaa Lol_dude– Aaa Lol_dude2021-10-13 23:40:19 +00:00Commented Oct 13, 2021 at 23:40
$\begingroup$
$\endgroup$
1
First, if $x=0$ then $$\frac{x}{x+y}=\frac{0}{y}=0$$ provided $y\neq 0$. As $x,y\in[0,1]$, it follows that we exclude the case $x=y=0$. More specially, if $x=0$ then $y\neq 0$ and if $y=0$ then $x\neq 0$. Next, if $x+y\neq 0$ then we may let $$A=\frac{x}{x+y}.$$ Taking the reciprocal of $A$ gives
$$\frac{1}{A}=\frac{x+y}{x}=1+\frac{y}{x}.$$
Then since the reciprocal of the reciprocal is our original function we have
$$A=\frac{1}{\left(\frac{1}{A}\right)}=\frac{1}{1+\frac{y}{x}}.$$ By our analysis above, we see that this is true provided $x\neq 0$ and $x \neq -y$. Since $x,y\in[0,1]$, the second condition is only valid when $x=y=0$ which we excluded.
-
1$\begingroup$ Indeed it's a simple check for near zero within a tolerance to avoid floating point underflow. :) $\endgroup$Chris– Chris2021-10-31 14:29:52 +00:00Commented Oct 31, 2021 at 14:29