Can the product $AB$ be computed using only $+, -,$ and reciprocal operators?

Edit: previous answer was wrong. Posted new answer. Hopefully right this time

1. We can compute and store $A^2$ using $$ \frac{1}{A} - \frac{1}{A+1} = \frac{1}{A^2 + A} $$ We can extract $A^2$ using only $+, -, ^{-1}.$ Similarly we can compute and store $B^2.$

2. Then

$$\frac{1}{A+B-1} - \frac{1}{A+B} = \frac{1}{(A+B)(A+B-1)} = \frac{1}{A^2 + B^2 + 2AB - A - B} $$

where we can extract $2AB,$ again, using only $+, -, ^{-1}$ and the values for $A^2, B^2$ we computed in step $1$ above.

Thanks to joriki, now to get $AB$ from $2AB$, add $\frac{1}{2AB} + \frac{1}{2AB},$ and take the reciprocal.

J.D. and N. S. have shown how to do it if a constant $1$ is allowed. Here's a proof that it can't be done if only $A$ and $B$ can be entered, no constants.

We can show by structural induction that all expressions we can generate change sign if both $A$ and $B$ change sign.

Base case: The two atomic expressions $A$ and $B$ change sign when both $A$ and $B$ change sign.

Induction step: $x+y$ changes sign when both $x$ and $y$ change sign, $x-y$ changes sign when both $x$ and $y$ change sign, and $x^{-1}$ changes sign when $x$ changes sign.

Since $AB$ doesn't change sign when both $A$ and $B$ change sign, it follows that it can't be generated from $A$ and $B$ using only these operations.

user2468's solution ends up requiring quite a lot of operations -- three reciprocals to get $A^2$, then the same to get $B^2$, then another three to get $A^2+B^2+2AB-A-B$. Then to get rid of that pesky factor of 2 you need another three reciprocals, for a total of 12 reciprocal operations (and 31 binary operations).

It's possible to do better by employing some tricks:

• Instead of $(A+B)^2 - A^2 - B^2$, which involves three terms, use $(A+B)^2 - (A-B)^2$, which involves only two terms.
• Instead of needing to divide by a constant near the end of the process, which involves a lot more reciprocals, identify where in the formula you actually introduced the constant, and scale your other constants to match.

Using those tricks, I was able to reduce this down to 6 reciprocal operations and 11 binary operations:

$$ \frac{1}{\frac{1}{A+B-2} - \frac{1}{A+B+2}}-\frac{1}{\frac{1}{A-B-2} - \frac{1}{A-B+2}} = AB $$

I suspect this is probably the most beautiful solution to this challenge.

We have $$\frac{A^2}4=\frac1{\dfrac1{A}-\dfrac1{A+4}}-A\tag1$$ and $$AB=\frac14\,(A+B)^2-\frac14\,(A-B)^2\tag2.$$