Does there exist a function $f_{\Box,\Box}(\Box)$ making the formula $a + (b \oplus c) = (f_{b,c}(a)+b) \oplus (f_{c,b}(a)+c)$ true?

Question

Let $a$ and $b$ denote the resistances of two resistors. If they're put in series, the total resistance is $a+b$. If they're put in parallel, the total resistance is $$a \oplus b := \frac{1}{\frac{1}{a}+\frac{1}{b}} = \frac{ab}{a+b}.$$

I suspect there's a formula describing how $+$ "distributes over" $\oplus$.

It should be of the form

$$a + (b \oplus c) = (f_{b,c}(a)+b) \oplus (f_{c,b}(a)+c)$$

for an appropriate choice of $f$. I haven't been able to find an $f$ that works, however.

Question. Does there exist an $f$ making the above formula true? If not, why not?

Remark. Unpacking the definitions, we're looking for a function $f$ such that

$$(ab+ac+bc)(f_{b,c}(a)+b + f_{c,b}(a)+c)$$

and

$$(b+c)(f_{b,c}(a)+b) (f_{c,b}(a)+c)$$

are equal.

If you assume $f_{b,c}(a)=f_{c,b}(a)$, then the equation in question is quadratic in $f_{b,c}(a)$, has negative zero order term, and there exists a positive square root. — daw, Jan 10 '20 at 09:33

score 5 · Accepted Answer · answered Jan 10 '20 at 09:47

Let me assume that $b\ge0,c\ge0, b+c>0$. I am looking for $f$ satisfying $f_{b,c}(a) = f_{c,b}(a)$. The equation in the OP is then equivalent to $$ (b+c) f_{bc}(a)^2 + ((b+c)^2-2ab-2bc-2ca)f_{bc}(a)- (b+c)(ab+bc+ca)=0. $$ Setting $d:=ab+bc+ca$ and dividing by $b+c$ this is equivalent to $$ f_{bc}(a)^2 + (b+c -\frac{2d}{b+c}) f_{bc}(a)- d=0. $$ The roots of this equation are $$\begin{split} f_{bc}(a) &= -\frac{ (b+c)^2 - 2d}{2(b+c)} \pm \frac1{b+c} \sqrt{ \frac14((b+c)^2 - 2d)^2+d(b+c)^2}\\ &=-\frac{ (b+c)^2 - 2d \pm\sqrt{ (b+c)^4 +d^2}}{2(b+c)}, \end{split} $$ one of them is positive, which gives you a formula for $f$.

You can simplify this somewhat further: you can also write it as $a + (b \oplus c) - \frac{b+c}{2} + \frac12 \sqrt{(b+c)^2 + (a + (b \oplus c))^2}$, and then if we agree that $0 \oplus 0 = 0$ you have a definition for all non-negative $a, b, c$. — Mees de Vries, Jan 10 '20 at 10:20

score 4 · Answer 2 · answered Jan 12 '20 at 00:02

There are many different possible definitions of $f_{b,c}(a)$. One approach is based on the identity $$(kb)\oplus (kc) = k(b\oplus c)$$

Since we would like to get $$a + \left(b\oplus c\right)=\left(f_{b,c}(a) + b\right)\oplus\left(f_{c,b}(a) + c\right)$$ it would be nice if the right-hand side can transformed into to contain $\left(b\oplus c\right)$. It can be rewritten as $$\left(f_{b,c}(a) + b\right)\oplus\left(f_{c,b}(a) + c\right) = \left(\left(\frac{f_{b,c}(a)}{b}+1\right)\cdot b\right)\oplus \left(\left(\frac{f_{c,b}(a)}{c}+1\right)\cdot c\right)$$ Equating $\frac{f_{b,c}(a)}{b} = \frac{f_{c,b}(a)}{c}$ and setting both equal to $g(a,b,c)$ yields $$a + \left(b\oplus c\right) = \left(g(a,b,c)+1\right)\left(b\oplus c\right)$$ Thus, we can set $g(a,b,c)=a/\left(b\oplus c\right)$, we get $$f_{b,c}(a)=\frac{ab}{b\oplus c}$$ and the equality $$a + \left(b\oplus c\right)=\left(\frac{ab}{b\oplus c} + b\right)\oplus\left(\frac{ac}{b\oplus c} + c\right)$$ Alternatively, we can also expand $\left(b\oplus c\right)=\frac{bc}{b+c}$ and get $$f_{b,c}(a)=\left(1+\frac{b}{c}\right)a$$

Cesareo · Answer 3 · 2020-01-12T01:20:28.773

3

Defining

$$ f_{b,c}(a) = 2a+b-\frac{2b^2}{b+c}\ \ \text{or}\ \ \ f_{c,b}(a) = 2a+c-\frac{2c^2}{b+c} $$

with $f_{b,c}(a)\ne f_{c,b}(a)$, we have

$$ a + (b \oplus c) = (f_{b,c}(a)+b) \oplus (f_{c,b}(a)+c) $$

edited Jan 12 '20 at 01:20

answered Jan 11 '20 at 21:54

Cesareo

36,341

It seems the formulas are not correct. Consider $a=b=c=1$ which yields $f_{1,1}(1)=(-1/3)$. On the left-hand side, we get $1+\frac{1}{2}$, while the right-hand side is equal to $1/3$. – Peter Košinár Jan 12 '20 at 00:21
1

Yes. A problem already fixed. Thanks. – Cesareo Jan 12 '20 at 01:21

Does there exist a function $f_{\Box,\Box}(\Box)$ making the formula $a + (b \oplus c) = (f_{b,c}(a)+b) \oplus (f_{c,b}(a)+c)$ true?

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