# Calculate an abstract algebra “sum” $\frac 12 * \frac13 *…*\frac 1{1000}$.

Let $x*y= \frac {x+y}{1+xy}$, $x,y\in(-1,1)$. Calculate $\frac 12 * \frac13 *...*\frac 1{1000}$.

My attempt:

First I tried to find some inductive formula but I get something like this:

$\frac 12*\frac 13=\frac 57$

$\frac 57*\frac 14=\frac 9{11}$

$\frac 9{11}*\frac 15=\frac 78$ didn't got anywhere...

Then I thought maybe if I let $G=(-1,1)$ then $(G,*)$ is an abelian group. That means I have to find another group that can be an isomorphism with this one and try to calculate using that composition. So I found that: $$f:(0,\infty)\to (-1,1),f(x)=\frac {x-1}{x+1}$$

is an isomorphism with $G$ from $(\mathbb{R},\times)$ to $(G,*).$

Then: $f(xy)=f(x)*f(y)$. So: $$f^{-1}:(-1,1)\to(0,\infty), f^{-1}(x)=-\frac {x+1}{x-1}.$$

Is is true that $f^{-1}(x*y)=f^{-1}(x)f^{-1}(y)?$ If for $f$ happens this then it happens for $f^{-1}$ too? If this is true then I think I solved my exercise because:

$$f^{-1}\left(\frac 12*\frac 13*\ldots*\frac 1{1000}\right)=f\left(\frac {1}{2}\right) f\left(\frac{1}{3}\right)\ldots f\left(\frac{1}{1000}\right)= \frac {\frac 32}{-\frac 12}\frac {\frac 43}{-\frac 23}...\frac {\frac {1000}{999}}{-\frac {998}{999}}\frac {\frac {1001}{1000}}{-\frac {999}{1000}}=500\cdot 1001=500500.$$

then $\frac 12*\frac 13*...*\frac 1{1000}=f(500500)=\frac {500499}{500501}$

• $f$ is an isomorphism, and so its inverse is a homomorphism. Well done! – lhf May 18 '18 at 12:08
• @lhf Oh, okay. Thank you so much! – C. Cristi May 18 '18 at 12:11
• @Jean-ClaudeArbaut I'm only a $12th$ grader so I haven't learned the hyperbolic functions yet but my first try of finding the function was something like $(\arctan(x)+\arctan(y))(\arctan(x)-\arctan(y))=\frac {x+y}{1+xy}\frac {x-y}{1-xy}$ if $x,y$ are actually some tangents – C. Cristi May 18 '18 at 12:12
• Check the sign of the last product. – Darth Geek May 18 '18 at 12:21
• I removed my initial comment because I don't think it helps much here. But for those interested, it may be noted that $\tanh u *\tanh v=\tanh (u+v)$, where $\tanh$ is the hyperbolic tangent. – Jean-Claude Arbaut May 18 '18 at 12:25

Yes: if $f$ is a bijective group homomorphism, then $f^{-1}$ is a homomorphism.