-1

How to find the root of $$(1-x)^{a} - {a \choose 2} x^2=0. $$ Here, $a$ is an integer. Or can we at least have an approximation of it ?

flag
-1 because it looks like homework : a question and no information about where the problem comes from or leads to – Julien Puydt Jun 25 2011 at 15:22
Maybe we can try Newton's method or Newton-Raphson method, right ? Is there any other possible ways ? – nana Jun 25 2011 at 15:27
Hey, Snark, Thanks for your time and attention. Maybe I should do some research online first. It is not a homework problem. It is from a project I am working on. It takes time to explain the whole stuff. If anyone think it is inappropriate to post it here, please delete it. Anyway, I am sorry for that. – nana Jun 25 2011 at 15:30
3 
 Which root? There are $a$ roots (except $a=2$), and at least if $2|a$ there can't even be a unique real root. Anyway there's probably no closed form solution: for $3 \leq a \leq 11$ the Galois group is either the full symmetric group (if $a$ is odd) or the imprimitive subgroup of size $2(a/2)!^2$ (if $a$ is even -- can this be explained?); already not solvable in radicals for $a=5$. – Noam D. Elkies Jun 25 2011 at 15:50
1 
 There is obviously a unique root between 0 and 1, and I am almost willing to make a bet that is the root that is wanted. Any standard numerical method will have no trouble finding this root. Another piece of information is that the root behaves like 0.901201/a when a is large (the factor here is the root of $e^{-y}=y^2/2$). – Michael Renardy Jun 25 2011 at 16:42
show 3 more comments

closed as off topic by Bruce Westbury, Simon Thomas, Felipe Voloch, Andres Caicedo, Gjergji Zaimi Jun 25 2011 at 21:46

Browse other questions tagged or ask your own question.