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Rational roots

melese

Member
Feb 24, 2012
27
(BGR,1989) Prove that for any integer $n>1$ the equation $\displaystyle \frac{x^n}{n!}+\frac{x^{n-1}}{(n-1)!}\cdots+\frac{x^2}{2!}+\frac{x^1}{1!}+1=0$ has no rational roots.
 

Albert

Well-known member
Jan 25, 2013
1,225
(BGR,1989) Prove that for any integer $n>1$ the equation $\displaystyle \frac{x^n}{n!}+\frac{x^{n-1}}{(n-1)!}\cdots+\frac{x^2}{2!}+\frac{x^1}{1!}+1=0$ has no rational roots.

$\displaystyle \frac{x^n}{n!}+\frac{x^{n-1}}{(n-1)!}\cdots+\frac{x^2}{2!}+\frac{x^1}{1!}+1=0----(*)$
if n=2 then we have :$x^2+2x+2=0$ no real solution
if m is a solution of original equation then m<0
multiply both sides with n ! we obtain :
$x^n+nx^{n-1}+-----+n! x+ n!=0$
using "the rational zero theorem"
if the original equation has a rational solution m<0 then n! must be a multiple of it(m is a negative integer)
replacing x with any negative factor of n! to (*)will not be zero ,so no rational root exists
in fact :
---+$\displaystyle \frac{x^n}{n!}+\frac{x^{n-1}}{(n-1)!}\cdots+\frac{x^2}{2!}+\frac{x^1}{1!}+1=e^x$
[FONT=&#32048](Maclaurin expasion of $e^x$)[/FONT]
 
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melese

Member
Feb 24, 2012
27
I agreed with you up to
replacing x with any negative factor of n! to (*)will not be zero

This part is a little vauge to me. To see what I mean, what if my orginal question was to show that there are no integer solutions? - Then your step appears hasty.

መለሰ
 

Albert

Well-known member
Jan 25, 2013
1,225
$\displaystyle \frac{x^n}{n!}+\frac{x^{n-1}}{(n-1)!}\cdots+\frac{x^2}{2!}+\frac{x^1}{1!}+1=0----(*)$

if n is even and m (a negative integer ) is a root then (*) becomes
$\displaystyle \frac{m^n}{n!}+\frac{m^{n-2}}{(n-2)!}\cdots+\frac{m^2}{2!}+1$
$- \dfrac{k^{n-1}}{(n-1)!}- \dfrac{k^{n-3}}{(n-3)!}\cdots-\dfrac{k}{1!}$
will not be zero,(the calculation is very tedious)
here k=-m is a positive integer
likewise if n is odd then (*) becomes
$\displaystyle -\frac{k^n}{n!}-\frac{k^{n-2}}{(n-2)!}\cdots-\frac{k^2}{2!}+1$
$+ \dfrac{m^{n-1}}{(n-1)!}+ \dfrac{m^{n-3}}{(n-3)!}\cdots+\dfrac{m}{1!}$
also will not be zero
so there is no integer root for original equation
 
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