341This is a well known problem from USAMO in 70's
http://www.goiit.com/posts/list/algebra-polynomial-prob-39369.htm
1)find number of real roots of \sum_{r=0}^{r=2n}{}x^r/r! =0 ,(n\epsilon N)
2)Find the number of cubic polynomial p(x) with integral coefficients such that for three distinct integers a,b and c, p(a)=b,p(b)=c and p(c)=a
source TMH
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19 Answers
341
21i have got another solution (simple?) for the 1st one
we have f(x) = f'(x) + x2n/ 2n!
clearly f(x) has a minimum
let z corresponds to the minimum value,z≠0
so f(z)= f'(z) + z2n/2n! f'(z)= 0
or f(z)>0 no solution
341In fact i just recalled reading recently that IMO 2006 problem #5 was based on this problem. You can read it up at
www.artofproblemsolving.com/Forum/viewtopic.php?p=572821&sid=97614cee98271130e32a440de98d4cfb#p572821
341That would be true if P(x)≥x for all x like say if P(x) = x2-x+1
As a counterexample if P(x) = x2-3x+2, then
P(2) = 0<2.
But P(P(2)) = P(0) = 2>0=P(2).
edit: in the post #9, the last line should read S2n>e-y>0
21#12 correction P(a)-P(b) is divisible by a-b where P(x) is a polynomial with integer co efficients
21just took the concept P(a)-P(b)=(a-b)k from there..(where P(x) is polynomial)
here goes the proof
P(a)-P(b)= (a-b)L=(b-c)
P(b)-P(c)=(b-c)M=(c-a) (L,M,N are positive integer)
P(c)-P(a)=(c-a)N=(a-b)
we have LMN=1 possible if and only if L=M=N=1
so a-b=b-c or a+c=2b ----1)
and a-b=c-a or b+c=2a------2)
from 1) and 2) we have a=b=c ,a contradiction so proved
@prophet sir by the way sir what was wrong in my proof #7
341Here's a solution at jee level:
As you said for x≥0, the expression is +ve
If x<0 then let x=-y so that y>0. Then we have
S = 1 - y + \frac{y^2}{2!}-...+\frac{y^{2n}}{2n!}
If 0<y<1 then we have
S = (1 - y) + \left(\frac{y^2}{2!}-\frac{y^3}{3!} \right)...+\frac{y^{2n}}{2n!}>0
If y≥2n, then
S = 1 + \left(\frac{y^2}{2!}-y \right)...+\left(\frac{y^{2n}}{2n!}-\frac{y^{2n-1}}{(2n-1)!}\right)>0
If 1<y<2n, then let
S_{2n}=1-y+\frac{y^2}{2!}-...+\frac{y^{2n}}{2n!}
S_{2n+1}=S_{2n} - \frac{y^{2n+1}}{(2n+1)!}<S_{2n}
Also
S_{2n+2}=S_{2n} + \frac{y^{2n+2}}{(2n+2)!} - \frac{y^{2n+1}}{(2n+1)!}<S_{2n}
This means we have for 1<y<2n
S_{2n}>S_k \forall k>2n and hence S_{2n}>e^{2y}>0
Hence no real roots
212)we can assume a>b>c without any loss of generality (as they are distinct)
given p(a)=b
or, p(a)<a
or, p(p(a))<p(a)<a
or, p(p(p(a)))<p(p(a))<p(a)<a
given p(a)=b p(b)=c p(c)=a
so p(p(p(a)))=a which is a contradiction so no such polynomial exists
21here goes the first one...
clearly no solution for x>0
(at x=0 value of the function is 1, thereafter it increases only(monotonic))
in order to consider the case when x<0 let us apply theorem of remainder of taylor formula
acc. to this
e^x=\sum_{r=0}^{r=2n}{}x^r/r! + R(n)
where R(n)= [x(2n+1)/(2n+1)!]ekx where 0<k<1
now for all x<0 R(n) is negative
so \sum_{r=0}^{r=2n}{}x^r/r! =ex-R(n)
or \sum_{r=0}^{r=2n}{}x^r/r! >0
for all x<0
so no solution
1to be frank , i dont have a solid proof for both
for the second one i think only function satisfying f(f(f(x)))=x....
is f(x)=xi know its very crude..