## this is interesting!!!

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• Let P(x)=ax^3+bx^2+cx+d from R[X] with a 0.If the coeficients of this polinomial are in arithmecical progression, and all of its roots are real then the
Message 1 of 6 , Jun 2 2:07 PM
Let P(x)=ax^3+bx^2+cx+d from R[X] with a<>0.If the coeficients of
this polinomial are in arithmecical progression, and all of its roots
are real then the following relations holds:

arctan(1/x1)+arctan(1/x2)+1/(arctan(x3))=-pi/4.

where x1,x2,x3 are its roots.

Pls help me!!

bye! thank you!
• P(x)=ax^2+bx^2+cx+d. (a,b,c,d) is a arithmetical progression, so P(x)=ax^3+(a+r)x^2+(a+2r)^x+(a+3r) let arctan(1/x1)=m arctan(1/x2)=n arctan(1/x3)=p so
Message 2 of 6 , Jun 3 1:57 AM
P(x)=ax^2+bx^2+cx+d. (a,b,c,d) is a arithmetical progression, so
P(x)=ax^3+(a+r)x^2+(a+2r)^x+(a+3r)

let
arctan(1/x1)=m
arctan(1/x2)=n
arctan(1/x3)=p

so tg(m+n+p)= -1. tg(a+b)=tga+tgb/(1-tga.tgb),
so tg(m+n+p)=x1x2+x1x3+x2x3 - 1/[x1x2x3 - (x1+x2+x3)]= -1

by girard relations:
[(a+2r)/a - 1]/[-(a+3r)/a + (a+r)/a]= -1 ,what is true.

Ricardo Filho
Fortaleza - CE - Brasil
ICQ 23260673

"Procure dividir-se em alguém."
Marcelo Camelo
----- Original Message -----
From: michaelballackachtung
To: mathforfun@yahoogroups.com
Sent: Monday, June 02, 2003 6:07 PM
Subject: [MATH for FUN] this is interesting!!!

Let P(x)=ax^3+bx^2+cx+d from R[X] with a<>0.If the coeficients of
this polinomial are in arithmecical progression, and all of its roots
are real then the following relations holds:

arctan(1/x1)+arctan(1/x2)+1/(arctan(x3))=-pi/4.

where x1,x2,x3 are its roots.
• Well the relation is actually: arctan(1/x1)+arctan(1/x2)+arctan(1/x3)=-pi/4 cause other wise it is impossible. try to write it:
Message 3 of 6 , Jun 3 12:11 PM
Well the relation is actually:
arctan(1/x1)+arctan(1/x2)+arctan(1/x3)=-pi/4
cause other wise it is impossible.
try to write it:
arctan(1/x1)+arctan(1/x2)=-pi/4-arctan(1/x3) and then
apply tan to both members and then use Viete
relationships and the fact that the
coeficients of this polinomial are in arithmecical
progression and u have the problem.

Bye!
Louis.

--- michaelballackachtung
<michaelballackachtung@...> wrote:
> Let P(x)=ax^3+bx^2+cx+d from R[X] with a<>0.If the
> coeficients of
> this polinomial are in arithmecical progression, and
> all of its roots
> are real then the following relations holds:
>
>
>
>
> arctan(1/x1)+arctan(1/x2)+1/(arctan(x3))=-pi/4.
>
>
>
>
> where x1,x2,x3 are its roots.
>
>
>
>
> Pls help me!!
>
>
> bye! thank you!
>
>
>

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• Valeu =] Ricardo Filho Fortaleza - CE - Brasil ICQ 23260673 Procure dividir-se em alguém. Marcelo Camelo ... From: Carlos Assale To:
Message 4 of 6 , Jun 3 4:31 PM
Valeu =]

Ricardo Filho
Fortaleza - CE - Brasil
ICQ 23260673

"Procure dividir-se em alguém."
Marcelo Camelo
----- Original Message -----
From: Carlos Assale
To: mathforfun@yahoogroups.com
Sent: Tuesday, June 03, 2003 8:57 PM
Subject: Re: [MATH for FUN] this is interesting!!!

Dá-lhe patrício!!!!
Carlos

[Non-text portions of this message have been removed]
• Dá-lhe patrício!!!! Carlos
Message 5 of 6 , Jun 3 4:57 PM
Dá-lhe patrício!!!!
Carlos

----- Original Message -----
> P(x)=ax^2+bx^2+cx+d. (a,b,c,d) is a arithmetical progression, so
> P(x)=ax^3+(a+r)x^2+(a+2r)^x+(a+3r)
>
> let
> arctan(1/x1)=m
> arctan(1/x2)=n
> arctan(1/x3)=p
>
> so tg(m+n+p)= -1. tg(a+b)=tga+tgb/(1-tga.tgb),
> so tg(m+n+p)=x1x2+x1x3+x2x3 - 1/[x1x2x3 - (x1+x2+x3)]= -1
>
> by girard relations:
> [(a+2r)/a - 1]/[-(a+3r)/a + (a+r)/a]= -1 ,what is true.
>
>
> Ricardo Filho
> Fortaleza - CE - Brasil
> ICQ 23260673
>
> "Procure dividir-se em alguém."
> Marcelo Camelo
> ----- Original Message -----
> From: michaelballackachtung
> To: mathforfun@yahoogroups.com
> Sent: Monday, June 02, 2003 6:07 PM
> Subject: [MATH for FUN] this is interesting!!!
>
>
> Let P(x)=ax^3+bx^2+cx+d from R[X] with a<>0.If the coeficients of
> this polinomial are in arithmecical progression, and all of its roots
> are real then the following relations holds:
>
>
>
>
> arctan(1/x1)+arctan(1/x2)+1/(arctan(x3))=-pi/4.
>
>
>
>
> where x1,x2,x3 are its roots.
>
>
>
>
> To unsubscribe from this group, send an email to:
> mathforfun-unsubscribe@yahoogroups.com
>
>
>
> Your use of Yahoo! Groups is subject to http://docs.yahoo.com/info/terms/
>
>
>
>
• Let P(x)=ax^3+bx^2+cx+d from R[X] with a 0.If the coeficients of ... roots ... where x1,x2,x3 are its roots. Suggested solution:
Message 6 of 6 , Jun 4 8:37 PM
Let P(x)=ax^3+bx^2+cx+d from R[X] with a<>0.If the coeficients of
> this polinomial are in arithmecical progression, and all of its
roots
> are real then the following relations holds:

> arctan(1/x1)+arctan(1/x2)+1/(arctan(x3))=-pi/4.
where x1,x2,x3 are its roots.

Suggested solution:

arctan(1/x1)+arctan(1/x2)=arctan[(1/x1+1/x2)/(1-1/x1x2)]=

arctan[(x1+x2)/(x1x2-1)]
arctan(1/x1)+arctan(1/x2)+arctan(1/x3)=
arctan[(x1x3+x2x3+x2x1-1)/(x1x2x3-x3-x1-x2)](1)

From the theory of equations:

x1+x2+x3=-b/a

x1x2+x2x3+x1x3=c/a

x1x2x3=-d/a

thus replacing we have in (1):

x1x3+x2x3+x2x1-1=c/a-1=(c-a)/a

x1x2x3-x3-x1-x2=-d/a+b/a=(b-d)/a

But since a,b,c, d are in AP

a=A-3D
b=A-D
c=A+D
d=A+3D

c-a=4D
b-d=-4D

Thus (1) reduces to

arctan[(c-a)/(b-d)]=arctan[4D/-4D]=-pi/4

Note that the solution of arctan must lie within -pi/2 to pi/2

Saurabh
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