dimanche 6 septembre 2009
Inscription à :
Publier les commentaires (Atom)
Qui êtes-vous ?
- zeraoulia rafik
- etudiant universitaires 3èmè anné math et aussi j'étude 2èmè biologie 1er informatique je suis actuellement un enseignant des eleve s terminal science experimentale et de 4ème anné moyenne
1 commentaire:
hello, i have honor to writing for you my article that explaine my new basic(inequality),and all thanks for ur encourage of publisher mathematicien
this
is my article ( new inequality)
In the first I search for true hypothese to solved this
Probleme, we have for n=1,we have x+y=z
For x,y,z€R POSITIF, and
x≤z and,y≤z quel que soit x and y positive number reel
so we can get |x|≤|z|,and |y|≤|z |
so -|z|≤ x≤|z|……1
-|z|≤ y≤|z|…….2
I add the too i thus get
-2α≤x+y≤2α that α=|z| so α≥0…..3
In this cas x+y mayb will 2α-k that k≥0
Or 2α+k , k≤0 ,so
{x+y=2α-k if k≥0 or
{x+y=2α+k if k≤0
So -2α≤2α-k≤2α, so we can get
-3α≤α-k≤α but we have α=|z| ,and z≤ |z| ,z≥0
So z=α-k ,k≥0 , now we find x+y,and z, so
y=α+k
now I find the only hypothese to solve the fermat equation,
for n=1,I put x=α,y= α+k, z= α-k i remplaced with this value in the equation we find ; α-2k=0, so α=2k, and x,y,z after substitution we can find , x=2k,y=k,z=3k , this solution of x+y=z
and now I solve the equation for n=2, so it must be to solved this equation pythagorthien
x²+y²=z²
so I put x=α,y =α+k,and z= α-k after substitution in the equation of second degree we find, α=4k,so
x=4k,y=3k,z=5k, k€R
now I solve the equation with n=3,but its equation 3 degre it diffuclt
to solved,in this cas it must be used the perception
-whene we compare with precident solution for n=1,n=2, we find good idea, the defference of the solution for n=1,and n=2 is constant and equal 2k so,I can find the result and solution for n=3
So x=6k,y=5k,z=7k, but this solution verified by 0 yet,and whene we put k=1,so 6^3+5^3 not equal 7^3, the defference between tow is 2,so 2k, in this cas I propose opened sphere ,such with r=1/n, n defferent for 0, and 1/n € ]0,1[ this interval ouvert belong in v, the voisinage of sphere , and we can doing the subdivision of ]0,1[, in lot area
Such; it was find r'=j*1/n, but with observation I look the number
0.1 verified the equation x^3+y^3=z^3 and whene we substate in the general solution for n=3, we find 0.6^3+0.5^3=0.7 ^3
And its equal 0.34 in the tow tom of equation,in add to i chose
k= r'=j*1=0.1/n.
-for n=4 we find x=8k,y=7k,z=9k,so for n we can obtien the general
Relation, so for n ≥4, we find the solution like this
X=2nk,y=2nk-k,z=2nk+k,k belong in R number
So the solution of probleme fermat like this
And x^n+y^n=z^n it equivalent for this equation or solution
(0.2)^n+[(0.2) -(0.1/n)] ^n ≤ (0.2) +(0.1/n)] ^n
For n≥4, but 0.1 is it solution for n=3, so this is a last my solution
this is general basic
quel que soit n≥1
a, b ;reel number no nul and
a €]0,1[ b=a/2n
a^n+(a-b)^n ≤ (a+b)^n ......(5)
this is last general basic(5)
pleas, i i can't send it with pdf
all thanks, and i like ur reply
Enregistrer un commentaire