GRITINZ TO ALL THE MTH GENERAL
AM BACK
HAPPY!!!!
can some one simplify 4 +4 +4 +4+ 4+...+

cong ! ,* MATHZ GURU ! LOLZ

STATE HAPPY END THEOREM
HAPPY

akpos4uall:

This is known as the magic square. I came across this in MATLAB. A special type of matrix known as magic square matrix http://en.m.wikipedia.org/wiki/Magic_square
I don't know how it is solved sha but this is the square for a 3 by 3
2 7 6
9 5 1
4 3 8

GOOD IT CAN SO BE ;
8 3 4
1 5 9
6 7 2


HAPPY

SOLVE FOR h in the equation : 6(raise to power h) -31(2)raise to power h+32 is equal to zero ( HINT : USE LINEAR APPROX)
:* HAPPY!

akpos4uall:

Let me use equations to make it clearer to you. Hope you know that I have two sets? The first one agrees with the question while the second one is the square roots of the perfect squares from the question.

According to the question, if (a, b, c) agree with the condition,
ab + 1 = X²
ac + 1 = Y²
bc + 1 = Z²
(a, b, c) & (X, Y, Z) are the two sets that I posted. Let M_n = (a_n, b_n, c_n) & P_n = (X_n, Y_n, Z_n)

We were able to get M₁ = (1, 8, 15) but I noticed that the subsequent ones that were gotten, had a relationship with M₁ as follows
M₁ = (a₁, b₁, c₁)
M₂ = (a₂, b₂, c₂)
M₃ = (a₃, b₃, c₃
Where a₂ = 0.5b₁, b₂ = 2c₁, c₂ = 2b₂ - a₂
Likewise a₃ = 0.5b₂, b₃ = 2c₂, c₃ = 2b₃ - a₃
Generally
M₁ = (a₁, b₁, c₁) = (1, 8, 15)
M_(n+1) = (0.5b_n, 2c_n, 2b_(n+1) - a_(n+1))
As for P_n
P_(n+1) = (Z_n, Z_n + Y_n, 2Y_(n+1) + X_(n+1)
P₁ = (X₁, Y₁, Z₁) = (3, 4, 11)


M₁ = 1, 8, 15
M₂ = 4, 30, 56
M₃ = 15, 112, 209
M₄ = 56, 418, 780
M₅ = 209, 1560, 2911
M₆ = 780, 5822, 10864
M₇ = 2911, 21728, 40545
M₈ = 10864, 81090, 151316

P₁ = 3, 4, 11
P₂ = 11, 15, 41
P₃ = 41, 56, 153
P₄ = 153, 209, 571
P₅ = 571, 780, 2131
P₆ = 2131, 2911, 7953
P₇ = 7953, 10864, 29681
P₈ = 29681, 40545, 110771

9ICE-ONE

Consider the digit from 1to 9 , select any three digit form (1,2,3,4,5,6,7,8,9) with out repetition the sum of the digit's will give 15 in the column , diagonal and row
HAPPY!

Laplacian:
...yea, i 've corrected that, u posted ur comment just as i was about to modify it... Ur result is very impressive...pls i need complete details (very elaborate) on how u generated those programs on ur PC...are u on Facebook?...

@ akpos ; your sequence did not satisfy laplacian question! that is three non- zero integers are in AP, the product of any two when increase by one is a perfect square.
:*HAPPY

akpos4uall:








I guess I know why @Laplacian said there are infinite number of triples that agree with this condition. From what I observed, I also agree with him.

1, 8, 15
4, 30, 56
15, 112, 127
56, 418, 780
2911, 21728, 40545

Take a look again, you'll notice that you can form another triple from the previous one because, half of the second term in the first triple is the first term in the next one while twice the 3rd term is the same as the second term in the next term. Since you already know the first and second term, you can get the third term. Following this trend, you can get an infinite number of triples.

This can be written as
X, 2Y, Z
Y, 2Z, 4Z - 2Y
Z, 8Z - 4Y, 15Z - 8Y
4Z - 2Y, 30Z - 16Y, 56Z - 30Y
15Z - 8Y, 112Z - 60Y, 209Z - 112Y
56Z - 30Y, 418Z -224Y,
209Z - 112Y,

take a look at the third sequence it is not an AP OK.
it ( 15,112,209) ,
:*@ HAPPY/

akpos4uall:








I guess I know why @Laplacian said there are infinite number of triples that agree with this condition. From what I observed, I also agree with him.

1, 8, 15
4, 30, 56
15, 112, 127
56, 418, 780
2911, 21728, 40545

Take a look again, you'll notice that you can form another triple from the previous one because, half of the second term in the first triple is the first term in the next one while twice the 3rd term is the same as the second term in the next term. Since you already know the first and second term, you can get the third term. Following this trend, you can get an infinite number of triples.

This can be written as
X, 2Y, Z
Y, 2Z, 4Z - 2Y
Z, 8Z - 4Y, 15Z - 8Y
4Z - 2Y, 30Z - 16Y, 56Z - 30Y
15Z - 8Y, 112Z - 60Y, 209Z - 112Y
56Z - 30Y, 418Z -224Y,
209Z - 112Y,

take a look at the third sequence it is not an AP OK.

akpos4uall:





Yes you are right but only for when n is greater than or equal to 9
What about when n < 9
For n = 0, X = 9 = 9 * 8! / 8!
n = 1, X = 9 * 8 = 9 * 8 * 7! / 7!
n = 2, X = 9 * 8 * 7 = 9 * 8 * 7 * 6! / 6!
This is the same as X = 9! / (8 - n)!

Hence X = (9-0)(9-1)(9-2)(9-3)...(9-n) can be written as
For 0 < n < 9, X = 9! / (8 - n)!
n greater than or equal to 9, X = 0
Well as for n < 0, that one pass me o.

why are you disturbing your self on (9-0)(9-1)(9-2)(9-3)(9-4)(9-5)(9-6)(9-7)(9-(9-9)(9-10)(9-11)... at the point (9-9) We have allthe product affected , which result to ZERO.
HAPPY

PROVE THE FIRST , SECOND,AND THIRD ISOMORPHISM THOREM ( ABSTRACT ALGEBRA)
HINT: SHOW THAT A FUNCTION IS ONE-ONE AND ALSO SHOW THAT IS A UNTO .MAPPING.
HAPPY DAY TO ALL THE MATHZ GURU.!

Laplacian:
proofs:

PROPOSITION 1:

proof: let xy be a two digit HAPPY number, then the followin sequenc holds;
x²+y²=abcd

a²+b²+c²+d²=mnpqsr
.....
.....
..... =10000....0

Suppose we have xy*10^k for some integer k, then
xy*10^k=xy00000...0 with the zeros in k places, hence
x²+y²+0²+0²+...+0²=abcd, and the above sequenc is repeatd and the result follows.

COROLLARY 1: there are infinitely many HAPPY numbers.

Proof; because k is a natural number and there are infinitely many natural numbers, the result follows

PROPOSITION 2:

proof: let xyz be HAPPY, then
x²+y²+z²=abcd, where abcd is HAPPY, now the permutation yxz is shows that y²+x²+z²=x²+y²+z²=abcd, {because addition is cummutative}, and the result follows..

COROLLARY 2: if any number of zeros is either struck out or fixed in between the digits of a HAPPY number, the new number formed is also HAPPY.

Proof:
case 1: let the number be
xyz0000...0 havin k zeros, then by the above proposition, the number can be permuted to take m zeros to the left of the natural numbers, i.e
000...0xyz000..0=xyz000...0, where the zeros to the right of xyz is k-m, for k>m. e.g, 2300 is HAPPY (by proposition 1), so 0230=230 is also HAPPY.
case 2: if xyz is HAPPY, then by proposition 1, xyz*10^k =xyz000...0 is HAPPY, so permutin we have, x000...0y00..0z000...0

PROPOSITION 3:

proof: obvious

PROPOSITION 4:

proof; the number 111...000, with m ones and n zeros follows the undelisted sequenc;
1²+1²+1²+...+0²+0²+0²=1+1+1+...+0+0+0=m, hence if m is HAPPY, the number is also happy

COROLLARY 3: there are infinitely many binary numbers

proof: from above, for any HAPPY number m, there exist a binary number containin m ones, or equivalently, there exist a one to one correspondenc between HAPPY numbers and binary numbers, but from COROLLARY 1 above, there are infinitely many HAPPY numbers, and the result follows.

PROPOSITION 5:

proof: obviously follows from 3.)

PROPOSITION 6:

proof: let xy and ab be two numbers such that x+a<=9, y+b<=9 , suppose both numbers are HAPPY, then x²+y²=pqrs, is HAPPY and a²+b²=uvwz is also HAPPY.
Now xy+ab=(x+a)(y+b) so the sequenc of HAPPINESS for the sum is; (x+a)²+(y+b)²=x²+a²+2ax+y²+b²+2by=x²+y²+a²+b²+2(ax+by), now the number 2(ax+by) is not alwaz HAPPY (why?, because the product of HAPPY or SAD numbers is not alwaz happy, by proposition 3 above), and the result follows. In particular, if ab and xy are HAPPY then ab+xy is HAPPY iff 2(ax+by) is HAPPY. The same reasonin goes for a subtraction.
PROPOSITIO 7: Loading...

Anyone who proves the last gets a reward from me...promise...
Hey!, my proofs above are open 2 criticism

GOOD PROOF

solve for h in the equation : 6(raise to power h) -31(2 raise to power h ) +32 is equal to zero.

what are HAPPY AND SAD NUMBERS

A MONIC EQUATION IS TYPE OF EQUATION THAT EXIST WHEN THE leading coefficient of a polynomial is one. example x raise to power 4 + (xy)raise to power 3 + x raise to power 2 equal to zero.

rhydex 247:

Do u mean harmonic equation?

MONIC EQUATION (*ABSTRACT ALGEBRA)

jackpot: Story!!!

Remember when I told you earlier that there are many answers to this question?

What rule-satisfication are you talking about?

I hope you are not referring to the discombobulated crap of a hint you supposedly ga

To the original question, my answer is on point.

Are you treasuring your answer? Oh well, I treasure my answer too as better.

we know by defination a sequence must have a common term (*chain) .and to the question what kind of sequence do you produce. Alright give me the rules that governs yr results

@ ,* MADAM JACKPOT I WAITED FOR YOUR SOLUTION TO THE PROBLEM 5,__ 1205,__ 271205.
WELL ALL IS WELL EVEN IN THE WELL . ( :hint multiply 15 to the previous +5 to get the next term) ! HAPPY DAY TO MY QUEEN SALMI!

WHAT ARE HAPPY AND SAD NUMBERS?

SLAMI

WHAT ARE ALGEBRAIC NUMBERS? :*@HAPPY

jackpot:
The formula for the nth term of your sequence may be given by

x_n = (8426n²-33209n+24813)/₅.

It is easy to verify that:
x₁ = 6
x₃= 204
x₆=25779

and as such, we have

x₂= - ⁷⁹⁰¹/₅
x₄=²⁶⁷⁹³/⁵
x₅=⁶⁹⁴¹⁸/₅

Use these numbers to fill in the gaps you . There you have it, bro.

.NICE , (:THE SEQUENCE IS A MULTIPLE OF 3 WE HAVE THE SEQUENCE AS 6,39,204,1029,5154,25779. ( RULE MULTIPLY 5 TO THE PREVIOUS TERM AND ADD 9 , TO GET THE NEXT TERM @ ,MADAM JACKPOT YOUR FORMULA DID NOT SATISFY THE RULE THAT GOVERNS THE SEQUENCE) :HAPPY

rhydex 247:
hmmmm.
dis is my best course
here the soln goes.
langrange theorem states that for any finite group G, the order number of element of every sub group H of G divides the order of G. The theorem is named after joseph louis langrange.

now the prove.
this can be shown using the concepts of left cosets of H in G. the left cosets are the equivalence classes of a certain equivalence relation on G and therefore form a partition of G. Specifically, x and y in G are related if and only if there exists h in H such that x=yh . If we can show dat all cosets of H av d same number of elements, then each coset of H has precisely /H/ elements. We are den done since the order of H times the number of cosets is equal to the no of elements in G, thereby proving that the order of H divides the order of G. Now, if aH and bH are two left cosets of H. We can define a map. f:aH--->bH. by setting f(x)=ba^-1x. This map is bijective because its inverse is given by f^-1 y=ab^-1(y).

GREAT! WHAT ARE MONIC EQUATION?

doubleDx:

Her formula explains everything bruv!

WE HAVE 6,39,204,1029,5154,25779. MULTIPLE OF 3 ; JACKPOT FORMULA IS WRONG

rhydex 247:
hmmmm.
dis is my best course
here the soln goes.
langrange theorem states that for any finite group G, the order number of element of every sub group H of G divides the order of G. The theorem is named after joseph louis langrange.

now the prove.
this can be shown using the concepts of left cosets of H in G. the left cosets are the equivalence classes of a certain equivalence relation on G and therefore form a partition of G. Specifically, x and y in G are related if and only if there exists h in H such that x=yh . If we can show dat all cosets of H av d same number of elements, then each coset of H has precisely /H/ elements. We are den done since the order of H times the number of is equal to the no of elements in G, thereby proving that the order of H divides the order of G. Now, if aH and bH are two left cosets of H. We can define a map. f:aH--->bH. by setting f(x)=ba^-1x. This map is bijective because its inverse is given by f^-1 y=ab^-1(y).

GOOD

@ ,MADAM jackpot the solution to the sequence ; 6,__,204,__,__,25779 - IS 6,39,204, 1029,5154, 25779.
THE RULE IS : muitiply the previous number by5 and add 9. HAPPY

doubleDx:

Her formula explains everything bruv!

APPLICATION
THE SEQUENCE 6,39,204,1029,5154,25779 ARE MULTIPLE OF 3 . :*HAPPY

MY QUEEN GAVE ME THIS QUESTION YESTERDAY AND AS ME TO COMPLETE IT : I said to my self what am i going to do ? told her that am going to post it to my oga on top : complete these : 6,__,204__,__,25779. (hint they are multiple of 3) : HAPPY

FIND THE VAULE OF K:6"K -31(2"K)+32 IS EQUAL TO ZERO ) : HAPPY ! ** DAY TO MY QUEEN

THE EXPRESSION (9-0)(9-1)(9-2)(9-3)(9-4)... IS EQAUL TO ZERO

HAPPY

Laplacian:
pls simplify it lets c @humphrey

HERE IT
SOLUTION
(9-0)(9-1)(9-2)(9-3)(9-4)(9-5)(9-6)(9-7)(9-(9-9)(9-10)... AT THE POINT (9-9) WE HAVE ZERO . CLEARLY , (: THE PRODUCT EXPANSION AS IT VAULE EQUAL TO ZERO WITH WHICHTHE BEAKING POINTEXIST AT (9-9) )

SIMPLIFY (9-0)(9-1)(9-3)...
AND STATE THE BREAKING POINT

:HAPPY DAY

state and prove LAGRANGE THEOREM ( ABSTRACT ALGEBRA)