YLL討論網

scsnake 寫到:[(√3+1)^k]=
(√3+1)^k+(√3-1)^k-1_______if k is even
(√3+1)^k-(√3-1)^k_________if k is odd

k is even
(√3+1)^k+(√3-1)^k為自然數，且0<(√3-1)^k<1，所以(√3+1)^k+(√3-1)^k=[(√3+1)^k]+1

k is odd
(√3+1)^k-(√3-1)^k是自然數，且0<(√3-1)^k<1，所以(√3+1)^k-(√3-1)^k=[(√3+1)^k]


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突然想到，如上面所說，如果0<√a-√b<1，則x[k]=(√a+√b)^k，k愈大時，|round(x[k])-x[k]|會趨近於0，也就是和最近的整數之差愈來愈小！
不知meowth能不能用連分數(或其他)解釋？   

Kang, where did you get these QQ? Did you see the answer?

[u^k] congruent to k-1 (mod 2).

can i do it in this way?

Let u be a root of x^2-ax-b=0
let a,b E Z, and v be the another root of this equation.
Let S(k)=u^k+v^k
It represents a recurrence relation, therefore S(k) is integer for k E N.
When a=0(mod 2), u=v(mod 2), u^k+v^k=0(mod 2), S(k)=0(mod 2)

Take a,b such that -1<v<0, then
S(k)<u^k<S(k+1), if k is odd.
S(k-1)<u^k<S(k), if k is even.
So [u^k]=s(k) is even, if k is odd.
=S(k-1) is odd, if k is even.

v= [a-sqrt(a^2-4b)]/2 ,
-a-sqrt(a^2-4b)>-2
a+sqrt(a^2-4b)<2
a^2-4b<a^2-4a+4
-b<-a+1
a<b+1
a=0(mod 2)
Say a=-2, b=-1
Then
x^2-2x-1=0
u = (2 + sqrt(8)/2
u = 1+sqrt2

Can I do it in this way?

Meowth 寫到:Kang, where did you get these QQ? Did you see the answer?

i'm learning more in 牛頓恆等式... i can solve the typical ones, i think. But i have difficulty in this one.

請問魔界大大大大大費瑪，牛頓恆等式是什麼？？

scsnake 寫到:請問魔界大大大大大費瑪，牛頓恆等式是什麼？？

f(x)=c0 x^n + c1 x^(n-1) +...+ cn

Roots of f(x)=0 are x1....xn

Denote S(k) = (x1)^k + (x2)^k +...+(xn)^k

c0 S(k) + c1 S(k-1) +... + cn S(k-n) = 0 if k>n
c0 S(k) + c1 S(k-1) +... + c k-1 S(1) + k ck =0 if 1<=k<=n

原來這就是牛頓恆等式∼
這是meowth幾個月前教我的..

純數書是叫

根之冪和的牛頓公式

scsnake 寫到:原來這就是牛頓恆等式∼
這是meowth幾個月前教我的..

I know a little bit, but not familiar.... since last year, i know a little bit as far as I learnt recurrence relations, they are so similar in certain aspect. But since I meet this problem, I found that I've still very poor in using recurrence relations, while originally I think I can use it to solve basic problem...

但在中六純數課中實際應用不多

Raceleader 寫到:但在中六純數課中實際應用不多

but quite useful in competitions

siuhochung 寫到:

scsnake 寫到:請問魔界大大大大大費瑪，牛頓恆等式是什麼？？

f(x)=c0 x^n + c1 x^(n-1) +...+ cn

Roots of f(x)=0 are x1....xn

Denote S(k) = (x1)^k + (x2)^k +...+(xn)^k

c0 S(k) + c1 S(k-1) +... + cn S(k-n) = 0 if k>n
c0 S(k) + c1 S(k-1) +... + c k-1 S(1) + k ck =0 if 1<=k<=n

.... 有看沒有懂 ........... 跟不上 Little Kangaroo ...........

書本課程跟競賽是不同程度

Meowth 寫到:

siuhochung 寫到:

scsnake 寫到:請問魔界大大大大大費瑪，牛頓恆等式是什麼？？

f(x)=c0 x^n + c1 x^(n-1) +...+ cn

Roots of f(x)=0 are x1....xn

Denote S(k) = (x1)^k + (x2)^k +...+(xn)^k

c0 S(k) + c1 S(k-1) +... + cn S(k-n) = 0 if k>n
c0 S(k) + c1 S(k-1) +... + c k-1 S(1) + k ck =0 if 1<=k<=n

.... 有看沒有懂 ........... 跟不上 Little Kangaroo ...........

teasing me?!

siuhochung 寫到:[u^k] congruent to k-1 (mod 2).

can i do it in this way?

Let u be a root of x^2-ax-b=0
let a,b E Z, and v be the another root of this equation.
Let S(k)=u^k+v^k
It represents a recurrence relation, therefore S(k) is integer for k E N.
When a=0(mod 2), u=v(mod 2), u^k+v^k=0(mod 2), S(k)=0(mod 2)

Take a,b such that -1<v<0, then
S(k)<u^k<S(k+1), if k is odd.
S(k-1)<u^k<S(k), if k is even.
So [u^k]=s(k) is even, if k is odd.
=S(k-1) is odd, if k is even.

v= [a-sqrt(a^2-4b)]/2 ,
-a-sqrt(a^2-4b)>-2
a+sqrt(a^2-4b)<2
a^2-4b<a^2-4a+4
-b<-a+1
a<b+1
a=0(mod 2)
Say a=-2, b=-1
Then
x^2-2x-1=0
u = (2 + sqrt(8)/2
u = 1+sqrt2

Can I do it in this way?

...
Take a,b such that -1<v<0, then
S(k)<u^k<S(k)+1, if k is odd.
S(k)-1<u^k<S(k), if k is even.
...

-1<v= [a-sqrt(a^2+4b)]/2 <0,
a+2>sqrt(a^2+4b) >a

if a=2k>=0,
aa+4a+4> aa+4b> aa
2k>= b> =1
x^2-2kx-b=0
x=k+sqrt(kk+b)
--------
when k =1:  b=1 or 2, x=1+sqrt(2) or 1+sqrt(3)
when k=2: b=1,2,3, or 4, x=2+sqrt(5) ,2+sqrt(6), 2+sqrt(7) or 2+sqrt(8)
...

That means I'm right?
Is this method applicable to no.2?

I don't think Q2 can be solved in this way.

Could Mathematica get some values of solutions?

test values near 2.11631819100246 ?? 8.49837073 ??

求實數u，使得對任何k E N，有
[u^k] congruent to 0 (mod 2) if k is not divisible by 3
congruent to 1 (mod 2) if k is divisible by 3.
---------------------------------
consider the series mod 2: F(n+1)=a*F(n)-b*F(n-1)
where a, b, F(1), F(2) are all odds.
then  
F(k) == 1 (mod 2) if k is not divisible by 3
== 0 (mod 2) if k is divisible by 3.
---------------------------------
Let u be a root of x^2-ax+b=0
let a,b are odds, and v be the another root of this equation.
Let F(k)=u^k+v^k
It represents a recurrence relation, therefore F(k) is integer for k E N.
F(1)=a is odd.
F(2)=aF(1)-b*2 is odd
F(3)=a*F(2)-b*F(1) is even

Take a,b such that 1>v>0, then
F(k)-1<u^k<F(k)
...

1>v= [a-sqrt(a^2-4b)]/2 >0,
a-2<sqrt(a^2-4b) <a
aa-4a+4< aa-4b< aa
a-1> b> 0

a>=b+2>=3 where a, b are all positive odds.
x=u=(a+sqrt(aa-4b))/2 ##

--------
when a =3: b=1 , x=(3+sqrt(5))/2
when a=5: b=1,3, x=(5+sqrt(21))/2 , or (5+sqrt(13))/2
...

clever meow!!!