Matematické Fórum

stuart clark · 09. 11. 2011 18:47

find value of $\left(%202+\sqrt{1+\sqrt{2+\sqrt{1+\sqrt{2+...}}}}\right)-\left(1+\sqrt{2+\sqrt{1+\sqrt{2+\sqrt{1+...}}}}%20\right)=$

vanok · 10. 11. 2011 12:05 — Editoval vanok (10. 11. 2011 14:01)

Hi ↑ stuart clark:,
First of all your expression has a sense because terms can be considered as limits of two sequences increasing and limited by 4.
Let us note M.
The second term
$1+\sqrt{2+\sqrt{1+\sqrt{2+\sqrt{1+...}}}}$
is the limit K of the sequences
$k_n= 1+ \sqrt {2+\sqrt{k_{n-1}}}$
So K verify
$K=1+\sqrt {2+\sqrt K}$
I confidedthe calculation of $K$ to wolframalpha
Which gives the value approached( and exact form also)
look here
http://www.wolframalpha.com/input/?i=K- … t+K%29%3D0

On the other hand the first term L
$2+\sqrt{1+\sqrt{2+\sqrt{1+\sqrt{2+...}}}}$
Give $L=2 +\sqrt K$
So $M=L- K= 2 +\sqrt K -K$
the approached value of which is: look to wolframalpha
http://www.wolframalpha.com/input/?i=2- … sqrt2.9263

Sincerely

Honzc · 10. 11. 2011 13:25 — Editoval Honzc (10. 11. 2011 13:27)

↑ vanok:
Zdravím, já teda nevím, ale výraz $1+\sqrt{2+\sqrt{1+\sqrt{2+\sqrt{1+...}}}}$ dá větší součet než $1.80927$ už pro první dva "členy". Tedy $1+\sqrt{2}>1.80927$
Navíc $M=L- K= 2 +\sqrt K -K$ se rozhodně nerovná $0.07163$

Autor skryl část textu. Zobrazit/skrýt!

vanok · 10. 11. 2011 13:45 — Editoval vanok (10. 11. 2011 14:09)

Hi ↑ Honzc:,

Thank you.
Look directly at the result on wolpframalpha (An error of typing that I corrected)

Otherwise, that you think of this fast solution drafted in 5 minutes.

Honzc · 10. 11. 2011 14:08

↑ vanok:
Totiž ty jsi asi do Wolframalpha asi zadal špatnou rovnici.
Asi to má být:
Takto

vanok · 10. 11. 2011 14:11

↑ Honzc:
Read my messages previous ones

Honzc · 10. 11. 2011 14:14

↑ vanok:
Já vím, oba jsme to napsali ve stejný čas.

Pavel · 11. 11. 2011 14:31 — Editoval Pavel (11. 11. 2011 16:40)

↑ stuart clark:

Using the theory of symetric polynomials I found a polynomial such that one of its root is the given expression. Thus, the expression fulfils the equation

$x^4-4x^3+3x^2+6x-5=0.$

However, the roots cannot be expressed in a nice form, see.

check_drummer · 11. 11. 2011 21:34

Pavel napsal(a):
↑ stuart clark:
Using the theory of symetric polynomials I found a polynomial such that one of its root is the given expression.

Hi, can you please write the solution using symetric polynomials - or at least the idea of your reasoning? Thank you.

Pavel · 13. 11. 2011 14:35 — Editoval Pavel (13. 11. 2011 14:39)

↑ check_drummer:

Hi, let us denote

$S:=\left(%202+\sqrt{1+\sqrt{2+\sqrt{1+\sqrt{2+...}}}}\right)-\left(1+\sqrt{2+\sqrt{1+\sqrt{2+\sqrt{1+...}}}}%20\right).$

The convergence of nested radicals was discussed here. So we can assume that $S$ is defined correctly and it is a finite real number.

Let us define

$\alpha:=\sqrt{2+\sqrt{1+\sqrt{2+\sqrt{1+...}}}}$

It is easy to see that $(\alpha^2-2)^2-1=\alpha$ and $S=\alpha^2-\alpha-1$ . It means that $\alpha$ is a root of the algebraic equation

$x^4-4x^2-x+3=0.$

Suppose that $x_1,x_2,x_3,x_4$ are all the roots of the equation. Then using Vieta's formulas, we obtain that

$A&:=-(x_1+x_2+x_3+x_4)=0,\\ B&:=x_1x_2+x_1x_3+x_1x_4+x_2x_3+x_2x_4+x_3x_4=-4,\\ C&:=-(x_1x_2x_3+x_1x_2x_4+x_1x_3x_4+x_2x_3x_4)=-1,\\ D&:=x_1x_2x_3x_4=3.$

Now we will derive a polynomial $x^4+Ux^3+Vx^2+Wx+Z$ from the given one such that all its roots are the numbers $x_1^2-x_1-1,x_2^2-x_2-1,x_3^2-x_3-1,x_4^2-x_4-1$ and therefore also $S$ is a root. It is not necessary solve the equation $x^4-4x^2-x+3=0$ , it suffices to find the coefficents $U,V,W,Z$ using $A,B,C,D$ , Vieta's formulas and symmetric polynomial identities that can be proved by elementary techniques. We will use some of them as:

$m_{(1,0,0,0)}(x_1,x_2,x_3,x_4)&:=x_1+x_2+x_3+x_4=-A=0,\\ m_{(1,1,0,0)}(x_1,x_2,x_3,x_4)&:=x_1x_2+x_1x_3+x_1x_4+x_2x_3+x_2x_4+x_3x_4=B=-4,\\ m_{(2,0,0,0)}(x_1,x_2,x_3,x_4)&:=x_1^2+x_2^2+x_3^2+x_4^2=A^2-2B=8,\\ m_{(1,1,1,0)}(x_1,x_2,x_3,x_4)&:=x_1x_2x_3+x_1x_2x_4+x_1x_3x_4+x_2x_3x_4=-C=1,\\ m_{(2,1,0,0)}(x_1,x_2,x_3,x_4)&: =x_1^2x_2+x_1^2x_3+x_1^2x_4+x_1x_2^2+x_2^2x_3+x_2^2x_4+x_1x_3^2+x_2x_3^2+x_3^2x_4\\ &\quad+x_1x_4^2+x_2x_4^2+x_3x_4^2=-AB+3C=-3,\\ m_{(1,1,1,1)}(x_1,x_2,x_3,x_4)&:=x_1x_2x_3x_4=D=3,\\ m_{(2,1,1,0)}(x_1,x_2,x_3,x_4)&: =x_1^2x_2x_3+x_1^2x_2x_4+x_1^2x_3x_4+x_1x_2^2x_3+x_1x_2^2x_4+x_2^2x_3x_4+x_1x_2x_3^2\\ &\quad+x_1x_3^2x_4+x_2x_3^2x_4+x_1x_2x_4^2+x_1x_3x_4^2+x_2x_3x_4^2=AC-4D=-12,\\ m_{(2,2,0,0)}(x_1,x_2,x_3,x_4)&: =x_1^2x_2^2+x_1^2x_3^2+x_1^2x_4^2+x_2^2x_3^2+x_2^2x_4^2+x_3^2x_4^2=B^2-2AC+2D=22,\\ m_{(2,1,1,1)}(x_1,x_2,x_3,x_4)&: =x_1^2x_2x_3x_4+x_1x_2^2x_3x_4+x_1x_2x_3^2x_4+x_1x_2x_3x_4^2=-AD=0,\\ m_{(2,2,1,0)}(x_1,x_2,x_3,x_4)&: =x_1^2x_2^2x_3+x_1^2x_2^2x_4+x_1^2x_2x_3^2+x_1^2x_3^2x_4+x_1^2x_2x_4^2+x_1^2x_3x_4^2+x_1x_2^2x_3^2\\ &\quad+x_2^2x_3^2x_4+x_1x_2^2x_4^2+x_2^2x_3x_4^2+x_1x_3^2x_4^2+x_2x_3^2x_4^2=-BC+3AD=-4,\\ m_{(2,2,1,1)}(x_1,x_2,x_3,x_4)&: =x_1^2x_2^2x_3x_4+x_1^2x_2x_3^2x_4+x_1^2x_2x_3x_4^2+x_1x_2^2x_3^2x_4+x_1x_2^2x_3x_4^2+x_1x_2x_3^2x_4^2=BD=-12,\\ m_{(2,2,2,0)}(x_1,x_2,x_3,x_4)&: =x_1^2x_2^2x_3^2+x_1^2x_2^2x_4^2+x_1^2x_3^2x_4^2+x_2^2x_3^2x_4^2=C^2-2BD=25,\\ m_{(2,2,2,1)}(x_1,x_2,x_3,x_4)&: =x_1^2x_2^2x_3^2x_4+x_1^2x_2^2x_3x_4^2+x_1^2x_2x_3^2x_4^2+x_1x_2^2x_3^2x_4^2=-CD=3,\\ m_{(2,2,2,2)}(x_1,x_2,x_3,x_4)&:=x_1^2x_2^2x_3^2x_4^2=D^2=9.$

If $x^4+Ux^3+Vx^2+Wx+Z$ is supposed to have roots $x_1^2-x_1-1,x_2^2-x_2-1,x_3^2-x_3-1,x_4^2-x_4-1$ then the following identities have to be true (due to Vieta's formulas):

$U&=-(x_1^2-x_1-1+x_2^2-x_2-1+x_3^2-x_3-1+x_4^2-x_4-1)\\ &=-\bigl(m_{(2,0,0,0)}(x_1,x_2,x_3,x_4)-m_{(1,0,0,0)}(x_1,x_2,x_3,x_4)-4\bigr)=-4,\\ V&=(x_1^2-x_1-1)(x_2^2-x_2-1)+(x_1^2-x_1-1)(x_3^2-x_3-1)+(x_1^2-x_1-1)(x_4^2-x_4-1)\\ &\quad+(x_2^2-x_2-1)(x_3^2-x_3-1)+(x_2^2-x_2-1)(x_4^2-x_4-1)+(x_3^2-x_3-1)(x_4^2-x_4-1)\\ &=m_{(2,2,0,0)}(x_1,x_2,x_3,x_4)-m_{(2,1,0,0)}(x_1,x_2,x_3,x_4)-3m_{(2,0,0,0)}(x_1,x_2,x_3,x_4)\\ &\quad+m_{(1,1,0,0)}(x_1,x_2,x_3,x_4)+3m_{(1,0,0,0)}(x_1,x_2,x_3,x_4)+6=3,\\ W&=-\bigl((x_1^2-x_1-1)(x_2^2-x_2-1)(x_3^2-x_3-1)+(x_1^2-x_1-1)(x_2^2-x_2-1)(x_4^2-x_4-1)\\ &\quad+(x_1^2-x_1-1)(x_3^2-x_3-1)(x_4^2-x_4-1)+(x_2^2-x_2-1)(x_3^2-x_3-1)(x_4^2-x_4-1)\bigr)\\ &=-\bigl(m_{(2,2,2,0)}(x_1,x_2,x_3,x_4)-m_{(2,2,1,0)}(x_1,x_2,x_3,x_4)-2m_{(2,2,0,0)}(x_1,x_2,x_3,x_4)\\ &\quad+m_{(2,1,1,0)}(x_1,x_2,x_3,x_4)+2m_{(2,1,0,0)}(x_1,x_2,x_3,x_4)+3m_{(2,0,0,0)}(x_1,x_2,x_3,x_4)\\ &\quad-m_{(1,1,1,0)}(x_1,x_2,x_3,x_4)-2m_{(1,1,0,0)}(x_1,x_2,x_3,x_4)-3m_{(1,0,0,0)}(x_1,x_2,x_3,x_4)-4\bigr)=6,\\ Z&=(x_1^2-x_1-1)(x_2^2-x_2-1)(x_3^2-x_3-1)(x_4^2-x_4-1)=\\ &=m_{(2,2,2,2)}(x_1,x_2,x_3,x_4)-m_{(2,2,2,1)}(x_1,x_2,x_3,x_4)-m_{(2,2,2,0)}(x_1,x_2,x_3,x_4)\\ &\quad+m_{(2,2,1,1)}(x_1,x_2,x_3,x_4)+m_{(2,2,1,0)}(x_1,x_2,x_3,x_4)+m_{(2,2,0,0)}(x_1,x_2,x_3,x_4)\\ &\quad-m_{(2,1,1,1)}(x_1,x_2,x_3,x_4)-m_{(2,1,1,0)}(x_1,x_2,x_3,x_4)-m_{(2,1,0,0)}(x_1,x_2,x_3,x_4)\\ &\quad-m_{(2,0,0,0)}(x_1,x_2,x_3,x_4)+m_{(1,1,1,1)}(x_1,x_2,x_3,x_4)+m_{(1,1,1,0)}(x_1,x_2,x_3,x_4)\\ &\quad+m_{(1,1,0,0)}(x_1,x_2,x_3,x_4)+m_{(1,0,0,0)}(x_1,x_2,x_3,x_4)+1=-5.$

We have found the polynomial $P(x)=x^4-4x^3+3x^2+6x-5$ such that

$P(S)=0,$

where

$S=\left(%202+\sqrt{1+\sqrt{2+\sqrt{1+\sqrt{2+...}}}}\right)-\left(1+\sqrt{2+\sqrt{1+\sqrt{2+\sqrt{1+...}}}}%20\right).$

check_drummer · 13. 11. 2011 16:25

Thank you for your solution, Pavel. Nevertheless:

Pavel napsal(a):
However, the roots cannot be expressed in a nice form

Why not? Roots of the polynomial of degree 4 can be expessed in nice (=closed) form.

Pavel · 13. 11. 2011 16:56

↑ check_drummer:

Yes, you are right. Despite of expressing in a closed form, it is necessary to use a lot of radicals. If we knew
another way of solving the problem, it might be expressed in a more simple form. However, I doubt about it.

byk7 · 18. 09. 2013 22:25

↑ Pavel:

Can we say, which one of the is solution of the exercise?

Matematické Fórum

#1 09. 11. 2011 18:47

infinite sum

#2 10. 11. 2011 12:05 — Editoval vanok (10. 11. 2011 14:01)

Re: infinite sum

#3 10. 11. 2011 13:25 — Editoval Honzc (10. 11. 2011 13:27)

Re: infinite sum

#4 10. 11. 2011 13:45 — Editoval vanok (10. 11. 2011 14:09)

Re: infinite sum

#5 10. 11. 2011 14:08

Re: infinite sum

#6 10. 11. 2011 14:11

Re: infinite sum

#7 10. 11. 2011 14:14

Re: infinite sum

#8 11. 11. 2011 14:31 — Editoval Pavel (11. 11. 2011 16:40)

Re: infinite sum

#9 11. 11. 2011 21:34

Re: infinite sum

Pavel napsal(a):

#10 13. 11. 2011 14:35 — Editoval Pavel (13. 11. 2011 14:39)

Re: infinite sum

#11 13. 11. 2011 16:25

Re: infinite sum

Pavel napsal(a):

#12 13. 11. 2011 16:56

Re: infinite sum

#13 18. 09. 2013 22:25

Re: infinite sum

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