Matematické Fórum

stuart clark · 14. 07. 2012 06:14

Comput all real solution of the equation $\{x^2\}+\{x\} = 1$

where $\{x\} =$ fractional part of $x$

vanok · 14. 07. 2012 12:03

Hi ↑ stuart clark:,
$\{x^2\}+\{x\} = 1$ is equivalent with $x^{2}+x=[x^{2}]+[x]+1=m$
Hence $x^{2}+x-m=0$ ( $m$ is a not zero natural number )
and
$x=\frac{-1\pm\sqrt{1+4m}}{2}$ .

Marian · 13. 09. 2012 13:08

↑ vanok:

This can not be an effective form of the solution. Note that $m=m(x)$ .

vanok · 14. 09. 2012 15:25

Hi ↑ Marian:,
Have you some examples?
Thank you.

Marian · 14. 09. 2012 18:32

↑ vanok:

Examples or ideas ...?

vanok · 14. 09. 2012 19:49

↑ Marian:,
Examples of the solutions of shape found, which do not suit.
If you have the other ideas, why not.
Pleasant evening

check_drummer · 14. 09. 2012 22:13

↑ vanok:
Hi, how should I understand the equality $x=\frac{-1\pm\sqrt{1+4m}}{2}$ ? Does it mean that for every m there exists solution x? But for e.g. m=2 we get x=1 or x=-2 which do not satisfy the equation that has to be solved.

vanok · 14. 09. 2012 22:32

↑ check_drummer:,
Thank you.

check_drummer

↑ stuart clark:
I am hiding because of error in the second row... But in fact the text below can serve as a solution to the
$\{x\}^2 + 2.[x].\{x\}+\{x\} -1 = 0$ .

Autor skryl část textu. Zobrazit/skrýt!

vanok · 15. 09. 2012 08:01 — Editoval vanok (15. 09. 2012 08:06)

↑ vanok:,
A correction to raise quite badly - waited:

$\{x^2\}+\{x\} = 1$ is equivalent with $x^{2}+x=[x^{2}]+[x]+1$
Thus it is sufficient, to solve $x^{2}+x-m=0$ , where $m$ is a not zero natural number suitable (such as $1+4m$ is a not squared number )
and thus
$x=\frac{-1\pm\sqrt{1+4m}}{2}$ is a solution of the problem posed (even although with the sign -).

Pavel Brožek

↑ vanok:

It really seems that your solution gives the right answer. But I don't think your reasoning is correct (at least it is not complete), so (considering ↑ Marian's note:) it seems rather like a miracle that your solution actually works.

I will try to explain why it works.

1) If some x satisfies $\{x^2\}+\{x\} = 1$ then

$x^{2}+x&=([x^2]+\{x^2\})+([x]+\{x\})=\\ &=[x^2]+[x]+\{x^2\}+\{x\}=\\ &=[x^2]+[x]+1\in\mathbb{Z}$ .

so there exists some $m\in\mathbb{Z}$ such that $x^2+x=m$ . In other words every solution can be written in this form.

2) If there exists some $m\in\mathbb{Z}$ such that $x^2+x=m$ , then

$\{x^{2}\}+\{x\}&=(x^2-[x^2])+(x-[x])=\\ &=x^2+x-[x^2]-[x]=\\ &=m-[x^2]-[x]\in\mathbb{Z}$ .

But it is obvious that $0\le\{x^{2}\}+\{x\}<2$ , so $\{x^{2}\}+\{x\}=0$ or $\{x^{2}\}+\{x\}=1$ . The last thing that needs to be done is to show which m give $\{x^{2}\}+\{x\}=0$ (since fractional part of a number is possitive, it is equivallent to $\{x\}=0$ , which means that x is integer), every other m will automatically give $\{x^{2}\}+\{x\}=1$ . By solving the quadratic equation $x^2+x=m$ for x we obtain

$x=\frac{-1\pm\sqrt{1+4m}}{2}.$

It is obvious that x is integer iff $\sqrt{1+4m}$ is odd number. $\sqrt{1+4m}$ is odd number iff $1+4m$ is square of some integer (because $1+4m$ is odd number) and $1+4m\ge0$ (which gives $m\ge0$ ).

So every solution of the original equation can be written in form $x=\frac{-1\pm\sqrt{1+4m}}{2}$ where $m\in\mathbb{N}$ and $1+4m$ is not square of some integer. Also every x that can be written in this form is solutions of the problem.

vanok · 15. 09. 2012 11:46

Hi dear ↑ Pavel Brožek:,

I thank you for your collaboration and to have detailed all the points of my solution. It is true that my initial writing, although it contains the key idea of the solution was drafted a little bit quickly and with some gaps.

But thanks to this forum we all arrive towards a completely satisfactory for many colleagues.

I take advantage these lines to thank all the colleagues who contributed to improve this solution of this problem.

And, I add a reflection (almost Olympic): the most important are to participate in the activities of the forum, and thanks to the interaction, we can improve our contributions thanks to the common collaboration.

vanok · 15. 09. 2012 19:33

Still some questions having one relations with the problem posed.
Comput all real solution of the equation $\{x^2\}+\{x\} = 0,99$ .
Comput all real solution of the equation $\{x^2\}+\{x\} = r$ , $r$ a real number.
Study variations of the function $f(x)=\{x^2\}+\{x\}$ .

vanok · 17. 09. 2012 12:58

Dear ↑ Marian:,
Have you the other ideas and the other observations to complete this subject?
It interests me, as also, certainly the other colleagues.
Thank you in advance

Matematické Fórum

#1 14. 07. 2012 06:14

real solution of x

#2 14. 07. 2012 12:03

Re: real solution of x

#3 13. 09. 2012 13:08

Re: real solution of x

#4 14. 09. 2012 15:25

Re: real solution of x

#5 14. 09. 2012 18:32

Re: real solution of x

#6 14. 09. 2012 19:49

Re: real solution of x

#7 14. 09. 2012 22:13

Re: real solution of x

#8 14. 09. 2012 22:32

Re: real solution of x

#9 15. 09. 2012 04:16 — Editoval check_drummer (15. 09. 2012 09:56)

Re: real solution of x

#10 15. 09. 2012 08:01 — Editoval vanok (15. 09. 2012 08:06)

Re: real solution of x

#11 15. 09. 2012 10:41 — Editoval Pavel Brožek (15. 09. 2012 11:55)

Re: real solution of x

#12 15. 09. 2012 11:46

Re: real solution of x

#13 15. 09. 2012 19:33

Re: real solution of x

#14 17. 09. 2012 12:58

Re: real solution of x

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