Matematické Fórum

stuart clark

Find Real values of $x$ in $\mid x \mid+\sqrt{[x]+\sqrt{1+\{x\}}} = 1$

vanok · 14. 07. 2012 12:34 — Editoval vanok (17. 07. 2012 10:52)

Hi ↑ stuart clark:,
First case:
$x\geq 0$
It is clear that $\sqrt{[x]+\sqrt{1+\{x\}}} \in \{0; 1;2;... \}$ .
Also $[x]+\sqrt{1+\{x\}} \in \{0; 1;2;... \}$ and thus $\sqrt{1+\{x\}} \in \{0; 1;2;... \}$ .
As $0 \leq \{x\} < 1$ , thus $\{x\}=0$ .
So $1+\{x\} =1$ and $[x] =0$ (because we have $\mid x \mid+\sqrt{[x]+\sqrt{1+\{x\}}}= x+\sqrt {x+1}= 1$ ).
Finally $x=0$ .

Edit:
Solution of the missing case. Cf ↑ anes:

The case $x <0$
In this case
$\mid x \mid+\sqrt{[x]+\sqrt{1+\{x\}}}= -x+\sqrt{[x]+\sqrt{1+x-[x]}}= 1$
We deduct so early that from it $[x]=-1$
So we have to solve
$-x +\sqrt{-1+\sqrt{1+x+1}}=1$
Where still
$-x +\sqrt{-1+\sqrt{x+2}}=1$
This implies:
$x^4+4x^3+8x^2+7x+2=0$
$(x+1)(x^3+3x^2+5x+2)=0$
So $x_1=1$ and $x_2 \approx -0.5466$
are 2 real solutions of the last one equation.
( cf. http://www.wolframalpha.com/input/?i=x^ … 7x%2B2%3D0 )

A check shows that it also verify the given equation.

Conclusion: The problem posed has 3 solutions $0; -1; \approx -0.5466$

anes · 14. 07. 2012 19:43 — Editoval anes (14. 07. 2012 19:47)

↑ vanok:

Could you explain the first two statements a bit closer? I suspect that you've misread $|x|$ as $[x]$ , but if not, I'd be interested in the ideas behind. I also suppose that at line 4, you forgot to consider $x < 0$ .

Anyway, $\sqrt{[x]+\sqrt{1+\{x\}}} \geq 0$ , so clearly $|x| \leq 1$ .
For $x > 0$ , we'd have $\sqrt{[x]+\sqrt{1+\{x\}}} > 1$ , so we are left with $-1 \leq x \leq 0$ .

We see, that $x=0$ and $x=-1$ are two solutions.
For all the other cases, it depends on what you mean by the integer & fractional part in negative numbers.

Understanding $[x]=-1$ , $\{x\} = x - [x] = x + 1$ we get $-x +\sqrt{-1+\sqrt{x+2}} = 1$ , which (according to wolframalpha, Odkaz) gives one additional solution of approx. $-0.547$ .

Understanding $[x]=0$ , $\{x\} = x$ , we come to $-x +\sqrt{\sqrt{1+x}} = 1$ , which clearly will not give any additional solution from $(-1,0)$ .

vanok · 14. 07. 2012 20:55 — Editoval vanok (14. 07. 2012 23:33)

Hi ↑ anes:,

Thank you, for your warning.

Effectively, I badly looked at the statement.
So my sol:ution considers only the case, when $x \geq 0$

Remark:
You written:

Understanding $[x]=0$ , $\{x\} = x$ , we come to $-x +\sqrt{\sqrt{1+x}} = 1$ , which clearly will not give any additional solution from $(-1,0)$ .

This is not exact , because in the case $[x]=0$ , we have $1>x \geq 0$ , and then $|x|=x$ and not $-x$ .
Reminder
Usually $\[ x \] \le x < \[x \]+1$ and $x= \[x\]+\{x\}$ .

anes · 14. 07. 2012 21:07

↑ vanok:

What I meant by the two cases actually is that sometimes people choose to define the integer part of a negative number as its ceil instead of floor value (effectively working eith truncation). And since this is an english thread on a czech forum, it seemed safer to check. Maybe it was not so clear, but at that stage, I've already referred only to the case $-1<x<0$ .

It's however possible that I might have a mistake somewhere too. But I don't see any atm.

Matematické Fórum

#1 08. 07. 2012 07:59 — Editoval stuart clark (08. 07. 2012 08:12)

real values of x

#2 14. 07. 2012 12:34 — Editoval vanok (17. 07. 2012 10:52)

Re: real values of x

#3 14. 07. 2012 19:43 — Editoval anes (14. 07. 2012 19:47)

Re: real values of x

#4 14. 07. 2012 20:55 — Editoval vanok (14. 07. 2012 23:33)

Re: real values of x

#5 14. 07. 2012 21:07

Re: real values of x

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