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Show without derivatives that the equation $x^{n+1}-2x+1=0,n>1,n\in\mathbb{R}$ has at least one solution $x_0\in (0, 1)$

I have thought that I need to use Bolzano for the equation as $f(x)$. So, we have: $$f(0)=1>0$$ $$f\left( \frac{1}{2}\right) =2^{-n-1}>0$$ $$f(1)=0$$

There is a problem here, as I cannot find any number or somehow use limit theorems in order for some $a\in (0, 1)$ to be $f(a)<0$. Any hint?

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$f(0)=1$ and $f(1)=0$ for all $n$, so it suffices to show, for example, that $f(2/3)\lt0$. But if $n\gt1$ then

$$f(2/3)=(2/3)^{n+1}-(4/3)+1\lt(2/3)^3-(1/3)=-1/27$$

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  • $\begingroup$ Yet, it is assumed $n\ge 2$ $\endgroup$ – Svetoslav Sep 28 '15 at 15:58
  • $\begingroup$ But still, the most elegant answer ! $\endgroup$ – Svetoslav Sep 28 '15 at 15:59
  • $\begingroup$ $n>1$ and $n\ge2$ is the same thing. Agree this is the most elegant answer ! $\endgroup$ – Michael Medvinsky Sep 29 '15 at 9:14
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if $x^{n+1} = 2x-1$, then $x^n = 2 - \frac{1}{x}$

Does this help?

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Hint:

For $x \rightarrow 1^-$ write the function as $f(x)=x\left(x^n-2+\dfrac{1}{x}\right)$ and you can see that its value approximate $0$ by negative numbers.

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  • $\begingroup$ How do I see that it is approaching 0 by negative numbers? $\endgroup$ – Jason Sep 28 '15 at 15:21
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Let's take $x = 1-\frac{1}{n+1}$ and prove that

$$ x^{n+1}-2x+1<0 \Leftrightarrow \left(1-\frac{1}{n+1}\right)^{n+1}<2 \left(1-\frac{1}{n+1}\right)-1\Leftrightarrow \left(\frac{n}{n+1}\right)^n<1-\frac{1}{n}.$$ The last inequality can be proved if we notice that for $n=1$ it is true and the left side is decreasing function and the right side is increasing function.

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  • $\begingroup$ But $n\in\mathbb R$ $\endgroup$ – Svetoslav Sep 28 '15 at 15:48
  • $\begingroup$ Yes, Svetoslav you are right, but any way all claims are true for real $n$ too. $\endgroup$ – pointer Sep 28 '15 at 15:51
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You want some $a \in(0,1)$ such that $f(a)<0$. Writing $b =\frac{1}{a}$ this is equivalent to finding a $b >1$ such that $$ 1-2b^n+b^{n+1} <0 $$ or $$2b^n > 1+ b^{n+1}$$

Now by Bernoulli we have $$b^n >1+n(b-1)$$

So by choosing $n(b-1)>1$ you have $b^n>2$ and hence $$\frac{1}{2}b^n>1 \,.$$

Also, as long as $b <\frac{3}{2}$ you have $$\frac{3}{2}b^n >b^{n+1}$$

Combining the two you get that for all $\frac{1}{n}+1< b < \frac{3}{2}$ we have $$2b^n >1+b^{n+1}$$

This solution works for $n \geq 3$, as we need $\frac{1}{n}+1< b < \frac{3}{2}$. For $n=2$ the inequalities are not strong enough, but in this case the problem is very easy to solve.

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  • $\begingroup$ Maybe I need to remind that $n\in\mathbb{R}$ not always a natural.... $\endgroup$ – Jason Sep 28 '15 at 19:53
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    $\begingroup$ @Jason Ups, I am used with $n$ always denoting an integer. Note that I am only using $n$ integer for Bernoulli. You can easely change the proof the following way: $$b^n \geq b^{\lfloor n \rfloor} \geq 1+\lfloor n \rfloor (b-1) \geq 1+ (n-1)(b-1)$$ and then the same proof should work for $n \geq 3 $ or $n \geq 4$. Small values of $n$ are easy to handle. $\endgroup$ – N. S. Sep 28 '15 at 23:24
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Hint: If we recall that $a_{n+1}=\left(1-\frac{1}{n+1}\right)^{n+1}$ gives an increasing sequence converging towards $\frac{1}{e}$ and define $p_{n+1}(x)$ as $x^{n+1}-2x+1$, we have: $$ p_{n+1}\left(1-\frac{1}{n+1}\right) < \left(\frac{1}{e}-1\right)+\frac{2}{n+1},$$ hence $p_{n+1}(x)$ has a root in $\left(0,1-\frac{1}{n+1}\right)$ for any $n\geq 2$.

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Try to show that $f(a)<0$ for $a\approx1$ for example $a=\frac{9}{10}$

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