Let $a_n$ be the only positive root of the equation $x^n+x=1$, for each $n\in \Bbb N$.

Show that $\lim \limits_{n\to \infty}a_{n}$ exists,and find its value.

My guess is that $$\lim \limits_{n\to \infty}a_{n}=1$$

But How prove that it exists, and how find it?

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    $\begingroup$ @GitGud the equation is $x^n+x=1$ which always has a root in $(0,1)$ $\endgroup$ – L. F. Sep 5 '13 at 17:18
  • $\begingroup$ You didn't read the question properly. It is $x^n + x - 1 = 0$ $\endgroup$ – Parth Thakkar Sep 5 '13 at 17:19
  • $\begingroup$ Right. I'll edit my comment. $\endgroup$ – Git Gud Sep 5 '13 at 17:19
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    $\begingroup$ What GitGud is saying is it should read "The equation $x^n+x=1$ has a positive root $a_n$ for each $n$." $\endgroup$ – Pedro Tamaroff Sep 5 '13 at 17:20
  • $\begingroup$ I guess we have to calculate that root and let n go to infinity. But to calculate that root for a large degree polynomial is not easy, so there has to be something different here to figure that out. $\endgroup$ – imranfat Sep 5 '13 at 17:21

It is sufficient to show that, for all $0<\varepsilon<1$, the function $f$, defined by $f_n(x)=x^n+x-1$, has a root in the interval $(1-\varepsilon,1)$ for all sufficiently large $n$.

We know $f_n(1)=1>0$. Thus, since $f_n$ is continuous, it sufficient to show that $f(1-\varepsilon)<0$ for all sufficiently large $n$.

We observe \begin{align*} f_n(1-\varepsilon) &= (1-\varepsilon)^n+1-\varepsilon-1 \\ &=(1-\varepsilon)^n-\varepsilon \\ & \rightarrow -\varepsilon \\ & < 0 \end{align*} since $c^n \rightarrow 0$ whenever $0 < c < 1$.

Hence, for all $0<\varepsilon<1$, we have shown that $f_n(1-\varepsilon)<0$ for all sufficiently large $n$.

  • $\begingroup$ Thank you, But you don't prove $\displaystyle\lim_{n\to\infty}a_{n}$ is exsit. $\endgroup$ – math110 Sep 6 '13 at 1:11
  • $\begingroup$ For all $0<\varepsilon<1$ there exists $N \geq 1$ such that $|a_n-1| \leq \varepsilon$ for all $n \geq N$. $\endgroup$ – Rebecca J. Stones Sep 6 '13 at 1:22
  • $\begingroup$ oh, my methods, I prove $$a_{n+1}>a_{n},0<a_{n}<1$$ $\endgroup$ – math110 Sep 6 '13 at 1:35
  • $\begingroup$ That works, using the monotone convergence theorem, but it means more work in showing that $a_{n+1}>a_n$. It's easier to show that sufficiently large $a_k$ with $k>n$ satisfies $a_k>a_n$, since we can take an interval $(1-\varepsilon,1)$ that does not contain $a_n$. $\endgroup$ – Rebecca J. Stones Sep 6 '13 at 1:38
  • $\begingroup$ I guess must $$g(n)<a_{n}<\dfrac{n}{n+1}$$?How find the simple form of $$g(n)$$? $\endgroup$ – math110 Sep 6 '13 at 2:26

You may show the sequence of roots is increasing, and bounded above by $1$, say. Rebecca's approach then shows the limit of the roots cannot be $<1$, so it must be $\geqslant 1$. Having shown it is $\leqslant 1$; we get it must be $1$.

If you want to be sketchy, consider the following. The functions $f_n=x^n$ are always increasing for $x>0$, with $f_n(0)=0$ and $f_n(1)=1$. The function $f(x)=1-x$ is monotone decreasing and $f(0)=1$; $f(1)=0$. You can show the $f_n=f$ has a root, and it must be $<1$. As $n$ grows, $f_n$ decreases over $0<x<1$ (so the ordinate of the root increases); with pointwise limit $=0$, so $1-x=0\implies x=1$.

  • $\begingroup$ Is there somehting about the Rule of Descartes that shows the roots is between 0 and 1? $\endgroup$ – imranfat Sep 5 '13 at 17:24
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    $\begingroup$ @imranfat No idea what the Rule of Descartes is, but you 'just' need the IMT. $\endgroup$ – Git Gud Sep 5 '13 at 17:25
  • $\begingroup$ Could you elaborate "so the ordinate of the root increases"? $\endgroup$ – Parth Thakkar Sep 5 '13 at 17:27
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    $\begingroup$ @ParthThakkar Since $f_{n+1}<f_n$, the root of $f_n=f$ must be closer to $0$ than that of $f_{n+1}=f$. $\endgroup$ – Pedro Tamaroff Sep 5 '13 at 17:29
  • $\begingroup$ "Having shown it is $\le 1$" should be "Having shown it is $\ge 1$", right? $\endgroup$ – Daniel R Sep 5 '13 at 18:10

Technically, one should first verify that $a_n$ is indeed unique. This can be done easiest by Descartes' Rule of Signs, a less-known but in my opinion underrated result which is used as follows. Note that the polynomial $x^n + x - 1$ has coefficients $1, 1, -1$ in that order (ignore $0$ coefficients), so the coefficients signs go $+,+, -$ (positive, positive, negative). The coefficients change sign exactly once, so the number of positive real roots of the polynomial $x^n + x - 1$ is one. (In general, if the coefficients change sign $n$ times, the number of positive real roots will be one of $n$, $n - 2$, $n - 4$, $n - 6$, etc.)

Once you have established that $a_n$ is unique, Rebecca's solution completes the proof.

  • $\begingroup$ Why should it be not unique? In the problem is also stated "the only positive root of..." $\endgroup$ – user92797 Sep 14 '13 at 15:50
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    $\begingroup$ I know how the problem is stated. But "the problem was stated that way" is not a rigorous reason to assume something. If the goal was to provide an answer to the question (i.e. prove only what you are asked to prove) then you don't need this. But if the goal is to provide a full mathematical proof then you do. $\endgroup$ – 6005 Sep 14 '13 at 15:57

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