Exercise on sequence of a function

Question

Subject: Seeking Help with a Mathematics Exercise

Hello everyone,

I hope this message finds you well. I'm reaching out to seek assistance with a mathematics exercise that has been posing some challenges for me. The exercise is as follows:

Let $n$ be an integer, $n \geq 1$. We denote $f_n$ as the function defined on $\mathbb{R}$ by $f_n(x) = x^n + x^{(n-1)} + \ldots + x - 1$.

Consider the equation $E(n): x^n + x^{(n-1)} + \ldots + x - 1 = 0$. I am required to:

1. Prove that the equation $E(n)$ has a unique strictly positive solution, denoted as $a_n$.
2. Calculate $a_1$ and $a_2$.
3. Show that the sequence $a_n$ is decreasing.
4. Demonstrate that the sequence $(a_n)^n$ tends towards zero. Caution: the reasoning involved is delicate!
5. Prove that $(a_n)^{n+1} - 1 = 2(a_{n-1})$ and deduce that the sequence $(a_n)$ tends towards $1/2$.

I've already made some progress with the exercise, but I'm facing difficulties in certain parts of the proof. If someone could help me articulate the necessary justifications and fill in the missing steps, I would greatly appreciate it.

Thank you in advance for your invaluable assistance!

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What I try to do :

1. Prove that the equation $E(n)$ has a unique strictly positive solution, denoted as $a_n$: I think we can consider the function $f_n(x)$. And we notice that $f_n(0) = -1$ and $f_n(1) = n$. Since $f_n(x)$ is continuous, by the Intermediate Value Theorem, there exists at least one root in the interval $(0, 1)$. To show uniqueness and strict positivity, we can use the fact that $f_n'(x) = nx^{(n-1)} + (n-1)x^{(n-2)} + \ldots + 1 > 0$ for $0 < x < 1$. Hence, there is only one root in the interval, and it is strictly positive.

2. Calculate $a_1$ and $a_2$: For $n = 1$, $E(1): x - 1 = 0$ implies $a_1 = 1$.

   For $n = 2$, $E(2): x^2 + x - 1 = 0$ maybe this can be solved using the quadratic formula, and the positive root is $a_2 = \frac{-1 + \sqrt{5}}{2}$.

3. Show that the sequence $(a_n)$ is decreasing:

I think we might observe that $f_n'(x) > 0$ for $0 < x < 1$, implying that $f_n(x)$ is strictly increasing on $(0,1)$. Therefore, $a_{n+1} < a_n$.

4. Demonstrate that the sequence $(a_n)^n$ tends towards zero:

Maybe I can say : As $0 < a_n < 1$, $(a_n)^n$ approaches zero as $n$ goes to infinity.

5. Prove that $(a_n)^{n+1} - 1 = 2(a_{n}-1)$ and deduce that the sequence $(a_n)$ tends towards $1/2$: I don't know how to do it

Answer 1

Since this is an exercise, I don't want to give you a full solution, but rather some sketches for you to expand on.

1. What you did is almost fine: notice that $f'_n(x)>0$ for all $x>0$, thus $f_n$ is strictly increasing on all $(0,+\infty)$ and it has a unique positive solution since $f_n(0)=-1<0$ (also notice that $f_n(1)=n-1)$.
2. Correct.
3. What you did here is not so clear: $a_{n+1}$ is the positive solution to $f_{n+1}(x)=0$, hence saying that $f_n(x)$ is increasing isn't enough. Perhaps notice that for $x>0$ you have $f_n(x)<f_{n+1}(x)$, thus $$f_{n+1}(a_{n+1})=0=f_n(a_n)< f_{n+1}(a_n)$$ and conclude from the fact that $f_{n+1}$ is strictly increasing.
4. You should add more details: by the previous points you know $0<a_n<a_2<1$ for all $n\geq 2$, thus $$0\leq a_n^n<a_2^n\to 0 \qquad \text{for } n\to \infty.$$
5. There must be some typo since this is false for $n=2$. Assuming this is $$a_n^{n+1}=2a_n-1 \qquad (\star)$$ we can proceed as follows: notice that $f_{n+1}(x)=x^{n+1}+f_n(x)$ and $f_{n+1}(x)-xf_n(x)=2x-1$, thus $$2a_n-1=f_{n+1}(a_n)-a_n f_n(a_n)=a_n^{n+1} + f_n(a_n)-0=a_n^{n+1}$$ for every $n\geq 1$. Since $a_n$ is strictly decreasing and bounded from below we know that $a_n\to L\geq 0$, thus taking the limit on both sides of $(\star)$ we get $$2L-1=\lim_{n\to\infty} 2a_n-1=\lim_{n\to\infty} a_n^{n+1}=0$$ by the previous point.

Answer 2

As the previous comment mentioned, the last question is false. However, note that for $0<x<1$, it holds $\sum_{k=1}^nx^k=\frac{x-x^{n+1}}{1-x}$ by the geometric series formula. Thus, since we know that $0<a_n<1$, we have that $a_n$ satisfies $$ a_n^{n+1}+1=2a_n $$ By part 4, the left hand side tends to 1, and so we conclude that $a_n \to \frac{1}{2}$.