A question that combines Darboux's theorem, Mean Value Theorem, and possibly Intermediate Value Theorem.
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I encounter this question in an exam paper of my uni, and to my amazement it somehow combines three important theorem in elementary analysis into a single question. My problem is with the a part of the question.
Statement of the question
Let $a,b$ $in mathbb{R}$ with $a<b$. And let $f:mathbb{R} to mathbb{R}$ be a differentiable function on $mathbb{R}$. Now suppose that $f(a)<0, f(b)>0$ and the derivative $f'$ is strictly decreasing. Prove that for the following sequence:
$$ x_{n+1} = x_n - frac{f(x_n)}{f'(x_n)} ; x_1 = a $$
that
$$1. aleq x_n<b$$
$$2. f(x_n)<0$$
$$3. f'(x_n)>0$$
I think I already succeed in proving that first and the third part of the question by induction. The problem is:
How can the second part be proved?
Here is my attempt for the remaining two parts:
First note that $f'(a)>0$, and for $n=1$ all the statements are correct. Then assume that they are true for $n=k$, so $f(x_k)<0$ and $f'(x_k)>0$. Then assume to the contrary that $x_{k+1} >b$, we have
$$x_k - frac{f(x_k)}{f'(x_k)} >b$$
$$- frac{f(x_k)}{b-x_k} > f'(x_k)$$
This implies:
$$frac{f(b)}{b-x_k}- frac{f(x_k)}{b-x_k} > f'(x_k)$$
Then by the Mean Value Theorem, there exists a point $cin (x_k,b)$ such that $f'(c)>f'(x_k)$ but $c>x_k$, this is a contradiction. Thus the first statement is proved. For the third statement, assume to the contrary that $f'(x_{k+1}) leq 0$, then by Darboux's theorem we have a point $c in (x_k, x_{k+1})$ such that $f'(c)=0$, this combined with the fact that the derivatives are decreasing implies that $f(b)<0$, a contradiction again.
real-analysis sequences-and-series derivatives
asked Dec 6 '18 at 16:25
hephaeshephaes
1709
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add a comment |
$begingroup$
I encounter this question in an exam paper of my uni, and to my amazement it somehow combines three important theorem in elementary analysis into a single question. My problem is with the a part of the question.
Statement of the question
Let $a,b$ $in mathbb{R}$ with $a<b$. And let $f:mathbb{R} to mathbb{R}$ be a differentiable function on $mathbb{R}$. Now suppose that $f(a)<0, f(b)>0$ and the derivative $f'$ is strictly decreasing. Prove that for the following sequence:
$$ x_{n+1} = x_n - frac{f(x_n)}{f'(x_n)} ; x_1 = a $$
that
$$1. aleq x_n<b$$
$$2. f(x_n)<0$$
$$3. f'(x_n)>0$$
I think I already succeed in proving that first and the third part of the question by induction. The problem is:
How can the second part be proved?
Here is my attempt for the remaining two parts:
First note that $f'(a)>0$, and for $n=1$ all the statements are correct. Then assume that they are true for $n=k$, so $f(x_k)<0$ and $f'(x_k)>0$. Then assume to the contrary that $x_{k+1} >b$, we have
$$x_k - frac{f(x_k)}{f'(x_k)} >b$$
$$- frac{f(x_k)}{b-x_k} > f'(x_k)$$
This implies:
$$frac{f(b)}{b-x_k}- frac{f(x_k)}{b-x_k} > f'(x_k)$$
Then by the Mean Value Theorem, there exists a point $cin (x_k,b)$ such that $f'(c)>f'(x_k)$ but $c>x_k$, this is a contradiction. Thus the first statement is proved. For the third statement, assume to the contrary that $f'(x_{k+1}) leq 0$, then by Darboux's theorem we have a point $c in (x_k, x_{k+1})$ such that $f'(c)=0$, this combined with the fact that the derivatives are decreasing implies that $f(b)<0$, a contradiction again.
real-analysis sequences-and-series derivatives
asked Dec 6 '18 at 16:25
hephaeshephaes
1709
$endgroup$
add a comment |
$begingroup$
I encounter this question in an exam paper of my uni, and to my amazement it somehow combines three important theorem in elementary analysis into a single question. My problem is with the a part of the question.
Statement of the question
Let $a,b$ $in mathbb{R}$ with $a<b$. And let $f:mathbb{R} to mathbb{R}$ be a differentiable function on $mathbb{R}$. Now suppose that $f(a)<0, f(b)>0$ and the derivative $f'$ is strictly decreasing. Prove that for the following sequence:
$$ x_{n+1} = x_n - frac{f(x_n)}{f'(x_n)} ; x_1 = a $$
that
$$1. aleq x_n<b$$
$$2. f(x_n)<0$$
$$3. f'(x_n)>0$$
I think I already succeed in proving that first and the third part of the question by induction. The problem is:
How can the second part be proved?
Here is my attempt for the remaining two parts:
First note that $f'(a)>0$, and for $n=1$ all the statements are correct. Then assume that they are true for $n=k$, so $f(x_k)<0$ and $f'(x_k)>0$. Then assume to the contrary that $x_{k+1} >b$, we have
$$x_k - frac{f(x_k)}{f'(x_k)} >b$$
$$- frac{f(x_k)}{b-x_k} > f'(x_k)$$
This implies:
$$frac{f(b)}{b-x_k}- frac{f(x_k)}{b-x_k} > f'(x_k)$$
Then by the Mean Value Theorem, there exists a point $cin (x_k,b)$ such that $f'(c)>f'(x_k)$ but $c>x_k$, this is a contradiction. Thus the first statement is proved. For the third statement, assume to the contrary that $f'(x_{k+1}) leq 0$, then by Darboux's theorem we have a point $c in (x_k, x_{k+1})$ such that $f'(c)=0$, this combined with the fact that the derivatives are decreasing implies that $f(b)<0$, a contradiction again.
real-analysis sequences-and-series derivatives
asked Dec 6 '18 at 16:25
hephaeshephaes
1709
$endgroup$
I encounter this question in an exam paper of my uni, and to my amazement it somehow combines three important theorem in elementary analysis into a single question. My problem is with the a part of the question.
Statement of the question
Let $a,b$ $in mathbb{R}$ with $a<b$. And let $f:mathbb{R} to mathbb{R}$ be a differentiable function on $mathbb{R}$. Now suppose that $f(a)<0, f(b)>0$ and the derivative $f'$ is strictly decreasing. Prove that for the following sequence:
$$ x_{n+1} = x_n - frac{f(x_n)}{f'(x_n)} ; x_1 = a $$
that
$$1. aleq x_n<b$$
$$2. f(x_n)<0$$
$$3. f'(x_n)>0$$
I think I already succeed in proving that first and the third part of the question by induction. The problem is:
How can the second part be proved?
Here is my attempt for the remaining two parts:
First note that $f'(a)>0$, and for $n=1$ all the statements are correct. Then assume that they are true for $n=k$, so $f(x_k)<0$ and $f'(x_k)>0$. Then assume to the contrary that $x_{k+1} >b$, we have
$$x_k - frac{f(x_k)}{f'(x_k)} >b$$
$$- frac{f(x_k)}{b-x_k} > f'(x_k)$$
This implies:
$$frac{f(b)}{b-x_k}- frac{f(x_k)}{b-x_k} > f'(x_k)$$
Then by the Mean Value Theorem, there exists a point $cin (x_k,b)$ such that $f'(c)>f'(x_k)$ but $c>x_k$, this is a contradiction. Thus the first statement is proved. For the third statement, assume to the contrary that $f'(x_{k+1}) leq 0$, then by Darboux's theorem we have a point $c in (x_k, x_{k+1})$ such that $f'(c)=0$, this combined with the fact that the derivatives are decreasing implies that $f(b)<0$, a contradiction again.
real-analysis sequences-and-series derivatives
real-analysis sequences-and-series derivatives
asked Dec 6 '18 at 16:25
hephaeshephaes
1709
asked Dec 6 '18 at 16:25
hephaeshephaes
1709
asked Dec 6 '18 at 16:25
hephaeshephaes
1709
asked Dec 6 '18 at 16:25
hephaeshephaes
1709
asked Dec 6 '18 at 16:25
hephaeshephaes
1709
1709
add a comment |
add a comment |
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