Prove $exists xi, eta in (0,1)$, such that $f(xi)+f'(eta)=xi+eta$.
Suppose $f(x)$ is continuous in $[0,1]$, derivable in $(0,1)$, $f(0)=0, f(1)=frac 12$, prove
$exists xi, eta in (0,1)$, such that
$$
f(xi)+f'(eta)=xi+eta.
$$
I have tried Cauchy mean value theorem:
let $F(x)=int_0^x f(t)-t ,dt, G(x)=frac{1}{2}x^2-f(x)$, then $exists xi in (0,1), eta in (0,1)$, such that
$$
frac{F(1)-F(0)}{1-0} = F'(xi)
Leftrightarrow f(xi)-xi = int_0^1 f(t) ,dt - frac 12.
$$
$$
frac{G(1)-G(0)}{1-0} = G'(eta)
Leftrightarrow eta-f'(eta) = frac 12 - f(1) = 0.
$$
but failed.
Any hints are appreciated, thanks for your help.
The conditions of this question are wrong, but if i rectify them, the question will contradict the anwser of Mr. Robert Z to some extent, so please be sure to see the anwser after skimming this question. I am sorry for the inconvenience.
calculus real-analysis
asked Nov 29 at 6:05
gcfsjfcus
596
add a comment |
Suppose $f(x)$ is continuous in $[0,1]$, derivable in $(0,1)$, $f(0)=0, f(1)=frac 12$, prove
$exists xi, eta in (0,1)$, such that
$$
f(xi)+f'(eta)=xi+eta.
$$
I have tried Cauchy mean value theorem:
let $F(x)=int_0^x f(t)-t ,dt, G(x)=frac{1}{2}x^2-f(x)$, then $exists xi in (0,1), eta in (0,1)$, such that
$$
frac{F(1)-F(0)}{1-0} = F'(xi)
Leftrightarrow f(xi)-xi = int_0^1 f(t) ,dt - frac 12.
$$
$$
frac{G(1)-G(0)}{1-0} = G'(eta)
Leftrightarrow eta-f'(eta) = frac 12 - f(1) = 0.
$$
but failed.
Any hints are appreciated, thanks for your help.
The conditions of this question are wrong, but if i rectify them, the question will contradict the anwser of Mr. Robert Z to some extent, so please be sure to see the anwser after skimming this question. I am sorry for the inconvenience.
calculus real-analysis
asked Nov 29 at 6:05
gcfsjfcus
596
1
It feels like a place where we want to use Rolle's somehow...
– Mason
Nov 29 at 6:23
@gcfsjfcus Do you mind to write something about the origin of this problem? I am curious about it!
– Robert Z
Nov 29 at 12:51
@RobertZ This is a postgraduate examination question of the Chinese Academy of Sciences.
– gcfsjfcus
Nov 30 at 4:05
@RobertZ May be recalled by candidates.
– gcfsjfcus
Nov 30 at 4:14
@gcfsjfcus Thanks!
– Robert Z
Nov 30 at 5:12
add a comment |
Suppose $f(x)$ is continuous in $[0,1]$, derivable in $(0,1)$, $f(0)=0, f(1)=frac 12$, prove
$exists xi, eta in (0,1)$, such that
$$
f(xi)+f'(eta)=xi+eta.
$$
I have tried Cauchy mean value theorem:
let $F(x)=int_0^x f(t)-t ,dt, G(x)=frac{1}{2}x^2-f(x)$, then $exists xi in (0,1), eta in (0,1)$, such that
$$
frac{F(1)-F(0)}{1-0} = F'(xi)
Leftrightarrow f(xi)-xi = int_0^1 f(t) ,dt - frac 12.
$$
$$
frac{G(1)-G(0)}{1-0} = G'(eta)
Leftrightarrow eta-f'(eta) = frac 12 - f(1) = 0.
$$
but failed.
Any hints are appreciated, thanks for your help.
The conditions of this question are wrong, but if i rectify them, the question will contradict the anwser of Mr. Robert Z to some extent, so please be sure to see the anwser after skimming this question. I am sorry for the inconvenience.
calculus real-analysis
asked Nov 29 at 6:05
gcfsjfcus
596
Suppose $f(x)$ is continuous in $[0,1]$, derivable in $(0,1)$, $f(0)=0, f(1)=frac 12$, prove
$exists xi, eta in (0,1)$, such that
$$
f(xi)+f'(eta)=xi+eta.
$$
I have tried Cauchy mean value theorem:
let $F(x)=int_0^x f(t)-t ,dt, G(x)=frac{1}{2}x^2-f(x)$, then $exists xi in (0,1), eta in (0,1)$, such that
$$
frac{F(1)-F(0)}{1-0} = F'(xi)
Leftrightarrow f(xi)-xi = int_0^1 f(t) ,dt - frac 12.
$$
$$
frac{G(1)-G(0)}{1-0} = G'(eta)
Leftrightarrow eta-f'(eta) = frac 12 - f(1) = 0.
$$
but failed.
Any hints are appreciated, thanks for your help.
The conditions of this question are wrong, but if i rectify them, the question will contradict the anwser of Mr. Robert Z to some extent, so please be sure to see the anwser after skimming this question. I am sorry for the inconvenience.
calculus real-analysis
calculus real-analysis
asked Nov 29 at 6:05
gcfsjfcus
596
asked Nov 29 at 6:05
gcfsjfcus
596
edited Nov 30 at 23:03
asked Nov 29 at 6:05
gcfsjfcus
596
asked Nov 29 at 6:05
gcfsjfcus
596
asked Nov 29 at 6:05
gcfsjfcus
596
596
1
It feels like a place where we want to use Rolle's somehow...
– Mason
Nov 29 at 6:23
@gcfsjfcus Do you mind to write something about the origin of this problem? I am curious about it!
– Robert Z
Nov 29 at 12:51
@RobertZ This is a postgraduate examination question of the Chinese Academy of Sciences.
– gcfsjfcus
Nov 30 at 4:05
@RobertZ May be recalled by candidates.
– gcfsjfcus
Nov 30 at 4:14
@gcfsjfcus Thanks!
– Robert Z
Nov 30 at 5:12
add a comment |
1
It feels like a place where we want to use Rolle's somehow...
– Mason
Nov 29 at 6:23
@gcfsjfcus Do you mind to write something about the origin of this problem? I am curious about it!
– Robert Z
Nov 29 at 12:51
@RobertZ This is a postgraduate examination question of the Chinese Academy of Sciences.
– gcfsjfcus
Nov 30 at 4:05
@RobertZ May be recalled by candidates.
– gcfsjfcus
Nov 30 at 4:14
@gcfsjfcus Thanks!
– Robert Z
Nov 30 at 5:12
1
1
It feels like a place where we want to use Rolle's somehow...
– Mason
Nov 29 at 6:23
It feels like a place where we want to use Rolle's somehow...
– Mason
Nov 29 at 6:23
@gcfsjfcus Do you mind to write something about the origin of this problem? I am curious about it!
– Robert Z
Nov 29 at 12:51
@gcfsjfcus Do you mind to write something about the origin of this problem? I am curious about it!
– Robert Z
Nov 29 at 12:51
@RobertZ This is a postgraduate examination question of the Chinese Academy of Sciences.
– gcfsjfcus
Nov 30 at 4:05
@RobertZ This is a postgraduate examination question of the Chinese Academy of Sciences.
– gcfsjfcus
Nov 30 at 4:05
@RobertZ May be recalled by candidates.
– gcfsjfcus
Nov 30 at 4:14
@RobertZ May be recalled by candidates.
– gcfsjfcus
Nov 30 at 4:14
@gcfsjfcus Thanks!
– Robert Z
Nov 30 at 5:12
@gcfsjfcus Thanks!
– Robert Z
Nov 30 at 5:12
add a comment |
1 Answer
1
active
oldest
votes
The statement is false, I have a counterexample: if $f(x)=x^2/2$ then $f(0)=0$, $f(1)=frac 12$, and
$$f(xi)+f'(eta)=frac{xi^2}{2}+eta<xi+eta$$
for any $xi,etain(0,1)$.
Note that the proposition holds with $xiin [0,1)$ and $etain(0,1)$. In fact, let $F(x)=f(x)-x^2/2$ then $F(0)=F(1)=0$ and therefore, by Rolle's theorem, there is $etain(0,1)$ such that $F'(eta)=f'(eta)-eta=0$. Then by letting $xi=0$ we get
$$f(xi)+f'(eta)=xi+eta.$$
answered Nov 29 at 6:44
Robert Z
93.1k1060131
Nice. Is it true that there is always $eta, xi in [0,1]$ that does the job?
– Mason
Nov 29 at 6:48
2
@Mason Yes, see my edit.
– Robert Z
Nov 29 at 6:48
@RobertZ many thanks!
– gcfsjfcus
Nov 29 at 6:52
add a comment |
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1 Answer
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1 Answer
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oldest
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active
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votes
The statement is false, I have a counterexample: if $f(x)=x^2/2$ then $f(0)=0$, $f(1)=frac 12$, and
$$f(xi)+f'(eta)=frac{xi^2}{2}+eta<xi+eta$$
for any $xi,etain(0,1)$.
Note that the proposition holds with $xiin [0,1)$ and $etain(0,1)$. In fact, let $F(x)=f(x)-x^2/2$ then $F(0)=F(1)=0$ and therefore, by Rolle's theorem, there is $etain(0,1)$ such that $F'(eta)=f'(eta)-eta=0$. Then by letting $xi=0$ we get
$$f(xi)+f'(eta)=xi+eta.$$
answered Nov 29 at 6:44
Robert Z
93.1k1060131
Nice. Is it true that there is always $eta, xi in [0,1]$ that does the job?
– Mason
Nov 29 at 6:48
2
@Mason Yes, see my edit.
– Robert Z
Nov 29 at 6:48
@RobertZ many thanks!
– gcfsjfcus
Nov 29 at 6:52
add a comment |
The statement is false, I have a counterexample: if $f(x)=x^2/2$ then $f(0)=0$, $f(1)=frac 12$, and
$$f(xi)+f'(eta)=frac{xi^2}{2}+eta<xi+eta$$
for any $xi,etain(0,1)$.
Note that the proposition holds with $xiin [0,1)$ and $etain(0,1)$. In fact, let $F(x)=f(x)-x^2/2$ then $F(0)=F(1)=0$ and therefore, by Rolle's theorem, there is $etain(0,1)$ such that $F'(eta)=f'(eta)-eta=0$. Then by letting $xi=0$ we get
$$f(xi)+f'(eta)=xi+eta.$$
answered Nov 29 at 6:44
Robert Z
93.1k1060131
Nice. Is it true that there is always $eta, xi in [0,1]$ that does the job?
– Mason
Nov 29 at 6:48
2
@Mason Yes, see my edit.
– Robert Z
Nov 29 at 6:48
@RobertZ many thanks!
– gcfsjfcus
Nov 29 at 6:52
add a comment |
The statement is false, I have a counterexample: if $f(x)=x^2/2$ then $f(0)=0$, $f(1)=frac 12$, and
$$f(xi)+f'(eta)=frac{xi^2}{2}+eta<xi+eta$$
for any $xi,etain(0,1)$.
Note that the proposition holds with $xiin [0,1)$ and $etain(0,1)$. In fact, let $F(x)=f(x)-x^2/2$ then $F(0)=F(1)=0$ and therefore, by Rolle's theorem, there is $etain(0,1)$ such that $F'(eta)=f'(eta)-eta=0$. Then by letting $xi=0$ we get
$$f(xi)+f'(eta)=xi+eta.$$
answered Nov 29 at 6:44
Robert Z
93.1k1060131
The statement is false, I have a counterexample: if $f(x)=x^2/2$ then $f(0)=0$, $f(1)=frac 12$, and
$$f(xi)+f'(eta)=frac{xi^2}{2}+eta<xi+eta$$
for any $xi,etain(0,1)$.
Note that the proposition holds with $xiin [0,1)$ and $etain(0,1)$. In fact, let $F(x)=f(x)-x^2/2$ then $F(0)=F(1)=0$ and therefore, by Rolle's theorem, there is $etain(0,1)$ such that $F'(eta)=f'(eta)-eta=0$. Then by letting $xi=0$ we get
$$f(xi)+f'(eta)=xi+eta.$$
answered Nov 29 at 6:44
Robert Z
93.1k1060131
edited Nov 29 at 12:48
answered Nov 29 at 6:44
Robert Z
93.1k1060131
answered Nov 29 at 6:44
Robert Z
93.1k1060131
answered Nov 29 at 6:44
Robert Z
93.1k1060131
93.1k1060131
Nice. Is it true that there is always $eta, xi in [0,1]$ that does the job?
– Mason
Nov 29 at 6:48
2
@Mason Yes, see my edit.
– Robert Z
Nov 29 at 6:48
@RobertZ many thanks!
– gcfsjfcus
Nov 29 at 6:52
add a comment |
Nice. Is it true that there is always $eta, xi in [0,1]$ that does the job?
– Mason
Nov 29 at 6:48
2
@Mason Yes, see my edit.
– Robert Z
Nov 29 at 6:48
@RobertZ many thanks!
– gcfsjfcus
Nov 29 at 6:52
Nice. Is it true that there is always $eta, xi in [0,1]$ that does the job?
– Mason
Nov 29 at 6:48
Nice. Is it true that there is always $eta, xi in [0,1]$ that does the job?
– Mason
Nov 29 at 6:48
2
2
@Mason Yes, see my edit.
– Robert Z
Nov 29 at 6:48
@Mason Yes, see my edit.
– Robert Z
Nov 29 at 6:48
@RobertZ many thanks!
– gcfsjfcus
Nov 29 at 6:52
@RobertZ many thanks!
– gcfsjfcus
Nov 29 at 6:52
add a comment |
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1
It feels like a place where we want to use Rolle's somehow...
– Mason
Nov 29 at 6:23
@gcfsjfcus Do you mind to write something about the origin of this problem? I am curious about it!
– Robert Z
Nov 29 at 12:51
@RobertZ This is a postgraduate examination question of the Chinese Academy of Sciences.
– gcfsjfcus
Nov 30 at 4:05
@RobertZ May be recalled by candidates.
– gcfsjfcus
Nov 30 at 4:14
@gcfsjfcus Thanks!
– Robert Z
Nov 30 at 5:12