Mixing
If $f(0)=0$ and $f(1)=1$ there exists $(x_i)$ such that $sumlimits_{i=1}^nfrac{1}{f'(x_i)}=n$
1
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Let $f:[0,1] tomathbb R$ be differentiable, with $f(0)=0$ and $f(1)=1 $. Prove that, for every $nin mathbb N$, there exists $x_1,x_2,ldots,x_nin[0,1]$ such that $$sum_{i=1}^nfrac{1}{f'(x_i)}=n$$
sequences-and-series analysis
edited Jan 11 at 21:58
Did
248k23223460
asked Feb 2 '13 at 17:22
M.HM.H
7,28711554
$endgroup$
add a comment |
1
$begingroup$
Let $f:[0,1] tomathbb R$ be differentiable, with $f(0)=0$ and $f(1)=1 $. Prove that, for every $nin mathbb N$, there exists $x_1,x_2,ldots,x_nin[0,1]$ such that $$sum_{i=1}^nfrac{1}{f'(x_i)}=n$$
sequences-and-series analysis
edited Jan 11 at 21:58
Did
248k23223460
asked Feb 2 '13 at 17:22
M.HM.H
7,28711554
$endgroup$
$begingroup$
i use Mean value theorem to find f'.yes i prove for n=1
$endgroup$
– M.H
Feb 2 '13 at 17:31
add a comment |
1
1
1
$begingroup$
Let $f:[0,1] tomathbb R$ be differentiable, with $f(0)=0$ and $f(1)=1 $. Prove that, for every $nin mathbb N$, there exists $x_1,x_2,ldots,x_nin[0,1]$ such that $$sum_{i=1}^nfrac{1}{f'(x_i)}=n$$
sequences-and-series analysis
edited Jan 11 at 21:58
Did
248k23223460
asked Feb 2 '13 at 17:22
M.HM.H
7,28711554
$endgroup$
Let $f:[0,1] tomathbb R$ be differentiable, with $f(0)=0$ and $f(1)=1 $. Prove that, for every $nin mathbb N$, there exists $x_1,x_2,ldots,x_nin[0,1]$ such that $$sum_{i=1}^nfrac{1}{f'(x_i)}=n$$
sequences-and-series analysis
sequences-and-series analysis
edited Jan 11 at 21:58
Did
248k23223460
asked Feb 2 '13 at 17:22
M.HM.H
7,28711554
edited Jan 11 at 21:58
Did
248k23223460
asked Feb 2 '13 at 17:22
M.HM.H
7,28711554
edited Jan 11 at 21:58
Did
248k23223460
edited Jan 11 at 21:58
Did
248k23223460
edited Jan 11 at 21:58
Did
248k23223460
248k23223460
asked Feb 2 '13 at 17:22
M.HM.H
7,28711554
asked Feb 2 '13 at 17:22
M.HM.H
7,28711554
asked Feb 2 '13 at 17:22
M.HM.H
7,28711554
7,28711554
$begingroup$
i use Mean value theorem to find f'.yes i prove for n=1
$endgroup$
– M.H
Feb 2 '13 at 17:31
add a comment |
$begingroup$
i use Mean value theorem to find f'.yes i prove for n=1
$endgroup$
– M.H
Feb 2 '13 at 17:31
$begingroup$
i use Mean value theorem to find f'.yes i prove for n=1
$endgroup$
– M.H
Feb 2 '13 at 17:31
$begingroup$
i use Mean value theorem to find f'.yes i prove for n=1
$endgroup$
– M.H
Feb 2 '13 at 17:31
add a comment |
1 Answer
1
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By the mean value theorem, there exists $x$ such that $f(1)-f(0)=f'(x)(1-0)$. Take $x_i=x$ for every $igeqslant1$.
answered Feb 2 '13 at 17:25
DidDid
248k23223460
$endgroup$
1
$begingroup$
You mean the mean value theorem, not the intermediate value theorem.
$endgroup$
– Chris Eagle
Feb 2 '13 at 17:27
$begingroup$
@ChrisEagle Indeed I do. Thanks.
$endgroup$
– Did
Feb 2 '13 at 17:28
$begingroup$
@did:thanks sir
$endgroup$
– M.H
Feb 2 '13 at 17:36
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
4
$begingroup$
By the mean value theorem, there exists $x$ such that $f(1)-f(0)=f'(x)(1-0)$. Take $x_i=x$ for every $igeqslant1$.
answered Feb 2 '13 at 17:25
DidDid
248k23223460
$endgroup$
1
$begingroup$
You mean the mean value theorem, not the intermediate value theorem.
$endgroup$
– Chris Eagle
Feb 2 '13 at 17:27
$begingroup$
@ChrisEagle Indeed I do. Thanks.
$endgroup$
– Did
Feb 2 '13 at 17:28
$begingroup$
@did:thanks sir
$endgroup$
– M.H
Feb 2 '13 at 17:36
add a comment |
4
$begingroup$
By the mean value theorem, there exists $x$ such that $f(1)-f(0)=f'(x)(1-0)$. Take $x_i=x$ for every $igeqslant1$.
answered Feb 2 '13 at 17:25
DidDid
248k23223460
$endgroup$
1
$begingroup$
You mean the mean value theorem, not the intermediate value theorem.
$endgroup$
– Chris Eagle
Feb 2 '13 at 17:27
$begingroup$
@ChrisEagle Indeed I do. Thanks.
$endgroup$
– Did
Feb 2 '13 at 17:28
$begingroup$
@did:thanks sir
$endgroup$
– M.H
Feb 2 '13 at 17:36
add a comment |
4
4
4
$begingroup$
By the mean value theorem, there exists $x$ such that $f(1)-f(0)=f'(x)(1-0)$. Take $x_i=x$ for every $igeqslant1$.
answered Feb 2 '13 at 17:25
DidDid
248k23223460
$endgroup$
By the mean value theorem, there exists $x$ such that $f(1)-f(0)=f'(x)(1-0)$. Take $x_i=x$ for every $igeqslant1$.
answered Feb 2 '13 at 17:25
DidDid
248k23223460
answered Feb 2 '13 at 17:25
DidDid
248k23223460
answered Feb 2 '13 at 17:25
DidDid
248k23223460
answered Feb 2 '13 at 17:25
DidDid
248k23223460
248k23223460
1
$begingroup$
You mean the mean value theorem, not the intermediate value theorem.
$endgroup$
– Chris Eagle
Feb 2 '13 at 17:27
$begingroup$
@ChrisEagle Indeed I do. Thanks.
$endgroup$
– Did
Feb 2 '13 at 17:28
$begingroup$
@did:thanks sir
$endgroup$
– M.H
Feb 2 '13 at 17:36
add a comment |
1
$begingroup$
You mean the mean value theorem, not the intermediate value theorem.
$endgroup$
– Chris Eagle
Feb 2 '13 at 17:27
$begingroup$
@ChrisEagle Indeed I do. Thanks.
$endgroup$
– Did
Feb 2 '13 at 17:28
$begingroup$
@did:thanks sir
$endgroup$
– M.H
Feb 2 '13 at 17:36
1
1
$begingroup$
You mean the mean value theorem, not the intermediate value theorem.
$endgroup$
– Chris Eagle
Feb 2 '13 at 17:27
$begingroup$
You mean the mean value theorem, not the intermediate value theorem.
$endgroup$
– Chris Eagle
Feb 2 '13 at 17:27
$begingroup$
@ChrisEagle Indeed I do. Thanks.
$endgroup$
– Did
Feb 2 '13 at 17:28
$begingroup$
@ChrisEagle Indeed I do. Thanks.
$endgroup$
– Did
Feb 2 '13 at 17:28
$begingroup$
@did:thanks sir
$endgroup$
– M.H
Feb 2 '13 at 17:36
$begingroup$
@did:thanks sir
$endgroup$
– M.H
Feb 2 '13 at 17:36
add a comment |
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$begingroup$
i use Mean value theorem to find f'.yes i prove for n=1
$endgroup$
– M.H
Feb 2 '13 at 17:31