Mixing
integration inequality over a unit interval [duplicate]
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This question already has an answer here:
Prove that $ln x leq x - 1$
4 answers
Let $mgeq 1$ be a natural number. I want to show that $$int_{m}^{m+1}log(frac{t}{m})dt leq int_{m}^{m+1}(frac{t}{m}-1)dt.$$ I have graphed it and can see the result, but don't know how to show it rigorously.
integration inequality
edited Jan 8 at 13:03
amWhy
1
asked Jan 8 at 12:34
Sam.SSam.S
649
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marked as duplicate by Martin R, Community♦ Jan 8 at 14:36
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
add a comment |
$begingroup$
This question already has an answer here:
Prove that $ln x leq x - 1$
4 answers
Let $mgeq 1$ be a natural number. I want to show that $$int_{m}^{m+1}log(frac{t}{m})dt leq int_{m}^{m+1}(frac{t}{m}-1)dt.$$ I have graphed it and can see the result, but don't know how to show it rigorously.
integration inequality
edited Jan 8 at 13:03
amWhy
1
asked Jan 8 at 12:34
Sam.SSam.S
649
$endgroup$
marked as duplicate by Martin R, Community♦ Jan 8 at 14:36
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
add a comment |
$begingroup$
This question already has an answer here:
Prove that $ln x leq x - 1$
4 answers
Let $mgeq 1$ be a natural number. I want to show that $$int_{m}^{m+1}log(frac{t}{m})dt leq int_{m}^{m+1}(frac{t}{m}-1)dt.$$ I have graphed it and can see the result, but don't know how to show it rigorously.
integration inequality
edited Jan 8 at 13:03
amWhy
1
asked Jan 8 at 12:34
Sam.SSam.S
649
$endgroup$
This question already has an answer here:
Prove that $ln x leq x - 1$
4 answers
Let $mgeq 1$ be a natural number. I want to show that $$int_{m}^{m+1}log(frac{t}{m})dt leq int_{m}^{m+1}(frac{t}{m}-1)dt.$$ I have graphed it and can see the result, but don't know how to show it rigorously.
This question already has an answer here:
Prove that $ln x leq x - 1$
4 answers
integration inequality
integration inequality
edited Jan 8 at 13:03
amWhy
1
asked Jan 8 at 12:34
Sam.SSam.S
649
edited Jan 8 at 13:03
amWhy
1
asked Jan 8 at 12:34
Sam.SSam.S
649
edited Jan 8 at 13:03
amWhy
1
edited Jan 8 at 13:03
amWhy
1
edited Jan 8 at 13:03
amWhy
1
1
asked Jan 8 at 12:34
Sam.SSam.S
649
asked Jan 8 at 12:34
Sam.SSam.S
649
asked Jan 8 at 12:34
Sam.SSam.S
649
649
marked as duplicate by Martin R, Community♦ Jan 8 at 14:36
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
marked as duplicate by Martin R, Community♦ Jan 8 at 14:36
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
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An answer without calculating the integrals is the following:
We have
$$log(frac{t}{m})leq frac{t}{m}-1;;;for ; all;mleq tleq m+1,$$
hence by the monotonicity of the integral we get the claim.
(One can see this inequality for example by considering
$$f(t)=log(frac{t}{m})-frac{t}{m}+1:$$
Since $f(m)=0$ and $f'(t)<0$ on $[m,m+1]$ (that's an easy calculation), we get
$$f(t)leq 0;;;on;[m,m+1]$$
and hence the needed inequality.)
answered Jan 8 at 13:00
Student7Student7
2089
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$begingroup$
ok thanks, this is now clear, although when you define f(t) should you not add 1 at the end, rather than subtract? I can't edit since its only 1 character.
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– Sam.S
Jan 8 at 14:33
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You're right, of course, thank you! I edited it.
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– Student7
Jan 8 at 14:35
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
An answer without calculating the integrals is the following:
We have
$$log(frac{t}{m})leq frac{t}{m}-1;;;for ; all;mleq tleq m+1,$$
hence by the monotonicity of the integral we get the claim.
(One can see this inequality for example by considering
$$f(t)=log(frac{t}{m})-frac{t}{m}+1:$$
Since $f(m)=0$ and $f'(t)<0$ on $[m,m+1]$ (that's an easy calculation), we get
$$f(t)leq 0;;;on;[m,m+1]$$
and hence the needed inequality.)
answered Jan 8 at 13:00
Student7Student7
2089
$endgroup$
$begingroup$
ok thanks, this is now clear, although when you define f(t) should you not add 1 at the end, rather than subtract? I can't edit since its only 1 character.
$endgroup$
– Sam.S
Jan 8 at 14:33
$begingroup$
You're right, of course, thank you! I edited it.
$endgroup$
– Student7
Jan 8 at 14:35
add a comment |
$begingroup$
An answer without calculating the integrals is the following:
We have
$$log(frac{t}{m})leq frac{t}{m}-1;;;for ; all;mleq tleq m+1,$$
hence by the monotonicity of the integral we get the claim.
(One can see this inequality for example by considering
$$f(t)=log(frac{t}{m})-frac{t}{m}+1:$$
Since $f(m)=0$ and $f'(t)<0$ on $[m,m+1]$ (that's an easy calculation), we get
$$f(t)leq 0;;;on;[m,m+1]$$
and hence the needed inequality.)
answered Jan 8 at 13:00
Student7Student7
2089
$endgroup$
$begingroup$
ok thanks, this is now clear, although when you define f(t) should you not add 1 at the end, rather than subtract? I can't edit since its only 1 character.
$endgroup$
– Sam.S
Jan 8 at 14:33
$begingroup$
You're right, of course, thank you! I edited it.
$endgroup$
– Student7
Jan 8 at 14:35
add a comment |
$begingroup$
An answer without calculating the integrals is the following:
We have
$$log(frac{t}{m})leq frac{t}{m}-1;;;for ; all;mleq tleq m+1,$$
hence by the monotonicity of the integral we get the claim.
(One can see this inequality for example by considering
$$f(t)=log(frac{t}{m})-frac{t}{m}+1:$$
Since $f(m)=0$ and $f'(t)<0$ on $[m,m+1]$ (that's an easy calculation), we get
$$f(t)leq 0;;;on;[m,m+1]$$
and hence the needed inequality.)
answered Jan 8 at 13:00
Student7Student7
2089
$endgroup$
An answer without calculating the integrals is the following:
We have
$$log(frac{t}{m})leq frac{t}{m}-1;;;for ; all;mleq tleq m+1,$$
hence by the monotonicity of the integral we get the claim.
(One can see this inequality for example by considering
$$f(t)=log(frac{t}{m})-frac{t}{m}+1:$$
Since $f(m)=0$ and $f'(t)<0$ on $[m,m+1]$ (that's an easy calculation), we get
$$f(t)leq 0;;;on;[m,m+1]$$
and hence the needed inequality.)
answered Jan 8 at 13:00
Student7Student7
2089
edited Jan 8 at 14:35
answered Jan 8 at 13:00
Student7Student7
2089
answered Jan 8 at 13:00
Student7Student7
2089
answered Jan 8 at 13:00
Student7Student7
2089
2089
$begingroup$
ok thanks, this is now clear, although when you define f(t) should you not add 1 at the end, rather than subtract? I can't edit since its only 1 character.
$endgroup$
– Sam.S
Jan 8 at 14:33
$begingroup$
You're right, of course, thank you! I edited it.
$endgroup$
– Student7
Jan 8 at 14:35
add a comment |
$begingroup$
ok thanks, this is now clear, although when you define f(t) should you not add 1 at the end, rather than subtract? I can't edit since its only 1 character.
$endgroup$
– Sam.S
Jan 8 at 14:33
$begingroup$
You're right, of course, thank you! I edited it.
$endgroup$
– Student7
Jan 8 at 14:35
$begingroup$
ok thanks, this is now clear, although when you define f(t) should you not add 1 at the end, rather than subtract? I can't edit since its only 1 character.
$endgroup$
– Sam.S
Jan 8 at 14:33
$begingroup$
ok thanks, this is now clear, although when you define f(t) should you not add 1 at the end, rather than subtract? I can't edit since its only 1 character.
$endgroup$
– Sam.S
Jan 8 at 14:33
$begingroup$
You're right, of course, thank you! I edited it.
$endgroup$
– Student7
Jan 8 at 14:35
$begingroup$
You're right, of course, thank you! I edited it.
$endgroup$
– Student7
Jan 8 at 14:35
add a comment |
