Log-sum inequality
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Let $(a^n)_{n in mathbb{N}} subset mathbb{R}^k$ be a sequence of vectors with non-negative components and with $sum_{i=1}^k a^n_i=1$ for all $ninmathbb{N}$. Then we have for each $ninmathbb{N}$ by the log-sum inequality:
$sum_{i=1}^k a_i^{n+1} logdfrac{a_i^{n+1}}{a_i^n} geq left(sum_{i=1}^k a_i^{n+1}right) logdfrac{sum_{i=1}^k a_i^{n+1}}{sum_{i=1}^k a_i^{n}}=0,$
where the equality holds if and only if $a_i^{n+1}/a_i^n$ is constant for any $i=1,...,k$.
Now assume that
$limlimits_{nrightarrow infty}sum_{i=1}^k a_i^{n+1} logdfrac{a_i^{n+1}}{a_i^n} = 0.$
Can we conclude that $limlimits_{nrightarrow infty} dfrac{a_i^{n+1}}{a_i^n} = 1$? Intuitively yes, but how can this be proven rigorously?
Thanks in advance!
real-analysis analysis inequality
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add a comment |
$begingroup$
Let $(a^n)_{n in mathbb{N}} subset mathbb{R}^k$ be a sequence of vectors with non-negative components and with $sum_{i=1}^k a^n_i=1$ for all $ninmathbb{N}$. Then we have for each $ninmathbb{N}$ by the log-sum inequality:
$sum_{i=1}^k a_i^{n+1} logdfrac{a_i^{n+1}}{a_i^n} geq left(sum_{i=1}^k a_i^{n+1}right) logdfrac{sum_{i=1}^k a_i^{n+1}}{sum_{i=1}^k a_i^{n}}=0,$
where the equality holds if and only if $a_i^{n+1}/a_i^n$ is constant for any $i=1,...,k$.
Now assume that
$limlimits_{nrightarrow infty}sum_{i=1}^k a_i^{n+1} logdfrac{a_i^{n+1}}{a_i^n} = 0.$
Can we conclude that $limlimits_{nrightarrow infty} dfrac{a_i^{n+1}}{a_i^n} = 1$? Intuitively yes, but how can this be proven rigorously?
Thanks in advance!
real-analysis analysis inequality
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I find your notation $a_i^n$ confusing and keep wanting to see the $n$ as an exponent rather than an index. I would write this as $a_{i, n}$ or $a_{n, i}$.
$endgroup$
– marty cohen
Dec 17 '18 at 18:18
add a comment |
$begingroup$
Let $(a^n)_{n in mathbb{N}} subset mathbb{R}^k$ be a sequence of vectors with non-negative components and with $sum_{i=1}^k a^n_i=1$ for all $ninmathbb{N}$. Then we have for each $ninmathbb{N}$ by the log-sum inequality:
$sum_{i=1}^k a_i^{n+1} logdfrac{a_i^{n+1}}{a_i^n} geq left(sum_{i=1}^k a_i^{n+1}right) logdfrac{sum_{i=1}^k a_i^{n+1}}{sum_{i=1}^k a_i^{n}}=0,$
where the equality holds if and only if $a_i^{n+1}/a_i^n$ is constant for any $i=1,...,k$.
Now assume that
$limlimits_{nrightarrow infty}sum_{i=1}^k a_i^{n+1} logdfrac{a_i^{n+1}}{a_i^n} = 0.$
Can we conclude that $limlimits_{nrightarrow infty} dfrac{a_i^{n+1}}{a_i^n} = 1$? Intuitively yes, but how can this be proven rigorously?
Thanks in advance!
real-analysis analysis inequality
$endgroup$
Let $(a^n)_{n in mathbb{N}} subset mathbb{R}^k$ be a sequence of vectors with non-negative components and with $sum_{i=1}^k a^n_i=1$ for all $ninmathbb{N}$. Then we have for each $ninmathbb{N}$ by the log-sum inequality:
$sum_{i=1}^k a_i^{n+1} logdfrac{a_i^{n+1}}{a_i^n} geq left(sum_{i=1}^k a_i^{n+1}right) logdfrac{sum_{i=1}^k a_i^{n+1}}{sum_{i=1}^k a_i^{n}}=0,$
where the equality holds if and only if $a_i^{n+1}/a_i^n$ is constant for any $i=1,...,k$.
Now assume that
$limlimits_{nrightarrow infty}sum_{i=1}^k a_i^{n+1} logdfrac{a_i^{n+1}}{a_i^n} = 0.$
Can we conclude that $limlimits_{nrightarrow infty} dfrac{a_i^{n+1}}{a_i^n} = 1$? Intuitively yes, but how can this be proven rigorously?
Thanks in advance!
real-analysis analysis inequality
real-analysis analysis inequality
asked Dec 17 '18 at 16:14
Max93Max93
319210
319210
$begingroup$
I find your notation $a_i^n$ confusing and keep wanting to see the $n$ as an exponent rather than an index. I would write this as $a_{i, n}$ or $a_{n, i}$.
$endgroup$
– marty cohen
Dec 17 '18 at 18:18
add a comment |
$begingroup$
I find your notation $a_i^n$ confusing and keep wanting to see the $n$ as an exponent rather than an index. I would write this as $a_{i, n}$ or $a_{n, i}$.
$endgroup$
– marty cohen
Dec 17 '18 at 18:18
$begingroup$
I find your notation $a_i^n$ confusing and keep wanting to see the $n$ as an exponent rather than an index. I would write this as $a_{i, n}$ or $a_{n, i}$.
$endgroup$
– marty cohen
Dec 17 '18 at 18:18
$begingroup$
I find your notation $a_i^n$ confusing and keep wanting to see the $n$ as an exponent rather than an index. I would write this as $a_{i, n}$ or $a_{n, i}$.
$endgroup$
– marty cohen
Dec 17 '18 at 18:18
add a comment |
1 Answer
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$begingroup$
Not necessarily. For instance, put $k=2$ and for each $ninBbb N$ put $a^n_1=frac 1{2^n}$ and $a^n_2=1-frac 1{2^n}$. Then
$$lim_{ntoinfty}sum_{i=1}^k a_i^{n+1} logdfrac{a_i^{n+1}}{a_i^n}=
lim_{ntoinfty}frac 1{2^{n+1}}logdfrac{frac 1{2^{n+1}}}{frac 1{2^{n}}}+
left(1-frac 1{2^{n+1}}right)logdfrac{1-frac 1{2^{n+1}}}{1-frac 1{2^{n}}}=0,$$
but $$limlimits_{nrightarrow infty} dfrac{a_1^{n+1}}{a_1^n} = frac 12.$$
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1 Answer
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1 Answer
1
active
oldest
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active
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$begingroup$
Not necessarily. For instance, put $k=2$ and for each $ninBbb N$ put $a^n_1=frac 1{2^n}$ and $a^n_2=1-frac 1{2^n}$. Then
$$lim_{ntoinfty}sum_{i=1}^k a_i^{n+1} logdfrac{a_i^{n+1}}{a_i^n}=
lim_{ntoinfty}frac 1{2^{n+1}}logdfrac{frac 1{2^{n+1}}}{frac 1{2^{n}}}+
left(1-frac 1{2^{n+1}}right)logdfrac{1-frac 1{2^{n+1}}}{1-frac 1{2^{n}}}=0,$$
but $$limlimits_{nrightarrow infty} dfrac{a_1^{n+1}}{a_1^n} = frac 12.$$
$endgroup$
add a comment |
$begingroup$
Not necessarily. For instance, put $k=2$ and for each $ninBbb N$ put $a^n_1=frac 1{2^n}$ and $a^n_2=1-frac 1{2^n}$. Then
$$lim_{ntoinfty}sum_{i=1}^k a_i^{n+1} logdfrac{a_i^{n+1}}{a_i^n}=
lim_{ntoinfty}frac 1{2^{n+1}}logdfrac{frac 1{2^{n+1}}}{frac 1{2^{n}}}+
left(1-frac 1{2^{n+1}}right)logdfrac{1-frac 1{2^{n+1}}}{1-frac 1{2^{n}}}=0,$$
but $$limlimits_{nrightarrow infty} dfrac{a_1^{n+1}}{a_1^n} = frac 12.$$
$endgroup$
add a comment |
$begingroup$
Not necessarily. For instance, put $k=2$ and for each $ninBbb N$ put $a^n_1=frac 1{2^n}$ and $a^n_2=1-frac 1{2^n}$. Then
$$lim_{ntoinfty}sum_{i=1}^k a_i^{n+1} logdfrac{a_i^{n+1}}{a_i^n}=
lim_{ntoinfty}frac 1{2^{n+1}}logdfrac{frac 1{2^{n+1}}}{frac 1{2^{n}}}+
left(1-frac 1{2^{n+1}}right)logdfrac{1-frac 1{2^{n+1}}}{1-frac 1{2^{n}}}=0,$$
but $$limlimits_{nrightarrow infty} dfrac{a_1^{n+1}}{a_1^n} = frac 12.$$
$endgroup$
Not necessarily. For instance, put $k=2$ and for each $ninBbb N$ put $a^n_1=frac 1{2^n}$ and $a^n_2=1-frac 1{2^n}$. Then
$$lim_{ntoinfty}sum_{i=1}^k a_i^{n+1} logdfrac{a_i^{n+1}}{a_i^n}=
lim_{ntoinfty}frac 1{2^{n+1}}logdfrac{frac 1{2^{n+1}}}{frac 1{2^{n}}}+
left(1-frac 1{2^{n+1}}right)logdfrac{1-frac 1{2^{n+1}}}{1-frac 1{2^{n}}}=0,$$
but $$limlimits_{nrightarrow infty} dfrac{a_1^{n+1}}{a_1^n} = frac 12.$$
answered Dec 18 '18 at 18:05
Alex RavskyAlex Ravsky
42.6k32383
42.6k32383
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$begingroup$
I find your notation $a_i^n$ confusing and keep wanting to see the $n$ as an exponent rather than an index. I would write this as $a_{i, n}$ or $a_{n, i}$.
$endgroup$
– marty cohen
Dec 17 '18 at 18:18