Why is $alpha^nbeta^m+alpha^mbeta^nlealpha^sbeta^t+alpha^tbeta^s$ if $(n,m)preceq(s,t)$?
$begingroup$
I've recently come across the following statement:
$$alpha^nbeta^m+alpha^mbeta^nlealpha^sbeta^t+alpha^tbeta^s$$
for $alpha,betage0$, $n,m,s,tinmathbb N_+$, $n+m=s+t$, and $(n,m)preceq(s,t)$.
If anyone is interested (though it's not particularly relevant for the question itself), I found this stated in a slightly different form in this paper, page 2, second column. They reference a book on majorization in which I wasn't able to find an easy answer to this particular point (both paper and book are paywalled, sorry about that).
I can prove the case $n=m$ (see below), but I'm at a loss for the more general statement. What is an easy way to prove this?
Also, can this sort of statement be generalized to products of more than two numbers? The direct generalization would seem to be
$$sum_{sigmainmathcal P_N}prod_{i=1}^N alpha_{sigma(i)}^{n_i}lesum_{sigmainmathcal P_N}prod_{i=1}^N alpha_{sigma(i)}^{m_i}$$
when $(n_1,..,n_N)preceq(m_1,...,m_N)$.
If this holds, can the same technique used for the first statement be applied to this more general case?
For $n=m$ we write $s=mu+Delta, t=mu-Delta$, where $$muequivfrac{s+t}{2}=frac{n+m}{2}=n.$$
We thus have
begin{align}
(alpha^nbeta^m+alpha^mbeta^n)-(alpha^sbeta^t+alpha^tbeta^s)
&=2(alphabeta)^n-(alphabeta)^{mu-Delta}(alpha^{2Delta}+beta^{2Delta}) \
&=(alphabeta)^{mu-Delta}left[2alpha^Deltabeta^Delta-alpha^{2Delta}-beta^{2Delta}right]le0,
end{align}
where the last step follows from
$$2alpha^Deltabeta^Delta-alpha^{2Delta}-beta^{2Delta}
=-(alpha^Delta-beta^Delta)^2.$$
The sign of the inequality is consistent with the original statement because
$(n,n)preceq(n+Delta,n-Delta)$ (provided $n-Deltage0$).
convex-analysis products convex-geometry
$endgroup$
|
show 3 more comments
$begingroup$
I've recently come across the following statement:
$$alpha^nbeta^m+alpha^mbeta^nlealpha^sbeta^t+alpha^tbeta^s$$
for $alpha,betage0$, $n,m,s,tinmathbb N_+$, $n+m=s+t$, and $(n,m)preceq(s,t)$.
If anyone is interested (though it's not particularly relevant for the question itself), I found this stated in a slightly different form in this paper, page 2, second column. They reference a book on majorization in which I wasn't able to find an easy answer to this particular point (both paper and book are paywalled, sorry about that).
I can prove the case $n=m$ (see below), but I'm at a loss for the more general statement. What is an easy way to prove this?
Also, can this sort of statement be generalized to products of more than two numbers? The direct generalization would seem to be
$$sum_{sigmainmathcal P_N}prod_{i=1}^N alpha_{sigma(i)}^{n_i}lesum_{sigmainmathcal P_N}prod_{i=1}^N alpha_{sigma(i)}^{m_i}$$
when $(n_1,..,n_N)preceq(m_1,...,m_N)$.
If this holds, can the same technique used for the first statement be applied to this more general case?
For $n=m$ we write $s=mu+Delta, t=mu-Delta$, where $$muequivfrac{s+t}{2}=frac{n+m}{2}=n.$$
We thus have
begin{align}
(alpha^nbeta^m+alpha^mbeta^n)-(alpha^sbeta^t+alpha^tbeta^s)
&=2(alphabeta)^n-(alphabeta)^{mu-Delta}(alpha^{2Delta}+beta^{2Delta}) \
&=(alphabeta)^{mu-Delta}left[2alpha^Deltabeta^Delta-alpha^{2Delta}-beta^{2Delta}right]le0,
end{align}
where the last step follows from
$$2alpha^Deltabeta^Delta-alpha^{2Delta}-beta^{2Delta}
=-(alpha^Delta-beta^Delta)^2.$$
The sign of the inequality is consistent with the original statement because
$(n,n)preceq(n+Delta,n-Delta)$ (provided $n-Deltage0$).
convex-analysis products convex-geometry
$endgroup$
$begingroup$
Could possibly the rearrangement formula, which I asked about a few days ago, have some bearing on this? It looks a kind-of similarish thing.
$endgroup$
– AmbretteOrrisey
Nov 24 '18 at 16:03
$begingroup$
@AmbretteOrrisey never heard of it. Link?
$endgroup$
– glS
Nov 24 '18 at 16:04
$begingroup$
How do you link to another post? Or I could give the Wikipedia link. I"ll do that anyway. But I've not been doing this long ... please tell me how to link to another post.
$endgroup$
– AmbretteOrrisey
Nov 24 '18 at 16:07
$begingroup$
I'm using an app. I'm beginning to think theae apps are very deficient in many ways. It doean't display the link. Hang on ... I'll sort both one way or another.
$endgroup$
– AmbretteOrrisey
Nov 24 '18 at 16:09
$begingroup$
Sum of Itemwise Products of Two Sets of Real Numbers.
$endgroup$
– AmbretteOrrisey
Nov 24 '18 at 16:13
|
show 3 more comments
$begingroup$
I've recently come across the following statement:
$$alpha^nbeta^m+alpha^mbeta^nlealpha^sbeta^t+alpha^tbeta^s$$
for $alpha,betage0$, $n,m,s,tinmathbb N_+$, $n+m=s+t$, and $(n,m)preceq(s,t)$.
If anyone is interested (though it's not particularly relevant for the question itself), I found this stated in a slightly different form in this paper, page 2, second column. They reference a book on majorization in which I wasn't able to find an easy answer to this particular point (both paper and book are paywalled, sorry about that).
I can prove the case $n=m$ (see below), but I'm at a loss for the more general statement. What is an easy way to prove this?
Also, can this sort of statement be generalized to products of more than two numbers? The direct generalization would seem to be
$$sum_{sigmainmathcal P_N}prod_{i=1}^N alpha_{sigma(i)}^{n_i}lesum_{sigmainmathcal P_N}prod_{i=1}^N alpha_{sigma(i)}^{m_i}$$
when $(n_1,..,n_N)preceq(m_1,...,m_N)$.
If this holds, can the same technique used for the first statement be applied to this more general case?
For $n=m$ we write $s=mu+Delta, t=mu-Delta$, where $$muequivfrac{s+t}{2}=frac{n+m}{2}=n.$$
We thus have
begin{align}
(alpha^nbeta^m+alpha^mbeta^n)-(alpha^sbeta^t+alpha^tbeta^s)
&=2(alphabeta)^n-(alphabeta)^{mu-Delta}(alpha^{2Delta}+beta^{2Delta}) \
&=(alphabeta)^{mu-Delta}left[2alpha^Deltabeta^Delta-alpha^{2Delta}-beta^{2Delta}right]le0,
end{align}
where the last step follows from
$$2alpha^Deltabeta^Delta-alpha^{2Delta}-beta^{2Delta}
=-(alpha^Delta-beta^Delta)^2.$$
The sign of the inequality is consistent with the original statement because
$(n,n)preceq(n+Delta,n-Delta)$ (provided $n-Deltage0$).
convex-analysis products convex-geometry
$endgroup$
I've recently come across the following statement:
$$alpha^nbeta^m+alpha^mbeta^nlealpha^sbeta^t+alpha^tbeta^s$$
for $alpha,betage0$, $n,m,s,tinmathbb N_+$, $n+m=s+t$, and $(n,m)preceq(s,t)$.
If anyone is interested (though it's not particularly relevant for the question itself), I found this stated in a slightly different form in this paper, page 2, second column. They reference a book on majorization in which I wasn't able to find an easy answer to this particular point (both paper and book are paywalled, sorry about that).
I can prove the case $n=m$ (see below), but I'm at a loss for the more general statement. What is an easy way to prove this?
Also, can this sort of statement be generalized to products of more than two numbers? The direct generalization would seem to be
$$sum_{sigmainmathcal P_N}prod_{i=1}^N alpha_{sigma(i)}^{n_i}lesum_{sigmainmathcal P_N}prod_{i=1}^N alpha_{sigma(i)}^{m_i}$$
when $(n_1,..,n_N)preceq(m_1,...,m_N)$.
If this holds, can the same technique used for the first statement be applied to this more general case?
For $n=m$ we write $s=mu+Delta, t=mu-Delta$, where $$muequivfrac{s+t}{2}=frac{n+m}{2}=n.$$
We thus have
begin{align}
(alpha^nbeta^m+alpha^mbeta^n)-(alpha^sbeta^t+alpha^tbeta^s)
&=2(alphabeta)^n-(alphabeta)^{mu-Delta}(alpha^{2Delta}+beta^{2Delta}) \
&=(alphabeta)^{mu-Delta}left[2alpha^Deltabeta^Delta-alpha^{2Delta}-beta^{2Delta}right]le0,
end{align}
where the last step follows from
$$2alpha^Deltabeta^Delta-alpha^{2Delta}-beta^{2Delta}
=-(alpha^Delta-beta^Delta)^2.$$
The sign of the inequality is consistent with the original statement because
$(n,n)preceq(n+Delta,n-Delta)$ (provided $n-Deltage0$).
convex-analysis products convex-geometry
convex-analysis products convex-geometry
edited Nov 24 '18 at 22:56
glS
asked Nov 24 '18 at 14:30
glSglS
716520
716520
$begingroup$
Could possibly the rearrangement formula, which I asked about a few days ago, have some bearing on this? It looks a kind-of similarish thing.
$endgroup$
– AmbretteOrrisey
Nov 24 '18 at 16:03
$begingroup$
@AmbretteOrrisey never heard of it. Link?
$endgroup$
– glS
Nov 24 '18 at 16:04
$begingroup$
How do you link to another post? Or I could give the Wikipedia link. I"ll do that anyway. But I've not been doing this long ... please tell me how to link to another post.
$endgroup$
– AmbretteOrrisey
Nov 24 '18 at 16:07
$begingroup$
I'm using an app. I'm beginning to think theae apps are very deficient in many ways. It doean't display the link. Hang on ... I'll sort both one way or another.
$endgroup$
– AmbretteOrrisey
Nov 24 '18 at 16:09
$begingroup$
Sum of Itemwise Products of Two Sets of Real Numbers.
$endgroup$
– AmbretteOrrisey
Nov 24 '18 at 16:13
|
show 3 more comments
$begingroup$
Could possibly the rearrangement formula, which I asked about a few days ago, have some bearing on this? It looks a kind-of similarish thing.
$endgroup$
– AmbretteOrrisey
Nov 24 '18 at 16:03
$begingroup$
@AmbretteOrrisey never heard of it. Link?
$endgroup$
– glS
Nov 24 '18 at 16:04
$begingroup$
How do you link to another post? Or I could give the Wikipedia link. I"ll do that anyway. But I've not been doing this long ... please tell me how to link to another post.
$endgroup$
– AmbretteOrrisey
Nov 24 '18 at 16:07
$begingroup$
I'm using an app. I'm beginning to think theae apps are very deficient in many ways. It doean't display the link. Hang on ... I'll sort both one way or another.
$endgroup$
– AmbretteOrrisey
Nov 24 '18 at 16:09
$begingroup$
Sum of Itemwise Products of Two Sets of Real Numbers.
$endgroup$
– AmbretteOrrisey
Nov 24 '18 at 16:13
$begingroup$
Could possibly the rearrangement formula, which I asked about a few days ago, have some bearing on this? It looks a kind-of similarish thing.
$endgroup$
– AmbretteOrrisey
Nov 24 '18 at 16:03
$begingroup$
Could possibly the rearrangement formula, which I asked about a few days ago, have some bearing on this? It looks a kind-of similarish thing.
$endgroup$
– AmbretteOrrisey
Nov 24 '18 at 16:03
$begingroup$
@AmbretteOrrisey never heard of it. Link?
$endgroup$
– glS
Nov 24 '18 at 16:04
$begingroup$
@AmbretteOrrisey never heard of it. Link?
$endgroup$
– glS
Nov 24 '18 at 16:04
$begingroup$
How do you link to another post? Or I could give the Wikipedia link. I"ll do that anyway. But I've not been doing this long ... please tell me how to link to another post.
$endgroup$
– AmbretteOrrisey
Nov 24 '18 at 16:07
$begingroup$
How do you link to another post? Or I could give the Wikipedia link. I"ll do that anyway. But I've not been doing this long ... please tell me how to link to another post.
$endgroup$
– AmbretteOrrisey
Nov 24 '18 at 16:07
$begingroup$
I'm using an app. I'm beginning to think theae apps are very deficient in many ways. It doean't display the link. Hang on ... I'll sort both one way or another.
$endgroup$
– AmbretteOrrisey
Nov 24 '18 at 16:09
$begingroup$
I'm using an app. I'm beginning to think theae apps are very deficient in many ways. It doean't display the link. Hang on ... I'll sort both one way or another.
$endgroup$
– AmbretteOrrisey
Nov 24 '18 at 16:09
$begingroup$
Sum of Itemwise Products of Two Sets of Real Numbers.
$endgroup$
– AmbretteOrrisey
Nov 24 '18 at 16:13
$begingroup$
Sum of Itemwise Products of Two Sets of Real Numbers.
$endgroup$
– AmbretteOrrisey
Nov 24 '18 at 16:13
|
show 3 more comments
2 Answers
2
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$begingroup$
The inequality you are interested in is wrong. To see this, take $m=3, n=1, s=t=2$. Then it is easy to see that the following inequality $alpha^3beta + alpha beta^3 leq 2 alpha^2 beta^2$ does not hold.
In fact, it must have an opposite sign. In this case it is a particular case of a powerful Muirhead's inequality.
Let $a = (a_1,dots, a_n)$ and $x=(x_1,dots, x_n)$ be real nonnegative numbers. Define a symmetric polynomial $amapstoPhi(a)$ with variables $x_1,dots, x_n$ to be an average of all possible terms $x_{sigma_1}^{a_1}...x_{sigma_n}^{a_n}$ over all possible permutations $sigma $ of ${1,dots, n}$. In other words, it is
$$Phi (a) = frac{1}{n!}sum_{sigma}x_{sigma_1}^{a_1}...x_{sigma_n}^{a_n}.$$
For example, $Phi(3,2) = frac 1 2 (x_1^3x_2^2 + x_1^2x_2^3)$, $Phi(3,2,0) = frac 1 6 (x_1^3x_2^2 + x_1^3 x_3^2 + x_1^2x_2^3+x_1^2x_3^3 + x_2^3x_3^2+x_2^2x_3^3)$.
Then Muirhead's inequality says that if $a$ majorizes $b$, then $Phi(a)geq Phi(b)$.
In your case $Phi(m,n) = frac 1 2 (alpha^m beta^n +alpha^n beta^m)$ and similarly for $Phi(s,t)$. Now since $(n,m)$ majorizes $(s,t)$, then $Phi (n,m)geq Phi(s,t)$.
The generalization works in the same way.
$endgroup$
$begingroup$
ah, you are totally right! I actually did write it correctly in the first version of the post, then for some reason convinced myself it was in the other direction. I see my proofs were in the right direction nonetheless though. This is exactly what I was hoping for, thanks.
$endgroup$
– glS
Nov 24 '18 at 23:00
add a comment |
$begingroup$
I eventually figured out how to prove the simple case, though I'm not sure how to generalize the idea to the more general one.
The main point is to "parametrize" $m,n,s,t$ via their averages, which the assumption $n+m=s+t$ implies must be equal:
begin{align}
n&=mu+Delta,\
m&=mu-Delta,\
s&=mu+delta,\
t&=mu-delta.
end{align}
With these definitions, $(n,m)succeq(s,t)$ is equivalent to $Deltagedelta$.
With these definitions we have
begin{align}
Cequiv (alpha^nbeta^m+alpha^mbeta^n)-(alpha^tbeta^s+alpha^sbeta^t)
=(alphabeta)^{mu}left[
(alpha/beta)^{Delta} +
(alpha/beta)^{-Delta} -
(alpha/beta)^{delta} -
(alpha/beta)^{-delta}
right].
end{align}
Let me know define $gammaequivalpha/beta$ for notational convenience. After some algebraic manipulation we get to
$$C=(alphabeta)^mugamma^{-Delta}(1-gamma^{Delta-delta})(1-gamma^{Delta+delta}).$$
It then follows that for $Deltagedelta$ we have $Cge0$.
The problem with this is that it relies to a "parametrisation" of $n,m,s,t$, which is likely hard to do in a more general case with sequences with more than two numbers.
$endgroup$
add a comment |
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2 Answers
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2 Answers
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$begingroup$
The inequality you are interested in is wrong. To see this, take $m=3, n=1, s=t=2$. Then it is easy to see that the following inequality $alpha^3beta + alpha beta^3 leq 2 alpha^2 beta^2$ does not hold.
In fact, it must have an opposite sign. In this case it is a particular case of a powerful Muirhead's inequality.
Let $a = (a_1,dots, a_n)$ and $x=(x_1,dots, x_n)$ be real nonnegative numbers. Define a symmetric polynomial $amapstoPhi(a)$ with variables $x_1,dots, x_n$ to be an average of all possible terms $x_{sigma_1}^{a_1}...x_{sigma_n}^{a_n}$ over all possible permutations $sigma $ of ${1,dots, n}$. In other words, it is
$$Phi (a) = frac{1}{n!}sum_{sigma}x_{sigma_1}^{a_1}...x_{sigma_n}^{a_n}.$$
For example, $Phi(3,2) = frac 1 2 (x_1^3x_2^2 + x_1^2x_2^3)$, $Phi(3,2,0) = frac 1 6 (x_1^3x_2^2 + x_1^3 x_3^2 + x_1^2x_2^3+x_1^2x_3^3 + x_2^3x_3^2+x_2^2x_3^3)$.
Then Muirhead's inequality says that if $a$ majorizes $b$, then $Phi(a)geq Phi(b)$.
In your case $Phi(m,n) = frac 1 2 (alpha^m beta^n +alpha^n beta^m)$ and similarly for $Phi(s,t)$. Now since $(n,m)$ majorizes $(s,t)$, then $Phi (n,m)geq Phi(s,t)$.
The generalization works in the same way.
$endgroup$
$begingroup$
ah, you are totally right! I actually did write it correctly in the first version of the post, then for some reason convinced myself it was in the other direction. I see my proofs were in the right direction nonetheless though. This is exactly what I was hoping for, thanks.
$endgroup$
– glS
Nov 24 '18 at 23:00
add a comment |
$begingroup$
The inequality you are interested in is wrong. To see this, take $m=3, n=1, s=t=2$. Then it is easy to see that the following inequality $alpha^3beta + alpha beta^3 leq 2 alpha^2 beta^2$ does not hold.
In fact, it must have an opposite sign. In this case it is a particular case of a powerful Muirhead's inequality.
Let $a = (a_1,dots, a_n)$ and $x=(x_1,dots, x_n)$ be real nonnegative numbers. Define a symmetric polynomial $amapstoPhi(a)$ with variables $x_1,dots, x_n$ to be an average of all possible terms $x_{sigma_1}^{a_1}...x_{sigma_n}^{a_n}$ over all possible permutations $sigma $ of ${1,dots, n}$. In other words, it is
$$Phi (a) = frac{1}{n!}sum_{sigma}x_{sigma_1}^{a_1}...x_{sigma_n}^{a_n}.$$
For example, $Phi(3,2) = frac 1 2 (x_1^3x_2^2 + x_1^2x_2^3)$, $Phi(3,2,0) = frac 1 6 (x_1^3x_2^2 + x_1^3 x_3^2 + x_1^2x_2^3+x_1^2x_3^3 + x_2^3x_3^2+x_2^2x_3^3)$.
Then Muirhead's inequality says that if $a$ majorizes $b$, then $Phi(a)geq Phi(b)$.
In your case $Phi(m,n) = frac 1 2 (alpha^m beta^n +alpha^n beta^m)$ and similarly for $Phi(s,t)$. Now since $(n,m)$ majorizes $(s,t)$, then $Phi (n,m)geq Phi(s,t)$.
The generalization works in the same way.
$endgroup$
$begingroup$
ah, you are totally right! I actually did write it correctly in the first version of the post, then for some reason convinced myself it was in the other direction. I see my proofs were in the right direction nonetheless though. This is exactly what I was hoping for, thanks.
$endgroup$
– glS
Nov 24 '18 at 23:00
add a comment |
$begingroup$
The inequality you are interested in is wrong. To see this, take $m=3, n=1, s=t=2$. Then it is easy to see that the following inequality $alpha^3beta + alpha beta^3 leq 2 alpha^2 beta^2$ does not hold.
In fact, it must have an opposite sign. In this case it is a particular case of a powerful Muirhead's inequality.
Let $a = (a_1,dots, a_n)$ and $x=(x_1,dots, x_n)$ be real nonnegative numbers. Define a symmetric polynomial $amapstoPhi(a)$ with variables $x_1,dots, x_n$ to be an average of all possible terms $x_{sigma_1}^{a_1}...x_{sigma_n}^{a_n}$ over all possible permutations $sigma $ of ${1,dots, n}$. In other words, it is
$$Phi (a) = frac{1}{n!}sum_{sigma}x_{sigma_1}^{a_1}...x_{sigma_n}^{a_n}.$$
For example, $Phi(3,2) = frac 1 2 (x_1^3x_2^2 + x_1^2x_2^3)$, $Phi(3,2,0) = frac 1 6 (x_1^3x_2^2 + x_1^3 x_3^2 + x_1^2x_2^3+x_1^2x_3^3 + x_2^3x_3^2+x_2^2x_3^3)$.
Then Muirhead's inequality says that if $a$ majorizes $b$, then $Phi(a)geq Phi(b)$.
In your case $Phi(m,n) = frac 1 2 (alpha^m beta^n +alpha^n beta^m)$ and similarly for $Phi(s,t)$. Now since $(n,m)$ majorizes $(s,t)$, then $Phi (n,m)geq Phi(s,t)$.
The generalization works in the same way.
$endgroup$
The inequality you are interested in is wrong. To see this, take $m=3, n=1, s=t=2$. Then it is easy to see that the following inequality $alpha^3beta + alpha beta^3 leq 2 alpha^2 beta^2$ does not hold.
In fact, it must have an opposite sign. In this case it is a particular case of a powerful Muirhead's inequality.
Let $a = (a_1,dots, a_n)$ and $x=(x_1,dots, x_n)$ be real nonnegative numbers. Define a symmetric polynomial $amapstoPhi(a)$ with variables $x_1,dots, x_n$ to be an average of all possible terms $x_{sigma_1}^{a_1}...x_{sigma_n}^{a_n}$ over all possible permutations $sigma $ of ${1,dots, n}$. In other words, it is
$$Phi (a) = frac{1}{n!}sum_{sigma}x_{sigma_1}^{a_1}...x_{sigma_n}^{a_n}.$$
For example, $Phi(3,2) = frac 1 2 (x_1^3x_2^2 + x_1^2x_2^3)$, $Phi(3,2,0) = frac 1 6 (x_1^3x_2^2 + x_1^3 x_3^2 + x_1^2x_2^3+x_1^2x_3^3 + x_2^3x_3^2+x_2^2x_3^3)$.
Then Muirhead's inequality says that if $a$ majorizes $b$, then $Phi(a)geq Phi(b)$.
In your case $Phi(m,n) = frac 1 2 (alpha^m beta^n +alpha^n beta^m)$ and similarly for $Phi(s,t)$. Now since $(n,m)$ majorizes $(s,t)$, then $Phi (n,m)geq Phi(s,t)$.
The generalization works in the same way.
answered Nov 24 '18 at 22:41
cheypcheyp
30117
30117
$begingroup$
ah, you are totally right! I actually did write it correctly in the first version of the post, then for some reason convinced myself it was in the other direction. I see my proofs were in the right direction nonetheless though. This is exactly what I was hoping for, thanks.
$endgroup$
– glS
Nov 24 '18 at 23:00
add a comment |
$begingroup$
ah, you are totally right! I actually did write it correctly in the first version of the post, then for some reason convinced myself it was in the other direction. I see my proofs were in the right direction nonetheless though. This is exactly what I was hoping for, thanks.
$endgroup$
– glS
Nov 24 '18 at 23:00
$begingroup$
ah, you are totally right! I actually did write it correctly in the first version of the post, then for some reason convinced myself it was in the other direction. I see my proofs were in the right direction nonetheless though. This is exactly what I was hoping for, thanks.
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– glS
Nov 24 '18 at 23:00
$begingroup$
ah, you are totally right! I actually did write it correctly in the first version of the post, then for some reason convinced myself it was in the other direction. I see my proofs were in the right direction nonetheless though. This is exactly what I was hoping for, thanks.
$endgroup$
– glS
Nov 24 '18 at 23:00
add a comment |
$begingroup$
I eventually figured out how to prove the simple case, though I'm not sure how to generalize the idea to the more general one.
The main point is to "parametrize" $m,n,s,t$ via their averages, which the assumption $n+m=s+t$ implies must be equal:
begin{align}
n&=mu+Delta,\
m&=mu-Delta,\
s&=mu+delta,\
t&=mu-delta.
end{align}
With these definitions, $(n,m)succeq(s,t)$ is equivalent to $Deltagedelta$.
With these definitions we have
begin{align}
Cequiv (alpha^nbeta^m+alpha^mbeta^n)-(alpha^tbeta^s+alpha^sbeta^t)
=(alphabeta)^{mu}left[
(alpha/beta)^{Delta} +
(alpha/beta)^{-Delta} -
(alpha/beta)^{delta} -
(alpha/beta)^{-delta}
right].
end{align}
Let me know define $gammaequivalpha/beta$ for notational convenience. After some algebraic manipulation we get to
$$C=(alphabeta)^mugamma^{-Delta}(1-gamma^{Delta-delta})(1-gamma^{Delta+delta}).$$
It then follows that for $Deltagedelta$ we have $Cge0$.
The problem with this is that it relies to a "parametrisation" of $n,m,s,t$, which is likely hard to do in a more general case with sequences with more than two numbers.
$endgroup$
add a comment |
$begingroup$
I eventually figured out how to prove the simple case, though I'm not sure how to generalize the idea to the more general one.
The main point is to "parametrize" $m,n,s,t$ via their averages, which the assumption $n+m=s+t$ implies must be equal:
begin{align}
n&=mu+Delta,\
m&=mu-Delta,\
s&=mu+delta,\
t&=mu-delta.
end{align}
With these definitions, $(n,m)succeq(s,t)$ is equivalent to $Deltagedelta$.
With these definitions we have
begin{align}
Cequiv (alpha^nbeta^m+alpha^mbeta^n)-(alpha^tbeta^s+alpha^sbeta^t)
=(alphabeta)^{mu}left[
(alpha/beta)^{Delta} +
(alpha/beta)^{-Delta} -
(alpha/beta)^{delta} -
(alpha/beta)^{-delta}
right].
end{align}
Let me know define $gammaequivalpha/beta$ for notational convenience. After some algebraic manipulation we get to
$$C=(alphabeta)^mugamma^{-Delta}(1-gamma^{Delta-delta})(1-gamma^{Delta+delta}).$$
It then follows that for $Deltagedelta$ we have $Cge0$.
The problem with this is that it relies to a "parametrisation" of $n,m,s,t$, which is likely hard to do in a more general case with sequences with more than two numbers.
$endgroup$
add a comment |
$begingroup$
I eventually figured out how to prove the simple case, though I'm not sure how to generalize the idea to the more general one.
The main point is to "parametrize" $m,n,s,t$ via their averages, which the assumption $n+m=s+t$ implies must be equal:
begin{align}
n&=mu+Delta,\
m&=mu-Delta,\
s&=mu+delta,\
t&=mu-delta.
end{align}
With these definitions, $(n,m)succeq(s,t)$ is equivalent to $Deltagedelta$.
With these definitions we have
begin{align}
Cequiv (alpha^nbeta^m+alpha^mbeta^n)-(alpha^tbeta^s+alpha^sbeta^t)
=(alphabeta)^{mu}left[
(alpha/beta)^{Delta} +
(alpha/beta)^{-Delta} -
(alpha/beta)^{delta} -
(alpha/beta)^{-delta}
right].
end{align}
Let me know define $gammaequivalpha/beta$ for notational convenience. After some algebraic manipulation we get to
$$C=(alphabeta)^mugamma^{-Delta}(1-gamma^{Delta-delta})(1-gamma^{Delta+delta}).$$
It then follows that for $Deltagedelta$ we have $Cge0$.
The problem with this is that it relies to a "parametrisation" of $n,m,s,t$, which is likely hard to do in a more general case with sequences with more than two numbers.
$endgroup$
I eventually figured out how to prove the simple case, though I'm not sure how to generalize the idea to the more general one.
The main point is to "parametrize" $m,n,s,t$ via their averages, which the assumption $n+m=s+t$ implies must be equal:
begin{align}
n&=mu+Delta,\
m&=mu-Delta,\
s&=mu+delta,\
t&=mu-delta.
end{align}
With these definitions, $(n,m)succeq(s,t)$ is equivalent to $Deltagedelta$.
With these definitions we have
begin{align}
Cequiv (alpha^nbeta^m+alpha^mbeta^n)-(alpha^tbeta^s+alpha^sbeta^t)
=(alphabeta)^{mu}left[
(alpha/beta)^{Delta} +
(alpha/beta)^{-Delta} -
(alpha/beta)^{delta} -
(alpha/beta)^{-delta}
right].
end{align}
Let me know define $gammaequivalpha/beta$ for notational convenience. After some algebraic manipulation we get to
$$C=(alphabeta)^mugamma^{-Delta}(1-gamma^{Delta-delta})(1-gamma^{Delta+delta}).$$
It then follows that for $Deltagedelta$ we have $Cge0$.
The problem with this is that it relies to a "parametrisation" of $n,m,s,t$, which is likely hard to do in a more general case with sequences with more than two numbers.
edited Nov 24 '18 at 22:57
answered Nov 24 '18 at 16:40
glSglS
716520
716520
add a comment |
add a comment |
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Could possibly the rearrangement formula, which I asked about a few days ago, have some bearing on this? It looks a kind-of similarish thing.
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– AmbretteOrrisey
Nov 24 '18 at 16:03
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@AmbretteOrrisey never heard of it. Link?
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– glS
Nov 24 '18 at 16:04
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How do you link to another post? Or I could give the Wikipedia link. I"ll do that anyway. But I've not been doing this long ... please tell me how to link to another post.
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– AmbretteOrrisey
Nov 24 '18 at 16:07
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I'm using an app. I'm beginning to think theae apps are very deficient in many ways. It doean't display the link. Hang on ... I'll sort both one way or another.
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– AmbretteOrrisey
Nov 24 '18 at 16:09
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Sum of Itemwise Products of Two Sets of Real Numbers.
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– AmbretteOrrisey
Nov 24 '18 at 16:13