How to apply the Cauchy-Schwartz inequality on the following inner product?
Reading a research article, I came across the following statement:
The following function (where $x_i$ and $p_j$ are two vectors in $mathbb{R}^n$ and $mu_{i,j}$ is a constant) - it doesn't matter where it came from, is just an inner product actually:
$sum_{i} mu_{i,j} - left< sum_{i} mu_{i,j} frac{x_i}{||x_i||}, frac{p_j}{||p_j||} right>$
is minimized by using the Cauchy-Schwartz inequality iff
$p_j propto sum_{i} mu_{i,j} frac{x_i}{||x_i||}$.
I want to apply the Cauchy-Schwartz inequality on $left< sum_{i} mu_{i,j} frac{x_i}{||x_i||}, frac{p_j}{||p_j||} right>$ in order to understand where the proportionality thing came from.
So basically the Cauchy-Schwartz inequality states that:
$left< x, yright> le ||x|| cdot ||y||$. But how can I apply this formula on that monstrous term?
vectors cauchy-schwarz-inequality
asked Nov 24 at 17:57
Hello Lili
1104
add a comment |
Reading a research article, I came across the following statement:
The following function (where $x_i$ and $p_j$ are two vectors in $mathbb{R}^n$ and $mu_{i,j}$ is a constant) - it doesn't matter where it came from, is just an inner product actually:
$sum_{i} mu_{i,j} - left< sum_{i} mu_{i,j} frac{x_i}{||x_i||}, frac{p_j}{||p_j||} right>$
is minimized by using the Cauchy-Schwartz inequality iff
$p_j propto sum_{i} mu_{i,j} frac{x_i}{||x_i||}$.
I want to apply the Cauchy-Schwartz inequality on $left< sum_{i} mu_{i,j} frac{x_i}{||x_i||}, frac{p_j}{||p_j||} right>$ in order to understand where the proportionality thing came from.
So basically the Cauchy-Schwartz inequality states that:
$left< x, yright> le ||x|| cdot ||y||$. But how can I apply this formula on that monstrous term?
vectors cauchy-schwarz-inequality
asked Nov 24 at 17:57
Hello Lili
1104
1
begin{align*}leftlangle sum_imu_{i,j}frac{x_i}{|x_i|},frac{p_j}{|p_j|}rightrangle &= sum_imu_{i,j}leftlanglefrac{x_i}{|x_i|},frac{p_j}{|p_j|}rightrangle \&= sum_imu_{i,j}frac{langle x_i,p_jrangle}{|x_i||p_j|}\&leqsum_imu_{i,j} end{align*} Does that help?
– user3482749
Nov 24 at 18:02
@user3482749 no, this is actually where the formula I stated came from (like going backwards), and it doesn't use the Cauchy-Schwartz inequality, but thanks for your contribution anyway
– Hello Lili
Nov 24 at 18:12
It does use the Cauchy-Schwartz inequality: that's precisely what that inequality is.
– user3482749
Nov 24 at 18:18
@user3482749 I don't think this is the correct answer because as I said, in the article the equations you stated are used to generate the formula in my question. So it's like going backwards.
– Hello Lili
Nov 24 at 18:22
@user3482749 I can give you a link if this helps: google.com/…
– Hello Lili
Nov 24 at 18:24
add a comment |
Reading a research article, I came across the following statement:
The following function (where $x_i$ and $p_j$ are two vectors in $mathbb{R}^n$ and $mu_{i,j}$ is a constant) - it doesn't matter where it came from, is just an inner product actually:
$sum_{i} mu_{i,j} - left< sum_{i} mu_{i,j} frac{x_i}{||x_i||}, frac{p_j}{||p_j||} right>$
is minimized by using the Cauchy-Schwartz inequality iff
$p_j propto sum_{i} mu_{i,j} frac{x_i}{||x_i||}$.
I want to apply the Cauchy-Schwartz inequality on $left< sum_{i} mu_{i,j} frac{x_i}{||x_i||}, frac{p_j}{||p_j||} right>$ in order to understand where the proportionality thing came from.
So basically the Cauchy-Schwartz inequality states that:
$left< x, yright> le ||x|| cdot ||y||$. But how can I apply this formula on that monstrous term?
vectors cauchy-schwarz-inequality
asked Nov 24 at 17:57
Hello Lili
1104
Reading a research article, I came across the following statement:
The following function (where $x_i$ and $p_j$ are two vectors in $mathbb{R}^n$ and $mu_{i,j}$ is a constant) - it doesn't matter where it came from, is just an inner product actually:
$sum_{i} mu_{i,j} - left< sum_{i} mu_{i,j} frac{x_i}{||x_i||}, frac{p_j}{||p_j||} right>$
is minimized by using the Cauchy-Schwartz inequality iff
$p_j propto sum_{i} mu_{i,j} frac{x_i}{||x_i||}$.
I want to apply the Cauchy-Schwartz inequality on $left< sum_{i} mu_{i,j} frac{x_i}{||x_i||}, frac{p_j}{||p_j||} right>$ in order to understand where the proportionality thing came from.
So basically the Cauchy-Schwartz inequality states that:
$left< x, yright> le ||x|| cdot ||y||$. But how can I apply this formula on that monstrous term?
vectors cauchy-schwarz-inequality
vectors cauchy-schwarz-inequality
asked Nov 24 at 17:57
Hello Lili
1104
asked Nov 24 at 17:57
Hello Lili
1104
asked Nov 24 at 17:57
Hello Lili
1104
asked Nov 24 at 17:57
Hello Lili
1104
asked Nov 24 at 17:57
Hello Lili
1104
1104
1
begin{align*}leftlangle sum_imu_{i,j}frac{x_i}{|x_i|},frac{p_j}{|p_j|}rightrangle &= sum_imu_{i,j}leftlanglefrac{x_i}{|x_i|},frac{p_j}{|p_j|}rightrangle \&= sum_imu_{i,j}frac{langle x_i,p_jrangle}{|x_i||p_j|}\&leqsum_imu_{i,j} end{align*} Does that help?
– user3482749
Nov 24 at 18:02
@user3482749 no, this is actually where the formula I stated came from (like going backwards), and it doesn't use the Cauchy-Schwartz inequality, but thanks for your contribution anyway
– Hello Lili
Nov 24 at 18:12
It does use the Cauchy-Schwartz inequality: that's precisely what that inequality is.
– user3482749
Nov 24 at 18:18
@user3482749 I don't think this is the correct answer because as I said, in the article the equations you stated are used to generate the formula in my question. So it's like going backwards.
– Hello Lili
Nov 24 at 18:22
@user3482749 I can give you a link if this helps: google.com/…
– Hello Lili
Nov 24 at 18:24
add a comment |
1
begin{align*}leftlangle sum_imu_{i,j}frac{x_i}{|x_i|},frac{p_j}{|p_j|}rightrangle &= sum_imu_{i,j}leftlanglefrac{x_i}{|x_i|},frac{p_j}{|p_j|}rightrangle \&= sum_imu_{i,j}frac{langle x_i,p_jrangle}{|x_i||p_j|}\&leqsum_imu_{i,j} end{align*} Does that help?
– user3482749
Nov 24 at 18:02
@user3482749 no, this is actually where the formula I stated came from (like going backwards), and it doesn't use the Cauchy-Schwartz inequality, but thanks for your contribution anyway
– Hello Lili
Nov 24 at 18:12
It does use the Cauchy-Schwartz inequality: that's precisely what that inequality is.
– user3482749
Nov 24 at 18:18
@user3482749 I don't think this is the correct answer because as I said, in the article the equations you stated are used to generate the formula in my question. So it's like going backwards.
– Hello Lili
Nov 24 at 18:22
@user3482749 I can give you a link if this helps: google.com/…
– Hello Lili
Nov 24 at 18:24
1
1
begin{align*}leftlangle sum_imu_{i,j}frac{x_i}{|x_i|},frac{p_j}{|p_j|}rightrangle &= sum_imu_{i,j}leftlanglefrac{x_i}{|x_i|},frac{p_j}{|p_j|}rightrangle \&= sum_imu_{i,j}frac{langle x_i,p_jrangle}{|x_i||p_j|}\&leqsum_imu_{i,j} end{align*} Does that help?
– user3482749
Nov 24 at 18:02
begin{align*}leftlangle sum_imu_{i,j}frac{x_i}{|x_i|},frac{p_j}{|p_j|}rightrangle &= sum_imu_{i,j}leftlanglefrac{x_i}{|x_i|},frac{p_j}{|p_j|}rightrangle \&= sum_imu_{i,j}frac{langle x_i,p_jrangle}{|x_i||p_j|}\&leqsum_imu_{i,j} end{align*} Does that help?
– user3482749
Nov 24 at 18:02
@user3482749 no, this is actually where the formula I stated came from (like going backwards), and it doesn't use the Cauchy-Schwartz inequality, but thanks for your contribution anyway
– Hello Lili
Nov 24 at 18:12
@user3482749 no, this is actually where the formula I stated came from (like going backwards), and it doesn't use the Cauchy-Schwartz inequality, but thanks for your contribution anyway
– Hello Lili
Nov 24 at 18:12
It does use the Cauchy-Schwartz inequality: that's precisely what that inequality is.
– user3482749
Nov 24 at 18:18
It does use the Cauchy-Schwartz inequality: that's precisely what that inequality is.
– user3482749
Nov 24 at 18:18
@user3482749 I don't think this is the correct answer because as I said, in the article the equations you stated are used to generate the formula in my question. So it's like going backwards.
– Hello Lili
Nov 24 at 18:22
@user3482749 I don't think this is the correct answer because as I said, in the article the equations you stated are used to generate the formula in my question. So it's like going backwards.
– Hello Lili
Nov 24 at 18:22
@user3482749 I can give you a link if this helps: google.com/…
– Hello Lili
Nov 24 at 18:24
@user3482749 I can give you a link if this helps: google.com/…
– Hello Lili
Nov 24 at 18:24
add a comment |
1 Answer
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Beyond the inequality itself, the full Cauchy-Schwarz theorem states that $|langle u, vrangle|$ is maximized exactly when $u$ and $v$ are parallel or anti-parallel. I.e., when $v propto u$. For your inner product, this is
$$frac{p_j}{|p_j|} propto sum_{i} mu_{i,j} frac{x_i}{||x_i||}$$
Since $1/|p_j|$ is just a scalar multiplier, it can be absorbed into the constant of proportionality, leaving
$$p_j propto sum_{i} mu_{i,j} frac{x_i}{||x_i||}$$
answered Nov 25 at 4:22
Paul Sinclair
19.2k21441
Thank you very much!
– Hello Lili
Nov 25 at 9:20
add a comment |
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1 Answer
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Beyond the inequality itself, the full Cauchy-Schwarz theorem states that $|langle u, vrangle|$ is maximized exactly when $u$ and $v$ are parallel or anti-parallel. I.e., when $v propto u$. For your inner product, this is
$$frac{p_j}{|p_j|} propto sum_{i} mu_{i,j} frac{x_i}{||x_i||}$$
Since $1/|p_j|$ is just a scalar multiplier, it can be absorbed into the constant of proportionality, leaving
$$p_j propto sum_{i} mu_{i,j} frac{x_i}{||x_i||}$$
answered Nov 25 at 4:22
Paul Sinclair
19.2k21441
Thank you very much!
– Hello Lili
Nov 25 at 9:20
add a comment |
Beyond the inequality itself, the full Cauchy-Schwarz theorem states that $|langle u, vrangle|$ is maximized exactly when $u$ and $v$ are parallel or anti-parallel. I.e., when $v propto u$. For your inner product, this is
$$frac{p_j}{|p_j|} propto sum_{i} mu_{i,j} frac{x_i}{||x_i||}$$
Since $1/|p_j|$ is just a scalar multiplier, it can be absorbed into the constant of proportionality, leaving
$$p_j propto sum_{i} mu_{i,j} frac{x_i}{||x_i||}$$
answered Nov 25 at 4:22
Paul Sinclair
19.2k21441
Thank you very much!
– Hello Lili
Nov 25 at 9:20
add a comment |
Beyond the inequality itself, the full Cauchy-Schwarz theorem states that $|langle u, vrangle|$ is maximized exactly when $u$ and $v$ are parallel or anti-parallel. I.e., when $v propto u$. For your inner product, this is
$$frac{p_j}{|p_j|} propto sum_{i} mu_{i,j} frac{x_i}{||x_i||}$$
Since $1/|p_j|$ is just a scalar multiplier, it can be absorbed into the constant of proportionality, leaving
$$p_j propto sum_{i} mu_{i,j} frac{x_i}{||x_i||}$$
answered Nov 25 at 4:22
Paul Sinclair
19.2k21441
Beyond the inequality itself, the full Cauchy-Schwarz theorem states that $|langle u, vrangle|$ is maximized exactly when $u$ and $v$ are parallel or anti-parallel. I.e., when $v propto u$. For your inner product, this is
$$frac{p_j}{|p_j|} propto sum_{i} mu_{i,j} frac{x_i}{||x_i||}$$
Since $1/|p_j|$ is just a scalar multiplier, it can be absorbed into the constant of proportionality, leaving
$$p_j propto sum_{i} mu_{i,j} frac{x_i}{||x_i||}$$
answered Nov 25 at 4:22
Paul Sinclair
19.2k21441
answered Nov 25 at 4:22
Paul Sinclair
19.2k21441
answered Nov 25 at 4:22
Paul Sinclair
19.2k21441
answered Nov 25 at 4:22
Paul Sinclair
19.2k21441
19.2k21441
Thank you very much!
– Hello Lili
Nov 25 at 9:20
add a comment |
Thank you very much!
– Hello Lili
Nov 25 at 9:20
Thank you very much!
– Hello Lili
Nov 25 at 9:20
Thank you very much!
– Hello Lili
Nov 25 at 9:20
add a comment |
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1
begin{align*}leftlangle sum_imu_{i,j}frac{x_i}{|x_i|},frac{p_j}{|p_j|}rightrangle &= sum_imu_{i,j}leftlanglefrac{x_i}{|x_i|},frac{p_j}{|p_j|}rightrangle \&= sum_imu_{i,j}frac{langle x_i,p_jrangle}{|x_i||p_j|}\&leqsum_imu_{i,j} end{align*} Does that help?
– user3482749
Nov 24 at 18:02
@user3482749 no, this is actually where the formula I stated came from (like going backwards), and it doesn't use the Cauchy-Schwartz inequality, but thanks for your contribution anyway
– Hello Lili
Nov 24 at 18:12
It does use the Cauchy-Schwartz inequality: that's precisely what that inequality is.
– user3482749
Nov 24 at 18:18
@user3482749 I don't think this is the correct answer because as I said, in the article the equations you stated are used to generate the formula in my question. So it's like going backwards.
– Hello Lili
Nov 24 at 18:22
@user3482749 I can give you a link if this helps: google.com/…
– Hello Lili
Nov 24 at 18:24