Mixing
Approximate scalar dot product with a vector's sum
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I have two vectors $u$ and $v$ of size $n$. The $u$ vector is a linear increasing function. For the $v$ vector the individual elements are not known, only its sum. Is it possible to approximate the $u cdot v$ somehow? I have noticed that in my case $u cdot v approx textmean( u) cdot sum v$ but it might be coincidence for one case. Is there any theorem about that? or any other better approximation I could do?
linear-algebra vector-spaces vectors vector-analysis approximation
edited Jul 26 at 10:12
Kenta S
1,1371418
asked Jul 26 at 10:10
AlphaBeta
236
add a comment |Â
up vote
0
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favorite
I have two vectors $u$ and $v$ of size $n$. The $u$ vector is a linear increasing function. For the $v$ vector the individual elements are not known, only its sum. Is it possible to approximate the $u cdot v$ somehow? I have noticed that in my case $u cdot v approx textmean( u) cdot sum v$ but it might be coincidence for one case. Is there any theorem about that? or any other better approximation I could do?
linear-algebra vector-spaces vectors vector-analysis approximation
edited Jul 26 at 10:12
Kenta S
1,1371418
asked Jul 26 at 10:10
AlphaBeta
236
What exactly do you mean by $u$ being a "linear increasing function"? And by "individual elements" of $v$, do you mean its components?
– Shirish Kulhari
Jul 26 at 10:30
@ShirishKulhari I mean that $u = (a, a + delta, dotsc, a + (n - 1)delta)$ for some $a$ and $delta$. Yes, its components.
– AlphaBeta
Jul 26 at 13:42
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
I have two vectors $u$ and $v$ of size $n$. The $u$ vector is a linear increasing function. For the $v$ vector the individual elements are not known, only its sum. Is it possible to approximate the $u cdot v$ somehow? I have noticed that in my case $u cdot v approx textmean( u) cdot sum v$ but it might be coincidence for one case. Is there any theorem about that? or any other better approximation I could do?
linear-algebra vector-spaces vectors vector-analysis approximation
edited Jul 26 at 10:12
Kenta S
1,1371418
asked Jul 26 at 10:10
AlphaBeta
236
I have two vectors $u$ and $v$ of size $n$. The $u$ vector is a linear increasing function. For the $v$ vector the individual elements are not known, only its sum. Is it possible to approximate the $u cdot v$ somehow? I have noticed that in my case $u cdot v approx textmean( u) cdot sum v$ but it might be coincidence for one case. Is there any theorem about that? or any other better approximation I could do?
linear-algebra vector-spaces vectors vector-analysis approximation
edited Jul 26 at 10:12
Kenta S
1,1371418
asked Jul 26 at 10:10
AlphaBeta
236
edited Jul 26 at 10:12
Kenta S
1,1371418
edited Jul 26 at 10:12
Kenta S
1,1371418
edited Jul 26 at 10:12
Kenta S
1,1371418
1,1371418
asked Jul 26 at 10:10
AlphaBeta
236
asked Jul 26 at 10:10
AlphaBeta
236
asked Jul 26 at 10:10
AlphaBeta
236
236
What exactly do you mean by $u$ being a "linear increasing function"? And by "individual elements" of $v$, do you mean its components?
– Shirish Kulhari
Jul 26 at 10:30
@ShirishKulhari I mean that $u = (a, a + delta, dotsc, a + (n - 1)delta)$ for some $a$ and $delta$. Yes, its components.
– AlphaBeta
Jul 26 at 13:42
add a comment |Â
What exactly do you mean by $u$ being a "linear increasing function"? And by "individual elements" of $v$, do you mean its components?
– Shirish Kulhari
Jul 26 at 10:30
@ShirishKulhari I mean that $u = (a, a + delta, dotsc, a + (n - 1)delta)$ for some $a$ and $delta$. Yes, its components.
– AlphaBeta
Jul 26 at 13:42
What exactly do you mean by $u$ being a "linear increasing function"? And by "individual elements" of $v$, do you mean its components?
– Shirish Kulhari
Jul 26 at 10:30
What exactly do you mean by $u$ being a "linear increasing function"? And by "individual elements" of $v$, do you mean its components?
– Shirish Kulhari
Jul 26 at 10:30
@ShirishKulhari I mean that $u = (a, a + delta, dotsc, a + (n - 1)delta)$ for some $a$ and $delta$. Yes, its components.
– AlphaBeta
Jul 26 at 13:42
@ShirishKulhari I mean that $u = (a, a + delta, dotsc, a + (n - 1)delta)$ for some $a$ and $delta$. Yes, its components.
– AlphaBeta
Jul 26 at 13:42
add a comment |Â
1 Answer
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You are asking for the difference $sum_k=1^n u_k v_k-frac1n sum_k=1^n u_k sum_k=1^n v_k$, where $u_k=ak+b$ so $frac1n sum u_k = b+a fracn+12$.
This means $$sum_k=1^n u_k v_k-frac1n sum_k=1^n u_k sum_k=1^n v_k=asum_k=1^n (k-fracn+12) v_k= asum_k=1^n (k-fracn+12) left(v_k-frac1nsum_j=1^n v_jright).$$
This will for example be small if $v$ is close to its mean for all $k$. Using various inequalities for $sum a_k b_k$, you can derive estimates such as
$$left|sum_k=1^n u_k v_k-frac1n sum_k=1^n u_k sum_k=1^n v_kright|le |a|sqrtsum_k=1^n (k-fracn+12)^2 sqrtsum_k=1^n(v_k-frac1nsum_k=1^n v_k)^2,$$
which tells you your approximation is good if the variance of $v$ is small.
answered Jul 26 at 10:44
Kusma
1,097111
could you elaborate how you substituted $v_k$ with $left(v_k-frac1nsum_k=1^n v_kright)$ between second and third equality?
– AlphaBeta
Jul 26 at 14:24
We have $sum_k=1^n (k-fracn+12)=0$, so $sum_k=1^n (k-fracn+12)frac1n sum_j=1^n v_j=0$ as well.
– Kusma
Jul 26 at 14:31
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
You are asking for the difference $sum_k=1^n u_k v_k-frac1n sum_k=1^n u_k sum_k=1^n v_k$, where $u_k=ak+b$ so $frac1n sum u_k = b+a fracn+12$.
This means $$sum_k=1^n u_k v_k-frac1n sum_k=1^n u_k sum_k=1^n v_k=asum_k=1^n (k-fracn+12) v_k= asum_k=1^n (k-fracn+12) left(v_k-frac1nsum_j=1^n v_jright).$$
This will for example be small if $v$ is close to its mean for all $k$. Using various inequalities for $sum a_k b_k$, you can derive estimates such as
$$left|sum_k=1^n u_k v_k-frac1n sum_k=1^n u_k sum_k=1^n v_kright|le |a|sqrtsum_k=1^n (k-fracn+12)^2 sqrtsum_k=1^n(v_k-frac1nsum_k=1^n v_k)^2,$$
which tells you your approximation is good if the variance of $v$ is small.
answered Jul 26 at 10:44
Kusma
1,097111
could you elaborate how you substituted $v_k$ with $left(v_k-frac1nsum_k=1^n v_kright)$ between second and third equality?
– AlphaBeta
Jul 26 at 14:24
We have $sum_k=1^n (k-fracn+12)=0$, so $sum_k=1^n (k-fracn+12)frac1n sum_j=1^n v_j=0$ as well.
– Kusma
Jul 26 at 14:31
add a comment |Â
up vote
0
down vote
You are asking for the difference $sum_k=1^n u_k v_k-frac1n sum_k=1^n u_k sum_k=1^n v_k$, where $u_k=ak+b$ so $frac1n sum u_k = b+a fracn+12$.
This means $$sum_k=1^n u_k v_k-frac1n sum_k=1^n u_k sum_k=1^n v_k=asum_k=1^n (k-fracn+12) v_k= asum_k=1^n (k-fracn+12) left(v_k-frac1nsum_j=1^n v_jright).$$
This will for example be small if $v$ is close to its mean for all $k$. Using various inequalities for $sum a_k b_k$, you can derive estimates such as
$$left|sum_k=1^n u_k v_k-frac1n sum_k=1^n u_k sum_k=1^n v_kright|le |a|sqrtsum_k=1^n (k-fracn+12)^2 sqrtsum_k=1^n(v_k-frac1nsum_k=1^n v_k)^2,$$
which tells you your approximation is good if the variance of $v$ is small.
answered Jul 26 at 10:44
Kusma
1,097111
could you elaborate how you substituted $v_k$ with $left(v_k-frac1nsum_k=1^n v_kright)$ between second and third equality?
– AlphaBeta
Jul 26 at 14:24
We have $sum_k=1^n (k-fracn+12)=0$, so $sum_k=1^n (k-fracn+12)frac1n sum_j=1^n v_j=0$ as well.
– Kusma
Jul 26 at 14:31
add a comment |Â
up vote
0
down vote
up vote
0
down vote
You are asking for the difference $sum_k=1^n u_k v_k-frac1n sum_k=1^n u_k sum_k=1^n v_k$, where $u_k=ak+b$ so $frac1n sum u_k = b+a fracn+12$.
This means $$sum_k=1^n u_k v_k-frac1n sum_k=1^n u_k sum_k=1^n v_k=asum_k=1^n (k-fracn+12) v_k= asum_k=1^n (k-fracn+12) left(v_k-frac1nsum_j=1^n v_jright).$$
This will for example be small if $v$ is close to its mean for all $k$. Using various inequalities for $sum a_k b_k$, you can derive estimates such as
$$left|sum_k=1^n u_k v_k-frac1n sum_k=1^n u_k sum_k=1^n v_kright|le |a|sqrtsum_k=1^n (k-fracn+12)^2 sqrtsum_k=1^n(v_k-frac1nsum_k=1^n v_k)^2,$$
which tells you your approximation is good if the variance of $v$ is small.
answered Jul 26 at 10:44
Kusma
1,097111
You are asking for the difference $sum_k=1^n u_k v_k-frac1n sum_k=1^n u_k sum_k=1^n v_k$, where $u_k=ak+b$ so $frac1n sum u_k = b+a fracn+12$.
This means $$sum_k=1^n u_k v_k-frac1n sum_k=1^n u_k sum_k=1^n v_k=asum_k=1^n (k-fracn+12) v_k= asum_k=1^n (k-fracn+12) left(v_k-frac1nsum_j=1^n v_jright).$$
This will for example be small if $v$ is close to its mean for all $k$. Using various inequalities for $sum a_k b_k$, you can derive estimates such as
$$left|sum_k=1^n u_k v_k-frac1n sum_k=1^n u_k sum_k=1^n v_kright|le |a|sqrtsum_k=1^n (k-fracn+12)^2 sqrtsum_k=1^n(v_k-frac1nsum_k=1^n v_k)^2,$$
which tells you your approximation is good if the variance of $v$ is small.
answered Jul 26 at 10:44
Kusma
1,097111
edited Jul 26 at 14:30
answered Jul 26 at 10:44
Kusma
1,097111
answered Jul 26 at 10:44
Kusma
1,097111
answered Jul 26 at 10:44
Kusma
1,097111
1,097111
could you elaborate how you substituted $v_k$ with $left(v_k-frac1nsum_k=1^n v_kright)$ between second and third equality?
– AlphaBeta
Jul 26 at 14:24
We have $sum_k=1^n (k-fracn+12)=0$, so $sum_k=1^n (k-fracn+12)frac1n sum_j=1^n v_j=0$ as well.
– Kusma
Jul 26 at 14:31
add a comment |Â
could you elaborate how you substituted $v_k$ with $left(v_k-frac1nsum_k=1^n v_kright)$ between second and third equality?
– AlphaBeta
Jul 26 at 14:24
We have $sum_k=1^n (k-fracn+12)=0$, so $sum_k=1^n (k-fracn+12)frac1n sum_j=1^n v_j=0$ as well.
– Kusma
Jul 26 at 14:31
could you elaborate how you substituted $v_k$ with $left(v_k-frac1nsum_k=1^n v_kright)$ between second and third equality?
– AlphaBeta
Jul 26 at 14:24
could you elaborate how you substituted $v_k$ with $left(v_k-frac1nsum_k=1^n v_kright)$ between second and third equality?
– AlphaBeta
Jul 26 at 14:24
We have $sum_k=1^n (k-fracn+12)=0$, so $sum_k=1^n (k-fracn+12)frac1n sum_j=1^n v_j=0$ as well.
– Kusma
Jul 26 at 14:31
We have $sum_k=1^n (k-fracn+12)=0$, so $sum_k=1^n (k-fracn+12)frac1n sum_j=1^n v_j=0$ as well.
– Kusma
Jul 26 at 14:31
add a comment |Â
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What exactly do you mean by $u$ being a "linear increasing function"? And by "individual elements" of $v$, do you mean its components?
– Shirish Kulhari
Jul 26 at 10:30
@ShirishKulhari I mean that $u = (a, a + delta, dotsc, a + (n - 1)delta)$ for some $a$ and $delta$. Yes, its components.
– AlphaBeta
Jul 26 at 13:42