Relative error of division
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How can I proof that $Rel(frac{x}{y})$ $leq$ $Rel(x)+Rel(y)$ where $Rel(x)$ is relative error of $x$
error-propagation
asked Nov 15 at 23:39
Mohamad Abasi
62
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up vote
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down vote
favorite
How can I proof that $Rel(frac{x}{y})$ $leq$ $Rel(x)+Rel(y)$ where $Rel(x)$ is relative error of $x$
error-propagation
asked Nov 15 at 23:39
Mohamad Abasi
62
What have you tried? Where are you stuck? Also, note that this is only true if $Rel(y)$ is small (or if $y$ has been rounded up). If you try out a couple of examples with $Rel(y)=0.5$, for instance comparing $frac11$ to $frac{1}{0.5}$, you will find that the actual bound is quite a bit larger.
– Arthur
Nov 16 at 0:00
I completly have no idea about this,I proved something like this for addition and multiplication but I can't do anything about division,and also all errors are small
– Mohamad Abasi
Nov 16 at 1:15
How did you do multiplication, then? Why doesn't that work for division?
– Arthur
Nov 16 at 2:34
$Rel(xy)= frac{| xy-(x-a)(y-b) | }{ | xy |} = frac{ |ay+bx-ab | }{ | xy | } leq frac{ | ay |+ | bx | + | ab |}{ | xy | } = frac{|a|}{|x|} + frac{|b|}{|y|}+frac{|a|}{|x|}frac{|b|}{|y|}=Rel(x)+Rel(y)+Rel(x)Rel(y)$
– Mohamad Abasi
Nov 16 at 10:42
Cool. And if you try that for division, what happens?
– Arthur
Nov 16 at 10:55
|
show 2 more comments
up vote
0
down vote
favorite
up vote
0
down vote
favorite
How can I proof that $Rel(frac{x}{y})$ $leq$ $Rel(x)+Rel(y)$ where $Rel(x)$ is relative error of $x$
error-propagation
asked Nov 15 at 23:39
Mohamad Abasi
62
How can I proof that $Rel(frac{x}{y})$ $leq$ $Rel(x)+Rel(y)$ where $Rel(x)$ is relative error of $x$
error-propagation
error-propagation
asked Nov 15 at 23:39
Mohamad Abasi
62
asked Nov 15 at 23:39
Mohamad Abasi
62
edited Nov 15 at 23:47
asked Nov 15 at 23:39
Mohamad Abasi
62
asked Nov 15 at 23:39
Mohamad Abasi
62
asked Nov 15 at 23:39
Mohamad Abasi
62
62
What have you tried? Where are you stuck? Also, note that this is only true if $Rel(y)$ is small (or if $y$ has been rounded up). If you try out a couple of examples with $Rel(y)=0.5$, for instance comparing $frac11$ to $frac{1}{0.5}$, you will find that the actual bound is quite a bit larger.
– Arthur
Nov 16 at 0:00
I completly have no idea about this,I proved something like this for addition and multiplication but I can't do anything about division,and also all errors are small
– Mohamad Abasi
Nov 16 at 1:15
How did you do multiplication, then? Why doesn't that work for division?
– Arthur
Nov 16 at 2:34
$Rel(xy)= frac{| xy-(x-a)(y-b) | }{ | xy |} = frac{ |ay+bx-ab | }{ | xy | } leq frac{ | ay |+ | bx | + | ab |}{ | xy | } = frac{|a|}{|x|} + frac{|b|}{|y|}+frac{|a|}{|x|}frac{|b|}{|y|}=Rel(x)+Rel(y)+Rel(x)Rel(y)$
– Mohamad Abasi
Nov 16 at 10:42
Cool. And if you try that for division, what happens?
– Arthur
Nov 16 at 10:55
|
show 2 more comments
What have you tried? Where are you stuck? Also, note that this is only true if $Rel(y)$ is small (or if $y$ has been rounded up). If you try out a couple of examples with $Rel(y)=0.5$, for instance comparing $frac11$ to $frac{1}{0.5}$, you will find that the actual bound is quite a bit larger.
– Arthur
Nov 16 at 0:00
I completly have no idea about this,I proved something like this for addition and multiplication but I can't do anything about division,and also all errors are small
– Mohamad Abasi
Nov 16 at 1:15
How did you do multiplication, then? Why doesn't that work for division?
– Arthur
Nov 16 at 2:34
$Rel(xy)= frac{| xy-(x-a)(y-b) | }{ | xy |} = frac{ |ay+bx-ab | }{ | xy | } leq frac{ | ay |+ | bx | + | ab |}{ | xy | } = frac{|a|}{|x|} + frac{|b|}{|y|}+frac{|a|}{|x|}frac{|b|}{|y|}=Rel(x)+Rel(y)+Rel(x)Rel(y)$
– Mohamad Abasi
Nov 16 at 10:42
Cool. And if you try that for division, what happens?
– Arthur
Nov 16 at 10:55
What have you tried? Where are you stuck? Also, note that this is only true if $Rel(y)$ is small (or if $y$ has been rounded up). If you try out a couple of examples with $Rel(y)=0.5$, for instance comparing $frac11$ to $frac{1}{0.5}$, you will find that the actual bound is quite a bit larger.
– Arthur
Nov 16 at 0:00
What have you tried? Where are you stuck? Also, note that this is only true if $Rel(y)$ is small (or if $y$ has been rounded up). If you try out a couple of examples with $Rel(y)=0.5$, for instance comparing $frac11$ to $frac{1}{0.5}$, you will find that the actual bound is quite a bit larger.
– Arthur
Nov 16 at 0:00
I completly have no idea about this,I proved something like this for addition and multiplication but I can't do anything about division,and also all errors are small
– Mohamad Abasi
Nov 16 at 1:15
I completly have no idea about this,I proved something like this for addition and multiplication but I can't do anything about division,and also all errors are small
– Mohamad Abasi
Nov 16 at 1:15
How did you do multiplication, then? Why doesn't that work for division?
– Arthur
Nov 16 at 2:34
How did you do multiplication, then? Why doesn't that work for division?
– Arthur
Nov 16 at 2:34
$Rel(xy)= frac{| xy-(x-a)(y-b) | }{ | xy |} = frac{ |ay+bx-ab | }{ | xy | } leq frac{ | ay |+ | bx | + | ab |}{ | xy | } = frac{|a|}{|x|} + frac{|b|}{|y|}+frac{|a|}{|x|}frac{|b|}{|y|}=Rel(x)+Rel(y)+Rel(x)Rel(y)$
– Mohamad Abasi
Nov 16 at 10:42
$Rel(xy)= frac{| xy-(x-a)(y-b) | }{ | xy |} = frac{ |ay+bx-ab | }{ | xy | } leq frac{ | ay |+ | bx | + | ab |}{ | xy | } = frac{|a|}{|x|} + frac{|b|}{|y|}+frac{|a|}{|x|}frac{|b|}{|y|}=Rel(x)+Rel(y)+Rel(x)Rel(y)$
– Mohamad Abasi
Nov 16 at 10:42
Cool. And if you try that for division, what happens?
– Arthur
Nov 16 at 10:55
Cool. And if you try that for division, what happens?
– Arthur
Nov 16 at 10:55
|
show 2 more comments
1 Answer
1
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oldest
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0
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If the rounded value of $x$ is $x(1+e_x)$ and the rounded value of $y$ is $y(1+e_y)$, then this means $|e_x|=Rel(x)$ and $|e_y|=Re(y)$. We get
$$
frac{x(1+e_x)}{y(1+e_y)} =frac xycdot(1+e_x)cdotfrac1{1+e_y}
$$
Since $e_y$ is small (specifically, between $-1$ and $1$), we have
$$
frac1{1+e_y}=1-e_y+e_y^2-e_y^3+cdots
$$
(Usually you go the other way, from the infinite geometric series to the fraction, but the equality is just as valid here.)
This gives us
$$
xycdot(1+e_x)cdotfrac1{1+e_y}=frac xycdot(1+e_x)(1-e_y+e_y^2-cdots)\
=frac xy(1+e_x-e_y-e_xe_y+e_y^2+e_xe_y^2-cdots)
$$
This means that
$$
Relleft(frac xyright)=|e_x-e_y-e_xe_y+e_y^2+e_xe_y^2-cdots|
$$
which, after using the triangle inequality, and also removing all higher degree terms, becomes the expression you were after.
answered Nov 16 at 11:49
Arthur
108k7103186
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
If the rounded value of $x$ is $x(1+e_x)$ and the rounded value of $y$ is $y(1+e_y)$, then this means $|e_x|=Rel(x)$ and $|e_y|=Re(y)$. We get
$$
frac{x(1+e_x)}{y(1+e_y)} =frac xycdot(1+e_x)cdotfrac1{1+e_y}
$$
Since $e_y$ is small (specifically, between $-1$ and $1$), we have
$$
frac1{1+e_y}=1-e_y+e_y^2-e_y^3+cdots
$$
(Usually you go the other way, from the infinite geometric series to the fraction, but the equality is just as valid here.)
This gives us
$$
xycdot(1+e_x)cdotfrac1{1+e_y}=frac xycdot(1+e_x)(1-e_y+e_y^2-cdots)\
=frac xy(1+e_x-e_y-e_xe_y+e_y^2+e_xe_y^2-cdots)
$$
This means that
$$
Relleft(frac xyright)=|e_x-e_y-e_xe_y+e_y^2+e_xe_y^2-cdots|
$$
which, after using the triangle inequality, and also removing all higher degree terms, becomes the expression you were after.
answered Nov 16 at 11:49
Arthur
108k7103186
add a comment |
up vote
0
down vote
If the rounded value of $x$ is $x(1+e_x)$ and the rounded value of $y$ is $y(1+e_y)$, then this means $|e_x|=Rel(x)$ and $|e_y|=Re(y)$. We get
$$
frac{x(1+e_x)}{y(1+e_y)} =frac xycdot(1+e_x)cdotfrac1{1+e_y}
$$
Since $e_y$ is small (specifically, between $-1$ and $1$), we have
$$
frac1{1+e_y}=1-e_y+e_y^2-e_y^3+cdots
$$
(Usually you go the other way, from the infinite geometric series to the fraction, but the equality is just as valid here.)
This gives us
$$
xycdot(1+e_x)cdotfrac1{1+e_y}=frac xycdot(1+e_x)(1-e_y+e_y^2-cdots)\
=frac xy(1+e_x-e_y-e_xe_y+e_y^2+e_xe_y^2-cdots)
$$
This means that
$$
Relleft(frac xyright)=|e_x-e_y-e_xe_y+e_y^2+e_xe_y^2-cdots|
$$
which, after using the triangle inequality, and also removing all higher degree terms, becomes the expression you were after.
answered Nov 16 at 11:49
Arthur
108k7103186
add a comment |
up vote
0
down vote
up vote
0
down vote
If the rounded value of $x$ is $x(1+e_x)$ and the rounded value of $y$ is $y(1+e_y)$, then this means $|e_x|=Rel(x)$ and $|e_y|=Re(y)$. We get
$$
frac{x(1+e_x)}{y(1+e_y)} =frac xycdot(1+e_x)cdotfrac1{1+e_y}
$$
Since $e_y$ is small (specifically, between $-1$ and $1$), we have
$$
frac1{1+e_y}=1-e_y+e_y^2-e_y^3+cdots
$$
(Usually you go the other way, from the infinite geometric series to the fraction, but the equality is just as valid here.)
This gives us
$$
xycdot(1+e_x)cdotfrac1{1+e_y}=frac xycdot(1+e_x)(1-e_y+e_y^2-cdots)\
=frac xy(1+e_x-e_y-e_xe_y+e_y^2+e_xe_y^2-cdots)
$$
This means that
$$
Relleft(frac xyright)=|e_x-e_y-e_xe_y+e_y^2+e_xe_y^2-cdots|
$$
which, after using the triangle inequality, and also removing all higher degree terms, becomes the expression you were after.
answered Nov 16 at 11:49
Arthur
108k7103186
If the rounded value of $x$ is $x(1+e_x)$ and the rounded value of $y$ is $y(1+e_y)$, then this means $|e_x|=Rel(x)$ and $|e_y|=Re(y)$. We get
$$
frac{x(1+e_x)}{y(1+e_y)} =frac xycdot(1+e_x)cdotfrac1{1+e_y}
$$
Since $e_y$ is small (specifically, between $-1$ and $1$), we have
$$
frac1{1+e_y}=1-e_y+e_y^2-e_y^3+cdots
$$
(Usually you go the other way, from the infinite geometric series to the fraction, but the equality is just as valid here.)
This gives us
$$
xycdot(1+e_x)cdotfrac1{1+e_y}=frac xycdot(1+e_x)(1-e_y+e_y^2-cdots)\
=frac xy(1+e_x-e_y-e_xe_y+e_y^2+e_xe_y^2-cdots)
$$
This means that
$$
Relleft(frac xyright)=|e_x-e_y-e_xe_y+e_y^2+e_xe_y^2-cdots|
$$
which, after using the triangle inequality, and also removing all higher degree terms, becomes the expression you were after.
answered Nov 16 at 11:49
Arthur
108k7103186
edited Nov 16 at 14:08
answered Nov 16 at 11:49
Arthur
108k7103186
answered Nov 16 at 11:49
Arthur
108k7103186
answered Nov 16 at 11:49
Arthur
108k7103186
108k7103186
add a comment |
add a comment |
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What have you tried? Where are you stuck? Also, note that this is only true if $Rel(y)$ is small (or if $y$ has been rounded up). If you try out a couple of examples with $Rel(y)=0.5$, for instance comparing $frac11$ to $frac{1}{0.5}$, you will find that the actual bound is quite a bit larger.
– Arthur
Nov 16 at 0:00
I completly have no idea about this,I proved something like this for addition and multiplication but I can't do anything about division,and also all errors are small
– Mohamad Abasi
Nov 16 at 1:15
How did you do multiplication, then? Why doesn't that work for division?
– Arthur
Nov 16 at 2:34
$Rel(xy)= frac{| xy-(x-a)(y-b) | }{ | xy |} = frac{ |ay+bx-ab | }{ | xy | } leq frac{ | ay |+ | bx | + | ab |}{ | xy | } = frac{|a|}{|x|} + frac{|b|}{|y|}+frac{|a|}{|x|}frac{|b|}{|y|}=Rel(x)+Rel(y)+Rel(x)Rel(y)$
– Mohamad Abasi
Nov 16 at 10:42
Cool. And if you try that for division, what happens?
– Arthur
Nov 16 at 10:55