Terminology for half the difference
$begingroup$
Is there a term for half of the difference between two numbers
$$x=dfrac{a-b}{2}$$
Analogous to say the mean?
terminology
asked Jan 15 at 9:13
ArchimedesprincipleArchimedesprinciple
34418
$endgroup$
add a comment |
$begingroup$
Is there a term for half of the difference between two numbers
$$x=dfrac{a-b}{2}$$
Analogous to say the mean?
terminology
asked Jan 15 at 9:13
ArchimedesprincipleArchimedesprinciple
34418
$endgroup$
2
$begingroup$
I sincerely doubt it
$endgroup$
– El borito
Jan 15 at 9:16
$begingroup$
I do not think that there is a terminlog better than "distance from the mid-point"
$endgroup$
– Peter
Jan 15 at 9:20
3
$begingroup$
In Italian, this is called semidifferenza, analogous to semisomma (which is a translation of semi-sum, a term that exists also in English). You could use it as an extension - semidifference (maybe)?
$endgroup$
– Easymode44
Jan 15 at 9:20
$begingroup$
Or simply "the mean of $;a,,,-b;$ ...
$endgroup$
– DonAntonio
Jan 15 at 9:34
$begingroup$
MAD (mean absolute derivation)?
$endgroup$
– achille hui
Jan 15 at 9:51
add a comment |
$begingroup$
Is there a term for half of the difference between two numbers
$$x=dfrac{a-b}{2}$$
Analogous to say the mean?
terminology
asked Jan 15 at 9:13
ArchimedesprincipleArchimedesprinciple
34418
$endgroup$
Is there a term for half of the difference between two numbers
$$x=dfrac{a-b}{2}$$
Analogous to say the mean?
terminology
terminology
asked Jan 15 at 9:13
ArchimedesprincipleArchimedesprinciple
34418
asked Jan 15 at 9:13
ArchimedesprincipleArchimedesprinciple
34418
asked Jan 15 at 9:13
ArchimedesprincipleArchimedesprinciple
34418
asked Jan 15 at 9:13
ArchimedesprincipleArchimedesprinciple
34418
asked Jan 15 at 9:13
ArchimedesprincipleArchimedesprinciple
34418
34418
2
$begingroup$
I sincerely doubt it
$endgroup$
– El borito
Jan 15 at 9:16
$begingroup$
I do not think that there is a terminlog better than "distance from the mid-point"
$endgroup$
– Peter
Jan 15 at 9:20
3
$begingroup$
In Italian, this is called semidifferenza, analogous to semisomma (which is a translation of semi-sum, a term that exists also in English). You could use it as an extension - semidifference (maybe)?
$endgroup$
– Easymode44
Jan 15 at 9:20
$begingroup$
Or simply "the mean of $;a,,,-b;$ ...
$endgroup$
– DonAntonio
Jan 15 at 9:34
$begingroup$
MAD (mean absolute derivation)?
$endgroup$
– achille hui
Jan 15 at 9:51
add a comment |
2
$begingroup$
I sincerely doubt it
$endgroup$
– El borito
Jan 15 at 9:16
$begingroup$
I do not think that there is a terminlog better than "distance from the mid-point"
$endgroup$
– Peter
Jan 15 at 9:20
3
$begingroup$
In Italian, this is called semidifferenza, analogous to semisomma (which is a translation of semi-sum, a term that exists also in English). You could use it as an extension - semidifference (maybe)?
$endgroup$
– Easymode44
Jan 15 at 9:20
$begingroup$
Or simply "the mean of $;a,,,-b;$ ...
$endgroup$
– DonAntonio
Jan 15 at 9:34
$begingroup$
MAD (mean absolute derivation)?
$endgroup$
– achille hui
Jan 15 at 9:51
2
2
$begingroup$
I sincerely doubt it
$endgroup$
– El borito
Jan 15 at 9:16
$begingroup$
I sincerely doubt it
$endgroup$
– El borito
Jan 15 at 9:16
$begingroup$
I do not think that there is a terminlog better than "distance from the mid-point"
$endgroup$
– Peter
Jan 15 at 9:20
$begingroup$
I do not think that there is a terminlog better than "distance from the mid-point"
$endgroup$
– Peter
Jan 15 at 9:20
3
3
$begingroup$
In Italian, this is called semidifferenza, analogous to semisomma (which is a translation of semi-sum, a term that exists also in English). You could use it as an extension - semidifference (maybe)?
$endgroup$
– Easymode44
Jan 15 at 9:20
$begingroup$
In Italian, this is called semidifferenza, analogous to semisomma (which is a translation of semi-sum, a term that exists also in English). You could use it as an extension - semidifference (maybe)?
$endgroup$
– Easymode44
Jan 15 at 9:20
$begingroup$
Or simply "the mean of $;a,,,-b;$ ...
$endgroup$
– DonAntonio
Jan 15 at 9:34
$begingroup$
Or simply "the mean of $;a,,,-b;$ ...
$endgroup$
– DonAntonio
Jan 15 at 9:34
$begingroup$
MAD (mean absolute derivation)?
$endgroup$
– achille hui
Jan 15 at 9:51
$begingroup$
MAD (mean absolute derivation)?
$endgroup$
– achille hui
Jan 15 at 9:51
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
You can interpret it as $1over sqrt 2$ times as much as the Standard deviation of $a$ and $b$. According to definition $$sigma=sqrt{(a-m)^2+(a-m)^2}$$where $m={a+bover 2}$ therefore $$sigma=sqrt{{(a-b)^2}over 2}=sqrt 2{left|{a-bover 2}right|}$$and we obtain$$left|{a-bover 2}right|={sigma(a,b)over sqrt 2}$$but I don't know whether any other interpretation exists.
answered Jan 15 at 9:39
Mostafa AyazMostafa Ayaz
18.1k31040
$endgroup$
add a comment |
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1 Answer
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active
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1 Answer
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active
oldest
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active
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votes
$begingroup$
You can interpret it as $1over sqrt 2$ times as much as the Standard deviation of $a$ and $b$. According to definition $$sigma=sqrt{(a-m)^2+(a-m)^2}$$where $m={a+bover 2}$ therefore $$sigma=sqrt{{(a-b)^2}over 2}=sqrt 2{left|{a-bover 2}right|}$$and we obtain$$left|{a-bover 2}right|={sigma(a,b)over sqrt 2}$$but I don't know whether any other interpretation exists.
answered Jan 15 at 9:39
Mostafa AyazMostafa Ayaz
18.1k31040
$endgroup$
add a comment |
$begingroup$
You can interpret it as $1over sqrt 2$ times as much as the Standard deviation of $a$ and $b$. According to definition $$sigma=sqrt{(a-m)^2+(a-m)^2}$$where $m={a+bover 2}$ therefore $$sigma=sqrt{{(a-b)^2}over 2}=sqrt 2{left|{a-bover 2}right|}$$and we obtain$$left|{a-bover 2}right|={sigma(a,b)over sqrt 2}$$but I don't know whether any other interpretation exists.
answered Jan 15 at 9:39
Mostafa AyazMostafa Ayaz
18.1k31040
$endgroup$
add a comment |
$begingroup$
You can interpret it as $1over sqrt 2$ times as much as the Standard deviation of $a$ and $b$. According to definition $$sigma=sqrt{(a-m)^2+(a-m)^2}$$where $m={a+bover 2}$ therefore $$sigma=sqrt{{(a-b)^2}over 2}=sqrt 2{left|{a-bover 2}right|}$$and we obtain$$left|{a-bover 2}right|={sigma(a,b)over sqrt 2}$$but I don't know whether any other interpretation exists.
answered Jan 15 at 9:39
Mostafa AyazMostafa Ayaz
18.1k31040
$endgroup$
You can interpret it as $1over sqrt 2$ times as much as the Standard deviation of $a$ and $b$. According to definition $$sigma=sqrt{(a-m)^2+(a-m)^2}$$where $m={a+bover 2}$ therefore $$sigma=sqrt{{(a-b)^2}over 2}=sqrt 2{left|{a-bover 2}right|}$$and we obtain$$left|{a-bover 2}right|={sigma(a,b)over sqrt 2}$$but I don't know whether any other interpretation exists.
answered Jan 15 at 9:39
Mostafa AyazMostafa Ayaz
18.1k31040
answered Jan 15 at 9:39
Mostafa AyazMostafa Ayaz
18.1k31040
answered Jan 15 at 9:39
Mostafa AyazMostafa Ayaz
18.1k31040
answered Jan 15 at 9:39
Mostafa AyazMostafa Ayaz
18.1k31040
18.1k31040
add a comment |
add a comment |
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2
$begingroup$
I sincerely doubt it
$endgroup$
– El borito
Jan 15 at 9:16
$begingroup$
I do not think that there is a terminlog better than "distance from the mid-point"
$endgroup$
– Peter
Jan 15 at 9:20
3
$begingroup$
In Italian, this is called semidifferenza, analogous to semisomma (which is a translation of semi-sum, a term that exists also in English). You could use it as an extension - semidifference (maybe)?
$endgroup$
– Easymode44
Jan 15 at 9:20
$begingroup$
Or simply "the mean of $;a,,,-b;$ ...
$endgroup$
– DonAntonio
Jan 15 at 9:34
$begingroup$
MAD (mean absolute derivation)?
$endgroup$
– achille hui
Jan 15 at 9:51