Terminology for half the difference












2












$begingroup$


Is there a term for half of the difference between two numbers
$$x=dfrac{a-b}{2}$$



Analogous to say the mean?










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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
















2












$begingroup$


Is there a term for half of the difference between two numbers
$$x=dfrac{a-b}{2}$$



Analogous to say the mean?










share|cite|improve this question









$endgroup$








  • 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














2












2








2





$begingroup$


Is there a term for half of the difference between two numbers
$$x=dfrac{a-b}{2}$$



Analogous to say the mean?










share|cite|improve this question









$endgroup$




Is there a term for half of the difference between two numbers
$$x=dfrac{a-b}{2}$$



Analogous to say the mean?







terminology






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










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














  • 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










1 Answer
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$begingroup$

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.






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    0












    $begingroup$

    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.






    share|cite|improve this answer









    $endgroup$


















      0












      $begingroup$

      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.






      share|cite|improve this answer









      $endgroup$
















        0












        0








        0





        $begingroup$

        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.






        share|cite|improve this answer









        $endgroup$



        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.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Jan 15 at 9:39









        Mostafa AyazMostafa Ayaz

        18.1k31040




        18.1k31040






























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