weights in multiplication












0












$begingroup$


I understand I have an equation of two fractions $0 < A, B < 1$.



$$R = dfrac{A+B}2,$$ where $0 < R, A, B <1$. I understand how to bias towards A or B by inserting weights k1,k2 to the equation, like



$$R = dfrac{k_1A + k_2B}2,$$ where $k_1+k_2=1$.



how do I bias towards a variable in a multiplication scenario, i.e., $R = AB$ ? I sense it should be something like $R = (A+k_1)(B+k_2)$ but I cannot find the boundary conditions and relationship for $k_1,k_2$. Would appreciate your help!



By the way my range for A, B is 1/7 -> 1, with a step of 1/7th.
So A*B should be 1/49 (or as close as possible)



Thanks,
Andreas










share|cite|improve this question











$endgroup$












  • $begingroup$
    Not too sure what you are asking. Do you have any context for this? I guess you can try $$R=A^{k_1}B^{k_2}$$ with $k_1+k_2=1$.
    $endgroup$
    – Karn Watcharasupat
    Dec 26 '18 at 11:59












  • $begingroup$
    Hi, its 0< R, A, B <1 , so this solution cannot be applied. Would I be looking at something like R = A^{1-k1}B^{1-k2} ?
    $endgroup$
    – Andreas Prs
    Dec 26 '18 at 12:26












  • $begingroup$
    The weighted arithmetic mean is simply $R = k_1A + k_2B,$ without dividing by $2.$ The equation $R = frac{A+B}2$ then is what you get when $k_1=k_2=frac12.$
    $endgroup$
    – David K
    Dec 26 '18 at 19:55
















0












$begingroup$


I understand I have an equation of two fractions $0 < A, B < 1$.



$$R = dfrac{A+B}2,$$ where $0 < R, A, B <1$. I understand how to bias towards A or B by inserting weights k1,k2 to the equation, like



$$R = dfrac{k_1A + k_2B}2,$$ where $k_1+k_2=1$.



how do I bias towards a variable in a multiplication scenario, i.e., $R = AB$ ? I sense it should be something like $R = (A+k_1)(B+k_2)$ but I cannot find the boundary conditions and relationship for $k_1,k_2$. Would appreciate your help!



By the way my range for A, B is 1/7 -> 1, with a step of 1/7th.
So A*B should be 1/49 (or as close as possible)



Thanks,
Andreas










share|cite|improve this question











$endgroup$












  • $begingroup$
    Not too sure what you are asking. Do you have any context for this? I guess you can try $$R=A^{k_1}B^{k_2}$$ with $k_1+k_2=1$.
    $endgroup$
    – Karn Watcharasupat
    Dec 26 '18 at 11:59












  • $begingroup$
    Hi, its 0< R, A, B <1 , so this solution cannot be applied. Would I be looking at something like R = A^{1-k1}B^{1-k2} ?
    $endgroup$
    – Andreas Prs
    Dec 26 '18 at 12:26












  • $begingroup$
    The weighted arithmetic mean is simply $R = k_1A + k_2B,$ without dividing by $2.$ The equation $R = frac{A+B}2$ then is what you get when $k_1=k_2=frac12.$
    $endgroup$
    – David K
    Dec 26 '18 at 19:55














0












0








0





$begingroup$


I understand I have an equation of two fractions $0 < A, B < 1$.



$$R = dfrac{A+B}2,$$ where $0 < R, A, B <1$. I understand how to bias towards A or B by inserting weights k1,k2 to the equation, like



$$R = dfrac{k_1A + k_2B}2,$$ where $k_1+k_2=1$.



how do I bias towards a variable in a multiplication scenario, i.e., $R = AB$ ? I sense it should be something like $R = (A+k_1)(B+k_2)$ but I cannot find the boundary conditions and relationship for $k_1,k_2$. Would appreciate your help!



By the way my range for A, B is 1/7 -> 1, with a step of 1/7th.
So A*B should be 1/49 (or as close as possible)



Thanks,
Andreas










share|cite|improve this question











$endgroup$




I understand I have an equation of two fractions $0 < A, B < 1$.



$$R = dfrac{A+B}2,$$ where $0 < R, A, B <1$. I understand how to bias towards A or B by inserting weights k1,k2 to the equation, like



$$R = dfrac{k_1A + k_2B}2,$$ where $k_1+k_2=1$.



how do I bias towards a variable in a multiplication scenario, i.e., $R = AB$ ? I sense it should be something like $R = (A+k_1)(B+k_2)$ but I cannot find the boundary conditions and relationship for $k_1,k_2$. Would appreciate your help!



By the way my range for A, B is 1/7 -> 1, with a step of 1/7th.
So A*B should be 1/49 (or as close as possible)



Thanks,
Andreas







calculus






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 26 '18 at 15:42







Andreas Prs

















asked Dec 26 '18 at 10:00









Andreas PrsAndreas Prs

11




11












  • $begingroup$
    Not too sure what you are asking. Do you have any context for this? I guess you can try $$R=A^{k_1}B^{k_2}$$ with $k_1+k_2=1$.
    $endgroup$
    – Karn Watcharasupat
    Dec 26 '18 at 11:59












  • $begingroup$
    Hi, its 0< R, A, B <1 , so this solution cannot be applied. Would I be looking at something like R = A^{1-k1}B^{1-k2} ?
    $endgroup$
    – Andreas Prs
    Dec 26 '18 at 12:26












  • $begingroup$
    The weighted arithmetic mean is simply $R = k_1A + k_2B,$ without dividing by $2.$ The equation $R = frac{A+B}2$ then is what you get when $k_1=k_2=frac12.$
    $endgroup$
    – David K
    Dec 26 '18 at 19:55


















  • $begingroup$
    Not too sure what you are asking. Do you have any context for this? I guess you can try $$R=A^{k_1}B^{k_2}$$ with $k_1+k_2=1$.
    $endgroup$
    – Karn Watcharasupat
    Dec 26 '18 at 11:59












  • $begingroup$
    Hi, its 0< R, A, B <1 , so this solution cannot be applied. Would I be looking at something like R = A^{1-k1}B^{1-k2} ?
    $endgroup$
    – Andreas Prs
    Dec 26 '18 at 12:26












  • $begingroup$
    The weighted arithmetic mean is simply $R = k_1A + k_2B,$ without dividing by $2.$ The equation $R = frac{A+B}2$ then is what you get when $k_1=k_2=frac12.$
    $endgroup$
    – David K
    Dec 26 '18 at 19:55
















$begingroup$
Not too sure what you are asking. Do you have any context for this? I guess you can try $$R=A^{k_1}B^{k_2}$$ with $k_1+k_2=1$.
$endgroup$
– Karn Watcharasupat
Dec 26 '18 at 11:59






$begingroup$
Not too sure what you are asking. Do you have any context for this? I guess you can try $$R=A^{k_1}B^{k_2}$$ with $k_1+k_2=1$.
$endgroup$
– Karn Watcharasupat
Dec 26 '18 at 11:59














$begingroup$
Hi, its 0< R, A, B <1 , so this solution cannot be applied. Would I be looking at something like R = A^{1-k1}B^{1-k2} ?
$endgroup$
– Andreas Prs
Dec 26 '18 at 12:26






$begingroup$
Hi, its 0< R, A, B <1 , so this solution cannot be applied. Would I be looking at something like R = A^{1-k1}B^{1-k2} ?
$endgroup$
– Andreas Prs
Dec 26 '18 at 12:26














$begingroup$
The weighted arithmetic mean is simply $R = k_1A + k_2B,$ without dividing by $2.$ The equation $R = frac{A+B}2$ then is what you get when $k_1=k_2=frac12.$
$endgroup$
– David K
Dec 26 '18 at 19:55




$begingroup$
The weighted arithmetic mean is simply $R = k_1A + k_2B,$ without dividing by $2.$ The equation $R = frac{A+B}2$ then is what you get when $k_1=k_2=frac12.$
$endgroup$
– David K
Dec 26 '18 at 19:55










2 Answers
2






active

oldest

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0












$begingroup$

From your first example, it looks like you want a formula for $R = f(k,A,B)$ where $lim_{k to 1} R = A$, and $lim_{k to 0} R = B$. The following is one possibility: $R = A^k B^{1-k}$, but I am not sure if this will behave as you want for any $0<k<1$.



Since R is the mean of A,B in your first example, perhaps you are looking for a weighted geometric mean?



$R = e^{k ln(A)+(1-k)ln(B)}$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Hi, so you propose instead of using two weights just use one, and find the neutral solution (bias-wise) at k=0.5? So k=0.6 biases towards B and k=0.4 biases towards A?
    $endgroup$
    – Andreas Prs
    Dec 26 '18 at 12:24










  • $begingroup$
    Yup. Sticking with $k_2=1-k_1$ I just wrote $k$ and $1-k$ instead of $k_1$ and $k_2$, same thing.
    $endgroup$
    – Samadin
    Dec 26 '18 at 12:52










  • $begingroup$
    Note for k=0.5, this doesn't give you back AB. So you'd have to do some tweaking.
    $endgroup$
    – Samadin
    Dec 26 '18 at 12:56










  • $begingroup$
    Thanks a lot for your reply! The weighted mean does not give me AB rather than the mean of the two values for k=0.5, right? What do you mean by tweaking? On the weighted geometric mean or for the R=A^k*B^k ? I would need an equation that is close to that product when k=0.5, but I am not sure of how to approach this.
    $endgroup$
    – Andreas Prs
    Dec 26 '18 at 15:26












  • $begingroup$
    By the way my range for A, B is 1/7 -> 1, with a step of 1/7th. So A*B should be 1/49 (or as close as possible)
    $endgroup$
    – Andreas Prs
    Dec 26 '18 at 15:42



















0












$begingroup$

Because you started your question by asking about a sum which is really an arithmetic mean, people who responded have tended to interpret your "multiplication" question as a question about the geometric mean,
because the geometric mean has the same relationship to multiplication that the arithmetic mean has to addition.



For a weighted geometric mean, you would choose $k_1$ and $k_2$ such that
$k_1+k_2=1,$ $0 leq k_1 leq 1,$ and $0 leq k_2 leq 1,$ and then the weighted geometric mean is
$$ R=A^{k_1}B^{k_2}.$$
Note that I wrote more conditions on $k_1$ and $k_2$ than you really need; in effect, I wrote four inequalities (two for $k_1,$ two for $k_2$) where only two are actually needed. For example, $0 leq k_1$ and $0 leq k_2$ would have been enough.



Alternatively, with the same conditions on $k_1$ and $k_2$ you could write
$$ R=A^{1 - k_1}B^{1 - k_2},$$
and this also would give a weighted geometric mean of the same kind, because the conditions imply that
$(1-k_1)+(1-k_2)=1,$ $0 leq (1-k_1) leq 1,$ and $0 leq (1-k_2) leq 1.$



The fact that $0 < R, A, B leq 1$ is not a problem. These formulas work for any positive numbers.



Also observe that when $k_1 = k_2 = frac12,$ you get the ordinary geometric mean,
$$ R = A^{1/2}B^{1/2} = sqrt{AB}.$$



If you actually want a weighted product, so that equal weighting would give you
$R = AB,$ then you can simply double the exponents in the geometric mean.
For example, you can set
$$ R=A^{2k_1}B^{2k_2},$$
which gives you $R = AB$ when $k_1 = k_2 = frac12.$



If you would prefer not to write the $2$ in each exponent, you can change the conditions so that $k_1+k_2=2,$ $0 leq k_1 leq 2,$ and $0 leq k_2 leq 2,$
and write
$$ R=A^{k_1}B^{k_2},$$
which gives you $R = AB$ when $k_1 = k_2 = 1$ and otherwise gives you a product weighted toward whichever factor has the greater exponent.






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    2 Answers
    2






    active

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    2 Answers
    2






    active

    oldest

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    active

    oldest

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    0












    $begingroup$

    From your first example, it looks like you want a formula for $R = f(k,A,B)$ where $lim_{k to 1} R = A$, and $lim_{k to 0} R = B$. The following is one possibility: $R = A^k B^{1-k}$, but I am not sure if this will behave as you want for any $0<k<1$.



    Since R is the mean of A,B in your first example, perhaps you are looking for a weighted geometric mean?



    $R = e^{k ln(A)+(1-k)ln(B)}$






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      Hi, so you propose instead of using two weights just use one, and find the neutral solution (bias-wise) at k=0.5? So k=0.6 biases towards B and k=0.4 biases towards A?
      $endgroup$
      – Andreas Prs
      Dec 26 '18 at 12:24










    • $begingroup$
      Yup. Sticking with $k_2=1-k_1$ I just wrote $k$ and $1-k$ instead of $k_1$ and $k_2$, same thing.
      $endgroup$
      – Samadin
      Dec 26 '18 at 12:52










    • $begingroup$
      Note for k=0.5, this doesn't give you back AB. So you'd have to do some tweaking.
      $endgroup$
      – Samadin
      Dec 26 '18 at 12:56










    • $begingroup$
      Thanks a lot for your reply! The weighted mean does not give me AB rather than the mean of the two values for k=0.5, right? What do you mean by tweaking? On the weighted geometric mean or for the R=A^k*B^k ? I would need an equation that is close to that product when k=0.5, but I am not sure of how to approach this.
      $endgroup$
      – Andreas Prs
      Dec 26 '18 at 15:26












    • $begingroup$
      By the way my range for A, B is 1/7 -> 1, with a step of 1/7th. So A*B should be 1/49 (or as close as possible)
      $endgroup$
      – Andreas Prs
      Dec 26 '18 at 15:42
















    0












    $begingroup$

    From your first example, it looks like you want a formula for $R = f(k,A,B)$ where $lim_{k to 1} R = A$, and $lim_{k to 0} R = B$. The following is one possibility: $R = A^k B^{1-k}$, but I am not sure if this will behave as you want for any $0<k<1$.



    Since R is the mean of A,B in your first example, perhaps you are looking for a weighted geometric mean?



    $R = e^{k ln(A)+(1-k)ln(B)}$






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      Hi, so you propose instead of using two weights just use one, and find the neutral solution (bias-wise) at k=0.5? So k=0.6 biases towards B and k=0.4 biases towards A?
      $endgroup$
      – Andreas Prs
      Dec 26 '18 at 12:24










    • $begingroup$
      Yup. Sticking with $k_2=1-k_1$ I just wrote $k$ and $1-k$ instead of $k_1$ and $k_2$, same thing.
      $endgroup$
      – Samadin
      Dec 26 '18 at 12:52










    • $begingroup$
      Note for k=0.5, this doesn't give you back AB. So you'd have to do some tweaking.
      $endgroup$
      – Samadin
      Dec 26 '18 at 12:56










    • $begingroup$
      Thanks a lot for your reply! The weighted mean does not give me AB rather than the mean of the two values for k=0.5, right? What do you mean by tweaking? On the weighted geometric mean or for the R=A^k*B^k ? I would need an equation that is close to that product when k=0.5, but I am not sure of how to approach this.
      $endgroup$
      – Andreas Prs
      Dec 26 '18 at 15:26












    • $begingroup$
      By the way my range for A, B is 1/7 -> 1, with a step of 1/7th. So A*B should be 1/49 (or as close as possible)
      $endgroup$
      – Andreas Prs
      Dec 26 '18 at 15:42














    0












    0








    0





    $begingroup$

    From your first example, it looks like you want a formula for $R = f(k,A,B)$ where $lim_{k to 1} R = A$, and $lim_{k to 0} R = B$. The following is one possibility: $R = A^k B^{1-k}$, but I am not sure if this will behave as you want for any $0<k<1$.



    Since R is the mean of A,B in your first example, perhaps you are looking for a weighted geometric mean?



    $R = e^{k ln(A)+(1-k)ln(B)}$






    share|cite|improve this answer









    $endgroup$



    From your first example, it looks like you want a formula for $R = f(k,A,B)$ where $lim_{k to 1} R = A$, and $lim_{k to 0} R = B$. The following is one possibility: $R = A^k B^{1-k}$, but I am not sure if this will behave as you want for any $0<k<1$.



    Since R is the mean of A,B in your first example, perhaps you are looking for a weighted geometric mean?



    $R = e^{k ln(A)+(1-k)ln(B)}$







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered Dec 26 '18 at 12:12









    SamadinSamadin

    42117




    42117












    • $begingroup$
      Hi, so you propose instead of using two weights just use one, and find the neutral solution (bias-wise) at k=0.5? So k=0.6 biases towards B and k=0.4 biases towards A?
      $endgroup$
      – Andreas Prs
      Dec 26 '18 at 12:24










    • $begingroup$
      Yup. Sticking with $k_2=1-k_1$ I just wrote $k$ and $1-k$ instead of $k_1$ and $k_2$, same thing.
      $endgroup$
      – Samadin
      Dec 26 '18 at 12:52










    • $begingroup$
      Note for k=0.5, this doesn't give you back AB. So you'd have to do some tweaking.
      $endgroup$
      – Samadin
      Dec 26 '18 at 12:56










    • $begingroup$
      Thanks a lot for your reply! The weighted mean does not give me AB rather than the mean of the two values for k=0.5, right? What do you mean by tweaking? On the weighted geometric mean or for the R=A^k*B^k ? I would need an equation that is close to that product when k=0.5, but I am not sure of how to approach this.
      $endgroup$
      – Andreas Prs
      Dec 26 '18 at 15:26












    • $begingroup$
      By the way my range for A, B is 1/7 -> 1, with a step of 1/7th. So A*B should be 1/49 (or as close as possible)
      $endgroup$
      – Andreas Prs
      Dec 26 '18 at 15:42


















    • $begingroup$
      Hi, so you propose instead of using two weights just use one, and find the neutral solution (bias-wise) at k=0.5? So k=0.6 biases towards B and k=0.4 biases towards A?
      $endgroup$
      – Andreas Prs
      Dec 26 '18 at 12:24










    • $begingroup$
      Yup. Sticking with $k_2=1-k_1$ I just wrote $k$ and $1-k$ instead of $k_1$ and $k_2$, same thing.
      $endgroup$
      – Samadin
      Dec 26 '18 at 12:52










    • $begingroup$
      Note for k=0.5, this doesn't give you back AB. So you'd have to do some tweaking.
      $endgroup$
      – Samadin
      Dec 26 '18 at 12:56










    • $begingroup$
      Thanks a lot for your reply! The weighted mean does not give me AB rather than the mean of the two values for k=0.5, right? What do you mean by tweaking? On the weighted geometric mean or for the R=A^k*B^k ? I would need an equation that is close to that product when k=0.5, but I am not sure of how to approach this.
      $endgroup$
      – Andreas Prs
      Dec 26 '18 at 15:26












    • $begingroup$
      By the way my range for A, B is 1/7 -> 1, with a step of 1/7th. So A*B should be 1/49 (or as close as possible)
      $endgroup$
      – Andreas Prs
      Dec 26 '18 at 15:42
















    $begingroup$
    Hi, so you propose instead of using two weights just use one, and find the neutral solution (bias-wise) at k=0.5? So k=0.6 biases towards B and k=0.4 biases towards A?
    $endgroup$
    – Andreas Prs
    Dec 26 '18 at 12:24




    $begingroup$
    Hi, so you propose instead of using two weights just use one, and find the neutral solution (bias-wise) at k=0.5? So k=0.6 biases towards B and k=0.4 biases towards A?
    $endgroup$
    – Andreas Prs
    Dec 26 '18 at 12:24












    $begingroup$
    Yup. Sticking with $k_2=1-k_1$ I just wrote $k$ and $1-k$ instead of $k_1$ and $k_2$, same thing.
    $endgroup$
    – Samadin
    Dec 26 '18 at 12:52




    $begingroup$
    Yup. Sticking with $k_2=1-k_1$ I just wrote $k$ and $1-k$ instead of $k_1$ and $k_2$, same thing.
    $endgroup$
    – Samadin
    Dec 26 '18 at 12:52












    $begingroup$
    Note for k=0.5, this doesn't give you back AB. So you'd have to do some tweaking.
    $endgroup$
    – Samadin
    Dec 26 '18 at 12:56




    $begingroup$
    Note for k=0.5, this doesn't give you back AB. So you'd have to do some tweaking.
    $endgroup$
    – Samadin
    Dec 26 '18 at 12:56












    $begingroup$
    Thanks a lot for your reply! The weighted mean does not give me AB rather than the mean of the two values for k=0.5, right? What do you mean by tweaking? On the weighted geometric mean or for the R=A^k*B^k ? I would need an equation that is close to that product when k=0.5, but I am not sure of how to approach this.
    $endgroup$
    – Andreas Prs
    Dec 26 '18 at 15:26






    $begingroup$
    Thanks a lot for your reply! The weighted mean does not give me AB rather than the mean of the two values for k=0.5, right? What do you mean by tweaking? On the weighted geometric mean or for the R=A^k*B^k ? I would need an equation that is close to that product when k=0.5, but I am not sure of how to approach this.
    $endgroup$
    – Andreas Prs
    Dec 26 '18 at 15:26














    $begingroup$
    By the way my range for A, B is 1/7 -> 1, with a step of 1/7th. So A*B should be 1/49 (or as close as possible)
    $endgroup$
    – Andreas Prs
    Dec 26 '18 at 15:42




    $begingroup$
    By the way my range for A, B is 1/7 -> 1, with a step of 1/7th. So A*B should be 1/49 (or as close as possible)
    $endgroup$
    – Andreas Prs
    Dec 26 '18 at 15:42











    0












    $begingroup$

    Because you started your question by asking about a sum which is really an arithmetic mean, people who responded have tended to interpret your "multiplication" question as a question about the geometric mean,
    because the geometric mean has the same relationship to multiplication that the arithmetic mean has to addition.



    For a weighted geometric mean, you would choose $k_1$ and $k_2$ such that
    $k_1+k_2=1,$ $0 leq k_1 leq 1,$ and $0 leq k_2 leq 1,$ and then the weighted geometric mean is
    $$ R=A^{k_1}B^{k_2}.$$
    Note that I wrote more conditions on $k_1$ and $k_2$ than you really need; in effect, I wrote four inequalities (two for $k_1,$ two for $k_2$) where only two are actually needed. For example, $0 leq k_1$ and $0 leq k_2$ would have been enough.



    Alternatively, with the same conditions on $k_1$ and $k_2$ you could write
    $$ R=A^{1 - k_1}B^{1 - k_2},$$
    and this also would give a weighted geometric mean of the same kind, because the conditions imply that
    $(1-k_1)+(1-k_2)=1,$ $0 leq (1-k_1) leq 1,$ and $0 leq (1-k_2) leq 1.$



    The fact that $0 < R, A, B leq 1$ is not a problem. These formulas work for any positive numbers.



    Also observe that when $k_1 = k_2 = frac12,$ you get the ordinary geometric mean,
    $$ R = A^{1/2}B^{1/2} = sqrt{AB}.$$



    If you actually want a weighted product, so that equal weighting would give you
    $R = AB,$ then you can simply double the exponents in the geometric mean.
    For example, you can set
    $$ R=A^{2k_1}B^{2k_2},$$
    which gives you $R = AB$ when $k_1 = k_2 = frac12.$



    If you would prefer not to write the $2$ in each exponent, you can change the conditions so that $k_1+k_2=2,$ $0 leq k_1 leq 2,$ and $0 leq k_2 leq 2,$
    and write
    $$ R=A^{k_1}B^{k_2},$$
    which gives you $R = AB$ when $k_1 = k_2 = 1$ and otherwise gives you a product weighted toward whichever factor has the greater exponent.






    share|cite|improve this answer









    $endgroup$


















      0












      $begingroup$

      Because you started your question by asking about a sum which is really an arithmetic mean, people who responded have tended to interpret your "multiplication" question as a question about the geometric mean,
      because the geometric mean has the same relationship to multiplication that the arithmetic mean has to addition.



      For a weighted geometric mean, you would choose $k_1$ and $k_2$ such that
      $k_1+k_2=1,$ $0 leq k_1 leq 1,$ and $0 leq k_2 leq 1,$ and then the weighted geometric mean is
      $$ R=A^{k_1}B^{k_2}.$$
      Note that I wrote more conditions on $k_1$ and $k_2$ than you really need; in effect, I wrote four inequalities (two for $k_1,$ two for $k_2$) where only two are actually needed. For example, $0 leq k_1$ and $0 leq k_2$ would have been enough.



      Alternatively, with the same conditions on $k_1$ and $k_2$ you could write
      $$ R=A^{1 - k_1}B^{1 - k_2},$$
      and this also would give a weighted geometric mean of the same kind, because the conditions imply that
      $(1-k_1)+(1-k_2)=1,$ $0 leq (1-k_1) leq 1,$ and $0 leq (1-k_2) leq 1.$



      The fact that $0 < R, A, B leq 1$ is not a problem. These formulas work for any positive numbers.



      Also observe that when $k_1 = k_2 = frac12,$ you get the ordinary geometric mean,
      $$ R = A^{1/2}B^{1/2} = sqrt{AB}.$$



      If you actually want a weighted product, so that equal weighting would give you
      $R = AB,$ then you can simply double the exponents in the geometric mean.
      For example, you can set
      $$ R=A^{2k_1}B^{2k_2},$$
      which gives you $R = AB$ when $k_1 = k_2 = frac12.$



      If you would prefer not to write the $2$ in each exponent, you can change the conditions so that $k_1+k_2=2,$ $0 leq k_1 leq 2,$ and $0 leq k_2 leq 2,$
      and write
      $$ R=A^{k_1}B^{k_2},$$
      which gives you $R = AB$ when $k_1 = k_2 = 1$ and otherwise gives you a product weighted toward whichever factor has the greater exponent.






      share|cite|improve this answer









      $endgroup$
















        0












        0








        0





        $begingroup$

        Because you started your question by asking about a sum which is really an arithmetic mean, people who responded have tended to interpret your "multiplication" question as a question about the geometric mean,
        because the geometric mean has the same relationship to multiplication that the arithmetic mean has to addition.



        For a weighted geometric mean, you would choose $k_1$ and $k_2$ such that
        $k_1+k_2=1,$ $0 leq k_1 leq 1,$ and $0 leq k_2 leq 1,$ and then the weighted geometric mean is
        $$ R=A^{k_1}B^{k_2}.$$
        Note that I wrote more conditions on $k_1$ and $k_2$ than you really need; in effect, I wrote four inequalities (two for $k_1,$ two for $k_2$) where only two are actually needed. For example, $0 leq k_1$ and $0 leq k_2$ would have been enough.



        Alternatively, with the same conditions on $k_1$ and $k_2$ you could write
        $$ R=A^{1 - k_1}B^{1 - k_2},$$
        and this also would give a weighted geometric mean of the same kind, because the conditions imply that
        $(1-k_1)+(1-k_2)=1,$ $0 leq (1-k_1) leq 1,$ and $0 leq (1-k_2) leq 1.$



        The fact that $0 < R, A, B leq 1$ is not a problem. These formulas work for any positive numbers.



        Also observe that when $k_1 = k_2 = frac12,$ you get the ordinary geometric mean,
        $$ R = A^{1/2}B^{1/2} = sqrt{AB}.$$



        If you actually want a weighted product, so that equal weighting would give you
        $R = AB,$ then you can simply double the exponents in the geometric mean.
        For example, you can set
        $$ R=A^{2k_1}B^{2k_2},$$
        which gives you $R = AB$ when $k_1 = k_2 = frac12.$



        If you would prefer not to write the $2$ in each exponent, you can change the conditions so that $k_1+k_2=2,$ $0 leq k_1 leq 2,$ and $0 leq k_2 leq 2,$
        and write
        $$ R=A^{k_1}B^{k_2},$$
        which gives you $R = AB$ when $k_1 = k_2 = 1$ and otherwise gives you a product weighted toward whichever factor has the greater exponent.






        share|cite|improve this answer









        $endgroup$



        Because you started your question by asking about a sum which is really an arithmetic mean, people who responded have tended to interpret your "multiplication" question as a question about the geometric mean,
        because the geometric mean has the same relationship to multiplication that the arithmetic mean has to addition.



        For a weighted geometric mean, you would choose $k_1$ and $k_2$ such that
        $k_1+k_2=1,$ $0 leq k_1 leq 1,$ and $0 leq k_2 leq 1,$ and then the weighted geometric mean is
        $$ R=A^{k_1}B^{k_2}.$$
        Note that I wrote more conditions on $k_1$ and $k_2$ than you really need; in effect, I wrote four inequalities (two for $k_1,$ two for $k_2$) where only two are actually needed. For example, $0 leq k_1$ and $0 leq k_2$ would have been enough.



        Alternatively, with the same conditions on $k_1$ and $k_2$ you could write
        $$ R=A^{1 - k_1}B^{1 - k_2},$$
        and this also would give a weighted geometric mean of the same kind, because the conditions imply that
        $(1-k_1)+(1-k_2)=1,$ $0 leq (1-k_1) leq 1,$ and $0 leq (1-k_2) leq 1.$



        The fact that $0 < R, A, B leq 1$ is not a problem. These formulas work for any positive numbers.



        Also observe that when $k_1 = k_2 = frac12,$ you get the ordinary geometric mean,
        $$ R = A^{1/2}B^{1/2} = sqrt{AB}.$$



        If you actually want a weighted product, so that equal weighting would give you
        $R = AB,$ then you can simply double the exponents in the geometric mean.
        For example, you can set
        $$ R=A^{2k_1}B^{2k_2},$$
        which gives you $R = AB$ when $k_1 = k_2 = frac12.$



        If you would prefer not to write the $2$ in each exponent, you can change the conditions so that $k_1+k_2=2,$ $0 leq k_1 leq 2,$ and $0 leq k_2 leq 2,$
        and write
        $$ R=A^{k_1}B^{k_2},$$
        which gives you $R = AB$ when $k_1 = k_2 = 1$ and otherwise gives you a product weighted toward whichever factor has the greater exponent.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 26 '18 at 19:53









        David KDavid K

        55.1k344120




        55.1k344120






























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