Limits of a parametric equation












1












$begingroup$


I have the following equation which states that $σ$ is an explicit function of $k$:$$σ=frac b{1-cleft(1+left(fracβαk^A-Bright)^{-1}right)},$$
where $A=dfrac{b+c-1}{b}, ;B=dfrac{c}{b+c-1},;;dfracβα$ is a constant and the value of $σ$ has to be positive.



The author found that:




  • If $b+c>1$, the value of $sigma$ declines with increased $k$ and approaches $dfrac{b}{1-c}$ which is greater than $1$ as $k$ increases without limit.

  • If $b+c<1$, the value of $sigma$ increases from $dfrac{b}{1-c}$ which is less than $1$, to $1$ as $k$ increases from $0$ to infinity.


How can I derive these results mathematically?



Further I have tried to simulate the function. While I can replicate the result when $b+c<1$ with the function that tends to the limit $dfrac{b}{1-c}$, I see strange behaviors not corresponding to the result described above when $b+c>1$. So, are there any restrictions so that the values of $b$ and $c$ whose sum is greater than one produce a decreasing function and tending to the $dfrac{b}{1-c}$ limit which this time should be greater than $1$?



Thanks a lot in advance to anyone who wants to help me :)






calculus limits







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$endgroup$

















    1












    $begingroup$


    I have the following equation which states that $σ$ is an explicit function of $k$:$$σ=frac b{1-cleft(1+left(fracβαk^A-Bright)^{-1}right)},$$
    where $A=dfrac{b+c-1}{b}, ;B=dfrac{c}{b+c-1},;;dfracβα$ is a constant and the value of $σ$ has to be positive.



    The author found that:




    • If $b+c>1$, the value of $sigma$ declines with increased $k$ and approaches $dfrac{b}{1-c}$ which is greater than $1$ as $k$ increases without limit.

    • If $b+c<1$, the value of $sigma$ increases from $dfrac{b}{1-c}$ which is less than $1$, to $1$ as $k$ increases from $0$ to infinity.


    How can I derive these results mathematically?



    Further I have tried to simulate the function. While I can replicate the result when $b+c<1$ with the function that tends to the limit $dfrac{b}{1-c}$, I see strange behaviors not corresponding to the result described above when $b+c>1$. So, are there any restrictions so that the values of $b$ and $c$ whose sum is greater than one produce a decreasing function and tending to the $dfrac{b}{1-c}$ limit which this time should be greater than $1$?



    Thanks a lot in advance to anyone who wants to help me :)






    calculus limits







    share|cite|improve this question











    $endgroup$















      1












      1








      1





      $begingroup$


      I have the following equation which states that $σ$ is an explicit function of $k$:$$σ=frac b{1-cleft(1+left(fracβαk^A-Bright)^{-1}right)},$$
      where $A=dfrac{b+c-1}{b}, ;B=dfrac{c}{b+c-1},;;dfracβα$ is a constant and the value of $σ$ has to be positive.



      The author found that:




      • If $b+c>1$, the value of $sigma$ declines with increased $k$ and approaches $dfrac{b}{1-c}$ which is greater than $1$ as $k$ increases without limit.

      • If $b+c<1$, the value of $sigma$ increases from $dfrac{b}{1-c}$ which is less than $1$, to $1$ as $k$ increases from $0$ to infinity.


      How can I derive these results mathematically?



      Further I have tried to simulate the function. While I can replicate the result when $b+c<1$ with the function that tends to the limit $dfrac{b}{1-c}$, I see strange behaviors not corresponding to the result described above when $b+c>1$. So, are there any restrictions so that the values of $b$ and $c$ whose sum is greater than one produce a decreasing function and tending to the $dfrac{b}{1-c}$ limit which this time should be greater than $1$?



      Thanks a lot in advance to anyone who wants to help me :)






      calculus limits







      share|cite|improve this question











      $endgroup$




      I have the following equation which states that $σ$ is an explicit function of $k$:$$σ=frac b{1-cleft(1+left(fracβαk^A-Bright)^{-1}right)},$$
      where $A=dfrac{b+c-1}{b}, ;B=dfrac{c}{b+c-1},;;dfracβα$ is a constant and the value of $σ$ has to be positive.



      The author found that:




      • If $b+c>1$, the value of $sigma$ declines with increased $k$ and approaches $dfrac{b}{1-c}$ which is greater than $1$ as $k$ increases without limit.

      • If $b+c<1$, the value of $sigma$ increases from $dfrac{b}{1-c}$ which is less than $1$, to $1$ as $k$ increases from $0$ to infinity.


      How can I derive these results mathematically?



      Further I have tried to simulate the function. While I can replicate the result when $b+c<1$ with the function that tends to the limit $dfrac{b}{1-c}$, I see strange behaviors not corresponding to the result described above when $b+c>1$. So, are there any restrictions so that the values of $b$ and $c$ whose sum is greater than one produce a decreasing function and tending to the $dfrac{b}{1-c}$ limit which this time should be greater than $1$?



      Thanks a lot in advance to anyone who wants to help me :)






      calculus limits




      calculus limits






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      edited Jan 9 at 13:22









      user376343

      3,3833826




      3,3833826










      asked Jan 8 at 15:47









      AlessandroAlessandro

      237




      237






















          1 Answer
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          $begingroup$

          It is not given, though I assume $b,c>0.$




          • If $;b+c>1;$ then $A=frac{b+c-1}{b}>0,$ and consequently $k^A;$ increases, $;k^A to infty;$ as $;kto infty.$

            Then $left(fracβαk^A-Bright)^{-1};$ decreases to $0$ and finally $sigma;$ decreases to $$lim_{ktoinfty}frac b{1-cleft(1+left(fracβαk^A-Bright)^{-1}right)}=frac{b}{1-ccdot (1+0)}=frac{b}{1-c}$$


          • If $;b+c<1;$ then $A=frac{b+c-1}{b}<0,$ and $k^A;$ decreases to $0$ as $;kto infty.$

            Note that $limlimits_{kto 0^{+}}k^A=infty.$

            Further, $left(fracβαk^A-Bright)=left(fracβαk^A+frac{c}{1-(b+c)}right);$ decreases to $frac{c}{1-(b+c)};$ from where we conclude that $left(fracβαk^A-Bright)^{-1};$ increases to $;frac{1-(b+c)}{c}.$

            Put together, $sigma$ increases from the value$$lim_{kto 0^{+}}sigma(k)=lim_{kto 0^{+}}frac b{1-cleft(1+left(fracβαk^A-Bright)^{-1}right)}=frac{b}{1-cleft(1+0right)}=frac{b}{1-c}$$
            to
            $$lim_{ktoinfty}sigma(k)=lim_{ktoinfty}frac b{1-cleft(1+left(fracβαk^A-Bright)^{-1}right)}=frac{b}{1-cleft(1+frac{1-(b+c)}{c}right)}=1.$$







          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Thank you for your answer! :) If I try to give some numerical values to the function on Excel for the second case no problem, but when b+c are greater than 1, the block $k^{A}-B$ has a strange behaviour and assume negative values. I attach a pictures of the two cases here below: drive.google.com/open?id=11HPJcunadtpBUzmjFDCrbpGldMPW3BEL
            $endgroup$
            – Alessandro
            Jan 9 at 15:53












          • $begingroup$
            Only when the b is "high enough" and the value of c is "low enough" the right behaviour of sigma took place. drive.google.com/file/d/1hzIRYBuFufCJ1f0dCBKWuQl9JDCfHYEm/…
            $endgroup$
            – Alessandro
            Jan 9 at 16:44






          • 1




            $begingroup$
            Is there any relation between $alpha, beta$ and other constants?
            $endgroup$
            – user376343
            Jan 9 at 18:09






          • 1




            $begingroup$
            Indeed yep: $alpha = (1-delta )gamma ^{-rho }$ and $beta= delta gamma ^{-rho }$. Where $rho = frac{1}{b}-1$ and $0<delta<1$. But since in my case $gamma =1$ always their ratio should always be constant. Thank u so much for u help! :)
            $endgroup$
            – Alessandro
            Jan 9 at 18:29













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          0












          $begingroup$

          It is not given, though I assume $b,c>0.$




          • If $;b+c>1;$ then $A=frac{b+c-1}{b}>0,$ and consequently $k^A;$ increases, $;k^A to infty;$ as $;kto infty.$

            Then $left(fracβαk^A-Bright)^{-1};$ decreases to $0$ and finally $sigma;$ decreases to $$lim_{ktoinfty}frac b{1-cleft(1+left(fracβαk^A-Bright)^{-1}right)}=frac{b}{1-ccdot (1+0)}=frac{b}{1-c}$$


          • If $;b+c<1;$ then $A=frac{b+c-1}{b}<0,$ and $k^A;$ decreases to $0$ as $;kto infty.$

            Note that $limlimits_{kto 0^{+}}k^A=infty.$

            Further, $left(fracβαk^A-Bright)=left(fracβαk^A+frac{c}{1-(b+c)}right);$ decreases to $frac{c}{1-(b+c)};$ from where we conclude that $left(fracβαk^A-Bright)^{-1};$ increases to $;frac{1-(b+c)}{c}.$

            Put together, $sigma$ increases from the value$$lim_{kto 0^{+}}sigma(k)=lim_{kto 0^{+}}frac b{1-cleft(1+left(fracβαk^A-Bright)^{-1}right)}=frac{b}{1-cleft(1+0right)}=frac{b}{1-c}$$
            to
            $$lim_{ktoinfty}sigma(k)=lim_{ktoinfty}frac b{1-cleft(1+left(fracβαk^A-Bright)^{-1}right)}=frac{b}{1-cleft(1+frac{1-(b+c)}{c}right)}=1.$$







          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Thank you for your answer! :) If I try to give some numerical values to the function on Excel for the second case no problem, but when b+c are greater than 1, the block $k^{A}-B$ has a strange behaviour and assume negative values. I attach a pictures of the two cases here below: drive.google.com/open?id=11HPJcunadtpBUzmjFDCrbpGldMPW3BEL
            $endgroup$
            – Alessandro
            Jan 9 at 15:53












          • $begingroup$
            Only when the b is "high enough" and the value of c is "low enough" the right behaviour of sigma took place. drive.google.com/file/d/1hzIRYBuFufCJ1f0dCBKWuQl9JDCfHYEm/…
            $endgroup$
            – Alessandro
            Jan 9 at 16:44






          • 1




            $begingroup$
            Is there any relation between $alpha, beta$ and other constants?
            $endgroup$
            – user376343
            Jan 9 at 18:09






          • 1




            $begingroup$
            Indeed yep: $alpha = (1-delta )gamma ^{-rho }$ and $beta= delta gamma ^{-rho }$. Where $rho = frac{1}{b}-1$ and $0<delta<1$. But since in my case $gamma =1$ always their ratio should always be constant. Thank u so much for u help! :)
            $endgroup$
            – Alessandro
            Jan 9 at 18:29


















          0












          $begingroup$

          It is not given, though I assume $b,c>0.$




          • If $;b+c>1;$ then $A=frac{b+c-1}{b}>0,$ and consequently $k^A;$ increases, $;k^A to infty;$ as $;kto infty.$

            Then $left(fracβαk^A-Bright)^{-1};$ decreases to $0$ and finally $sigma;$ decreases to $$lim_{ktoinfty}frac b{1-cleft(1+left(fracβαk^A-Bright)^{-1}right)}=frac{b}{1-ccdot (1+0)}=frac{b}{1-c}$$


          • If $;b+c<1;$ then $A=frac{b+c-1}{b}<0,$ and $k^A;$ decreases to $0$ as $;kto infty.$

            Note that $limlimits_{kto 0^{+}}k^A=infty.$

            Further, $left(fracβαk^A-Bright)=left(fracβαk^A+frac{c}{1-(b+c)}right);$ decreases to $frac{c}{1-(b+c)};$ from where we conclude that $left(fracβαk^A-Bright)^{-1};$ increases to $;frac{1-(b+c)}{c}.$

            Put together, $sigma$ increases from the value$$lim_{kto 0^{+}}sigma(k)=lim_{kto 0^{+}}frac b{1-cleft(1+left(fracβαk^A-Bright)^{-1}right)}=frac{b}{1-cleft(1+0right)}=frac{b}{1-c}$$
            to
            $$lim_{ktoinfty}sigma(k)=lim_{ktoinfty}frac b{1-cleft(1+left(fracβαk^A-Bright)^{-1}right)}=frac{b}{1-cleft(1+frac{1-(b+c)}{c}right)}=1.$$







          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Thank you for your answer! :) If I try to give some numerical values to the function on Excel for the second case no problem, but when b+c are greater than 1, the block $k^{A}-B$ has a strange behaviour and assume negative values. I attach a pictures of the two cases here below: drive.google.com/open?id=11HPJcunadtpBUzmjFDCrbpGldMPW3BEL
            $endgroup$
            – Alessandro
            Jan 9 at 15:53












          • $begingroup$
            Only when the b is "high enough" and the value of c is "low enough" the right behaviour of sigma took place. drive.google.com/file/d/1hzIRYBuFufCJ1f0dCBKWuQl9JDCfHYEm/…
            $endgroup$
            – Alessandro
            Jan 9 at 16:44






          • 1




            $begingroup$
            Is there any relation between $alpha, beta$ and other constants?
            $endgroup$
            – user376343
            Jan 9 at 18:09






          • 1




            $begingroup$
            Indeed yep: $alpha = (1-delta )gamma ^{-rho }$ and $beta= delta gamma ^{-rho }$. Where $rho = frac{1}{b}-1$ and $0<delta<1$. But since in my case $gamma =1$ always their ratio should always be constant. Thank u so much for u help! :)
            $endgroup$
            – Alessandro
            Jan 9 at 18:29
















          0












          0








          0





          $begingroup$

          It is not given, though I assume $b,c>0.$




          • If $;b+c>1;$ then $A=frac{b+c-1}{b}>0,$ and consequently $k^A;$ increases, $;k^A to infty;$ as $;kto infty.$

            Then $left(fracβαk^A-Bright)^{-1};$ decreases to $0$ and finally $sigma;$ decreases to $$lim_{ktoinfty}frac b{1-cleft(1+left(fracβαk^A-Bright)^{-1}right)}=frac{b}{1-ccdot (1+0)}=frac{b}{1-c}$$


          • If $;b+c<1;$ then $A=frac{b+c-1}{b}<0,$ and $k^A;$ decreases to $0$ as $;kto infty.$

            Note that $limlimits_{kto 0^{+}}k^A=infty.$

            Further, $left(fracβαk^A-Bright)=left(fracβαk^A+frac{c}{1-(b+c)}right);$ decreases to $frac{c}{1-(b+c)};$ from where we conclude that $left(fracβαk^A-Bright)^{-1};$ increases to $;frac{1-(b+c)}{c}.$

            Put together, $sigma$ increases from the value$$lim_{kto 0^{+}}sigma(k)=lim_{kto 0^{+}}frac b{1-cleft(1+left(fracβαk^A-Bright)^{-1}right)}=frac{b}{1-cleft(1+0right)}=frac{b}{1-c}$$
            to
            $$lim_{ktoinfty}sigma(k)=lim_{ktoinfty}frac b{1-cleft(1+left(fracβαk^A-Bright)^{-1}right)}=frac{b}{1-cleft(1+frac{1-(b+c)}{c}right)}=1.$$







          share|cite|improve this answer









          $endgroup$



          It is not given, though I assume $b,c>0.$




          • If $;b+c>1;$ then $A=frac{b+c-1}{b}>0,$ and consequently $k^A;$ increases, $;k^A to infty;$ as $;kto infty.$

            Then $left(fracβαk^A-Bright)^{-1};$ decreases to $0$ and finally $sigma;$ decreases to $$lim_{ktoinfty}frac b{1-cleft(1+left(fracβαk^A-Bright)^{-1}right)}=frac{b}{1-ccdot (1+0)}=frac{b}{1-c}$$


          • If $;b+c<1;$ then $A=frac{b+c-1}{b}<0,$ and $k^A;$ decreases to $0$ as $;kto infty.$

            Note that $limlimits_{kto 0^{+}}k^A=infty.$

            Further, $left(fracβαk^A-Bright)=left(fracβαk^A+frac{c}{1-(b+c)}right);$ decreases to $frac{c}{1-(b+c)};$ from where we conclude that $left(fracβαk^A-Bright)^{-1};$ increases to $;frac{1-(b+c)}{c}.$

            Put together, $sigma$ increases from the value$$lim_{kto 0^{+}}sigma(k)=lim_{kto 0^{+}}frac b{1-cleft(1+left(fracβαk^A-Bright)^{-1}right)}=frac{b}{1-cleft(1+0right)}=frac{b}{1-c}$$
            to
            $$lim_{ktoinfty}sigma(k)=lim_{ktoinfty}frac b{1-cleft(1+left(fracβαk^A-Bright)^{-1}right)}=frac{b}{1-cleft(1+frac{1-(b+c)}{c}right)}=1.$$








          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Jan 9 at 14:18









          user376343user376343

          3,3833826




          3,3833826












          • $begingroup$
            Thank you for your answer! :) If I try to give some numerical values to the function on Excel for the second case no problem, but when b+c are greater than 1, the block $k^{A}-B$ has a strange behaviour and assume negative values. I attach a pictures of the two cases here below: drive.google.com/open?id=11HPJcunadtpBUzmjFDCrbpGldMPW3BEL
            $endgroup$
            – Alessandro
            Jan 9 at 15:53












          • $begingroup$
            Only when the b is "high enough" and the value of c is "low enough" the right behaviour of sigma took place. drive.google.com/file/d/1hzIRYBuFufCJ1f0dCBKWuQl9JDCfHYEm/…
            $endgroup$
            – Alessandro
            Jan 9 at 16:44






          • 1




            $begingroup$
            Is there any relation between $alpha, beta$ and other constants?
            $endgroup$
            – user376343
            Jan 9 at 18:09






          • 1




            $begingroup$
            Indeed yep: $alpha = (1-delta )gamma ^{-rho }$ and $beta= delta gamma ^{-rho }$. Where $rho = frac{1}{b}-1$ and $0<delta<1$. But since in my case $gamma =1$ always their ratio should always be constant. Thank u so much for u help! :)
            $endgroup$
            – Alessandro
            Jan 9 at 18:29




















          • $begingroup$
            Thank you for your answer! :) If I try to give some numerical values to the function on Excel for the second case no problem, but when b+c are greater than 1, the block $k^{A}-B$ has a strange behaviour and assume negative values. I attach a pictures of the two cases here below: drive.google.com/open?id=11HPJcunadtpBUzmjFDCrbpGldMPW3BEL
            $endgroup$
            – Alessandro
            Jan 9 at 15:53












          • $begingroup$
            Only when the b is "high enough" and the value of c is "low enough" the right behaviour of sigma took place. drive.google.com/file/d/1hzIRYBuFufCJ1f0dCBKWuQl9JDCfHYEm/…
            $endgroup$
            – Alessandro
            Jan 9 at 16:44






          • 1




            $begingroup$
            Is there any relation between $alpha, beta$ and other constants?
            $endgroup$
            – user376343
            Jan 9 at 18:09






          • 1




            $begingroup$
            Indeed yep: $alpha = (1-delta )gamma ^{-rho }$ and $beta= delta gamma ^{-rho }$. Where $rho = frac{1}{b}-1$ and $0<delta<1$. But since in my case $gamma =1$ always their ratio should always be constant. Thank u so much for u help! :)
            $endgroup$
            – Alessandro
            Jan 9 at 18:29


















          $begingroup$
          Thank you for your answer! :) If I try to give some numerical values to the function on Excel for the second case no problem, but when b+c are greater than 1, the block $k^{A}-B$ has a strange behaviour and assume negative values. I attach a pictures of the two cases here below: drive.google.com/open?id=11HPJcunadtpBUzmjFDCrbpGldMPW3BEL
          $endgroup$
          – Alessandro
          Jan 9 at 15:53






          $begingroup$
          Thank you for your answer! :) If I try to give some numerical values to the function on Excel for the second case no problem, but when b+c are greater than 1, the block $k^{A}-B$ has a strange behaviour and assume negative values. I attach a pictures of the two cases here below: drive.google.com/open?id=11HPJcunadtpBUzmjFDCrbpGldMPW3BEL
          $endgroup$
          – Alessandro
          Jan 9 at 15:53














          $begingroup$
          Only when the b is "high enough" and the value of c is "low enough" the right behaviour of sigma took place. drive.google.com/file/d/1hzIRYBuFufCJ1f0dCBKWuQl9JDCfHYEm/…
          $endgroup$
          – Alessandro
          Jan 9 at 16:44




          $begingroup$
          Only when the b is "high enough" and the value of c is "low enough" the right behaviour of sigma took place. drive.google.com/file/d/1hzIRYBuFufCJ1f0dCBKWuQl9JDCfHYEm/…
          $endgroup$
          – Alessandro
          Jan 9 at 16:44




          1




          1




          $begingroup$
          Is there any relation between $alpha, beta$ and other constants?
          $endgroup$
          – user376343
          Jan 9 at 18:09




          $begingroup$
          Is there any relation between $alpha, beta$ and other constants?
          $endgroup$
          – user376343
          Jan 9 at 18:09




          1




          1




          $begingroup$
          Indeed yep: $alpha = (1-delta )gamma ^{-rho }$ and $beta= delta gamma ^{-rho }$. Where $rho = frac{1}{b}-1$ and $0<delta<1$. But since in my case $gamma =1$ always their ratio should always be constant. Thank u so much for u help! :)
          $endgroup$
          – Alessandro
          Jan 9 at 18:29






          $begingroup$
          Indeed yep: $alpha = (1-delta )gamma ^{-rho }$ and $beta= delta gamma ^{-rho }$. Where $rho = frac{1}{b}-1$ and $0<delta<1$. But since in my case $gamma =1$ always their ratio should always be constant. Thank u so much for u help! :)
          $endgroup$
          – Alessandro
          Jan 9 at 18:29




















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