Differential of iterative function












0












$begingroup$


I previously got great help with understanding that



$$C = -Afrac{frac{K1}{f1^2H}+2frac{K1K2}{f1^2f2H^2}+3frac{K1K2K3}{f1^3f3H^3}}{1+frac{K1}{f1^2H}+frac{K1K2}{f1^2f2H^2}+frac{K1K2K3}{f1^3f3H^3}}$$



always had a positive differential in H (differentiation of difficult function with unknown constants)



Now I wonder if we can generalize - the general next entry in the numerator would be



$$ifrac{K1K2...Ki}{f1^ifiH^i}$$ and the next entry in the denominator would be



$$frac{K1K2...Ki}{f1^ifiH^i}$$ - it seems also from actual numeric experimenttion that it is indeed true that the general differential is still positive - but can we prove that?



I should add that all K and f values are positive and H (proton concentration) is also positive. K values are true thermodynamic constants, and f values, activity coefficients are assumed fixed for each value of total buffer concentration (A) as a first approximation.






calculus







share|cite|improve this question











$endgroup$

















    0












    $begingroup$


    I previously got great help with understanding that



    $$C = -Afrac{frac{K1}{f1^2H}+2frac{K1K2}{f1^2f2H^2}+3frac{K1K2K3}{f1^3f3H^3}}{1+frac{K1}{f1^2H}+frac{K1K2}{f1^2f2H^2}+frac{K1K2K3}{f1^3f3H^3}}$$



    always had a positive differential in H (differentiation of difficult function with unknown constants)



    Now I wonder if we can generalize - the general next entry in the numerator would be



    $$ifrac{K1K2...Ki}{f1^ifiH^i}$$ and the next entry in the denominator would be



    $$frac{K1K2...Ki}{f1^ifiH^i}$$ - it seems also from actual numeric experimenttion that it is indeed true that the general differential is still positive - but can we prove that?



    I should add that all K and f values are positive and H (proton concentration) is also positive. K values are true thermodynamic constants, and f values, activity coefficients are assumed fixed for each value of total buffer concentration (A) as a first approximation.






    calculus







    share|cite|improve this question











    $endgroup$















      0












      0








      0





      $begingroup$


      I previously got great help with understanding that



      $$C = -Afrac{frac{K1}{f1^2H}+2frac{K1K2}{f1^2f2H^2}+3frac{K1K2K3}{f1^3f3H^3}}{1+frac{K1}{f1^2H}+frac{K1K2}{f1^2f2H^2}+frac{K1K2K3}{f1^3f3H^3}}$$



      always had a positive differential in H (differentiation of difficult function with unknown constants)



      Now I wonder if we can generalize - the general next entry in the numerator would be



      $$ifrac{K1K2...Ki}{f1^ifiH^i}$$ and the next entry in the denominator would be



      $$frac{K1K2...Ki}{f1^ifiH^i}$$ - it seems also from actual numeric experimenttion that it is indeed true that the general differential is still positive - but can we prove that?



      I should add that all K and f values are positive and H (proton concentration) is also positive. K values are true thermodynamic constants, and f values, activity coefficients are assumed fixed for each value of total buffer concentration (A) as a first approximation.






      calculus







      share|cite|improve this question











      $endgroup$




      I previously got great help with understanding that



      $$C = -Afrac{frac{K1}{f1^2H}+2frac{K1K2}{f1^2f2H^2}+3frac{K1K2K3}{f1^3f3H^3}}{1+frac{K1}{f1^2H}+frac{K1K2}{f1^2f2H^2}+frac{K1K2K3}{f1^3f3H^3}}$$



      always had a positive differential in H (differentiation of difficult function with unknown constants)



      Now I wonder if we can generalize - the general next entry in the numerator would be



      $$ifrac{K1K2...Ki}{f1^ifiH^i}$$ and the next entry in the denominator would be



      $$frac{K1K2...Ki}{f1^ifiH^i}$$ - it seems also from actual numeric experimenttion that it is indeed true that the general differential is still positive - but can we prove that?



      I should add that all K and f values are positive and H (proton concentration) is also positive. K values are true thermodynamic constants, and f values, activity coefficients are assumed fixed for each value of total buffer concentration (A) as a first approximation.






      calculus




      calculus






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Jan 18 at 16:02







      user37217

















      asked Jan 14 at 8:39









      user37217user37217

      12




      12






















          1 Answer
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          oldest

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          0












          $begingroup$

          This is not a proof.



          Let $H=frac 1x$ and consider the function
          $$f=frac{sum _{i=1}^n i, a_i, x^i}{1+sum _{i=1}^n a_i, x^i}$$ Compute the derivative to get
          $$left(1+sum _{i=1}^n a_i, x^i right)^2, f'=left(sum _{i=1}^n i^2 a_i x^{i-1}right) left(1+sum _{i=1}^n a_i
          x^iright)-left(sum _{i=1}^n i a_i x^{i-1}right)left( sum _{i=1}^n i a_i
          x^iright)$$           The rhs is then a polynomial of degree $2(n-1)$ that is to say



          $$text{rhs}=sum_{i=0}^{2(n-1)} b_i,x^i$$ in which the coefficients are all positive since all $a_i$'s are non negative. For example, using $n=10$, we should get
          $$left(
          begin{array}{cc}
          i & b_i \
          0 & a_1 \
          1 & 4 a_2 \
          2 & a_1 a_2+9 a_3 \
          3 & 4 a_1 a_3+16 a_4 \
          4 & a_2 a_3+9 a_1 a_4+25 a_5 \
          5 & 4 a_2 a_4+16 a_1 a_5+36 a_6 \
          6 & a_3 a_4+9 a_2 a_5+25 a_1 a_6+49 a_7 \
          7 & 4 a_3 a_5+16 a_2 a_6+36 a_1 a_7+64 a_8 \
          8 & a_4 a_5+9 a_3 a_6+25 a_2 a_7+49 a_1 a_8+81 a_9 \
          9 & 4 a_4 a_6+16 a_3 a_7+36 a_2 a_8+64 a_1 a_9+100 a_{10} \
          10 & a_5 a_6+9 a_4 a_7+25 a_3 a_8+49 a_2 a_9+81 a_1 a_{10} \
          11 & 4 a_5 a_7+16 a_4 a_8+36 a_3 a_9+64 a_2 a_{10} \
          12 & a_6 a_7+9 a_5 a_8+25 a_4 a_9+49 a_3 a_{10} \
          13 & 4 a_6 a_8+16 a_5 a_9+36 a_4 a_{10} \
          14 & a_7 a_8+9 a_6 a_9+25 a_5 a_{10} \
          15 & 4 a_7 a_9+16 a_6 a_{10} \
          16 & a_8 a_9+9 a_7 a_{10} \
          17 & 4 a_8 a_{10} \
          18 & a_9 a_{10}
          end{array}
          right)$$






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Thanks a lot - but what is then the interpretation in regard tothe question of whether we caan know if thederivativeis positive?
            $endgroup$
            – user37217
            Jan 26 at 8:06










          • $begingroup$
            @user37217. You have sums of only positive terms.
            $endgroup$
            – Claude Leibovici
            Jan 26 at 8:43










          • $begingroup$
            Thanks a lot - so this is indeed a proof? - nice!!!
            $endgroup$
            – user37217
            Jan 26 at 13:35











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          1 Answer
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          0












          $begingroup$

          This is not a proof.



          Let $H=frac 1x$ and consider the function
          $$f=frac{sum _{i=1}^n i, a_i, x^i}{1+sum _{i=1}^n a_i, x^i}$$ Compute the derivative to get
          $$left(1+sum _{i=1}^n a_i, x^i right)^2, f'=left(sum _{i=1}^n i^2 a_i x^{i-1}right) left(1+sum _{i=1}^n a_i
          x^iright)-left(sum _{i=1}^n i a_i x^{i-1}right)left( sum _{i=1}^n i a_i
          x^iright)$$           The rhs is then a polynomial of degree $2(n-1)$ that is to say



          $$text{rhs}=sum_{i=0}^{2(n-1)} b_i,x^i$$ in which the coefficients are all positive since all $a_i$'s are non negative. For example, using $n=10$, we should get
          $$left(
          begin{array}{cc}
          i & b_i \
          0 & a_1 \
          1 & 4 a_2 \
          2 & a_1 a_2+9 a_3 \
          3 & 4 a_1 a_3+16 a_4 \
          4 & a_2 a_3+9 a_1 a_4+25 a_5 \
          5 & 4 a_2 a_4+16 a_1 a_5+36 a_6 \
          6 & a_3 a_4+9 a_2 a_5+25 a_1 a_6+49 a_7 \
          7 & 4 a_3 a_5+16 a_2 a_6+36 a_1 a_7+64 a_8 \
          8 & a_4 a_5+9 a_3 a_6+25 a_2 a_7+49 a_1 a_8+81 a_9 \
          9 & 4 a_4 a_6+16 a_3 a_7+36 a_2 a_8+64 a_1 a_9+100 a_{10} \
          10 & a_5 a_6+9 a_4 a_7+25 a_3 a_8+49 a_2 a_9+81 a_1 a_{10} \
          11 & 4 a_5 a_7+16 a_4 a_8+36 a_3 a_9+64 a_2 a_{10} \
          12 & a_6 a_7+9 a_5 a_8+25 a_4 a_9+49 a_3 a_{10} \
          13 & 4 a_6 a_8+16 a_5 a_9+36 a_4 a_{10} \
          14 & a_7 a_8+9 a_6 a_9+25 a_5 a_{10} \
          15 & 4 a_7 a_9+16 a_6 a_{10} \
          16 & a_8 a_9+9 a_7 a_{10} \
          17 & 4 a_8 a_{10} \
          18 & a_9 a_{10}
          end{array}
          right)$$






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Thanks a lot - but what is then the interpretation in regard tothe question of whether we caan know if thederivativeis positive?
            $endgroup$
            – user37217
            Jan 26 at 8:06










          • $begingroup$
            @user37217. You have sums of only positive terms.
            $endgroup$
            – Claude Leibovici
            Jan 26 at 8:43










          • $begingroup$
            Thanks a lot - so this is indeed a proof? - nice!!!
            $endgroup$
            – user37217
            Jan 26 at 13:35
















          0












          $begingroup$

          This is not a proof.



          Let $H=frac 1x$ and consider the function
          $$f=frac{sum _{i=1}^n i, a_i, x^i}{1+sum _{i=1}^n a_i, x^i}$$ Compute the derivative to get
          $$left(1+sum _{i=1}^n a_i, x^i right)^2, f'=left(sum _{i=1}^n i^2 a_i x^{i-1}right) left(1+sum _{i=1}^n a_i
          x^iright)-left(sum _{i=1}^n i a_i x^{i-1}right)left( sum _{i=1}^n i a_i
          x^iright)$$           The rhs is then a polynomial of degree $2(n-1)$ that is to say



          $$text{rhs}=sum_{i=0}^{2(n-1)} b_i,x^i$$ in which the coefficients are all positive since all $a_i$'s are non negative. For example, using $n=10$, we should get
          $$left(
          begin{array}{cc}
          i & b_i \
          0 & a_1 \
          1 & 4 a_2 \
          2 & a_1 a_2+9 a_3 \
          3 & 4 a_1 a_3+16 a_4 \
          4 & a_2 a_3+9 a_1 a_4+25 a_5 \
          5 & 4 a_2 a_4+16 a_1 a_5+36 a_6 \
          6 & a_3 a_4+9 a_2 a_5+25 a_1 a_6+49 a_7 \
          7 & 4 a_3 a_5+16 a_2 a_6+36 a_1 a_7+64 a_8 \
          8 & a_4 a_5+9 a_3 a_6+25 a_2 a_7+49 a_1 a_8+81 a_9 \
          9 & 4 a_4 a_6+16 a_3 a_7+36 a_2 a_8+64 a_1 a_9+100 a_{10} \
          10 & a_5 a_6+9 a_4 a_7+25 a_3 a_8+49 a_2 a_9+81 a_1 a_{10} \
          11 & 4 a_5 a_7+16 a_4 a_8+36 a_3 a_9+64 a_2 a_{10} \
          12 & a_6 a_7+9 a_5 a_8+25 a_4 a_9+49 a_3 a_{10} \
          13 & 4 a_6 a_8+16 a_5 a_9+36 a_4 a_{10} \
          14 & a_7 a_8+9 a_6 a_9+25 a_5 a_{10} \
          15 & 4 a_7 a_9+16 a_6 a_{10} \
          16 & a_8 a_9+9 a_7 a_{10} \
          17 & 4 a_8 a_{10} \
          18 & a_9 a_{10}
          end{array}
          right)$$






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Thanks a lot - but what is then the interpretation in regard tothe question of whether we caan know if thederivativeis positive?
            $endgroup$
            – user37217
            Jan 26 at 8:06










          • $begingroup$
            @user37217. You have sums of only positive terms.
            $endgroup$
            – Claude Leibovici
            Jan 26 at 8:43










          • $begingroup$
            Thanks a lot - so this is indeed a proof? - nice!!!
            $endgroup$
            – user37217
            Jan 26 at 13:35














          0












          0








          0





          $begingroup$

          This is not a proof.



          Let $H=frac 1x$ and consider the function
          $$f=frac{sum _{i=1}^n i, a_i, x^i}{1+sum _{i=1}^n a_i, x^i}$$ Compute the derivative to get
          $$left(1+sum _{i=1}^n a_i, x^i right)^2, f'=left(sum _{i=1}^n i^2 a_i x^{i-1}right) left(1+sum _{i=1}^n a_i
          x^iright)-left(sum _{i=1}^n i a_i x^{i-1}right)left( sum _{i=1}^n i a_i
          x^iright)$$           The rhs is then a polynomial of degree $2(n-1)$ that is to say



          $$text{rhs}=sum_{i=0}^{2(n-1)} b_i,x^i$$ in which the coefficients are all positive since all $a_i$'s are non negative. For example, using $n=10$, we should get
          $$left(
          begin{array}{cc}
          i & b_i \
          0 & a_1 \
          1 & 4 a_2 \
          2 & a_1 a_2+9 a_3 \
          3 & 4 a_1 a_3+16 a_4 \
          4 & a_2 a_3+9 a_1 a_4+25 a_5 \
          5 & 4 a_2 a_4+16 a_1 a_5+36 a_6 \
          6 & a_3 a_4+9 a_2 a_5+25 a_1 a_6+49 a_7 \
          7 & 4 a_3 a_5+16 a_2 a_6+36 a_1 a_7+64 a_8 \
          8 & a_4 a_5+9 a_3 a_6+25 a_2 a_7+49 a_1 a_8+81 a_9 \
          9 & 4 a_4 a_6+16 a_3 a_7+36 a_2 a_8+64 a_1 a_9+100 a_{10} \
          10 & a_5 a_6+9 a_4 a_7+25 a_3 a_8+49 a_2 a_9+81 a_1 a_{10} \
          11 & 4 a_5 a_7+16 a_4 a_8+36 a_3 a_9+64 a_2 a_{10} \
          12 & a_6 a_7+9 a_5 a_8+25 a_4 a_9+49 a_3 a_{10} \
          13 & 4 a_6 a_8+16 a_5 a_9+36 a_4 a_{10} \
          14 & a_7 a_8+9 a_6 a_9+25 a_5 a_{10} \
          15 & 4 a_7 a_9+16 a_6 a_{10} \
          16 & a_8 a_9+9 a_7 a_{10} \
          17 & 4 a_8 a_{10} \
          18 & a_9 a_{10}
          end{array}
          right)$$






          share|cite|improve this answer









          $endgroup$



          This is not a proof.



          Let $H=frac 1x$ and consider the function
          $$f=frac{sum _{i=1}^n i, a_i, x^i}{1+sum _{i=1}^n a_i, x^i}$$ Compute the derivative to get
          $$left(1+sum _{i=1}^n a_i, x^i right)^2, f'=left(sum _{i=1}^n i^2 a_i x^{i-1}right) left(1+sum _{i=1}^n a_i
          x^iright)-left(sum _{i=1}^n i a_i x^{i-1}right)left( sum _{i=1}^n i a_i
          x^iright)$$           The rhs is then a polynomial of degree $2(n-1)$ that is to say



          $$text{rhs}=sum_{i=0}^{2(n-1)} b_i,x^i$$ in which the coefficients are all positive since all $a_i$'s are non negative. For example, using $n=10$, we should get
          $$left(
          begin{array}{cc}
          i & b_i \
          0 & a_1 \
          1 & 4 a_2 \
          2 & a_1 a_2+9 a_3 \
          3 & 4 a_1 a_3+16 a_4 \
          4 & a_2 a_3+9 a_1 a_4+25 a_5 \
          5 & 4 a_2 a_4+16 a_1 a_5+36 a_6 \
          6 & a_3 a_4+9 a_2 a_5+25 a_1 a_6+49 a_7 \
          7 & 4 a_3 a_5+16 a_2 a_6+36 a_1 a_7+64 a_8 \
          8 & a_4 a_5+9 a_3 a_6+25 a_2 a_7+49 a_1 a_8+81 a_9 \
          9 & 4 a_4 a_6+16 a_3 a_7+36 a_2 a_8+64 a_1 a_9+100 a_{10} \
          10 & a_5 a_6+9 a_4 a_7+25 a_3 a_8+49 a_2 a_9+81 a_1 a_{10} \
          11 & 4 a_5 a_7+16 a_4 a_8+36 a_3 a_9+64 a_2 a_{10} \
          12 & a_6 a_7+9 a_5 a_8+25 a_4 a_9+49 a_3 a_{10} \
          13 & 4 a_6 a_8+16 a_5 a_9+36 a_4 a_{10} \
          14 & a_7 a_8+9 a_6 a_9+25 a_5 a_{10} \
          15 & 4 a_7 a_9+16 a_6 a_{10} \
          16 & a_8 a_9+9 a_7 a_{10} \
          17 & 4 a_8 a_{10} \
          18 & a_9 a_{10}
          end{array}
          right)$$







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Jan 22 at 13:00









          Claude LeiboviciClaude Leibovici

          124k1158135




          124k1158135












          • $begingroup$
            Thanks a lot - but what is then the interpretation in regard tothe question of whether we caan know if thederivativeis positive?
            $endgroup$
            – user37217
            Jan 26 at 8:06










          • $begingroup$
            @user37217. You have sums of only positive terms.
            $endgroup$
            – Claude Leibovici
            Jan 26 at 8:43










          • $begingroup$
            Thanks a lot - so this is indeed a proof? - nice!!!
            $endgroup$
            – user37217
            Jan 26 at 13:35


















          • $begingroup$
            Thanks a lot - but what is then the interpretation in regard tothe question of whether we caan know if thederivativeis positive?
            $endgroup$
            – user37217
            Jan 26 at 8:06










          • $begingroup$
            @user37217. You have sums of only positive terms.
            $endgroup$
            – Claude Leibovici
            Jan 26 at 8:43










          • $begingroup$
            Thanks a lot - so this is indeed a proof? - nice!!!
            $endgroup$
            – user37217
            Jan 26 at 13:35
















          $begingroup$
          Thanks a lot - but what is then the interpretation in regard tothe question of whether we caan know if thederivativeis positive?
          $endgroup$
          – user37217
          Jan 26 at 8:06




          $begingroup$
          Thanks a lot - but what is then the interpretation in regard tothe question of whether we caan know if thederivativeis positive?
          $endgroup$
          – user37217
          Jan 26 at 8:06












          $begingroup$
          @user37217. You have sums of only positive terms.
          $endgroup$
          – Claude Leibovici
          Jan 26 at 8:43




          $begingroup$
          @user37217. You have sums of only positive terms.
          $endgroup$
          – Claude Leibovici
          Jan 26 at 8:43












          $begingroup$
          Thanks a lot - so this is indeed a proof? - nice!!!
          $endgroup$
          – user37217
          Jan 26 at 13:35




          $begingroup$
          Thanks a lot - so this is indeed a proof? - nice!!!
          $endgroup$
          – user37217
          Jan 26 at 13:35


















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          Human spaceflight

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          Human spaceflight From Wikipedia, the free encyclopedia Jump to navigation Jump to search "Space traveler" redirects here. For other uses, see Space traveler (disambiguation). @media all and (max-width:720px){.mw-parser-output .tmulti>.thumbinner{width:100%!important;max-width:none!important}.mw-parser-output .tmulti .tsingle{float:none!important;max-width:none!important;width:100%!important;text-align:center}} Apollo 11 crewmember Buzz Aldrin on the Moon, 1969 International Space Station crewmember Tracy Caldwell Dyson views the Earth, 2010 Space Shuttle Discovery heads into space with a crew aboard, STS-121 in 2006 Inside a space suit on the Canadarm, 1993 Human spaceflight (also referred to as crewed spaceflight or manned spaceflight ) is space travel with a crew or passengers aboard the spacecraft. Spacecraft carrying people may be operated directly, by human crew, or it may be e...
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          Can not write log (Is /dev/pts mounted?) - openpty in Ubuntu-on-Windows?

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          0 1 What exactly does the error message beginning with E: mean? I assume it has to do with the file structure on Ubuntu-on-Windows, but what exactly? The following package was automatically installed and is no longer required: os-prober Use 'apt-get autoremove' to remove it. The following extra packages will be installed: libxslt1.1 The following NEW packages will be installed: libxslt1.1 xmlstarlet 0 upgraded, 2 newly installed, 0 to remove and 0 not upgraded. Need to get 435 kB of archives. After this operation, 1023 kB of additional disk space will be used. Get:1 http://archive.ubuntu.com/ubuntu/ trusty/main libxslt1.1 amd64 1.1.28-2build1 [145 kB] Get:2 http://archive.ubuntu.com/ubuntu/ trusty/universe xmlstarlet amd64 1.5.0-1 [290 kB] ...
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          張江高科駅

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          張江高科駅 ちょうこうこうか Zhangjiang Hi-Tech Park 所在地 中国上海市浦東新区 所属事業者 上海地鉄運営有限公司（上海軌道交通・駅詳細） 上海浦東現代有軌交通有限公司（駅詳細） テンプレートを表示 張江高科駅 各種表記 繁体字： 張江高科站 簡体字： 张江高科站 拼音 ： Zhàngjiāng Gāokē Zhàn 発音： ヂャンジャン ガオクー ヂャン 英文： Zhangjiang Hi-Tech Park Station テンプレートを表示 張江高科駅 （ちょうこうこうか-えき）は中華人民共和国上海市浦東新区に位置する、上海軌道交通2号線の駅である。 なお、張江高科駅前にある張江軌道電車の 張江地鉄駅駅 についてもここで述べる。 目次 1 利用可能な鉄道路線 2 沿革 3 駅構造 3.1 上海軌道交通 3.2 張江軌道電車 4 駅周辺 5 隣の駅 6 注釈 7 関連項目 利用可能な鉄道路線 上海軌道交通 ■ 2号線 上海浦東現代有軌交通有限公司 ■ 張江軌道電車 沿革 高架駅時代の姿 2000年12月27日 - 2号線の駅が高架駅として開業。 2010年 1月1日 - 張江軌道電車の開業に伴い、張江地鉄駅駅（停留所）を設置。 2月13日 - 駅地下化に伴い高架駅の運用停止。 2月24日 - 10日間の閉鎖を経て地下新駅の開業。 2017年2月9日 - リニューアル工事を経て再開業。乗降客の増加に対応し、ホームへのエレベーター、エスカレーターを増設、券売機エリアを拡張した [1] 。 駅構造 上海軌道交通 上海軌道交通 張江高科駅 5号口 ちょうこうこうか Zhangjiang Hi-Tech Park ◄ 竜陽路 (2.60km) (1.61km) 金科路 ► 所在地 中国上海市浦東新区松涛路 祖冲之路 （中国語版） 北緯31度12分11秒 東経121度34分50秒  /  北緯31.2...
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