Uncertainties in Logarithms












7












$begingroup$


I have plotted data of temperature against time for a cooling cup of coffee and gotten a nice curve. I have then linearised this data by taking the log of the temperature for the purposes of gatting a straight line. My question now is how do I deal with the uncertainty in the temperature measurement? If the absolute uncertainty was +/-0.5 degrees C what is it now that I've taken the log of the temperature? I found something else here that MIGHT be what I'm looking for but I'm afraid I don't understand that either :(



Many thanks



David






error-analysis







share|cite|improve this question









$endgroup$

















    7












    $begingroup$


    I have plotted data of temperature against time for a cooling cup of coffee and gotten a nice curve. I have then linearised this data by taking the log of the temperature for the purposes of gatting a straight line. My question now is how do I deal with the uncertainty in the temperature measurement? If the absolute uncertainty was +/-0.5 degrees C what is it now that I've taken the log of the temperature? I found something else here that MIGHT be what I'm looking for but I'm afraid I don't understand that either :(



    Many thanks



    David






    error-analysis







    share|cite|improve this question









    $endgroup$















      7












      7








      7





      $begingroup$


      I have plotted data of temperature against time for a cooling cup of coffee and gotten a nice curve. I have then linearised this data by taking the log of the temperature for the purposes of gatting a straight line. My question now is how do I deal with the uncertainty in the temperature measurement? If the absolute uncertainty was +/-0.5 degrees C what is it now that I've taken the log of the temperature? I found something else here that MIGHT be what I'm looking for but I'm afraid I don't understand that either :(



      Many thanks



      David






      error-analysis







      share|cite|improve this question









      $endgroup$




      I have plotted data of temperature against time for a cooling cup of coffee and gotten a nice curve. I have then linearised this data by taking the log of the temperature for the purposes of gatting a straight line. My question now is how do I deal with the uncertainty in the temperature measurement? If the absolute uncertainty was +/-0.5 degrees C what is it now that I've taken the log of the temperature? I found something else here that MIGHT be what I'm looking for but I'm afraid I don't understand that either :(



      Many thanks



      David






      error-analysis




      error-analysis






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Nov 23 '18 at 4:03









      greenplasticdavegreenplasticdave

      903




      903






















          2 Answers
          2






          active

          oldest

          votes


















          3












          $begingroup$

          The general rule is that when you have a value $g$ that depends of another value $f$ then if you write $u(g)$ for the incertitude of $g$ then :
          $$ u(g(f)) = frac{partial g}{partial f}(f) times u(f) $$
          With here $ g = mathrm{log}(T) $ and so :
          $$ u(mathrm{log}(T)) = frac{u(T)}{T} $$



          So for a temperature $T$ the incertitude on its log will be of $frac{0.5}{T}$






          share|cite|improve this answer









          $endgroup$





















            8












            $begingroup$

            This question is a little bit on the homework-like side of things for us, but there's so many heuristic rules about propagating uncertainties floating around that it seems worth stating correctly again.



            The right way to propagate uncertainties in a single variable is to use calculus to decide what a small variation in the input does to the output.



            In your case, you have $y = ln x$, so



            begin{align}
            delta y &= frac{partial y}{partial x} delta x
            = frac{delta x}{x}
            end{align}



            So the absolute error in $ln x$ is the same as the fractional error in $x$. Beware of temperature units (like centigrade) with non-physical zeros.






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              Note that this is only an approximate (first-order) formula, as many others that involve uncertainties.
              $endgroup$
              – Federico Poloni
              Nov 23 '18 at 13:28










            • $begingroup$
              To account for second-order corrections, you can either (a) Taylor-expand around the central value, or (b) repeat your experiment with better precision.
              $endgroup$
              – rob♦
              Nov 23 '18 at 15:08











            Your Answer





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            2 Answers
            2






            active

            oldest

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            2 Answers
            2






            active

            oldest

            votes









            active

            oldest

            votes






            active

            oldest

            votes









            3












            $begingroup$

            The general rule is that when you have a value $g$ that depends of another value $f$ then if you write $u(g)$ for the incertitude of $g$ then :
            $$ u(g(f)) = frac{partial g}{partial f}(f) times u(f) $$
            With here $ g = mathrm{log}(T) $ and so :
            $$ u(mathrm{log}(T)) = frac{u(T)}{T} $$



            So for a temperature $T$ the incertitude on its log will be of $frac{0.5}{T}$






            share|cite|improve this answer









            $endgroup$


















              3












              $begingroup$

              The general rule is that when you have a value $g$ that depends of another value $f$ then if you write $u(g)$ for the incertitude of $g$ then :
              $$ u(g(f)) = frac{partial g}{partial f}(f) times u(f) $$
              With here $ g = mathrm{log}(T) $ and so :
              $$ u(mathrm{log}(T)) = frac{u(T)}{T} $$



              So for a temperature $T$ the incertitude on its log will be of $frac{0.5}{T}$






              share|cite|improve this answer









              $endgroup$
















                3












                3








                3





                $begingroup$

                The general rule is that when you have a value $g$ that depends of another value $f$ then if you write $u(g)$ for the incertitude of $g$ then :
                $$ u(g(f)) = frac{partial g}{partial f}(f) times u(f) $$
                With here $ g = mathrm{log}(T) $ and so :
                $$ u(mathrm{log}(T)) = frac{u(T)}{T} $$



                So for a temperature $T$ the incertitude on its log will be of $frac{0.5}{T}$






                share|cite|improve this answer









                $endgroup$



                The general rule is that when you have a value $g$ that depends of another value $f$ then if you write $u(g)$ for the incertitude of $g$ then :
                $$ u(g(f)) = frac{partial g}{partial f}(f) times u(f) $$
                With here $ g = mathrm{log}(T) $ and so :
                $$ u(mathrm{log}(T)) = frac{u(T)}{T} $$



                So for a temperature $T$ the incertitude on its log will be of $frac{0.5}{T}$







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Nov 23 '18 at 4:13









                Q.ReindeersonQ.Reindeerson

                3396




                3396























                    8












                    $begingroup$

                    This question is a little bit on the homework-like side of things for us, but there's so many heuristic rules about propagating uncertainties floating around that it seems worth stating correctly again.



                    The right way to propagate uncertainties in a single variable is to use calculus to decide what a small variation in the input does to the output.



                    In your case, you have $y = ln x$, so



                    begin{align}
                    delta y &= frac{partial y}{partial x} delta x
                    = frac{delta x}{x}
                    end{align}



                    So the absolute error in $ln x$ is the same as the fractional error in $x$. Beware of temperature units (like centigrade) with non-physical zeros.






                    share|cite|improve this answer









                    $endgroup$













                    • $begingroup$
                      Note that this is only an approximate (first-order) formula, as many others that involve uncertainties.
                      $endgroup$
                      – Federico Poloni
                      Nov 23 '18 at 13:28










                    • $begingroup$
                      To account for second-order corrections, you can either (a) Taylor-expand around the central value, or (b) repeat your experiment with better precision.
                      $endgroup$
                      – rob♦
                      Nov 23 '18 at 15:08
















                    8












                    $begingroup$

                    This question is a little bit on the homework-like side of things for us, but there's so many heuristic rules about propagating uncertainties floating around that it seems worth stating correctly again.



                    The right way to propagate uncertainties in a single variable is to use calculus to decide what a small variation in the input does to the output.



                    In your case, you have $y = ln x$, so



                    begin{align}
                    delta y &= frac{partial y}{partial x} delta x
                    = frac{delta x}{x}
                    end{align}



                    So the absolute error in $ln x$ is the same as the fractional error in $x$. Beware of temperature units (like centigrade) with non-physical zeros.






                    share|cite|improve this answer









                    $endgroup$













                    • $begingroup$
                      Note that this is only an approximate (first-order) formula, as many others that involve uncertainties.
                      $endgroup$
                      – Federico Poloni
                      Nov 23 '18 at 13:28










                    • $begingroup$
                      To account for second-order corrections, you can either (a) Taylor-expand around the central value, or (b) repeat your experiment with better precision.
                      $endgroup$
                      – rob♦
                      Nov 23 '18 at 15:08














                    8












                    8








                    8





                    $begingroup$

                    This question is a little bit on the homework-like side of things for us, but there's so many heuristic rules about propagating uncertainties floating around that it seems worth stating correctly again.



                    The right way to propagate uncertainties in a single variable is to use calculus to decide what a small variation in the input does to the output.



                    In your case, you have $y = ln x$, so



                    begin{align}
                    delta y &= frac{partial y}{partial x} delta x
                    = frac{delta x}{x}
                    end{align}



                    So the absolute error in $ln x$ is the same as the fractional error in $x$. Beware of temperature units (like centigrade) with non-physical zeros.






                    share|cite|improve this answer









                    $endgroup$



                    This question is a little bit on the homework-like side of things for us, but there's so many heuristic rules about propagating uncertainties floating around that it seems worth stating correctly again.



                    The right way to propagate uncertainties in a single variable is to use calculus to decide what a small variation in the input does to the output.



                    In your case, you have $y = ln x$, so



                    begin{align}
                    delta y &= frac{partial y}{partial x} delta x
                    = frac{delta x}{x}
                    end{align}



                    So the absolute error in $ln x$ is the same as the fractional error in $x$. Beware of temperature units (like centigrade) with non-physical zeros.







                    share|cite|improve this answer












                    share|cite|improve this answer



                    share|cite|improve this answer










                    answered Nov 23 '18 at 4:18









                    rob♦rob

                    40.7k972167




                    40.7k972167












                    • $begingroup$
                      Note that this is only an approximate (first-order) formula, as many others that involve uncertainties.
                      $endgroup$
                      – Federico Poloni
                      Nov 23 '18 at 13:28










                    • $begingroup$
                      To account for second-order corrections, you can either (a) Taylor-expand around the central value, or (b) repeat your experiment with better precision.
                      $endgroup$
                      – rob♦
                      Nov 23 '18 at 15:08


















                    • $begingroup$
                      Note that this is only an approximate (first-order) formula, as many others that involve uncertainties.
                      $endgroup$
                      – Federico Poloni
                      Nov 23 '18 at 13:28










                    • $begingroup$
                      To account for second-order corrections, you can either (a) Taylor-expand around the central value, or (b) repeat your experiment with better precision.
                      $endgroup$
                      – rob♦
                      Nov 23 '18 at 15:08
















                    $begingroup$
                    Note that this is only an approximate (first-order) formula, as many others that involve uncertainties.
                    $endgroup$
                    – Federico Poloni
                    Nov 23 '18 at 13:28




                    $begingroup$
                    Note that this is only an approximate (first-order) formula, as many others that involve uncertainties.
                    $endgroup$
                    – Federico Poloni
                    Nov 23 '18 at 13:28












                    $begingroup$
                    To account for second-order corrections, you can either (a) Taylor-expand around the central value, or (b) repeat your experiment with better precision.
                    $endgroup$
                    – rob♦
                    Nov 23 '18 at 15:08




                    $begingroup$
                    To account for second-order corrections, you can either (a) Taylor-expand around the central value, or (b) repeat your experiment with better precision.
                    $endgroup$
                    – rob♦
                    Nov 23 '18 at 15:08


















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