Using Newton-Raphson method, find the solution for $e^{frac{x^2}{4vt}} = 1+frac{x^2}{2vt}$












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I need help with solving this difficult fluid dynamic expression. I have tried using rules of logs, symbolab algebra calculator and Wolfram Alpha calculator, and I have got no solution.




How would you solve the following expression USING the NEWTON-RAPHSON method for $x$?
$$e^{frac{x^2}{4vt}} = 1+frac{x^2}{2vt}$$




When solving this USING the NEWTON-RAPHSON method, the solution is: $x=2.2418sqrt{vt}$





I want to know how you could solve the first expression using the NEWTON-RAPHSON method to get the solution. So could someone please provide a step-by-step solution, by using this method please?





Note: This question was answered, however it was NOT answered using NEWTON-RAPHSON method. It was answered using the Lambert W function, which is a very long and complicated process as compared to the Newton-Raphson method.










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    1












    $begingroup$


    I need help with solving this difficult fluid dynamic expression. I have tried using rules of logs, symbolab algebra calculator and Wolfram Alpha calculator, and I have got no solution.




    How would you solve the following expression USING the NEWTON-RAPHSON method for $x$?
    $$e^{frac{x^2}{4vt}} = 1+frac{x^2}{2vt}$$




    When solving this USING the NEWTON-RAPHSON method, the solution is: $x=2.2418sqrt{vt}$





    I want to know how you could solve the first expression using the NEWTON-RAPHSON method to get the solution. So could someone please provide a step-by-step solution, by using this method please?





    Note: This question was answered, however it was NOT answered using NEWTON-RAPHSON method. It was answered using the Lambert W function, which is a very long and complicated process as compared to the Newton-Raphson method.










    share|cite|improve this question









    $endgroup$















      1












      1








      1


      1



      $begingroup$


      I need help with solving this difficult fluid dynamic expression. I have tried using rules of logs, symbolab algebra calculator and Wolfram Alpha calculator, and I have got no solution.




      How would you solve the following expression USING the NEWTON-RAPHSON method for $x$?
      $$e^{frac{x^2}{4vt}} = 1+frac{x^2}{2vt}$$




      When solving this USING the NEWTON-RAPHSON method, the solution is: $x=2.2418sqrt{vt}$





      I want to know how you could solve the first expression using the NEWTON-RAPHSON method to get the solution. So could someone please provide a step-by-step solution, by using this method please?





      Note: This question was answered, however it was NOT answered using NEWTON-RAPHSON method. It was answered using the Lambert W function, which is a very long and complicated process as compared to the Newton-Raphson method.










      share|cite|improve this question









      $endgroup$




      I need help with solving this difficult fluid dynamic expression. I have tried using rules of logs, symbolab algebra calculator and Wolfram Alpha calculator, and I have got no solution.




      How would you solve the following expression USING the NEWTON-RAPHSON method for $x$?
      $$e^{frac{x^2}{4vt}} = 1+frac{x^2}{2vt}$$




      When solving this USING the NEWTON-RAPHSON method, the solution is: $x=2.2418sqrt{vt}$





      I want to know how you could solve the first expression using the NEWTON-RAPHSON method to get the solution. So could someone please provide a step-by-step solution, by using this method please?





      Note: This question was answered, however it was NOT answered using NEWTON-RAPHSON method. It was answered using the Lambert W function, which is a very long and complicated process as compared to the Newton-Raphson method.







      algebra-precalculus newton-raphson






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      asked Jan 3 at 18:03









      Alan GlennAlan Glenn

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          5 Answers
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          1












          $begingroup$

          Let $u = frac{x^2}{4 v t}$ so that
          $$ e^u - 1 - 2u = 0. $$



          enter image description here



          Defining $g(u) = e^u - 1 - 2u, $ we iterate
          $$ u mapsto u - frac{g(u)}{g'(u)} ,$$
          $$ u mapsto frac{1 +(u-1)e^u}{e^u - 2} $$



          enter image description here



          ? u = 1.25
          %9 = 1.250000000000000000000000000
          ? u = ( 1 + (u-1) * exp(u) ) / (exp(u)-2 )
          %10 = 1.256479745141752637179209827
          ? u = ( 1 + (u-1) * exp(u) ) / (exp(u)-2 )
          %11 = 1.256431211361022818343348929
          ? u = ( 1 + (u-1) * exp(u) ) / (exp(u)-2 )
          %12 = 1.256431208626169685666336003
          ? u = ( 1 + (u-1) * exp(u) ) / (exp(u)-2 )
          %13 = 1.256431208626169676982737617
          ? u = ( 1 + (u-1) * exp(u) ) / (exp(u)-2 )
          %14 = 1.256431208626169676982737617
          ?


          We find that
          $$ frac{x^2}{4vt} = 1.256431208626169676982737617 $$ so
          $$ x^2 = 5.025724834504678707930950466 ; vt $$
          $$ x = 2.241812845557068063953533471 ; sqrt {vt} $$






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Thank you very much @Will
            $endgroup$
            – Alan Glenn
            Jan 3 at 22:45










          • $begingroup$
            My final question is, when choosing the initial value 'u', do we always plot the '$u mapsto u - frac{g(u)}{g'(u)}$' expression and also plot the 'y=u' expression and wherever they intercept we choose that as the initial value of u?
            $endgroup$
            – Alan Glenn
            Jan 3 at 22:51












          • $begingroup$
            @AlanGlenn I added in a picture of $y=e^u - 1 - 2u,$ and I do recommend graphing that. That is where I got the initial estimate 1.25, looking at that picture.
            $endgroup$
            – Will Jagy
            Jan 3 at 23:18





















          3












          $begingroup$

          Well, we have:



          $$expleft(frac{x^2}{4cdottext{v}cdot t}right)=1+frac{x^2}{2cdottext{v}cdot t}tag1$$



          Now, we know that we can write:



          $$expleft(alpharight)=sum_{text{n}=0}^inftyfrac{alpha^text{n}}{text{n}!}=frac{alpha^0}{0!}+frac{alpha^1}{1!}+frac{alpha^2}{2!}+dots=$$
          $$1+alpha+frac{alpha^2}{2}+dotstag2$$



          So, for equation $(1)$ we can write:



          $$1+frac{x^2}{4cdottext{v}cdot t}+frac{1}{2}cdotleft(frac{x^2}{4cdottext{v}cdot t}right)^2+dots=1+frac{x^2}{2cdottext{v}cdot t}tag3$$



          Using the aproximation of three terms we have:



          $$1+frac{x^2}{4cdottext{v}cdot t}+frac{1}{2}cdotleft(frac{x^2}{4cdottext{v}cdot t}right)^2approx1+frac{x^2}{2cdottext{v}cdot t}spaceLongleftrightarrowspace$$
          $$xapprox0spaceveespace xapproxpm2sqrt{2}cdotsqrt{text{v}cdottext{t}}tag4$$






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          • $begingroup$
            That is the wrong answer. $2sqrt{2}=2.828$ which is not the same as 2.2418.
            $endgroup$
            – Alan Glenn
            Jan 3 at 18:50










          • $begingroup$
            @AlanGlenn It is not wrong!
            $endgroup$
            – Jan
            Jan 3 at 19:00



















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          $begingroup$

          First express this in terms of a single variable: letting $s = x/sqrt{vt}$, the equation becomes
          $$ e^{s^2/4} = 1 + s^2/2$$
          Now with $f(s) = exp(s^2/4) - (1 + s^2/2)$, $f'(s) = s exp(s^2/4)/2 - s$, and the Newton iteration is
          $$ s_{n+1} = s_n - frac{f(s_n)}{f'(s_n)}$$
          Note that $s=0$ is also a solution, so you don't want to start too close to that.
          Starting with, say, $s_0 = 2$, you just iterate until the numbers get close enough to each other.



          $s_1 = 2 - f(2)/f'(2) = 2.392211192$



          $s_2 = 2.392211192 - f(2.392211192)/f'(2.392211192) = 2.269512712$



          etc.



          I find that $s_5$ and $s_6$ differ only in the $9$'th decimal place.






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            0












            $begingroup$

            We put $0 le y=x^2/(4vt)$ and look for the non-negative zeros of the function $f(y)$
            $$
            left{ matrix{
            f(y) = e^{,y} - 2y - 1 = 0 hfill cr
            f'(y) = e^{,y} - 2quad Rightarrow quad min f(y):;y = ln 2 hfill cr
            0 < f''(y) = e^{,y} hfill cr} right.
            $$



            Clearly, $f(y)$ is convex, has a negative minimum at $y=ln2$, thus it has two zeros.

            One of them is at $y=0$ and the other will be past the minimum.

            Since for $ln2 < y$ the function is increasing we can apply Newton-Raphson
            method to find the second zero, provided that the starting point $y_0$ be to the right of the minimum.



            Exp_Newton_1



            We can choose $y_0=2$ for instance, and then start the recursion
            $$
            eqalign{
            & {{f(y_0 )} over {y_1 - y_0 }} = f'(y_0 )quad Rightarrow quad cr
            & Rightarrow quad y_1 = y_0 + {{f(y_0 )} over {f'(y_0 )}} = y_0 + {{e^{,y_0 } - 2y_0 - 1} over {e^{,y_0 } - 2}}quad Rightarrow cr
            & Rightarrow quad y_{n + 1} = y_n + {{e^{,y_{,n} } - 2y_n - 1} over {e^{,y_{,n} } - 2}} cr}
            $$



            Of course, once found a satisfactory value for $y$, you can easily
            convert it back to $x$






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              0












              $begingroup$

              Just as Will Jagy did, let$u = frac{x^2}{4 v t}$ to make the equation $e^u - 1 - 2u = 0$.



              So, let consider that you look for the zero's of function
              $$f(u)=e^u - 1 - 2u $$ for which
              $$f'(u)=e^u - 2 qquad text{and} qquad f''(u)=e^u > ,,forall u$$



              The first derivative cancels when $u=log(2)$. You can get an estimate of the root builiding the Taylor series at this point. This would give
              $$e^u - 1 - 2u =(1-2 log (2))+(u-log (2))^2+Oleft((u-log (2))^3right)$$ Ignoring the gigher order terms, you then have as an estimate
              $$u_0=log(2)+sqrt{2 log (2)-1}approx 1.31467$$ With this estimate, you can now use Newton method
              $$u_{n+1}=u_n-frac{f(u_n)}{f'(u_n)}=frac{e^{u_n} (u_n-1)+1}{e^{u_n}-2}$$ and get, for twelve significant figures, the following iterates
              $$left(
              begin{array}{cc}
              n & u_n \
              0 & 1.31467301359 \
              1 & 1.26002526328 \
              2 & 1.25644611685 \
              3 & 1.25643120888 \
              4 & 1.25643120863
              end{array}
              right)$$



              One of the important points when you use Newton method is to get a "reasonable" estimates.



              You will notice in the table that, at no time, we overshoot the solution because we started at a point whe $f(x_0) times f''(x_0) > 0$ (Darboux theorem).



              Starting instead with $x_0=1$, the iterates would have been
              $$left(
              begin{array}{cc}
              n & u_n \
              0 & 1.00000000000 \
              1 & color{red}{1.39221119118} \
              2 & 1.27395717022 \
              3 & 1.25677778598 \
              4 & 1.25643134800 \
              5 & 1.25643120863
              end{array}
              right)$$






              share|cite|improve this answer









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                5 Answers
                5






                active

                oldest

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                5 Answers
                5






                active

                oldest

                votes









                active

                oldest

                votes






                active

                oldest

                votes









                1












                $begingroup$

                Let $u = frac{x^2}{4 v t}$ so that
                $$ e^u - 1 - 2u = 0. $$



                enter image description here



                Defining $g(u) = e^u - 1 - 2u, $ we iterate
                $$ u mapsto u - frac{g(u)}{g'(u)} ,$$
                $$ u mapsto frac{1 +(u-1)e^u}{e^u - 2} $$



                enter image description here



                ? u = 1.25
                %9 = 1.250000000000000000000000000
                ? u = ( 1 + (u-1) * exp(u) ) / (exp(u)-2 )
                %10 = 1.256479745141752637179209827
                ? u = ( 1 + (u-1) * exp(u) ) / (exp(u)-2 )
                %11 = 1.256431211361022818343348929
                ? u = ( 1 + (u-1) * exp(u) ) / (exp(u)-2 )
                %12 = 1.256431208626169685666336003
                ? u = ( 1 + (u-1) * exp(u) ) / (exp(u)-2 )
                %13 = 1.256431208626169676982737617
                ? u = ( 1 + (u-1) * exp(u) ) / (exp(u)-2 )
                %14 = 1.256431208626169676982737617
                ?


                We find that
                $$ frac{x^2}{4vt} = 1.256431208626169676982737617 $$ so
                $$ x^2 = 5.025724834504678707930950466 ; vt $$
                $$ x = 2.241812845557068063953533471 ; sqrt {vt} $$






                share|cite|improve this answer











                $endgroup$













                • $begingroup$
                  Thank you very much @Will
                  $endgroup$
                  – Alan Glenn
                  Jan 3 at 22:45










                • $begingroup$
                  My final question is, when choosing the initial value 'u', do we always plot the '$u mapsto u - frac{g(u)}{g'(u)}$' expression and also plot the 'y=u' expression and wherever they intercept we choose that as the initial value of u?
                  $endgroup$
                  – Alan Glenn
                  Jan 3 at 22:51












                • $begingroup$
                  @AlanGlenn I added in a picture of $y=e^u - 1 - 2u,$ and I do recommend graphing that. That is where I got the initial estimate 1.25, looking at that picture.
                  $endgroup$
                  – Will Jagy
                  Jan 3 at 23:18


















                1












                $begingroup$

                Let $u = frac{x^2}{4 v t}$ so that
                $$ e^u - 1 - 2u = 0. $$



                enter image description here



                Defining $g(u) = e^u - 1 - 2u, $ we iterate
                $$ u mapsto u - frac{g(u)}{g'(u)} ,$$
                $$ u mapsto frac{1 +(u-1)e^u}{e^u - 2} $$



                enter image description here



                ? u = 1.25
                %9 = 1.250000000000000000000000000
                ? u = ( 1 + (u-1) * exp(u) ) / (exp(u)-2 )
                %10 = 1.256479745141752637179209827
                ? u = ( 1 + (u-1) * exp(u) ) / (exp(u)-2 )
                %11 = 1.256431211361022818343348929
                ? u = ( 1 + (u-1) * exp(u) ) / (exp(u)-2 )
                %12 = 1.256431208626169685666336003
                ? u = ( 1 + (u-1) * exp(u) ) / (exp(u)-2 )
                %13 = 1.256431208626169676982737617
                ? u = ( 1 + (u-1) * exp(u) ) / (exp(u)-2 )
                %14 = 1.256431208626169676982737617
                ?


                We find that
                $$ frac{x^2}{4vt} = 1.256431208626169676982737617 $$ so
                $$ x^2 = 5.025724834504678707930950466 ; vt $$
                $$ x = 2.241812845557068063953533471 ; sqrt {vt} $$






                share|cite|improve this answer











                $endgroup$













                • $begingroup$
                  Thank you very much @Will
                  $endgroup$
                  – Alan Glenn
                  Jan 3 at 22:45










                • $begingroup$
                  My final question is, when choosing the initial value 'u', do we always plot the '$u mapsto u - frac{g(u)}{g'(u)}$' expression and also plot the 'y=u' expression and wherever they intercept we choose that as the initial value of u?
                  $endgroup$
                  – Alan Glenn
                  Jan 3 at 22:51












                • $begingroup$
                  @AlanGlenn I added in a picture of $y=e^u - 1 - 2u,$ and I do recommend graphing that. That is where I got the initial estimate 1.25, looking at that picture.
                  $endgroup$
                  – Will Jagy
                  Jan 3 at 23:18
















                1












                1








                1





                $begingroup$

                Let $u = frac{x^2}{4 v t}$ so that
                $$ e^u - 1 - 2u = 0. $$



                enter image description here



                Defining $g(u) = e^u - 1 - 2u, $ we iterate
                $$ u mapsto u - frac{g(u)}{g'(u)} ,$$
                $$ u mapsto frac{1 +(u-1)e^u}{e^u - 2} $$



                enter image description here



                ? u = 1.25
                %9 = 1.250000000000000000000000000
                ? u = ( 1 + (u-1) * exp(u) ) / (exp(u)-2 )
                %10 = 1.256479745141752637179209827
                ? u = ( 1 + (u-1) * exp(u) ) / (exp(u)-2 )
                %11 = 1.256431211361022818343348929
                ? u = ( 1 + (u-1) * exp(u) ) / (exp(u)-2 )
                %12 = 1.256431208626169685666336003
                ? u = ( 1 + (u-1) * exp(u) ) / (exp(u)-2 )
                %13 = 1.256431208626169676982737617
                ? u = ( 1 + (u-1) * exp(u) ) / (exp(u)-2 )
                %14 = 1.256431208626169676982737617
                ?


                We find that
                $$ frac{x^2}{4vt} = 1.256431208626169676982737617 $$ so
                $$ x^2 = 5.025724834504678707930950466 ; vt $$
                $$ x = 2.241812845557068063953533471 ; sqrt {vt} $$






                share|cite|improve this answer











                $endgroup$



                Let $u = frac{x^2}{4 v t}$ so that
                $$ e^u - 1 - 2u = 0. $$



                enter image description here



                Defining $g(u) = e^u - 1 - 2u, $ we iterate
                $$ u mapsto u - frac{g(u)}{g'(u)} ,$$
                $$ u mapsto frac{1 +(u-1)e^u}{e^u - 2} $$



                enter image description here



                ? u = 1.25
                %9 = 1.250000000000000000000000000
                ? u = ( 1 + (u-1) * exp(u) ) / (exp(u)-2 )
                %10 = 1.256479745141752637179209827
                ? u = ( 1 + (u-1) * exp(u) ) / (exp(u)-2 )
                %11 = 1.256431211361022818343348929
                ? u = ( 1 + (u-1) * exp(u) ) / (exp(u)-2 )
                %12 = 1.256431208626169685666336003
                ? u = ( 1 + (u-1) * exp(u) ) / (exp(u)-2 )
                %13 = 1.256431208626169676982737617
                ? u = ( 1 + (u-1) * exp(u) ) / (exp(u)-2 )
                %14 = 1.256431208626169676982737617
                ?


                We find that
                $$ frac{x^2}{4vt} = 1.256431208626169676982737617 $$ so
                $$ x^2 = 5.025724834504678707930950466 ; vt $$
                $$ x = 2.241812845557068063953533471 ; sqrt {vt} $$







                share|cite|improve this answer














                share|cite|improve this answer



                share|cite|improve this answer








                edited Jan 3 at 23:15

























                answered Jan 3 at 20:32









                Will JagyWill Jagy

                104k5102201




                104k5102201












                • $begingroup$
                  Thank you very much @Will
                  $endgroup$
                  – Alan Glenn
                  Jan 3 at 22:45










                • $begingroup$
                  My final question is, when choosing the initial value 'u', do we always plot the '$u mapsto u - frac{g(u)}{g'(u)}$' expression and also plot the 'y=u' expression and wherever they intercept we choose that as the initial value of u?
                  $endgroup$
                  – Alan Glenn
                  Jan 3 at 22:51












                • $begingroup$
                  @AlanGlenn I added in a picture of $y=e^u - 1 - 2u,$ and I do recommend graphing that. That is where I got the initial estimate 1.25, looking at that picture.
                  $endgroup$
                  – Will Jagy
                  Jan 3 at 23:18




















                • $begingroup$
                  Thank you very much @Will
                  $endgroup$
                  – Alan Glenn
                  Jan 3 at 22:45










                • $begingroup$
                  My final question is, when choosing the initial value 'u', do we always plot the '$u mapsto u - frac{g(u)}{g'(u)}$' expression and also plot the 'y=u' expression and wherever they intercept we choose that as the initial value of u?
                  $endgroup$
                  – Alan Glenn
                  Jan 3 at 22:51












                • $begingroup$
                  @AlanGlenn I added in a picture of $y=e^u - 1 - 2u,$ and I do recommend graphing that. That is where I got the initial estimate 1.25, looking at that picture.
                  $endgroup$
                  – Will Jagy
                  Jan 3 at 23:18


















                $begingroup$
                Thank you very much @Will
                $endgroup$
                – Alan Glenn
                Jan 3 at 22:45




                $begingroup$
                Thank you very much @Will
                $endgroup$
                – Alan Glenn
                Jan 3 at 22:45












                $begingroup$
                My final question is, when choosing the initial value 'u', do we always plot the '$u mapsto u - frac{g(u)}{g'(u)}$' expression and also plot the 'y=u' expression and wherever they intercept we choose that as the initial value of u?
                $endgroup$
                – Alan Glenn
                Jan 3 at 22:51






                $begingroup$
                My final question is, when choosing the initial value 'u', do we always plot the '$u mapsto u - frac{g(u)}{g'(u)}$' expression and also plot the 'y=u' expression and wherever they intercept we choose that as the initial value of u?
                $endgroup$
                – Alan Glenn
                Jan 3 at 22:51














                $begingroup$
                @AlanGlenn I added in a picture of $y=e^u - 1 - 2u,$ and I do recommend graphing that. That is where I got the initial estimate 1.25, looking at that picture.
                $endgroup$
                – Will Jagy
                Jan 3 at 23:18






                $begingroup$
                @AlanGlenn I added in a picture of $y=e^u - 1 - 2u,$ and I do recommend graphing that. That is where I got the initial estimate 1.25, looking at that picture.
                $endgroup$
                – Will Jagy
                Jan 3 at 23:18













                3












                $begingroup$

                Well, we have:



                $$expleft(frac{x^2}{4cdottext{v}cdot t}right)=1+frac{x^2}{2cdottext{v}cdot t}tag1$$



                Now, we know that we can write:



                $$expleft(alpharight)=sum_{text{n}=0}^inftyfrac{alpha^text{n}}{text{n}!}=frac{alpha^0}{0!}+frac{alpha^1}{1!}+frac{alpha^2}{2!}+dots=$$
                $$1+alpha+frac{alpha^2}{2}+dotstag2$$



                So, for equation $(1)$ we can write:



                $$1+frac{x^2}{4cdottext{v}cdot t}+frac{1}{2}cdotleft(frac{x^2}{4cdottext{v}cdot t}right)^2+dots=1+frac{x^2}{2cdottext{v}cdot t}tag3$$



                Using the aproximation of three terms we have:



                $$1+frac{x^2}{4cdottext{v}cdot t}+frac{1}{2}cdotleft(frac{x^2}{4cdottext{v}cdot t}right)^2approx1+frac{x^2}{2cdottext{v}cdot t}spaceLongleftrightarrowspace$$
                $$xapprox0spaceveespace xapproxpm2sqrt{2}cdotsqrt{text{v}cdottext{t}}tag4$$






                share|cite|improve this answer











                $endgroup$













                • $begingroup$
                  That is the wrong answer. $2sqrt{2}=2.828$ which is not the same as 2.2418.
                  $endgroup$
                  – Alan Glenn
                  Jan 3 at 18:50










                • $begingroup$
                  @AlanGlenn It is not wrong!
                  $endgroup$
                  – Jan
                  Jan 3 at 19:00
















                3












                $begingroup$

                Well, we have:



                $$expleft(frac{x^2}{4cdottext{v}cdot t}right)=1+frac{x^2}{2cdottext{v}cdot t}tag1$$



                Now, we know that we can write:



                $$expleft(alpharight)=sum_{text{n}=0}^inftyfrac{alpha^text{n}}{text{n}!}=frac{alpha^0}{0!}+frac{alpha^1}{1!}+frac{alpha^2}{2!}+dots=$$
                $$1+alpha+frac{alpha^2}{2}+dotstag2$$



                So, for equation $(1)$ we can write:



                $$1+frac{x^2}{4cdottext{v}cdot t}+frac{1}{2}cdotleft(frac{x^2}{4cdottext{v}cdot t}right)^2+dots=1+frac{x^2}{2cdottext{v}cdot t}tag3$$



                Using the aproximation of three terms we have:



                $$1+frac{x^2}{4cdottext{v}cdot t}+frac{1}{2}cdotleft(frac{x^2}{4cdottext{v}cdot t}right)^2approx1+frac{x^2}{2cdottext{v}cdot t}spaceLongleftrightarrowspace$$
                $$xapprox0spaceveespace xapproxpm2sqrt{2}cdotsqrt{text{v}cdottext{t}}tag4$$






                share|cite|improve this answer











                $endgroup$













                • $begingroup$
                  That is the wrong answer. $2sqrt{2}=2.828$ which is not the same as 2.2418.
                  $endgroup$
                  – Alan Glenn
                  Jan 3 at 18:50










                • $begingroup$
                  @AlanGlenn It is not wrong!
                  $endgroup$
                  – Jan
                  Jan 3 at 19:00














                3












                3








                3





                $begingroup$

                Well, we have:



                $$expleft(frac{x^2}{4cdottext{v}cdot t}right)=1+frac{x^2}{2cdottext{v}cdot t}tag1$$



                Now, we know that we can write:



                $$expleft(alpharight)=sum_{text{n}=0}^inftyfrac{alpha^text{n}}{text{n}!}=frac{alpha^0}{0!}+frac{alpha^1}{1!}+frac{alpha^2}{2!}+dots=$$
                $$1+alpha+frac{alpha^2}{2}+dotstag2$$



                So, for equation $(1)$ we can write:



                $$1+frac{x^2}{4cdottext{v}cdot t}+frac{1}{2}cdotleft(frac{x^2}{4cdottext{v}cdot t}right)^2+dots=1+frac{x^2}{2cdottext{v}cdot t}tag3$$



                Using the aproximation of three terms we have:



                $$1+frac{x^2}{4cdottext{v}cdot t}+frac{1}{2}cdotleft(frac{x^2}{4cdottext{v}cdot t}right)^2approx1+frac{x^2}{2cdottext{v}cdot t}spaceLongleftrightarrowspace$$
                $$xapprox0spaceveespace xapproxpm2sqrt{2}cdotsqrt{text{v}cdottext{t}}tag4$$






                share|cite|improve this answer











                $endgroup$



                Well, we have:



                $$expleft(frac{x^2}{4cdottext{v}cdot t}right)=1+frac{x^2}{2cdottext{v}cdot t}tag1$$



                Now, we know that we can write:



                $$expleft(alpharight)=sum_{text{n}=0}^inftyfrac{alpha^text{n}}{text{n}!}=frac{alpha^0}{0!}+frac{alpha^1}{1!}+frac{alpha^2}{2!}+dots=$$
                $$1+alpha+frac{alpha^2}{2}+dotstag2$$



                So, for equation $(1)$ we can write:



                $$1+frac{x^2}{4cdottext{v}cdot t}+frac{1}{2}cdotleft(frac{x^2}{4cdottext{v}cdot t}right)^2+dots=1+frac{x^2}{2cdottext{v}cdot t}tag3$$



                Using the aproximation of three terms we have:



                $$1+frac{x^2}{4cdottext{v}cdot t}+frac{1}{2}cdotleft(frac{x^2}{4cdottext{v}cdot t}right)^2approx1+frac{x^2}{2cdottext{v}cdot t}spaceLongleftrightarrowspace$$
                $$xapprox0spaceveespace xapproxpm2sqrt{2}cdotsqrt{text{v}cdottext{t}}tag4$$







                share|cite|improve this answer














                share|cite|improve this answer



                share|cite|improve this answer








                edited Jan 3 at 19:00

























                answered Jan 3 at 18:14









                JanJan

                22k31440




                22k31440












                • $begingroup$
                  That is the wrong answer. $2sqrt{2}=2.828$ which is not the same as 2.2418.
                  $endgroup$
                  – Alan Glenn
                  Jan 3 at 18:50










                • $begingroup$
                  @AlanGlenn It is not wrong!
                  $endgroup$
                  – Jan
                  Jan 3 at 19:00


















                • $begingroup$
                  That is the wrong answer. $2sqrt{2}=2.828$ which is not the same as 2.2418.
                  $endgroup$
                  – Alan Glenn
                  Jan 3 at 18:50










                • $begingroup$
                  @AlanGlenn It is not wrong!
                  $endgroup$
                  – Jan
                  Jan 3 at 19:00
















                $begingroup$
                That is the wrong answer. $2sqrt{2}=2.828$ which is not the same as 2.2418.
                $endgroup$
                – Alan Glenn
                Jan 3 at 18:50




                $begingroup$
                That is the wrong answer. $2sqrt{2}=2.828$ which is not the same as 2.2418.
                $endgroup$
                – Alan Glenn
                Jan 3 at 18:50












                $begingroup$
                @AlanGlenn It is not wrong!
                $endgroup$
                – Jan
                Jan 3 at 19:00




                $begingroup$
                @AlanGlenn It is not wrong!
                $endgroup$
                – Jan
                Jan 3 at 19:00











                1












                $begingroup$

                First express this in terms of a single variable: letting $s = x/sqrt{vt}$, the equation becomes
                $$ e^{s^2/4} = 1 + s^2/2$$
                Now with $f(s) = exp(s^2/4) - (1 + s^2/2)$, $f'(s) = s exp(s^2/4)/2 - s$, and the Newton iteration is
                $$ s_{n+1} = s_n - frac{f(s_n)}{f'(s_n)}$$
                Note that $s=0$ is also a solution, so you don't want to start too close to that.
                Starting with, say, $s_0 = 2$, you just iterate until the numbers get close enough to each other.



                $s_1 = 2 - f(2)/f'(2) = 2.392211192$



                $s_2 = 2.392211192 - f(2.392211192)/f'(2.392211192) = 2.269512712$



                etc.



                I find that $s_5$ and $s_6$ differ only in the $9$'th decimal place.






                share|cite|improve this answer









                $endgroup$


















                  1












                  $begingroup$

                  First express this in terms of a single variable: letting $s = x/sqrt{vt}$, the equation becomes
                  $$ e^{s^2/4} = 1 + s^2/2$$
                  Now with $f(s) = exp(s^2/4) - (1 + s^2/2)$, $f'(s) = s exp(s^2/4)/2 - s$, and the Newton iteration is
                  $$ s_{n+1} = s_n - frac{f(s_n)}{f'(s_n)}$$
                  Note that $s=0$ is also a solution, so you don't want to start too close to that.
                  Starting with, say, $s_0 = 2$, you just iterate until the numbers get close enough to each other.



                  $s_1 = 2 - f(2)/f'(2) = 2.392211192$



                  $s_2 = 2.392211192 - f(2.392211192)/f'(2.392211192) = 2.269512712$



                  etc.



                  I find that $s_5$ and $s_6$ differ only in the $9$'th decimal place.






                  share|cite|improve this answer









                  $endgroup$
















                    1












                    1








                    1





                    $begingroup$

                    First express this in terms of a single variable: letting $s = x/sqrt{vt}$, the equation becomes
                    $$ e^{s^2/4} = 1 + s^2/2$$
                    Now with $f(s) = exp(s^2/4) - (1 + s^2/2)$, $f'(s) = s exp(s^2/4)/2 - s$, and the Newton iteration is
                    $$ s_{n+1} = s_n - frac{f(s_n)}{f'(s_n)}$$
                    Note that $s=0$ is also a solution, so you don't want to start too close to that.
                    Starting with, say, $s_0 = 2$, you just iterate until the numbers get close enough to each other.



                    $s_1 = 2 - f(2)/f'(2) = 2.392211192$



                    $s_2 = 2.392211192 - f(2.392211192)/f'(2.392211192) = 2.269512712$



                    etc.



                    I find that $s_5$ and $s_6$ differ only in the $9$'th decimal place.






                    share|cite|improve this answer









                    $endgroup$



                    First express this in terms of a single variable: letting $s = x/sqrt{vt}$, the equation becomes
                    $$ e^{s^2/4} = 1 + s^2/2$$
                    Now with $f(s) = exp(s^2/4) - (1 + s^2/2)$, $f'(s) = s exp(s^2/4)/2 - s$, and the Newton iteration is
                    $$ s_{n+1} = s_n - frac{f(s_n)}{f'(s_n)}$$
                    Note that $s=0$ is also a solution, so you don't want to start too close to that.
                    Starting with, say, $s_0 = 2$, you just iterate until the numbers get close enough to each other.



                    $s_1 = 2 - f(2)/f'(2) = 2.392211192$



                    $s_2 = 2.392211192 - f(2.392211192)/f'(2.392211192) = 2.269512712$



                    etc.



                    I find that $s_5$ and $s_6$ differ only in the $9$'th decimal place.







                    share|cite|improve this answer












                    share|cite|improve this answer



                    share|cite|improve this answer










                    answered Jan 3 at 18:55









                    Robert IsraelRobert Israel

                    330k23219473




                    330k23219473























                        0












                        $begingroup$

                        We put $0 le y=x^2/(4vt)$ and look for the non-negative zeros of the function $f(y)$
                        $$
                        left{ matrix{
                        f(y) = e^{,y} - 2y - 1 = 0 hfill cr
                        f'(y) = e^{,y} - 2quad Rightarrow quad min f(y):;y = ln 2 hfill cr
                        0 < f''(y) = e^{,y} hfill cr} right.
                        $$



                        Clearly, $f(y)$ is convex, has a negative minimum at $y=ln2$, thus it has two zeros.

                        One of them is at $y=0$ and the other will be past the minimum.

                        Since for $ln2 < y$ the function is increasing we can apply Newton-Raphson
                        method to find the second zero, provided that the starting point $y_0$ be to the right of the minimum.



                        Exp_Newton_1



                        We can choose $y_0=2$ for instance, and then start the recursion
                        $$
                        eqalign{
                        & {{f(y_0 )} over {y_1 - y_0 }} = f'(y_0 )quad Rightarrow quad cr
                        & Rightarrow quad y_1 = y_0 + {{f(y_0 )} over {f'(y_0 )}} = y_0 + {{e^{,y_0 } - 2y_0 - 1} over {e^{,y_0 } - 2}}quad Rightarrow cr
                        & Rightarrow quad y_{n + 1} = y_n + {{e^{,y_{,n} } - 2y_n - 1} over {e^{,y_{,n} } - 2}} cr}
                        $$



                        Of course, once found a satisfactory value for $y$, you can easily
                        convert it back to $x$






                        share|cite|improve this answer









                        $endgroup$


















                          0












                          $begingroup$

                          We put $0 le y=x^2/(4vt)$ and look for the non-negative zeros of the function $f(y)$
                          $$
                          left{ matrix{
                          f(y) = e^{,y} - 2y - 1 = 0 hfill cr
                          f'(y) = e^{,y} - 2quad Rightarrow quad min f(y):;y = ln 2 hfill cr
                          0 < f''(y) = e^{,y} hfill cr} right.
                          $$



                          Clearly, $f(y)$ is convex, has a negative minimum at $y=ln2$, thus it has two zeros.

                          One of them is at $y=0$ and the other will be past the minimum.

                          Since for $ln2 < y$ the function is increasing we can apply Newton-Raphson
                          method to find the second zero, provided that the starting point $y_0$ be to the right of the minimum.



                          Exp_Newton_1



                          We can choose $y_0=2$ for instance, and then start the recursion
                          $$
                          eqalign{
                          & {{f(y_0 )} over {y_1 - y_0 }} = f'(y_0 )quad Rightarrow quad cr
                          & Rightarrow quad y_1 = y_0 + {{f(y_0 )} over {f'(y_0 )}} = y_0 + {{e^{,y_0 } - 2y_0 - 1} over {e^{,y_0 } - 2}}quad Rightarrow cr
                          & Rightarrow quad y_{n + 1} = y_n + {{e^{,y_{,n} } - 2y_n - 1} over {e^{,y_{,n} } - 2}} cr}
                          $$



                          Of course, once found a satisfactory value for $y$, you can easily
                          convert it back to $x$






                          share|cite|improve this answer









                          $endgroup$
















                            0












                            0








                            0





                            $begingroup$

                            We put $0 le y=x^2/(4vt)$ and look for the non-negative zeros of the function $f(y)$
                            $$
                            left{ matrix{
                            f(y) = e^{,y} - 2y - 1 = 0 hfill cr
                            f'(y) = e^{,y} - 2quad Rightarrow quad min f(y):;y = ln 2 hfill cr
                            0 < f''(y) = e^{,y} hfill cr} right.
                            $$



                            Clearly, $f(y)$ is convex, has a negative minimum at $y=ln2$, thus it has two zeros.

                            One of them is at $y=0$ and the other will be past the minimum.

                            Since for $ln2 < y$ the function is increasing we can apply Newton-Raphson
                            method to find the second zero, provided that the starting point $y_0$ be to the right of the minimum.



                            Exp_Newton_1



                            We can choose $y_0=2$ for instance, and then start the recursion
                            $$
                            eqalign{
                            & {{f(y_0 )} over {y_1 - y_0 }} = f'(y_0 )quad Rightarrow quad cr
                            & Rightarrow quad y_1 = y_0 + {{f(y_0 )} over {f'(y_0 )}} = y_0 + {{e^{,y_0 } - 2y_0 - 1} over {e^{,y_0 } - 2}}quad Rightarrow cr
                            & Rightarrow quad y_{n + 1} = y_n + {{e^{,y_{,n} } - 2y_n - 1} over {e^{,y_{,n} } - 2}} cr}
                            $$



                            Of course, once found a satisfactory value for $y$, you can easily
                            convert it back to $x$






                            share|cite|improve this answer









                            $endgroup$



                            We put $0 le y=x^2/(4vt)$ and look for the non-negative zeros of the function $f(y)$
                            $$
                            left{ matrix{
                            f(y) = e^{,y} - 2y - 1 = 0 hfill cr
                            f'(y) = e^{,y} - 2quad Rightarrow quad min f(y):;y = ln 2 hfill cr
                            0 < f''(y) = e^{,y} hfill cr} right.
                            $$



                            Clearly, $f(y)$ is convex, has a negative minimum at $y=ln2$, thus it has two zeros.

                            One of them is at $y=0$ and the other will be past the minimum.

                            Since for $ln2 < y$ the function is increasing we can apply Newton-Raphson
                            method to find the second zero, provided that the starting point $y_0$ be to the right of the minimum.



                            Exp_Newton_1



                            We can choose $y_0=2$ for instance, and then start the recursion
                            $$
                            eqalign{
                            & {{f(y_0 )} over {y_1 - y_0 }} = f'(y_0 )quad Rightarrow quad cr
                            & Rightarrow quad y_1 = y_0 + {{f(y_0 )} over {f'(y_0 )}} = y_0 + {{e^{,y_0 } - 2y_0 - 1} over {e^{,y_0 } - 2}}quad Rightarrow cr
                            & Rightarrow quad y_{n + 1} = y_n + {{e^{,y_{,n} } - 2y_n - 1} over {e^{,y_{,n} } - 2}} cr}
                            $$



                            Of course, once found a satisfactory value for $y$, you can easily
                            convert it back to $x$







                            share|cite|improve this answer












                            share|cite|improve this answer



                            share|cite|improve this answer










                            answered Jan 3 at 19:21









                            G CabG Cab

                            20.4k31341




                            20.4k31341























                                0












                                $begingroup$

                                Just as Will Jagy did, let$u = frac{x^2}{4 v t}$ to make the equation $e^u - 1 - 2u = 0$.



                                So, let consider that you look for the zero's of function
                                $$f(u)=e^u - 1 - 2u $$ for which
                                $$f'(u)=e^u - 2 qquad text{and} qquad f''(u)=e^u > ,,forall u$$



                                The first derivative cancels when $u=log(2)$. You can get an estimate of the root builiding the Taylor series at this point. This would give
                                $$e^u - 1 - 2u =(1-2 log (2))+(u-log (2))^2+Oleft((u-log (2))^3right)$$ Ignoring the gigher order terms, you then have as an estimate
                                $$u_0=log(2)+sqrt{2 log (2)-1}approx 1.31467$$ With this estimate, you can now use Newton method
                                $$u_{n+1}=u_n-frac{f(u_n)}{f'(u_n)}=frac{e^{u_n} (u_n-1)+1}{e^{u_n}-2}$$ and get, for twelve significant figures, the following iterates
                                $$left(
                                begin{array}{cc}
                                n & u_n \
                                0 & 1.31467301359 \
                                1 & 1.26002526328 \
                                2 & 1.25644611685 \
                                3 & 1.25643120888 \
                                4 & 1.25643120863
                                end{array}
                                right)$$



                                One of the important points when you use Newton method is to get a "reasonable" estimates.



                                You will notice in the table that, at no time, we overshoot the solution because we started at a point whe $f(x_0) times f''(x_0) > 0$ (Darboux theorem).



                                Starting instead with $x_0=1$, the iterates would have been
                                $$left(
                                begin{array}{cc}
                                n & u_n \
                                0 & 1.00000000000 \
                                1 & color{red}{1.39221119118} \
                                2 & 1.27395717022 \
                                3 & 1.25677778598 \
                                4 & 1.25643134800 \
                                5 & 1.25643120863
                                end{array}
                                right)$$






                                share|cite|improve this answer









                                $endgroup$


















                                  0












                                  $begingroup$

                                  Just as Will Jagy did, let$u = frac{x^2}{4 v t}$ to make the equation $e^u - 1 - 2u = 0$.



                                  So, let consider that you look for the zero's of function
                                  $$f(u)=e^u - 1 - 2u $$ for which
                                  $$f'(u)=e^u - 2 qquad text{and} qquad f''(u)=e^u > ,,forall u$$



                                  The first derivative cancels when $u=log(2)$. You can get an estimate of the root builiding the Taylor series at this point. This would give
                                  $$e^u - 1 - 2u =(1-2 log (2))+(u-log (2))^2+Oleft((u-log (2))^3right)$$ Ignoring the gigher order terms, you then have as an estimate
                                  $$u_0=log(2)+sqrt{2 log (2)-1}approx 1.31467$$ With this estimate, you can now use Newton method
                                  $$u_{n+1}=u_n-frac{f(u_n)}{f'(u_n)}=frac{e^{u_n} (u_n-1)+1}{e^{u_n}-2}$$ and get, for twelve significant figures, the following iterates
                                  $$left(
                                  begin{array}{cc}
                                  n & u_n \
                                  0 & 1.31467301359 \
                                  1 & 1.26002526328 \
                                  2 & 1.25644611685 \
                                  3 & 1.25643120888 \
                                  4 & 1.25643120863
                                  end{array}
                                  right)$$



                                  One of the important points when you use Newton method is to get a "reasonable" estimates.



                                  You will notice in the table that, at no time, we overshoot the solution because we started at a point whe $f(x_0) times f''(x_0) > 0$ (Darboux theorem).



                                  Starting instead with $x_0=1$, the iterates would have been
                                  $$left(
                                  begin{array}{cc}
                                  n & u_n \
                                  0 & 1.00000000000 \
                                  1 & color{red}{1.39221119118} \
                                  2 & 1.27395717022 \
                                  3 & 1.25677778598 \
                                  4 & 1.25643134800 \
                                  5 & 1.25643120863
                                  end{array}
                                  right)$$






                                  share|cite|improve this answer









                                  $endgroup$
















                                    0












                                    0








                                    0





                                    $begingroup$

                                    Just as Will Jagy did, let$u = frac{x^2}{4 v t}$ to make the equation $e^u - 1 - 2u = 0$.



                                    So, let consider that you look for the zero's of function
                                    $$f(u)=e^u - 1 - 2u $$ for which
                                    $$f'(u)=e^u - 2 qquad text{and} qquad f''(u)=e^u > ,,forall u$$



                                    The first derivative cancels when $u=log(2)$. You can get an estimate of the root builiding the Taylor series at this point. This would give
                                    $$e^u - 1 - 2u =(1-2 log (2))+(u-log (2))^2+Oleft((u-log (2))^3right)$$ Ignoring the gigher order terms, you then have as an estimate
                                    $$u_0=log(2)+sqrt{2 log (2)-1}approx 1.31467$$ With this estimate, you can now use Newton method
                                    $$u_{n+1}=u_n-frac{f(u_n)}{f'(u_n)}=frac{e^{u_n} (u_n-1)+1}{e^{u_n}-2}$$ and get, for twelve significant figures, the following iterates
                                    $$left(
                                    begin{array}{cc}
                                    n & u_n \
                                    0 & 1.31467301359 \
                                    1 & 1.26002526328 \
                                    2 & 1.25644611685 \
                                    3 & 1.25643120888 \
                                    4 & 1.25643120863
                                    end{array}
                                    right)$$



                                    One of the important points when you use Newton method is to get a "reasonable" estimates.



                                    You will notice in the table that, at no time, we overshoot the solution because we started at a point whe $f(x_0) times f''(x_0) > 0$ (Darboux theorem).



                                    Starting instead with $x_0=1$, the iterates would have been
                                    $$left(
                                    begin{array}{cc}
                                    n & u_n \
                                    0 & 1.00000000000 \
                                    1 & color{red}{1.39221119118} \
                                    2 & 1.27395717022 \
                                    3 & 1.25677778598 \
                                    4 & 1.25643134800 \
                                    5 & 1.25643120863
                                    end{array}
                                    right)$$






                                    share|cite|improve this answer









                                    $endgroup$



                                    Just as Will Jagy did, let$u = frac{x^2}{4 v t}$ to make the equation $e^u - 1 - 2u = 0$.



                                    So, let consider that you look for the zero's of function
                                    $$f(u)=e^u - 1 - 2u $$ for which
                                    $$f'(u)=e^u - 2 qquad text{and} qquad f''(u)=e^u > ,,forall u$$



                                    The first derivative cancels when $u=log(2)$. You can get an estimate of the root builiding the Taylor series at this point. This would give
                                    $$e^u - 1 - 2u =(1-2 log (2))+(u-log (2))^2+Oleft((u-log (2))^3right)$$ Ignoring the gigher order terms, you then have as an estimate
                                    $$u_0=log(2)+sqrt{2 log (2)-1}approx 1.31467$$ With this estimate, you can now use Newton method
                                    $$u_{n+1}=u_n-frac{f(u_n)}{f'(u_n)}=frac{e^{u_n} (u_n-1)+1}{e^{u_n}-2}$$ and get, for twelve significant figures, the following iterates
                                    $$left(
                                    begin{array}{cc}
                                    n & u_n \
                                    0 & 1.31467301359 \
                                    1 & 1.26002526328 \
                                    2 & 1.25644611685 \
                                    3 & 1.25643120888 \
                                    4 & 1.25643120863
                                    end{array}
                                    right)$$



                                    One of the important points when you use Newton method is to get a "reasonable" estimates.



                                    You will notice in the table that, at no time, we overshoot the solution because we started at a point whe $f(x_0) times f''(x_0) > 0$ (Darboux theorem).



                                    Starting instead with $x_0=1$, the iterates would have been
                                    $$left(
                                    begin{array}{cc}
                                    n & u_n \
                                    0 & 1.00000000000 \
                                    1 & color{red}{1.39221119118} \
                                    2 & 1.27395717022 \
                                    3 & 1.25677778598 \
                                    4 & 1.25643134800 \
                                    5 & 1.25643120863
                                    end{array}
                                    right)$$







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                                    answered Jan 4 at 8:48









                                    Claude LeiboviciClaude Leibovici

                                    125k1158136




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