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Fast method to solve transcendental equation for a range of parameters?


Limit solution to a transcendental equationNewton Raphson Method Overestimating ParametersNewton-Raphson 2 variable method: Finding complex solutionsNewton Raphson when we can calculate second derivativeNewton-Raphson method with $c_0=3$ to calculate $c_1$.Newton-Raphson method for a vector function with root bracketing / root constraint?Multiple roots for nonlinear algebraic equationsOrder of convergence of the Newton-Raphson methodNewton-Raphson method for complex numbersUsing Newton-Raphson method, find the solution for $e^fracx^24vt = 1+fracx^22vt$













1












$begingroup$


I have an equation of the form



$t + e^Ax + e^Bx = 0$



This is a transcendental equation, and I would use a Newton-Raphson algorithm or uniroot in R, to solve for $x$.



I am interested in finding the solution to this equation for a range of values of $t$.



Do I need to repeat the same algorithm for every value of $t$, or is there any way in which I can use the solution for a particular $t$ to calculate the solution for other values of $t$?




Edit: In my specific problem I have $A > 0$ and $B<0$.










share|cite|improve this question











$endgroup$











  • $begingroup$
    Can you say something about the range of values that can be taken by parmaeter $A$ and $B$, positive, negative, $[0,1]$, any value ?
    $endgroup$
    – Jean Marie
    Mar 13 at 22:30










  • $begingroup$
    Sure, I added the restrictions on $A$ and $B$ from my specific problem to the question.
    $endgroup$
    – robust
    Mar 14 at 0:33















1












$begingroup$


I have an equation of the form



$t + e^Ax + e^Bx = 0$



This is a transcendental equation, and I would use a Newton-Raphson algorithm or uniroot in R, to solve for $x$.



I am interested in finding the solution to this equation for a range of values of $t$.



Do I need to repeat the same algorithm for every value of $t$, or is there any way in which I can use the solution for a particular $t$ to calculate the solution for other values of $t$?




Edit: In my specific problem I have $A > 0$ and $B<0$.










share|cite|improve this question











$endgroup$











  • $begingroup$
    Can you say something about the range of values that can be taken by parmaeter $A$ and $B$, positive, negative, $[0,1]$, any value ?
    $endgroup$
    – Jean Marie
    Mar 13 at 22:30










  • $begingroup$
    Sure, I added the restrictions on $A$ and $B$ from my specific problem to the question.
    $endgroup$
    – robust
    Mar 14 at 0:33













1












1








1


1



$begingroup$


I have an equation of the form



$t + e^Ax + e^Bx = 0$



This is a transcendental equation, and I would use a Newton-Raphson algorithm or uniroot in R, to solve for $x$.



I am interested in finding the solution to this equation for a range of values of $t$.



Do I need to repeat the same algorithm for every value of $t$, or is there any way in which I can use the solution for a particular $t$ to calculate the solution for other values of $t$?




Edit: In my specific problem I have $A > 0$ and $B<0$.










share|cite|improve this question











$endgroup$




I have an equation of the form



$t + e^Ax + e^Bx = 0$



This is a transcendental equation, and I would use a Newton-Raphson algorithm or uniroot in R, to solve for $x$.



I am interested in finding the solution to this equation for a range of values of $t$.



Do I need to repeat the same algorithm for every value of $t$, or is there any way in which I can use the solution for a particular $t$ to calculate the solution for other values of $t$?




Edit: In my specific problem I have $A > 0$ and $B<0$.







numerical-methods newton-raphson transcendental-equations programming






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Mar 14 at 0:32







robust

















asked Mar 13 at 18:08









robustrobust

163




163











  • $begingroup$
    Can you say something about the range of values that can be taken by parmaeter $A$ and $B$, positive, negative, $[0,1]$, any value ?
    $endgroup$
    – Jean Marie
    Mar 13 at 22:30










  • $begingroup$
    Sure, I added the restrictions on $A$ and $B$ from my specific problem to the question.
    $endgroup$
    – robust
    Mar 14 at 0:33
















  • $begingroup$
    Can you say something about the range of values that can be taken by parmaeter $A$ and $B$, positive, negative, $[0,1]$, any value ?
    $endgroup$
    – Jean Marie
    Mar 13 at 22:30










  • $begingroup$
    Sure, I added the restrictions on $A$ and $B$ from my specific problem to the question.
    $endgroup$
    – robust
    Mar 14 at 0:33















$begingroup$
Can you say something about the range of values that can be taken by parmaeter $A$ and $B$, positive, negative, $[0,1]$, any value ?
$endgroup$
– Jean Marie
Mar 13 at 22:30




$begingroup$
Can you say something about the range of values that can be taken by parmaeter $A$ and $B$, positive, negative, $[0,1]$, any value ?
$endgroup$
– Jean Marie
Mar 13 at 22:30












$begingroup$
Sure, I added the restrictions on $A$ and $B$ from my specific problem to the question.
$endgroup$
– robust
Mar 14 at 0:33




$begingroup$
Sure, I added the restrictions on $A$ and $B$ from my specific problem to the question.
$endgroup$
– robust
Mar 14 at 0:33










1 Answer
1






active

oldest

votes


















2












$begingroup$

Here is a solution that transforms the initial problem with 2 parameters into a 1 parameter issue, easier to tackle.



Set $C=B/A$ (a negative number because $A>0$ and $B<0$).



Let us make a (bijective !) change of variable



$$u=e^Ax iff x=frac1Atextln(u)tag1$$



Important remark : $u>0$.



The initial equation



$$x+e^Ax+e^Bx=0 tag2$$



is thus transformed into



$$underbracet+u+u^C_f(u)=0tag3$$



or, geometrically, into the intersection of



$$begincasesy&=&u^C& text(curve with asymptotes) \y&=&-u-t& textline with slope -1endcasestag4$$



Fig. 1 illustrates (4).



Three cases can occur (due to the fact that the straight line hits the $y$ axis in $-t$) :



  • either zero solutions when $t<t_0$ for a certain $t_0$.


  • (exceptionally) one solution when $t=t_0$,


  • 2 solutions when $t=t_0$,



and in this case which one do you desire ?



I have assumed here that we are looking for the root with the highest magnitude.




The choice of the initial value is directed by the diagram : if we take the initial value $u=-t$, for big values of $-t$, we will be already rather close to the final value.



The following Matlab program applying Newton iteration is very efficient :



A=1.;B=-2.5;C=B/A;
t=-3;
u=-t, % initialization
for k=1:4, % 4 iterations are usually enough
u=u-(t+u+u^C)/(1+C*u^(C-1)), % Newton iteration step u = u-f(u)/f'(u)
end;
x=(1/A)*log(u), % solution x=x(t)
t+exp(A*x)+exp(B*x), % checking that this value is (almost !) zero


enter image description here



Fig. 1 : (case $A=1, B=-2.5, C=B/A=-2.5$. Horizontal axis is for $u$ variable). Curve with equation $y=u^C$ and two straight lines with resp. equations $y=-u-t_0$ and $y=-u-t$ for two different values of $t$ ($t=-3$ and $t=t_0 approx 1.818$); Newton recurrence relationship with initial value $u_0=3$ gives final value $u_5=2.9321...$ out of which $x=1.0757...$.






share|cite|improve this answer











$endgroup$












  • $begingroup$
    I have completely re-written my text that is now a full-fleshed solution. It can be improved, of course. Can you tell me if it agrees you ?
    $endgroup$
    – Jean Marie
    Mar 15 at 4:40






  • 1




    $begingroup$
    +1. I appreciate the reduction to a one parameter problem. It occurred to me that usual error formula for Newton's method $e_n+1^2 = - frac12 fracf''(xi_n)f'(x_n)$ allows us to approximate the distribution of the initial guesses needed to quickly solve any for any $t$ in a bounded interval. This suggests to me a whole set of nice practice problems of the non-linear-solve, table-look-up, interpolation type.
    $endgroup$
    – Carl Christian
    Mar 15 at 9:44










  • $begingroup$
    @Carl Christian Yes, this issue could be extended as a nice student's project for example for a recap of a certain number of Numerical Analysis methods...
    $endgroup$
    – Jean Marie
    Mar 15 at 10:33










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1 Answer
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active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









2












$begingroup$

Here is a solution that transforms the initial problem with 2 parameters into a 1 parameter issue, easier to tackle.



Set $C=B/A$ (a negative number because $A>0$ and $B<0$).



Let us make a (bijective !) change of variable



$$u=e^Ax iff x=frac1Atextln(u)tag1$$



Important remark : $u>0$.



The initial equation



$$x+e^Ax+e^Bx=0 tag2$$



is thus transformed into



$$underbracet+u+u^C_f(u)=0tag3$$



or, geometrically, into the intersection of



$$begincasesy&=&u^C& text(curve with asymptotes) \y&=&-u-t& textline with slope -1endcasestag4$$



Fig. 1 illustrates (4).



Three cases can occur (due to the fact that the straight line hits the $y$ axis in $-t$) :



  • either zero solutions when $t<t_0$ for a certain $t_0$.


  • (exceptionally) one solution when $t=t_0$,


  • 2 solutions when $t=t_0$,



and in this case which one do you desire ?



I have assumed here that we are looking for the root with the highest magnitude.




The choice of the initial value is directed by the diagram : if we take the initial value $u=-t$, for big values of $-t$, we will be already rather close to the final value.



The following Matlab program applying Newton iteration is very efficient :



A=1.;B=-2.5;C=B/A;
t=-3;
u=-t, % initialization
for k=1:4, % 4 iterations are usually enough
u=u-(t+u+u^C)/(1+C*u^(C-1)), % Newton iteration step u = u-f(u)/f'(u)
end;
x=(1/A)*log(u), % solution x=x(t)
t+exp(A*x)+exp(B*x), % checking that this value is (almost !) zero


enter image description here



Fig. 1 : (case $A=1, B=-2.5, C=B/A=-2.5$. Horizontal axis is for $u$ variable). Curve with equation $y=u^C$ and two straight lines with resp. equations $y=-u-t_0$ and $y=-u-t$ for two different values of $t$ ($t=-3$ and $t=t_0 approx 1.818$); Newton recurrence relationship with initial value $u_0=3$ gives final value $u_5=2.9321...$ out of which $x=1.0757...$.






share|cite|improve this answer











$endgroup$












  • $begingroup$
    I have completely re-written my text that is now a full-fleshed solution. It can be improved, of course. Can you tell me if it agrees you ?
    $endgroup$
    – Jean Marie
    Mar 15 at 4:40






  • 1




    $begingroup$
    +1. I appreciate the reduction to a one parameter problem. It occurred to me that usual error formula for Newton's method $e_n+1^2 = - frac12 fracf''(xi_n)f'(x_n)$ allows us to approximate the distribution of the initial guesses needed to quickly solve any for any $t$ in a bounded interval. This suggests to me a whole set of nice practice problems of the non-linear-solve, table-look-up, interpolation type.
    $endgroup$
    – Carl Christian
    Mar 15 at 9:44










  • $begingroup$
    @Carl Christian Yes, this issue could be extended as a nice student's project for example for a recap of a certain number of Numerical Analysis methods...
    $endgroup$
    – Jean Marie
    Mar 15 at 10:33















2












$begingroup$

Here is a solution that transforms the initial problem with 2 parameters into a 1 parameter issue, easier to tackle.



Set $C=B/A$ (a negative number because $A>0$ and $B<0$).



Let us make a (bijective !) change of variable



$$u=e^Ax iff x=frac1Atextln(u)tag1$$



Important remark : $u>0$.



The initial equation



$$x+e^Ax+e^Bx=0 tag2$$



is thus transformed into



$$underbracet+u+u^C_f(u)=0tag3$$



or, geometrically, into the intersection of



$$begincasesy&=&u^C& text(curve with asymptotes) \y&=&-u-t& textline with slope -1endcasestag4$$



Fig. 1 illustrates (4).



Three cases can occur (due to the fact that the straight line hits the $y$ axis in $-t$) :



  • either zero solutions when $t<t_0$ for a certain $t_0$.


  • (exceptionally) one solution when $t=t_0$,


  • 2 solutions when $t=t_0$,



and in this case which one do you desire ?



I have assumed here that we are looking for the root with the highest magnitude.




The choice of the initial value is directed by the diagram : if we take the initial value $u=-t$, for big values of $-t$, we will be already rather close to the final value.



The following Matlab program applying Newton iteration is very efficient :



A=1.;B=-2.5;C=B/A;
t=-3;
u=-t, % initialization
for k=1:4, % 4 iterations are usually enough
u=u-(t+u+u^C)/(1+C*u^(C-1)), % Newton iteration step u = u-f(u)/f'(u)
end;
x=(1/A)*log(u), % solution x=x(t)
t+exp(A*x)+exp(B*x), % checking that this value is (almost !) zero


enter image description here



Fig. 1 : (case $A=1, B=-2.5, C=B/A=-2.5$. Horizontal axis is for $u$ variable). Curve with equation $y=u^C$ and two straight lines with resp. equations $y=-u-t_0$ and $y=-u-t$ for two different values of $t$ ($t=-3$ and $t=t_0 approx 1.818$); Newton recurrence relationship with initial value $u_0=3$ gives final value $u_5=2.9321...$ out of which $x=1.0757...$.






share|cite|improve this answer











$endgroup$












  • $begingroup$
    I have completely re-written my text that is now a full-fleshed solution. It can be improved, of course. Can you tell me if it agrees you ?
    $endgroup$
    – Jean Marie
    Mar 15 at 4:40






  • 1




    $begingroup$
    +1. I appreciate the reduction to a one parameter problem. It occurred to me that usual error formula for Newton's method $e_n+1^2 = - frac12 fracf''(xi_n)f'(x_n)$ allows us to approximate the distribution of the initial guesses needed to quickly solve any for any $t$ in a bounded interval. This suggests to me a whole set of nice practice problems of the non-linear-solve, table-look-up, interpolation type.
    $endgroup$
    – Carl Christian
    Mar 15 at 9:44










  • $begingroup$
    @Carl Christian Yes, this issue could be extended as a nice student's project for example for a recap of a certain number of Numerical Analysis methods...
    $endgroup$
    – Jean Marie
    Mar 15 at 10:33













2












2








2





$begingroup$

Here is a solution that transforms the initial problem with 2 parameters into a 1 parameter issue, easier to tackle.



Set $C=B/A$ (a negative number because $A>0$ and $B<0$).



Let us make a (bijective !) change of variable



$$u=e^Ax iff x=frac1Atextln(u)tag1$$



Important remark : $u>0$.



The initial equation



$$x+e^Ax+e^Bx=0 tag2$$



is thus transformed into



$$underbracet+u+u^C_f(u)=0tag3$$



or, geometrically, into the intersection of



$$begincasesy&=&u^C& text(curve with asymptotes) \y&=&-u-t& textline with slope -1endcasestag4$$



Fig. 1 illustrates (4).



Three cases can occur (due to the fact that the straight line hits the $y$ axis in $-t$) :



  • either zero solutions when $t<t_0$ for a certain $t_0$.


  • (exceptionally) one solution when $t=t_0$,


  • 2 solutions when $t=t_0$,



and in this case which one do you desire ?



I have assumed here that we are looking for the root with the highest magnitude.




The choice of the initial value is directed by the diagram : if we take the initial value $u=-t$, for big values of $-t$, we will be already rather close to the final value.



The following Matlab program applying Newton iteration is very efficient :



A=1.;B=-2.5;C=B/A;
t=-3;
u=-t, % initialization
for k=1:4, % 4 iterations are usually enough
u=u-(t+u+u^C)/(1+C*u^(C-1)), % Newton iteration step u = u-f(u)/f'(u)
end;
x=(1/A)*log(u), % solution x=x(t)
t+exp(A*x)+exp(B*x), % checking that this value is (almost !) zero


enter image description here



Fig. 1 : (case $A=1, B=-2.5, C=B/A=-2.5$. Horizontal axis is for $u$ variable). Curve with equation $y=u^C$ and two straight lines with resp. equations $y=-u-t_0$ and $y=-u-t$ for two different values of $t$ ($t=-3$ and $t=t_0 approx 1.818$); Newton recurrence relationship with initial value $u_0=3$ gives final value $u_5=2.9321...$ out of which $x=1.0757...$.






share|cite|improve this answer











$endgroup$



Here is a solution that transforms the initial problem with 2 parameters into a 1 parameter issue, easier to tackle.



Set $C=B/A$ (a negative number because $A>0$ and $B<0$).



Let us make a (bijective !) change of variable



$$u=e^Ax iff x=frac1Atextln(u)tag1$$



Important remark : $u>0$.



The initial equation



$$x+e^Ax+e^Bx=0 tag2$$



is thus transformed into



$$underbracet+u+u^C_f(u)=0tag3$$



or, geometrically, into the intersection of



$$begincasesy&=&u^C& text(curve with asymptotes) \y&=&-u-t& textline with slope -1endcasestag4$$



Fig. 1 illustrates (4).



Three cases can occur (due to the fact that the straight line hits the $y$ axis in $-t$) :



  • either zero solutions when $t<t_0$ for a certain $t_0$.


  • (exceptionally) one solution when $t=t_0$,


  • 2 solutions when $t=t_0$,



and in this case which one do you desire ?



I have assumed here that we are looking for the root with the highest magnitude.




The choice of the initial value is directed by the diagram : if we take the initial value $u=-t$, for big values of $-t$, we will be already rather close to the final value.



The following Matlab program applying Newton iteration is very efficient :



A=1.;B=-2.5;C=B/A;
t=-3;
u=-t, % initialization
for k=1:4, % 4 iterations are usually enough
u=u-(t+u+u^C)/(1+C*u^(C-1)), % Newton iteration step u = u-f(u)/f'(u)
end;
x=(1/A)*log(u), % solution x=x(t)
t+exp(A*x)+exp(B*x), % checking that this value is (almost !) zero


enter image description here



Fig. 1 : (case $A=1, B=-2.5, C=B/A=-2.5$. Horizontal axis is for $u$ variable). Curve with equation $y=u^C$ and two straight lines with resp. equations $y=-u-t_0$ and $y=-u-t$ for two different values of $t$ ($t=-3$ and $t=t_0 approx 1.818$); Newton recurrence relationship with initial value $u_0=3$ gives final value $u_5=2.9321...$ out of which $x=1.0757...$.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Mar 15 at 10:29

























answered Mar 14 at 1:14









Jean MarieJean Marie

30.8k42154




30.8k42154











  • $begingroup$
    I have completely re-written my text that is now a full-fleshed solution. It can be improved, of course. Can you tell me if it agrees you ?
    $endgroup$
    – Jean Marie
    Mar 15 at 4:40






  • 1




    $begingroup$
    +1. I appreciate the reduction to a one parameter problem. It occurred to me that usual error formula for Newton's method $e_n+1^2 = - frac12 fracf''(xi_n)f'(x_n)$ allows us to approximate the distribution of the initial guesses needed to quickly solve any for any $t$ in a bounded interval. This suggests to me a whole set of nice practice problems of the non-linear-solve, table-look-up, interpolation type.
    $endgroup$
    – Carl Christian
    Mar 15 at 9:44










  • $begingroup$
    @Carl Christian Yes, this issue could be extended as a nice student's project for example for a recap of a certain number of Numerical Analysis methods...
    $endgroup$
    – Jean Marie
    Mar 15 at 10:33
















  • $begingroup$
    I have completely re-written my text that is now a full-fleshed solution. It can be improved, of course. Can you tell me if it agrees you ?
    $endgroup$
    – Jean Marie
    Mar 15 at 4:40






  • 1




    $begingroup$
    +1. I appreciate the reduction to a one parameter problem. It occurred to me that usual error formula for Newton's method $e_n+1^2 = - frac12 fracf''(xi_n)f'(x_n)$ allows us to approximate the distribution of the initial guesses needed to quickly solve any for any $t$ in a bounded interval. This suggests to me a whole set of nice practice problems of the non-linear-solve, table-look-up, interpolation type.
    $endgroup$
    – Carl Christian
    Mar 15 at 9:44










  • $begingroup$
    @Carl Christian Yes, this issue could be extended as a nice student's project for example for a recap of a certain number of Numerical Analysis methods...
    $endgroup$
    – Jean Marie
    Mar 15 at 10:33















$begingroup$
I have completely re-written my text that is now a full-fleshed solution. It can be improved, of course. Can you tell me if it agrees you ?
$endgroup$
– Jean Marie
Mar 15 at 4:40




$begingroup$
I have completely re-written my text that is now a full-fleshed solution. It can be improved, of course. Can you tell me if it agrees you ?
$endgroup$
– Jean Marie
Mar 15 at 4:40




1




1




$begingroup$
+1. I appreciate the reduction to a one parameter problem. It occurred to me that usual error formula for Newton's method $e_n+1^2 = - frac12 fracf''(xi_n)f'(x_n)$ allows us to approximate the distribution of the initial guesses needed to quickly solve any for any $t$ in a bounded interval. This suggests to me a whole set of nice practice problems of the non-linear-solve, table-look-up, interpolation type.
$endgroup$
– Carl Christian
Mar 15 at 9:44




$begingroup$
+1. I appreciate the reduction to a one parameter problem. It occurred to me that usual error formula for Newton's method $e_n+1^2 = - frac12 fracf''(xi_n)f'(x_n)$ allows us to approximate the distribution of the initial guesses needed to quickly solve any for any $t$ in a bounded interval. This suggests to me a whole set of nice practice problems of the non-linear-solve, table-look-up, interpolation type.
$endgroup$
– Carl Christian
Mar 15 at 9:44












$begingroup$
@Carl Christian Yes, this issue could be extended as a nice student's project for example for a recap of a certain number of Numerical Analysis methods...
$endgroup$
– Jean Marie
Mar 15 at 10:33




$begingroup$
@Carl Christian Yes, this issue could be extended as a nice student's project for example for a recap of a certain number of Numerical Analysis methods...
$endgroup$
– Jean Marie
Mar 15 at 10:33

















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