Maple Finding the solution of this long equation using Maple

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SUMMARY

The discussion focuses on solving a complex equation using Maple, specifically for very small values of x (##x \ll 1##). Users suggest employing the 'mtaylor' command in Maple to perform a Taylor expansion, simplifying the equation to a polynomial form for easier resolution. The equation involves constants such as 434972871000000000.0 and 1.488388992×10^-36, and participants identify potential roots for x within specific ranges. The conversation emphasizes the importance of careful formatting and verification of the equation.

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jamesbrazil
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TL;DR
Command to solve a strange equation in Maple for ##x \ll 1##
I need help solving an equation. I started using Maple, but had no success. Could someone explain to me which command to use? I need to find a very small value of ##x##, that is, ##x \ll 1##. The equation is

$$434972871000000000.0+{\frac {\sqrt {6} \left( { 1.488388992\times 10^
{-36}}\,\ln \left( 1/12\,{\frac {12\,\sqrt {6}{x}^{2}+{
1.488388992\times 10^{-36}}\,\sqrt {6}+12\,\sqrt {6\,{x}^{2}+{
1.488388992\times 10^{-36}}}x}{x}} \right) \sqrt {6}-12\,
\sqrt {6\,{x}^{2}+{ 1.488388992\times 10^{-36}}}x \right) }{
72\,{x}^{3}}}-{\frac { \left( 0.001704000000\,{x}^{2}+{
1.488388992\times 10^{-36}} \right) \left( { 1.488388992\times 10^{
-36}}\,\ln \left( 1/12\,{\frac { 0.003408000000\,\sqrt {6}{x}^{2
}+{ 1.488388992\times 10^{-36}}\,\sqrt {6}+12\,\sqrt {
0.0000004839360000\,{x}^{2}+{ 4.227024737\times 10^{-40}}}x
}{x}} \right) \sqrt {6}-12\,\sqrt { 0.0000004839360000\,{x}^
{2}+{ 4.227024737\times 10^{-40}}}x \right) \sqrt {6}}{72\,\sqrt
{ 6.000000001\,{x}^{2}+{ 5.240806311\times 10^{-33}}}\sqrt {
0.0000004839360000\,{x}^{2}+{ 4.227024737\times 10^{-40}}}{
x}^{3}}}=0$$
 
Last edited by a moderator:
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jamesbrazil said:
TL;DR Summary: Command to solve a strange equation in Maple for ##x \ll 1##

I need help solving an equation. I started using Maple, but had no success. Could someone explain to me which command to use? I need to find a very small value of ##x##, that is, ##x \ll 1##. The equation is

$$434972871000000000.0+{\frac {\sqrt {6} \left( { 1.488388992\times 10^
{-36}}\,\ln \left( 1/12\,{\frac {12\,\sqrt {6}{x}^{2}+{
1.488388992\times 10^{-36}}\,\sqrt {6}+12\,\sqrt {6\,{x}^{2}+{
1.488388992\times 10^{-36}}}x}{x}} \right) \sqrt {6}-12\,
\sqrt {6\,{x}^{2}+{ 1.488388992\times 10^{-36}}}x \right) }{
72\,{x}^{3}}}-{\frac { \left( 0.001704000000\,{x}^{2}+{
1.488388992\times 10^{-36}} \right) \left( { 1.488388992\times 10^{
-36}}\,\ln \left( 1/12\,{\frac { 0.003408000000\,\sqrt {6}{x}^{2
}+{ 1.488388992\times 10^{-36}}\,\sqrt {6}+12\,\sqrt {
0.0000004839360000\,{x}^{2}+{ 4.227024737\times 10^{-40}}}x
}{x}} \right) \sqrt {6}-12\,\sqrt { 0.0000004839360000\,{x}^
{2}+{ 4.227024737\times 10^{-40}}}x \right) \sqrt {6}}{72\,\sqrt
{ 6.000000001\,{x}^{2}+{ 5.240806311\times 10^{-33}}}\sqrt {
0.0000004839360000\,{x}^{2}+{ 4.227024737\times 10^{-40}}}{
x}^{3}}}=0$$
Can you repost this equation, using symbols for all of the lengthy decimal numbers, followed with a separate list of all the symbols and their respective decimal values? That would make everything much easier to read and work with!
 
If anyone might still be interested in this then I've tried to reformat it as requested by jamesbrazil

a=434972871000000000.0;
b=1.488388992*10^-36;
c=0.001704000000;
d=0.003408000000;
e=0.0000004839360000;
f=4.227024737*10^-40;
g=5.240806311*10^-33;
a+
(sqrt(6)*(b*ln(1/12*
12*sqrt(6)*x^2+b*sqrt(6)+
12*sqrt(6*x^2+b)*x)/x))*sqrt(6)-
12*sqrt(6*x^2+b)*x)/(72*x^3)-
((c*x^2+b)*
(b*ln(1/12*(d*sqrt(6)*x^2+b*sqrt(6)+
12*sqrt(e*x^2+f)*x)/x)*sqrt(6)-
12*sqrt(e*x^2+f)*x)*sqrt(6))/
(72*sqrt(6*x^2+g)*
sqrt(e*x^2+f)*x^3)

I believe there is a root for x<<1 that lies between
8.071385266*10^-19 and 8.071385267*10^-19

It is certainly possible that I've made a mistake when trying to reformat this.
And I suppose it might be possible that the OP made a mistake when trying to format this.

Please check this carefully to try to find any of my mistakes before depending on this.

When I look at a plot of this it seems that the expression is approximately 4.34973*10^17
for a range of modest positive x.
It also seems that it is approximately that same value for many small negative x values,
but it seems there may be a number of points which are indeterminant.
 
Last edited:
jamesbrazil said:
TL;DR Summary: Command to solve a strange equation in Maple for ##x \ll 1##

I need help solving an equation. I started using Maple, but had no success. Could someone explain to me which command to use? I need to find a very small value of ##x##, that is, ##x \ll 1##. The equation is

$$434972871000000000.0+{\frac {\sqrt {6} \left( { 1.488388992\times 10^
{-36}}\,\ln \left( 1/12\,{\frac {12\,\sqrt {6}{x}^{2}+{
1.488388992\times 10^{-36}}\,\sqrt {6}+12\,\sqrt {6\,{x}^{2}+{
1.488388992\times 10^{-36}}}x}{x}} \right) \sqrt {6}-12\,
\sqrt {6\,{x}^{2}+{ 1.488388992\times 10^{-36}}}x \right) }{
72\,{x}^{3}}}-{\frac { \left( 0.001704000000\,{x}^{2}+{
1.488388992\times 10^{-36}} \right) \left( { 1.488388992\times 10^{
-36}}\,\ln \left( 1/12\,{\frac { 0.003408000000\,\sqrt {6}{x}^{2
}+{ 1.488388992\times 10^{-36}}\,\sqrt {6}+12\,\sqrt {
0.0000004839360000\,{x}^{2}+{ 4.227024737\times 10^{-40}}}x
}{x}} \right) \sqrt {6}-12\,\sqrt { 0.0000004839360000\,{x}^
{2}+{ 4.227024737\times 10^{-40}}}x \right) \sqrt {6}}{72\,\sqrt
{ 6.000000001\,{x}^{2}+{ 5.240806311\times 10^{-33}}}\sqrt {
0.0000004839360000\,{x}^{2}+{ 4.227024737\times 10^{-40}}}{
x}^{3}}}=0$$
If you know that x is small, I'd suggest a Taylor expansion first (use mtaylor in Maple), then solve it. That way it's just a polynomial equation.

-Dan
 
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If anyone might still be interested in this then I've tried to reformat it as requested by jamesbrazil.

In experimenting with that I see that I dropped a couple of ( ) in my post above.

I apologize for that and I can't seem to edit that now to correct those mistakes.

This will hopefully correct my ( ) mistakes.

a=434972871000000000.0;
b=1.488388992*10^-36;
c=0.001704000000;
d=0.003408000000;
e=0.0000004839360000;
f=4.227024737*10^-40;
g=5.240806311*10^-33;
a+
(sqrt(6)*(b*ln(1/12*
(12*sqrt(6)*x^2+b*sqrt(6)+
12*sqrt(6*x^2+b)*x)/x))*sqrt(6)-
12*sqrt(6*x^2+b)*x)/(72*x^3)-
((c*x^2+b)*
(b*ln(1/12*(d*sqrt(6)*x^2+b*sqrt(6)+
12*sqrt(e*x^2+f)*x)/x)*sqrt(6)-
12*sqrt(e*x^2+f)*x)*sqrt(6))/
(72*sqrt(6*x^2+g)*
sqrt(e*x^2+f)*x^3)

Strangely enough, I now believe there may be two roots for x<<1

One may lie between 10^-37 and 10^-36 and may lie near 3.0381613366*10^-37

One may lie between 10^-28 and 10^-27 and may lie near 6.8635202356*10^-28

But with the size of the numbers involved this problem is very delicate and needs some care.

If anyone wants to try to check this then I would urge you to begin with checking that I have reformatted the original expression correctly and then begin carefully looking for the roots. I would greatly appreciate if someone would do that independently, I am always happy when someone finds and points out any of my mistakes.

Thank you and again I apologize for any and all of my errors.
 
Last edited:
Bill Simpson said:
If anyone might still be interested in this then I've tried to reformat it as requested by jamesbrazil.
Frankly, until @jamesbrazil returns to show that he's still interested (and to explain the problem background and show the derivation his equation) I'm don't think this is worth pursuing.
 

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