Transcendental equation problem

[dji3214]

i have a transcendental equation and i have not a mathematique superieur formation ( I'm an hydraulic engeneer) and i want to resolve it but i can't so if you can help me with it !

the equation is : 2*x*n*ctg(2x)= x² - n² or (same equation) : (n*ctg(x)-x)*(n*tg(x)+x) =0

n= constante ( i have this n... in example n = 0.5 )
x_p = are the roots of equation above

 

[berkeman]

i have a transcendental equation and i have not a mathematique superieur formation ( I'm an hydraulic engeneer) and i want to resolve it but i can't so if you can help me with it !

the equation is : 2*x*n*ctg(2x)= x² - n² or (same equation) : (n*ctg(x)-x)*(n*tg(x)+x) =0

n= constante ( i have this n... in example n = 0.5 )
x_p = are the roots of equation above

Welcome to the PF.

Did you mean to put this in the schoolwork forums? If it's for your work, it could be posted in the general Math forums...

 

[SteamKing]

i have a transcendental equation and i have not a mathematique superieur formation ( I'm an hydraulic engeneer) and i want to resolve it but i can't so if you can help me with it !

the equation is : 2*x*n*ctg(2x)= x² - n² or (same equation) : (n*ctg(x)-x)*(n*tg(x)+x) =0

n= constante ( i have this n... in example n = 0.5 )
x_p = are the roots of equation above

Such equations are normally solved by iteration, also known as trial and error.

 

[Ray Vickson]

i have a transcendental equation and i have not a mathematique superieur formation ( I'm an hydraulic engeneer) and i want to resolve it but i can't so if you can help me with it !

the equation is : 2*x*n*ctg(2x)= x² - n² or (same equation) : (n*ctg(x)-x)*(n*tg(x)+x) =0

n= constante ( i have this n... in example n = 0.5 )
x_p = are the roots of equation above

First: the ##\cot(2x)## function "blows up" (goes to ##\pm \infty##) at (infinitely many) finite values of ##x##, so that will give trouble in any numerical solution scheme. It is better to eliminate that problem by re-writing the equation as
[tex] 2 n x \cos(2x) = (x^2 - n^2) \sin(2x) [/tex]
Now a plot of both sides will reveal the approximate positions of roots, and these can be refined by various methods to get accurate values. There are infinitely many positive and negative roots.

 

[dji3214]

First: the ##\cot(2x)## function "blows up" (goes to ##\pm \infty##) at (infinitely many) finite values of ##x##, so that will give trouble in any numerical solution scheme. It is better to eliminate that problem by re-writing the equation as
[tex] 2 n x \cos(2x) = (x^2 - n^2) \sin(2x) [/tex]
Now a plot of both sides will reveal the approximate positions of roots, and these can be refined by various methods to get accurate values. There are infinitely many positive and negative roots.

DEAR ,, RAY thnx for helping me,, if i understand u this infitely positive and negative roots are for this n in our example 0.5 ... AND if i tell you that x_p the roots are depending to this index p like in series Σ ...because in my formula they wrote in it X_p like this ; X is h ( h_n = are the roots of equation above) ,, the index _p is _n and n (n=0.5) in the equation is mew !

the GENERAL EQUATION is ;

where in the left side constant i have it

the right side this serie and THIS INDEX n batherd me a lot and complicate my equation because i don't know in which number of n i stop my serie and also stop the root of the transcendental equation

 

[Ray Vickson]

DEAR ,, RAY thnx for helping me,, if i understand u this infitely positive and negative roots are for this n in our example 0.5 ... AND if i tell you that x_p the roots are depending to this index p like in series Σ ...because in my formula they wrote in it X_p like this ; X is h ( h_n = are the roots of equation above) ,, the index _p is _n and n (n=0.5) in the equation is mew !
View attachment 87287

the GENERAL EQUATION is ;
View attachment 87288
where in the left side constant i have it

the right side this serie and THIS INDEX n batherd me a lot and complicate my equation because i don't know in which number of n i stop my serie and also stop the root of the transcendental equation

I cannot understand anything you have written.

 

[dji3214]

I cannot understand anything you have written.

lol ,, ok ,, my general formula is : my problem is hn roots of transcendental equation

in PHI _n i have h_n ( hn is roots of the first transcendental equation that i wrote in first time )

this is PHI_n

look at h_n ,, now h_n are roots of the transcendental equation

that u transform to 2nxcos(2x)=(x2−n2)sin(2x) i would tell you that the greek lettre eta (

) is n in first post that's all because i change variables lettre

 

[SteamKing]

lol ,, ok ,, my general formula is : my problem is hn roots of transcendental equation

in PHI _n i have h_n ( hn is roots of the first transcendental equation that i wrote in first time )

this is PHI_n

look at h_n ,, now h_n are roots of the transcendental equation View attachment 87304 that u transform to 2nxcos(2x)=(x2−n2)sin(2x) i would tell you that the greek lettre mew ( View attachment 87305 ) is n in first post that's all because i change variables lettre

η is called "eta"

μ is called "mu", or "mew", as you prefer.

 

[dji3214]

η is called "eta"

μ is called "mu", or "mew", as you prefer.

lol see i forget very think lol