Items tagged with roots

Hi guys,

I would like to compute the complex roots of the following equations

u*(BesselJ(0,u)^2 + BesselJ(1,u)^2) = 2 BesselJ(0,u)*BesselJ(1,u)

The function fsolve in Maple gives only 0. I was wondering whether other complex solutions could be obtained as well.

Your help is highly appreciated.

Thanks

Federiko

I am solving "Fx=0" for geting "roots:x" using "solve(Fx,x)". Solution is in the form of

"a+sqrt(b)", "a-sqrt(b)"

Here my question is how to extract "a", "b" separately (a, b are complex and very big expressions).

 

Thank you in advance for your help.

 

 

Hi,

I have to find the root of an equation corresponding to the maximum absolute value. I am using root finding package to get all the roots. But after getting all the roots i am not able to apply abs function. Maple sheet is attached.

restart

with(plots):

with(LinearAlgebra):

with(DEtools):

with(ColorTools):

Digits := 30

30

(1)

x := proc (t) options operator, arrow; x0*exp(lambda*t) end proc:

phi := proc (t) options operator, arrow; phi0*exp(lambda*t) end proc:

eqm1 := collect(simplify(coeff(expand(diff(x(t), `$`(t, 2))+(2*0)*beta*(diff(x(t), t))+0*x(t)+n*psi*(-v*(phi(t)-phi(t-2*Pi/(n*omega0)))+x(t)-x(t-2*Pi/(n*omega0)))), exp(lambda*t))), {phi0, x0})

(-n*psi*v+n*psi*v*exp(-2*lambda*Pi/(n*omega0)))*phi0+(lambda^2+n*psi-n*psi*exp(-2*lambda*Pi/(n*omega0)))*x0

(2)

eqm2 := collect(simplify(coeff(expand(diff(phi(t), `$`(t, 2))+(2*0)*(diff(phi(t), t))+phi(t)+n*(-v*(phi(t)-phi(t-2*Pi/(n*omega0)))+x(t)-x(t-2*Pi/(n*omega0)))), exp(lambda*t))), {phi0, x0})

(-n*v+n*v*exp(-2*lambda*Pi/(n*omega0))+lambda^2+1)*phi0+(n-n*exp(-2*lambda*Pi/(n*omega0)))*x0

(3)

mode := simplify(evalc(Re(evalc(subs(lambda = I*Omega, solve(subs(x0 = m*phi0, eqm1), m)))))^2+evalc(Im(evalc(subs(lambda = I*Omega, solve(subs(x0 = m*phi0, eqm1), m)))))^2)

-2*n^2*psi^2*v^2*(-1+cos(2*Omega*Pi/(n*omega0)))/(Omega^4-2*Omega^2*n*psi+2*Omega^2*n*psi*cos(2*Omega*Pi/(n*omega0))+2*n^2*psi^2-2*n^2*psi^2*cos(2*Omega*Pi/(n*omega0)))

(4)

A, b := GenerateMatrix([eqm1, eqm2], [x0, phi0])

A, b := Matrix(2, 2, {(1, 1) = lambda^2+n*psi-n*psi*exp(-2*lambda*Pi/(n*omega0)), (1, 2) = -n*psi*v+n*psi*v*exp(-2*lambda*Pi/(n*omega0)), (2, 1) = n-n*exp(-2*lambda*Pi/(n*omega0)), (2, 2) = -n*v+n*v*exp(-2*lambda*Pi/(n*omega0))+lambda^2+1}), Vector(2, {(1) = 0, (2) = 0})

(5)

with(RootFinding):

eq := subs(n = 6, psi = 1000, omega0 = 1.15, v = 0.1e-1, Determinant(A))

6000.94*lambda^2-5999.94*exp(-.289855072463768115942028985507*lambda*Pi)*lambda^2+lambda^4+6000-6000*exp(-.289855072463768115942028985507*lambda*Pi)

(6)

zeros := RootFinding:-Analytic(eq, lambda, re = 0 .. 400, im = -200 .. 200)

0.899769545162895563524511282265e-56, 0.813609592584011756247655681635e-1-20.6993361029378520006643410260*I, .242743338419727199544214811606-34.4961764258358768825593120288*I, .440964962950043888796944083291-100.074138054178692973033664525*I, .107710271188082726666762251538-106.954651646879437684160623413*I, 1.12290283496379505456476079030-62.0290638297730162295171014475*I, .879463466045683309032252293625-93.2168861049771086211729407830*I, 2.54860869821265794971735119535-80.1919866273564551209847942490*I, 1.52678990439144770439544731898-86.4450560720567958301493690195*I, 2.62945288424037545703549470125-75.0161229879790946191171617450*I, 1.68779005203728587549371003511-68.8012471850312399391042105550*I, .776570081405504740452992339900-55.1681878011205261920670466495*I, 0.851171007270465178285429398270e-9+1.00000500045406723708450960132*I, 0.851171007270465178285445699470e-9-1.00000500045406723708450960133*I, 0.874874719902730972066854301075e-2-6.89997772561385443312823760560*I, 0.354201863215292148351069041542e-1-13.7998152076043523748759861636*I, .369195444156713173497807954493-41.3921704506707022569621870947*I, .540047057129385026999638567235-48.2843908783769449582520027744*I, .149078330738225743331408017894-27.5982749361891156626731068484*I, .369195444156713173497807954500+41.3921704506707022569621870948*I, .440964962950043888796944083291+100.074138054178692973033664525*I, .107710271188082726666762251538+106.954651646879437684160623413*I, 1.12290283496379505456476079030+62.0290638297730162295171014475*I, .879463466045683309032252293625+93.2168861049771086211729407830*I, 2.54860869821265794971735119535+80.1919866273564551209847942490*I, 1.52678990439144770439544731898+86.4450560720567958301493690195*I, 2.62945288424037545703549470125+75.0161229879790946191171617450*I, 1.68779005203728587549371003511+68.8012471850312399391042105550*I, .776570081405504740452992339900+55.1681878011205261920670466495*I, 0.813609592584011756247655681660e-1+20.6993361029378520006643410260*I, 0.354201863215292148351069041261e-1+13.7998152076043523748759861634*I, 0.874874719902730972066854301075e-2+6.89997772561385443312823760560*I, .540047057129385026999638567235+48.2843908783769449582520027744*I, .242743338419727199544214811602+34.4961764258358768825593120288*I, .149078330738225743331408017894+27.5982749361891156626731068484*I

(7)

"zeros.select(int 1)"

Error, missing operation

"zeros.select(int 1)"

 

``


Download question.mw

I will be really thankful for the help.

Regards

Sunit

I am having 26th degree polynomial univariate equation , I used Isolate to get the roots. but I am getting some extra roots which are not true they I even tried to substitute those roots in original equation then I got non zero answer instead of getting nearly zero answer.How is it possible??

 

equation looks like:

-12116320194738194778134937600000000*t^26+167589596741213731838990745600000000*t^24+1058345691529498270472972795904000000*t^22-4276605572538658673086219419648000000*t^20-23240154739806540070988490473472000000*t^18-5442849111209103187871341215744000000*t^16+49009931453396028716875310432256000000*t^14+74247033158233643322704589225984000000*t^12-2762178990802317464801412907008000000*t^10-25947900993773120244883450232832000000*t^8-7468990043547273070742668836864000000*t^6-567730116675454293925108383744000000*t^4+3703566799705707258760396800000000*t^2-4742330812072533924249600000000

Solutions i got:

[t = -4.162501845, t = -2.295186769, t = -1.300314688, t = -.8048430445, t = -0.6596008501e-1, t = -0.4212510777e-1, t = 0.4212510777e-1, t = 0.6596008501e-1, t = .8048430445, t = 1.300314688, t = 2.295186769, t = 4.162501845]

t=4.162501845 give me non zero answer when I substitute it in the equation given above:

I got this answer: 4.750212083*10^39

 

Hi there,

            Recently, I encountered a problem. I have a function( omega as its variable)  (18)

 gamma*sqrt(4)*sqrt(omega^2*C2^2*R4^2/(C2^4*R4^4*beta^2*gamma^2*omega^4+C2^2*(1+gamma^2*(beta+1)^2-2*gamma)*R4^2*omega^2+1))

I tried to find a point where its first derivative equals 0. In this case, Maple returned four solutions. In my

question, both beta, gamma, R4 and C2 >0, I want it to return a real positive solution, the first term

in (19) (i.e. 1/(sqrt(beta *gamma) *1/R4 C2).

 

I know it is easy to find out the positive real roots in this case. This question seems to make no sense.

However, sometime I came across an expression complicated enough that I cannot tell whether it is real

positive.

Is there a approach to find a real positive solution of an symbolic eqution?

Thanks in advance!

 

                                                                     A University student in BeiHang University, Beijing

 

 

how i can calculate roots of the characteristic polynomial equations {dsys and dsys2}
and dsolve them with arbitrary initial condition for differennt amont of m and n?
thanks
Kr.mw

restart; a := 1; b := 2; Number := 10; q := 1; omega := 0.2e-1
``

Q1 := besselj(0, xi*b)*(eval(diff(bessely(0, xi*r), r), r = a))-(eval(diff(besselj(0, xi*r), r), r = a))*bessely(0, xi*b):

J := 0:

m := 0:

U1 := (int(r*K1[m]*(diff(K_01[m], r)+K_01[m]/r), r = a .. b))/(int(r*K1[m]^2, r = a .. b)); -1; U2 := -(int(r*K_01[m]*(diff(K1[m], r)), r = a .. b))/(int(r*K_01[m]^2, r = a .. b)); -1; U3 := (int(r^2*omega^2*K_01[m], r = a .. b))/(int(r*K_01[m]^2, r = a .. b))

0.6222222222e-3/K_01[12]

(1)

Q2 := besselj(1, eta*b)*(eval(diff(bessely(1, eta*r), r), r = a))-(eval(diff(besselj(1, eta*r), r), r = a))*bessely(1, eta*b):

E2 := unapply(Q2, eta):

m := 0:

 
dsys := {diff(S_mn(t), t, t, t)+xi[m]^2*(diff(S_mn(t), t, t))+(-U1*U2+eta__n^2)*(diff(S_mn(t), t))+xi[m]^2*eta__n^2*S_mn(t) = -(2*U2*b_m/(Pi*xi[m])*(-besselj(0, xi[m]*b)/besselj(1, xi[m]*a)))*q+xi[m]^2*U3}; 1; dsolve(dsys)

{S_mn(t) = (3111111111/5000000000000)/(K_01[12]*eta__n^2)+_C1*cos(eta__n*t)+_C2*sin(eta__n*t)+_C3*exp(-xi[12]^2*t)}

(2)

dsys2 := {diff(Q_mn(t), t, t, t)+xi[m]^2*(diff(Q_mn(t), t, t))+(-U1*U2+eta__n^2)*(diff(Q_mn(t), t))+xi[m]^2*eta__n^2*Q_mn(t) = -2*besselj(0, xi[m]*b)*U1*U2*b_m*(1-exp(-xi[m]^2*t))/(besselj(1, xi[m]*a)*Pi*xi[m]^3)}; 1; dsolve(dsys2)

{Q_mn(t) = _C1*exp(-xi[12]^2*t)+_C2*sin(eta__n*t)+_C3*cos(eta__n*t)}

(3)

``

 

``



Download Kr.mw

 

Dear all,

I have a question: how to compute the roots of exp(z) = -1 with z in C? 

I tried: 

fsolve( exp(z) = -1, z, complex );

But it only gives one root (0.1671148658e-3+4.934802220*10^9*I) which does not even seem to be correct. I would prefere smth like z_n = I*(2*n-1)*pi or at least multiple roots...

By using

solve(exp(x) = -1, x);

it returns I*Pi.

 

MATLAB MuPAD gives the desired result:


solve(exp(x) = -1, x)

(PI*I + 2*PI*k*I, k in Z)

 

 

Thanks!

when the kummerM function equal to 0?????

or

-(((beta*eta^2-(1/2)*beta+1)*p^2-(1/2)*beta^3)*KummerM((1/4)*((-beta+2)*p^2-beta^3)/p^2, 1, beta*eta^2)+(1/2)*KummerM((1/4)*((-beta+6)*p^2-beta^3)/p^2, 1, beta*eta^2)*((beta-2)*p^2+beta^3))*exp(-(1/2)*beta*eta^2)/(p^2*beta)=0

how solve this equation unitage beta?

What is the easiest way to ask roots of a polynomial on a finite field. For example asking roots of x^2+xy+y on GF(8)? I was thinking to run a two for on members of GF(8) and ask to check it but I couldn't do it using Galois package or maybe I couldn't use that package. Thanks for any help.

Hello,

I would like to plot a vertical line each time my function is null and is increasing.

Here an extract of my code:

v:=unapply(H*sin(w*t),t);
L:=0.080;
H:=0.020;
Vf:=0.3;
w:=10;
fsolve(v(t)=0,t=0.5..2);
zeros:=Roots(v(t),t=0.5..2);

vdot:=unapply(diff(v(t),t),t);
map(vdot,zeros);
LignesVerticales=implicitplot(x=To complete, colour=yellow,linestyle=3, thickness=2):

With Roots function, I could obtain the zero-crossings.

With the derative of the functions, I could obtain a vector which gives me when the derivative is positive or not.

I would like now to obtain the list of the zero-crossing values which are increasing but I have difficulties to obtain it. 

By list manipulation, it should be easy to do. 

May you help to plot obtain this list so that I can plot the desired vertical lines?

Thank you for your help

I'm trying to create a routine to perform the test of rational roots , but I'm having some problems. Below is the routine I created :

But the program is only printing " aux = -24 " . I don't know what it can be .

I need to modify my code , but I don't know where. Can someone help me? Thank you!

hw2_unfinished.mw

There is something wroung with the t0.

How to correct it?

can anyone please help me to find roots of (2*cos(0.5*x)*sin(0.5*x)*cos(3.775*x)+2.2075*((cos(0.5*x))^2)*sin(3.775*x)-0.453*((sin(0.5*x))^2)*sin(3.775*x)=0 ?

i type on maple like this:

solve(2*cos(0.5*x)*sin(0.5*x)*cos(3.775*x)+2.2075*((cos(0.5*x))^2)*sin(3.775*x)-0.453*((sin(0.5*x))^2)*sin(3.775*x)=0);

but it said the solutions may have been lost

thanks before

Hi,

 

I am trying to find the roots of Hankel function H1(2, z)?

 

j := 1;
for i from 0 to 10 do z0 := i*step+z_min; x[j] := fsolve(f = 0, z = z0, complex) end do;


for p to 10 do for j from p+1 to 9 do if `and`(Re(x[j])-Re(x[p]) < 0.1e-4, Im(x[j])-Im(x[p]) < 0.1e-4) then for i from j to 10 do x[i] := x[i+1] end do; p := p-1; break end if end do end do;

 

the first of these two processes works fine however the second does not. The second on is to get rid of same value solutions! I am not sure if I have missed anything and also is there a way to determine a max value in the complex domain and use it in a for loop?

 

 

any help would be great 

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