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Dear all,

I have the following problem: Maple does not simplify the denominator in the following example:

which gives

16*a^8*B/((dz*L*sqrt(s)*sqrt(s+c)*sqrt(L^2*s*(s+c)*dz^2+4*a^2)+L^2*s*(s+c)*dz^2+2*a^2)^2*(-dz*L*sqrt(s)*sqrt(s+c)*sqrt(L^2*s*(s+c)*dz^2+4*a^2)+L^2*s*(s+c)*dz^2+2*a^2)^2)

However, the result should be B. If only the denomiator is expanded it works: 

gives

16*a^8

which equals the nominator except for the B...

How can I use simplify in order to yield the desired result? 

Thanks a lot!

 

 

 

 

hi.please help me for gain result with out root of form....

thanks

root_of.mw

digite := 20; -1; L := 20*R; -1; varepsilon := solve(tan(sigma*L/(2*R))+tanh(sigma*L/(2*R)), sigma)

(1/10)*RootOf(tan(_Z)+tanh(_Z))

(1)

``


Download root_of.mw

hi.how convert root of to explicit form.

w is a imaginary..

thanks

123.mw

restart; w := (1/2)*(2*d-5+I*sqrt(4*d-9))/(d-2)

(1/2)*(2*d-5+I*(4*d-9)^(1/2))/(d-2)

(1)

with(LinearAlgebra):

{Q1 = RootOf((2*I)*(4*d-9)^(1/2)*_Z*d-((2*d-5+I*(4*d-9)^(1/2))/(d-2).(d*(I*(4*d-9)^(1/2)+1)*_Z/((d-2)*b)))*b*d+2*((2*d-5+I*(4*d-9)^(1/2))/(d-2).(d*(I*(4*d-9)^(1/2)+1)*_Z/((d-2)*b)))*b-2*d*_Z), Q2 = -(1/2)*d*(I*(4*d-9)^(1/2)+1)*RootOf((2*I)*(4*d-9)^(1/2)*_Z*d-((2*d-5+I*(4*d-9)^(1/2))/(d-2).(d*(I*(4*d-9)^(1/2)+1)*_Z/((d-2)*b)))*b*d+2*((2*d-5+I*(4*d-9)^(1/2))/(d-2).(d*(I*(4*d-9)^(1/2)+1)*_Z/((d-2)*b)))*b-2*d*_Z)/((d-2)*b)}

(2)

``

 

Download 123.mw

solve.mw

Hi all, I want to determine the roots of this figure that attached here. But the function has 5 parameters so the code doesn't work for it! Help me.

I have another question: The code that attached, determine the roots on horizontal axis, how could I find the values of root on vertical axis?In this figure I want to know the value of F(0) that cut the vertical axis?

Regards

 

``

``

``

plot(KummerM(1/2-(1/4)*sqrt(2*Nu), 1, sqrt(2*Nu)), Nu = 0 .. 120)

 

``

j := 1

"for i from 0 to 100000 do  Nu[i]= i/(1000); end do;    for i from 0 to 100000 do  if (KummerM(1/(2)-1/(4)*sqrt(2*Nu[i]),1,sqrt(2*Nu[i]))=0)  x[j]=Nu[i]; j= j+1;  end do if ;  end;   I know the answers for   x    are: 3.6568,  22.3047,  56.9605,  107.6203,  174.2820,  256.9450,  355.6088,  470.2730...     I    can    solve    only    one    of    them:           "

evalf(solve(KummerM(1/2-(1/4)*sqrt(2*Nu), 1, sqrt(2*Nu)) = 0))

3.656793458

(1)

PLEASE*HELP*ME

 

Download loop.mw

To motivate some ideas in my research, I've been looking at the expected number of real roots of random polynomials (and their derivatives).  In doing so I have noticed an issue/bug with fsolve and RootFinding[Isolate].  One of the polynomials I came upon was

f(x) = -32829/50000-(9277/50000)*x-(37251/20000)*x^2-(6101/6250)*x^3-(47777/20000)*x^4+(291213/50000)*x^5.

We know that f(x) has at least 1 real root and, in fact, graphing shows that f(x) has exactly 1 real root (~1.018).  However, fsolve(f) and Isolate(f) both return no real roots.  On the other hand, Isolate(f,method=RC) correctly returns the root near 1.018.  I know that fsolve's details page says "It may not return all roots for exceptionally ill-conditioned polynomials", though this system does not seem especially ill-conditioned.  Moreover, Isolate's help page says confidently "All significant digits returned by the program are correct, and unlike purely numerical methods no roots are ever lost, although repeated roots are discarded" which is clearly not the case here.  It also seems interesting that the RealSolving package used by Isolate(f,method=RS) (default method) misses the root while the RegularChains package used by Isolate(f,method=RC) correctly finds the root.

 All-in-all, I am not sure what to make of this.  Is this an issue which has been fixed in more recent incarnations of fsolve or Isolate?  Is this a persistent problem?  Is there a theoretical reason why the root is being missed, particularly for Isolate?

Any help or insight would be greatly appreciated.

Hi all,

By solving cubic equation in maple (version 17), I got

restart

``

-0.363700352e-2*y^3-.4041941000*y^2+3.397775673*y-2.377540486 = 0

-0.363700352e-2*y^3-.4041941000*y^2+3.397775673*y-2.377540486 = 0

(1)

"(->)"

[[y = .7709248124], [y = 7.123944371], [y = -119.0286907]]

(2)

``

Now I want to find these roots through the formula.

 

I solve it generally in Maple.. 

 

``# Suppose

A*y^3+B*y^2+C*y+E = 0

A*y^3+B*y^2+C*y+E = 0

(3)

NULL

A := -0.363700352e-2:

B := -.4041941000:

C := 3.397775673:

E := -2.377540486:

``

A*y^3+B*y^2+C*y+E = 0

 

A*y^3+B*y^2+C*y+E = 0

(4)

``

y1 := (1/6)*(-108*E*A^2+36*A*B*C+12*sqrt(3)*sqrt(27*A^2*E^2-18*A*B*C*E+4*A*C^3+4*B^3*E-B^2*C^2)*A-8*B^3)^(1/3)/A-(2/3)*(3*A*C-B^2)/(A*(-108*E*A^2+36*A*B*C+12*sqrt(3)*sqrt(27*A^2*E^2-18*A*B*C*E+4*A*C^3+4*B^3*E-B^2*C^2)*A-8*B^3)^(1/3))-(1/3)*B/A

-45.82526955*(.7114884222-(0.5542993294e-1*I)*3^(1/2))^(1/3)-36.74197467/(.7114884222-(0.5542993294e-1*I)*3^(1/2))^(1/3)-37.04460717

(5)

"(=)"

-119.0286907-0.1e-8*I

(6)

y2 := y = -(1/12)*(-108*E*A^2+36*A*B*C+12*sqrt(3)*sqrt(27*A^2*E^2-18*A*B*C*E+4*A*C^3+4*B^3*E-B^2*C^2)*A-8*B^3)^(1/3)/A+(1/3)*(3*A*C-B^2)/(A*(-108*E*A^2+36*A*B*C+12*sqrt(3)*sqrt(27*A^2*E^2-18*A*B*C*E+4*A*C^3+4*B^3*E-B^2*C^2)*A-8*B^3)^(1/3))-(1/3)*B/A+(1/2*I)*sqrt(3)*((1/6)*(-108*E*A^2+36*A*B*C+12*sqrt(3)*sqrt(27*A^2*E^2-18*A*B*C*E+4*A*C^3+4*B^3*E-B^2*C^2)*A-8*B^3)^(1/3)/A+(2/3)*(3*A*C-B^2)/(A*(-108*E*A^2+36*A*B*C+12*sqrt(3)*sqrt(27*A^2*E^2-18*A*B*C*E+4*A*C^3+4*B^3*E-B^2*C^2)*A-8*B^3)^(1/3)))

y = 22.91263477*(.7114884222-(0.5542993294e-1*I)*3^(1/2))^(1/3)+18.37098733/(.7114884222-(0.5542993294e-1*I)*3^(1/2))^(1/3)-37.04460717+((1/2)*I)*3^(1/2)*(-45.82526955*(.7114884222-(0.5542993294e-1*I)*3^(1/2))^(1/3)+36.74197467/(.7114884222-(0.5542993294e-1*I)*3^(1/2))^(1/3))

(7)

"(=)"

y = .770924807+0.1772050808e-7*I

(8)

y3 := y = -(1/12)*(-108*E*A^2+36*A*B*C+12*sqrt(3)*sqrt(27*A^2*E^2-18*A*B*C*E+4*A*C^3+4*B^3*E-B^2*C^2)*A-8*B^3)^(1/3)/A+(1/3)*(3*A*C-B^2)/(A*(-108*E*A^2+36*A*B*C+12*sqrt(3)*sqrt(27*A^2*E^2-18*A*B*C*E+4*A*C^3+4*B^3*E-B^2*C^2)*A-8*B^3)^(1/3))-(1/3)*B/A-(1/2*I)*sqrt(3)*((1/6)*(-108*E*A^2+36*A*B*C+12*sqrt(3)*sqrt(27*A^2*E^2-18*A*B*C*E+4*A*C^3+4*B^3*E-B^2*C^2)*A-8*B^3)^(1/3)/A+(2/3)*(3*A*C-B^2)/(A*(-108*E*A^2+36*A*B*C+12*sqrt(3)*sqrt(27*A^2*E^2-18*A*B*C*E+4*A*C^3+4*B^3*E-B^2*C^2)*A-8*B^3)^(1/3)))

y = 22.91263477*(.7114884222-(0.5542993294e-1*I)*3^(1/2))^(1/3)+18.37098733/(.7114884222-(0.5542993294e-1*I)*3^(1/2))^(1/3)-37.04460717-((1/2)*I)*3^(1/2)*(-45.82526955*(.7114884222-(0.5542993294e-1*I)*3^(1/2))^(1/3)+36.74197467/(.7114884222-(0.5542993294e-1*I)*3^(1/2))^(1/3))

(9)

"(=)"

y = 7.123944373-0.1692050808e-7*I

(10)

``


y1, y2, y3 formulas are computed by Maple by solving it for general formula.
But, now I got answers in real and imaginery parts, i.e

 

y1 = -119.0286907-1.*10^(-9)*I

y2 = .770924807+1.772050808*10^(-8)*I

y3 = 7.123944373-1.692050808*10^(-8)*I

 

Why, is it so?

 

 

I want answers in simple forum directly only by using these formulas. As i have to show the proof!

Thanks in advance

 

Download qstn.mw

x^4-1equivalent 0 mod 29

after find root of inverse?

I want to simplify (x+1)/sqrt(2*x+2) into sqrt((x+1)/2). Is there a way to do this?

Helo every one,

I'm trying to find the positive real root from:

and substitute it in:

my prog is:

eq5:=u->-3*b*k*u^2-3*b*k^2*u^4-k^3*u^6*b+k*u+5*k^2*u^3-b:  

                   (solve(eq5(u),u)): S:=array([...

underline is a simple procedure:
> s := solve({5*A+3*B = 5, 8*A-9*B = 94}, [A, B]);
s := [[A = 109/23, B = -430/69]]
> assign(s);
> A; B;
109/23
-430/69

But, now what I want to use is not 109/23 or -430/69, but A or B.
That is to say, firstly, I need use root of equation, so Execute Command
> assign(s);
> A; B;
then, I can apply 109/23 or -430/69. Then later on, I want to use A or B, which now are viewed asariable? Now how can I do?

I am having trouble identifying all the possible roots existing in the range specified.

In the file attached i have a non-linear set of equations that are solved for "w" value specified...but fsolve provides solutions only at certain values of "w" , does it mean there is no solution for other data points ?...or is there a way to find out all possible solutions for each value of "w"

Question_primes.mw

I have defined a function as follows:

f:=alpha->1-RombInt(10*tan(x)^alpha,0,Pi/4,1,2);

and now i need to plot this function to see where the root lies, i have tried to do this using the following command:

plot(10*tan(x)^alpha);

and i have tried putting in boundaries but it doesnt seem to work. Can anyone help me please. 

Dear Primers

I have an exponantial equation of the form below:

eq161 := 1/2*alpha^4*(-2*exp(-1+1/2*(4*alpha^2+1)^(1/2)+1/2*(-4*alpha^2+1)^(1/2))*(4*alpha^2+1)^(1/2)+2*exp(-1+1/2*(4*alpha^2+1)^(1/2)-1/2*(-4*alpha^2+1)^(1/2))*(4*alpha^2+1)^(1/2)-2*exp(-1+1/2*(4*alpha^2+1)^(1/2)-1/2*(-4*alpha^2+1)^(1/2))*(-4*alpha^2+1)^(1/2)+2*exp(-1-1/2*(-4*alpha^2+1)^(1/2)-1/2*(4*alpha^2+1)^(1/2))*(-4*alpha^2+1)^(1/2)+2*exp(-1+1/2*(-4*alpha^2+1)^(1/2)-1/2*(4*alpha^2+1)^(1/2)...

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