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Hi, I'm trying to find the first ten roots of the equation:

sin(a*Pi)*sqrt(4*a^2+3)*cos(sqrt(4*a^2+3)*Pi)+sin(sqrt(4*a^2+3)*Pi)*a*cos(a*Pi) = 0

But fsolve only finds one root, and not necessarily the first one. So I'm having a difficult time getting the first 10 roots without preknowledge of the intervals to search in. (I can obviously plot it, but there are some other parameters which affect this equation and I can't manually find the intervals every time I change them).

Hi,

This is for a code that analyzes composite laminates with many plys of varying orientation stacked on top of each other, and this is the first semester I've used Maple.

I'm solving for the roots of an equation which will always have a positive and a negative root, and I do this for each ply (could be 100 plies in a laminate).  I want to collect the positive root for each ply, put them in a list or array, and then pick out the smallest value of all the plies ...

In my complex root finding problem it is essential to obtain complex roots in each period of the function. to do this i have write the following code with functional periods but it doesnt work. would you please tell me why? and how can i correct the code?

I am trying to find the complex roots of a nonlinear function within a region using Rootfinding:- Analytic. The problem is that with exactly the same function, the same parameters, etc. it sometimes finds all the roots and sometimes does not! Does anyone know why?

hi all,

m := 1+(-21+.7*I)*sqrt((2.3*6.494)*10^13-x^2)/(2.3*sqrt(((-21+.7*I)*6.494)*10^13-x^2))

fsolve(m = 0, x, {x = 10^7+26500*I .. 2*10^7+26600*I}, complex)

the above characteristic equation m=0 will give complex roots if we solve it. but fsolve is not able to solve this equation evenif i provided the range.through some light for my problem.

thanks

Hi everyone.

Does anyone know what engine is behind the fsolve command? As we know fsolve can solve the sets of equations even without any initial guesses. Is it working based on GA or another global optimization techniques?

Appreciate any help.

MJ

Hi,

     I'm trying to find the zero's of a 'nearly parabolic' trajectory, so narurally there are two zeros' of course I don't care about the first zero only the second one. Is there a way to have fsolve find both, or just skip the first?

     Restricting the range of fsolve won't work because I don't know where my first of second zero will be.  
Thanks!

Hello, I have this question:

I have to plot the roots of a polynomial of the 6th degree, a*t⁶ + b*t⁵ + c*t⁴ + d*t³ + e*t² + f*t = 0, where

a,b,c,d,e,f = g(sigma), then I use implicitplot function to plot t as function of sigma, I want to generate an array, vector or list of pairs as (sigma,t).

Thanks,

Benjamin.

Find, to 10 significant figures, the unique turning point x0 of f(x)=3sin(x^4/4)-sin(x^4/2)in the interval [1,2] and enter it in the box below.x0=？

How to write this in maple？

Hello, 

I am trying to solve a simple equation 

x(0.0178){tan([2.10^-5](x)^0.5)}^2=+2.32.10^12-x, which I want to be solved over the range 1.10^10......2.10^12.

I have tried the following program (which I have got help on):

restart; 

eq := x(0.0178)*tan²(Sqrt(x)*2E-5) = 2.32e12-x;

Use Newton’s Method to approximate the indicated root of the equation to correct six decimal places.

The root of 2.2x⁵ – 4.4x³ + 1.3x²-0.9x-4.0=0 in the interval [-2, -1].

The rest of the assignment states : "

Start by plotting the function in
Maple to get a reasonably good initial approximation. You may use a
“while” loop, but do not use existing Maple commands for Newton’s

The question is to use a program of modified newton raphson , incorporating the Romberg intergal procedure which i have already created, to create a new program which evaluates the integral f(alpha) = 1-10*int(tan(x)^alpha),from 0 to Pi/4.
The following needs to be incorporated in your program:

-Let ci be the approximation of alpha* on iterate i of your "modified newton raphson" method, then the program should run until:
             |ci-c(i-1)|<10^(-6)

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)...

I came across this problem while helping another user find the maxima of an expression for 
various values of a parameter here:

http://www.mapleprimes.com/questions/124104-Maximum-Points--Of-Function-With-More

For various values of ga, it was required to find the maxima in a range of 0<delta<2. 
Plots of the expression indicate that one such maxima exists for each value of the parameter.

My approach was to find the zeros of the first...

I want to find the first positive and first negative root of a bessel function

j2 := (x, a) -> k(x)^2*a^2*(diff((diff(1000000000*sin(1/1000000000*k(x)*a)/(k(x)*a), x))/(k(x)*a), x))/(k(x)*a)

I couldn't use the built in bessel function with two variables, so I had to enter it by hand.

I want to find the values of x where j2 is zero as a function of a:

f := (a) -> RootOf(j2(x, a), x, 0 .. 2)

plot(f(a), a = 2 .. 2.25, y = -3 .. 3)