Hello all,

I'm trying to solve nine non-linear equations for various input. I am using a for loop to vary the input, and for few input combinations there is no solution. I am trying to exclude these and print the values only if there is a solution. After looking at some of the forms I used if (not type(sol[7], numeric) then print("no solution") else print(sol[7]) end if but this is the output I am getting for all the input.

Attached is the maple...

In it's recent edition of Mathematics Today (in print and online), the UK-based Institute of Mathematics and it's Applications, compared 4 symbolic solvers: Maple 15, Mathcad 15, the student edition of Matlab v5 and the Casio CFX-9970G calculator, concluding that "Maple would be the natural choice for research mathematicians, theortetical physicists, those working in any area where mathematics is demanding or for mathematics undergraduates for whom costs are lower"

Hello, 

I am trying to analyse a nonlinear ode. To see the behaviour,

I am using implicit plot (suggested by @Preben Alsholm).

The problem is that every time I need to search for the value of N

(=infinity) to have some sort of graph (needed for analysis).

The other thing is, if we have an output of the implicitplot, How we can be sure

that it is the correct one?

Hello,

Can anyone help me with this error in Maple while using prcNewton to find local extrema: 

> prcNewton := proc () 

local ftn, strpt, epsilon, maxlps, i, xn, dftn; 

if 4 < nargs then 

elif nargs < 2 then end if; 

if nargs = 2 then 

epsilon := 1/10000000; 

maxlps := 1000 

elif nargs = 3 then 

> G := proc (N)

local s, n;

s := 0.;

n := 0;

while n < N do

n := n+1;

s := s+1/n

end do;

[s, n] end proc;

    Comput  G(1.0),G(2.0), .....G(7.0)   try to figure out what
    the procdure G is doing

Anyone knows how to use an expression like , where t is a polynomial variable and n is an intager variable, in such a way that ? Actually, I would like to use it inside some package like polynomial tools, in order to normalize such expressions.

restart:with(plots):with(plottools):with(stats):
unprotect(D):
a[1]:=0:a[2]:=0.2:a[3]:=-0.15:a[4]:=0.85:
b[1]:=0:b[2]:=-0.26:b[3]:=0.28:b[4]:=0.04:
c[1]:=0:c[2]:=0.23:c[3]:=0.26:c[4]:=-0.04:
d[1]:=0.16:d[2]:=0.22:d[3]:=0.24:d[4]:=0.85:
e[1]:=0:e[2]:=0:e[3]:=0:e[4]:=0:
f[1]:=0:f[2]:=0.2:f[3]:=0.2:f[4]:=0.2:
p[1]:=0.01:p[2]:=0.08:p[3]:=0.15:
N:=10000:x[0]:=0:y[0]:=0:
randomize():
for n from 0 to N do
r[n]:=rand()/(10)^(12);
if r[n]then i:=2 elif r[n...

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!

> limit(n*int((1-(3/26)*x-(37/49)*x^2)^n, x = 0 .. 1), n = infinity); with Maple? The calculations suggest its existence. I don't have  in mind the usage of known asymptotic formulas.

fprime0 := diff(BfT[0]*z^ls*(ls*p[0]*z^lk+1-ls*p[0])/(.99*(1-(1-.989*f[0])^5)), z); 'fprime0(1)' = subs(z = 1, fprime0)

 je veux savoir comment je calcule les valeurs de fprime0 s'il vous plait je suis coincé depuis une semaine et j'ai pas pu terminer mon travail le resultat me donne 

"Large sortie de plus de 1000000 noeuds" 

I am presently using Tally to read through a list of numbers to try and calculate their frequency, I end up with a list as follows:

frequency := [1 = 6, 2 = 7, 3 = 91, 4 = 10], etc.

Now, I've been given a function that should turn this function into a "list of lists":

[code]
list_of_lists := n -> [ op(n)[1],  op(n)[2] ]
[/code]

As can be expected from looking at this function, it only returns the first two elements from the original list

is there a simple maple package template that i can look at?

I wrote a module with about 30 procs, i want to make it into a package but I don't really know where to start.

all i need is an existing package that i can simply

• rename the title
• swap the code with my own code
• swap the example with my own example

Input: solve(O/N=k^x*e^(-x)/x!,x);

Output: 

            /                    _Z  (-_Z)  \
           RootOf\-O factorial(_Z) + k   e      N/
What is this _Z? 

In the 20-29 age group , the heigts were normally ditributed with a mean 0f 64.3 inches and a standard deviation of 2.6 inches

find probability that her height is less than 56.5 inches.