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Hello! How can I find extremes of numeric solution of ODE system obtained using "dsolve"? Can I use something like "extrema" function?

If I use a decimal like 0.569840290998053 into a fraction?

I notice that when I tried to factor the polynomial

x^3+x^2+2*x+1 I did not get the rational roots of the polynomial, then I tried using synthetic division to solve for the roots but I could not find a root. So I believe that the roots of the polynomial equation are complex numbers...

How could I use synthetic division to find the rational roots of this integral so that I could do a Partial Fraction Decomposition for the integral...

Here is my code & the error mesage.  What's wrong?


X := Vector*([0, 5, 10, 15, 20, 25, 30], datatype = float);
Y := Vector*([38.8, 53.8, 82.4, 107.6, 130.7, 152.4, 173.2], datatype = float);
NonlinearFit(av^2+bv+c, X, Y, v);
Error, (in Statistics:-NonlinearFit) invalid input: PostProcessData expects its 1st argument, x, to be of type {array, list, rtable}, but received w

Hello everybody,

I want to find all of roots of the complex variables functions in two ways.

(1) find the value which can make the function equals 0

(2) find the real value and imaginary value which make real part and imaginary part of function equal 0

(I know answers of these two case is not equal completely.)


The function is a non-linear function, including sin, cos and Bessel function, such as:


And, I used Analytic and fsolve to do case (1) and (2), but failure. The follow result is how I tried to find the real value answer:


It seems that both of two commands can only find some of roots. 

How to find all of roots of these cases? The related .mw file is attached.


Thanks a lot.


Issues with Bases ...

August 16 2014 casalcantara 5

I am trying to take out the intersection between two bases from one of the original basis. I have two matrices (A and B) and want to find the intesection between the range (or column space) of A and the null space of B. The range of A is 


For some reason, the column space is not presented as a list, but with square brakes, so I convert this into a list of vectors:

X2:=SumBasis([X1[1],X1[2]])    %If I use the command Basis, it returns again square brakets, not sure why... 

The null space of B



X5:= X2\cap X4 % I am using latex code for the intersection symbol...

The result is the empty set! Evidently, X2 and X3 are differnt bases.

Any help would be really welcome! Many thanks!



How do I interpolate using polar coordinates?  I have a series with non-uniform spacing, that I want to map onto a uniform space polar grid.

Any assistance would be appreciated.

let A be a matrix=


[  7        7      9    -17

   6        6      1    -2

 -12    -12    -27    1

   7       7      17   -15 ]

What is the reduced row echelon form of A?

What is the rank of A?

A consistent system of linear equations in 14 unknowns is reduced to row echelon form. There are then 10 non-zero rows (i.e. 10 pivots). How many parameters (free variables) will occur in the solution?

I failed to solve the ODE system shown as follows, where y1(x) and y2(x) are functions of x, ranging from -L/2 to L/2. All the other parameters are constants (A,B,C,F,G). The analytic or numeric solution of y1(x) and y2(x) are wanted.Really appreciate for you experts' help and time!!!


boundary conditions:y1(0)=0, diff(y2(L/2),x$2)=0, D(y2)(0)=0, y2(L/2)=0



RandomCompositions:= module()
Compositions, Rand,

RandomCompositions:= module()
Compositions, Rand,
ModuleApply:= proc(n::posint, k::posint)
local C;
Compositions:= [seq(C-~1, C= combinat:-composition(n+k, k))];
Rand:= rand(1..nops(Compositions));
()-> Compositions[Rand()]
end proc;

end module:
R:= RandomCompositions(8,6):

[0, 0, 2, 6, 0, 0], [1, 0, 0, 3, 4, 0], [0, 3, 3, 2, 0, 0],[1, 2, 4, 0, 1, 0], [0, 4, 0, 1, 3, 0],

[2, 0, 1, 4, 1, 0],[2, 0, 1, 1, 3, 1], [1, 0, 4, 2, 1, 0], [1, 3, 0, 2, 0, 2],[2, 0, 3, 2, 1, 0]

[0 = 1, 1 = 4, 4 = 1]

I want to count the 0 to 8 respectively summation,and divide by i.

 But the  seq command  R() isn't conform  with Tally command R() .




I am left wondering why evalf as well as right clicking and selecting approximate both are in error.

Try 1/2*sqrt(109) and find a solution with 5 digits (english isnt my first language so the wording may not be correct).

If you calculate it as 1/2*sqrt(109.0) you correctly get 5.2202

However, if you use evalf(1/2*sqrt(109)) or 1/2*sqrt(109) then right click and approximate with 5 digits you incorrectly get 5.2200.

I assume this is because both incorrectly round to 5 digits on sqrt(109) before dividing by 2 - a common mistake for very inexperienced students but surely not something I would expect from Maple.

Please explain and better yet fix. Thanks.

i'm going to solve the two eq. and plot the answer point

i try to do two group but the first succeed the second get some wrong

the first i named 1 ,and second named 2 as below

the 1 plot the correct point that i want

Although the 2 solve eq too but it didn't plot the point on(-0.36,0.29),

the code of this two is same ,

anyone can help me?



I am trying to extract all the coefficients of a linear equation by calling coeffs function.

eq := 1.2*a[2]+1.3*a[3]+1.1*a[1]+1.0




vars := [seq(a[i], i = 1 .. 3)]

[a[1], a[2], a[3]]


coeffs(eq, vars)

1.0, 1.2, 1.3, 1.1



I expected I receive the coefficients in the same order as in vars variable, but I was wrong. How to extract coefficients in the order I need?




in Kurzweil und Stellmacher "Theorie der endlichen Gruppen" page 97, Exercice 3 I have read

Is G nilpotent then G is cyclic.

I have found that if G is nilpotent it must not be cyclic.

Can anybody tell me whether I am right or not?


Best Regards

Kurt Ewald

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