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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?

 

with(Statistics);
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.

Cannot_find_all_of_roots.mw

 

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 

X1:=ColumnSpace(A)

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

X3:=NullSpace(B) 

X4:=IntersectionBasis([X2,X3])

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!

Joaquin

Hello,

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!!!

dsys:={diff(y1(x),x$2)-A*x^2*y1=B*diff(y2(x),x$3),diff(y2(x),x$2)-C*diff(y1(x),x)=F*x^2+G}

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

 

 

RandomCompositions:= module()
local
Compositions, Rand,

RandomCompositions:= module()
local
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):
seq(R(),i=1..10);

[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]

with(Statistics):
Tally(R());
[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() .

Thanks.

 

Greetings

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?

1.mw

 

 

 

 

2.mw

Hello.

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

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

(1)

 

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

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

(2)

coeffs(eq, vars)

1.0, 1.2, 1.3, 1.1

(3)

``


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?

Download test.mw

 

Hello,

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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