OrbitA

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These are questions asked by OrbitA

hello i startet using subs. and i got a problem, my expression is only finished and valid if it stands in the factorised version. i am calculating some koefficients and then with subs i am trying to insert it back into the original expression. but it always expand and simplyfy. looking at it, i can see it is a oneway arrow => after the simplicifaction, so it is impossible to rearrange.

what i would like it to happen is the expression i am asking subs to insert a koefficient into is insertet only, and not evaluated and rearrange. can i use a tecknique er something to stop the evaluation and only perform an insertion.

my equation are

a*(2*A + 2*B) + b*(C) = c*(2*A + B) + d*(B*2) + e*(C + 4*B)

i have calculated the koefficients as

{a = 6, b = 1, c = 6, d = 1, e = 1}

i would like it to look like this when i use subs

6*(2*A + 2*B) + 1(C) = 6*(2*A + B) + 1*(B*2) + 1(C + 4*B)      

but it looks like this

12*A + 12*B + C = 12*A + 12*B + C

 

hello i am trying to remove "[ ]" from an expression so it can be used inside a solve function.

the problem is (eqs) has an "[ ]" outside the expression so it lookslike this "[a = c, a = d, 4*a = e, b = 2*e, b = c + 2*d + 2*f]"

how can i remove [] from this expression?

 

code

eq := a*(K + Mn + 4*O) + b*(H + Cl) = c*(K + Cl) + d*(Mn + 2*Cl) + e*(2*H + O) + 2*f*Cl;
elems := [K, Mn, O, H, Cl];
eqr := collect(expand(rhs(eq)), elems);
eql := collect(expand(lhs(eq)), elems);
eqs := zip(`=`, map2(coeff, eql, elems), map2(coeff, eqr, elems));
solve([eq, eqs, a = 1]);

 

thanks

hello how do i solve these equations and what type exactly are they ? so i can find more information

(i dont understand if this is a partial-something)

for the following function

f(y,z)=y*sin(z) + 2*cos(y)

i need to evaluate it at the following point

fy(pi/2,pi)  and  fz(pi/2,pi)

this is the part i dont understand how to do, i keep getting 0 in the equation.

 

thanks for you! and your time

 

i have a task who is like this:

let h(x)=2/f(x)

and the function f meets the following criteria

 

f(-1)=4 and f '(-1)=2

 

what is h'(-1)

 

how to i solve this, i am thinking of ordinary differential equations (ODEs)

and dsolve funktion in maple. but how do i do this.

or is there another way to solve it

 

 

hello this is so simple, and so essentiel in mathprograms, but i cannot get it to work.

how do i run two list through a function that needs both variables.

i am gessing a "for loop" is the solution, but i dont really care if its a loop or not, just i am getting a solution to the problem.

the two list is of equal lengt.

example:

my variables

A:=[1,2,3,4,5,6,7,8,9,10]:

B:=[10,20,30,40,50,60,70,80,90,100]:

my function

C:=A*E/4*pi*B

E is a konstant, in this case E=5

Need

so i want another list object. D to contain all the results from the calculation. as a list.

so D:=[]  needs to be filled with result.

for every run/calculation it takes, one of each of the variables F.ex.

C:=1*E/4*pi*10 and insert into the list D,

then C:=2*E/4*pi*20 and insert into the list D

then C:=3*E/4*pi*30 and insert into the list D.

and so on.

 

thx in advance

 

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