Maple Questions and Posts

These are Posts and Questions associated with the product, Maple

I am try to program a proc that will plot the kinematic equations.

kinematics := proc(a::procedure, t::name, s0::numeric, v0::numeric, t0::numeric, t1:: numeric)

local v:=t->v0+int(a(t),t);

local s:=t->s0+int(v(t),t);

plot([a(t),v(t),s(t)], t=t0..t1);

end proc;

the above is my code, which works ok. 

Note: a is acceleration function, v is velocity function, s is position function, t is independent variable,...

There seems to be patterns for sin(10^-k) for rational k;

Here we have the "floats."

n sin(10^(-n-1/2))

1 0.03161750640

2 0.003162272390

3 0.0003162277607

4 0.00003162277660

5 0.000003162277660

6 0.0000003162277660

7 0.00000003162277660

 

More later on using the mantissa. You're welcome to join me.

(i)develop an algorithm for computing f¡ÊF[x,y],F a field,where the degree of f in y is less than n and and f(x,ui)=vi; for i=0,1......,n-1, for distinct ui∈F,and arbitrary Vi∈F[x].showthat f is unique.

(ii) assuming that the degree of each Vi is less than m, what is the computing time of your algorithm (in term of m and n)?

(iii) computer f∈ F[11][x,y]such thatf(x,0)=x^2+7,f(x,1)=x^3+2*x+3,f(x,2)=x^3+5.

sys := [1.8*a+.4000000000*b+103.9*c-.5000000000*e-102.6*g = 0, .5000000000*b+102.6*c+.3*d+.3500000000*e-102.1*g = 0, -3.3*a-1.050000000*b+102.4*c+.7*d+.2500000000*e-105.1*g = 0, -2.5*a+1.450000000*b+102.6*c+3.3*d+1.050000000*e-102.4*g = 0, -1.6*a+1.100000000*b+105.8*c+.3*d+1.750000000*e-105.5*g = 0, -.7*a-.2500000000*b+105.1*c-.2*d+.2000000000*e-105.7*g = 0, -.3*a-.3500000000*b+102.1*c+2.5*d-1.450000000*e-102.6*g = 0, .2*a-.2000000000*b+105.7*c+1.6*d-1.100000000*e-105.8*g = 0...

hello. how can I translate a graph in the x axis 3 times and then again in the y axis 2 times?

Find a(n)of sequence1,8,1,4,7,0,7,0,....   (n=1,2,3,4,...) ,

'1.8*a+.4000000000*b+103.9*c-.5000000000*e-102.6*g = 0', '.5000000000*b+102.6*c+.3*d+.3500000000*e-102.1*g = 0', '-.3*a-.3500000000*b+102.1*c+2.5*d-1.450000000*e-102.6*g = 0', '-2.5*a+1.450000000*b+102.6*c+3.3*d+1.050000000*e-102.4*g = 0', '-3.3*a-1.050000000*b+102.4*c+.7*d+.2500000000*e-105.1*g = 0', '-.7*a-.2500000000*b+105.1*c-.2*d+.2000000000*e-105.7*g = 0'

i use solve, it return a = a 

> 1) letJ subset F[7][x] be the set of all polynomials h in F[7][x] sloving the interpolation problem h(0);
=1,h(1)=5,h(6)=2. computer the unique polynomial f∈J of least degree.

2) find a surjective ring homomorphism ×:F[7][x]->F[7]^(3) such that Ker×=={rm:r∈F[7][x]};and computer ×(f)and×(x^2+3*x+2).

3) show thatJ=f+Ker×={f+rm:r∈ F[7][x]}.

> find*a*polynomial*f;

 in F[7][x] ofdegree less than 4 solving the congruence(x^(2)-1)∗f≡x^(3)+2*x+5 mod x^(4)+2*x+1 inF[7][x] .;

show that the residue class ring F[343]=F[7][x]/is a field,and computer the inverse of x^(2) mod x^(3)+x+1in F[343].

Hello everyone,

I saw a question on a different forum related to the stability of the numerical solution of a coupled ode system. The original question was about the convergence of the far field condition (infinity) for some specific values of the different parameters involved.  For the details please see the link below

http://mathematica...

Hi,I want to compute some expressions as binomial(d-2,r)*binomial(v-d-1,r-1) and something like that. In the process, there are many such expressions but with differernt parameters. Some of them can pass successfully, but some of them will generate a mistake. That is: Error, (in gcd/LinZip) variable in input expression must be in either the evaluate or retained variables.Could you please what happened to it? Thanks.

hi i have one probleam:i want solve one pde

(1/r)*(d/dr(r*dT/dr))=(1/ß)*dT/dt

ß is cte.

T=T(r,t)

Hi,

I need help for the following. I have system of ODE as follows

 

Hello

 I wish to solve nonlinear PDE equation numerically  in u & v coordinate . where u=t-x, v=t+x . we note x=(v-u)/2.. 

pde := diff(u(t,x),t,t) - diff(u(t,x),x,x) + x^2*u(t,x);  

 

 We need only initial condition let f(u)=sech(u) & g(v)=sech(v). So u((i,0)=f(ih), u(0,j)=g(jh) 

we summe step size of time j =  step size of spatial i .

the stencil we use , 

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