MaplePrimes Questions

We put A = {1,2,3,4} and B is the set of lower cubes
to 100 of positive integers.Determine A ∪ B  A ∩ B , A ∩ {6,8} and A \ B 

Write the list of numbers of prime numbers that are less than 10000

As always, thank you all in advanced.

I found this challenge by chance.

solve 615+x^2=2^y over integers.

I rushed to Maple and tried to solve it  with “solve” and "assuming" but I did not get results.

solve(615+x^2=2^y) assuming x::integer,y::integer   did not work.

How could this equation be suitably formulated for Maple to solve it?
 

 

Hi,

i am trying to solve a differential equation numercially but since it is second degree,I can not achieve the proper answer.What should I do?

thank you
diff1.mw
 

restart

h := 1-.8*x+.5*(x^2-x)

1-1.3*x+.5*x^2

(1)

z := 3^(3/2)*(((2+k*De^2*(diff(p(x), x))^2*h^2)^3*h^2+108*k*De^2)/k)^(1/2)

3*3^(1/2)*(((2+k*De^2*(diff(p(x), x))^2*(1-1.3*x+.5*x^2)^2)^3*(1-1.3*x+.5*x^2)^2+108*k*De^2)/k)^(1/2)

(2)

c := (1/6)*((-54*De+z)*k^2*h^2)^(1/3)/(k*De*h)-(1/2)*h*(2+k*De^2*(diff(p(x), x))^2*h^2)/(De*((-54*De+z)*k^2*h^2)^(1/3))-(1/2)*h*(diff(p(x), x))

(1/6)*((-54*De+3*3^(1/2)*(((2+k*De^2*(diff(p(x), x))^2*(1-1.3*x+.5*x^2)^2)^3*(1-1.3*x+.5*x^2)^2+108*k*De^2)/k)^(1/2))*k^2*(1-1.3*x+.5*x^2)^2)^(1/3)/(k*De*(1-1.3*x+.5*x^2))-(1/2)*(1-1.3*x+.5*x^2)*(2+k*De^2*(diff(p(x), x))^2*(1-1.3*x+.5*x^2)^2)/(De*((-54*De+3*3^(1/2)*(((2+k*De^2*(diff(p(x), x))^2*(1-1.3*x+.5*x^2)^2)^3*(1-1.3*x+.5*x^2)^2+108*k*De^2)/k)^(1/2))*k^2*(1-1.3*x+.5*x^2)^2)^(1/3))-(1/2)*(1-1.3*x+.5*x^2)*(diff(p(x), x))

(3)

ode1 := ((1/2)*c*(diff(p(x), x))^2*h^4+(diff(p(x), x))*c^2*h^3+(1/10)*(diff(p(x), x))*h^5+c^3*h^2)*k*De^2+(1/2)*c*h^2+(1/6)*(diff(p(x), x))*h^3+h = 0

((1/2)*((1/6)*((-54*De+3*3^(1/2)*(((2+k*De^2*(diff(p(x), x))^2*(1-1.3*x+.5*x^2)^2)^3*(1-1.3*x+.5*x^2)^2+108*k*De^2)/k)^(1/2))*k^2*(1-1.3*x+.5*x^2)^2)^(1/3)/(k*De*(1-1.3*x+.5*x^2))-(1/2)*(1-1.3*x+.5*x^2)*(2+k*De^2*(diff(p(x), x))^2*(1-1.3*x+.5*x^2)^2)/(De*((-54*De+3*3^(1/2)*(((2+k*De^2*(diff(p(x), x))^2*(1-1.3*x+.5*x^2)^2)^3*(1-1.3*x+.5*x^2)^2+108*k*De^2)/k)^(1/2))*k^2*(1-1.3*x+.5*x^2)^2)^(1/3))-(1/2)*(1-1.3*x+.5*x^2)*(diff(p(x), x)))*(diff(p(x), x))^2*(1-1.3*x+.5*x^2)^4+(diff(p(x), x))*((1/6)*((-54*De+3*3^(1/2)*(((2+k*De^2*(diff(p(x), x))^2*(1-1.3*x+.5*x^2)^2)^3*(1-1.3*x+.5*x^2)^2+108*k*De^2)/k)^(1/2))*k^2*(1-1.3*x+.5*x^2)^2)^(1/3)/(k*De*(1-1.3*x+.5*x^2))-(1/2)*(1-1.3*x+.5*x^2)*(2+k*De^2*(diff(p(x), x))^2*(1-1.3*x+.5*x^2)^2)/(De*((-54*De+3*3^(1/2)*(((2+k*De^2*(diff(p(x), x))^2*(1-1.3*x+.5*x^2)^2)^3*(1-1.3*x+.5*x^2)^2+108*k*De^2)/k)^(1/2))*k^2*(1-1.3*x+.5*x^2)^2)^(1/3))-(1/2)*(1-1.3*x+.5*x^2)*(diff(p(x), x)))^2*(1-1.3*x+.5*x^2)^3+(1/10)*(diff(p(x), x))*(1-1.3*x+.5*x^2)^5+((1/6)*((-54*De+3*3^(1/2)*(((2+k*De^2*(diff(p(x), x))^2*(1-1.3*x+.5*x^2)^2)^3*(1-1.3*x+.5*x^2)^2+108*k*De^2)/k)^(1/2))*k^2*(1-1.3*x+.5*x^2)^2)^(1/3)/(k*De*(1-1.3*x+.5*x^2))-(1/2)*(1-1.3*x+.5*x^2)*(2+k*De^2*(diff(p(x), x))^2*(1-1.3*x+.5*x^2)^2)/(De*((-54*De+3*3^(1/2)*(((2+k*De^2*(diff(p(x), x))^2*(1-1.3*x+.5*x^2)^2)^3*(1-1.3*x+.5*x^2)^2+108*k*De^2)/k)^(1/2))*k^2*(1-1.3*x+.5*x^2)^2)^(1/3))-(1/2)*(1-1.3*x+.5*x^2)*(diff(p(x), x)))^3*(1-1.3*x+.5*x^2)^2)*k*De^2+(1/2)*((1/6)*((-54*De+3*3^(1/2)*(((2+k*De^2*(diff(p(x), x))^2*(1-1.3*x+.5*x^2)^2)^3*(1-1.3*x+.5*x^2)^2+108*k*De^2)/k)^(1/2))*k^2*(1-1.3*x+.5*x^2)^2)^(1/3)/(k*De*(1-1.3*x+.5*x^2))-(1/2)*(1-1.3*x+.5*x^2)*(2+k*De^2*(diff(p(x), x))^2*(1-1.3*x+.5*x^2)^2)/(De*((-54*De+3*3^(1/2)*(((2+k*De^2*(diff(p(x), x))^2*(1-1.3*x+.5*x^2)^2)^3*(1-1.3*x+.5*x^2)^2+108*k*De^2)/k)^(1/2))*k^2*(1-1.3*x+.5*x^2)^2)^(1/3))-(1/2)*(1-1.3*x+.5*x^2)*(diff(p(x), x)))*(1-1.3*x+.5*x^2)^2+(1/6)*(diff(p(x), x))*(1-1.3*x+.5*x^2)^3+1-1.3*x+.5*x^2 = 0

(4)

ivp := {ode1, p(0) = 0}

{((1/2)*((1/6)*((-54*De+3*3^(1/2)*(((2+k*De^2*(diff(p(x), x))^2*(1-1.3*x+.5*x^2)^2)^3*(1-1.3*x+.5*x^2)^2+108*k*De^2)/k)^(1/2))*k^2*(1-1.3*x+.5*x^2)^2)^(1/3)/(k*De*(1-1.3*x+.5*x^2))-(1/2)*(1-1.3*x+.5*x^2)*(2+k*De^2*(diff(p(x), x))^2*(1-1.3*x+.5*x^2)^2)/(De*((-54*De+3*3^(1/2)*(((2+k*De^2*(diff(p(x), x))^2*(1-1.3*x+.5*x^2)^2)^3*(1-1.3*x+.5*x^2)^2+108*k*De^2)/k)^(1/2))*k^2*(1-1.3*x+.5*x^2)^2)^(1/3))-(1/2)*(1-1.3*x+.5*x^2)*(diff(p(x), x)))*(diff(p(x), x))^2*(1-1.3*x+.5*x^2)^4+(diff(p(x), x))*((1/6)*((-54*De+3*3^(1/2)*(((2+k*De^2*(diff(p(x), x))^2*(1-1.3*x+.5*x^2)^2)^3*(1-1.3*x+.5*x^2)^2+108*k*De^2)/k)^(1/2))*k^2*(1-1.3*x+.5*x^2)^2)^(1/3)/(k*De*(1-1.3*x+.5*x^2))-(1/2)*(1-1.3*x+.5*x^2)*(2+k*De^2*(diff(p(x), x))^2*(1-1.3*x+.5*x^2)^2)/(De*((-54*De+3*3^(1/2)*(((2+k*De^2*(diff(p(x), x))^2*(1-1.3*x+.5*x^2)^2)^3*(1-1.3*x+.5*x^2)^2+108*k*De^2)/k)^(1/2))*k^2*(1-1.3*x+.5*x^2)^2)^(1/3))-(1/2)*(1-1.3*x+.5*x^2)*(diff(p(x), x)))^2*(1-1.3*x+.5*x^2)^3+(1/10)*(diff(p(x), x))*(1-1.3*x+.5*x^2)^5+((1/6)*((-54*De+3*3^(1/2)*(((2+k*De^2*(diff(p(x), x))^2*(1-1.3*x+.5*x^2)^2)^3*(1-1.3*x+.5*x^2)^2+108*k*De^2)/k)^(1/2))*k^2*(1-1.3*x+.5*x^2)^2)^(1/3)/(k*De*(1-1.3*x+.5*x^2))-(1/2)*(1-1.3*x+.5*x^2)*(2+k*De^2*(diff(p(x), x))^2*(1-1.3*x+.5*x^2)^2)/(De*((-54*De+3*3^(1/2)*(((2+k*De^2*(diff(p(x), x))^2*(1-1.3*x+.5*x^2)^2)^3*(1-1.3*x+.5*x^2)^2+108*k*De^2)/k)^(1/2))*k^2*(1-1.3*x+.5*x^2)^2)^(1/3))-(1/2)*(1-1.3*x+.5*x^2)*(diff(p(x), x)))^3*(1-1.3*x+.5*x^2)^2)*k*De^2+(1/2)*((1/6)*((-54*De+3*3^(1/2)*(((2+k*De^2*(diff(p(x), x))^2*(1-1.3*x+.5*x^2)^2)^3*(1-1.3*x+.5*x^2)^2+108*k*De^2)/k)^(1/2))*k^2*(1-1.3*x+.5*x^2)^2)^(1/3)/(k*De*(1-1.3*x+.5*x^2))-(1/2)*(1-1.3*x+.5*x^2)*(2+k*De^2*(diff(p(x), x))^2*(1-1.3*x+.5*x^2)^2)/(De*((-54*De+3*3^(1/2)*(((2+k*De^2*(diff(p(x), x))^2*(1-1.3*x+.5*x^2)^2)^3*(1-1.3*x+.5*x^2)^2+108*k*De^2)/k)^(1/2))*k^2*(1-1.3*x+.5*x^2)^2)^(1/3))-(1/2)*(1-1.3*x+.5*x^2)*(diff(p(x), x)))*(1-1.3*x+.5*x^2)^2+(1/6)*(diff(p(x), x))*(1-1.3*x+.5*x^2)^3+1-1.3*x+.5*x^2 = 0, p(0) = 0}

(5)

dsolve(ivp, p(x), numeric, parameters = [k, De])

Error, (in dsolve/numeric/make_proc) Could not convert to an explicit first order system due to 'RootOf'

 

``


 

Download diff1.mw

 

 


 

Download diff1.mw

Simple, but racking my brains, what is the formula for the sequence 2,3,4,6,8,9,10,12,14,15,16,18,...?

I have problem: Let  A(2,0,0), B(0,3,0), C(0,0,6) and D(1,1,1) be four points and Delta is the line passing through the point D so that sum of distances from the points A, B, C to Delta maximum. Find a direction vector of Delta.

I tried. Let v(a,b,c) where a^2 + b^2 + c^2 = 1 and my code
 

with(Student:-MultivariateCalculus):
with(Optimization):
A := [2, 0, 0]: 
B := [0, 3, 0]: 
C := [0, 0, 6]: 
DD := [1, 1, 1]: 
v := <a, b, c>: 
d := Line(DD, v): 
d1 := Distance(A, d): 
d2 := Distance(B, d): 
d3 := Distance(C, d): 
S := d1+d2+d3: 
Maximize(S, {a^2+b^2+c^2 = 1});

I didn't get the result. How to get the numbers a, b, c?

Hello,

Is it possible to expand vector calculus identities directly in Maple without taking them to basis form (i.e. their component partial derivatives).

For example: 

restart

with(Physics[Vectors]):

%Curl(u*%Gradient(v)) = u*%Curl(%Gradient(v)+`&x`(%Gradient(u), %Gradient(v)))

 

Given the Left Hand Side, can Maple come up with the RHS on its own?

Many Thanks.

Let (u (n)) be the sequence defined by u (n + 1) = 3.5u(n) (1 - u (n)) and u (0) = 0.4.
1. Create the sequence s whose elements are [k, u (k]) for k varying from 0 to 100.
2. Graph the list of points with the plot function and the style = point option.

1. Write the list of numbers that are the sum of two squares of integers a ^ 2 and b ^ 2 with 0 <= a, b <= 5.
2. Select the odd numbers from the previous list. What seems to be their general form

How 2 argumented matrices can be written in one matrix 

Hello,

I currently have

And was wondering if there's a way to declare x as a real so that this evaluates to 1?

Why, when parsing Pi/6, is it displayed differently in a textplot situation?

parse("Pi/6")
               

with(plots):
textplot([1,1,parse("Pi/6")])

Hellow,I use maple 13 (linux)

 

How can I get a output data file solution of my ODE? For example, the maple resolved the harmonic equation and got a u(t) function, but I want manipulated the data in a external programm, like gnuplot ou xmgrace.

 

Sorry my bad english!

 

wave.mw

 

i wnat to take this export numbers from for loops in the matrix
restart;
with(LinearAlgebra);
f(x) :=  3*x^2+1 :
for i by .5 to 3.5 do print(i, f(i)) end do;

same this pic

I want to get solutions (approximate solutions) of the inequality g(x) > 0. I tried. 
 

restart; 
fprime := x-> (x-1)*(x-2)^2*(x-3)*(x-4); 
f := unapply(simplify(int(fprime(x), x)), x); 
g := unapply(expand(f(3*x+1)-x^3+3*x), x); 
solve(g(x) > 0, x);

I got

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