Absolutely amazing. Absolutely every single time I go to Maple to invert a matrix,

the syntax changes. Sometimes it can recognize M:=[[1,3],[2,5]] as a matrix. Then the next time it never will. It requires M:=Matrix([[1,3],[2,5]]); Sometimes it is MatrixInverse(A). Other times it is MatrixInverse[A]. Other times a matrix must be written as <<1,3>|<2,5>>. Sometimes the command is with(LinearAlgebra). Other times with(LinearAlgebra[Generic]). Some of these commands on in the Help menu. Other times they cannot be found.

Hi, I'm trying to find the first few roots of the Bessel function (1st kind) for a few different values of v, but I'm new to Maple and am having some trouble.

This is what I've tried:

------

u:= v->BesselJ(v,x);

solve({u(0)=0},{x});

                            {x=RootOf(BesselJ(0,_Z))}

evalf(%)

I am brand new to Maple and am having difficulty with the syntax for entering an equation and plotting.

Can anyone assist me with the syntax for the following (the equation syntax and plot syntax):

Given that the slope of a line is -3 and passes through the point (1,4), find the equation of the line in slope- intercept form and plot the line.

 

I have already worked the problem by hand; x=2.333, y=7 and I need to plot the line through points 1,4.

 I need to get the surface area and volume of a tetrahedron - coords: a=(2,0,0), b=(0,3,0),c=(3,6,-1),d=(1,2,3). Also, it's height starting from point 'd'.

 The second problem is: how to create an animation of a torus that spins around the Oz axis.

How do I do that? Any help appreciated :*

Also g1 and g2 are costants. I need the expression of  T,that solves the equation,  in function of  x1, P, g1, g2 because later I've to use this expression in Fortran. However I find some mistakes in the equation. The correct equation is funz=0

Thanks

Suppose that I define a unit step function:

u:=t->piecewise(t<0,0,t>=0,1):

How can you periodiclly extend this functions so that it has a period of p ?

Hello, I have found a difference in the handling of an integral between Maple versions 10 and 12. It bothers me since it was working in version 10, but now it gives a numeric exception... I'm not sure whether the replacement is valid or not, so I hope someone can validate it for me. Suppose I have the function definition > f := (x,a,b) -> int( t^(a-1) * (1-t)^(b-1), t=0..x ); As an example, consider > f(2.,3.,.4); This gives -0.8 both in Maple version 10 and 12. However, if I replace the previous statement with > f(x,a,b); > subs( x=2., a=3., b=4., % );

restart;
with( plots ):
P := Array(1..51);
P1 := Array(1..51);
P2 := Array(1..51);
P3 := Array(1..51);
R := proc (s) options operator, arrow; 0.6e-1*(arctan(10*s/1.3-3.0)/Pi+1/2) end proc;
y := proc (s) options operator, arrow; evalf((-1)*0.6e-1*(arctan(10*s/1.3-3.0)/Pi+1/2)*cos(6*Pi*s/1.3)) end proc;
z := proc (s) options operator, arrow; evalf(0.6e-1*(arctan(10*s/1.3-3.0)/Pi+1/2)*sin(6*Pi*s/1.3)) end proc;
f := unapply( simplify(sqrt(1-((6*Pi/(2.6)*0.6e-1)*R(s))^2-(D(R))(s)^2)), s ):
for i from 1 to 51 do

for what values of a does the equation:

ax^3+x^2-ax+1

have exactly two distinct real roots?

I've to solve this equation f(T)=0:

 

f:=T->P*y1*{exp[(0.00662*P*(0.1696-3154.69/T^1.6-(1.58*10^9)/T^4.2))/T]-exp[(0.0022*P*(0.131-5523.13/T^1.6-(3.79*10^9)/T^4.2))/T]}+P*{exp[(0.0022*P*(0.131-5523.13/T^1.6-(3.79*10^9)/T^4.2))/T]}-{x1*g1*1.197*log10(7.74439-1473.686/(T+198.463))*[P-log(7.74439-1473.686/(T+198.463))]*exp[(0.00662*P*(0.1696-3154.69/T^1.6-(1.58*10^9)/T^4.2))/T]}/T-{(1-x1)*g2*0.291*log10[8.07126-1730.63/(T+233.426)]*[P-log10[8.07126-1730.63/(T+233.426)]]*exp[(0.0022*P*(0.131-5523.13/T^1.6-(3.79*10^9)/T^4.2))/T]}/T

hi everyone,

i have study maple for many days .now i have two questions about DAE with bvp probelem . maple 11 doesn't seem to solve the problems.the two problems are similar.

the question one is :

k1'(t)= -1              (1)      

k1(t)-2*a(t)*k2(t)=0       (2)

a(t)=v'(t)                 (3 )    

I'm quite confused by this matter. The following equation (for example) is written into Maple 12: y=e^0.05x Then it is highlighted, right clicked -> copy. Pasting into for example notepad gives: y = exp(0.5e-1*x) This is not the original equation. 0.5e-1*x is not equal to 0.05x?? Any help greatly appreciated.

Hello there...

My name's Gilberto and i'm tryin' to make something out of one particular PDE...

Let me span it for you quick:

I'm trying to discover the solution of Cauchy's heat equation for the cirlce S1 = {x mod 2p}, given the following parameters:

ut = uxx, u(0,x) = phi(x), t > 0, x belongs to R

I'm currently thinking that my next step is solving that equation using the Poisson integral. Is that correct?

solve(exp(-x)-x = 0)

gives 'LambertW(1)' intead of '0.56' which i found by simple fixed pt iteration.