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.

Hi everyone,

For my research, I needed a procedure to calculate an interpolant respecting the monotonicity of the given data. The curve fitting package of Maple 11 didn't help.

I'm pasting my code below.  I hope it helps some of you too.

Cheers,

Ozgur

PS: Thanks goes to Joe Riel for his help.

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.

I have this commands:

> for i from 1 to 10 do x1:=1*i:  g:= x1*exp (T) -1 : T=fsolve( g,T) : y1=3*T end do;               

If i execute if i=1 the first lines will be:

x1 := 1

g := exp(T) - 1                  

T = 0.

y1 = 3 T

Now my questions are two:

 

how to covert function to expression???

View 8710_ZZZ.mw on MapleNet or Download 8710_ZZZ.mw
View file details

What are those 'Z' s represent?[File is attached]

hi all....how to find pseudo inverse in maple...like the one 'pinv' in matlab.

is there any command to generate identity matrix,zero matrix

thanks