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

It's been a while since I've updated my blog, but the recent Maple 12 release gives me a good opportunity to talk about some of the features I'd been working on for the past months. A few people on MaplePrimes had asked for more details about Maple 12, so I'll start by saying a bit about the new polar axes. A lot of this work was done by my colleagues in the GUI Group and they may have additional interesting things to say about the feature.

In previous versions of Maple, you could draw polar plots using the plots[polarplot] command or with the coords=polar option, but these were always displayed with Cartesian axes. In Maple 12, polar axes are displayed by default, as seen here.

plots[polarplot](1+cos(theta), theta=0..2*Pi, axis[radial]=[tickmarks=5])

 A number of new options were added to the polarplot command so that you can customize the axes.  The most useful ones are the axis[radial] and axis[angular] options. These work like the axis[1], axis[2] and axis[3] options available for general plots, and you can use them to control color, tickmarks and other properties of the radial and angular axes.

Typeset math on plots had been introduced in Maple 11, and now we can take advantage of this with nice axis labels, in multiples of Pi, on the angular axis. These labels appear by default, but of course, they can be customized with the axis options. The plot/typesetting help page provides information on how to add typeset math to plots through the command line. There are also interactive ways to do this, using the context menu.

You can add polar axes to plots created by commands other than plots[polarplot], by using the axiscoordinates=polar option. However, not all the options offered by plots[polarplot] are available generally. Here is an example using plots[implicitplot].

plots[implicitplot]([x^2+2*y^2 = 1, x^2+1.5*y^2 = 1], color = ["Blue", "Green"], x = -1 .. 1, y = -1 .. 1, axiscoordinates = polar);

It is also possible to get the pre-Maple 12 Cartesian axes back with polar plots, by adding the axiscoordinates=cartesian option.