gkokovidis

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20 years, 299 days
Draeger Medical Systems, Inc.

 

 

Regards,
Georgios Kokovidis
Dräger Medical

MaplePrimes Activity


These are answers submitted by gkokovidis

>restart: with(Student[Calculus1]):
>Roots( sin(3*x)=cos(5*x) ,0..5,numeric);

[.1963495408, .9817477042, 1.767145868, 2.356194490, 2.552544031, 3.337942194, 4.123340358, 4.908738521]

>identify(%);

[1/16*Pi, 5/16*Pi, 9/16*Pi, 3/4*Pi, 13/16*Pi, 17/16*Pi, 21/16*Pi, 25/16*Pi]

 

Regards,
Georgios Kokovidis
Dräger Medical

How about using the buillt-in version:

I - root of x^2 = -1
 

> z1 := -3*I*(2+I)*(3+2*I)*(1+4*I);
> zz1 := -18*I-93*I^2-90*I^3-24*I^4;
> z2 := (2+4*I)*(-1+3*I)/((1+I)*(4-I));
> zz2 := -2/((1+I)*(4-I))+2*I/((1+I)*(4-I))+12*I^2/((1+I)*(4-I));


                           z1 := 69 + 72 I


                           zz1 := 69 + 72 I


                                 -32   26
                           z2 := --- + -- I
                                 17    17


                                 -32   26
                          zz2 := --- + -- I
                                 17    17
 

 Are those the results you were looking for?

 

Regards,
Georgios Kokovidis
Dräger Medical

There are plenty of code examples on the web.  Here is a link.  There are more.  Search for "Maple  Aitoff Projection".

 

Regards,
Georgios Kokovidis
Dräger Medical

I tried the code above in both Maple 12 and Maple 9.5 and it works without error.  Here is a worksheet done in Maple 12.  When you open it, do you see the same error that you reported.  I used the logn form of the plot command, by loading the plots package first.

View 221_slope_field.mw on MapleNet or Download 221_slope_field.mw
View file details

 

Regards,
Georgios Kokovidis
Dräger Medical

I see the same error using Maple 12 .02.  Using Maple 9.52, I get the image below.

 

 

Regards,
Georgios Kokovidis
Dräger Medical

With Digits set to 10, you are probably running into roundoff error.  Try increasing Digits to 20 or higher, and see if you get more consistent results.

 

Regards,
Georgios Kokovidis
Dräger Medical

The following works for me using an earlier version of Maple.

restart:

Int(exp(-24358.0/x),x=295..1970)=int(exp(-24358.0/x),x=295..1970);

Int((exp(-24358.0/x)/x),x=295..1970)=int((exp(-24358.0/x)/x),x=295..1970);

I changed -24358 to -24358.0 to get the floating point result.  I also used exp to respresent "e". 

 

Regards,
Georgios Kokovidis
Dräger Medical

?plottools[line]

 

Regards,
Georgios Kokovidis
Dräger Medical

Here is one way to do this.  Replace the path with your own data file.  The example below reads in data that is 2 columns wide.  It is in a file called altitude.dat.  It can have any extension.  The data looks like this:

0.000    26.515
0.753    53.030
1.505    212.121
2.258    437.500
3.011    875.000
3.763    1405.303

 

restart:
with(plots):
a:=readdata("C:\\temp\\altitude.dat",2):
pointplot(a,axes=normal,symbol=box);

 

Regards,
Georgios Kokovidis
Dräger Medical

 

Here is a link to a previous post that looks similar to what you are doing.

 

Regards,
Georgios Kokovidis
Dräger Medical

Visit the Maple Applications Center, and take a look at some of the examples that are posted there.  Here is one example.

 

Regards,
Georgios Kokovidis
Dräger Medical

There are other ways to plot vectors in Maple.  Below is one of them.

> with(plottools):
> l1 := arrow([-3, 5], [-1, 8], .2, .4, .1, color=red):
> l2 := arrow([-1,8],[4, 6], .2, .4, .1, color=blue):
> l3 := arrow([4, 6], [8, -4], .2, .4, .1, color=green):
> l4 := arrow([8,-4],[-3, 5], .2, .4, .1, color=black):
> plots[display](l1, l2,l3,l4, axes=normal);

See also the help file for the arrow command in the plots package.

?plots[arrow]

 

Regards,
Georgios Kokovidis
Dräger Medical

See if the examples in this post are helpful.  The concept is the same.

 

Regards,
Georgios Kokovidis
Dräger Medical

The code below is a step in the right direction.  It bypasses the for loop.   The output is in Fortran format, but not exactly what you want.  Using printf might be a better choice.

> restart:
> with(CodeGeneration):
> ans:=[seq(exp(i),i=1.0..10.0)]:
> Fortran(ans):
 

 

Regards,
Georgios Kokovidis
Dräger Medical

Output of solve has 2 solutions for nstart.  By enclosing the output in square brackets, you are puting it into a list.  Then you can select which of the two answers you want to substitue values for.

>eq1 := 1/L = R*(1/nfinal^2-1/nstart^2);

>sol1:=[solve(eq1, nstart)];

>sol1[1];

>sol1[2];

>eval(subs(L = 6.561123702*10^(-7), nfinal = 2, R = 0.1097373e8, sol1[1]));

>eval(subs(L = 6.561123702*10^(-7), nfinal = 2, R = 0.1097373e8, sol1[2]));

 

 

Regards,
Georgios Kokovidis
Dräger Medical

 

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