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dharr

Dr. David Harrington

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15 years, 176 days
University of Victoria
Professor or university staff
Victoria, British Columbia, Canada

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Contact Dr. David Harrington
I am a professor of chemistry at the University of Victoria, BC, Canada, where my research areas are electrochemistry and surface science. I have been a user of Maple since about 1990.

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These are answers submitted by dharr

invlaplace...

dharr 953
Yesterday at 8:15 PM
1 6

Not entirely certain I know what you want, but this may be a start.

[Edit: I changed pi (just a symbol) to Pi (the circle constant thing)]

At the beginning I define i to be sqrt(-1) as you want and D to be a regular variable rather than the differential operator.

Maple gives the answer as a sum over roots of a quartic polynomial. To get a simpler formula, you will need to give values to some variables or parhaps tell Maple their signs (using assuming)

Download Linear_System.mw

also assignment...

dharr 953
December 07 2019
1 0

As well as Joe's corrections, you need to assign to y (y:=1). 
 

> 

r:=proc(x)
    local y;
    if (x=1)
    then    
    y:=1;
    else
        y:=0;
    end if;
    return y
    end proc;    
 

proc (x) local y; if x = 1 then y := 1 else y := 0 end if; return y end proc

> 

h:=proc(x);
    if (x= infinity)
    then    
    return 1 ;
    else
        return 0;
        end if;
    end proc

proc (x) if x = infinity then return 1 else return 0 end if end proc

> 

r(1);r(2);

1

0

> 

h(1);h(infinity);

0

1

> 

 


 

Download ww.mw

 

gcdex...

dharr 953
December 01 2019
2 0

I suspect you want gcdex(x^9-a, b*x^6-c, x); but the answer is 1. You may need to specify the numerical value of more of the coefficients.

delayed evaluation...

dharr 953
November 28 2019
0 2

Try

T := unapply('fsolve'(m(t, c) = -1, t = 0 .. (1/2)*Pi), c, numeric);

In general use unapply to make functions from expressions earlier in the worksheet. The numeric option means that plot and other functions evaluated with a symbolic value will not throw an error.

bc at infinity...

dharr 953
November 25 2019
0 0

So the constant _C3 means another condition is required. Presumably you want the solution to drop off at y=infinity.

pdetest suggests your solution does not solve the pde.
 

Download pde.mw

assignment...

dharr 953
October 27 2019
1 1

Use := to assign the right-hand side to the left. Some other issues also:

> 

v:= <1 |1| 0>;
w:=<1,1,0>;
M:=<1, 2, 3; 5, 4, 3; 7, 9, 0>;
c:=v.M;
d:=M.w;

v := Vector[row](3, {(1) = 1, (2) = 1, (3) = 0})

w := Vector(3, {(1) = 1, (2) = 1, (3) = 0})

M := Matrix(3, 3, {(1, 1) = 1, (1, 2) = 2, (1, 3) = 3, (2, 1) = 5, (2, 2) = 4, (2, 3) = 3, (3, 1) = 7, (3, 2) = 9, (3, 3) = 0})

c := Vector[row](3, {(1) = 6, (2) = 6, (3) = 6})

d := Vector(3, {(1) = 3, (2) = 9, (3) = 16})

> 

 


 

Download vectors.mw

code for less than or equal...

dharr 953
October 26 2019
0 0

Try entering in 1D Maple input, e.g. at a > prompt use ctrl-M then `&le;`

You will see that `&le;` is rendered as less-than-or-equals, as it would be in HTML.

missing * and extra spaces...

dharr 953
October 18 2019
0 1

Changes in red.
 

> 

 RK2skritt:=proc(FR::procedure,xo,yo,vxo,vyo,h)
 local x, y, vx, vy, r, kx1, kx2, kx3, kx4, ky1, ky2, ky3, ky4, lx1, lx2, lx3, lx4, ly1, ly2, ly3, ly4, tmp;
 r:=sqrt(xo*xo+yo*yo);
 tmp:=-(h*FR(r))/(r);
 lx1:=h*vxo;
 ly1:=h*vyo;
 kx1:=tmp*xo;
 ky1:=tmp*yo;
 lx2:=h*(vxo+0.5*kx1);
 ly2:=h*(vyo+0.5*ky1);
 r:=sqrt((xo+0.5*lx1)^(2)+(yo+0.5*ly1)^(2));
 tmp:=-(h*FR(r))/(r);
 kx2:=tmp*(xo+0.5*lx1);
 ky2:=tmp*(yo+0.5*ly1);
 lx3:=h*(vxo+0.5*kx2);
 ly3:=h*(vyo+0.5*ky2);
 r:=sqrt((xo+0.5*lx2)^(2)+(yo+0.5*ly2)^(2));
 tmp:=-(h*FR(r))/(r);
 kx3:=tmp*(xo+0.5*lx2);
 ky3:=tmp*(yo+0.5*ly2);
 lx4:=h*(vxo+kx3);
 ly4:=h*(vyo+ky3);
 r:=sqrt((xo+lx3)^(2)+(yo+ly3)^(2));
 tmp:=-(h*FR(r))/(r);
 kx4:=tmp*(xo+lx3);
 ky4:=tmp*(yo+ly3);  #edit - one more here
 x:=xo+(lx1+2*lx2+2*lx3+lx4)/(6);
 y:=yo+(ly1+2*ly2+2*ly3+ly4)/(6);
 vx:=vxo+(kx1+2*kx2+2*kx3+kx4)/(6);
 vy:=vyo+(ky1+2*ky2+2*ky3+ky4)/(6);
 [x,y,vx,vy]; end proc:
> 

 

 

Download proc.mw

Works...

dharr 953
October 17 2019
1 11

There is a missing *. The differentiation works in Maple 2017.
 

Download Diff.mw

 

 

a start...

dharr 953
October 02 2019
2 1
> 

restart

Case 1

> 

de1:=diff(U(x),x,x)=0;
bc1:=U(0)=10,U(L)=20;
ans1:=dsolve({de1,bc1},U(x));

diff(diff(U(x), x), x) = 0

U(0) = 10, U(L) = 20

U(x) = 10*x/L+10

Case 2

> 

de2:=diff(U(x),x,x)=-x^2;
bc2:=U(0)=T,D(U)(L)=0;
ans2:=dsolve({de2,bc2},U(x));

diff(diff(U(x), x), x) = -x^2

U(0) = T, (D(U))(L) = 0

U(x) = -(1/12)*x^4+(1/3)*L^3*x+T

> 

 

> 

 


 

Download rod.mw

use Vectors...

dharr 953
October 01 2019
0 2

Like this?

NULL

> 

restart
> 

n := 4; C := proc (K) options operator, arrow; Vector([seq(cos(2*Pi*K*l/n), l = 0 .. n-1)]) end proc; S := proc (K) options operator, arrow; Vector([seq(sin(2*Pi*K*l/n), l = 0 .. n-1)]) end proc

4

proc (K) options operator, arrow; Vector([seq(cos(2*Pi*K*l/n), l = 0 .. n-1)]) end proc

proc (K) options operator, arrow; Vector([seq(sin(2*Pi*K*l/n), l = 0 .. n-1)]) end proc

> 

C(2); S(3)

Vector[column]([[1], [-1], [1], [-1]])

Vector[column]([[0], [-1], [0], [1]])

These two are orthogonal

> 

C(2).S(3)

0

> 

C(2).C(2)

4

> 

 

NULL


 

Download orthogonal.mw

laplace and solve...

dharr 953
September 22 2019
1 0

I did G11; the others are similar

Transfer_function.mw

Document properties...

dharr 953
September 18 2019
2 1

In the File Menu under Document Properties, you should have an attribute named Active which is false for a regular help page and true for an active worksheet.

2-D math...

dharr 953
September 15 2019
2 0

in 1D input:

ode:=diff(y(x),x,x)=2*y(x)+1;
dsolve(ode);
 

does work, so this is probably incorrect entry of the derivative in 2-D math. [Edit: actually the error message confirms this.] Use the right-click menu 2-D math/convert to 1D math input and see if it looks like the above. If not, use 1-D input (ctrl-M) before the input, or use the expression palette for the derivative and use "^" to raise the powers of 2.

symmetry...

dharr 953
September 13 2019
0 0

I don't understand exactly what you mean - if you want real-only frequency domain, you need the time domain array to have the real part  symmetric about the centre and the imaginary part antisymmetric about the centre.

Edit: And here is a complex example that works pretty much as I would expect:

> 

restart;
> 

with(DiscreteTransforms):
> 

N:=2^10;df:=.2;dt:=1/df;T1:=500.;

1024

.2

5.000000000

500.

> 

tvec:=Vector(N,i->i*dt):
fvec:=Vector(N,i->i*df):
> 

fidr:=Vector(N,i->cos(2*Pi*df*i)*exp(-dt*i/T1)):
fidi:=Vector(N,i->sin(2*Pi*df*i)*exp(-dt*i/T1)):
fid:=fidr+I*fidi:
> 

plot(tvec,fidr);

> 

spec:=FourierTransform(fid):
> 

plot(fvec,map(abs,spec)); #magnitude

> 

plot(fvec,map(Im,spec));  #imaginary part

> 

plot(fvec,map(Re,spec));  #real part

> 

 


 

Download fid.mw

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