I am using Maple worksheets in a class. I have a student who finds it difficult to spend much time at the keyboard because of a joint disorder. I am looking for alternate input methods such as Dragon or the Dictate Facility on MacBooks which might help her out. Any suggestions would be appreciated. 

Thanks,

-Maury

f:=Intat(1.0000000000000000000*(1.7969454312181156991*_f^1.2+1.80)^1.2/sqrt(-1.4974545260150964159*(8.9847271560905784954*_f^3+14.640368911168931285*_f^2+30.220202497712627297)^1.2), _f = 0);

I tried to use  value(f);  eval(f); simplify(f); expand(f), but non provide an answer, but return an integral unevaluated.

Is there a command to produce a  numerical result ?

Can anyone tell me how to use dsolve to find the solution to the problem in the attachment.  It is faily easy to do using substitution for homogeneous coefficients, but dsolve seems to put out a very complicated solution to the problem.

Download DEprob.mw

I have 2 problem with my jacobian matrix:

first: i can not evaluate 11*11 jacobian matrix. at last i can evaluate 10*10 matrix. can i solve this?
second: i want to export my matrix for matlab but i see this error : {export matrix"cannot convert matrix element to float[8] data type"}
so how i can use this matrix in my matlab code?
 my jacobian matrix:

with(VectorCalculus); Jacobian([VectorCalculus:-`+`(VectorCalculus:-`+`(VectorCalculus:-`+`(VectorCalculus:-`*`(2.68, ex), VectorCalculus:-`-`(VectorCalculus:-`*`(2, vx))), VectorCalculus:-`*`(VectorCalculus:-`*`(VectorCalculus:-`*`(VectorCalculus:-`*`(VectorCalculus:-`*`(3.500000001, e^VectorCalculus:-`*`(1.666666667, sqrt(VectorCalculus:-`+`(VectorCalculus:-`-`(VectorCalculus:-`+`(sqrt(VectorCalculus:-`+`(rx^2, ry^2)), sqrt(VectorCalculus:-`+`(VectorCalculus:-`+`(rx, VectorCalculus:-`-`(VectorCalculus:-`*`(2, vb(ex))))^2, VectorCalculus:-`+`(ry, VectorCalculus:-`-`(VectorCalculus:-`*`(2, vb(ey))))^2)))^2), VectorCalculus:-`-`(VectorCalculus:-`*`(4, vb^2)))))), VectorCalculus:-`+`(sqrt(VectorCalculus:-`+`(rx^2, ry^2)), sqrt(VectorCalculus:-`+`(VectorCalculus:-`+`(rx, VectorCalculus:-`-`(VectorCalculus:-`*`(2, vb(ex))))^2, VectorCalculus:-`+`(ry, VectorCalculus:-`-`(VectorCalculus:-`*`(2, vb(ey))))^2)))), VectorCalculus:-`+`(VectorCalculus:-`*`(rx, 1/sqrt(VectorCalculus:-`+`(rx^2, ry^2))), VectorCalculus:-`*`(1/2, VectorCalculus:-`*`(VectorCalculus:-`+`(VectorCalculus:-`*`(2, rx), VectorCalculus:-`-`(VectorCalculus:-`*`(4, vb(ex)))), 1/sqrt(VectorCalculus:-`+`(VectorCalculus:-`+`(rx, VectorCalculus:-`-`(VectorCalculus:-`*`(2, vb(ex))))^2, VectorCalculus:-`+`(ry, VectorCalculus:-`-`(VectorCalculus:-`*`(2, vb(ey))))^2)))))), ln(e)), 1/sqrt(VectorCalculus:-`+`(VectorCalculus:-`-`(VectorCalculus:-`+`(sqrt(VectorCalculus:-`+`(rx^2, ry^2)), sqrt(VectorCalculus:-`+`(VectorCalculus:-`+`(rx, VectorCalculus:-`-`(VectorCalculus:-`*`(2, vb(ex))))^2, VectorCalculus:-`+`(ry, VectorCalculus:-`-`(VectorCalculus:-`*`(2, vb(ey))))^2)))^2), VectorCalculus:-`-`(VectorCalculus:-`*`(4, vb^2)))))), VectorCalculus:-`*`(VectorCalculus:-`*`(VectorCalculus:-`*`(VectorCalculus:-`*`(50.00000000, e^VectorCalculus:-`-`(VectorCalculus:-`*`(5.000000000, sqrt(VectorCalculus:-`+`(Rx^2, Ry^2))))), Rx), ln(e)), 1/sqrt(VectorCalculus:-`+`(Rx^2, Ry^2)))), VectorCalculus:-`+`(VectorCalculus:-`+`(VectorCalculus:-`+`(VectorCalculus:-`*`(2.68, ey), VectorCalculus:-`-`(VectorCalculus:-`*`(2, vy))), VectorCalculus:-`*`(VectorCalculus:-`*`(VectorCalculus:-`*`(VectorCalculus:-`*`(VectorCalculus:-`*`(3.500000001, e^VectorCalculus:-`*`(1.666666667, sqrt(VectorCalculus:-`+`(VectorCalculus:-`-`(VectorCalculus:-`+`(sqrt(VectorCalculus:-`+`(rx^2, ry^2)), sqrt(VectorCalculus:-`+`(VectorCalculus:-`+`(rx, VectorCalculus:-`-`(VectorCalculus:-`*`(2, vb(ex))))^2, VectorCalculus:-`+`(ry, VectorCalculus:-`-`(VectorCalculus:-`*`(2, vb(ey))))^2)))^2), VectorCalculus:-`-`(VectorCalculus:-`*`(4, vb^2)))))), VectorCalculus:-`+`(sqrt(VectorCalculus:-`+`(rx^2, ry^2)), sqrt(VectorCalculus:-`+`(VectorCalculus:-`+`(rx, VectorCalculus:-`-`(VectorCalculus:-`*`(2, vb(ex))))^2, VectorCalculus:-`+`(ry, VectorCalculus:-`-`(VectorCalculus:-`*`(2, vb(ey))))^2)))), VectorCalculus:-`+`(VectorCalculus:-`*`(ry, 1/sqrt(VectorCalculus:-`+`(rx^2, ry^2))), VectorCalculus:-`*`(1/2, VectorCalculus:-`*`(VectorCalculus:-`+`(VectorCalculus:-`*`(2, ry), VectorCalculus:-`-`(VectorCalculus:-`*`(4, vb(ey)))), 1/sqrt(VectorCalculus:-`+`(VectorCalculus:-`+`(rx, VectorCalculus:-`-`(VectorCalculus:-`*`(2, vb(ex))))^2, VectorCalculus:-`+`(ry, VectorCalculus:-`-`(VectorCalculus:-`*`(2, vb(ey))))^2)))))), ln(e)), 1/sqrt(VectorCalculus:-`+`(VectorCalculus:-`-`(VectorCalculus:-`+`(sqrt(VectorCalculus:-`+`(rx^2, ry^2)), sqrt(VectorCalculus:-`+`(VectorCalculus:-`+`(rx, VectorCalculus:-`-`(VectorCalculus:-`*`(2, vb(ex))))^2, VectorCalculus:-`+`(ry, VectorCalculus:-`-`(VectorCalculus:-`*`(2, vb(ey))))^2)))^2), VectorCalculus:-`-`(VectorCalculus:-`*`(4, vb^2)))))), VectorCalculus:-`*`(VectorCalculus:-`*`(VectorCalculus:-`*`(VectorCalculus:-`*`(50.00000000, e^VectorCalculus:-`-`(VectorCalculus:-`*`(5.000000000, sqrt(VectorCalculus:-`+`(Rx^2, Ry^2))))), Ry), ln(e)), 1/sqrt(VectorCalculus:-`+`(Rx^2, Ry^2)))), 1, 1, 1, 1, 1, 1, 1, 1, 1], [vx, vy, ex, ey, rx, ry, Ex, Ey, vb, Rx, Ry])

 Calculus problem S(n)=1^2+3^2+5^2+...+(2n-1)^2 ?

Hello everyone,

restart:with(plots):

xNData := ExcelTools:-Import("D:/a.xls"):

L := convert(xNData, listlist);

L := [[1.0, 0.75e-1], [2.0, .1], [3.0, .12], [4.0, .14], [5.0, .16], [6.0, .18], [7.0, .2], [8.0, .22], [9.0, .24], [10.0, .26]]

In the above L for each x we have N. I want to sub a single set of value [x,N] into THE equation NN to get a value for d and repeat the same process for each set of [x,N] to get d.  

NN := -N+(N0B-NB)*(erf((1/2)*x/(sqrt(t)*sqrt(d)))-sqrt(erf((1/2)*alpha/d)))/sqrt(erfc((1/2)*alpha/d))+NB;

alpha:=2*10^(-12):NB:=0.075:N0B:=0.2:t:=360000:

Cheers.

@Care Love, sorry for the trouble. I have edited the question.

a.xls

Hello,

restart:

N := N0-(1/2)*sqrt(2)*sqrt(Pi*Kc/d)*(sum(erfc((1/2)*(L*n+x)/sqrt(d*t))+erfc((1/2)*((n+1)*L-x)/sqrt(d*t)),

n = 0 .. infinity));

N0:=0.2:L:=0.25:Kc:=2*10^(-12):t:=360000:d:=2.010619298*10^(-10):

When I plot N vs x = 0..0.25 then there is no issue

plot(N,x=0..0.25,axes=box);

but when try to use a loop to get the data, Maple cannot evaluate

for x from 0 to 0.25 by 0.01 do

N[x]:=evalf(N0-sqrt(Pi*Kc/(2*d))*sum(((((erfc((n*L+x)/(2*sqrt(d*t))))+erfc(((n+1)*L-x)/(2*sqrt(d*t))))),n=0..infinity)));
end do;

Thanks

thanks

THE FOLLOWING CODE 

restart;

A:=Matrix([[ a , b ], [ c , d ]]);

a:=1; b:=0; c:=0; d:=1;

A; 

produces differents results under MAPLE 16  linux i386 and under MAPLE 16  linux amd64

in the first case the last evalution has the following printed output:

on the second machine the printed output is

Does anybody has an explication; I thought that the "coorect behaviour was the first one since tables use last-name evalutation. But now I am puzzled.

how to decompose a matrix into time invariant and time variant 

is it possible to make time invariant and time variant template and then decompose into it

i mean decomposition can be 

 time invariant matrix + time variant matrix

or

 time invariant matrix * time variant matrix

 dsolve([Diff(f, t) = f, Diff(f,t) + g = h], f);

 dsolve([Diff(f, t) = f, Diff(f,t)*g = h], f);

where h is orthogonal matrix, f,g,h are matrix

would like to find g and f from h

can dsolve solve differential equation of matrix ? how?

dsolve([Diff(f(t), t) = f(t), Diff(f(t),t) + g(t) = h1(x)*h2(x), int(h1(x)*h2(x),x=-1..1) = 0], [f(t),g(t),h1(x),h2(x)]);

dsolve([Diff(f(t), t) = f(t), Diff(f(t),t)*g(t) = h1(x)*h2(x), int(h1(x)*h2(x),x=-1..1) = 0], [f(t),g(t),h1(x),h2(x)]);

assume x^2 + 1 is from interpolation of polynomial

pdsolve([Diff(f(t), t) = f(t), Diff(f(t),t) + g(t) = h1(x,t)*h2(x,t), h1(x,t)*h2(x,t)= x^2+1], [f(t),g(t),h1(x,t),h2(x,t)]);
pdsolve([Diff(f(t), t) = f(t), Diff(f(t),t)*g(t) = h1(x,t)*h2(x,t), h1(x,t)*h2(x,t) = x^2+1], [f(t),g(t),h1(x,t),h2(x,t)]);

these system can not be solved

hope no real number any more after decomposition and only have iinteger in I time invariant function

The matrix:

<3,-2,-1,2,0>;

<11,4,-8,2,7>;

<0,0,2,0,0>;

<3,3,-4,3,3>;

<-8,4,5,-4,-1>;

has eigenvector:

<2,0,-1,0,1>

Find its corresponding eigenvalue.

(Hint: you don't need to find all the eigenvalues and eigenvectors to answer this question.)

Steps and the solution will be greatly appreciated. thanks!

number10:=`466d06ece998b7a2fb1d464fed2ced7641ddaa3cc31c9941cf110abbf409ed39598005b3399ccfafb61d0315fca0a314be138a9f32503bedac8067f03adbf3575c3b8edc9ba7f537530541ab0f9f3cd04ff50d66f1d559ba520e89a2cb2a83`:

number8:=`315c4eeaa8b5f8bffd11155ea506b56041c6a00c8a08854dd21a4bbde54ce56801d943ba708b8a3574f40c00fff9e00fa1439fd0654327a3bfc860b92f89ee04132ecb9298f5fd2d5e4b45e40ecc3b9d59e9417df7c

I first define

f:=x->convert(x, decimal, hex):

with(Bits):
str1:=convert( `Xor(f(number8), f(number10))`, bytes);

now how can I get back the alphabets, since again use of convert with bytes return the inital argument.

Moreover, I would really appreciate if someone could explain the difference between 

convert(`expr`, bytes)

convert( [expr], bytes)

Many regards!!

if kernel is solve(A*x, x);

then , what is cokernel of a numeric matrix ?

I cannot show that the following two sums(A and B) are equal to each other.

How can I simplify the difference(A-B)?

A := sum((-1)^(i-1)*factorial(n0)/(factorial(n1)*(n1+i)^2*factorial(i-1)*factorial(n0-n1-i)), i = 1 .. n0-n1);

B := sum(1/(n1+i), i = 1 .. n0-n1);

`assuming`([simplify(A-B)], [n0::nonnegint, n1::nonnegint, n1 <= n0]);

does not give zero. 

The result of a simple test: 

map(simplify, [seq(eval(A-B, n1 = 10), n0 = 20 .. 30)]);
print(`output redirected...`); 

is

[0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0]

I am trying to get a solution to the heat equation with multiple boundary conditions.

Most of them work but I am having trouble with two things: a Robin boundary condition and initial conditions.

First, here are my equations that work:

returns a solution (actually two including u(x,y,z,t)=0).

However, when I try to add:

or

I no longer get a solution.

Any guidance would be appreciated.

Regards.

I have uploaded a worksheet with the equations...

Download heat_equation_pde.mw