Hi all

Assume that we have folowings:

Assume that we devide [0,1] to N subintervals and in each subinterval we have:

Also we want to approximate arbitrary function x(t) with following manner:

How can we produce these basic polynomials?

Thanks in advance

Mahmood   Dadkhah

Ph.D Candidate

Applied Mathematics Department

Hello,

I would like to create a function from an equation.
I have define 4 functions u[1](t), ..., u[4](t).
From these 4 functions, I would like to define 4 equations. I would like to obtain this result:
Equ[1]:=u[1](t)=L/2*cos(w*t+phi[1])
Equ[2]:=u[2](t)=L/2*cos(w*t+phi[2])
Equ[3]:=u[3](t)=L/2*cos(w*t+phi[3])
Equ[4]:=u[4](t)=L/2*cos(w*t+phi[4])

My code is the following:

for i to 4
do
u[i]:=unapply(L/2*cos(w*t+phi[i]),t);
Equ[i]:='u[i](t)'=u[i](t);
end do;

These lines are not working because the left member is not incremented (u[i](t) stays at each iteration u[i](t))

Generally, how can I transform a function to an equation ?
And, in this specific case, how can I obtain the four equations mentioned above ?

Thanks a lot for your help.

Hello,

I would like to plot a vertical line each time my function is null and is increasing.

Here an extract of my code:

v:=unapply(H*sin(w*t),t);
L:=0.080;
H:=0.020;
Vf:=0.3;
w:=10;
fsolve(v(t)=0,t=0.5..2);
zeros:=Roots(v(t),t=0.5..2);

vdot:=unapply(diff(v(t),t),t);
map(vdot,zeros);
LignesVerticales=implicitplot(x=To complete, colour=yellow,linestyle=3, thickness=2):

With Roots function, I could obtain the zero-crossings.

With the derative of the functions, I could obtain a vector which gives me when the derivative is positive or not.

I would like now to obtain the list of the zero-crossing values which are increasing but I have difficulties to obtain it. 

By list manipulation, it should be easy to do. 

May you help to plot obtain this list so that I can plot the desired vertical lines?

Thank you for your help

Hello,

I would like to obtain all the solutions of a equation.

Here an extract of my code:

v:=unapply(H*sin(w*t),t);
L:=0.080;
H:=0.020;
Vf:=0.3;
w:=10;
fsolve(v(t)=0,t=0.5..2);

How can I do to obtain all the solutions of the equation in the wanted interval ?

I'm not fixed to use fsolve function.

Thank you for your help

Hello
I need to plot this region :

 Can anyone explain me how to use the next feature that you can find in ?dsolve,numeric,events,Round-off and simple triggers or refer me to a previous answer that explain this:

   This is primarily desired to be able to apply events for an ODE system that has been separated into disjoint cases dependent on the values of particular triggers (in which case you always want to use a form that provides the values just past the trigger point).

 Specially how to separate the ODE system into disjoint cases. Thanks in advanced.

Hello all,

I am trying to create a matrix throgh a procedure but unfortunately I could not find how to define an empty matrix in Maple. This is how my procedure looks like,

Initial_Matrix := proc (BC::list, n, BP::list)
local M::Matrix, i;
M:=[][];
for i from 1 to numelems(BC) do
M:= <M,function_coeffs(BC[i], n, BP)>;
end do;

where the procedure function_coeffs returns a row vector. Logically, it should keep adding rows function_coeffs(BC[i], n, BP) to the emty matrix, but its not happening. Please help me out.

Thank you for your time.

Is there a help page which explains why braces provide the partial text evaluation in this code?

RopeLen := 30;RopeAddLen := .5;

plots[textplot]([1, 1, typeset("%1", (({(1/2)*RopeLen}+{(1/2)*RopeAddLen})^2-{(1/2)*RopeLen}^2)^(1/2))])

I have the following equation:

Diff(W(t), t) = -q*V*(sin(Phi)-sin(Psi[s]))/(2*h*Pi);

I solve it for rhs() = 0:

soln := solve([rhs((2)) = 0, Phi < 2*Pi], [Phi], allsolutions = true,explicit);

This works and I get this result:

Now I want to get the first zeros, which occur for _Z1 and _Z2 equal to 0. So I substitute:

subs(_Z2=0,(3));

and get

In other words, the substitution did not work.

The original problem is embedded in a larger sheet created with Maple 15 and there it does work. It fails on Maple 2015.2. I then pulled out the relevant pieces to make this example demonstrating the problem (see the attached sheet, which has some of my other (unsuccessful) attempts to diagnose what is going on). It seems like the created variables _Z1 and _Z2 are somehow not recognized at all.

The only way I can get the _Z2 terms out is to substitute 2=0. This is really too icky to seriously consider, though.

Anyone has seen this before?

FWIW: Maple 2015.2 on Mac OS X 10.10.5.

Thanks,

Mac Dude

Assumptions_test.mw

This is not actually a question, but an interesting problem found in the recent book (2nd edition, 2015):
Mathematica®: A Problem-Centered Approach
by  Hazrat Roozbeh
Springer

I hope that you will enjoy the problem too.

Define the functions f, h : N --> N by
f(n) = the sum of the squares of the digits of n; e.g. f(25) = 2^2 + 5^2 = 29.
h(n) = min {f(n), f(f(n)), f(f(f(n))), ... };  e.g. h(7)  = min{49,97,130,10,1} = 1.

A natural number n is happy if h(n) = 1.
Find all the happy ages, i.e., happy numbers up to 100.
Conclude that happy ages are mostly before one gets a job or after retirement!

Hello,

I would like to plot coupled elliptic parametric curves.

I have a problem in the for loop in my code. The index i of the phi variable is not incremented.

Here my code:

for i to 4
do
u[i]:=t->L/2*cos(w*t);
v[i]:=t->H*sin(w*t+phi[i]);
end do;

The result looks like this :

u[1](t) -> - L cos(w t)
v[1](t) -> H sin(w t + phi[i])

u[2](t) -> - L cos(w t)
v[2](t) -> H sin(w t + phi[i])

u[3](t) -> - L cos(w t)
u[3](t) -> H sin(w t + phi[i])

u[4](t)  -> - L cos(w t)
u[4](t)  -> H sin(w t + phi[i])

And i would have prefered :

u[1](t) -> - L cos(w t)
v[1](t) -> H sin(w t + phi[1])

u[2](t) -> - L cos(w t)
v[2](t) -> H sin(w t + phi[2])

u[3](t) -> - L cos(w t)
v[3](t) -> H sin(w t + phi[3])

u[4](t)  -> - L cos(w t)
v[4](t)  -> H sin(w t + phi[4])

Do you have ideas how I can obtain iteration on my phi variable ?

Thank you for your help.

Can some one help me " How to conert a three or four coupled equation by using non dimensions variable "

Hi,

I have an issue calculating an electronics circuit with Maple, using units. I have a current source that I know, and I want to determin the voltage in a capacitor by solving an ODE (except that the current source is defined piecewise). And to make sure I have all the units and scales right, I use the standard unit package. All my variables have their units defined.

Except that Maple doesn't want to solve the equation. It seems to me that it assumes that the function I am trying to solve is unitless, and therefore refuses to solve. 

V__out := 3*Unit('kV');

C__out := 2*Unit('nF');
R__blead := 520*Unit('`k&Omega;`');

I__fly := proc (t) options operator, arrow; Unit('A')*piecewise(t < 3.25*Unit('us'), (1+(-1)*t/(3.25*Unit('us')))*.2, 0) end proc;

dsolve({I__fly(t*Unit('s'))-V__C(t*Unit('s'))/R__blead = C__out*(diff(V__C(t*Unit('s')), t)), V__C(0*Unit('s')) = V__out}, V__C(t*Unit('s')));
Error, (in Units:-Standard:-+) the units `A` and `S` have incompatible dimensions

Is there a way to make Maple assume the unit of what it's trying to solve ? I need it to understands that V__C is in Unit('V') ...

Thanks

Hi,

what i have to calculate in Maple for q=Mises-Stress and the third invariant of deviatoric stress as r???

It is for:

Link:
http://www.egr.msu.edu/software/abaqus/Documentation/docs/v6.7/books/stm/default.htm?startat=ch04s04ath116.html

Suppose we are given a matrix M[n*2n] of n linearly independent row vectors. Then I am trying to create a procedure to add n more linearly independent row vectors to this matix resulting in to a matrix M[2n*2n].

Consider this easy example, if the given matrix $M_{2*4}$ is

 
then we can add $[0,0,0,1]$ and $[0,0,1,0]$ to obtain the matrix $M_{4*4}$

Can I use rref form?Thank you for your help.