See help page on map by typing "?map;",specifically read the definition of "map".

You could get (x+y)^2 by

f:=(t)->t^2; then

f(x+y);

Just change the order of the dot product

ans:=Bra(b,0).Ket(b,0).Ket(a,0).Ket(c,1).Ket(d,1).

Otherwise you need a predifined relation between Bra(b,0) & Ket(a,0).You just can't force Maple to violet the laws of dot product in Quantum Mechanics.Here a & b are not the same basis.

 Well,I agree with Dave,you should try yourself first by taking help from the help pages,but still I'm going to give you some hints:

for part 2(b):

with(Student[Calculus1]):

expr:=your expression here,what is it : x^3-4*x^2-1 or, x^3-4*x^3-1=-3*x^3-1 ?  both have only one soln.

assume(n::posint)

f:=(n)->NewtonsMethod(expr,x=5,iterations=n,output=sequence)

now you can call  f for any value of n=1,2,....your wish & test them by

subs(x=the resulting values(one at a time) from calling f,expr);

I already gave you the solution of the of the previous problem,I thought you should solve this problem using the ideas of the

previous solution.I already told you that deceleration is constant,it doesn't depend on time,so it will be a straight line parallel to the

x-axis(here t-axis).

Why you are so worried about infinity?Your time isn't infinite,so the graph can't be infinite,you have to put the time interval from 0 to the time taken by the car to stop.

These are the hints I'm giving you now,try harder,it's very,very,very easy!This time I'm not giving the solution soon.

If you can't solve it after 1 or 2(maximum)  hours trying,then let me know,I will send the solution then. 

u:=3;

a:=9.81;

t:=2;

solve(v=u+a*t,v);

or,solve(v=u+a*t);

or,fsolve(v=u+a*t,v);          Any 1 of these would do.

or,fsolve(v=u+a*t);

or,solve({v=u+a*t},v);

or,solve({v=u+a*t});

or,fsolve({v=u+a*t},v);

or,fsolve({v=u+a*t});     In the last 4 square bracket instead of curly would do.

Alternatively,

v:=(u,a,t)->u+a*t;

v(3,9.81,2);

Open this   Ans_23.mw

Open this worksheet in standard mode,not in classical mode.

ANS.mw

Are you familier with any numerical root-finding method like bisection,secant,iteration or Newton-Rapson method?

Maple uses for fsolve() something like those,and those methods give one real root in the given interval(provided,at least one real root lies in the interval).So,in your 1st provided interval the root converges with the 1st answer,and for the 2nd with 2nd answer.By default it gives the 1st answer.

you can get help from:

?fsolve;

?solve;

?isolate;

?Isolate;

?RootFinding;

PS:Since I'm no expert,may be it is possible  to find them both at the same time,I don't know,what I've said,I said from my little experiance from my undergrad numerical-analysis course.

Could you plz explain your formulas more clearly?What are these?Euler method,improved Eular method?RK4?,or something else?

Here is a worksheet that solve your DE using maple DE solver.If you using maple you don't have to write your own procedure to solve it(this is the beauty of a CAS),though you can write it,if it is mandatory.Actually you use these formulas in C/C++,Fortran,BASIC ect.

DEsolve.mw

It seems to me that your equation is a transcendental one,i.e. it has no analytic solution,so you have to solve this numerically,instead of solve(),you use fsolve(),also you have to give the numerical values of h and b.Here is an example:

                                    h:=3;

                                    b:=5;

                     eqn:=h*sin(mu*b)+mu*cos(mu*b)=0; ..................................................(1)

                     fsolve(eqn,mu=0..5);  

Your answer should be 2.37917......You can change the mu interval,h and b values according to your need.

There is a second way to solve this kind of equations,plotting.Use the following commands:

                              f:=(mu)->h*sin(mu*b);

                              g:=(mu)->-mu*cos(mu*b); //see eqn(1) for the reason for - sign.

                              plot({f(mu),g(mu)},mu=-5..5);

From the plot you can find(actually approximate) the intersections of 2 curves,hence the root.

A equivalent way of doing this is explicitly evaluate the 2 functions for the same mu values,compare them and findout those mu values for which 2 functions are the same.