Rouben Rostamian

MaplePrimes Activity


These are replies submitted by Rouben Rostamian

@spalinowy

restart;

with(Student[MultivariateCalculus]):

Let's say you want to linearize a function F(x, y) about the point a, b.
Here is the equivalent of what you do in your worksheet:

TaylorApproximation(F(x,y), [x,y]=[a,b], 0);

F(a, b)

But that only gives you the value of the function F at the point a, b.

That's not what you mean by linearization, do you?

What you should be doing is:

TaylorApproximation(F(x,y), [x,y]=[a,b], 1);

F(a, b)+(-a+x)*(D[1](F))(a, b)+(-b+y)*(D[2](F))(a, b)

In this expression a, "b," F(a, b), (D__1(F))(a, b), and (D__2(F))(a, b) are

constants, while x and y are variables.  We may express the result equivalently,

but more neatly, as

A + B*(x-a) + C*(y-b);

A+B*(-a+x)+C*(-b+y)

or better yet, letting u = x-a and v = y-b, that takes the form

A + B*u + C*v;

B*u+C*v+A

There you have it.  This is what you may call the linearization of F.

 

You write: "For me linear equation is when I don't have sin(x)

or cos(x) in all equation".

 

That's not true.  The coefficients A, B, and C may have sine or cosine,

or all sorts of other things in them, as long as they don't involve the

linearization variables u and "v.  "For example, B may be cos(ln(a)).

There is nothing wrong with that.

 

Now look at the result shown at the end of my calculations, that is:
"sin(phi2) lk m1 ((ⅆ)^2)/(ⅆt^2) (sin(`psi__0`(t)) u(t))."

Note that this is linear in the linearization variable "u(t)."  The coefficient
sin(`ψ__0`(t)) has no u(t) in it, so it doesn't hurt.

 

That said, you need to know that Maple's TaylorApproximation

is meant to calculate the Taylor series of a function of several variables, such

as F(a, b).  Your expression, however, is not of the form F(a, b).  Rather,

due to the presence of terms involving the derivatives of psi1(t), it is a

differential operator.  Linearization of differential operators is studied in courses

beyond calculus.  Maple's TaylorApproximation is not the right tool for it.

I have already shown how to do that calculation properly, and vv has presented

a compact summary.

 

 

@spalinowy Where do you see a nonlinearity?  The expression
               sin(phi2)*lk*m1*(Diff(sin(psi__0(t))*u(t), t, t))
is linear in u(t).
That is what is meant by linearization.

As Edgardo has pointed out, your initial value problem is ill-posed in

that it has no unique solutions.  As a matter of fact, it has infinitely

many solutions given by

y(x) = piecewise(x < c, 0, 16.1*(x-c)^2);

y(x) = piecewise(x < c, 0, 16.1*(x-c)^2)

where c >= 0 is an arbitrary constant.  I'd be surprised if a computer

algebra system could discover that fact.

@MargoMaples You say xn+1 =xn- (f(xn)*d) / (f(xn+d)-f(xn)) and "d is a really small number".

Your problem of two intersecting circles leads to a nonlinear system of two equations in two unknowns. Consequently, the variable xn in that iteration scheme would be a point in R2.

This raises the question: What do you mean by xn+d where xn is in R2 and d is a number?

Do you see the problem?  You cannot add a point and a number!

That's why I am asking for either a precise description of iteration scheme that you wish to apply, or an easily accessible reference where that description can be found.

Or... ask the person who gave you this problem for clarification.

 

@MargoMaples What is d?

It will help if you either describe precisely the algorithm that you have in mind, or refer to a web page that describes it.  Offering various permutations of the words Newton, Raphson, Modified, Secant is not a good substitute.

@spalinowy I have already shown how to linearize about any ψ0(t).  In particular we see that if ψ0(t)≡ 0, then the linearized expression is just zero. This is analogous to linearizing the function f(x) = x*sin(x) at x=0 — you just get zero.

 

I know the meanings of the "Newton method", "modified Newton method", "secant method", but I don't know "modified Newton secant method".  What is it?  References?

 

@Joe Riel And here is a cruder way:

 y := 2*4^n;  

2*4^n

simplify(log[2](y)) assuming n>0;

2*n+1

2^%;

2^(2*n+1)

@brian bovril You essentially have the answer in the first four lines of your spline.mw.  Just change the first line from Y := [3,10,2,7,7] to Y := [3,10,2,7,7,42]. That is, the answer "42" is not arrived at by accident; it is built in!

The result may be cast into Horner's form through convert(f, horner).

 

@rallezet I can't make sense of what you are asking. Don't be afraid of using more words to explain what you have in mind.

I think you had better explain more clearly what it is that you are asking.

@Samir Khan I tried displaying your globe on 4 different computers, all running 64-bit Linux:
1. Thinkpad X301
2. Thinkpad X301
3. Thinkpad X1
4. ThinkStation C30

The X1 and C30 display the globe  perfectly.   The two identical X301 machines both display the globe with the artifacts that I have shown.  I suppose that this is due to the limitations of the graphics cards in X301.

What I really like about your software is how responsive the globe is to the mouse motion—there is no resistance to the motion at all.  In contrast, the globe produced by the previously noted plot3d() method with the image=… option is somewhat difficult to rotate; it appears that the globe is recalculated/repainted with each incremental rotation.  So all else being equal, I prefer your globe. 

 

@acer That code snippet does exactly what I wanted. It works particularly well with NASA's Earth images which are referenced in the old post that you cited.  Thanks!

@Samir Khan Thank you very much for the sample. It's a very good application.   The 3D globe, however, exhibits some strange artifacts.  I can't tell whether it is a software or hardware problem.  Here is what I see in the north Africa part of the globe:

This is with Maple 2017.3 on 64 bit Ubuntu Linux.

 

  1. In a couple of places you have a^().  What is that meant to be?
  2. You have Omega:=(0..376).  But later you do Omega[i]:=i*step which overwrites the previous definition in Omega.  Do you really mean that?

I have not looked at the rest of the code.

 

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