MaplePrimes Questions

Hello there, 

Would you please have a look at this question?

I got the answer in the 'eq_4_35', but what I wanted to see is the 'desired' expression. 

Somehow, the term '3*C' doesn't want to go away from the 'x__0' term, thus I wonder there is a way to cancel the term in the expression. 


 

restart;

eq_4_34 := x__alpha = C*x__a - C/2 * (x__b + x__c);

x__alpha = C*x__a-(1/2)*C*(x__b+x__c)

(1)

eq_4_33 := x__0 = 1/3*(x__a + x__b + x__c);

x__0 = (1/3)*x__a+(1/3)*x__b+(1/3)*x__c

(2)

eq_4_33aux := X__sub = (x__b + x__c);

X__sub = x__b+x__c

(3)

attempt2 := algsubs((x__b + x__c) = X__sub, eq_4_33);

x__0 = (1/3)*x__a+(1/3)*X__sub

(4)

attempt3 := X__sub = solve(attempt2, X__sub);

X__sub = -x__a+3*x__0

(5)

eq_4_34x := x__alpha = algsubs((x__b + x__c) = 3*x__0 - x__a, rhs(eq_4_34));

x__alpha = -(1/2)*C*(-x__a+3*x__0)+C*x__a

(6)

eq_4_35 := x__a = simplify(solve(eq_4_34x, x__a));

x__a = (1/3)*(3*C*x__0+2*x__alpha)/C

(7)

desired:= x__a = x__0 + (2*x__alpha)/(3*C);

x__a = x__0+(2/3)*x__alpha/C

(8)

 


 

Download Q20201005_2.mw

Hi

I have the following of fractional of ode system.

.How to solve it by maple.

Just curious about an output in an if statement.

Why do I need to use Print() in an if statement, when I don't need it otherwise? Why doesn't "hei" show in the output of the attached example?

Printinloop.mw

Charging a capacitor with a DC is well known and interaction with a sinusoidal signal is known to perfection.

The mathematical solutions behind an arbitrary waveform V1(t) is hard or impossible? to find .

To simplify, the arbitrary wave can be expressed with a 2-d degree polynom. This gives a fairly simple

expression with an integral. The problem however is that V2(t) - the capacitor voltage - appears on both sides and maple seems unable to resolv this into a symbolic expression.

If this can be solved it would be great ! The solution should be of the symbolic form V2(a,b,c,V2start,t1,t2,RC) and give the answer what voltage the capacitor V2 would reach arter charging between t1 and t2  with a V1(t) that is changing during charge.

Professor Robert J. Lopez is a Maple Fellow and the author of many Maple webinars and applications. He is also co-author of a book entitled Discovering Calculus With Maple, which I have recently acquired, and which turns out to be keyed to Maple V Release 3 on a Mac. Well, I have Maple 2020 on a Mac. What is the best way to get this book to work for me?  Thanks very much.

How do I plot the complex function |z+1||z-1|=1 in Maple? I know there is complexplot but don't quite understand how to use it.

I'm trying to readin a XML file, where some figures are attached a unit.

Here's part of the XML file.

When doing the readin and assigning the values to  a variable with AttributeValue, the type of the variable is a string.

How can I convert this string value ("10.*Units:-Unit(m)") to a number attached the defined units?

hello,

My equations are not getting solved.

Please tell me, is there any other command to solve these type of equations.

Q1.mw

 

 

Hi,

How to keep fractions and parentheses in the expression, and not calculate the product ?

Thanks

QuestionFraction.mw

For example this fraction?

 

Why Maple gives different looking solution when calling  

dsolve(sys);

vs.

dsolve(sys,deps);

? Both solution are actually correct. But look different.  Here is an example

restart;
sys:=[
diff(x(t),t) = 2*x(t)-z(t), 
diff(y(t),t) = 2*y(t)+z(t), 
diff(z(t),t)=2*z(t),
diff(w(t),t)=-z(t)+2*w(t)];
deps:=[x(t),y(t),z(t),w(t)];

dsolve(sys);

dsolve(sys,deps);

Notice the difference:

In first case, the x(t) and y(t) solutions have 2 constants of integrations, and in the second case, they have 3 constants of integration.

If we solve this using the matrix exponential method, the solution comes out to match the first one:

A:=Matrix([[2,0,-1,0],[0,2,1,0],[0,0,2,0],[0,0,-1,2]]):
sol:=Vector([x(t),y(t),z(t),w(t)])=LinearAlgebra:-MatrixExponential(A,t).Vector([_C1,_C2,_C3,_C4]);
simplify(sol)

Again, both solution verify to 0, using odetest, and I assume now it is correct (have no reason to think odetest is not correct).

But I'd like to understand more why when passing the unknowns to dsolve, the solution comes out different looking (3 constants of integrations, vs. two in this case). Should the solution be the same looking in both cases? 

This came out, since I thought my solution was wrong, since it did not look like Maple's. 

 

Maple 2020.1

 

In a recent question, the solution involved [...]::[list$3].

H:= proc(n, x, y)
    if n=0 then y+1
    elif n=1 and y=0 then x
    elif n=2 and y=0 then 0
    elif y=0 then 1
    elif not [n,x,y]::[nonnegint$3] then 
        'procname'(n-1, x, 'procname'(n, x, y-1))
    else
        thisproc(n-1, x, thisproc(n, x, y-1))
    fi
end proc

Where is  the  form [...]::[list$3] documented?

 

 

Please refer to this for the definition(s):

http://www.mrob.com/pub/math/largenum-3.html#hyper

The hy(a,n,b) recursion turns  into the following Maple code:

 

Unfortunately this code depends on x and y being explicit numeric values (x,y \in N), otherwise the recursion crashes (if I ask for example H(2,x,y) or H(2, x, 2))

Is there any way to transform the code so that the final construct can be shown symbolically?

I don't see anything obvious, especially since if y is not a specific natural number the recursion will crash. Can we maybe force Maple to not evaluate it for natural number arguments and return the final construct as either a sum, product or tower of the symbols for x and y (i.e. symbols of the digits of x and y like a tower of 2's 3's, etc)?

I do remember something about putting primes around functions prevents premature evaluation, but in this case it doesn't do anything like I'd want. This definition is not primitive recursive (like that of the Ackerman function), so I don't expect it to be able to be called abstractly (H(2,x,y) or something else such), but maybe we can turn it into something that shows the structure of the final construct as symbols of the digits of x and y?

Thanks.

Yannis

PS: One can improve the situation a bit, by implementing the extended definition as:

hy := proc (n, x, y) if n = 0 then y+1 elif n = 1 then x+y else if y = 1 then x else if n = 2 then x*y elif n = 3 then x^y elif n = 4 then x^hy(4, x, y-1) else hy(n-1, x, hy(n, x, y-1)) end if end if end if end proc

(which can be called abstractly for n=0,1,2,3,4 (hy(n,x,m) for natural m), but for higher n it crashes similar to H, since the definition falls back to the previous if n>4.

How to use spacecurve with two variables and how to use display? 

Thanks!!

I want to make a procedure that takes a function as an input, but everything I've tried has turned the function into a string.
This is what I want to do.

newtons := proc( f::function, x, n::integer)

But when I try and use the the function f that I defined, it just gets treated like a string or something. I can't put variables into it, I cant evaluate anything from it.

How do I get proc to see f as a function?

There are some answers up here, but they are for given known functions.
I need to place parentheses around functions that are automatically created.
For example, the function $f(x)$ takes many polynomial values during the execution
of a procedure. For examplle, it takes the values, 2x-1, x^2+3x+2, x^3-2x-3, etc.
How do I automatically place parentheses around these polynomials?
Thank you!
mapleatha

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