Hello everyone,

I have an issue that I don't know how to solve.
I would like to plot a part of a surface that is enclosed by another surface. I wrote a proc() function with an if statement and when the statement is statisfienied I returned desired function. The roblem is that if statement gets ploted as well..

Here is my proc()

yield := proc (sigma__x, sigma__y, tau__xy, sigma__cx, sigma__cy, f__45, f__cx, f__cy, tau__u, f__tx, f__ty, alpha)

if eval(evalf(f__r(sigma__x, sigma__y, tau__xy, f__tx, f__ty, alpha)) < 0) then

return f__h(sigma__x, sigma__y, tau__xy, sigma__cx, sigma__cy, f__45, f__cx, f__cy, tau__u)

else 9999999

end if

end proc;

Here is my plot request:

Explore(implicitplot3d('yield(sigma__x, sigma__y, tau__xy, sigma__cx, sigma__cy, f__45, f__cx, f__cy, tau__u, f__tx, f__ty, alpha)' = 0, sigma__x = -10 .. 10, sigma__y = -10 .. 10, tau__xy = 0 .. 10, style = surfacecontour, numpoints = 100000, axes = normal), parameters = [f__cx = 0.1e-3 .. 10, f__cy = 0.1e-3 .. 10, f__45 = 0.1e-3 .. 10, sigma__cx = 0.1e-3 .. 10, sigma__cy = 0.1e-3 .. 10, tau__u = 0.1e-3 .. 10, f__tx = 0.1e-3 .. 10, f__ty = 0.1e-3 .. 10, alpha = 0.1e-3 .. 10]

 and this is what i get:

The grainy surface should not be there. If it's not possible to remove it, it would be alright that surface would be smooth. Any ideas?

Maple evaluates:

but when I ask:

Maple will verify sin(u)/cos(u)=tan(u) but beyond that it does not seem to work too well. I suspect this has something to do with verify being a "semi-boolean" function but searching the docs for what that mean returns no results. It may mean "I am unable to decidee". But how should I know? :-(

This is Maple 2016.1 on Windows. There is a post on verifying very simple identities.

Edit As usual some very helpful answers for which many thanks. It is poor that Maple does not document the meaning of FAIL.It seems that for trig identities the expandverfier is the best.

Edit Maple documents FAIL under ? FAIL. I am not sure how I got a blank page when I searched.

Dear Community,

Is there an easy way to convert elapsed days since January 1 1900 to date, i.e. to year, month, day in Maple? E.g. 23021 days should convert to 1963.01.10. Also what is the opposite, i.e. from date to elapsed days?

tx in advance,

best regards

Andras

Hello Everyone

I have an expression which I would like to integrate from x=0 to x=L. The expression is 

x1:=(sin(beta*x)*cos(m*Pi*x/L))/(1+alpha*x)

Here, beta, m, alpha are constant. However I want the result in terms of these quantities.

I will be grateful if you could help mw in this regard.

Thanks a lot.

Ex: Give: Sum = 16;

Result: [1,3,5,7]

Dear All,

I am going to solve the following systems of ODEs but get the error: Newton iteration is not converging.
Could you please share your idea with me. In the case of AA=-0.2,0,0.2,0.4,...; I could get the solution.
Thank you in advance.

restart;
with(plots);
Pr := 2; Le := 2; nn := 2; Nb := .1; Nt := .1; QQ := .1; SS := .1; BB := .1; CC := .1; Ec := .1; MM := .2;AA:=-0.4;

Eq1 := diff(f(eta), `$`(eta, 3))+f(eta).(diff(f(eta), `$`(eta, 2)))-2.*nn/(nn+1).((diff(f(eta), eta))^2)-MM.(diff(f(eta), eta)) = 0; Eq2 := 1/Pr.(diff(theta(eta), `$`(eta, 2)))+f(eta).(diff(theta(eta), eta))-4.*nn/(nn+1).(diff(f(eta), eta)).theta(eta)+Nb.(diff(theta(eta), eta)).(diff(h(eta), eta))+Nt.((diff(theta(eta), eta))^2)+Ec.((diff(f(eta), `$`(eta, 2)))^2)-QQ.theta(eta) = 0;
Eq3 := diff(h(eta), `$`(eta, 2))+Le.f(eta).(diff(h(eta), eta))+Nt/Nb.(diff(theta(eta), `$`(eta, 2))) = 0;

bcs := f(0) = SS, (D(f))(0) = 1+AA.((D@@2)(f))(0), theta(0) = 1+BB.(D(theta))(0), phi(0) = 1+CC.(D(phi))(0), (D(f))(etainf) = 0, theta(etainf) = 0, phi(etainf) = 0

Error, (in dsolve/numeric/ComputeSolution) Newton iteration is not converging

Ibragimova Evelina, 6 class,
school № 57, Kazan

The manual with examples
( templates for the solution of )

The solution of problems on simple interest

> restart:
> with(finance);

[amortization, annuity, blackscholes, cashflows, effectiverate,

futurevalue, growingannuity, growingperpetuity, levelcoupon,

perpetuity, presentvalue, yieldtomaturity]

Team futurevalue (the first installment, rate, period) - the total calculation for a given down payment, interest rate, payments and number of periods.

Example 1. To the Bank account, the income of which is 15% per annum, has made 24 thousand rubles. How many thousands of rubles will be in this account after a year if no transactions on the account will not be carried out? (The answer: 27.60 thousand rubles.)

> futurevalue(260,0.40,1);

364.00

> evalf(1000/216);

> 364*3;

1092

> u:=fsolve(presentvalue(1e6,x,1250)=950,x)*950;

u := 5.303626495

>

Team presentvalue (future amount, rate, period) - the calculation of the initial input to obtain a specified final amount at an interest rate of charges and the number of periods.

Example 2. How much you need to put money in the Bank today, so that when the rate of 27% per annum have in the account after 10 years 100000 thousand rubles? (The answer: 9161.419934 rubles.)

> presentvalue(680,-0.20,1);

850.0000000

The solution of problems in compound interest

The solution of problems 
Using commands <futurevalue> и <presentvalue >
> restart;
> with(finance):
Direct task
> futurevalue(,0.,);
`,` unexpected
The inverse problem
> presentvalue(,0.,);
`,` unexpected

I. Case with the same interest rate every period

Using the universal formula F = P*(1+r)^n; , where:
F - the future value (final amount).
P - the initial payment (current amount).
r - the interest rate period.
n - the number of periods.
This formula for the case with the same interest rate every period

> restart:
The task of the formula
> y:=F=P*((1+r)^n):
> y;

n
F = P (1 + r)

The job parameters are known quantities
The interest rate

> r:=;
`;` unexpected
The number of years (periods)
> n:=3;

n := 3

The initial payment (present value)
> P:=;
`;` unexpected
The final amount
> F:=2.16;

F := 2.16

The solution of the equation - the calculation of unknown values (in decimal form)
> `Unknown`;fsolve(y);

Unknown

0

>

II. The case of different interest rates for each period

Formula An = A*(1+1/100*p1)*(1+1/100*p2)*(1+1/100*p3); ... %?(1+1/100*pn); , where
An - the final amount
A - the initial payment (current amount at the moment)
p1, p2, p3, .... pn - interest rate periods
n - the number of periods

> restart:
The task of the formula (need to be adjusted based on the number of periods)
> y:=An=A*(1+1/100*p1)*(1+1/100*p2)*(1+1/100*p3):
> y;

An = A (1 + 1/100 p1) (1 + 1/100 p2) (1 + 1/100 p3)

The task of the parameters of the known values
The initial payment (present value)
> A:=;
`;` unexpected
Interest rate periods
p1:=0.30;
p2:=0.10;
p3:=0.15;

p1 := .30

p2 := .10

p3 := .15

The final amount
> An:=;
`;` unexpected
The solution of the equation - the calculation of unknown values (in decimal form)
> `Unknown`;fsolve(y);

Unknown

0

>

 angl.FINANCE.mws

Ibragimova Evelina, 6th form,
school № 57, Kazan

     Matreshka.mws 

how to translate python code which use scipy, numpy to maple code

import numpy as np
from scipy.sparse.linalg import svds
from functools import partial


def emsvd(Y, k=None, tol=1E-3, maxiter=None):
    """
    Approximate SVD on data with missing values via expectation-maximization

    Inputs:
    -----------
    Y:          (nobs, ndim) data matrix, missing values denoted by NaN/Inf
    k:          number of singular values/vectors to find (default: k=ndim)
    tol:        convergence tolerance on change in trace norm
    maxiter:    maximum number of EM steps to perform (default: no limit)

    Returns:
    -----------
    Y_hat:      (nobs, ndim) reconstructed data matrix
    mu_hat:     (ndim,) estimated column means for reconstructed data
    U, s, Vt:   singular values and vectors (see np.linalg.svd and 
                scipy.sparse.linalg.svds for details)
    """

    if k is None:
        svdmethod = partial(np.linalg.svd, full_matrices=False)
    else:
        svdmethod = partial(svds, k=k)
    if maxiter is None:
        maxiter = np.inf

    # initialize the missing values to their respective column means
    mu_hat = np.nanmean(Y, axis=0, keepdims=1)
    valid = np.isfinite(Y)
    Y_hat = np.where(valid, Y, mu_hat)

    halt = False
    ii = 1
    v_prev = 0

    while not halt:

        # SVD on filled-in data
        U, s, Vt = svdmethod(Y_hat - mu_hat)

        # impute missing values
        Y_hat[~valid] = (U.dot(np.diag(s)).dot(Vt) + mu_hat)[~valid]

        # update bias parameter
        mu_hat = Y_hat.mean(axis=0, keepdims=1)

        # test convergence using relative change in trace norm
        v = s.sum()
        if ii >= maxiter or ((v - v_prev) / v_prev) < tol:
            halt = True
        ii += 1
        v_prev = v

    return Y_hat, mu_hat, U, s, Vt

hi.how convert root of to explicit form.

w is a imaginary..

thanks

123.mw

Download 123.mw

Dear Community,

How could I specify a list of random colors using some kind of an RGB function, which then could be used in another command for coloring? I think of something like this:

myColors := [ seq( RGB ( [rint(0,255) , rint(0,255) , rint(0,255)] ) , j = 1 .. 20 ) ] :

which does not work of course :-)  This should produce me a list of 20 random colors.  What would be the right RGB color function?

Tx for the kind help in advance

best regards

Andras

Dear Community,

I generate two vectors and try to plot them one vs. the other ( ZPLOT vs. PREDS )  with the pointplot command. Unfortunately I get the message "points are not in the correct format" ... Probably a minor error, but it is unclear what format Maple expects here? What do I do wrong?

Tx. for the kind help in advance,

best regards

Andras

P.S. Maple file attached

Z_DAK_PROC.mw

Hello people in mapleprimes,

I tried to solve y=x^3 for x, expecting of getting a result of x^(1/3),

through using restart;assume(x::real,y::real);
b:=y=x^3;
solve(b,x);

But, the result was:

Warning, solve may be ignoring assumptions on the input variables.
             (1/3)    1  (1/3)   1    (1/2)  (1/3)  
            y     , - - y      + - I 3      y     ,
                      2          2                  

                1  (1/3)   1    (1/2)  (1/3)
              - - y      - - I 3      y     
                2          2                

.

It means that solve couldn't use the assumption of x and y being real.

On the other hand, reading RealDomain package, y^(1/3) is returned properly:

with(RealDomain):
solve(b,x);
                              (1/3)
                             y     

What I want to ask you is

Aren't there ways other than using the RealDomain package, to obtain the solution of y^(1/3)?

I will be very glad if you give me answers.

Best wishes.

taro

Hi,

I'm new to the physics package - wondering if i can tweak it a bit to look like things i'm used to:

is there a way to make Christoffel symbols print as upper case gamma, instead of  'G'?

KroneckerDelta print as lower case delta, instead of 'd'?

can i make the Schwarzschild metric look like it does in Hartle, Carroll, and others:

-(1 - 2M/r)dt^2 + (1-2M/r)^-1 + r^2(dtheta^2 + sin(theta)^2 dphi^2)

i know about setting the signature in Setup.

i have tried the 'Coordinates' command, but when i give it X=[t,r,theta,phi] i always seem to get back

[t,r,q,f]

i am running maple 2016

many thanks,

larry

Hello everyone,
I would like to get a symbolic result of each variable x,y and z for the following 3 nonlinear equations. Maple does not respond to the following code at all. (Not even an error report.)

restart;

eq1 := x^2+y^2+z^2-134*x+800*y-360*z+31489, 2;
eq2 := x^2+y^2+z^2-934*x+900*y-370*z+321789, 2;
eq3 := x^2+y^2+z^2-614*x+1350*y-1110*z+70048, 97;
solve({eq1, eq2, eq3}, {x, y, z});

Thanks in advance.

P.S: Afterwards my intention is to solve these equaitons numerically for different variable values, and transfer to MatLab in order to plot animations and graphs.