Maxima add current loop iteration to filename - output

I have code similar to the one below, where a function with a parameter depending on the loop iteration is plotted after every iteration. I would like to save the plot with the name trigplot_i.ps where i is the iteration number, but don't know how.
I have tried trigplot_"i".ps but didn't work, and have not been able to find how to cast i to a string either.
I'm a beginner so any help is very welcome.
f(x) := sin(x);
g(x) := cos(x);
for i:1 thru 10 do
(plot2d([i*f(x), i*g(x)], [x,-5,5],[legend,"sin(x)","cos(x)"],
[xlabel,"x"],[ylabel,"y"],
[ps_file,"./trigplot_i.ps"],
[gnuplot_preamble,"set key box spacing 1.3 top right"])
);
code after edits gives an error:
f(x) := sin(x);
g(x) := cos(x);
for i:1 thru 10
do block([myfile],
myfile: sconcat("./trigplot_", i, ".ps"),
printf (true, "iteration ~d, myfile = ~a~%", myfile),
plot2d([i*f(x), i*g(x)], [x,-5,5],[legend,"sin(x)","cos(x)"],
[xlabel,"x"],[ylabel,"y"],
[ps_file, myfile],
[gnuplot_preamble,"set key box spacing 1.3 top right"])
);
error:
"declare: argument must be a symbol; found "./trigplot_1.ps
-- an error.
To debug this try: debugmode(true);"

Looks good. To construct a file name, try this: sconcat("./trigplot_", i, ".ps") or also you can try: printf(false, "./trigplot_~d.ps", i). My advice is to make that a separate step in the loop, and then you can use it in the call to plot2d, e.g.:
for i:1 thru 10
do block ([myfile],
myfile: sconcat("./trigplot_", i, ".ps"),
printf (true, "iteration ~d, myfile = ~a~%", i, myfile),
plot2d (<stuff goes here>, [ps_file, myfile], <more stuff>));
EDIT: Fixed a bug in printf (omitted argument i).

Related

Solving a system of equations in Maple

I have a system of n equations and n unknown variables under symbol sum. I want to create a loop to solve this system of equations when inputting n.
y := s -> 1/6cos(3s);
A := (k, s) -> piecewise(k <> 0, 1/2exp(ksI)/abs(k), k = 0, ln(2)exp(s0I) - sin(s));
s := (j, n) -> 2jPi/(2*n + 1);
n := 1;
for j from -n to n do
eqn[j] := sum((A(k, s(j, n))) . (a[k]), k = -n .. n) = y(s(j, n));
end do;
eqs := seq(eqn[i], i = -n .. n);
solve({eqs}, {a[i]});
enter image description here
Please help me out!
I added some missing multiplication symbols to your plaintext code, to reproduce it.
restart;
y:=s->1/6*cos(3*s):
A:=(k,s)->piecewise(k<>0,1/2*exp(k*s*I)/abs(k),
k=0,ln(2)*exp(s*I*0)-sin(s)):
s:=(j,n)->2*j*Pi/(2*n+1):
n:=1:
for j from -n to n do
eqn[j]:=add((A(k,s(j,n)))*a[k],k=-n..n)=y(s(j,n));
end do:
eqs:=seq(eqn[i],i=-n..n);
(-1/4+1/4*I*3^(1/2))*a[-1]+(ln(2)+1/2*3^(1/2))*a[0]+(-1/4-1/4*I*3^(1/2))*a[1] = 1/6,
1/2*a[-1]+ln(2)*a[0]+1/2*a[1] = 1/6,
(-1/4-1/4*I*3^(1/2))*a[-1]+(ln(2)-1/2*3^(1/2))*a[0]+(-1/4+1/4*I*3^(1/2))*a[1] = 1/6
You can pass the set of names (for which to solve) as an optional argument. But that has to contain the actual names, and not just the abstract placeholder a[i] as you tried it.
solve({eqs},{seq(a[i],i=-n..n)});
{a[-1] = 1/6*I/ln(2),
a[0] = 1/6/ln(2),
a[1] = -1/6*I/ln(2)}
You could also omit the indeterminate names here, as optional argument to solve (since you wish to solve for all of them, and no other names are present).
solve({eqs});
{a[-1] = 1/6*I/ln(2),
a[0] = 1/6/ln(2),
a[1] = -1/6*I/ln(2)}
For n:=3 and n:=4 it helps solve to get a result quicker here if exp calls are turned into trig calls. Ie,
solve(evalc({eqs}),{seq(a[i],i=-n..n)});
If n is higher than 4 you might have to wait long for an exact (symbolic) result. But even at n:=10 a floating-point result was fast for me. That is, calling fsolve instead of solve.
fsolve({eqs},{seq(a[i],i=-n..n)});
But even that might be unnecessary, as it seems that the following is a solution for n>=3. Here all the variables are set to zero, except a[-3] and a[3] which are both set to 1/2.
cand:={seq(a[i]=0,i=-n..-4),seq(a[i]=0,i=-2..2),
seq(a[i]=0,i=4..n),seq(a[i]=1/2,i=[-3,3])}:
simplify(eval((rhs-lhs)~({eqs}),cand));
{0}

Julia waits for function to finish before printing message in a loop

I have a function in Julia that requires to do things in a loop. A parameter is passed to the loop, and the bigger this parameter, the slower the function gets. I would like to have a message to know in which iteration it is, but it seems that Julia waits for the whole function to be finished before printing anything. This is Julia 1.4. That behaviour was not on Julia 1.3.
A example would be like this
function f(x)
rr=0.000:0.0001:x
aux=0
for r in rr
print(r, " ")
aux+=veryslowfunction(r)
end
return aux
end
As it is, f, when called, does not print anything until it has finished.
You need to add after the print command:
flush(stdout)
Explanation
The standard output of a process is usually buffered. The particular buffer size and behavior will depend on your system setting and perhaps the terminal type.
By flushing the buffer you make sure that the contents is actually sent to the terminal.
Alternatively, you can also use a library like ProgressLogging.jl (needs TerminalLoggers.jl to see actual output), or ProgressMeter.jl, which will automatically update a nicely formatted status bar during each step of the loop.
For example, with ProgressMeter, a call to
function f(x)
rr=0.000:0.0001:x
aux=0
#showprogress for r in rr
aux += veryslowfunction(r)
end
return aux
end
will show something like (in the end):
Progress: 100%|██████████████████████████████████████████████████████████████| Time: 0:00:10
Again I can't reproduce the behaviour in my terminal (it always prints), but I wanted to add that for these types of situations the #show macro is quite neat:
julia> function f(x)
rr=0.000:0.0001:x
aux=0
for r in rr
#show r
aux+=veryslowfunction(r)
end
return aux
end
f (generic function with 1 method)
julia> f(1)
r = 0.0
r = 0.0001
r = 0.0002
...
It uses println under the hood:
julia> using MacroTools
julia> a = 5
5
julia> prettify(#expand(#show a))
quote
Base.println("a = ", Base.repr($(Expr(:(=), :ibex, :a))))
ibex
end

MapleSoft: solutions to the inverse of a function puzzle

I have to find the inverse of a function which looks like:
T := ->x (x)^0.5/(x^0.5+(1-x)^0.5)^2.
As we can see from the polynomial, we have 4 solutions when solving y= f(x). In maple,I soled for the inverse of T(x)
V := x-> solve(t=T(x),x,useassumptions=true) assuming 0<=t<=1.
and I can evaluate V, i.e maple can do V(0)=0 V(1)=1 etc.
However, as discussed, there are four solutions to the inverse function, the output of V is an expressions sequence, which looks like (solution1, solution2, solution3, solution4).
In later part of the task, I have to find the derivative of V(x)and integrate it. When I apply diff(V(x),x), maple gives me an error, saying V(x) is not valid.As V(x) is an expression sequence. I tried to use the function D(V), but still no luck.
My questions is how would I be able to handle this V(x) as an expression sequence to finish the rest of the task. Is V(x) a piecewise function? If that's the case, how would I be able to convert this expression sequence to a piecewise function.
Regards,
restart:
T := proc (x) options operator, arrow; sqrt(x)/(sqrt(x)+sqrt(1-x))^2 end proc:
V := proc (x) options operator, arrow; solve(x = T(y), y) end proc:
sol := [allvalues(V(x))]:# Extract 4 solution, with command op(1, sol)->Only first solution is correct.
plot([x, T(x), op(1, sol)], x = 0 .. 2, legend = [typeset("Curve: ", "x"),
typeset("Curve: ", "T(x)"), typeset("Curve: ", "V(x)")]);
VV := proc (x) options operator, arrow; evalf(op(1, sol)) end proc;
eval(VV(x), x = 1/2); #Inverse function at point x=1/2
eval(diff(VV(x), x), x = 1/2);# Derivative of inverse function at point x=1/2
int(VV(x), x = 1/10 .. 1/2, numeric);# Integral of inverse function at range (1/10..1/2)
Mathematica 11.3 solution:

After one call to myfun, new parametrization does not affect the result, which conforms to the first call

I am new to Octave although I can say I am an expert Matlab user. I am running Octave on a Linux server (Red Hat) remotely through PuTTY, from a windows machine.
I am observing a very strange behavior in Octave. I call myfun(a) which performs as expected giving the sought results. Now, if I run, say, myfun(b) with b!=a, I get again myfun(a). Clear -f does not solve the problem. I need to reboot octave to change the parameters.
What am I doing wrong?
Thanks a lot
Francesco
This is the code for the function I mentioned:
function [a, v, obj, infos, iter] = mle_garch( p )
#{
% this function estimates the GARCH(1,1) parameters
% it is assumed we pass the adjusted price level p
#}
global y = (diff(log(p))-mean(diff(log(p))))*100;
global h = zeros(size(y));
a0 = [var(y)*0.9; 0.8; 0.1];
[a, obj, infos, iter] = sqp(a0, #loglike_garch, [], #loglike_con, [], [], 1000);
v = sqrt(h * 260);
endfunction
function g = loglike_garch( a )
global y h
n = length(y);
h(1) = var(y);
for i = 2 : n,
h(i) = a(1) + a(2) * h(i-1) + a(3) * y(i-1)^2;
endfor
g = 0.5 * ( sum(log(h)) + sum(y.^2./h) ) / n;
endfunction
function f = loglike_con( a )
f = [1;0;0;0] + [0 -1 -1;eye(3)] * a;
endfunction
I'm assuming the myfun you mentioned is mle_garch. The problem is the way you're initializing the global h and v variables (do you really need them to be global?). When you have a piece of code like this
global y = (diff(log(p))-mean(diff(log(p))))*100;
global h = zeros(size(y));
the values of y and h are defined the first time only. You can change their values later on, but this specific lines will never be ran again. Since your code only uses the input argument to define these two variables, the value which you use to run the function the first time will be used every single other time. If you really want to keep those variables global, replace it with the following:
global y;
global h;
y = (diff(log(p))-mean(diff(log(p))))*100;
h = zeros(size(y));
But I don't see any reason to keep them global so just don't make them global.
Also, you mentioned this code worked fine in Matlab. I was under the impression that you couldn't initialize global and persistent variables in Matlab which would make your code illegal in Matlab.

Program Works in Python 2.7, but not Python 3.3 (Syntax is compatible for both)

So I have the function integral(function, n=1000, start=0, stop=100) defined in nums.py:
def integral(function, n=1000, start=0, stop=100):
"""Returns integral of function from start to stop with 'n' rectangles"""
increment, num, x = float(stop - start) / n, 0, start
while x <= stop:
num += eval(function)
if x >= stop: break
x += increment
return increment * num
However, my teacher (for my Programming class) wants us to create a separate program that gets the input using input() and then returns it. So, I have:
def main():
from nums import integral # imports the function that I made in my own 'nums' module
f, n, a, b = get_input()
result = integral(f, n, a, b)
msg = "\nIntegration of " + f + " is: " + str(result)
print(msg)
def get_input():
f = str(input("Function (in quotes, eg: 'x^2'; use 'x' as the variable): ")).replace('^', '**')
# The above makes it Python-evaluable and also gets the input in one line
n = int(input("Numbers of Rectangles (enter as an integer, eg: 1000): "))
a = int(input("Start-Point (enter as an integer, eg: 0): "))
b = int(input("End-Point (enter as an integer, eg: 100): "))
return f, n, a, b
main()
When run in Python 2.7, it works fine:
>>>
Function (in quotes, eg: 'x^2'; use 'x' as the variable): 'x**2'
Numbers of Rectangles (enter as an integer, eg: 1000): 1000
Start-Point (enter as an integer, eg: 0): 0
End-Point (enter as an integer, eg: 100): 100
Integration of x**2 is: 333833.5
However, in Python 3.3 (which my teacher insists we use), it raises an error in my integral function, with the same exact input:
Traceback (most recent call last):
File "D:\my_stuff\Google Drive\documents\SCHOOL\Programming\Python\Programming Class\integration.py", line 20, in <module>
main()
File "D:\my_stuff\Google Drive\documents\SCHOOL\Programming\Python\Programming Class\integration.py", line 8, in main
result = integral(f, n, a, b)
File "D:\my_stuff\Google Drive\Modules\nums.py", line 142, in integral
num += eval(function)
TypeError: unsupported operand type(s) for +=: 'int' and 'str'
In addition, integral by itself (in Python 3.3) works fine:
>>> from nums import integral
>>> integral('x**2')
333833.4999999991
Because of that, I believe the fault is in my program for my class... Any and all help is appreciated. Thanks :)
The issue you're running into is that input works differently in Python 2 and Python 3. In Python 3, the input function works like the raw_input does in Python 2. Python 2's input function is equivalent to eval(input()) in Python 3.
You're getting into trouble because of the quoteation marks you're typing with the formula. When you type 'x**2' (with the quotes) as your formula when running on Python 2, the text gets evaled in the input function and you get a string with no quotation marks as the result. This works.
When you give the same string to Python 3's input function, it doesn't do an eval, so the quotation marks remain. When you later eval the formula as part of your integral calculation, you get the string x**2 (without any quotation marks) as the result, not the value of x squared. This results in an exception when you try the string to 0.
To fix this, I suggest either using just one version of Python, or putting the following code at the top of your file to get a Python 3 style input in both versions:
# ensure we have Python 3 semantics from input, even in Python 2
try:
input = raw_input
except NameError:
pass
Then just type in your formula without quotation marks and it should work correctly.