I have created a function with one procedure....
Func1[n_] := Table[a[i], {i, n}]
which returns
Func1[5]
{a[1], a[2], a[3], a[4], a[5]}
I also have created a function with a few parameters or with a few arguments, few variables!
Func1[x_, y_, z_] := (x + y)*z - 1
which returns
Func1[5, 2, 3]
20
But what about if I want to create a function with a several procedure which returns whatever I want?
I already know that when one procedure is done I have to type " ; " at the end of this procedure!
Like in for loops we do....
For[k = 2, k < 3, k++,
S := Table[a[i], {i, n}];
B := Dimensions[S][[1]]];
]
So I need to create a function with a several procedure!
How to do it?
Please help me!
A couple of examples here. Remember to use lower-case initial letters to avoid conflicting with built-in functions which all start with a capital letter.
s[n_] := Table[a[i], {i, n}]
b[s_] := Dimensions[s][[1]]
For[k = 2, k < 3, k++,
x = s[k];
Print[b[x]]]
2
For[k = 2, k < 3, k++,
Print[b[s[k]]]]
2
Use parentheses for grouping.
For example
set$s$b[n_Integer] := ($s = Table[a[i], {i, n}];
$b = Dimensions[$s][[1]];)
Now, after executing, e.g.,
set$s$b[5]
one will get
$s
{a[1], a[2], a[3], a[4], a[5]}
$b
5
However, making use of modularity might be a better design choice in situations where the execution of several procedures is needed.
Related
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}
I was trying to write a function that solves following;
persistence 39 = 3 // because 3*9 = 27, 2*7 = 14, 1*4=4
// and 4 has only one digit
persistence 999 = 4 // because 9*9*9 = 729, 7*2*9 = 126,
// 1*2*6 = 12, and finally 1*2 = 2
persistence 4 = 0 // because 4 is already a one-digit number
After I solved the question I tried to make all functions looks like Ramda.js function styles like this;
This code works;
let multiply = List.reduce (*)
let gt from input = input > from
let just input = fun _ -> input
let ifElse cond trueFn falseFn input =
if cond input then trueFn input else falseFn input
let digits n =
(string n) |> Seq.toList |> List.map (System.Char.GetNumericValue >> int)
let rec persRec iter current =
current
|> digits
|> multiply
|> ifElse (gt 9) (persRec (iter + 1)) (just iter)
let persistence n = if n > 9 then persRec 1 n else 0
But when I tried to modify persRec function with a curried composed version like following, it makes this stack overflow.
let rec persRec iter =
digits
>> multiply
>> ifElse (gt 9) (persRec (iter + 1)) (just iter)
What's wrong with this?
The function persRec is calling itself unconditionally. Here:
>> ifElse (gt 9) (persRec (iter + 1)) (just iter)
^^^^^^^^^^^^^^^^^^^^
|
unconditional recursive call
This happens always. Every time persRec is called by somebody, it immediately calls itself right away.
You may expect that the recursive call should only happen when gt 9, because, after all, it's inside an ifElse, right? But that doesn't matter: ifElse is not special, it's just a function. In order to call a function, F# has to compute all its parameter before the call (aka "applicative order of evaluation"), which means it has to call persRec (iter + 1) before it can call ifElse, and it has to call ifElse before it can call (>>), and it has to call (>>) in order to compute result of persRec. So ultimately, it needs to call persRec in order to compute the result of persRec. See where this is going?
The previous version works, because the body of persRec is not actually executed before the call to ifElse. The body of persRec will only be executed when all its parameters are supplied, and the last parameter will only be supplied inside the body of ifElse when the condition is true.
The way I see it, the confusion stems from the difference between denotational and operational semantics. Yes, mathematically, logically, the functions are equivalent. But execution also matters. Normal vs. applicative evaluation order. Memory concerns. Performance. Those are all outside of the domain of lambda-calculus.
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:
I would like to define a list using a for loop and I need to do it using a function of the n-iterate.
I have:
Initialization
In[176]: Subscript[y, 0] = {1, 2, 3}
Out[180]: {1,2,3}
The function:
In[181]: F[n_] := For[l = 1, l++, l <= 3, Subscript[y, n + 1][[l]] :=Subscript[y, n][[l]]+ n]
I call the function
F[0]
and I get:
In[183]: Subscript[y, 1]
Out[183]: Subscript[0, 1]
I should have {1,2,3}.
Anyone know why it isn't working as it should?
I have troubles recreating your error, problem.
I understand you want to add n to your vector, where n is the number of the subscript.
Here's another way to have a go at your question, avoiding the loop and the subscripts:
Clear#y;
y[0] = {1, 2, 3};
y[n_Integer] : =y[n - 1] + n
(as Plus is Listable, you can just add n to the vector, avoiding the For)
and then call it using, e.g.
y[0]
{1,2,3}
or
y[5]
{16,17,18}
Alternatively, using memoization, you could define y as follows:
y[n_Integer] := y[n] = y[n - 1] + n
This will then store already calculated values (check ?y after executing e.g. y[5]). Don't forget to Clear y, if y changes.
Obviously, for a function as this one, you might want to consider:
y[n_Integer] := y[0] + Total[Range[n]]
When I want to solve a set of linear equations for two functions, e.g.
solutions := solve({f(x)=x,g(x)=x},{f(x),g(x)});
what exactly can I do to work with the solutions as functions themselves in maple?
The only thing which I was able to do was
f_solution := x2 -> subs(x=x2, rhs(solutions[1]))
But that is ugly in many aspects. First, this trivial substitution x->x2 seems necessary, without it will not work. Second, the construct rhs(solutions[1]) is very bad, as it is not possible to control the order of the solutions. Consequently everytime I modify my equations, I would have to check manually, if the index [1] is still correct.
Is there a clean and standard way to extract the functions from the set?
solutions := solve({2*f(x)=sin(x),g(x)/3=cos(x)},{f(x),g(x)});
/ 1 \
{ f(x) = - sin(x), g(x) = 3 cos(x) }
\ 2 /
and now, with f_solution as an expression,
f_solution := eval(f(x), solutions);
1
- sin(x)
2
or with f_solution as a procedure,
f_solution := unapply( eval(f(x), solutions), x);
1
x -> - sin(x)
2
Have a look at assign. It can fix the solutions you obtain in your calculation
> restart:
> solutions := solve({f(x)=x,g(x)=x},{f(x),g(x)});
solutions := {f(x) = x, g(x) = x}
> assign(%);
> f(x);
x
You could also just try subs like this
> restart:
> solutions := solve({f(x)=x,g(x)=x},{f(x),g(x)});
solutions := {f(x) = x , g(x) = x}
> subs(solutions,f(x));
x