I want to simplify this symbolic expression, and then take the limit of it (this is not too hard on a paper) using Matlab
syms n x
sn = 8^n * n^2 * (sin(x))^(3*n)
simplify(sn^(1/n))
which results in
ans =
8*n^(2/n)*(sin(x)^(3*n))^(1/n)
This must be like 8 * n^(2/n) * (sin(x))^3. However, if I use
x = sym('x', 'positive'); n = sym('n', 'positive');
sn = 8^n * n^2 * (sin(x))^(3*n)
simplify(sn^(1/n))
to obtain a similar answer and then take limit, I get:
limit(ans, n, inf)
ans =
8*limit(n^(2/n)*(sin(x)^(3*n))^(1/n), n == Inf)
instead of 8*sin(x)^3.
If I simplify this on paper and then take limit, everything works:
limit(8*n^(2/n)*sin(x)^3, n, inf)
ans =
8*sin(x)^3
Is it possible to solve this using Matlab?
MuPAD doesn't simplify the expression, and thus can't take the limit, because you haven't provided the appropriate assumptions. It's not true that an expression such as (sin(x)^n)^(1/n) simplifies to sin(x) if n and x are positive. You need to fully restrict the range of x as it the argument of a periodic function:
x = sym('x','positive');
n = sym('n','real');
assumeAlso(x<=sym(pi));
sn = 8^n * n^2 * (sin(x))^(3*n);
sn2 = simplify(sn^(1/n))
limit(sn2, n, Inf)
which returns the answer you were expecting
ans =
8*sin(x)^3
If you have access to Mathematica, this sort of thing can be accomplished very easily:
Limit[(8^n n^2 Sin[x]^(3 n))^(1/n), n -> \[Infinity], Assumptions -> {n \[Element] Reals, x >= 0, x <= \[Pi]}]