subs does not substitutes in the derivation (maura.mon​ville@gmai​l.com)

2 ビュー (過去 30 日間)
mauede
mauede 2015 年 11 月 16 日
回答済み: Walter Roberson 2015 年 11 月 16 日
syms s k mu2(s,k) mu3(s,k) Scap(s,k)
Scap = mu3/mu2^(3/2);
dsScap = diff(Scap,s);
dsScap(s, k) =
diff(mu3(s, k), s)/mu2(s, k)^(3/2) - (3*diff(mu2(s, k), s)*mu3(s, k))/(2*mu2(s, k)^(5/2))
mu2(s,k) = 1 + 6*k^2 - 24*k*s^2 + 25*s^4;
mu3(s,k) = 6*s - 76*s^3 + 510*s^5 + 36*s*k - 468*k*s^3 + 108*s*k^2;
subs(dsScap)
ans(s, k) =
diff(mu3(s, k), s)/mu2(s, k)^(3/2) - (3*diff(mu2(s, k), s)*mu3(s, k))/(2*mu2(s, k)^(5/2))

回答 (1 件)

Walter Roberson
Walter Roberson 2015 年 11 月 16 日
subs(dsScap) is not a command. subs() tells MATLAB to process the input expression and return the expression with substitutions. The input variable is not changed. You can assign the result of subs() to a variable.
Because the input variable is not changed, if you have used the input variable elsewhere (such as taking its derivative) then those other locations are not changed.
If you were to assign the result of the subs() on top of the original variable, then other places that used the variable before would also not be changed -- not until you used subs() on them. When you use a symbolic variable in an expression, the symbolic variable is "captured" in the state it was at the time it was used. For example if you have
syms x y z
z = x + y
y = 2 * x
then z will still be x + y, and would not become 3 * x until you did a subs(z)

カテゴリ

Help Center および File ExchangeSymbolic Math Toolbox についてさらに検索

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by