# インデックスが配列要​素数を超えています。​インデックスは 0 を超えてはなりません。

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Masaaki Yanagihara on 28 Mar 2024
Commented: Dyuman Joshi on 28 Mar 2024
syms s
for kq = -10:1:10
eqnqroop = 1 + (kq*(-((4483*s^3)/5000 + (1915281*s^2)/5000000 - (551896897*s)/5000000000)/(s^4 + (52*s^3)/125 ...
+ (764541*s^2)/1000000 + (4137177603*s)/500000000 - 427810011/250000000))) == 0;
q1 = solve(eqnqroop,s); %%求解
kq
plot(real(q1(1)),imag(q1(1)),'-s','MarkerSize',5,'MarkerEdgeColor','red', 'MarkerFaceColor',[1 .6 .6])
hold on
end
kq = -10
kq = -9
kq = -8
kq = -7
kq = -6
kq = -5
kq = -4
kq = -3
kq = -2
kq = -1
kq = 0
Index exceeds the number of array elements. Index must not exceed 0.

Error in indexing (line 962)
R_tilde = builtin('subsref',L_tilde,Idx);
kqを-10~10まで1刻みで変えながら，eqnqroopという方程式の解を複素平面にプロットする，ということをしたいのですが，「インデックスが配列要素数を超えています。インデックスは 0 を超えてはなりません。」とでてきてしまい，-10~0までで計算がストップしてしまいます．

Dyuman Joshi on 28 Mar 2024
Edited: Dyuman Joshi on 28 Mar 2024
When kq is 0, the equation becomes 1==0, which is obviously not true, thus there is no solution for it. For that iteration, q1 is empty and you can't use indexing for it.
You can skip that value or plot a value manually for it. I have implemented the 1st option in my code below.
Note that the solve() returns the real solutions first if there are any. As the equation is a cubic polynomial in s, there will atleast be one real solution, so use the 2nd index for plotting.
syms s
figure
hold on
%Modify the loop indices to skip the value 0
for kq = setdiff(-10:10, 0)
eqnqroop = 1 + (kq*(-((4483*s^3)/5000 + (1915281*s^2)/5000000 - (551896897*s)/5000000000)/(s^4 + (52*s^3)/125 ...
+ (764541*s^2)/1000000 + (4137177603*s)/500000000 - 427810011/250000000))) == 0;
q1 = solve(eqnqroop,s);
%updated indexing
plot(real(q1(2)),imag(q1(2)),'-s','MarkerSize',5,'MarkerEdgeColor','red', 'MarkerFaceColor',[1 .6 .6])
end
Masaaki Yanagihara on 28 Mar 2024
ご回答ありがとうございます．

"Note"も有用で助かりました．
Dyuman Joshi on 28 Mar 2024
You're welcome!

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