Help with proper indexing for a programming assignment

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Lavorizia Vaughn
Lavorizia Vaughn 2021 年 11 月 11 日
コメント済み: Lavorizia Vaughn 2021 年 11 月 13 日
hi i have the code below. for the line that reads "y = y + (k1 + 2*k2 + 2*k3 + k4) / 6;' I would like to make it "y(i+1) = y(i) + (k1 + 2*k2 + 2*k3 + k4) / 6;" but that doesnt work. i should be computing y(i+1). could someone please tell me what to do so that i have the proper indexing, thanks. I would basically like to assign that y to the I+1 row of my matrix ‘computed’.btw i would have p0osted the assignments question but i dont think thats needed for such a small fix.
cases = [1, 1, 0, 0; pi, 0, 0, exp(-10); 2, 2, 0, 0; 2, 2+exp(-3), 0, 0];
%h=b-a/N;
a=0;
b=100;
N=2000; %steps
h=0.05;
t=(0:0.05:100)';
for k = 1:4 th1_2 = cases(k,1); th2_2 = cases(k,2); w1_2 = cases(k,3); w2_2 = cases(k,4);
y = [th1_2,th2_2,w1_2,w2_2];
computed = [y; zeros(N,4)];
for i = 1:N
k1 = h*fpend(y); %function only wants the coordinate in the y position
k2 = h*fpend(y + k1/2);
k3 = h*fpend(y + k2/2);
k4 = h*fpend(y + k3);
y = y + (k1 + 2*k2 + 2*k3 + k4) / 6; %i need help here. i should be computing y(i+1) terms.
????
end

採用された回答

Walter Roberson
Walter Roberson 2021 年 11 月 11 日
y = [th1_2,th2_2,w1_2,w2_2];
That is a row vector of length 4.
| would like to make it "y(i+1) = y(i) + (k1 + 2*k2 + 2*k3 + k4) / 6;"
y(1) is th1_2, y(2) is th2_2 -- are you sure you want to be looping extending y when you are using all of y in lines such as k1 = h*fpend(y); ?
If the idea is that you want to keep track of all of the different versions of your 1 x 4 y vector, then you would want a 1 x 4 for the first iteration, another 1 x 4 for the second iteration, and so on. You might, for example, want to use a 2D array, in which the row number was the number of iterations -- like assigning to y(i+1, :)
plot(t,approx(:,2));
Your code does not assign to approx and your code does not use y after it is calculated.
  7 件のコメント
Lavorizia Vaughn
Lavorizia Vaughn 2021 年 11 月 13 日
I will give this a try.
Lavorizia Vaughn
Lavorizia Vaughn 2021 年 11 月 13 日
that worked ty

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