Add to a vector from loop

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Matt Baron
Matt Baron 2021-3-6
评论： Matt Baron 2021-3-27
Team,
Just learning how to use MatLab here,
How do I insert the value n into a different variable so I can recall each number? Any other advice is much appreciated. I just started MATH240 so will need more help later ;)
-----------------------------------------------
totaln=0;
n=1;
for n=1:1000000;
if abs(cos(n))<1/n;
totaln=totaln+1;
display(n) %this displays n but then updates it with the next iteration, I want to insert the value of n into a vector so I can see recall them later.
n=n+1;
else;
n=n+1;
end;
end;
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Matt Baron
Matt Baron 2021-3-6
Ok, thank you. I will take a look! I appreciate your time.
Matt Baron
Matt Baron 2021-3-27
Mr. Cobeldick,
Just wanted to share with you how far I have come since this question with you and others' help;
syms e imp r
A=zeros([],2);%create an empty 0 by 2 matrix for later
display("Hello! Welcome to Matt Baron's numerical solver. Press enter to continue.")
pause
step = input('What step size would you like?');
initcondx = input('What is the initial condition for x?');
initcondy = input('What is the initial condition for y?');
n=1;%start at 1
x=initcondx;%start at the initial condition
y=initcondy;%start at the initial condition
dfeq=input('What differential equation do you want approximated? i.e. .2 * x * y','s');
f=str2func(sprintf('@(x,y)%s',dfeq));%the function to be evaluated
est=input('What value do you want the equation estimated to?');
%actual=@(x) (exp(.1 .* (x .^2) - .1));%the solved function
choice = input('Press e for Euler, imp for Improved Euler, or r for RK4.');
if choice == e
while x <= est%what value do you want it estimated to
A(n,:)=[x,y];
y = (A(n,2) + ((step) * f(A(n,1),A(n,2))));
x = (initcondx) + ((n) * (step));
%abs = actual(x) - y;
%rel = (abs / actual(x)) * 100;
n=n+1;
end
else
if choice == imp
while x <= est
A(n,:)=[x,y];
A(n+1,2) = y + ((step) * f(A(n,1),A(n,2)));
A(n+1,1) = (initcondx) + ((n) * (step));
y = A(n,2) + ((step) * ((f(A(n,1),A(n,2)))+f(A(n+1,1),A(n+1,2))))/(2);
x = (initcondx) + ((n) * (step));
n=n+1;
end
A(n,:)=[];
end
end
A

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采纳的回答

Stephen23
Stephen23 2021-3-6
编辑：Stephen23 2021-3-6
Ran in:
The MATLAB way:
N = 1:1000000;
X = abs(cos(N)) < 1./N;
T = nnz(X) % total
T = 7
Z = N(X) % those which pass your condition
Z = 1×7
1 2 11 33 52174 260515 573204

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