ratio using for loops

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Raushan 2023-10-19

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https://ww2.mathworks.cn/matlabcentral/answers/2035579-ratio-using-for-loops

评论： Walter Roberson 2023-10-20

在 MATLAB Online 中打开

Dear All,

I would like to calculate this and find out s^*. How can I calculate it using loop. I think reindexing doesn't help, since boundaries are different.

Thank you very much,

I tried in this way, but it doesn't work.

sum_x_s_ast_2 = 0;
sum_y_s_ast_2 = 0;
for n=((s_asterisk+1):1:(T-1))
    sum_x_s_asterisk_2=sum_x_s_ast_2+x_2^n;
    sum_y_s_asterisk_2=sum_y_s_ast_2+y_2^n;
    ratio_1(n+1)=(sum_x_s_ast_2/sum_y_s_ast_2)*epsilon_2^(s_asterisk);
end 
sum_x_s_ast_1 = 0;
sum_y_s_ast_1 = 0;
for l=(0:1:s_asterisk)
    sum_x_s_ast_1=sum_x_s_ast_1+x_1^l;
    sum_y_s_ast_1=sum_y_s_ast_1+y_1^l;
    ratio_2(l+1)=(sum_x_s_ast_1/(sum_y_s_ast_1+K^(1/gamma)*x_1^T))*epsilon_1^(s_asterisk);
end
z=ratio_1(n+1)/ratio_2(l+1);

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Walter Roberson 2023-10-20

在 MATLAB Online 中打开

syms s s__ast T integer

assumeAlso(s__ast >= 0)

assumeAlso(T - 1 >= s__ast + 1)

syms x_2 y_2 real

syms epsilon_2

part1 = symsum(y_2^s, s, s__ast + 1, T - 1)

part1 =

part2 = symsum(x_2^s, s, s__ast + 1, T - 1)

part2 =

part3 = simplify(part1 / part2 * epsilon_2^s__ast)

part3 =

This shows you how to construct the numerator for the expression. You follow similar steps to create the denominator for the expression.

Then you consider the expression P/Q >= 1 and multiply both sides by Q to get P >= Q . Subtract Q from both sides and you get P - Q >= 0 . Now you can solve() or vpasolve()

Walter Roberson 2023-10-20

Can you do better than that? Maybe. But it would requiire assumptions that we as outside observers are not permitted to make. As onlookers, we do not know whether the values involved are real-valued or complex-valued. If

are negative then we would want to know whether their absolute values are greater less than 1 . For that matter, we would also want to know about the absolute values of those variables even if the variables are positive in order to be able to reason about convergence.

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