syms k n Sn=symsum(1/k^2,1,n) thres=symsum(1/k^2,1,inf); tol=10^(-5); n=1; while Sn(n)<=thres-tol n=n+1; end [n,Sn(n)]
时间: 2024-06-04 13:12:31 浏览: 70
The code above is intended to find the smallest value of n such that the partial sum Sn of the series 1/k^2 from k=1 to n is within a tolerance of 10^(-5) of the sum of the series from k=1 to infinity (which is pi^2/6).
However, there are several syntax errors and logical errors in the code that prevent it from running correctly. Here is a corrected version:
syms k n
Sn = symsum(1/k^2, k, 1, n);
thres = symsum(1/k^2, k, 1, inf);
tol = 10^(-5);
n = 1;
while Sn < thres - tol
n = n + 1;
Sn = symsum(1/k^2, k, 1, n);
end
[n, double(Sn)]
Explanation:
- The "syms" command declares the variables k, n, Sn, and thres as symbolic variables.
- The "symsum" command computes the partial sum Sn and the sum of the series thres using symbolic summation.
- The "tol" variable stores the tolerance value.
- The "n" variable is initialized to 1.
- The while loop checks if Sn is less than thres minus tol. If it is, then n is incremented by 1 and Sn is recomputed. The loop continues until Sn is greater than or equal to thres minus tol.
- The "[n, double(Sn)]" command outputs the final values of n and Sn as a row vector. The "double" function converts Sn from symbolic to numeric format.
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