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

I'm sorry, but the code you provided has a syntax error. The line "n=n 1;" should be "n=n+1;". Additionally, the code does not output anything. Please let me know what you would like the code to do, and I can provide assistance.

相关问题

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)]

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.

thres=symsum(1/k^2,1,inf); tol=10^(-5); Snn=@(i)(1/i^2); i=1; while Snn(i)<=thres-tol i=i+1; end [i,Snn(i)]

There is an error in the code provided. The line "i=i 1;" should be "i=i+1;" to increment the value of i.

Here is the corrected code:

syms thres;
thres = symsum(1/k^2, 1, inf);
tol = 10^(-5);
Snn = @(i) (1/i^2);
i = 1;
while Snn(i) <= thres - tol
i = i + 1;
end
[i, Snn(i)]