Summation of Infinity Terms
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S = 0;
tol = eps;
term = inf;
n = 0;
while abs(term) > tol
term = exp(x*log10(n));
S = S + term;
n = n + 1;
end
It try by this way but I don't know what is wrong
لا
9 件のコメント
dpb
2019 年 12 月 9 日
編集済み: dpb
2019 年 12 月 9 日
I missed one thing in my first look...look at the summation limits carefully in the definition and what you initialized n as.
Actually, missed the biggie by not paying attention...dunno where you got the expression you're evaluatiing, but that isn't the expression in the assignment at all...
Again, you do need to define x (>1 for series to converge)
採用された回答
David Hill
2019 年 12 月 8 日
You need to create a function. The easiest is to use the existing function. It doesn't sound like your instructions prevent this.
function z = myzeta(x)
z=zeta(x);
end
Alternatively, it is easy to determine the stopping point but floating point inaccuracies will affect the result.
function z = myzeta(x)
n=ceil((1/eps)^(1/x));
z=sum((1:n).^(-x));
end
0 件のコメント
その他の回答 (1 件)
dpb
2019 年 12 月 9 日
S = 0;
tol = eps;
term = inf;
n = 1;
If (x<1)
'not valid starting points enter another points'
end
while abs(term) > tol
term = (1/power (n , x));
S = S + term;
n = n + 1;
end
Is not bad start for learner...I used similar with few changes and got answers that match those looked up...
S = 0;
tol = 1E-7; % run quicker; approx single precision accuracy eps;
term = bigvalue; % use finite values instead of inf
n = 1;
if (x<=1), error('not valid x--enter another points >1'), end
while abs(term) > tol
term = (1/power (n , x));
S = S + term;
n = n + 1;
end
0 件のコメント
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