Error calculation and using for loop
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The cosine function can be evaluated by the following infinite series: cos(x)=1-x^2/2!+x^4/4!-x^6/6!+... Create an M-file to compute cos(1.18 rad) for up to and including number of terms terms, which is up to the term x^8/8!. Your program should compute and display the values of cos x as each term in the series is added. then compute and display true relative error as a percentage..
attempted solution:
n=14;
x=pi;
terms=[0:2:n]
facts=factorial(terms);
x_power=x.^terms;
signs=(-1).^(terms/2);
quotes=x_power ./facts;
series=signs.*quotes;
m_cos_n=cumsum(series);
plot (m_cos_n);
true=-1.0;
p_error=(true-m_cos_n)/true*100;
plot(terms,(abs(p_error)));
回答 (2 件)
Abhishek Jain
2016 年 10 月 4 日
You can use the following code:
cosx=1;
x=1.18;
for i=1:4
cosx=cosx+(-1)^i.*x^(2*i)/factorial(2*i)
end
It prints the value of cos(x) after each step.
0 件のコメント
elias GR
2016 年 10 月 4 日
編集済み: elias GR
2016 年 10 月 4 日
- You need to put your code inside a function, in order x and n to be parameters.
- You can estimate the true value by just using MATLAB's cos function.
- Add figure commands in order to see both plots.
- Usually we take the absolute value of the relative error.
- You need to take care if true=0 (not to divide by zero) - this is not included in the code below
function cosAppr(x,n)
terms=[0:2:n];
facts=factorial(terms);
x_power=x.^terms;
signs=(-1).^(terms/2);
quotes=x_power ./facts;
series=signs.*quotes;
m_cos_n=cumsum(series);
figure
plot (m_cos_n);
true=cos(x);
p_error=abs((true-m_cos_n)/true*100);
figure
plot(terms,(abs(p_error)));
end
You can call the function like
cosAppr(1.18,8)
0 件のコメント
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