Taylor Series Expansions for sin(x)

25 次查看（过去 30 天）

显示更早的评论

Spaceman 2024-3-22

0
链接

此问题的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/2097781-taylor-series-expansions-for-sin-x

评论： Spaceman 2024-4-8，3:26

采纳的回答： Voss

在 MATLAB Online 中打开

Given: sin(x)=(-1)^i*x^(2*i=1)/(2i+1)!...or...

...

Find: Code that calculates the Taylor series for sin(pi/3) using the equation above

Issue: It just seems like my code isn't approximating right, anything I am overlooking here?

My Solution: I also am unsure of how to compare the CORRECT summation results to realsin which happens to be 0.866

x=pi/3;
sum1=0;
N=10;
for i=0:N
    a=(2*i+1);
    sum1=(-1).^i.*(x.^(2i+1))./factorial(a);
    fprintf('This is the estimate of sin(pi/3): %10e \n',sum1)
    % diff=sum1-sin(pi/3);
end
% Could also use the input function to have the user enter a number>10, factorial is accurate up to N<=21
% summation here is always negative every other value?
% realsin=pi/3=0.8660
% Need a way to display how close my estimate was, difference between,
% howclose=diff(sum1, realsin)
% fprintf('Estimate was ... off of actual value: %0.10e', howclose)

0 个评论
显示 -2更早的评论隐藏 -2更早的评论

请先登录，再进行评论。

请先登录，再回答此问题。

采纳的回答

Voss 2024-3-22

1
链接

此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/2097781-taylor-series-expansions-for-sin-x#answer_1429676

2i should be 2*i.

And you need to accumulate the sum; as it is now sum1 is only the current term and it's never added to anything.

11 个评论
显示 9更早的评论隐藏 9更早的评论

Voss 2024-3-22

在 MATLAB Online 中打开

"Is there a way to make it formatted such that it displays only the last value?"

Yes, move the fprintf command from inside the loop to after the loop:

x=pi/3;
s=0;
N=10;
for i=0:N
    a=(2*i+1);
    sum1=(-1).^i.*(x.^(2*i+1))./factorial(a);
    s=s+sum1;
end
fprintf('This is the estimate of sin(pi/3): %0.10e \n',s)
This is the estimate of sin(pi/3): 8.6602540378e-01 

If you also want the difference between the estimate and the actual value at every iteration, then that has to go inside the loop:

x=pi/3;
realsin=sin(x);
s=0;
N=10;
for i=0:N
    a=(2*i+1);
    sum1=(-1).^i.*(x.^(2*i+1))./factorial(a);
    s=s+sum1;
    acc = s-realsin;
    fprintf('This is the accuracy of the taylor series: %0.10e\n',acc)
end
This is the accuracy of the taylor series: 1.8117214741e-01
This is the accuracy of the taylor series: -1.0224622219e-02
This is the accuracy of the taylor series: 2.6988000240e-04
This is the accuracy of the taylor series: -4.1321280660e-06
This is the accuracy of the taylor series: 4.1315342481e-08
This is the accuracy of the taylor series: -2.9095592602e-10
This is the accuracy of the taylor series: 1.5211165660e-12
This is the accuracy of the taylor series: -6.2172489379e-15
This is the accuracy of the taylor series: -1.1102230246e-16
This is the accuracy of the taylor series: -1.1102230246e-16
This is the accuracy of the taylor series: -1.1102230246e-16
fprintf('This is the estimate of sin(pi/3): %0.10e \n',s)
This is the estimate of sin(pi/3): 8.6602540378e-01 