Taylor Series Expansions for sin(x)

22 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

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 

Voss 2024-4-8

You're welcome!

Spaceman 2024-4-8

You're the best, Voss!

请先登录，再进行评论。

类别

MATLAB Language Fundamentals Matrices and Arrays Matrix Indexing

在 Help Center 和 File Exchange 中查找有关 Matrix Indexing 的更多信息

产品

MATLAB

版本

R2023b

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by

Taylor Series Expansions for sin(x)

0 个评论
显示 -2更早的评论隐藏 -2更早的评论

采纳的回答

11 个评论
显示 9更早的评论隐藏 9更早的评论

更多回答（0 个）

另请参阅

类别

标签

产品

版本

Community Treasure Hunt

Taylor Series Expansions for sin(x)

0 个评论 显示 -2更早的评论隐藏 -2更早的评论

采纳的回答

11 个评论 显示 9更早的评论隐藏 9更早的评论

更多回答（0 个）

另请参阅

类别

标签

产品

版本

Community Treasure Hunt

0 个评论
显示 -2更早的评论隐藏 -2更早的评论

11 个评论
显示 9更早的评论隐藏 9更早的评论