最后的MATLAB代码故障 [英] MATLAB code glitch at the end
问题描述
我是新来的.
场景说明:
我已经编写了MATLAB代码以使用部分和S_n(x)=(-1)^ j * x ^(2j + 1)/(2j + 1)从j = 0到n的总和！近似于罪恶x.我已经编写了一个计算机程序，以通过方法
评估此部分和.LS:计算和总结从最大的术语到最小的术语
和
SL:以相反的顺序进行计算.
我已经使用我的程序通过LS和SL两种方法针对x = 0.1、1和10来计算S_n(x)，并将结果与sin x进行了比较.我用过n = 10、100和1000.
问题:我在代码的最后两行遇到问题.它说"NaN"(不是数字)，但在上一列中也显示了异常数字.我究竟做错了什么?谁能帮我吗?
这是我的代码:
第一个功能:
function ret = hw1_6_1(x,n)
ret=((-1)^n)*((x^((2*n)+1))/(factorial((2*n)+1)));
end
第二个功能:
function ret = partialsum(x,n,log)
%% Begin code
% j=n
% 1 to n+1
% LS 0 to n
% SL n to 0
%((-1)^n)*((x^((2*n)+1))/(factorial((2*n)+1)))
clear sum
ret = 0;
if log == 1
for i=0:1:n
ret= ret + hw1_6_1(x,i) ;
i=i+1;
end
elseif log == 0
for i=n:-1:0
ret= ret + hw1_6_1(x,i) ;
i=i+1;
end
end
end
最后，我的主要代码:
%% Hw 1 Problem 6
% Running approximation of sin x
% LS and SL
% LS == log=1, SL == log=0
% x = 0.1, 1, 10
% n = 10,100,1000
clear all
clc
format long %displays more decimal points
%% For x = 0.1
sin1 = sin(0.1);
% LS
% n = 10, 100, 1000 Generated in array format
LS1 = [partialsum(0.1,10,1);partialsum(0.1,100,1);partialsum(0.1,1000,1)];
% Now SL
SL1 = [partialsum(0.1,10,0);partialsum(0.1,100,0);partialsum(0.1,1000,0)];
%% For x = 1
sin2 = sin(1);
% LS
% n = 10, 100, 1000 Generated in array format
LS2 = [partialsum(1,10,1);partialsum(1,100,1);partialsum(1,1000,1)];
% Now SL
SL2 = [partialsum(1,10,0);partialsum(1,100,0);partialsum(1,1000,0)];
%% For x = 10
sin3 = sin(10);
% LS
% n = 10, 100, 1000 Generated in array format
LS3 = [partialsum(10,10,1);partialsum(10,100,1);partialsum(10,1000,1)];
% Now SL
SL3 = [partialsum(10,10,0);partialsum(10,100,0);partialsum(10,1000,0)];
%% Comparison stage
Sines = [sin1;sin2;sin3]
Approxs = [LS1 SL1 LS2 SL2 LS3 SL3]'
我的输出:
Sines =
0.099833416646828
0.841470984807897
-0.544021110889370
Approxs =
0.099833416646828 0.099833416646828 0.099833416646828
0.099833416646828 0.099833416646828 0.099833416646828
0.841470984807897 0.841470984807897 0.841470984807897
0.841470984807897 0.841470984807897 0.841470984807897
2.761090925979680 -0.544021110889270 NaN
2.761090925979687 -0.544021110889190 NaN
谢谢.
1)sin_taylor(10)== 2.76是因为在 x 较小的情况下，20度多项式逼近值与正弦值不同. /p>
2)序列x ^ n/n！对于| x |> 1，不是单调的，但是在大约n = 6处有最大值；因此，与| x |< = 1且一个正在迅速消失的序列求和的情况相比，会有一些舍入差异.
3)当 n 大时，分母和分母(10 ^ n，n！)都将表示为无穷大. Inf/Inf 被定义为 NaN ，而 NaN +任何== NaN . 这与partial_sum(10，100)不同，在partial_sum(10，100)中，一个项合计为(x/inf == 0).
I'm new here.
Description of scenario:
I have written MATLAB code to use the partial sum S_n(x)=Sum from j=0 to n of (-1)^j*x^(2j+1)/(2j+1)! to approximate sin x. I have written a computer program to evaluate this partial sum by methods
LS: computing and summing the terms from the largest term fi rst to the smallest term last
and
SL: doing the calculation in the reverse order.
I have used my program to calculate S_n(x) for x = 0.1, 1, and 10 by both methods LS and SL, and compared the results against sin x. I have used n = 10, 100, and 1000.
Question: I am running into a problem in the last two lines of my code. It says "NaN" (Not a number), but also shows deviant numbers in the previous column. What am I doing wrong? Can anyone help me?
Here is my code:
The first function:
function ret = hw1_6_1(x,n)
ret=((-1)^n)*((x^((2*n)+1))/(factorial((2*n)+1)));
end
The second function:
function ret = partialsum(x,n,log)
%% Begin code
% j=n
% 1 to n+1
% LS 0 to n
% SL n to 0
%((-1)^n)*((x^((2*n)+1))/(factorial((2*n)+1)))
clear sum
ret = 0;
if log == 1
for i=0:1:n
ret= ret + hw1_6_1(x,i) ;
i=i+1;
end
elseif log == 0
for i=n:-1:0
ret= ret + hw1_6_1(x,i) ;
i=i+1;
end
end
end
Finally, my main code:
%% Hw 1 Problem 6
% Running approximation of sin x
% LS and SL
% LS == log=1, SL == log=0
% x = 0.1, 1, 10
% n = 10,100,1000
clear all
clc
format long %displays more decimal points
%% For x = 0.1
sin1 = sin(0.1);
% LS
% n = 10, 100, 1000 Generated in array format
LS1 = [partialsum(0.1,10,1);partialsum(0.1,100,1);partialsum(0.1,1000,1)];
% Now SL
SL1 = [partialsum(0.1,10,0);partialsum(0.1,100,0);partialsum(0.1,1000,0)];
%% For x = 1
sin2 = sin(1);
% LS
% n = 10, 100, 1000 Generated in array format
LS2 = [partialsum(1,10,1);partialsum(1,100,1);partialsum(1,1000,1)];
% Now SL
SL2 = [partialsum(1,10,0);partialsum(1,100,0);partialsum(1,1000,0)];
%% For x = 10
sin3 = sin(10);
% LS
% n = 10, 100, 1000 Generated in array format
LS3 = [partialsum(10,10,1);partialsum(10,100,1);partialsum(10,1000,1)];
% Now SL
SL3 = [partialsum(10,10,0);partialsum(10,100,0);partialsum(10,1000,0)];
%% Comparison stage
Sines = [sin1;sin2;sin3]
Approxs = [LS1 SL1 LS2 SL2 LS3 SL3]'
My output:
Sines =
0.099833416646828
0.841470984807897
-0.544021110889370
Approxs =
0.099833416646828 0.099833416646828 0.099833416646828
0.099833416646828 0.099833416646828 0.099833416646828
0.841470984807897 0.841470984807897 0.841470984807897
0.841470984807897 0.841470984807897 0.841470984807897
2.761090925979680 -0.544021110889270 NaN
2.761090925979687 -0.544021110889190 NaN
Thanks in advance.
1) sin_taylor(10) == 2.76 is because 20th degree polynomial approximation diverges from sine at much smaller value of x.
2) The sequence x^n / n! for |x|>1, is not monotonic, but has a maximum at about n=6; thus there will be some rounding differences compared to cases, where |x|<= 1 and one is summing up a rapidly vanishing sequence.
3) Both the nominator and the denominator (10^n, n!) will be represented as infinities, when n is large. Inf / Inf is defined as NaN, and NaN + anything == NaN. This is different to partial_sum(10, 100), where one sums up terms (x / inf == 0).
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