Question

下面是我的MATLAB代码，用于生成wavepeed（c）的虚部，当雷诺数为249时使用函数oscalcpcf，我想以2为步长运行0.1到2之间的波数（alpha）。

c1holder = [];
c2holder = [];
reyholder = [];
alphaholder = [];
wantedrey = [];
wantedalpha = [];
for Rey = 249
    for alpha = 0.1:0.1:2
        c1=oscalcpcf(Rey,alpha,100);
        c2=oscalcpcf(Rey,alpha,200);
        c1holder = [c1holder c1];
        c2holder = [c2holder c2];    
        reyholder = [reyholder Rey];
        alphaholder = [alphaholder alpha];      
    end
end
vectors = [c1holder' c2holder' reyholder' alphaholder'];

上面的代码在我看来并不困难，但我的笔记本电脑却因一些雷诺数和alpha而变得暴躁。只需命名其中一个，Reynolds number = 249和alpha = 0.3。

当我运行上述代码时，我得到c1 = 6.06002472332094E-08和c2 = 0.0000010870344982811。

现在问题就在于此。如果我使用2步骤从0.1到-0.1，我会得到c1 = -0.337584041016646和c2 = 0.0000364854401656638。

如果我要使用oscalcpcf手动核对，即oscalcpcf(249,0.3,100)和oscalcpcf(249,0.3,200)，我会c1 = -0.337583911335139和c2 = -0.337577395716528。

我真的不知道这里发生了什么，有人可以帮忙！

修改

alpha: 2.000000000000000000000000000000
alpha: 1.899999999999999900000000000000
alpha: 1.800000000000000000000000000000
alpha: 1.700000000000000000000000000000
alpha: 1.600000000000000100000000000000
alpha: 1.500000000000000000000000000000
alpha: 1.399999999999999900000000000000
alpha: 1.299999999999999800000000000000
alpha: 1.200000000000000000000000000000
alpha: 1.100000000000000100000000000000
alpha: 1.000000000000000000000000000000
alpha: 0.899999999999999910000000000000
alpha: 0.799999999999999820000000000000
alpha: 0.699999999999999960000000000000
alpha: 0.599999999999999870000000000000
alpha: 0.500000000000000000000000000000
alpha: 0.399999999999999910000000000000
alpha: 0.299999999999999820000000000000
alpha: 0.199999999999999960000000000000
alpha: 0.099999999999999867000000000000

以及0.1至2

alpha: 0.100000000000000010000000000000
alpha: 0.200000000000000010000000000000
alpha: 0.300000000000000040000000000000
alpha: 0.400000000000000020000000000000
alpha: 0.500000000000000000000000000000
alpha: 0.599999999999999980000000000000
alpha: 0.700000000000000070000000000000
alpha: 0.800000000000000040000000000000
alpha: 0.900000000000000020000000000000
alpha: 1.000000000000000000000000000000
alpha: 1.100000000000000100000000000000
alpha: 1.200000000000000200000000000000
alpha: 1.300000000000000300000000000000
alpha: 1.400000000000000100000000000000
alpha: 1.500000000000000200000000000000
alpha: 1.600000000000000100000000000000
alpha: 1.700000000000000200000000000000
alpha: 1.800000000000000300000000000000
alpha: 1.900000000000000100000000000000
alpha: 2.000000000000000000000000000000

OMG，当我告诉它时，我的计算机为什么不给出0.1的精确步骤。函数oscalcpcf对alpha中的小变化非常敏感，当我检查我的脚本使用的这些值时，如果我由oscalcpcf手动执行，则匹配。您能为我的计算机建议一种方法来提供0.1的精确步骤吗？谢谢。

Answer 1

我相信你有一个浮点错误，因为你的冒号生成0.1:0.1:2向量。

alpha的值不准确，因为计算机无法准确表示固定存储大小（例如双精度）的所有数字。对colon operator尤其如此，这使得不准确性通过向量元素传播。

现在，我不确定这是否会改善您的结果，但基于how the COLON operator works，我建议您尝试按以下方式运行循环（类似于this answer of mine）：< / p>

for alpha = ((1:20) / 10)

此外，如果oscalcpcf()是一个允许被篡改的函数，我建议你研究它并提高其对输入中微小变化的鲁棒性/灵敏度。 10 ^-14％不准确不应对您的结果产生重大影响。

Answer 2

试试这个：

alpha = ((1:20) *1e-1). you should get :

0.100000000000000005551115123126 
0.200000000000000011102230246252 
0.300000000000000044408920985006 
0.400000000000000022204460492503 
0.500000000000000000000000000000 
0.600000000000000088817841970013 
0.700000000000000066613381477509 
0.800000000000000044408920985006 
0.900000000000000022204460492503 
1.000000000000000000000000000000 
1.100000000000000088817841970013 
1.200000000000000177635683940025 
1.300000000000000044408920985006 
1.400000000000000133226762955019 
1.500000000000000000000000000000 
1.600000000000000088817841970013 
1.700000000000000177635683940025 
1.800000000000000044408920985006 
1.900000000000000133226762955019 
2.000000000000000000000000000000

结果将精确到双精度标准。此外，matlab symbolic toolbox中的sym和vpa函数可能会有所帮助。

非常简单的MATLAB代码，结果不合逻辑

2 个答案: