一般平滑步方程
[英]General smoothstep equation
问题描述
I saw this in Wikipedia: Generalization of higher-order equations 
我在Wikipedia中看到了这一点： 高阶方程的一般化
My code when I tried to implement it: 我尝试执行的代码：
function generalSmoothStep(a, x) { //Generalized smoothstep var result = 0; for (var n = 0; n < a - 1; n ++) { result += binom(-a, n) * binom(2 * a - 1, a - n - 1) * Math.pow(x, a + n); } return result; } function smoothStep(x) { //Normal smoothstep return -2 * Math.pow(x, 3) + 3 * Math.pow(x, 2); //I know about x * x * (3 - 2 * x); } function binom(a, b) { //Binomial coefficient return Math.factorial(a) / (Math.factorial(ab) * Math.factorial(b)); } Math.factorial = function(value) { //Factorial var result = 1; if (value > 0) { for (var count = 1; count <= value; count ++) { result *= count; } } else if (value < 0) { for (var count = -1; count >= value; count --) { result *= count; } } else { result = 1; } return result; }; document.getElementById("normalStep").innerHTML = "smoothStep(0.2) = " + smoothStep(0.2); document.getElementById("generalStep").innerHTML = "generalSmoothStep(2, 0.2) = " + generalSmoothStep(2, 0.2);
<head> <script src="https://cdnjs.cloudflare.com/ajax/libs/mathjax/2.7.0/MathJax.js?config=TeX-MML-AM_CHTML"></script> <script> MathJax.Hub.Config({ tex2jax: {inlineMath: [['$','$'], ['\\\\(','\\\\)']]} }); </script> </head> <body> <p>The function: $\\operatorname{S}_a(x) = \\sum_{n=0}^{a-1} \\binom{-a}{n}\\binom{2a - 1}{a - n - 1}x^{a + n}$</p> <p>Here is the result when I run:</p> <p id="normalStep"></p> <p id="generalStep"></p> </body>
I know about binomial coefficient and factorial, but one problem is factorial can't use with negative number, so I tried to use tricks to bypass it, and failed... 我知道二项式系数和阶乘，但是一个问题是阶乘不能与负数一起使用，因此我尝试使用技巧绕过它，但失败了...
And also the part with ()() chained together, I think it is multiplication ()*() between binom, and multiply with x^(a+n), as above, but it still not work. 还有（）（）链接在一起的部分，我认为这是二项式之间的（）*（）乘以x ^（a + n），如上所述，但是仍然无法正常工作。
Used google with words like: "general smooth step", "smooth step sum" but still not return any good explanation about it... 用过google的词，例如：“一般平滑步长”，“平滑步长和”，但仍未返回有关此的任何良好解释...
Anyone have any idea why my code not work, and also how to implement general smoothStep function in Javascript 任何人都知道为什么我的代码不起作用，以及如何在Javascript中实现常规的smoothStep函数
2 个解决方案
解决方案1
2 已采纳 2016-12-17 07:41:50
You can use Pascal Triangle instead of binomial coefficient to solve the issue. 您可以使用Pascal Triangle而不是二项式系数来解决此问题。 Here is the link for more detail of Pascal Triangle. 这是有关Pascal Triangle的更多详细信息的链接。
https://trans4mind.com/personal_development/mathematics/series/pascalsTriangle.htm https://trans4mind.com/personal_development/mathematics/series/pascalGenCoefficients.gif https://trans4mind.com/personal_development/mathematics/series/pascalsTriangle.htm https://trans4mind.com/personal_development/mathematics/series/pascalGenCoefficients.gif
Here is the solution using Pascal Triangle. 这是使用Pascal Triangle的解决方案。
function generalSmoothStep(a, x) { //Generalized smoothstep var result = 0; for (var n = 0; n <= a - 1; n++) { result += (pascalTriangle(-a, n) * pascalTriangle(2 * a - 1, a - n - 1) * Math.pow(x, a + n)); } return result; } function pascalTriangle(a, b){ var result = 1; for(var i = 1; i <= b; i++){ result *= ((a - (i - 1)) / i); } return result; } function smoothStep(x) { //Normal smoothstep return -2 * Math.pow(x, 3) + 3 * Math.pow(x, 2); } console.log("smoothStep(0.3) = " + smoothStep(0.3)); console.log("generalSmoothStep(2, 0.3) = " + generalSmoothStep(2, 0.3));
解决方案2
0 2016-12-19 09:08:07
You can replace the negative binomial coefficient with a variant that uses positive numbers. 您可以将负二项式系数替换为使用正数的变体。 Based on the generalization of negative binomial coefficients as described on another Wikipedia article, the smoothstep equation using positive binomial coefficients would be: 根据另一篇Wikipedia文章所述的负二项式系数的一般化，使用正二项式系数的平滑步方程为：
So the resulting code would be: 因此，结果代码将是：
function generalSmoothStep(a, x) { //Generalized smoothstep var result = 0; for (var n = 0; n <= a - 1; n ++) { // Change negative binom coeff to positive equivalent result += Math.pow(-1, n) * binom(a + n - 1, n) * binom(2 * a - 1, a - n - 1) * Math.pow(x, a + n); } return result; } function binom(a, b) { //Binomial coefficient return Math.factorial(a) / (Math.factorial(ab) * Math.factorial(b)); } Math.factorial = function(value) { //Factorial var result = 1; if (value > 0) { for (var count = 1; count <= value; count ++) { result *= count; } } else if (value < 0) { for (var count = -1; count >= value; count --) { result *= count; } } else { result = 1; } return result; } function smoothStep(x) { //Normal smoothstep return -2 * Math.pow(x, 3) + 3 * Math.pow(x, 2); } console.log("smoothStep(0.3) = " + smoothStep(0.3)); console.log("generalSmoothStep(2, 0.3) = " + generalSmoothStep(2, 0.3));
However, since this version computes factorials, this function will begin to become less accurate a lot quicker than the code in Tushar Ghosh's answer for larger values of a . 但是，由于此版本计算阶乘，因此此函数的准确性将比Tushar Ghosh的答案中的a值更大的代码快得多。
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