sombody可以简化这个等式给我吗？

问题描述

我知道我可以使用数学，但遗憾的是我没有之一。
我只是想找到A，B，C，D
形矩阵

  | X1 ^ 3 ^ X1 X1 2 1 | | A | | Y0 |
| X2 ^ 3 ^ X2 X2 2 1 | | B | | Y1 |
| X3 ^ 3 ^ X3 X3 2 1 | X | B | = | Y2 |
| X4 ^ 3 ^ X4 X4 2 1 | | D | | Y3 |
 

我只是想找到A，B，C和D的简化公式。

其实我试图做一个程序的的Arduino ，需要使用曲线4分接头，这样我可以predict未来百分点。我见过this帖子，但抛物线是不是我需要足够的精确。

我已经尝试。

  linearSolve [{{（X1）^ 3，（X1）^ 2，X1，1}，
              {（X2）^ 3，（×2）^ 2，X 2，1}，
              {（X3）^ 3，（X3）^ 2，X3，1}，
              {（4个）^ 3，（X4）^ 2，×4，1}}，{{Y1}，{Y2}，{Y3}，{Y 4}}]
 

它返回一个长的结果，这可以被简化。但是，我不能进入全面结果，以简化搜索栏（它给我的错误：输入太长）。

任何想法？嗯，我想这将有可能在桌面版本。

甚至在此之后，如果结果是相当长的，请代X1 = 0，让我知道最简化的结果。

解决方案

  //输入数据：数组x []和y []
// X [1]，X [2]中，x [3]中，x [4]  - 的X值
// Y [1]，Y [2]，Y [3]，Y [4]  -  Y值//计算
A = 0
B = 0
C = 0
D = 0
S1 = X [1] + X [2] + X [3] + X [4]
S2 = X [1] * X [2] + X [1] * X [3] + X [1] * X [4] + X [2] * X [3] + X [2] * X [4 ] + X [3] * X [4]
S3 = X [1] * X [2] * X [3] + X [1] * X [2] * X [4] + X [1] * X [3] * X [4] + X [2 ] * X [3] * X [4]
对于i = 1至4环
   C0 =值Y [i] /（（（4 * X [i]于-3 * S 1）* X [I] + 2 * S 2）* X [i]于-S3）
   C1 = C0 *（S1  -  X [I]）
   C2 = S2 * C0  -  C1 * X [I]
   C3 = S3 * C0  -  C2 * X [I]
   A = A + C0
   B =  -  C1
   C = C + C2
   Ð= D  -  C3
结束循环//结果：A，B，C，D
 

I know I can use Mathematica, but sadly I dont have one.I just want to find the A,B,C,Dform matrix

| X1^3 x1^2 X1 1 |   |A|   |y0|
| X2^3 x2^2 X2 1 |   |B|   |y1|
| X3^3 x3^2 X3 1 | X |C| = |y2|
| X4^3 x4^2 X4 1 |   |D|   |y3|

I just want to find the simplified equations for A, B, C and D.

Actually I am trying to do a program in arduino that requires curve fitting using 4 points, so that I can predict the future points. I have seen this post , but parabola isn't accurate enough for my need.

I have already tried http://www.wolframalpha.com/.

linearSolve [{{(x1)^3, (x1)^2, x1, 1},
              {(x2)^3, (x2)^2, x2, 1},
              {(x3)^3, (x3)^2, x3, 1},
              {(x4)^3, (x4)^2, x4, 1}}, {{y1},{y2},{y3},{y4}}]

It returns a long result, which can be simplified. But, I cannot enter the full result in the search bar for simplifying (It gives me the error :Input Too Long!).

Any Ideas? Well I guess it would be possible in desktop versions.

Even after that, if the result is quite long, please substitute x1 = 0 and let me know the simplified result.

解决方案

// Input data: arrays x[] and y[]
// x[1],x[2],x[3],x[4] - X values
// y[1],y[2],y[3],y[4] - Y values

// Calculations
A = 0
B = 0
C = 0
D = 0
S1 = x[1] + x[2] + x[3] + x[4]
S2 = x[1]*x[2] + x[1]*x[3] + x[1]*x[4] + x[2]*x[3] + x[2]*x[4] + x[3]*x[4]
S3 = x[1]*x[2]*x[3] + x[1]*x[2]*x[4] + x[1]*x[3]*x[4] + x[2]*x[3]*x[4]
for i = 1 to 4 loop
   C0 = y[i]/(((4*x[i]-3*S1)*x[i]+2*S2)*x[i]-S3)
   C1 = C0*(S1 - x[i])
   C2 = S2*C0 - C1*x[i]
   C3 = S3*C0 - C2*x[i]
   A = A + C0
   B = B - C1
   C = C + C2
   D = D - C3
end-loop

// Result: A, B, C, D

这篇关于sombody可以简化这个等式给我吗？的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持！