In MATLAB, C = zeros (30,

Zeros () creates a matrix whose elements are all 0. Parameters 30 and 1 define the number of rows and columns of the matrix. The result of zeros (30,1) is a matrix with 30 rows and 1 column and all 0. Zeros (30,1) + 1 is to add 1 to each element, and the result is a matrix with 30 rows and 1 column and all 1 > > C = zeros (10,1) + 1C = 11

What does zeros (1: n) mean? I have encountered this problem. Please get zeros (1: 4) and why it is the result calculated by MATLAB

With the help of MATLAB, help zeros all help zeros all zeros zeros array. Zeros (n) is an n-by-n matrix of zeros. Zeros (m, n) or zeros ([M, n]) is an M-by-N matrix of zeros. Zeros (m, N, P,...) or zeros

Linear algebra must be calculated by Vandermonde determinant

Making auxiliary determinant D1=
1 1 1 1 1
a b c d x
a^2 b^2 c^2 d^2 x^2
a^3 b^3 c^3 d^3 x^3
a^4 b^4 c^4 d^4 x^4
This is the Vandermonde determinant
D1 = (b-a)(c-a)(d-a)(c-b)(d-b)(d-c)(x-a)(x-b)(x-c)(x-d).
Because the coefficient - M44 of x ^ 3 in determinant D1 is determinant D
therefore
D = -(b-a)(c-a)(d-a)(c-b)(d-b)(d-c)(-a-b-c-d)
= (a-b)(a-c)(a-d)(b-c)(b-d)(c-d)(a+b+c+d).

Determinant calculation and Vandermonde calculation

It's like "the king of the five halls". I can't see clearly!
1) First of all, the determinant "drop every one" is transformed into a standard Vandermonde. It needs n (n + 1) / 2 times [line by line exchange]
(because the determinant itself is a determinant of order n + 1)
Dn+1=(-1)^[n(n+1)/2] |1 1 1 . 1|
a a-1 a-2 . a-n
a² (a-1)²(a-2)²..(a-n)²
.
aⁿ(a-1)ⁿ(a-2)ⁿ...(a-n)ⁿ
2) According to Vandermonde expansion formula
=[(a-n)-(a-n+1)][(a-n)-(a-n+2)]...[(a-n)-a]*(-1)(-2)..[-(n-1)].(-2)(-1)(-1)[(-1)^n(n+1)/2]
=[(-n)^1]*[-(n-1)^2]*[-(n-2)^3]*...*[(-2)^(n-1)]*[(-1)^n]*{(-1)^[n(n+1)]/2}
=[(-1)^(1+2+3+...+n)]*{(-1)^[n(n+1)/2]}*∏ k! (k=1 to n)
={(-1)^[n(1+n)/2+n(1+n)/2]}* ∏ k! (k=1 to n)
=[(-1)^n(n+1)]*∏ k ! （k=1 to n)

How to calculate Vandermonde determinant
I see an example in the book. The determinant D is as follows
1 1 1 1
1 2 4 8
1 3 9 27
1 4 16 64
And then he's a Vandermonde determinant
Then we get the result d = 1 * 2 * 3 * 1 * 2 * 1 = 12
I want to say that Vandermonde's determinant is not the first line is 1, how this first column is also 1, and how to calculate what Vandermonde's formula is

To transpose the determinant, (according to the first article of the determinant nature), the determinant is the Vandermonde determinant: D = | 1 1 1 1 1 1 1 1 1 2 3 41 & # 178; 2 & # 178; 3 & # 178; 4 & # 178; 1 & # 179; 2 & # 179; 3 & # 179; 4 & # 179; = (4-3) (4-2) (4-1) (3-2) (3-1) (2-1) = 1 * 2 * 1 = 12

Concrete proof of finding Vandermonde determinant formula~

Proof of Vandermonde determinant
I want to ask if there is a recursive process in the proof of Vandermonde determinant? It directly comes out n, and there is no n + 1. If there is a recursive process, why? Thank you

If n-1 is proved by induction, then n-1 is proved

The proof of Vandermonde determinant and his conclusion

Use mathematical induction
When n = 2
Vandermonde determinant D2 = x2-x1 Vandermonde determinant holds
Let's assume that the Vandermonde determinant holds for order n-1
First of all, we need to reduce DN from the nth line, subtract the X1 times of the previous line from the next line, and then expand it according to the first line, so DN = (x2-x1) (x3-x1)... (xn-x1) DN-1, then DN = ||||||| (Xi XJ) (where |||||||||||||||||||||||||

How to understand Vandermonde's determinant?

For its proof, I won't write it. In fact, it's very simple. According to the final result, you can use mathematical induction to transform the determinant
For its application
1, you can use it to get some special determinant values
2. Use it to judge the uniqueness of interpolation polynomials; judge the linear independence of some special vector systems (especially when the application of independence of solutions of differential equations is obvious), and judge the existence and uniqueness of solutions of equations
3. Through the above, we can solve some special equations

To calculate the determinant, Vandermonde | 1 | a B C D | a ^ 2 B ^ 2 C ^ 2 D ^ 2 | a ^ 4 B ^ 4 C ^ 4 d ^ 4 is required|

Consider the determinant (*) 1 1A B C D XA ^ 2 B ^ 2 C ^ 2 D ^ 2 x ^ 2A ^ 3 B ^ 3 C ^ 3 D ^ 3 x ^ 3A ^ 4 B ^ 4 C ^ 4 d ^ 4 x ^ 4. Obviously, the determinant in the title is the opposite number of the coefficient of x ^ 3 of the determinant (*) (the coefficient of x ^ 3 is its algebraic cofactor)

In MATLAB, C = zeros (30,