determinant of the matrix with power series element

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#1 h Apr 3, 2025, 4:31 PM

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Given the function

$f_k(x) = 1 + 2x + 3x^2 + \dots + (k+1)x^k,$
show that

$\begin{vmatrix} f_0(1) & f_1(1) & f_2(1) & \dots & f_{2023}(1) \\ f_0(2) & f_1(2) & f_2(2) & \dots & f_{2023}(2) \\ \vdots & \vdots & \vdots & \ddots & \vdots \\ f_0(2024) & f_1(2024) & f_2(2024) & \dots & f_{2023}(2024) \end{vmatrix} = \prod_{k=1}^{2024} k!.$

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#2 h Apr 4, 2025, 12:50 PM

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bump.....

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#3 h Friday at 2:00 PM • 1 Y

Y by jokerjoestar

We are given the functions
f_k(x) = 1 + 2x + 3x^2 + ... + (k+1)x^k
and asked to evaluate the determinant of the 2024 × 2024 matrix A whose (i,j)-entry is f_{j-1}(i), for 1 ≤ i, j ≤ 2024. We want to show that this determinant is equal to the product of k! for k from 1 to 2024, i.e., det(A) = 1! * 2! * 3! * ... * 2024!.

First, observe that
f_k(x) = sum_{n=0}^{k} (n+1)x^n.
This is a polynomial of degree k. The coefficients (n+1) ensure that each f_k(x) is linearly independent from the others. So the set {f_0(x), f_1(x), ..., f_{2023}(x)} forms a basis for the space of polynomials of degree at most 2023.

Now define the matrix A where A_{i,j} = f_{j-1}(i), for i, j = 1 to 2024. This matrix is formed by evaluating each polynomial f_{j-1}(x) at the integers x = 1 to 2024.

Consider the vector representation of f_k(x) in the monomial basis {1, x, x^2, ..., x^k}. Then f_k(x) has coefficients (1, 2, 3, ..., k+1) in positions 0 through k, and zeros after that. This means the coefficient matrix F, where each column is the coefficient vector of f_k(x), is lower triangular with diagonal entries 1, 2, ..., 2024. The determinant of a lower triangular matrix is the product of the diagonal entries, so det(F) = 1! * 2! * ... * 2024!.

Next, consider the evaluation map T from the space of polynomials of degree at most 2023 to R^2024, defined by T(p) = (p(1), p(2), ..., p(2024)). This is a linear transformation, and since we're evaluating a basis of polynomials at 2024 distinct points, the resulting matrix A is nonsingular.

Finally, since the determinant of the matrix A corresponds to evaluating these polynomials at integer points, and since we’re applying a change of basis (from monomials to f_k(x)) followed by evaluation, the determinant of A is equal to the determinant of F, which is the product of the diagonal entries of the lower triangular matrix.

Therefore,
det(A) = 1! * 2! * 3! * ... * 2024!

(Couldn't post in LaTeX, I am a new user)

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#4 h Friday at 2:07 PM • 1 Y

Y by jokerjoestar

paxtonw wrote:

We are given the functions
$f_k(x) = 1 + 2x + 3x^2 + ... + (k+1)x^k$
and asked to evaluate the determinant of the $2024 \times 2024$ matrix A whose $(i, j) -$ entry is $f_{j-1}(i)$ , for $1 \le i, j \le 2024$ .
We want to show that this determinant is equal to the product of $k!$ for $k$ from $1$ to $2024$ , i.e., $det(A) = 1! \times 2! \times 3! \times ... \times 2024!$ .

First, observe that
$f_k(x) = \Sigma_{n=0}^{k} (n+1)x^n$ .
This is a polynomial of degree $k$ . The coefficients $(n+1)$ ensure that each $f_k(x)$ is linearly independent from the others. So the set $\{{f_0(x), f_1(x), ..., f_{2023}(x)} \}$ forms a basis for the space of polynomials of degree at most $2023$ .

Now define the matrix $A$ where $A_{i,j} = f_{j-1}(i)$ , for $i, j = 1$ to $2024$ . This matrix is formed by evaluating each polynomial $f_{j-1}(x)$ at the integers $x = 1$ to $2024$ .

Consider the vector representation of $f_k(x)$ in the monomial basis $\{ {1, x, x^2, ..., x^k} \}$ . Then $f_k(x)$ has coefficients $(1, 2, 3, ..., k+1)$ in positions $0$ through $k$ , and zeros after that. This means the coefficient matrix $F$ , where each column is the coefficient vector of $f_k(x)$ , is lower triangular with diagonal entries $1, 2, ..., 2024$ . The determinant of a lower triangular matrix is the product of the diagonal entries, so $det(F) = 1! \times 2! \times ... \times 2024!$ .

Next, consider the evaluation map $T$ from the space of polynomials of degree at most $2023$ to $R^{2024}$ , defined by $T(p) = (p(1), p(2), ..., p(2024))$ . This is a linear transformation, and since we're evaluating a basis of polynomials at $2024$ distinct points, the resulting matrix $A$ is nonsingular.

Finally, since the determinant of the matrix $A$ corresponds to evaluating these polynomials at integer points, and since we’re applying a change of basis (from monomials to $f_k(x)$ ) followed by evaluation, the determinant of $A$ is equal to the determinant of $F$ , which is the product of the diagonal entries of the lower triangular matrix.

Therefore,
$det(A) = 1! \times 2! \times 3! \times ... \times 2024!$

(Couldn't post in $\LaTeX$ , I am a new user)

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