daviddoji/CalculusWithJuliaNotes.jl
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some typos

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This commit is contained in:
Fang Liu
2025-05-04 15:08:47 +08:00
parent ed1d92197a
commit d55e4802fb
7 changed files with 17 additions and 17 deletions
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8
quarto/precalc/polynomials_package.qmd
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@@ -454,18 +454,18 @@ numericq(4)
The identification of a collection of coefficients with a polynomial depends on an understood **basis**. A basis for the polynomials of degree $n$ or less, consists of a minimal collection of polynomials for which all the polynomials of degree $n$ or less can be expressed through a combination of sums of terms, each of which is just a coefficient times a basis member. The typical basis is the $n+1$ polynomials $1, x, x^2, \dots, x^n$. However, though every basis must have $n+1$ members, they need not be these.
A basis used by [Lagrange](https://en.wikipedia.org/wiki/Lagrange_polynomial) is the following. Let there be $n+1$ points distinct points $x_0, x_1, \dots, x_n$. For each $i$ in $0$ to $n$ define
A basis used by [Lagrange](https://en.wikipedia.org/wiki/Lagrange_polynomial) is the following. Let there be $n+1$ distinct points $x_0, x_1, \dots, x_n$. For each $i$ in $0$ to $n$ define
$$
l_i(x) = \prod_{0 \leq j \leq n; j \ne i} \frac{x-x_j}{x_i - x_j} =
\frac{(x-x_1)\cdot(x-x_2)\cdot \cdots \cdot (x-x_{i-1}) \cdot (x-x_{i+1}) \cdot \cdots \cdot (x-x_n)}{(x_i-x_1)\cdot(x_i-x_2)\cdot \cdots \cdot (x_i-x_{i-1}) \cdot (x_i-x_{i+1}) \cdot \cdots \cdot (x_i-x_n)}.
\frac{(x-x_0)\cdot(x-x_1)\cdot \cdots \cdot (x-x_{i-1}) \cdot (x-x_{i+1}) \cdot \cdots \cdot (x-x_n)}{(x_i-x_0)\cdot(x_i-x_1)\cdot \cdots \cdot (x_i-x_{i-1}) \cdot (x_i-x_{i+1}) \cdot \cdots \cdot (x_i-x_n)}.
$$
That is $l_i(x)$ is a product of terms like $(x-x_j)/(x_i-x_j)$ *except* when $j=i$.
What is is the value of $l_0(x_0)$?
What is the value of $l_0(x_0)$?
```{julia}
@@ -578,7 +578,7 @@ choices = [
It is ``0\cdot T_0(x) + 1\cdot T_1(x) + 2\cdot T_2(x) + 3\cdot T_3(x) = 0``
"""
raw"""
It is ``0\cdot T_0(x) + 1\cdot T_1(x) + 2\cdot T_2(x) + 3\cdot T_3(x) = -2 - 8\cdot x + 4\cdot x^2 + 12\cdot x^3```
It is ``0\cdot T_0(x) + 1\cdot T_1(x) + 2\cdot T_2(x) + 3\cdot T_3(x) = -2 - 8\cdot x + 4\cdot x^2 + 12\cdot x^3``
"""]
radioq(choices, 3)
```