Higher Order Derivatives — Rodrigues-type Polynomial
DAILY_CHALLENGE
Grade None
Question:
Let $\mathbb{R}$ denote the set of all real numbers. Consider the polynomial function $f:\mathbb{R}\to\mathbb{R}$ defined by
$$f(x)=\frac{d^{10}}{dx^{10}}\left((x^2-1)^{10}\right), \quad \text{for all } x\in\mathbb{R}.$$
Here $\dfrac{d^{10}}{dx^{10}}\left((x^2-1)^{10}\right)$ is the $10^{\text{th}}$ order derivative of the function $(x^2-1)^{10}$.
Then which of the following statements is (are) TRUE?
The coefficient of $x^8$ in the polynomial $f(x)$ is $(-10)\left(\dfrac{18!}{8!}\right)$
The value of $f(1)+f(-1)$ is equal to $10!\cdot2^{11}$
The degree of the polynomial $f(x)$ is $10$
The constant term of the polynomial $f(x)$ is $-\left(\dfrac{10!}{5!}\right)$
Step-by-Step Solution
Key Concept: The 10th derivative of $(x^2-1)^{10}$ is a degree-10 polynomial (Legendre-type). Specific coefficients are found by identifying which binomial term contributes after differentiation.
**Step 1: Expand $(x^2-1)^{10}$**
$(x^2-1)^{10}=\sum_{k=0}^{10}\binom{10}{k}x^{2k}(-1)^{10-k}$. The 10th derivative of $x^n$ is $\frac{n!}{(n-10)!}x^{n-10}$ for $n\geq10$, else $0$.
**Step 2: Check (A): coefficient of $x^8$**
Comes from term $k=9$: coefficient of $x^{18}$ is $\binom{10}{9}(-1)^1=-10$. After 10 differentiations: $-10\cdot\frac{18!}{8!}x^8$. Coefficient $=-10\cdot\frac{18!}{8!}$. (A) TRUE ✓
**Step 3: Check (B): $f(1)+f(-1)$**
$(x^2-1)^{10}=(x-1)^{10}(x+1)^{10}$. By Leibniz at $x=1$: only $k=10$ term survives: $f(1)=\binom{10}{10}(10!)\cdot 2^{10}=10!\cdot2^{10}$. Since $(x^2-1)^{10}$ is even, $f(-1)=f(1)=10!\cdot2^{10}$. Sum $=10!\cdot2^{11}$. (B) TRUE ✓
**Step 4: Check (C): degree of $f(x)$**
$(x^2-1)^{10}$ has degree 20. Taking 10th derivative reduces degree by 10. Degree of $f=10$. (C) TRUE ✓
**Step 5: Check (D): constant term**
Comes from $k=5$: $\binom{10}{5}(-1)^5x^{10}=-252x^{10}$. 10th derivative: $-252\cdot10!$. Option D claims $-\frac{10!}{5!}=-10!/120$, but $252\cdot10!\neq\frac{10!}{5!}$. (D) FALSE ✗
Correct Answer: A, B, C