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pawlo392
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I am trying to prove the expression for Dickson polynomials:
$$D_n(x, a)=\sum_{i=0}^{\lfloor \frac{n}{2}\rfloor}d_{n,i}x^{n-2i}, \quad \text{where} \quad d_{n,i}=\frac{n}{n-i}{n-i\choose i}(-a)^i$$
I am supposed to use the recurrence relation:
$$D_n(x,a)=xD_{n-1}(x,a)-aD_{n-2}(x,a)$$
I have tried to use induction to prove this, but I am having trouble. Here's what I have so far:
Base case: For ##n=1##, we have:
$$D_1(x,a)=x$$
which matches the expression for ##D_n(x,a)## when ##n=1##. So the base case holds.
Inductive step: Assume that the expression for ##D_n(x,a)## holds for some ##n \geq 1##. We want to show that it also holds for ##n+1##.
From the recurrence relation, we have:
$$D_{n+1}(x,a) = xD_n(x,a) - aD_{n-1}(x,a)$$
Substituting the expression for ##D_n(x,a)## and ##D_{n-1}(x,a)##, we get:
\begin{align*}
D_{n+1}(x,a) &= xD_n(x,a) - aD_{n-1}(x,a) \\
&= x\sum_{i=0}^{\lfloor \frac{n}{2}\rfloor}d_{n,i}x^{n-2i} - a\sum_{i=0}^{\lfloor \frac{n-1}{2}\rfloor}d_{n-1,i}x^{n-1-2i} \\
&= \sum_{i=0}^{\lfloor \frac{n}{2}\rfloor}d_{n,i}x^{n-2i+1} - \sum_{i=0}^{\lfloor \frac{n-1}{2}\rfloor}d_{n-1,i}ax^{n-1-2i} \\
&= \sum_{i=0}^{\lfloor \frac{n}{2}\rfloor}d_{n,i}x^{n-2i+1} - \sum_{i=1}^{\lfloor \frac{n}{2}\rfloor}d_{n,i-1}ax^{n-2i+1} \\
\end{align*}
Now, I am stuck here and not sure how to proceed. Can someone please provide some guidance or hints on how to continue with the proof? Thanks in advance.
$$D_n(x, a)=\sum_{i=0}^{\lfloor \frac{n}{2}\rfloor}d_{n,i}x^{n-2i}, \quad \text{where} \quad d_{n,i}=\frac{n}{n-i}{n-i\choose i}(-a)^i$$
I am supposed to use the recurrence relation:
$$D_n(x,a)=xD_{n-1}(x,a)-aD_{n-2}(x,a)$$
I have tried to use induction to prove this, but I am having trouble. Here's what I have so far:
Base case: For ##n=1##, we have:
$$D_1(x,a)=x$$
which matches the expression for ##D_n(x,a)## when ##n=1##. So the base case holds.
Inductive step: Assume that the expression for ##D_n(x,a)## holds for some ##n \geq 1##. We want to show that it also holds for ##n+1##.
From the recurrence relation, we have:
$$D_{n+1}(x,a) = xD_n(x,a) - aD_{n-1}(x,a)$$
Substituting the expression for ##D_n(x,a)## and ##D_{n-1}(x,a)##, we get:
\begin{align*}
D_{n+1}(x,a) &= xD_n(x,a) - aD_{n-1}(x,a) \\
&= x\sum_{i=0}^{\lfloor \frac{n}{2}\rfloor}d_{n,i}x^{n-2i} - a\sum_{i=0}^{\lfloor \frac{n-1}{2}\rfloor}d_{n-1,i}x^{n-1-2i} \\
&= \sum_{i=0}^{\lfloor \frac{n}{2}\rfloor}d_{n,i}x^{n-2i+1} - \sum_{i=0}^{\lfloor \frac{n-1}{2}\rfloor}d_{n-1,i}ax^{n-1-2i} \\
&= \sum_{i=0}^{\lfloor \frac{n}{2}\rfloor}d_{n,i}x^{n-2i+1} - \sum_{i=1}^{\lfloor \frac{n}{2}\rfloor}d_{n,i-1}ax^{n-2i+1} \\
\end{align*}
Now, I am stuck here and not sure how to proceed. Can someone please provide some guidance or hints on how to continue with the proof? Thanks in advance.
Last edited: