Problem source

$\lozenge$ is an operation which take polynomials in $x$ to polynomials in $x$; that is, given a polynomial $\mathrm{h}(x)$ there is another polynomial called $\lozenge\mathrm{h}(x)$. It is given that, if $\mathrm{f}(x)$ and $\mathrm{g}(x)$ are any two polynomials in $x$, the following are always true: 
\begin{questionparts}
\item $\lozenge(\mathrm{f}(x)\mathrm{g}(x))=\mathrm{g}(x)\lozenge\mathrm{f}(x)+\mathrm{f}(x)\lozenge\mathrm{g}(x),$ 
\item $\lozenge(\mathrm{f}(x)+\mathrm{g}(x))=\lozenge\mathrm{f}(x)+\lozenge\mathrm{g}(x),$ 
\item $\lozenge x=1$
\item if $\lambda$ is a constant then $\lozenge(\lambda\mathrm{f}(x))=\lambda\lozenge\mathrm{f}(x).$
\end{questionparts}
Show that, if $\mathrm{f}(x)$ is a constant (i.e., a polynomial of degree zero), then $\lozenge\mathrm{f}(x)=0.$ 
Calculate $\lozenge x^{2}$ and $\lozenge x^{3}.$ Prove that $\lozenge\mathrm{h}(x)=\dfrac{\mathrm{d}}{\mathrm{d}x}(\mathrm{h}(x))$ for any polynomial $\mathrm{h}(x)$.

Solution source

Claim: If $f(x) = c$ then $\lozenge f(x) = 0$

Proof: Consider $g(x) = x$ then 

\begin{align*}
(1) && \lozenge(f(x)g(x)) &= g(x) \lozenge f(x) + f(x) \lozenge g(x) \\
\Rightarrow && \lozenge(c x) &= x \lozenge f(x) + c \lozenge x \\
(4) && \lozenge(c x) &= c \lozenge x \\
\Rightarrow && 0 &= x \lozenge f(x)  \\
\Rightarrow && \lozenge f(x) &= 0
\end{align*}

\begin{align*}
(1) && \lozenge(x^2) &= x \lozenge x + x \lozenge x \\
(3) &&&= 2 x \cdot 1  \\
&&&= 2x \\
\\
(1) && \lozenge (x^3) &= x^2 \lozenge x + x \lozenge (x^2) \\
&&&= x^2 \cdot \underbrace{1}_{(3)} + x \cdot\underbrace{ 2x}_{\text{previous part}} \\
&&&= 3x^2
\end{align*}

Claim: $\lozenge h(x) = \frac{\d }{\d x} ( h(x))$ for any polynomial $h$.

Proof: (By (strong) induction on the degree of $h$).

Base case: True, we proved this in the first part of the question.

Inductive step: Assume true for all polynomials of degree less than or equal to $k$. Then consider $n = k+1$.

We can write $h(x) = ax^{k+1} + h_k(x)$ where $h_k(x)$ is a polynomial of degree less than or equal to $k$. Then notice:

\begin{align*}
&& \lozenge (h(x)) &= \lozenge (ax^{k+1} + h_k(x)) \\
(2) &&&= \lozenge (ax^{k+1})+ \lozenge (h_k(x)) \\
&&&=\underbrace{a\lozenge (x^{k+1})}_{(4)}+ \underbrace{\frac{\d}{\d x} (h_k(x))}_{\text{inductive hypothesis}}\\ 
&&&= a \underbrace{\left (x \lozenge x^k + x^k \lozenge x \right)}_{(1)} + \frac{\d}{\d x} (h_k(x)) \\
&&&= a \left ( x \cdot \underbrace{k x^{k-1}}_{\text{inductive hyp.}} + x^k \cdot \underbrace{1}_{(3)} \right)  + \frac{\d}{\d x} (h_k(x)) \\
&&&= (k+1)a x^k + \frac{\d}{\d x} (h_k(x)) \\
&&&= \frac{\d }{\d x} \left ( ax^{k+1} + h_k(x) \right) \\
&&&= \frac{\d }{\d x} (h(x))
\end{align*} 

Therefore since our statement is true for $n=0$ and if it is true for $n=k$ it is true for $n=k+1$ by the principle of mathematical induction it is true for all $n \geq 0$