Problem source

$\triangle$ is an operation that takes polynomials in $x$ to polynomials in $x$; that is, given any polynomial $\h(x)$, there is a polynomial called $\triangle \h(x)$ which is obtained from $\h(x)$ using the rules that define $\triangle$. These rules are as follows:
\begin{questionparts}
\item $\triangle x = 1\,$;
\item $\triangle \big( \f(x)+\g(x)\big) = \triangle \f(x) + \triangle\g(x)\,$ for any polynomials $\f(x)$ and $\g(x)$;
\item $\triangle \big( \lambda \f(x)\big) =\lambda \triangle \f(x)$ for any constant $\lambda$ and any polynomial $\f(x)$;
\item $\triangle \big( \f(x)\g(x)\big) = \f(x) \triangle \g(x) +\g(x)\triangle \f(x)$ for any polynomials $\f(x)$ and $\g(x)$.
\end{questionparts}
Using these rules show that, if $\f(x)$ is a polynomial of degree zero (that is, a constant), then $\triangle \f(x) =0$. Calculate $\triangle x^2$ and $\triangle x^3$.
Prove that $\triangle \h(x) \equiv  \dfrac{\d \h(x)}{\d x \ \ \ }$ for any polynomial $\h(x)$. You should make it  clear whenever you use
one of the above rules in your proof. $\vphantom{\int}$

Solution source

Claim: If $f$ is a constant, then $\triangle f = 0$

Proof: First consider $f(x) = 1, g(x) = x$ then we must have:
\begin{align*}
&& \triangle (1x) &= 1 \triangle x + x \triangle 1 \tag{iv} \\
&&&= 1 \cdot 1 + x \triangle 1 \tag{i} \\
\Rightarrow && 1 &= 1 + x \triangle 1 \tag{i} \\
\Rightarrow && \triangle 1 &= 0 \\
\Rightarrow && \triangle c &= 0 \tag{iii}
\end{align*}

\begin{align*}
&& \triangle (x^2) &= x \triangle x + x \triangle x \tag{iv} \\
&&&= x \cdot 1 + x \cdot 1 \tag{i} \\
&&&= 2x \\
\\
&& \triangle (x^3) &= x^2 \triangle x + x \triangle (x^2) \tag{iv} \\
&&&= x^2 \cdot 1 + x \cdot 2x \tag{$\triangle x^2 = 2x$}\\
&&&= 3x^2
\end{align*}

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

Proof: Since both $\triangle$ and $\frac{\d}{\d x}$ are linear (properties $(ii)$ and $(iii)$) it suffices to prove that: $\triangle x^n = nx^{n-1}$. For this we proceed by induction.

Base cases (we've proved up to $n = 3$ so we're good).

Suppose it's true for some $n$, then consider $n + 1$,

\begin{align*}
&& \triangle (x^{n+1}) &= x \triangle (x^n) + x^n \triangle x \tag{iv} \\
&&&= x \cdot n x^{n-1} + x^n \triangle x \tag{Ind. hyp.} \\
&&&= nx^n + x^n \tag{i} \\
&&&= (n+1)x^{n}
\end{align*}

Therefore it's true for for $n+1$. Therefore by induction it's true for all $n$.