[proofplan] We prove the result by induction on the derivative order $m$. The base case is the standard fact that a differentiable real-valued function on an interval whose derivative vanishes everywhere is constant, obtained from the [Mean Value Theorem](/theorems/186). For the induction step, we apply the induction hypothesis to $f'$ to see that $f'$ is a polynomial of degree at most $m-2$, integrate that polynomial explicitly, and then use the base case on the difference between $f$ and the constructed antiderivative. [/proofplan] [step:Show that a function with zero derivative on an interval is constant] We first record the elementary consequence of the Mean Value Theorem needed below. [claim:Zero derivative on an interval forces constancy] Let $J \subset \mathbb{R}$ be an interval, and let $g: J \to \mathbb{R}$ be a [continuous function](/page/Continuous%20Function) on $J$ and differentiable at every interior point of $J$. If $g'(x)=0$ for every interior point $x$ of $J$, then $g$ is constant on $J$. [/claim] [proof] Let $x,y \in J$ with $x<y$. Since $J$ is an interval, $[x,y] \subset J$. The restriction $g|_{[x,y]}: [x,y] \to \mathbb{R}$ is continuous on $[x,y]$ and differentiable on $(x,y)$. By the Mean Value Theorem, there exists $c \in (x,y)$ such that \begin{align*} g(y)-g(x)=g'(c)(y-x). \end{align*} Since $g'(c)=0$, this gives $g(y)=g(x)$. Because any two distinct points of $J$ can be ordered in this way, and equal points have equal values by reflexivity of equality, $g$ is constant on $J$. [/proof] [guided] We need a precise way to pass from the pointwise condition $g'=0$ to a global statement about all values of $g$. Let $x,y \in J$ with $x<y$. The interval hypothesis is essential here: it gives $[x,y] \subset J$, so the restriction $g|_{[x,y]}: [x,y] \to \mathbb{R}$ is a legitimate function on a closed interval. The Mean Value Theorem applies to $g|_{[x,y]}$ because $g$ is continuous on $J$, hence continuous on $[x,y]$, and differentiable at every interior point of $J$, hence differentiable on $(x,y)$. Therefore there exists $c \in (x,y)$ satisfying \begin{align*} g(y)-g(x)=g'(c)(y-x). \end{align*} The hypothesis gives $g'(c)=0$, so \begin{align*} g(y)-g(x)=0. \end{align*} Thus $g(y)=g(x)$. Since this holds for every ordered pair $x<y$ in $J$, the function $g$ has the same value at every point of $J$. Hence $g$ is constant on $J$. [/guided] [/step] [step:Prove the polynomial representation by induction on the derivative order] We prove the theorem for every $m \in \mathbb{N}$ by induction on $m$. For $m=1$, the hypothesis is $f'(x)=0$ for every $x \in I$. Applying the claim with $J=I$ and $g=f$, there exists $a_0 \in \mathbb{R}$ such that \begin{align*} f(x)=a_0 \end{align*} for every $x \in I$. This is the required representation \begin{align*} f(x)=\sum_{i=0}^{0} a_i x^i. \end{align*} Assume now that $m \geq 2$ and that the result is known for $m-1$. Let $f \in C^m(I;\mathbb{R})$ satisfy $f^{(m)}(x)=0$ for every $x \in I$. Define the derivative function \begin{align*} h: I \to \mathbb{R}, \quad h(x)=f'(x). \end{align*} Since $f \in C^m(I;\mathbb{R})$, we have $h \in C^{m-1}(I;\mathbb{R})$, and \begin{align*} h^{(m-1)}(x)=f^{(m)}(x)=0 \end{align*} for every $x \in I$. By the induction hypothesis applied to $h$, there exist [real numbers](/page/Real%20Numbers) $b_0,\dots,b_{m-2} \in \mathbb{R}$ such that \begin{align*} h(x)=\sum_{i=0}^{m-2} b_i x^i \end{align*} for every $x \in I$. Define the polynomial antiderivative \begin{align*} Q: \mathbb{R} \to \mathbb{R}, \quad Q(x)=\sum_{i=0}^{m-2} \frac{b_i}{i+1}x^{i+1}. \end{align*} By the ordinary differentiation rule for monomials and finite sums, \begin{align*} Q'(x)=\sum_{i=0}^{m-2} b_i x^i \end{align*} for every $x \in \mathbb{R}$. Define \begin{align*} r: I \to \mathbb{R}, \quad r(x)=f(x)-Q(x). \end{align*} Since $f \in C^m(I;\mathbb{R})$ and $m \geq 2$, the function $f$ is continuous on $I$ and differentiable at every interior point of $I$. Since $Q$ is a polynomial function on $\mathbb{R}$, the function $Q|_I: I \to \mathbb{R}$ is continuous on $I$ and differentiable at every interior point of $I$. Hence $r=f-Q|_I$ is continuous on $I$ and differentiable at every interior point of $I$. For every interior point $x$ of $I$, \begin{align*} r'(x)=f'(x)-Q'(x)=h(x)-h(x)=0. \end{align*} Applying the claim to $r$, there exists $a_0 \in \mathbb{R}$ such that $r(x)=a_0$ for every $x \in I$. Therefore \begin{align*} f(x)=a_0+\sum_{i=0}^{m-2} \frac{b_i}{i+1}x^{i+1} \end{align*} for every $x \in I$. For each integer $j$ with $1 \leq j \leq m-1$, define \begin{align*} a_j=\frac{b_{j-1}}{j}. \end{align*} Together with the constant $a_0$ above, this gives real numbers $a_0,\dots,a_{m-1}$ satisfying \begin{align*} f(x)=\sum_{j=0}^{m-1} a_j x^j \end{align*} for every $x \in I$. This completes the induction. [guided] We now prove the theorem for every positive integer $m$ by induction. The statement has no $m=0$ case, because the formalized statement specifies $\mathbb{N}=\{1,2,3,\dots\}$. For $m=1$, the hypothesis says $f'(x)=0$ for every $x \in I$. The function $f$ belongs to $C^1(I;\mathbb{R})$, so it is continuous on $I$ and differentiable at every interior point of $I$. Therefore the zero-derivative claim applies with $J=I$ and $g=f$. It gives a real number $a_0 \in \mathbb{R}$ such that \begin{align*} f(x)=a_0 \end{align*} for every $x \in I$. Since the only summand in $\sum_{i=0}^{0} a_i x^i$ is $a_0x^0=a_0$, this is exactly the desired representation for $m=1$. Assume now that $m \geq 2$ and that the theorem has already been proved for the positive integer $m-1$. Let $f \in C^m(I;\mathbb{R})$ satisfy $f^{(m)}(x)=0$ for every $x \in I$. The natural object to which the induction hypothesis can be applied is the first derivative of $f$, because differentiating $f'$ exactly $m-1$ times gives the $m$th derivative of $f$. Define \begin{align*} h: I &\to \mathbb{R} \end{align*} by setting $h(x)=f'(x)$ for $x \in I$. Since $f \in C^m(I;\mathbb{R})$, its first derivative satisfies $h \in C^{m-1}(I;\mathbb{R})$. For every $x \in I$, \begin{align*} h^{(m-1)}(x)=f^{(m)}(x)=0. \end{align*} Thus the induction hypothesis applies to $h$. Hence there exist real numbers $b_0,\dots,b_{m-2} \in \mathbb{R}$ such that \begin{align*} h(x)=\sum_{i=0}^{m-2} b_i x^i \end{align*} for every $x \in I$. We now build an explicit polynomial antiderivative of this polynomial expression for $h$. Define \begin{align*} Q: \mathbb{R} &\to \mathbb{R} \end{align*} by \begin{align*} Q(x)=\sum_{i=0}^{m-2} \frac{b_i}{i+1}x^{i+1}. \end{align*} By the ordinary differentiation rule for monomials and finite sums, \begin{align*} Q'(x)=\sum_{i=0}^{m-2} b_i x^i \end{align*} for every $x \in \mathbb{R}$. In particular, for every $x \in I$, we have $Q'(x)=h(x)=f'(x)$. Define the remainder function \begin{align*} r: I &\to \mathbb{R} \end{align*} by setting $r(x)=f(x)-Q(x)$ for $x \in I$. Since $f \in C^m(I;\mathbb{R})$ and $m \geq 2$, the function $f$ is continuous on $I$ and differentiable at every interior point of $I$. Since $Q$ is a polynomial function on $\mathbb{R}$, its restriction $Q|_I: I \to \mathbb{R}$ is continuous on $I$ and differentiable at every interior point of $I$. Therefore $r=f-Q|_I$ is continuous on $I$ and differentiable at every interior point of $I$. For every interior point $x$ of $I$, \begin{align*} r'(x)=f'(x)-Q'(x)=h(x)-h(x)=0. \end{align*} The zero-derivative claim applies to $r$, so there exists $a_0 \in \mathbb{R}$ such that $r(x)=a_0$ for every $x \in I$. Hence \begin{align*} f(x)=a_0+\sum_{i=0}^{m-2} \frac{b_i}{i+1}x^{i+1} \end{align*} for every $x \in I$. It remains only to rewrite the coefficients with the indexing used in the theorem statement. For each integer $j$ with $1 \leq j \leq m-1$, define \begin{align*} a_j=\frac{b_{j-1}}{j}. \end{align*} Together with the constant $a_0$ obtained from the zero-derivative claim, these real numbers satisfy \begin{align*} f(x)=\sum_{j=0}^{m-1} a_j x^j \end{align*} for every $x \in I$. This proves the induction step, and therefore the theorem holds for every positive integer $m$. [/guided] [/step]