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The Rolle theorem states that if a real function is continuous on $[u,v]$, differentiable on $(u,v)$, and has equal endpoint values, then its derivative vanishes somewhere in $(u,v)$. Order the $k$ distinct real roots as $r_1<\cdots <r_k$. The theorem gives a root of $f^{\prime }$ in each interval $(r_j,r_{j+1})$. These intervals are disjoint, so the obtained roots are distinct:

We prove the requested bound for positive roots of a polynomial by mathematical induction on the number of nonzero terms. In fact, the proof also works with roots counted according to their multiplicity. Write $V(p)$ for the number of sign changes and $N_{+}(p)$ for the positive-root count.

A one-term polynomial $a_1{x}^{d_1}$ has no positive roots and no sign changes. For the induction step, factor out the smallest power of $x$:For $x>0$ this factor is nonzero, so $N_{+}(q)=N_{+}(p)$, including multiplicities, and $V(q)=V(p)$. Differentiation removes exactly the constant term:The exponents remain strictly increasing, and the positive multipliers $e_j$ do not change coefficient signs. The induction hypothesis therefore gives

First suppose $a_1{a}_2<0$. Removing $a_1$ removes exactly one sign change, so $V(q)=V(q^{\prime })+1$. If $q$ has positive roots $r_1<\cdots <r_k$ of multiplicities $m_1,\dots ,m_k$, its derivative has multiplicity $m_j-1$ at $r_j$ and, by the Rolle theorem, an additional root between each pair. Thus the Rolle root count with multiplicities giveswhen there is a positive root. ConsequentlyIf there is no positive root, the desired bound is automatic.

Now suppose $a_1{a}_2>0$, so $V(q)=V(q^{\prime })$. Again the zero-root case is immediate. Otherwise, multiplying $q$ by $-1$ if necessary lets us assume $a_1>0$ and $a_2>0$ without changing roots or sign changes. For sufficiently small $x>0$,since $q(0)=a_1$ and the lowest-power term of $q^{\prime }$ has positive coefficient. Choose such a point $u$ before the first positive root $r_1$. The mean value theorem on $[u,r_1]$ gives a point $v\in (u,r_1)$ withBut $q^{\prime }(u)>0$. The intermediate value theorem applied to the polynomial $q^{\prime }$ therefore gives an extra root in $(u,v)$, before $r_1$. It is distinct from the roots at $r_j$ and those between consecutive roots. HenceBoth sign cases complete the induction:An initial sign change allows one extra positive root; without that sign change, the derivative must already turn before the first positive root. This is the upper-bound part of the Descartes rule of signs. The multiplicity claim used above follows by differentiating $q(x)=(x-r{)}^{m}h(x)$ with $h(r)\ne 0$: the derivative has a factor $(x-r{)}^{m-1}$ and its remaining factor is nonzero at $r$.

The Lagrange cardinal polynomials areThus the Lagrange interpolation polynomial isIt has degree at most $n$ and takes the required values. Uniqueness follows because the difference of two such polynomials has $n+1$ distinct roots and degree at most $n$, so is zero. The source's phrase “degree $n$” must allow a smaller degree, for example when $f$ is constant.

Define the divided difference recursively by $f[x_i]=f_i$ andInduction on this recursion, or comparison of the leading coefficient in the Newton interpolation polynomial, givesFor completeness the induction works term by term: an interior term gets the coefficient $(1/(x_i-x_n)-1/(x_i-x_0))/(x_n-x_0)=1/((x_i-x_n)(x_i-x_0))$, multiplying the product over the other interior nodes; the two endpoint terms give the same formula. This is also the leading coefficient of $p$ directly from the cardinal formula, so $p^{(n)}=n!f[x_0,\dots ,x_n]$.

The function $f-p\in C^n[x_0,x_n]$ has $n+1$ distinct zeros. Repeated application of Rolle theorem gives $\xi \in (x_0,x_n)$ with $(f-p{)}^{(n)}(\xi )=0$ when $n\ge 1$. HenceFor $n=0$, this is simply $f[x_0]=f(x_0)$ and we take $\xi =x_0$.

Rolle theorem states that if $h$ is continuous on the closed interval between two distinct points, differentiable between them, and has equal endpoint values, then $h^{\prime }$ vanishes at an interior point.

Let $P_N(t)={\sum }_{j=0}^N{f}^{(j)}(0)t^j/j!$ and, for $x\ne 0$, putThen $h(x)=0$ and $h^{(j)}(0)=0$ for $0\le j\le N$. Applying Rolle's theorem successively $N+1$ times gives a point $\xi =\theta x$, $0<\theta <1$, where $h^{(N+1)}(\xi )=0$. Hencewhich is the stated Taylor theorem with Lagrange remainder.

If $f^{\prime }=0$, the $N=0$ case gives $f(x)=f(0)+x{f}^{\prime }(\theta x)=f(0)$. Thus $f$ is constant; equivalently this is the immediate corollary of the mean value theorem.

The Taylor theorem with Lagrange remainder states that if $f$ is $n$ times differentiable between $a$ and $x$, then some $\xi$ between $a$ and $x$ satisfies

Let $T_{n-1}$ denote the displayed polynomial and choose $K$ so thatsatisfies $F(x)=0$. By construction,Starting with the two zeros $a,x$ and applying Rolle theorem repeatedly, there is a $\xi$ between them with $F^{(n)}(\xi )=0$. Sincewe have $K=f^{(n)}(\xi )/n!$, proving the theorem.

Rolle theorem: if $f$ is continuous on $[a,b]$, differentiable on $(a,b)$, and $f(a)=f(b)$, then $f^{\prime }(c)=0$ for some $c\in (a,b)$. Indeed, the extreme value theorem gives a maximum and minimum. If both occur only at the endpoints then $f$ is constant; otherwise an interior extremum $c$ satisfies $f^{\prime }(c)=0$ by comparing the two-sided difference quotients.

Mean value theorem: if $f$ is continuous on $[a,b]$ and differentiable on $(a,b)$, then some $c\in (a,b)$ satisfiesApply Rolle's theorem towhich has $h(a)=h(b)=0$.

SetIts endpoint values are equal. Rolle theorem therefore supplies $\xi \in (a,b)$ withIf $g^{\prime }(\xi )=0$, then (1) and $g(b)\ne g(a)$ force $f^{\prime }(\xi )=0$, contrary to the hypothesis. Hence $g^{\prime }(\xi )\ne 0$, and division in (1) gives

The condition is necessary. On $[0,2\pi ]$, letHere $g(2\pi )-g(0)=\pi /2$ while $f(2\pi )-f(0)=0$, so the endpoint ratio is zero. ButWhere $g^{\prime }(x)\ne 0$, their ratio is $1/(1+\frac{1}{2}\cos x)$ and is never zero. At the remaining points both derivatives vanish, so the derivative ratio is undefined. Thus the conclusion fails when simultaneous zeros are allowed.

For any $n\ge 0$, take $f_n=T_{n+1}$, the Chebyshev polynomial of degree $n+1$. It has $n+2$ alternating extrema of magnitude one on $[-1,1]$. The Chebyshev alternation theorem shows that the zero polynomial is best from ${\mathcal{P}}_n$, with error one. Since it also lies in ${\mathcal{P}}_{n-1}$ when $n\ge 1$,

Now suppose $f^{(n)}(x)>0$ throughout $[-1,1]$ and, contrary to the claim, $E_{n-1}(f)=E_n(f)$. A best $p\in {\mathcal{P}}_{n-1}$ would then also be best in ${\mathcal{P}}_n$. Its error $e=f-p$ would have $n+2$ alternating extrema by the alternation theorem, hence at least $n+1$ distinct zeros. Applying the Rolle theorem $n$ times gives a zero ofcontradicting positivity. Therefore

If a nonzero real polynomial has $N$ roots in an open real interval, counted with multiplicity of a root, its derivative has at least $N-1$ roots in that interval, also counted with multiplicity, provided $N\ge 1$. Each root of multiplicity $m$ becomes a root of the derivative of multiplicity $m-1$. The Rolle theorem supplies one further root between each pair of distinct consecutive roots. Thus $k$ distinct roots of multiplicities $m_j$ give at least ${\sum }_j(m_j-1)+(k-1)=N-1$ derivative roots.