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Consider the original worked example $f(x)=x^3+x-1$ on $[0,1]$. First, existence: a polynomial function is a continuous function, and $f(0)=-1<0<1=f(1)$. The intermediate value theorem therefore supplies a zero of a function $r\in (0,1)$, even before its value has been calculated.

Second, uniqueness is a separate argument. The derivative is $f^{\prime }(x)=3x^2+1>0$. The derivative criterion for monotonicity, obtained from the mean value theorem, makes $f$ a strictly increasing function. Two distinct zeros of a function would contradict that strict increase. Thus this root bracket contains exactly one zero of a function; indeed the same derivative argument proves uniqueness on the entire real line.

Third, localization uses the existing bisection method. Set $[a_0,b_0]=[0,1]$; after $k$ completed halvings, retain $[a_k,b_k]$ with opposite endpoint signs. Each update tests the old midpoint $m_{k-1}=(a_{k-1}+b_{k-1})/2$. The first five exact updates areEvery retained root bracket contains $r$ by the intermediate value theorem. After ten halvings, exact arithmetic givesHence $0.681640625<r<0.6826171875$. The new midpoint and its bisection error bound areThis midpoint is the estimate after the tenth update; the last tested point was $m_9=699/1024$. In general the retained width is $2^{-k}$, so the bisection error bound is $2^{-(k+1)}$. Nine halvings already suffice for an error at most $10^{-3}$, since $2^{-10}=0.0009765625<10^{-3}$. This guarantee comes from the width of the root bracket, independently of any displayed decimal approximation to the unknown zero of a function. Endpoint signs must be established reliably if evaluations use floating-point arithmetic.

Let a polynomial $q$ of degree at most $n$ have errors $f(x_j)-q(x_j)=(-1{)}^{j}h$ at $n+2$ ordered distinct points. Then $|h|$ bounds below the error $E_n(f)$ of the best uniform approximation. Otherwise an improved polynomial $p$ with error strictly below $|h|$ would make $p-q$ alternate strictly in sign at all those points, giving at least $n+1$ distinct roots of a polynomial by the intermediate value theorem. Its degree is at most $n$, a contradiction. Moreover, if $q$ is not optimal, $|h|<E_n(f)$. In the equality case, an optimal $p$ gives weak alternating signs of $p-q$; its vanishing order-$(n+1)$ divided difference has same-sign summands, forcing every nodal value to vanish and hence $p=q$.

In exact arithmetic, after $k$ completed halvings the retained root bracket has width $W_k=b_k-a_k$ and midpoint $m_k$, whereThe absolute value bound holds for every zero of a function $x_{\ast }$ in the current root bracket, including the one approached by continued bisection method updates. A zero of a function discarded with an earlier half need not satisfy later bounds. Before update $k\ge 1$, the evaluated candidate is $m_{k-1}$, whose corresponding guarantee is $(b-a)/2^k$. After update $k$, the new midpoint $m_k$ has the smaller guarantee $(b-a)/2^{k+1}$, even without evaluating $f(m_k)$.

The guaranteed error halves at every step; the actual absolute value error need not decrease at every step or have successive ratio $1/2$. In the square root example it increases from $|m_6-\sqrt{2}|\approx 0.000151062$ to $|m_7-\sqrt{2}|\approx 0.003755188$, while both errors remain below their guarantees. An exact midpoint zero of a function instead gives zero error and immediate termination. For this example $\sqrt{2}$ is an irrational number, so no finite exact halving can produce it as a midpoint.

For a brief proof outline, each retained half has half the previous width and contains a zero of a function by the intermediate value theorem. Every point in the retained root bracket is at most half its width from its midpoint. If the process does not stop at a zero of a function, the nested interval theorem gives one common point as the widths tend to zero; continuity and the opposite endpoint signs force its function value to be zero. Uniqueness in the initial root bracket is unnecessary.

The bisection method finds a zero of a function by repeatedly halving a root bracket. Let $f$ be a real-valued continuous function on the closed interval $[a,b]$, with $a<b$. First check for endpoint zeros of a function; otherwise require strictly opposite signs at the endpoints. Evaluate the midpoint and retain the half with strictly opposite endpoint signs, stopping immediately if the midpoint is a zero of a function. The intermediate value theorem ensures that every retained root bracket still contains a zero of a function. This is the construction in MIT's root-finding notes.^[ref]

For a worked square root calculation, use the polynomial $f(x)=x^2-2$ on $[1,2]$. Write $[a_k,b_k]$ for the root bracket after $k$ halvings, with $[a_0,b_0]=[1,2]$. The candidate evaluated during update $k$ is the old midpoint $m_{k-1}$:The initial midpoint is $m_0=1.5$. After four halvings the new midpoint is $m_4=1.40625$, and the bisection error bound is $|m_4-\sqrt{2}|\le 0.03125$. Its actual error is approximately $0.00796356237310$. The last tested candidate $m_3=1.4375$ and the reported midpoint $m_4$ are different points.

The useful feature of the bisection method is that the retained width halves on every update, independently of the shape of the function graph. The bisection error bound therefore supplies an accuracy guarantee before the zero of a function is known. No derivative is needed, and a valid initial root bracket gives convergence without a close initial guess. This makes the bisection method a dependable fallback for the faster Newton root-finding iteration.

The Borsuk-Ulam theorem says that recording $n$ continuous real-valued measurements on an $n$-dimensional unit sphere must give identical records somewhere on opposite sides. The measurements can vary wildly; continuity alone forces a coincidence at an antipodal pair.

Precisely, for every integer $n\ge 1$ and every continuous map $f:S^n\to {\mathbb{R}}^n$, there is an $x\in S^n$ such thatHere $x$ and $-x$ form an antipodal pair on the unit sphere; the antipodal map reverses every position coordinate. The target is the Euclidean space ${\mathbb{R}}^n$. Both points have the same entire output, so every one of the $n$ measurements agrees at this common pair. See Hatcher, Algebraic Topology, Corollary 2B.7, p. 176.

Circle example. Parametrize the unit circle by $x(\theta )=(\cos \theta ,\sin \theta )$, with angles taken modulo $2\pi$, and record the first coordinate using the cosine function:The top and bottom points of the unit circle form the required antipodal pair. Their first coordinates agree, while their second coordinates are $1$ and $-1$.

Sphere example. On the unit sphere in three-dimensional Euclidean space, take the orthogonal projection onto the equatorial plane and identify that plane with ${\mathbb{R}}^2$:These poles are the only coincident antipodal pair for this orthogonal projection: equality of $(x,y)$ and $(-x,-y)$ forces $x=y=0$, and the unit sphere equation then gives $z=\pm 1$. The Borsuk-Ulam theorem guarantees an antipodal pair for every continuous map into ${\mathbb{R}}^2$, even when no convenient projection formula is available.

Equal output need not mean equal position coordinates. Keeping all three coordinates gives a continuous map $i:S^2\to {\mathbb{R}}^3$ defined by $i(x,y,z)=(x,y,z)$, the restriction of the identity map of ${\mathbb{R}}^3$. Its outputs at an antipodal pair are always distinct, since $u=-u$ would force $u=0$, which is outside the unit sphere. Thus replacing the target ${\mathbb{R}}^2$ by ${\mathbb{R}}^3$ destroys this guarantee.

Why the circle case follows from the intermediate value theorem. For an arbitrary continuous map $f:S^1\to \mathbb{R}$, write $F(\theta )=f(x(\theta ))$ using the unit circle parametrization above. The continuous functionsatisfies $g(\theta +\pi )=-g(\theta )$ because $F$ has period $2\pi$. If $g(0)=0$, the first antipodal pair already works. Otherwise $g(0)$ and $g(\pi )$ have opposite signs, and the intermediate value theorem supplies a zero between them. At that zero, the two antipodal values of $f$ agree. In the cosine function example, $g(\theta )=2\cos \theta$. This short argument handles only $n=1$. See Hatcher, Algebraic Topology, discussion of Theorem 1.10, p. 32.

An ideal weather interpretation. Model the surface of Earth as a sphere. At one fixed instant, suppose temperature $T$ and pressure $P$ are globally defined continuous scalar measurements on that sphere. Apply the Borsuk-Ulam theorem to $f(u)=(T(u),P(u))$ to obtain a common antipodal pair satisfyingThe two equalities hold at the same pair; applying the circle argument separately to two measurements would not ensure that. This is a conclusion about the ideal continuous model, with both measurements taken at the same instant, rather than an assertion that an exactly matching pair has been empirically measured. Matoušek gives this motivating interpretation and explicitly cautions that physical temperature need not be continuous.^[ref]

Equivalent formulations. A continuous map $h:S^n\to {\mathbb{R}}^n$ is odd when $h(-x)=-h(x)$, extending the symmetry of an odd function. The Borsuk-Ulam theorem equivalently says every such map has a zero. For a general $f$, the difference $h(x)=f(x)-f(-x)$ has this symmetry; conversely, equal antipodal outputs of an odd map must be zero. Another equivalent formulation is that no odd map between spheres $S^n\to S^{n-1}$ exists. Normalizing a nonvanishing odd map by its Euclidean norm would produce such an odd map between spheres. These formulations are useful starting points for algebraic topology; the general proof is beyond the circle argument here.^[ref]

For further study, Jiří Matoušek's Using the Borsuk-Ulam Theorem (2003) develops the theorem and its applications to combinatorics and Euclidean geometry. An early publication is Karol Borsuk's 1933 paper Drei Sätze über die n-dimensionale euklidische Sphäre.

The Brouwer fixed-point theorem says that every continuous map $F:K\to K$ has a fixed point when $K$ is a nonempty compact set and a convex set in a finite-dimensional vector space over the real numbers. Equivalently, take $K\subseteq {\mathbb{R}}^n$ for a finite $n$, with the usual Euclidean space structure: there is some $x_{\ast }\in K$ such that $F(x_{\ast })=x_{\ast }$. The conclusion guarantees at least one fixed point, without asserting uniqueness or telling us how quickly to find one. ^[ref]

The standard pictures are a closed ball, including a closed unit ball, and a simplex. A filled disk qualifies; its bounding circle alone does not. A probability simplex also qualifies even though its dimension is smaller than that of its ambient Euclidean space. The hypotheses have concrete meanings: being a convex set keeps every line segment between allowed points inside $K$, and compactness prevents escape or loss of a limiting point. In Euclidean space, the Heine-Borel theorem identifies compact sets with closed sets that are bounded sets. Crucially, every output must belong to the same $K$ as the input.

For a fully computable example, let $D=\{(u,v):u^2+v^2\le 1\}$ be the filled unit disk. Apply a counterclockwise quarter-turn rotation, shrink by $0.4$, and add a translation:This affine map is a continuous map. Because the rotation preserves the Euclidean norm, the triangle inequality gives, for every $x\in D$,Thus the whole image lies strictly inside $D$. More precisely, it is a disk of radius $0.4$ centred at $(0.3,0.1)$. This containment check is what makes the Brouwer fixed-point theorem applicable; a formula being a continuous map by itself would not suffice.

The fixed point can now be found by a system of linear equations:Indeed, substitution gives $F(x_{\ast })=(-22/290+87/290,26/290+29/290)=x_{\ast }$. Its Euclidean norm is $\sqrt{290}/58\approx 0.293610$, safely inside $D$. The figure shows arrows from sample inputs to their outputs; an arrow disappears at the fixed point because the displacement $F(x)-x$ vanishes there.

Repeated iterations of a map happen to succeed for this example. Starting at $x_0=(1,0)$ and setting $x_{k+1}=F(x_k)$ gives $x_1=(0.3,0.5)$, $x_2=(0.1,0.22)$, and $x_3=(0.212,0.14)$. Subtracting the fixed point equation yields $x_{k+1}-x_{\ast }=0.4R_{90}(x_k-x_{\ast })$, so the Euclidean distance to $x_{\ast }$ shrinks by exactly $0.4$ each step. This extra fact follows from the contraction mapping property and the Banach fixed-point theorem. It is additional information beyond the Brouwer fixed-point theorem.

The distinction becomes visible with two simpler continuous maps of the same disk. The pure rotation $G(u,v)=(-v,u)$ has exactly one fixed point, $(0,0)$, since $u=-v$ and $v=u$ force both to vanish. Yet iterations of a map starting at $(1,0)$ here cycle through $(0,1)$, $(-1,0)$, and $(0,-1)$ forever. Their Euclidean distance from $(0,0)$ stays one, so there is no convergence in a metric space. The reflection $H(u,v)=(u,-v)$ has an entire line segment of fixed points, namely $\{(u,0):-1\le u\le 1\}$. Starting off that line segment produces an alternating pair. Existence, uniqueness, and convergence of iterations of a map are separate questions.

Removing hypotheses changes the answer. Restrict the same quarter-turn rotation to the circle $S^1=\{(u,v):u^2+v^2=1\}$. It remains a continuous map into that compact set, but its only possible fixed point, $(0,0)$, has been removed. The circle is not a convex set. Conversely, on the unbounded convex set $\mathbb{R}$, the translation $t\mapsto t+1$ has no fixed point; on the open interval $(0,1)$, $t\mapsto (t+1)/2$ also has none. Neither domain is a compact set. The finite-dimensional closed ball statement should likewise not be extended to arbitrary infinite-dimensional vector spaces, where a closed ball need not be a compact set.

The one-dimensional version explains the connection to the intermediate value theorem. For a continuous map $f:[a,b]\to [a,b]$, put $h(t)=f(t)-t$. Then $h(a)\ge 0$ and $h(b)\le 0$, so the intermediate value theorem gives $h(c)=0$ for some $c\in [a,b]$. In higher dimensions, the Brouwer fixed-point theorem supplies a corresponding existence principle when this scalar sign argument is unavailable. Its practical value is ensuring that a proposed equation $x=F(x)$ has a solution once a suitable domain is found; it supplies no general promise of a fast algorithm.

A useful application is positive matrix eigenvector from simplex normalization: rescale the output of a strictly positive matrix to make its entries sum to one. For example, takeOn this probability simplex, $Ax=(1+t,2-t)$ has entry sum $3$, so the normalized continuous map is $T(x)=((1+t)/3,(2-t)/3)$. Both entries are positive and sum to one, keeping the image inside the probability simplex. Its fixed point equation is $t=(1+t)/3$, hence $t=1/2$. Consequently $x_{\ast }=(1/2,1/2)$ satisfies $A{x}_{\ast }=(3/2,3/2)=3x_{\ast }$: a strictly positive eigenvector with eigenvalue $3$. In larger dimensions, the Brouwer fixed-point theorem guarantees the corresponding positive eigenvector without requiring us to solve a system of linear equations in advance. The fixed point expresses preservation of relative weights, rather than preservation of their unnormalized size.

Proofs can be approached through the no-retraction theorem or Sperner's lemma. In the latter approach, increasingly fine triangulations locate approximate fixed points; compactness supplies a convergent subsequence, and continuous maps turn its limit into an exact fixed point. ^[ref]

A primary historical reference is L. E. J. Brouwer's 1911 paper Über Abbildung von Mannigfaltigkeiten, whose digitized original is hosted by Zenodo under DOI 10.1007/BF01456812. This gives a direct route to the early literature behind the Brouwer fixed-point theorem. ^[ref]

Every real interval is connected. Otherwise a separation into two nonempty relatively open sets defines a continuous real function equal to zero on one side and one on the other. Choose one point of each set. The intermediate value theorem on the closed segment between them requires the value $1/2$, a contradiction. The converse also holds: a connected subset of the real line contains every point between two of its points, since omitting an intermediate point separates it into its intersections with the two open half-lines.

A topological space $X$ is connected when it has no separation of a topological space: no partition of a set into two nonempty open subsets $U,V$ satisfyingHere the union covers all of $X$, and the intersection is the empty set, so no point belongs to both pieces. For a subset $X$ of a larger topological space, "open" means open in the subspace topology: each piece is the intersection of $X$ with an ambient open set. Both pieces must be open; merely dividing a drawing into two colours does not establish a separation of a topological space. ^[ref]

For example, the closed interval $I=[0,1]$ with its usual subspace topology from the real line is connected. The proposed split $[0,1/2)\cup [1/2,1]$ fails: the second subset is not open in $I$, since every neighbourhood of $1/2$ in $I$ contains points less than $1/2$. More generally, a connected subset of the real line is precisely a real interval. In contrast, for the unionthe open intervals $(-3,0)$ and $(0,3)$ exhibit $U,V$ as nonempty open subsets in the subspace topology of $D$. Their intersection is empty and their union is $D$, so $D$ is not connected. Although $U,V$ are closed intervals in the real line, they are also open within $D$.

Why is connectedness useful? A continuous image of a connected space is connected: a proposed separation of a topological space in the image would pull back through the continuous map to a forbidden separation of a topological space in its source. Thus a continuous map from a real interval to the real line cannot skip values between two attained values. This is the intermediate value theorem. For a computed example, the continuous function $f(x)=x^2$ on $[0,1]$ has image $[0,1]$; the value $1/2$ occurs at $x=1/\sqrt{2}\approx 0.707107$. Connectedness guarantees an input attaining that value even when we have no formula for it.

Path connectedness is a stronger condition: it supplies a continuous path between every pair of points. The topologist's sine curve below shows that a connected space need not be a path-connected space.

In the usual subspace topology from the real line, a subset $A\subseteq \mathbb{R}$ is connected exactly when it has the defining property of a real interval:Every real number between two members must be present. Thus open intervals, closed intervals, intervals with only one endpoint included, and unbounded real intervals are connected. Singleton sets are connected too. With the no-separation of a topological space convention used here, the empty set satisfies the same property; some texts reserve the word connected for nonempty topological spaces.^[ref]

Removing an interior point changes the answer:These two real intervals are exactly its connected components and its path components. Removing an endpoint instead leaves a real interval, for example $[-2,2]\setminus \{2\}=[-2,2)$, which stays connected. This is why an interior point of a closed interval is a cut point, whereas an endpoint is not. A nonempty finite set of points in the real line is connected exactly when it is a singleton set.

Proof outline: if $a,b\in A$ but $c\in (a,b)$ is missing, the subsetsgive a separation of a topological space, since the two open intervals make both pieces open in $A$. Conversely, a nonempty real interval contains the continuous path $t\mapsto (1-t)a+tb$ between any two of its points, so it is path-connected and hence connected. Together with continuous image of a connected space, this characterization gives the intermediate value theorem: a continuous function on a real interval cannot jump over a value lying between two attained values.

Suppose random continuous real functions have unique zeros. If pointwise convergence in probability gives opposite strict limiting signs at the two ends of every sufficiently small interval about the target, the intermediate value theorem traps their zeros in those intervals with probability tending to one. The limiting function need not be continuous and the random functions need not be monotone.

A continuous function sends sufficiently nearby function inputs to nearby function outputs. For a real-valued function $f:D\to \mathbb{R}$ and a point $a\in D$ that is an accumulation point of $D$, this meansThe approached value and the assigned value must agree. Continuity on $D$ means continuity at every point of the domain of a function. At an endpoint of an interval, only approaches inside that domain of a function count. The epsilon-delta definition of continuity below also applies at an isolated point, where continuity holds automatically.^[ref]

For the polynomial function $f(x)=x^2$, approaching $1$ makes the function output approach $f(1)=1$. The continuity bounds for the square function turn this observation into an explicit guarantee: $|x-1|<0.2$ implies $|x^2-1|<0.6$. This is useful whenever a small error in an function input must produce a controlled error in an function output. The required input tolerance can depend on the point and on the chosen output tolerance.

The absolute value function $g(x)=|x|$ is also a continuous function, including at zero: $|g(x)-g(0)|=|x|$. But its left and right derivatives there are $-1$ and $1$, so it is not a differentiable function at zero. Continuity controls values without requiring a tangent line. Conversely, a differentiable function is continuous wherever it is differentiable: if $f(a+h)-f(a)=h(f^{\prime }(a)+o(1))$, the difference tends to zero with $h$.

Several useful consequences reuse this one local idea. Addition, multiplication and function composition preserve continuity; a quotient of functions is continuous where its denominator is nonzero. On an interval, the intermediate value theorem turns continuity into a guarantee that no intermediate function output is skipped. On a compact set, the extreme value theorem guarantees attained maximum and minimum values. Those extra hypotheses matter: continuity alone on an unbounded domain of a function does not provide a global bound or an attained global maximum.

A rational function such as $x\mapsto 1/x$ is continuous on its domain of a function $\mathbb{R}\setminus \{0\}$. Zero is excluded from that domain of a function; the missing point prevents a continuous extension there, but it does not make the existing map discontinuous at one of its actual inputs. Always state the domain of a function before judging continuity.

A continuous function on any real interval has image a real interval. Given two image values and an intermediate value, choose preimages and apply the intermediate value theorem on the closed segment joining them, which stays inside the domain. Neither compactness nor boundedness of the original interval is needed. In particular $f(x)=\sin (1/x)/x$ on $(0,1]$ is a continuous surjection onto $\mathbb{R}$: its values are unbounded above and below, so its interval image contains every real number.

For a Lebesgue measurable set $E\subseteq \mathbb{R}$ of finite measure, the displayed function satisfies $|H_E(t)-H_E(s)|\le |t-s|$. Indeed, for $s<t$, finite additivity and monotonicity of a measure give $H_E(t)-H_E(s)=m(E\cap (s,t])\le t-s$. Thus $H_E$ is a nondecreasing Lipschitz function and is uniformly continuous even if $E$ is unbounded.

Continuity from above of a measure gives $H_E(t)\to 0$ as $t\to -\infty$, while continuity from below of a measure gives $H_E(t)\to m(E)$ as $t\to \infty$. Every strictly intermediate value is attained by the intermediate value theorem. The corresponding half-line cut of $E$ realizes any prescribed measure between zero and $m(E)$; the endpoints are realized by the empty set and $E$ itself.

If $E\subseteq [0,1]$ is a Lebesgue measurable set and $0\le b\le \lambda (E)$, some measurable subset $B\subseteq E$ has Lebesgue measure $b$. The function $F(t)=\lambda (E\cap [0,t])$ has Lipschitz continuity with constant $1$, starts at $0$ and ends at $\lambda (E)$; the intermediate value theorem proves the assertion. Applying the assertion successively to remaining portions of $E$ gives pairwise disjoint measurable pieces with any finite list of nonnegative measures whose sum is at most $\lambda (E)$. This is a concrete consequence of non-atomic measure structure, which fails for general measures containing an atom of a measure.

If $f,g:[a,b]\to \mathbb{R}$ are continuous functions, $g\ge 0$, and $f$ is a nondecreasing monotonic function, there is $d\in [a,b]$ satisfying the identity in the title. It expresses the weighted integral as a mixture of the two endpoint values of $f$, with the split occurring in the accumulated weight.

Put $G={\int }_a^{b}g$ and $I={\int }_a^{b}fg$. The monotonicity of the Riemann integral gives $f(a)G\le I\le f(b)G$. The function $H(x)={\int }_a^{x}g$ is continuous; indeed, $|H(y)-H(x)|\le {\max }_{[a,b]}|g||y-x|$ by the extreme value theorem and the triangle inequality for the Riemann integral. Thereforeis continuous, with $K(a)=f(b)G$ and $K(b)=f(a)G$. The intermediate value theorem gives $K(d)=I$. If the endpoint values coincide, the bounds force $I$ to have that common value and an endpoint works. The proof covers zero total weight and a degenerate interval without division.

Suppose $f$ has a fourth derivative that is a continuous function on the interval $[a,b]$, and $|f^{(4)}(x)|\le M$ there. With $I={\int }_a^{b}f(x)dx$, the signed quadrature error here is $I-S_n$. Then, for some point $\xi \in (a,b)$,In particular, a strictly positive fourth derivative makes $I-S_n<0$: composite Simpson's rule overestimates the integral.^[ref]

The bound's scale follows by adding the panel errors. A panel of width $2h$ has signed quadrature error $-h^5{f}^{(4)}({\xi }_j)/90$ and absolute error at most $M{h}^5/90$. There are $(b-a)/(2h)$ panels, so the triangle inequality gives $M(b-a)h^4/180$. The intermediate value theorem for the continuous fourth derivative replaces the average of the panel values $f^{(4)}({\xi }_j)$ by a single value $f^{(4)}(\xi )$, giving the signed formula.

The underlying single-panel calculation can be seen through the Peano kernel theorem. On the interval $[0,2]$, define the raw Peano kernel $K(t)=L[(x-t{)}_{+}^3]$, where $L$ is integral minus Simpson's rule and $(x-t{)}_{+}=\max (x-t,0)$ is the positive part. Then $K(t)=-t^3(4-3t)/12$ for $0\le t\le 1$ and $K(t)=-(2-t{)}^3(3t-2)/12$ for $1\le t\le 2$. It is strictly negative inside the interval and has integral $-1/15$. ThusThe uniform norm $\Vert f^{(4)}{\Vert }_{\infty }$ bounds the fourth derivative. The polynomial $f(x)=x^4/24$ attains equality. Scaling to endpoint spacing $h$ multiplies the coefficient by $h^5$, giving the sharp single-panel bound $h^5\Vert f^{(4)}{\Vert }_{\infty }/90$. The normalized Peano kernel is $K/3!$, so the external factorial must be omitted when using that normalization.

For the exponential function on $[0,1]$, $f^{(4)}(x)=e^x$, so $M=e$ and the bound is $e/(180n^4)$. Its leading actual absolute error is $(e-1)/(180n^4)$, smaller than the bound because a uniform upper bound on the fourth derivative does not use its variation across the interval. The theorem bounds truncation error with exact function evaluations; sample errors and roundoff error are additional contributions, as described by noise sensitivity of a quadrature rule.

For the finite square well $V=-V_0$ on $|x|<a$ and $V=0$ outside, with $a,V_0>0$, there is always an even bound state, however small $V_0$ is. Define $\rho =a\sqrt{2m{V}_0}/\hslash$, the dimensionless decay constant $\alpha =a\sqrt{-2mE}/\hslash$, and interior oscillation constant $\beta =a\sqrt{2m(E+V_0)}/\hslash$. Wavefunction matching at a finite potential step gives $\alpha =\beta \tan \beta$ and ${\alpha }^2+{\beta }^2={\rho }^2$.

On $0<\beta <\pi /2$, these equations reduce to $\beta /\cos \beta =\rho$. The left side is continuous, strictly increasing from zero to infinity, with derivative $\sec \beta (1+\beta \tan \beta )>0$. The intermediate value theorem gives exactly one root on this first branch for every $\rho >0$. It has $0<\beta <\rho$, hence $\alpha >0$ and $-V_0<E<0$. The matched cosine interior and exponential tails therefore give a nonzero even square-integrable function with continuous first derivative, which can be normalized. As the well becomes shallow the binding energy approaches zero but remains negative; see shallow finite-square-well energy.

For continuous $f:[0,1]\to [0,1]$, the function $f(x)-x$ is nonnegative at zero and nonpositive at one, so the intermediate value theorem gives a fixed point.

Let $G_N(B)=E(20N[1+2B(1-B)],21N)-B$. The proportional scaling limit of the Erlang loss formula makes its limits at $B=1/100$ and $B=1/8$ respectively $-1/100$ and $7/520$. Since $G_N(0)>0$ and $G_N(1)<0$, sufficiently large $N$ give alternating signs at these four points. The intermediate value theorem supplies three distinct zeros in the intervening intervals. These are multiple solutions of the Erlang fixed point approximation, not multiple stationary laws of an exact finite irreducible Markov chain.

If $f:[0,1]\to \mathbb{R}$ is continuous and $f(0)=f(1)$, then for every positive integer $n$ there is $\alpha \in [0,1-1/n]$ with $f(\alpha +1/n)=f(\alpha )$. For $n>1$, the continuous difference $h_n(x)=f(x+1/n)-f(x)$ satisfies ${\sum }_{j=0}^{n-1}{h}_n(j/n)=0$. Either one sampled difference is zero or two have opposite signs; the intermediate value theorem then supplies a zero between them. For $n=1$, the endpoint chord suffices.