nyuuzyou/wikis huggingface dataset:
Acceleration Due to Gravity/Historical Note

== Historical Note on Acceleration Due to Gravity ==
This result famously contradicts , who taught that heavier objects fall faster than light ones.
There exists a well-known story about how proved this law by dropping two objects of different weights from a height (supposedly the ).
Apparently this never actually happened.
However, did perform a similar experiment some time before started experimenting.
The fact that in the general case air resistance can not be ignored goes some way to explaining how the truth was not arrived at earlier.
Even children notice how leaves, for example, flutter gently down from on high, whereas stones tend to plummet.
== Sources ==
* : $\text {I}$: Motion, Sound and Heat: Chapter $2$: Falling Bodies: Inclined Planes
Category:Historical Notes
Acceleration Vector in Polar Coordinates

== Theorem ==
Consider a particle $p$ moving in the plane.
Let the position of $p$ at time $t$ be given in polar coordinates as $\polar {r, \theta}$.
Then the acceleration $\mathbf a$ of $p$ can be expressed as:
:$\mathbf a = \paren {r \dfrac {\d^2 \theta} {\d t^2} + 2 \dfrac {\d r} {\d t} \dfrac {\d \theta} {\d t} } \mathbf u_\theta + \paren {\dfrac {\d^2 r} {\d t^2} - r \paren {\dfrac {\d \theta} {\d t} }^2} \mathbf u_r$
where:
:$\mathbf u_r$ is the unit vector in the direction of the radial coordinate of $p$
:$\mathbf u_\theta$ is the unit vector in the direction of the angular coordinate of $p$
== Proof ==
Let the radius vector $\mathbf r$ from the origin to $p$ be expressed as:
:$(1): \quad \mathbf r = r \mathbf u_r$
:
From Derivatives of Unit Vectors in Polar Coordinates:
From Velocity Vector in Polar Coordinates:
:$\mathbf v = r \dfrac {\d \theta} {\d t} \mathbf u_\theta + \dfrac {\d r} {\d t} \mathbf u_r$
where $\mathbf v$ is the velocity of $p$.
The acceleration of $p$ is by definition the rate of change in its velocity:
== Sources ==
* : $\S 3.21$: Newton's Law of Gravitation: $(5)$
* 
* 
Category:Polar Coordinates
Category:Acceleration
Acceleration is Second Derivative of Displacement with respect to Time

== Theorem ==
The acceleration $\mathbf a$ of a body $M$ is the second derivative of the displacement $\mathbf s$ of $M$ from a given point of reference time $t$:
:$\mathbf a = \dfrac {\d^2 \mathbf s} {\d t^2}$
== Proof ==
By definition, the acceleration of a body $M$ is defined as the first derivative of the velocity $\mathbf v$ of $M$ relative to a given point of reference time:
:$\mathbf a = \dfrac {\d \mathbf v} {\d t}$
Also by definition, the velocity of $M$ is defined as the first derivative of the displacement of $M$ from a given point of reference time:
:$\mathbf v = \dfrac {\d \mathbf s} {\d t}$
That is:
:$\mathbf a = \map {\dfrac \d {\d t} } {\dfrac {\d \mathbf s} {\d t} }$
Hence the result by definition of the second derivative.
== Sources ==
* 
* 
* : Chapter $8$: The System of the World: Newton
* 
* 
Category:Mechanics
Category:Acceleration
Acceleration of Particle moving in Circle

== Theorem ==
Let $P$ be a particle moving in a circular path $C$.
Then the acceleration of $P$, denoted by $\mathbf a \in \R$, is given as:
:$\mathbf a = -\dfrac {\size {\mathbf v}^2 \mathbf r} {\size {\mathbf r}^2}$
where:
:$\mathbf v$ is the instantaneous velocity of $P$
:$\mathbf r$ is the vector whose magnitude equals the length of the radius of $C$ and whose direction is from the center of $C$ to $P$
:$\size {\, \cdot \,}$ denotes the magnitude of a vector.
== Proof ==
We use Lagrangian mechanics to derive the result.
Let $\mathbf r = \map {\mathbf r} t$ be a function of time $t \in \R_{\ge 0}$ .
Let $R = \size r \in \R_{>0}$ be the constant radius of the circle $C$.
The constraint can then be written as:
:$\forall t: \map f {\mathbf r} := \size {\mathbf r}^2 - R^2 = 0$
The Principle of Stationary Action states that the first variation of the action $S$ must vanish.
:$\ds \delta S = \delta \int L \rd t = 0$
The Lagrangian $L$ is given by the kinetic energy alone, because the constraint force is handled implicitly by the constraint, and there are no other forces acting on $P$.
:$L = \dfrac 1 2 m \size {\mathbf v}^2$
To incorporate the constraint, we use a scalar Lagrange multiplier $\lambda$.
Because we are dealing with functionals, and not functions, and the constraint is local and not global, $\lambda$ is also a function of time $t$, as is $\mathbf r$:
:$\lambda = \map \lambda t$ 
The augmented action $S^+$ is given by:
:$\ds S^+ = S + \int \lambda f \d t$
We have that:
:$\dfrac {\d f} {\d \mathbf r} = 2 \mathbf r$
Then:
:$\ds \delta S^+ = \int \paren {-\map {\dfrac \d {\d t} } {m \mathbf v} + \map \lambda {2 \mathbf r} } \cdot \delta \mathbf r \rd t + \int f \delta \lambda \rd t = 0$
This means that both the terms in the brackets as well as $f$ must vanish.
We eliminate $\lambda$ from the term in the brackets by taking its scalar product with $\mathbf r$ and simplifying $\map {\dfrac \d {\d t} } {m \mathbf v}$ to $m \mathbf a$:
:$-\map {\dfrac \d {\d t} } {m \mathbf v} \cdot \mathbf r + \map \lambda {2 \mathbf r} \cdot \mathbf r = 0$
This gives:
:$2 \lambda = \dfrac {m \mathbf a \cdot \mathbf r} {R^2}$
We must replace $\mathbf a \cdot \mathbf r$.
We take the second derivative $t$ of the constraint, which must vanish too:
:$\dfrac {\d^2 f} {\d t^2} = 2 \map {\dfrac \d {\d t} } {\mathbf r \cdot \mathbf v} = 2 \paren {\mathbf a \cdot \mathbf r + \size {\mathbf v}^2} = 0$
So:
:$2 \lambda = \dfrac {-m \size {\mathbf v}^2} {R^2}$
Substituting $2 \lambda$ in the term with the big brackets yields:
:$-m \mathbf a - \dfrac {m \size {\mathbf v}^2} {R^2} \mathbf r = 0$
Division by $m$ completes the proof.
== Sources ==
* 
* 
Category:Physics
Acceleration of Point in Plane in Intrinsic Coordinates

== Theorem ==
Let $\mathbf a$ be the acceleration of a particle $P$ in space.
Let $P$ be moving in a plane $\PP$.
Then the motion of $P$ can be expressed in intrinsic coordinates as:
:$\mathbf a = \dfrac {\d \map v t} {\d t} \mathbf s + \dfrac {\paren {\map v t}^2} \rho \bspsi$
where:
:$\mathbf s$ denotes the unit vector along the tangential direction of $P$
:$\bspsi$ denotes the unit vector toward the center of curvature of the motion of $P$
:$\map v t$ is the speed of $P$ at the time $t$
:$\rho$ is the radius of curvature of the motion of $P$ at time $t$.
== Proof ==
== Sources ==
* 
* 
Category:Acceleration
Acceleration of Point in Straight Line

== Theorem ==
Let $\mathbf a$ be the acceleration of a particle $P$ in space.
Let $P$ be moving along a straight line $\LL$ whose positive direction has been established.
Then the motion of $P$ can be defined by:
:$\mathbf a = \dfrac {\d^2 s} {\d t^2} \mathbf i$
where $\mathbf i$ denotes the unit vector in the positive direction of $\LL$.
== Proof ==
We have that $P$ moves along a straight line $\LL$.
Then the rate of change of displacement perpendicular to $\LL$ is zero.
Let the $\LL$ be embedded in a Cartesian space $\CC$.
From the components of acceleration vector, we have:
:$\mathbf a = \dfrac {\d \mathbf v} {\d t} = \dfrac {\d^2 \mathbf r} {\d t^2}$
where:
:$\mathbf v$ is the velocity of $P$ at time $t$
:$\mathbf r$ is the displacement of $P$ at time $t$ expressed as a position vector from a given origin.
Thus:
:$\mathbf a = \dfrac {\d^2 x} {\d t^2} \mathbf i + \dfrac {\d^2 x} {\d t^2} \mathbf j + \dfrac {\d^2 x} {\d t^2} \mathbf k$
where:
:$\mathbf r = x \mathbf i + y \mathbf j + z \mathbf k$
Let us define $\CC$ such that $\LL$ is parallel to the $x$-axis of $\CC$.
Then we have:
== Sources ==
* 
* 
* 
* 
Category:Acceleration
Acceleration of Rocket in Outer Space

== Theorem ==
Let $B$ be a rocket travelling in outer space.
Let the velocity of $B$ at time $t$ be $\mathbf v$.
Let the mass of $B$ at time $t$ be $m$.
Let the exhaust velocity of $B$ be constant at $\mathbf b$.
Then the acceleration of $B$ at time $t$ is given by:
:$\mathbf a = \dfrac 1 m \paren {-\mathbf b \dfrac {\d m} {\d t} }$
== Proof ==
From Motion of Rocket in Outer Space, the equation of motion of $B$ is given by:
:$(1): \quad m \dfrac {\d \mathbf v} {\d t} = -\mathbf b \dfrac {\d m} {\d t}$
By definition, the acceleration of $B$ is its rate of change of velocity:
:$\mathbf a = \dfrac {\d \mathbf v} {\d t}$
The result follows by substituting $\mathbf a$ for $\dfrac {\d \mathbf v} {\d t}$ and dividing by $m$.
== Sources ==
* : Chapter $\text {B}.8$: Rocket Propulsion in Outer Space
Category:Rocket Science
Acceleration with respect to Relative Acceleration

== Theorem ==
Let $A$ and $B$ be bodies in space.
Let $\mathbf a_A$ and $\mathbf a_B$ denote the accelerations of $A$ and $B$.
Let $\mathbf a_{AB}$ denote the acceleration of $A$ relative to $B$.
Then:
:$\mathbf a_A = \mathbf a_{AB} + \mathbf a_B$
== Proof ==
The result follows directly.
== Sources ==
* 
* 
Category:Relative Velocity
Acceptance Sampling/Examples/200

== Examples of Acceptance Sampling ==
Let the sample size be $200$.
Let the rule be to accept a batch for which the number of defectives $d$ is such that $d \le 3$.
Let the producer's risk $\alpha$ be the probability of the consumer rejecting a batch with $0 \cdotp 5 \%$ or less defectives.
Let the consumer's risk $\beta$ be the probability of the consumer accepting a batch with $5 \%$ or more defectives.
Then:
== Proof ==
== Sources ==
* 
Category:Acceptance Sampling
Accumulation Point of Infinite Sequence in First-Countable Space is Subsequential Limit

== Theorem ==
Let $\struct {X, \tau}$ be a first-countable topological space.
Let $\sequence {x_n}_{n \mathop \in \N}$ be an infinite sequence in $X$.
Let $x$ be an accumulation point of $\sequence {x_n}$.
Then $x$ is a subsequential limit of $\sequence {x_n}$.
== Proof ==
By the definition of a first-countable space, there exists a countable local basis $\BB$ at $x$.
By Surjection from Natural Numbers iff Countable, there exists a surjection $\phi: \N \to \BB$.
For all $n \in \N$, define the set:
:$\ds U_n = \bigcap_{k \mathop = 0}^n \map \phi k$
By General Intersection Property of Topological Space, it follows that $U_n$ is an open neighborhood of $x$.
Using the Principle of Recursive Definition, we construct a strictly increasing sequence $\sequence {n_k}_{k \mathop \in \N}$ in $\N$.
By the definition of an accumulation point, we can choose $n_0 \in \N$ such that $x_{n_0} \in U_0$.
For all $k \in \N$, let $n_{k + 1} > n_k$ be the (unique) smallest natural number such that $x_{n_{k + 1} } \in U_{k + 1}$.
Such an $n_{k + 1}$ exists by the definition of an accumulation point, and by the well-ordering principle.
We now show that $x$ is a limit of $\sequence {x_{n_k} }$.
Let $U$ be an open neighborhood of $x$.
By the definition of a local basis, there exists an $H \in \BB$ such that $H \subseteq U$.
By the definition of a surjection, there exists a natural number $m$ such that $H = \map \phi m$.
By construction, we have:
:$\forall k \in \N: k > m \implies x_{n_k} \in U_k \subseteq \map \phi m = H$
Hence the result, by the definition of a limit point.
Category:First-Countable Spaces
Category:Sequences
Accumulation Point of Sequence is not necessarily Limit Point

== Theorem ==
Let $T = \struct {S, \tau}$ be a topological space.
Let $\sequence {a_n}$ be a sequence in $T$.
Let $q \in S$ be an accumulation point of $\sequence {a_n}$.
Then it is not necessarily the case that $q$ is also a limit of $\sequence {a_n}$.
== Proof ==
Proof by Counterexample:
Let $\struct {\R, \tau_d}$ be the real number line with the usual (Euclidean) topology.
Let $\sequence {a_n}$ be the sequence defined as:
Then $\sequence {a_n}$ has exactly one accumulation point, that is $1$.
However, $1$ is not a limit of $\sequence {a_n}$, as $\sequence {a_n}$ has no limit.
Hence the result.
== Sources ==
* : Part $\text {II}$: Counterexamples: $28$. Euclidean Topology: $3$
Category:Accumulation Points
Category:Limit Points
Accumulation Point of Sequence of Distinct Terms is Omega-Accumulation Point of Range

== Theorem ==
Let $T = \struct {S, \tau}$ be a topological space.
Let $\sequence {x_n}$ be a sequence of distinct terms of $S$.
Let $\alpha$ be an accumulation point of $\sequence {x_n}$.
Then $\alpha$ is also an $\omega$-accumulation point of $\set {x_n: n \in \N}$.
== Proof ==
Let $U$ be an open set of $T$ containing $\alpha$.
By definition of accumulation point of $\sequence {x_n}$, $U$ contains infinitely many terms of $\sequence {x_n}$.
As $\sequence {x_n}$ is a sequence of distinct terms:
$U$ contains infinitely many elements of $\set {x_n: n \in \N}$.
Thus by definition $\alpha$ is an $\omega$-accumulation point of $\set {x_n: n \in \N}$.
== Sources ==
* : Part $\text I$: Basic Definitions: Section $1$: General Introduction: Limit Points
Category:Accumulation Points
Category:Omega-Accumulation Points
Accumulation Point of Sequence of Reciprocals and Reciprocals + 1

== Theorem ==
Let $\struct {\R, \tau}$ denote the real number line under the usual (Euclidean) topology.
Let $\sequence {a_n}$ denote the sequence in $\struct {\R, \tau}$ defined as:
Then $0$ is an accumulation point of $\sequence {a_n}$.
== Proof ==
Let $\epsilon \in \R_{>0}$ be a (strictly) positive real number.
Then the open interval $\openint {-\epsilon} \epsilon$ contains $0$ and all elements $a_m$ of $S$ such that $0 < \dfrac 1 m < \epsilon$.
We have that:
:$\forall n \in \N: n \ge m$
have the property that $0 < \dfrac 1 n < \epsilon$.
Hence there are a countably infinite number of terms of $\sequence {a_n}$ such that $a_n \in \openint {-\epsilon} \epsilon$.
Hence the result by definition of accumulation point of $\sequence {a_n}$.
== Also see ==
* Zero is not a Limit Point of Sequence of Reciprocals and Reciprocals + 1 where it is shown that $0$ is not a limit of $\sequence {a_n}$.
== Sources ==
* : Part $\text {II}$: Counterexamples: $32$. Special Subsets of the Real Line: $3$
Category:Sequences
Category:Accumulation Points
Accumulation Points for Sequence in Particular Point Space

== Theorem ==
Let $T = \struct {S, \tau_p}$ be a particular point space.
Let $\sequence {a_i}$ be an infinite sequence in $T$.
Let $\beta$ be an accumulation point of $\sequence {a_i}$.
Then $\beta$ is such that an infinite number of terms of $\sequence {a_i}$ are equal either to $\beta$ or to $p$.
== Proof ==
Let $\beta$ be an accumulation point of $\sequence {a_i}$.
Then by definition:
:all open sets of $T$ which contain $\beta$ also contain an infinite number of terms of $\sequence {a_i}$.
This condition applies to the open set $\set {\beta, p}$.
So $\set {\beta, p}$ contains an infinite number of terms of $\sequence {a_i}$.
The result follows.
== Sources ==
* : Part $\text {II}$: Counterexamples: $8 \text { - } 10$. Particular Point Topology: $1$
Category:Particular Point Topology
Category:Accumulation Points
Accumulation Points of Sequence of Distinct Terms in Infinite Particular Point Space

== Theorem ==
Let $T = \struct {S, \tau_p}$ be an infinite particular point space.
Let $U \in \tau_p$ be a countably infinite open set of $T$.
Let the elements of $U$ be arranged into a sequence $\sequence {a_i}$ of distinct terms in $T$.
Then while every element $x$ of $U$ such that $x \ne p$ is a limit point of $U$, there exists no $x \in U$ such that $x$ is an accumulation point of $\sequence {a_i}$.
== Proof ==
We note from Limit Points in Particular Point Space that every $x \in U$ such that $x \ne p$ is a limit point of $U$.
Let $\sequence {a_i}$ be the (infinite) sequence in $T$ whose terms are the elements of $U$.
 $\beta \in S$ is an accumulation point of $\sequence {a_i}$.
Then from Accumulation Points for Sequence in Particular Point Space, $\beta$ is such that an infinite number of terms of $\sequence {a_i}$ are equal either to $\beta$ or to $p$.
But this contradicts the assertion that $\sequence {a_i}$ is a sequence of distinct terms.
Hence no element of $U$ is an accumulation point of $\sequence {a_i}$.
== Sources ==
* : Part $\text {II}$: Counterexamples: $9 \text { - } 10$. Infinite Particular Point Topology: $1$
Category:Limit Points
Category:Particular Point Topology
Accumulation Points of Sequence of Distinct Terms in Infinite Particular Point Space/Mistake

== Source Work ==
:
:Part $\text {II}$: Counterexamples
::Section $9 \text { - } 10$: Infinite Particular Point Topology
:::Item $1$
== Mistake ==
:The sequences $\set {a_i}$ which converge are those for which the $a_i \ne p$ are equal for all but a finite number of indices. The only accumulation points for sequences are the points $b_j$ that the $a_i$ equal for infinitely many indices.
== Correction ==
It is essential to clarify that in the sentence:
:The only accumulation points for sequences are the points $b_j$ that the $a_i$ equal for infinitely many indices.
the sequences in question are specifically those where all $a_i \ne p$.
It is insufficient for complete understanding that the $a_i \ne p$ in the previous question (for which it is not even clear whether it should apply to all terms of $\set {a_i}$ or not) be an implicit condition on the sequences being referred to in the second sentence.
== Also see ==
* Accumulation Points of Sequence of Distinct Terms in Infinite Particular Point Space
== Sources ==
* : Part $\text {II}$: Counterexamples: $9 \text { - } 10$. Infinite Particular Point Topology: $1$
Category:Mistakes/Counterexamples in Topology
Accuracy of Convergents of Continued Fraction

== Theorem ==
=== Accuracy of Convergents of Continued Fraction Expansion of Irrational Number ===
=== Accuracy of Convergents of Convergent Simple Infinite Continued Fraction ===
== Caution ==
We have , by Correspondence between Irrational Numbers and Simple Infinite Continued Fractions, that the statements above coincide, or at least partially.
The point is that they are used when proving the correspondence.
== Also see ==
* Properties of Convergents of Continued Fractions
Category:Continued Fractions
Accuracy of Convergents of Continued Fraction Expansion of Irrational Number

== Theorem ==
Let $x$ be an irrational number.
Let $(a_0, a_1, \ldots)$ be its continued fraction expansion.
Let $\left \langle {C_n}\right \rangle_{n \geq 0}$ be its sequence of convergents.
Let $p_0, p_1, p_2, \ldots$ and $q_0, q_1, q_2, \ldots$ be its numerators and denominators.
Then:
:$\forall k \ge 1: \left\vert{x - \dfrac {p_{k + 1} } {q_{k + 1} } }\right\vert < \dfrac 1 {q_{k + 1} q_{k + 2} } \le \dfrac 1 {2 q_k q_{k + 1} } < \left\vert{x - \dfrac {p_k} {q_k} }\right\vert$
Thus:
:The of the inequality gives an indication of how close each convergent gets to its true value.
:The gives a bound that limits its accuracy.
=== Corollary ===
== Proof ==
The Continued Fraction Algorithm gives the following system of equations:
and
Now from the Continued Fraction Algorithm:
:$x_{n + 1} = \left[{a_{n + 1}, a_{n + 2}, a_{n + 3}, \ldots}\right]$
So:
: $a_{n + 1} < x_{n + 1} < a_{n + 1} + 1$
Therefore:
:$\left\vert{x - \dfrac {p_n} {q_n} }\right\vert < \dfrac 1 {q_n \left({a_{n + 1} q_n + q_{n - 1} }\right)} = \dfrac 1 {q_n q_{n + 1} }$
This gives the of the inequality when $n = k + 1$.
We also have:
:$\left\vert{x - \dfrac {p_n} {q_n} }\right\vert > \dfrac 1 {q_n \left({\left({a_{n + 1} + 1}\right) q_n + q_{n - 1} }\right)}$
This gives the of the inequality when $n = k$.
For the middle inequality, note that:
:$q_{k + 2} = a_{k + 2} q_{k + 1} + q_k > q_k + q_k = 2 q_k$
So:
:$\dfrac 1 {q_{k + 1} q_{k + 2} } \le \dfrac 1 {2 q_k q_{k + 1} }$
== Also see ==
* Accuracy of Convergents of Continued Fraction
Category:Continued Fractions
Accuracy of Convergents of Continued Fraction Expansion of Irrational Number/Corollary

== Corollary to Accuracy of Convergents of Continued Fraction ==
Let $x$ be an irrational number.
Let $\left \langle {C_n}\right \rangle$ be the sequence of convergents of $x$.
Let $p_1, p_2, p_3, \ldots$ and $q_1, q_2, q_3, \ldots$ be its numerators and denominators.
Then:
:$\forall k \ge 1: \dfrac 1 {q_k q_{k+1}} > \left\vert{x - \dfrac {p_k} {q_k}}\right\vert > \dfrac 1 {2 q_k q_{k+1}}$
== Proof ==
From Accuracy of Convergents of Continued Fraction:
:$\forall k \ge 1: \left\vert{x - \dfrac {p_{k+1}} {q_{k+1}}}\right\vert \le \dfrac 1 {q_{k+1} q_{k+2}} \le \dfrac 1 {2 q_k q_{k+1}} < \left\vert{x - \dfrac {p_k} {q_k}}\right\vert$
The last inequality immediately gives:
:$\left\vert{x - \dfrac {p_k} {q_k}}\right\vert > \dfrac 1 {2 q_k q_{k+1}}$
The first inequality with $k-1$ gives:
:$\dfrac 1 {q_k q_{k+1}} \ge \left\vert{x - \dfrac {p_k} {q_k}}\right\vert$
as desired.
Category:Continued Fractions
Accuracy of Convergents of Convergent Simple Infinite Continued Fraction

== Theorem ==
Let $C = \tuple {a_0, a_1, \ldots}$ be an simple infinite continued fraction in $\R$.
Let $C$ converge to $x \in \R$.
For $n \ge 0$, let $C_n = \dfrac {p_n} {q_n}$ be the $n$th convergent of $C$, where $p_n$ and $q_n$ are the $n$th numerator and denominator.
Then for all $n \ge 0$:
:$\size {x - \dfrac {p_n} {q_n} } < \dfrac 1 {q_n q_{n + 1} }$
== Proof ==
We show that either:
:$x \in \closedint {C_n} {C_{n + 1} }$
or:
:$x \in \closedint {C_{n + 1} } {C_n}$
so that the result follows from:
:Difference between Adjacent Convergents of Simple Continued Fraction
:Distance between Point of Real Interval and Endpoint is at most Length
==== Odd case ====
Let $n \ge 1$ be odd.
By Limit of Subsequence equals Limit of Sequence:
:$x = \ds \lim_{k \mathop \to \infty} C_{2 k}$
For all $2 k \ge n + 1$, by:
:Even Convergents of Simple Continued Fraction are Strictly Increasing
:Even Convergent of Simple Continued Fraction is Strictly Smaller than Odd Convergent
we have:
:$C_{n + 1} \le C_{2 k} < C_n$
By Lower and Upper Bounds for Sequences:
:$x \in \closedint {C_{n + 1} } {C_n}$
==== Even case ====
Let $n \ge 0$ be even.
By Limit of Subsequence equals Limit of Sequence:
:$x = \ds \lim_{k \mathop \to \infty} C_{2 k + 1}$
For all $2 k + 1 \ge n + 1$, by:
:Odd Convergents of Simple Continued Fraction are Strictly Decreasing
:Even Convergent of Simple Continued Fraction is Strictly Smaller than Odd Convergent
we have:
:$C_n < C_{2 k + 1} \le C_{n + 1}$
By Lower and Upper Bounds for Sequences:
:$x \in \closedint {C_n} {C_{n + 1} }$
== Also see ==
* Accuracy of Convergents of Continued Fraction
Category:Simple Continued Fractions
Accurate to n Decimal Places/Examples/Pi to 4 Places

== Example of Accurate to $n$ Decimal Places ==
The constant pi ($\pi$) expressed accurate to $4$ decimal places is:
:$\pi \approx 3 \cdotp 1415$
== Proof ==
We start with the decimal expansion of $\pi$:
and so $\pi$ expressed to the first $4$ decimal places is:
:$3.1415$
== Sources ==
* 
Category:Examples of Accuracy
Accurate to n Decimal Places/Examples/Root of 86.56 to 2 Places

== Example of Accurate to $n$ Decimal Places ==
The square root of $86 \cdotp 56$ expressed accurate to $2$ decimal places is:
:$\sqrt {86 \cdotp 56} \approx 9 \cdotp 30$
== Proof ==
We evaluate the square root of $86.56$ to arbitrary accuracy:
:$\sqrt {86 \cdotp 56} \approx 9 \cdotp 3037626...$
and so $\sqrt {86 \cdotp 56}$ expressed to the first $2$ decimal places is:
:$9 \cdotp 30$
where the $2$nd decimal place has been rounded down.
== Sources ==
* 
* 
Category:Examples of Accuracy
Accurate to n Decimal Places/Examples/e to 2 Places

== Example of Accurate to $n$ Decimal Places ==
Euler's number $e$ expressed accurate to $2$ decimal places is:
:$e \approx 2 \cdotp 72$
== Proof ==
We start with the decimal expansion of $e$:
and so $e$ expressed to the first $2$ decimal places is:
:$2.72$
where the $2$nd decimal place has been rounded up.
== Sources ==
* 
* 
Category:Examples of Accuracy
Accurate to n Decimal Places/Examples/e to 3 Places

== Example of Accurate to $n$ Decimal Places ==
Euler's number $e$ expressed accurate to $3$ decimal places is:
:$e \approx 2 \cdotp 718$
== Proof ==
We start with the decimal expansion of $e$:
and so $e$ expressed to the first $3$ decimal places is:
:$2.718$
== Sources ==
* 
* 
Category:Examples of Accuracy
Accurate to n Significant Figures/Examples/Pi to 5 Figures

== Example of Accurate to $n$ Significant Figures ==
The constant pi ($\pi$) expressed accurate to $5$ significant figures is:
:$\pi \approx 3 \cdotp 1416$
== Proof ==
We start with the decimal expansion of $\pi$:
We note that:
:$3.1416 - 3.14159 \ldots < 3.14159 \ldots - 3.1415$
and so $\pi$ rounded to $5$ significant figures is:
:$3.1416$
== Sources ==
* 
Category:Examples of Accuracy