nyuuzyou/wikis huggingface dataset:
Ladies' Diary/Hour and Minute Hand in Straight Line

== Puzzle ==
:"Being at so large a distance from the dial-plate of a great clock, that I could not distinguish the figures;
:but as the hour and minute hands were very bright and glaring,"
:the correspondent noted that they were in a straight line and pointing upwards to the right.
:It was evening.
:What was the time?
== Solution ==
:$5 \frac 5 {11}$ minutes, or approximately $5$ minutes and $27$ seconds, past $7$.
== Proof ==
Let it be assumed from the description of the general direction of the hands, and the fact that it is evening, that it is some short time after $7$ p.m.
Thus the hour hand is pointing a little way past the $7$.
Hence the minute hand is pointing a little way past the $1$.
Let $x$ be the number of minutes after $7$ that it is.
Let $\theta$ be the angle made by the hands of the clock from the vertical at that time.
The minute hand rotates at $1$ revolution every $60$ minutes.
That is, $\dfrac {360 \degrees} {60} = 6 \degrees$ per minute.
The hour hand rotates at $1$ revolution every $12$ hours.
That is, $\dfrac {360 \degrees} {12 \times 60} = \dfrac 1 2 \degrees$ per minute.
But at $7$, the hour hand is already at $\dfrac {360 \degrees} {12} = 30 \degrees$ from the vertical.
Thus:
So it is $5 \frac 5 {11}$ minutes past $7$.
$\dfrac 5 {11}$ minutes is $\dfrac {5 \times 60} {11}$ seconds, which evaluates to $27.\dot 2 \dot 7$ seconds.
== Sources ==
* : The Ladies' Diary or Woman's Almanac, : $142$
Category:Ladies' Diary
Category:Clock Puzzles
Ladies' Diary/Largest Cylinder from given Cone

== Puzzle ==
:From a given (right circular) cone to cut the greatest cylinder possible.
== Solution ==
The maximum volume of the cylinder when its height is $\dfrac 2 3$ of the height of the cone.
== Proof ==
It is assumed the base of the cylinder is on the base of the cone.
It remains to determine its height as a proportion of the height of the cone.
:
The volume of the cylinder is proportional to $DE^2 \times PQ$.
We have that $\dfrac {DE} {AP}$ is constant.
Therefore we are to ensure that $AP^2 \times PQ$ is a maximum.
That is, given any line $AQ$, find a point $P$ on it such that $AP^2 \times PQ$ is a maximum.
Let $AP$ be denoted $x$.
Let $AQ$ be denoted $L$.
Thus we need to make $x^2 \paren {L - x}$ a maximum.
That is the same as making $x^2 \paren {2 L - 2 x}$ a maximum.
But the latter is the product of $3$ factors whose sum is constant at $2 L$.
It will achieve that maximum when all $3$ are equal.
Therefore the maximum is when $x = 2 L - 2 x$ and $x = \dfrac {2 L} 3$.
So $AP$ is $\dfrac 2 3$ of $AQ$.
== Historical Note ==
== Sources ==
* 
* : Geometrical Problems relating to Maxima and Minima
* : The Ladies' Diary or Woman's Almanac, : $139$
Category:Ladies' Diary
Category:Right Circular Cylinders
Category:Cones
Ladies' Diary/Largest Cylinder from given Cone/Historical Note

== Historical Note on Largest Cylinder from given Cone ==
This often-cited problem was raised by , about whom little is known.
 informs us in his of $1992$ that the solution as given here is "something of a cheat", as it refers the reader to to one of 's texbooks.
This is puzzling, however, as this problem is cited as appearing in $1714$, and was not born until $1737$.
It was often the case that questions to were taken from already-published sources.
== Sources ==
* : The Ladies' Diary or Woman's Almanac, : $139$
Category:Historical Notes
Ladies' Diary/Lowest Common Multiple of 1 to 9

== Puzzle ==
:What is the least number which will divide by the nine digits without leaving a remainder?
== Solution ==
$2520$.
== Proof ==
This is the same as asking:
:What is the lowest common multiple of $\set {1, 2, 3, 4, 5, 6, 7, 8, 9}$?
This must be $5 \times 7 \times 8 \times 9$ which is $2520$.
== Sources ==
* : The Ladies' Diary or Woman's Almanac, : $140$
Category:Ladies' Diary
Category:Examples of Lowest Common Multiples of Integers
Category:2520
Ladies' Diary/Money Counting Problem

== Puzzle ==
:In how long a time would a million of millions of money be in counting,
:supposing one hundred pounds to be counted every minute without intermission,
:and the year to consist of $365$ days, $5$ hours, $45$ minutes?
== Solution ==
The number of minutes $m$ in the year is calculated as:
The number of $100$s $h$ in a million of millions is calculated as:
This gives the total number of minutes to count the money.
The total number of years $t$ needed to count the money is therefore:
== Historical Note ==
== Sources ==
* : The Ladies' Diary or Woman's Almanac, : $136$
Category:Ladies' Diary
Ladies' Diary/Ratios of Ages of Groom to Bride

== Puzzle ==
:A person remarked that upon his wedding day the proportion of his own age to that of his bride was $3$ to $1$,
:but $15$ years afterwards the proportion of their ages was $2$ to $1$.
:What were their ages upon the day of their marriage?
== Solution ==
The bride was $15$ and the groom was $45$ when they married.
Outrageous.
== Proof ==
Let $x$ and $y$ be the ages of the bride and groom at their marriage respectively.
We have:
== Sources ==
* : The Ladies' Diary or Woman's Almanac, : $138$
Category:Ladies' Diary
Lagrange's Formula

== Theorem ==
Let $\mathbf a$, $\mathbf b$ and $\mathbf c$ be vectors in a vector space $\mathbf V$ of $3$ dimensions.
Then:
:$\mathbf a \times \paren {\mathbf b \times \mathbf c} = \paren {\mathbf a \cdot \mathbf c} \mathbf b - \paren {\mathbf a \cdot \mathbf b} \mathbf c$
=== Corollary ===
== Proof ==
Let $\mathbf a$, $\mathbf b$ and $\mathbf c$ be embedded in a Cartesian $3$-space:
:$\mathbf a = \begin{bmatrix} a_x \\ a_y \\ a_z \end{bmatrix}$, $\mathbf b = \begin{bmatrix} b_x \\ b_y \\ b_z \end{bmatrix}$, $\mathbf c = \begin{bmatrix} c_x \\ c_y \\ c_z \end{bmatrix}$
== Sources ==
* : Chapter $\text {II}$: The Products of Vectors: $7$. Products of Three Vectors: $(2.22)$
* : Chapter $\text I$: Vector Algebra: $\S 5$
* : $\S 22$: Miscellaneous Formulas involving Dot and Cross Products: $22.18$
* 
* 
* 
Category:Vector Algebra
Category:Vector Triple Product
Lagrange's Formula/Corollary

== Theorem ==
Let $\mathbf a$, $\mathbf b$ and $\mathbf c$ be vectors in a vector space $\mathbf V$ of $3$ dimensions.
Then:
:$\paren {\mathbf a \times \mathbf b} \times \mathbf c = \paren {\mathbf a \cdot \mathbf c} \mathbf b - \paren {\mathbf b \cdot \mathbf c} \mathbf a$
== Proof ==
== Sources ==
* : $\S 22$: Miscellaneous Formulas involving Dot and Cross Products: $22.19$
Category:Vector Algebra
Category:Dot Product
Category:Vector Cross Product
Lagrange's Four Square Theorem

== Theorem ==
Every positive integer can be expressed as a sum of four squares.
== Proof 1 ==
== Proof 2 ==
== Also known as ==
== Examples ==
== Also see ==
* Waring's Problem
== Historical Note ==
== Sources ==
* 
* : $\text {3-5}$ The Use of Computers in Number Theory: Exercise $5$
* : $4$
* 
* : Chapter $\text {B}.2$: More about Numbers: Irrationals, Perfect Numbers and Mersenne Primes
* : $4$
* 
* 
* : Chapter $7$: Patterns in Numbers: Fermat
* 
* 
Category:Lagrange's Four Square Theorem
Category:Sums of Squares
Lagrange's Four Square Theorem/Also known as

== Lagrange's Four Square Theorem: Also known as == 
Some sources use the plural form for Lagrange's Four Square Theorem , as Lagrange's Four Squares Theorem.
Some omit 's name, and refer to this as just the Four Squares Theorem.
Some sources refer to this as just Lagrange's Theorem, but this is ambiguous as there is more than one theorem with 's name on it.
== Sources ==
* 
* : Chapter $\text {B}.2$: More about Numbers: Irrationals, Perfect Numbers and Mersenne Primes
* 
* 
* 
* 
* 
Category:Lagrange's Four Square Theorem
Lagrange's Four Square Theorem/Historical Note

== Historical Note on Lagrange's Four Square Theorem ==
It is suggested by some sources that the result of Lagrange's Four Square Theorem was known, at least empirically, by .
Some sources suggest that the theorem was originally stated formally by .
However, it appears that published the results of his having tested it thoroughly up to $120$, and stated the theorem in his $1621$ translation of the of .
 read about it in his copy of that work, and studied it, but appears to have failed to find a proof, as no proof of his can be found.
 investigated the problem, discussing it with in around $1730$, publishing a proof of a weaker version around that time.
It was finally proved by in $1770$. The proof was published in $1772$.
 subsequently submitted a proof of his own in $1772$, which was finally published in $1780$.
== Sources ==
* 
* : $4$
* : $\S 3$: Appendix $\text A$: Euler
* : $4$
* 
* 
* 
* 
Category:Historical Notes
Lagrange's Method of Multipliers/Examples/Arbitrary Example 1

== Examples of Use of Lagrange's Method of Multipliers ==
Find the maximum $M$ of the function $u: \R^2 \to \R$ defined as:
:$\forall \tuple {x, y} \in \R^2: \map u {x, y} = x y$
subject to the constraint:
:$\text C: \quad x + y = 1$
== Solution ==
The maximum $M$ is
:$M = \dfrac 1 4$
at $\tuple {x, y} = \tuple {\frac 1 2, \frac 1 2}$.
== Proof ==
We write:
:$L = x y + \lambda \paren {x + y - 1}$
Differentiation $x$, $y$ and $\lambda$ and equating to zero gives:
This leads to:
giving:
:$u = \dfrac 1 4$
This is the only stationary point for $L$.
We need to confirm it is indeed a maximum for this constraint:
:$\text C: \quad x + y = 1$
We check at $\tuple {1, 0}$ and $\tuple {0, 1}$, both of which obey $\text C$, and confirm that $u = 0$ at both points.
So it goes:
:up from $0$ at $\tuple {1, 0}$ to $\dfrac 1 4$ at $\tuple {\frac 1 2, \frac 1 2}$
:back down again from $\dfrac 1 4$ at $\tuple {\frac 1 2, \frac 1 2}$ to $0$ at $\tuple {0, 1}$.
There is only one stationary point for $L$.
Hence at $\dfrac 1 4$ it must be a maximum.
</onlyinclude>
== Sources ==
* 
* 
Category:Examples of Use of Lagrange's Method of Multipliers
Lagrange's Theorem (Group Theory)

== Theorem ==
Let $G$ be a finite group.
Let $H$ be a subgroup of $G$.
Then:
:$\order H$ divides $\order G$
where $\order G$ and $\order H$ are the order of $G$ and $H$ respectively.
In fact:
:$\index G H = \dfrac {\order G} {\order H}$
where $\index G H$ is the index of $H$ in $G$.
When $G$ is an infinite group, we can still interpret this theorem sensibly:
:A subgroup of finite index in an infinite group is itself an infinite group.
:A finite subgroup of an infinite group has infinite index.
== Proof 1 ==
== Proof 2 ==
== Proof 3 ==
== Examples ==
== Also see ==
* Group does not Necessarily have Subgroup of Order of Divisor of its Order, where it is shown that the converse is not true in general. 
== Historical Note ==
== Sources ==
* : Chapter $\text{II}$: Groups: Problem $\text {GG}$
* : $\S 5$: Groups $\text{I}$: Subgroups
* 
* 
* : $\S 2$: Some Basic Facts, Proof and Converse of Lagrange's Theorem.
Category:Order of Groups
Category:Index of Subgroups
Category:Subgroups
Category:Lagrange's Theorem (Group Theory)
Lagrange's Theorem (Group Theory)/Examples

== Examples of Use of Lagrange's Theorem ==
=== Intersection of Subgroups of Order $25$ and $36$ ===
=== Order of Group with Subgroups of Order $25$ and $36$ ===
=== Order of Union of Subgroups of Order $16$ ===
Category:Examples of Use of Lagrange's Theorem (Group Theory)
Lagrange's Theorem (Group Theory)/Examples/Intersection of Subgroups of Order 25 and 36

== Examples of Use of Lagrange's Theorem ==
Let $G$ be a group.
Let $H$ and $K$ be subgroups of $G$ such that:
:$\order H = 25$
:$\order K = 36$
where $\order {\, \cdot \,}$ denotes the order of the subgroup.
Then:
:$\order {H \cap K} = 1$ 
== Proof ==
From Intersection of Subgroups is Subgroup:
:$H \cap K \le H$
and:
:$H \cap K \le K$
where $\le$ denotes subgrouphood.
From Lagrange's theorem:
:$\order {H \cap K} \divides 25$
and:
:$\order {H \cap K} \divides 36$
As $25$ and $36$ are coprime, it follows that:
:$\order {H \cap K} = 1$
== Sources ==
* : Chapter $7$: Cosets and Lagrange's Theorem: Exercise $11 \ \text{(i)}$
Category:Examples of Use of Lagrange's Theorem (Group Theory)
Lagrange's Theorem (Group Theory)/Examples/Order of Group with Subgroups of Order 25 and 36

== Examples of Use of Lagrange's Theorem ==
Let $G$ be a group.
Let $H$ and $K$ be subgroups of $G$ such that:
:$\order H = 25$
:$\order K = 36$
where $\order {\, \cdot \,}$ denotes the order of the subgroup.
Then:
:$900 \divides \order G$ 
where $\divides$ denotes divisibility.
== Proof ==
From Lagrange's theorem:
:$25 \divides \order G$
and:
:$36 \divides \order G$
We have that:
:$\lcm \set {25, 36} = 900$
Hence the result.
== Sources ==
* : Chapter $7$: Cosets and Lagrange's Theorem: Exercise $11 \ \text{(ii)}$
Category:Examples of Use of Lagrange's Theorem (Group Theory)
Lagrange's Theorem (Group Theory)/Examples/Order of Union of Subgroups of Order 16

== Examples of Use of Lagrange's Theorem ==
Let $G$ be a group whose identity is $e$.
Let $H$ and $K$ be subgroups of $G$ such that:
:$\order H = \order K = 16$
:$H \ne K$
where $\order {\, \cdot \,}$ denotes the order of the subgroup.
Then:
:$24 \le \order {H \cup K} \le 31$ 
== Proof ==
As $H$ and $K$ are subgroups of $G$, they share at least $e$.
That is, $\order {H \cap K} \ge 1$.
On the other hand, we have that $H \ne K$.
Thus $\order {H \cap K} < 16$.
But from Intersection of Subgroups is Subgroup, $H \cap K$ is a subgroup of both $H$ and $K$.
From Lagrange's Theorem, that means:
:$\order {H \cap K} \divides 16$
and so:
:$\order {H \cap K} \le 8$
From Cardinality of Set Union:
:$\order {H \cup K} = \order H + \order K - \order {H \cap K}$
and so:
:$\order {H \cup K} = 16 + 16 - \order {H \cap K} = 32 - \order {H \cap K}$
As $\order {H \cap K}$ lies within the range of $1$ to $8$, the result follows.
== Sources ==
* : Chapter $7$: Cosets and Lagrange's Theorem: Exercise $12$
Category:Examples of Use of Lagrange's Theorem (Group Theory)
Lagrange's Theorem (Group Theory)/Historical Note

== Historical Note on Lagrange's Theorem in Group Theory ==
Lagrange's theorem was actually proved by .
's proof merely showed that a subgroup of the symmetric group $S_n$ has an order which is a divisor of $n!$.
== Sources ==
* : Chapter $5$: Cosets and Lagrange's Theorem: Summary for Chapter $5$
* : $\S 5$: Groups $\text{I}$: Subgroups
* : $\S2$: Some Basic Facts, Proof and Converse of Lagrange's Theorem.
Category:Historical Notes
Lagrange's Theorem (Group Theory)/Proof 1

== Theorem ==
== Proof ==
Let $G$ be finite.
Consider the mapping $\phi: G \to G / H^l$, defined as:
:$\phi: G \to G / H^l: \map \phi x = x H^l$
where $G / H^l$ is the left coset space of $G$ modulo $H$.
For every $y H \in G / H^l$, there exists a corresponding $y \in G$, so $\phi$ is a surjection.
From Cardinality of Codomain of Surjection it follows that $G / H^l$ is finite.
From Cosets are Equivalent, $G / H^l$ has the same number of elements as $H$.
We have that the $G / H^l$ is a partition of $G$.
It follows from Number of Elements in Partition that $\index G H = \dfrac {\order G} {\order H}$
== Sources ==
* : Chapter $\text {IV}$: Rings and Fields: $25$. Cyclic Groups and Lagrange's Theorem: Theorem $25.6$ and Corollary
* : $\S 1.9$
Category:Lagrange's Theorem (Group Theory)
Lagrange's Theorem (Group Theory)/Proof 2

== Theorem ==
== Proof ==
Let $G$ be a group.
Let $H$ be a subgroup of $G$.
From Cosets are Equivalent, a left coset $y H$ has the same number of elements as $H$, namely $\order H$.
Since left cosets are identical or disjoint, each element of $G$ belongs to exactly one left coset.
From the definition of index of subgroup, there are $\index G H$ left cosets, and therefore:
:$\order G = \index G H \order H$
Let $G$ be of finite order.
All three numbers are finite, and the result follows.
Now let $G$ be of infinite order.
If $\index G H = n$ is finite, then $\order G = n \order H$ and so $H$ is of infinite order.
If $H$ is of finite order such that $\order H = n$, then $\order G = \index G H \times n$ and so $\index G H$ is infinite.
== Sources ==
* : $\S 6.4$. Lagrange's theorem
* : Chapter $2$: Subgroups and Cosets: $\S 40$
* : $\S 43.1$ Lagrange's theorem
* : $0$: Some Conventions and some Basic Facts
* : $\S 3.3$: Group actions and coset decompositions: Theorem $1$
* : Chapter $5$: Cosets and Lagrange's Theorem: Theorem $5.9$
Category:Lagrange's Theorem (Group Theory)
Lagrange's Theorem (Number Theory)

== Theorem ==
Let $\struct {\Z_p, +_p, \times_p}$ be the ring of integers modulo $p$ for some prime $p$.
Let $f$ be a polynomial in one variable of degree $n$ over $\Z_p$.
Then $f$ has at most $n$ roots in $\Z_p$.
== Proof ==
Proof by induction on $n$:
=== Basis for the Induction ===
When $n = 1$, we have:
:$\map f x = a x + b$ for some $a, b \in \Z_p$ and $a \ne 0$
Suppose $x_1, x_2 \in \Z_p$ are two roots of $\map f x$.
Then:
Hence these two roots must be the same, implying that there is at most $1$ root.
This is our basis for the induction.
=== Induction Hypothesis ===
This is our induction hypothesis:
:A polynomial in one variable of degree $k$ has at most $k$ roots in $\Z_p$.
It is to be demonstrated that:
:A polynomial in one variable of degree $k + 1$ has at most $k + 1$ roots in $\Z_p$.
=== Induction Step ===
This is our induction step:
Consider $n = k + 1$, and let $f$ be a polynomial in one variable of degree $k + 1$.
If $f$ does not have a root in $\Z_p$, our claim is satisfied.
Otherwise, suppose $f$ does have a root $x_0$.
From Ring of Integers Modulo Prime is Field, $\Z_p$ is a field.
Applying the Polynomial Factor Theorem, since $\map f {x_0} = 0$:
:$\map f x = \paren {x - x_0} \map Q x$
where $Q$ is a polynomial of degree $k$.
By Euclid's Lemma for Prime Divisors:
:$\map f x = 0 \iff x - x_0 = 0 \text { or } \map Q x = 0$
By the induction hypothesis, $Q$ has at most $k$ roots.
Hence $f$ has at most $k + 1$ roots.
By the Principle of Mathematical Induction, the theorem holds for all $n$.
== Examples ==
Category:Lagrange's Theorem (Number Theory)
Category:Number Theory
Category:Polynomial Theory
Category:Proofs by Induction
Lagrange's Theorem (Number Theory)/Examples/Non-Prime Index

== Example of Use of Lagrange's Theorem (Number Theory) ==
This does not hold for composite number $p$.
For $p = 8$, the polynomial $x^2 - 1$ has $4$ roots $\eqclass 1 8, \eqclass 3 8, \eqclass 5 8, \eqclass 7 8$ in $\Z_p$ and $4 > n = 2$.
Category:Examples of Use of Lagrange's Theorem (Number Theory)
Lagrange's Trigonometric Identities/Cosine

== Theorem ==
where $x$ is not an integer multiple of $2 \pi$.
== Proof ==
By the Werner Formula for Cosine by Sine:
:$2 \cos \alpha \sin \beta = \map \sin {\alpha + \beta} - \map \sin {\alpha - \beta}$
Thus we establish the following sequence of identities:
Summing the above:
:$\ds 2 \sin \frac x 2 \paren {\frac 1 2 + \sum_{k \mathop = 1}^n \map \cos {k x} } = \sin \frac {\paren {2 n + 1} x} 2$
as the sums on the form a telescoping series.
The result follows by dividing both sides by $2 \sin \dfrac x 2$.
It is noted that when $x$ is a multiple of $2 \pi$ then:
:$\sin \dfrac x 2 = 0$
leaving the undefined.
== Also see ==
* Lagrange's Sine Identity:
** Sine Form of Lagrange's Sine Identity
** Cosine Form of Lagrange's Sine Identity
== Sources ==
* : $\S 19$: Miscellaneous Series: $19.39$
* : Chapter $\text {B}.15$: A Rigorous Proof of Euler's Formula $\ds \sum_1^\infty \frac 1 {n^2} = \frac {\pi^2} 6$
* : $\S 21$: Series of Constants: Miscellaneous Series: $21.39.$
Category:Lagrange's Cosine Identity
Category:Lagrange's Trigonometric Identities
Category:Cosine Function
Category:Telescoping Series
Lagrange's Trigonometric Identities/Sine

== Lagrange's Sine Identity ==
==== Cosine Form of Lagrange's Sine Identity ====
==== Sine Form of Lagrange's Sine Identity ====
== Also see ==
* Lagrange's Cosine Identity
Category:Lagrange's Sine Identity
Category:Lagrange's Trigonometric Identities
Category:Trigonometric Identities
Category:Sine Function
Lagrange's Trigonometric Identities/Sine/Cosine Form

== Theorem ==
:$\ds \sum_{k \mathop = 0}^n \sin k \theta = \dfrac {\map \cos {\frac 1 2 \theta} - \map \cos {n \theta + \frac 1 2 \theta} } {2 \map \sin {\frac 1 2 \theta} }$
== Proof ==
The is a telescoping sum. 
== Also see ==
* Sine Form of Lagrange's Sine Identity
* Lagrange's Cosine Identity
* Lagrange's Trigonometric Identities
Category:Cosine Form of Lagrange's Sine Identity
Category:Lagrange's Sine Identity
Category:Lagrange's Trigonometric Identities
Lagrange's Trigonometric Identities/Sine/Sine Form

== Theorem ==
where $x$ is not an integer multiple of $2 \pi$.
== Proof 1 ==
== Proof 2 ==
== Also see ==
* Lagrange's Cosine Identity
== Sources ==
* : $\S 19$: Miscellaneous Series: $19.40$
* : $\S 21$: Series of Constants: Miscellaneous Series: $21.40.$
Category:Sine Form of Lagrange's Sine Identity
Category:Lagrange's Sine Identity
Category:Lagrange's Trigonometric Identities
Category:Sine Function
Category:Telescoping Series
Lagrange's Trigonometric Identities/Sine/Sine Form/Proof 1

== Theorem ==
== Proof ==
By Werner Formula for Sine by Sine:
:$2 \sin \alpha \sin \beta = \map \cos {\alpha - \beta} - \map \cos {\alpha + \beta}$
Thus we establish the following sequence of identities:
Summing the above:
The result follows by dividing both sides by $2 \sin \dfrac x 2$.
It is noted that when $x$ is a multiple of $2 \pi$ then:
:$\sin \dfrac x 2 = 0$
leaving the undefined.
Category:Sine Form of Lagrange's Sine Identity
Lagrange's Trigonometric Identities/Sine/Sine Form/Proof 2

== Theorem ==
== Proof ==
Let $x$ be a real number that is not a integer multiple of $2 \pi$.
Let $k$ be a non-negative integer.
We have, from Euler's Formula:
:$\map \exp {i k x} = i \sin k x + \cos k x$
Summing from $k = 0$ to $k = n$, we have: 
:$\ds \sum_{k \mathop = 0}^n \map \exp {i k x} = i \sum_{k \mathop = 0}^n \sin k x + \sum_{k \mathop = 0}^n \cos k x$
As $\sin k x$ and $\cos k x$ are both real for real $k, x$, we have: 
Category:Sine Form of Lagrange's Sine Identity
Lagrange Interpolation Formula/Formulation 1

== Theorem ==
Let $\tuple {x_0, \ldots, x_n}$ and $\tuple {a_0, \ldots, a_n}$ be ordered tuples of real numbers such that $x_i \ne x_j$ for $i \ne j$.
Then there exists a unique polynomial $P \in \R \sqbrk X$ of degree at most $n$ such that:
:$\map P {x_i} = a_i$ for all $i \in \set {0, 1, \ldots, n}$
Moreover $P$ is given by the formula:
:$\ds \map P X = \sum_{j \mathop = 0}^n a_i \map {L_j} X$
where $\map {L_j} X$ is the $j$th Lagrange basis polynomial associated to the $x_i$.
== Proof ==
Recall the definition:
:$\ds \map {L_j} X = \prod_{\substack {0 \mathop \le i \mathop \le n \\ i \mathop \ne j}} \frac {X - x_i} {x_j - x_i} \in \R \sqbrk X$
From this we see that:
:$\map {L_j} {x_i} = \delta_{i j}$
Therefore:
:$\ds \map P{x_i} = \sum_{j \mathop = 0}^n a_i \delta_{i j} = a_i$
Moreover, by Degree of Product of Polynomials over Integral Domain and Degree of Sum of Polynomials, the degree of $P$ as defined above is at most $n$.
It remains to show that the choice of $P$ is unique.
 $\tilde P$ is another polynomial with the required properties.
Let $\Delta = P - \tilde P$.
By Degree of Sum of Polynomials, the degree of $\Delta$ is at most $n$.
Now we see that for $i = 0, \ldots, n$:
:$\map \Delta {x_i} = \map P {x_i} - \map {\tilde P} {x_i} = a_i - a_i = 0$
Since by hypothesis the $x_i$ are distinct, $\Delta$ has $n + 1$ distinct zeros in $\R$.
But by the corollary to the Polynomial Factor Theorem this shows that:
:$\ds \map \Delta X = \alpha \prod_{i \mathop = 0}^n \paren {X - x_i}$
If $\alpha \ne 0$, then this shows that the degree of $\Delta$ is $n+1$, a contradiction.
Therefore:
:$\Delta = 0$
and so:
:$P = \tilde P$
This establishes uniqueness.
== Also known as ==
== Also see ==
* Equivalence of Formulations of Lagrange Interpolation Formula
== Sources ==
* 
Category:Lagrange Interpolation Formula
Lagrange Interpolation Formula/Formulation 2

== Theorem ==
Let $f : \R \to \R$ be a real function.
Let $f$ have known values $y_i = \map f {x_i}$ for $n \in \set {0, 1, \ldots, n}$.
Let a value $y' = \map f {x'}$ be required to be estimated at some $x'$.
Then:
== Proof ==
== Also known as ==
== Also see ==
* Equivalence of Formulations of Lagrange Interpolation Formula
== Sources ==
* 
* 
* 
Category:Lagrange Interpolation Formula