nyuuzyou/wikis huggingface dataset:
Higher Dimensional Hausdorff Measure than Euclidean Space is Zero

== Theorem ==
Let $\R^n$ be the $n$-dimensional Euclidean space.
Let $s \in \R_{>n}$.
Then:
:$\map {\HH ^s} {\R^n} = 0$
where $\map {\HH^s} \cdot$ denotes the $s$-dimensional Hausdorff measure on $\R^n$
== Proof ==
Let $s \in \R_{>n}$.
Consider the $n$-cube:
:$Q := {\closedint 0 1}^n$
Then:
:$\ds \R^n = \bigcup _{x \mathop \in \Z^n} Q + x$
As $\map {\HH ^s} \cdot$ is countably subadditive and translation invariant, it suffices to show:
:$\map {\HH ^s} Q = 0$
Let $N \in \N_{>0}$.
Let $\CC_N$ be the set of $n$-cubes:
:$\ds \closedint {\dfrac {i_1 - 1} {2^N} } {\dfrac {i_1} {2^N} } \times \closedint {\dfrac {i_2 - 1} {2^N} } {\dfrac {i_2} {2^N} } \times \cdots \times \closedint {\dfrac {i_n - 1} {2^N} } {\dfrac {i_n} {2^N} }$
where $i_1,\ldots ,i_n \in \set {1, 2, 3, \ldots , 2^N}$.
By Diameter of N-Cube:
:$\forall Z \in \CC_N : \size Z = \dfrac {\sqrt n} {2^N}$
where $\size Z$ denotes the diameter of $Z$.
Thus:
Category:Measure Theory
Category:Fractals
Higher Homotopy Groups are Abelian

== Theorem ==
Let $T = \left({S, \tau}\right)$ be a topological space.
Let $x_0 \in S$.
Let $n \ge 2$ be a integer.
Let $\pi_n \left({T, x_0}\right)$ be the $n$th homotopy group with base point $x_0$.
Then $\pi_n \left({T, x_0}\right)$ is abelian.
== Proof ==
Category:Homotopy Theory
Category:Algebraic Topology
Higher Order Derivatives of Laplace Transform

== Theorem ==
Let $f: \R \to \R$ or $\R \to \C$ be a continuous function on any interval of the form $0 \le t \le A$.
Let $\dfrac {\partial f}{\partial s}$, the partial derivative of $f$ with respect to $s$, exist and be continuous on said intervals.
Let $\laptrans f = F$ denote the Laplace transform of $f$.
Then, everywhere that $\laptrans f$ exists and is $n$ times differentiable:
:$\dfrac {\d^n} {\d s^n} \laptrans {\map f t} = \paren {-1}^n \laptrans {t^n \, \map f t}$
== Proof ==
The proof proceeds by induction on $n$, the order of the derivative of $\laptrans f$.
The case $n = 0$ is verified as follows:
=== Basis for the Induction ===
The case $n = 1$ is demonstrated in Derivative of Laplace Transform:
:$\dfrac \d {\d s} \laptrans {\map f t} = -\laptrans {t \, \map f t}$
This is the basis for the induction.
=== Induction Hypothesis ===
Fix $n \in \N$ with $n \ge 0$.
Assume:
:$\dfrac {\d^n} {\d s^n} \laptrans {\map f t} = \paren {-1}^n \laptrans {t^n \, \map f t}$
This is our induction hypothesis.
=== Induction Step ===
This is our induction step:
The result follows by the Principle of Mathematical Induction.
== Examples ==
== Also see ==
* Laplace Transform of Higher Order Derivatives
== Sources ==
* : Chapter $1$: The Laplace Transform: Some Important Properties of Laplace Transforms: $7$. Multiplication by $t^n$: Theorem $1 \text{-} 12$
* : Chapter $1$: The Laplace Transform: Solved Problems: Multiplication by Powers of $t$: $19$
* : $\S 32$: Table of General Properties of Laplace Transforms: $32.12$
Category:Laplace Transforms
Higher Order Derivatives of Laplace Transform/Examples/Example 1

== Examples of Use of Laplace Transform of Integral ==
Let $\laptrans f$ denote the Laplace transform of the real function $f$.
:$\laptrans {t^2 e^{2 t} } = \dfrac 2 {\paren {s - 2}^3}$
== Proof ==
== Sources ==
* : Chapter $1$: The Laplace Transform: Some Important Properties of Laplace Transforms: $7$. Multiplication by $t^n$
Category:Laplace Transforms
Highest Power of 2 Dividing Numerator of Sum of Odd Reciprocals

== Theorem ==
Let:
: $S = \dfrac p q = \ds \sum_{k \mathop = 1}^n \dfrac 1 {2 k - 1}$
where $\dfrac p q$ is the canonical form of $S$.
Let $n = 2^k m$ where $m$ is odd.
Then the largest power of $2$ that divides $p$ is $2^{2 k}$.
== Proof ==
We have that:
:$\ds \sum_{k \mathop = 1}^n \dfrac 1 {2 k - 1} = \sum_{i \mathop = 0}^{M - 1} \paren {\dfrac 1 {i \times 2^{r + 1} + 1} + \dfrac 1 {i \times 2^{r + 1} + 3} + \cdots + \dfrac 1 {\paren {i + 1} \times 2^{r + 1} - 1} }$
where $k = 2^r M$ where $M$ is odd.
Collect the $r$ terms in the parenthesis on the of the $i$th term under a single common denominator $P_i$.
Then the resulting numerators will each consist of $2^r$ terms, each of the form $\dfrac {P_i} {R_i}$, where $R_i$ is a distinct odd residue of $2^{r + 1}$.
These $\dfrac {P_i} {R_i}$ must themselves also be the distinct odd residue of $2^{r + 1}$, in some order.
The odd residues of $2^{r + 1}$ are:
:$1, 3, 5, \dots, 2^{r + 1} - 1$
Their sum is:
Thus each of the $M$ numerators is of the form $2^{2 r} M_i$, where $M_i$ is odd.
Thus the numerator of $S$ is in the form $2^{2 r} Q$ where $Q$ is the sum of an odd number of odd terms.
Therefore $Q$ is itself odd.
Hence, if $2^r$ is the largest power of $2$ which divides $k$, then $2^{2 r}$ is the largest power of $2$ dividing the numerator of $S$.
== Historical Note ==
== Sources ==
* 
* : $\S 1.2.7$: Harmonic Numbers: Exercise $18$
Category:Harmonic Numbers
Highest Power of 2 Dividing Numerator of Sum of Odd Reciprocals/Historical Note

== Historical Note on Highest Power of 2 Dividing Numerator of Sum of Odd Reciprocals ==
This result was posed as an elementary problem by in March $1960$: .
He notes that and leave the problem unsolved in an article in a $1959$ commercial publication, where they suggest that:
:an estimate [of this] power of $2$ ... seems in general to be a difficult number theoretic problem.
In the event, $7$ contributors are reported as having submitted a solution, of which that by was the one published.
Among the solvers was , who included the problem as an exercise of difficulty level $M33$ in his .
== Sources ==
* 
* 
* : $\S 1.2.7$: Harmonic Numbers: Exercise $18$
Category:Historical Notes
Highly Composite Number/Examples/1

== Example of Highly Composite Number ==
$1$ is a highly composite number, being the smallest positive integer with $1$ divisor or more.
== Proof ==
From :
:$\map {\sigma_0} 1 = 1$
The positive integer $1$ has $1$ divisor, that is, $1$ itself.
Vacuously, no smaller positive integer has a greater number of divisors.
Thus, despite not actually being composite, $1$ is a highly composite number.
Category:Highly Composite Numbers
Highly Composite Number/Examples/2

== Example of Highly Composite Number ==
$2$ is a highly composite number, being the smallest positive integer with $2$ divisors or more.
== Proof ==
From :
:$\map {\sigma_0} 2 = 2$
From :
:$\map {\sigma_0} 1 = 1$
That is, the only positive integer smaller than $2$ has a smaller number of divisors.
Thus, despite not actually being composite, $2$ is a highly composite number.
Category:Highly Composite Numbers
Highly Composite Number/Examples/60

== Example of Highly Composite Number ==
$60$ is a highly composite number, being the smallest positive integer with $12$ divisors or more.
== Proof ==
From :
:$\map {\sigma_0} {60} = 12$
For a positive integer $n$ to have a $\sigma_0$ of $12$, the product of $1$ greater than the multiplicities of all the prime factors must be $12$.
The divisors of $12$ are $1, 2, 3, 4, 6, 12$.
The smallest $n$ with one prime factor such that $\map {\sigma_0} n = 12$ is $2^{11} = 2048$.
The smallest $n$ with two prime factors such that $\map {\sigma_0} n = 12$ are:
:$2^3 3^2 = 72$
:$2^5 3^1 = 96$
The smallest $n$ with three prime factors such that $\map {\sigma_0} n = 12$ is:
:$2^2 3^1 5^1 = 60$
== Sources ==
* : $60$
* : $60$
Category:Highly Composite Numbers
Hilbert's Basis Theorem

== Theorem ==
Let $A$ be a Noetherian ring.
Let $A \sqbrk x$ be the ring of polynomial forms over $A$ in the single indeterminate $x$.
Then $A \sqbrk x$ is also a Noetherian ring.
=== Corollary ===
== Proof ==
From the definition, a Noetherian ring is also a commutative ring with unity.
Let $f = a_n x^n + \cdots + a_1 x + a_0 \in A \sqbrk x$ be a polynomial over $x$.
Let $I \subseteq A \sqbrk x$ be an ideal of $A \sqbrk x$.
We will show that $I$ is finitely generated. 
Let $f_1$ be an element of least degree in $I$, and let $\ideal {g_1, \ldots, g_r}$ denote the ideal generated by the polynomials $g_1, \dotsc, g_r$.
For $i \ge 1$, if $\ideal {f_1, \ldots, f_i} \ne I$, then choose $f_{i + 1}$ to be an element of minimal degree in $I \setminus \ideal {f_1, \ldots, f_i}$.
If $\ideal {f_1, \ldots, f_i} = I$ then stop choosing elements.
Let $a_j$ be the leading coefficient of $f_j$.
Since $A$ is Noetherian, the ideal $\ideal {a_1, a_2, \ldots} \subseteq A$ is generated by $a_1, a_2, \ldots, a_m$ for some $m \in \N$.
We claim that $f_1, f_2, \ldots, f_m$ generate $I$.
 not.
Then our process chose an element $f_{m + 1}$, and $\ds a_{m + 1} = \sum_{j \mathop = 1}^m u_j a_j$ for some $u_j \in A$.
Since the degree of $f_{m + 1}$ is greater than or equal to the degree of $f_j$ for $j = 1, \ldots, m$, the polynomial:
:$\ds g = \sum_{j \mathop = 1}^m u_j f_j x^{\deg f_{m + 1} - \deg f_j} \in \ideal {f_1, \ldots, f_m}$
has the same leading coefficient and degree as $f_{m + 1}$.
The difference $f_{m + 1} - g$ is not in $\ideal {f_1, \ldots, f_m}$ and has degree strictly less than $f_{m + 1}$, a contradiction of our choice of $f_{m + 1}$.
Thus $I = \ideal {f_1, \ldots, f_m}$ is finitely generated, and the proof is complete.
Category:Noetherian Rings
Category:Commutative Algebra
Hilbert's Basis Theorem/Corollary

== Theorem ==
Let $A$ be a Noetherian ring.
Let $n \ge 1$ be an integer.
Let $A \sqbrk {x_1, \ldots, x_n}$ be the ring of polynomial forms over $A$ in the indeterminates $x_1, \ldots, x_n$.
Then $A \sqbrk {x_1, \ldots, x_n}$ is also a Noetherian ring.
== Proof ==
We proceed by induction over $n \ge 1$.
=== Basis for the induction ===
Suppose that $n = 1$.
Then $A \sqbrk {x_1, \ldots, x_n} = A \sqbrk {x_1}$ is the ring of polynomial forms in a single indeterminate $x_1$.
Thus by Hilbert's Basis Theorem, $A \sqbrk {x_1}$ is Noetherian.
=== Induction step ===
Suppose that $A \sqbrk {x_1, \ldots, x_{n - 1} }$ is Noetherian.
We wish to show that $A \sqbrk {x_1, \ldots, x_n}$ is Noetherian.
By Adjunction of Indeterminates to Ring is Associative:
:$A \sqbrk {x_1, \ldots, x_n} = A \sqbrk {x_1, \ldots, x_{n - 1} } \sqbrk {x_n}$
is the polynomial ring in a single indeterminate $x_n$ over the Noetherian ring $A \sqbrk {x_1, \ldots, x_{n - 1} }$.
Thus by Hilbert's Basis Theorem, $A \sqbrk {x_1, \ldots, x_n}$ is Noetherian.
Category:Commutative Algebra
Hilbert's Basis Theorem for Finitely Generated Algebras

== Theorem ==
Let $A$ be a Noetherian ring.
Let $B$ be a finitely generated algebra over $A$.
Then $B$ is a Noetherian ring.
== Proof ==
Let $\set {b_1, \ldots , b_n} \subseteq B$ be a generator of $B$.
Let $A \sqbrk {X_1, \ldots, X_n}$ be the ring of polynomial forms over $A$ in $\sequence {X_1, \ldots ,X_n}$.
By Hilbert's basis theorem for polynomial rings, $A \sqbrk {X_1, \ldots, X_n}$ is Noetherian.
Let $\phi : A \sqbrk {X_1, \ldots, X_n} \to B$ be the homomorphism defined by:
:$\map f {X_1, \ldots , X_n} \mapsto \map f {b_1, \ldots , b_n}$
Then, by :
:$\Img \phi = B$
Thus by First Ring Isomorphism Theorem:
:$B \cong A \sqbrk {X_1, \ldots, X_n} / \map \ker \phi$
Now the is Noetherian, since Quotient Ring of Noetherian Ring is Noetherian.
Thus $B$ is Noetherian.
Category:Noetherian Rings
Category:Commutative Algebra
Hilbert's Nullstellensatz

== Theorem ==
Let $k$ be an algebraically closed field.
Let $n \geq 0$ be an natural number.
Let $k \sqbrk {x_1, \ldots, x_n}$ be the polynomial ring in $n$ variables over $k$.
Then for every ideal $J \subseteq k \sqbrk {x_1, \ldots, x_n}$, the associated ideal of its zero-locus equals its radical:
:$\map I {\map Z J} = \map \Rad J$
== Proof ==
Note first that the operations $\map I {\, \cdot \,}$ and $\map Z {\, \cdot \,}$ are inclusion reversing.
That is:
:$X \subseteq Y \subseteq k^n \implies \map I X \supseteq \map I Y$
:$I \subseteq J \implies \map Z I \supseteq \map Z J$
Let $m_a$ be the ideal $\ideal {x_1 - a_1, \ldots, x_n - a_n}$ with $a \in k^n$.
=== Step 1: Maximal Ideals ===
It is demonstrated that $m_a$ are the only maximal ideals.
Let $a \in k^n$.
Define now:
:$\pi_a: k \sqbrk {x_1, \ldots, x_n} \to k: f \mapsto \map f {a_1, \ldots, a_n}$
and note that is an epimorphism of $k$-algebras with kernel:
:$\map I {\set a} = m_a$
Let now $m$ be a maximal ideal of $k \sqbrk {x_1, \ldots, x_n}$.
Now $\dfrac {k \sqbrk {x_1, \ldots, x_n} } m$ is a field extension of $k$, which is finitely generated.
Hence by a corollary of the Noether Normalization Lemma, we find that $\dfrac {k \sqbrk {x_1, \ldots, x_n} } m$ is a finite field extension of $k$.
Since $k$ is algebraically closed, there is an isomorphism of $k$-algebras:
:$\dfrac {k \sqbrk {x_1, \ldots, x_n} } m \to k$
Let $a_i$ denote the image $x_i$. Hence we find that $m_a \subseteq m$, which implies an equality since the first one is a maximal ideal.
=== Step 2: Radical is Intersection of Maximum Ideals ===
It is to be demonstrated that the radical of an ideal $J$ in a finitely generated $k$-algebra $A$ is equal to the intersection of the maximal ideals that contain $J$.
Note that the projection morphism:
:$\pi: A \to \dfrac A J$
induces a bijection $I \mapsto \map {\pi^{-1} } I$ from the sets of radical, prime and maximal ideals of $\dfrac A J$ to the sets radical, prime and maximal ideals of $A$ that contain $J$.
Hence we need to prove this only if $J = \ideal 0$.
It is clear that $\map \Rad {\ideal 0}$ is contained in every maximal ideal.
Hence we need to prove that every element that is not in $\map \Rad {\ideal 0}$ is not contained in some maximal ideal.
Let $f \in A$ such that it is not nilpotent, that is:
:$f \notin \map \Rad {\ideal 0}$
Hence:
:$A_f \cong \dfrac {A \sqbrk x} {\paren {f x - 1} }$
is a non-trivial $k$-algebra, which thus has a maximal ideal $\MM$.
Consider now the morphism
:$\phi : A \to A_f$
which is a morphism of finitely generated $k$-algebras.
Hence by a corollary of the Noether Normalization Lemma, $\map {\phi^{-1} } \MM$ must also be maximal.
This is a maximal ideal $A$ that does not contain $f$.
=== Step 3 ===
Note now that a point $a \in k^n$ belongs to $\map Z J$ $J \subseteq m_a$.
This implies that the maximal ideals containing $J$ are just the maximal ideals $m_a$ with $a \in \map Z J$.
From Step 2:
:$\ds \bigcap_{a \mathop \in \map Z J} m_a = \map \Rad J$
Note now also that:
which implies the required result.
== Linguistic Note ==
Category:Ideal Theory
Category:Algebraic Geometry
Hilbert-Waring Theorem

== Theorem ==
For each $k \in \Z: k \ge 2$, there exists a positive integer $\map g k$ such that every positive integer can be expressed as a sum of at most $\map g k$ positive $k$th powers.
=== Sequence ===
== Particular Cases ==
== Partial Resolution ==
== Proof ==
== Also known as ==
== Historical Note ==
== Sources ==
* 
* 
* 
* 
Category:Hilbert-Waring Theorem
Category:Number Theory
Hilbert-Waring Theorem/Also known as

== Hilbert-Waring Theorem ==
The Hilbert-Waring Theorem is often referred to as Waring's problem, which was how it was named before proved it in $1909$.
However, Waring's problem is properly used for the particular case $3$ and the particular case $4$.
== Sources ==
* 
* 
Category:Hilbert-Waring Theorem
Hilbert-Waring Theorem/Historical Note

== Historical Note on Hilbert-Waring Theorem ==
The Hilbert-Waring Theorem was conjectured for $k = 3$ and $k = 4$ by in $1770$, in his , and was generally referred to as Waring's problem.
It was proved by in $1909$.
The assertion is that for each $k$ there exist such a number $\map g k$.
The problem remains to determine what that $\map g k$ actually is.
=== Partial Resolution ===
== Sources ==
* : $9$
* : $9$
* 
* 
* 
Category:Historical Notes
Hilbert-Waring Theorem/Partial Resolution

== Theorem ==
== Partial Resolution ==
It was determined in $1990$ by and that for every $k \le 471 \, 600 \, 000$, the value of $\map g k$ is given by the formula:
:$\map g k = \floor {\paren {\dfrac 3 2}^k} + 2^k - 2$
It is suspected that it is true for all $k$, but this still remains to be proved.
== Proof ==
== Examples ==
== Sources ==
* 
Category:Hilbert-Waring Theorem
Hilbert-Waring Theorem/Partial Resolution/Examples/5

== Example of Partial Resolution of Hilbert-Waring Theorem ==
The Hilbert-Waring Theorem states that:
== Proof ==
From the partial resolution of the Hilbert-Waring theorem:
For $k = 5$, we have:
which is the result established by in $1964$.
== Sources ==
* 
Category:Hilbert-Waring Theorem
Hilbert-Waring Theorem/Particular Cases

== Particular Cases of the Hilbert-Waring Theorem ==
The Hilbert-Waring Theorem states that:
=== Hilbert-Waring Theorem: $k = 2$ ===
=== Hilbert-Waring Theorem: $k = 3$ ===
=== Hilbert-Waring Theorem: $k = 4$ ===
=== Hilbert-Waring Theorem: $k = 5$ ===
=== Hilbert-Waring Theorem: $k = 6$ ===
=== Hilbert-Waring Theorem: $k = 7$ ===
Category:Hilbert-Waring Theorem
Hilbert-Waring Theorem/Particular Cases/2

== Particular Case of the Hilbert-Waring Theorem: $k = 2$ ==
The Hilbert-Waring Theorem states that:
The case where $k = 2$ is proved by Lagrange's Four Square Theorem‎:
:$\map g 2 = 4$
That is, every positive integer can be expressed as the sum of at most $4$ squares.
== Historical Note ==
== Sources ==
* 
* 
Category:Hilbert-Waring Theorem
Hilbert-Waring Theorem/Particular Cases/2/Mistake

== Source Work ==
:
:Waring's problem
== Mistake ==
:Thus, if $n = 2$ then $N = 4$; i.e. any positive integer can be expressed by a sum of up to four squared terms -- a result proved by .
== Correction ==
While it is true that did indeed publish such a proof (in $1780$, having written it in $1772$), the fact is that got there first with his own proof in $1770$.
These errors had both been corrected by the time of the publication of the $4$th edition.
== Sources ==
* 
Category:Mistakes/The Penguin Dictionary of Mathematics
Hilbert-Waring Theorem/Particular Cases/3

== Particular Case of the Hilbert-Waring Theorem: $k = 3$ ==
The Hilbert-Waring Theorem states that:
The case where $k = 3$ is:
Every positive integer can be expressed as the sum of at most $9$ positive cubes.
That is:
:$\map g 3 = 9$
== Proof ==
== Historical Note ==
== Sources ==
* 
* 
* 
* 
* 
Category:Hilbert-Waring Theorem
Category:9
Hilbert-Waring Theorem/Particular Cases/3/Historical Note

== Particular Case of the Hilbert-Waring Theorem: $k = 3$: Historical Note ==
 knew that some integers required at least $9$ positive cubes to represent them as a sum:
The fact that $\map g 3 = 9$ was mostly established by $1909$.
The remaining gap in the argument was resolved in $1912$ by .
== Sources ==
* : $9$
* : $9$
* 
Category:Historical Notes
Hilbert-Waring Theorem/Particular Cases/4

== Particular Case of the Hilbert-Waring Theorem: $k = 4$ ==
The Hilbert-Waring Theorem states that:
The case where $k = 4$ is:
Every positive integer can be expressed as the sum of at most $19$ powers of $4$.
That is:
:$\map g 4 = 19$
== Proof ==
== Also see ==
* Positive Integers not Expressible as Sum of Fewer than 19 Fourth Powers
== Historical Note ==
== Sources ==
* : $19$
* : $19$
* 
* 
* 
Category:Hilbert-Waring Theorem
Hilbert-Waring Theorem/Particular Cases/4/Historical Note

== Particular Case of the Hilbert-Waring Theorem: $k = 4$: Historical Note ==
It is clear that some integers require at least $19$ powers of $4$ to represent them as a sum, as $79$ requires $19$ of them:
Thus knew that:
:$\map g 4 \ge 19$
 subsequently showed that:
:$\map g 4 \le 53$
 and showed that all sufficiently large numbers are the sum of at most $19$ fourth powers.
Finally it was shown in $1986$ by , and that $\map g 4 \le 19$.
Thus:
:$\map g 4 = 19$
and so 's claim has been shown to be true.
== Sources ==
* : $9$
* : $9$
* 
Category:Historical Notes
Hilbert-Waring Theorem/Particular Cases/4/Mistake

== Source Work ==
:
:The Dictionary
::$53$
== Mistake ==
:Every positive integer is the sum of at most $53$ $4$th powers.
The number $53$ is the number as determined by .
Since then, it has been ascertained that the number is in fact $19$.
== Sources ==
* : $53$
Category:Mistakes/Curious and Interesting Numbers
Hilbert-Waring Theorem/Particular Cases/5

== Particular Case of the Hilbert-Waring Theorem: $k = 5$ ==
The Hilbert-Waring Theorem states that:
The case where $k = 5$ is:
Every positive integer can be expressed as the sum of at most $37$ positive fifth powers.
That is:
:$\map g 5 = 37$
== Proof ==
== Also see ==
* Largest Number not Expressible as Sum of Less than 37 Positive Fifth Powers
* Largest Number not Expressible as Sum of Less than 32 Positive Fifth Powers
== Historical Note ==
== Sources ==
* : $37$
* : $37$
* 
Category:Hilbert-Waring Theorem
Category:37
Category:Fifth Powers
Hilbert-Waring Theorem/Particular Cases/5/Mistake

== Source Work ==
:
:The Dictionary
::$37$
:
:The Dictionary
::$37$
== Mistake ==
:Every number is the sum of at most $37$ $5$th powers.
It needs to be stated that they should all be positive $5$th powers.
== Sources ==
* : $37$
* : $37$
Category:Mistakes/Curious and Interesting Numbers
Hilbert-Waring Theorem/Particular Cases/6

== Particular Case of the Hilbert-Waring Theorem: $k = 6$ ==
The Hilbert-Waring Theorem states that:
The case where $k = 6$ is:
Every positive integer can be expressed as the sum of at most $73$ positive sixth powers.
That is:
:$\map g 6 = 73$
== Proof ==
== Sources ==
* : $73$
* : $73$
Category:Hilbert-Waring Theorem
Category:73
Category:Sixth Powers
Hilbert-Waring Theorem/Particular Cases/7

== Particular Case of the Hilbert-Waring Theorem: $k = 7$ ==
The Hilbert-Waring Theorem states that:
The case where $k = 7$ is:
Every positive integer can be expressed as the sum of at most $143$ positive seventh powers.
That is:
:$\map g 7 = 143$
== Proof ==
Category:Hilbert-Waring Theorem
Category:143
Category:Seventh Powers
Hilbert-Waring Theorem/Particular Cases/Mistake

== Source Work ==
:
:Waring's problem
== Mistake ==
:The first part, on cubes, was proved by in $1909$. The second part is still unsolved.
== Correction ==
The first part, that is:
:Every positive integer can be expressed as the sum of at most $9$ positive cubes
was in fact resolved by and between $1909$ and $1921$.
The second part, that is:
:Every positive integer can be expressed as the sum of at most $19$ powers of $4$
had been resolved in $1986$ by , and , some $12$ years before the publication of this edition of this dictionary.
What showed in $1909$ was that for every $n$ there exists a $\map g n$ such that every positive integer can be expressed by no more than $\map g n$ positive integers raised to the power of $n$.
These errors had both been corrected by the time of the publication of the $4$th edition.
== Sources ==
* 
Category:Mistakes/The Penguin Dictionary of Mathematics
Hilbert-Waring Theorem/Sequence

== Sequence ==
The integer sequence of values of $\map g k$ begins:
:$1, 4, 9, 19, 37, 73, 143, 279, 548, 1079, 2132, 4223, 8384, 16673, 33203, 66190, 132055, \ldots$
== Historical Note ==
Category:Hilbert-Waring Theorem
Hilbert-Waring Theorem/Variant Form

== Theorem ==
For each $k \in \Z: k \ge 2$, there exists a positive integer $G \left({k}\right)$ such that every sufficiently large positive integer can be expressed as a sum of at most $G \left({k}\right)$ $k$th powers.
== Particular Cases ==
== Proof ==
== Also known as ==
== Historical Note ==
Category:Number Theory
Category:Hilbert-Waring Theorem
Hilbert-Waring Theorem/Variant Form/Historical Note

== Historical Note on Waring's Problem ==
The Hilbert-Waring Theorem was conjectured by in $1770$ in , and was generally referred to as Waring's problem.
It was proved by in $1909$.
Its variant form was investigated by and .
The assertion is that for each $k$ there exist such a number $G \left({k}\right)$.
The problem remains to determine what that $G \left({k}\right)$ actually is.
Category:Historical Notes
Hilbert-Waring Theorem/Variant Form/Particular Cases

== Particular Cases of the Hilbert-Waring Theorem -- Variant Form ==
The Hilbert-Waring Theorem -- Variant Form states that:
=== Hilbert-Waring Theorem -- Variant Form: $k = 2$ ===
=== Hilbert-Waring Theorem -- Variant Form: $k = 3$ ===
=== Hilbert-Waring Theorem -- Variant Form: $k = 4$ ===
=== Hilbert-Waring Theorem -- Variant Form: $k = 7$ ===
Category:Hilbert-Waring Theorem