avewright/proofwiki-math huggingface dataset:
Theorem: Product of Sums

Let $\ds \sum_{n \mathop \in A} a_n$ and $\ds \sum_{n \mathop \in B} b_n$ be absolutely convergent sequences.
Then:
:$\ds \paren {\sum_{i \mathop \in A} a_i} \paren {\sum_{j \mathop \in B} b_j} = \sum_{\tuple {i, j} \mathop \in A \times B} a_i b_j$

Proof:
We have that both series are absolutely convergent.
Thus by Manipulation of Absolutely Convergent Series, it is permitted to expand the product as:
:$\ds \paren {\sum_{i \mathop \in A} a_i} \paren {\sum_{j \mathop \in B} b_j} = \sum_{i \mathop \in A} \paren {a_i \sum_{j \mathop \in B} b_j}$
But since $a_i$ is a constant, it may be brought into the summation, to obtain:
:$\ds \sum_{i \mathop \in A} \sum_{j \mathop \in B} a_i b_j$
Hence the result.
{{qed}}
Category:Number Theory
Category:Analysis
87kvg12tgx2bj9hugrreicuyu5cm3ri
Theorem: Product of Sums

Let $\ds \sum_{n \mathop \in A} a_n$ and $\ds \sum_{n \mathop \in B} b_n$ be absolutely convergent sequences.
Then:
:$\ds \paren {\sum_{i \mathop \in A} a_i} \paren {\sum_{j \mathop \in B} b_j} = \sum_{\tuple {i, j} \mathop \in A \times B} a_i b_j$

Proof:
Taking each of the squares on the {{RHS}} and multiplying them out in turn:
{{begin-eqn}}
{{eqn | l = \paren {a w + b x + c y + d z}^2
      | r = a^2 w^2 + b^2 x^2 + c^2 y^2 + d^2 z^2
      | c = 
}}
{{eqn | o = 
      | ro= +
      | r = 2 \paren {a b w x + a c w y + a d w z + b c x y + b d x z + c d y z}
      | c = 
}}
{{end-eqn}}
{{begin-eqn}}
{{eqn | l = \paren {a x - b w + c z - d y}^2
      | r = a^2 x^2 + b^2 w^2 + c^2 z^2 + d^2 y^2
      | c = 
}}
{{eqn | o = 
      | ro= +
      | r = 2 \paren {-a b w x + a c x z - a d x y - b c w z + b d w y - c d y z}
      | c = 
}}
{{end-eqn}}
{{begin-eqn}}
{{eqn | l = \paren {a y - b z - c w + d x}^2
      | r = a^2 y^2 + b^2 z^2 + c^2 w^2 + d^2 x^2
      | c = 
}}
{{eqn | o = 
      | ro= +
      | r = 2 \paren {-a b y z - a c w y + a d x y + b c w z - b d x z - c d w x}
      | c = 
}}
{{end-eqn}}
{{begin-eqn}}
{{eqn | l = \paren {a z + b y - c x - d w}^2
      | r = a^2 z^2 + b^2 y^2 + c^2 x^2 + d^2 w^2
      | c = 
}}
{{eqn | o = 
      | ro= +
      | r = 2 \paren {a b y z - a c x z - a d w z - b c x y - b d w y + c d w x}
      | c = 
}}
{{end-eqn}}
All the non-square terms cancel out with each other, leaving:
{{begin-eqn}}
{{eqn | o = 
      | r = \paren {a w + b x + c y + d z}^2
      | c = 
}}
{{eqn | o = 
      | ro= +
      | r = \paren {a x - b w + c z - d y}^2
      | c = 
}}
{{eqn | o = 
      | ro= +
      | r = \paren {a y - b z - c w + d x}^2
      | c = 
}}
{{eqn | o = 
      | ro= +
      | r = \paren {a z + b y - c x - d w}^2
      | c = 
}}
{{eqn | r = a^2 w^2 + b^2 x^2 + c^2 y^2 + d^2 z^2
      | c = 
}}
{{eqn | o = 
      | ro= +
      | r = a^2 x^2 + b^2 w^2 + c^2 z^2 + d^2 y^2
      | c = 
}}
{{eqn | o = 
      | ro= +
      | r = a^2 y^2 + b^2 z^2 + c^2 w^2 + d^2 x^2
      | c = 
}}
{{eqn | o = 
      | ro= +
      | r = a^2 z^2 + b^2 y^2 + c^2 x^2 + d^2 w^2
      | c = 
}}
{{eqn | r = \paren {a^2 + b^2 + c^2 + d^2} \paren {w^2 + x^2 + y^2 + z^2}
      | c = 
}}
{{end-eqn}}
{{qed}}
Theorem: Product of Sums

Let $\ds \sum_{n \mathop \in A} a_n$ and $\ds \sum_{n \mathop \in B} b_n$ be absolutely convergent sequences.
Then:
:$\ds \paren {\sum_{i \mathop \in A} a_i} \paren {\sum_{j \mathop \in B} b_j} = \sum_{\tuple {i, j} \mathop \in A \times B} a_i b_j$

Proof:
Let:
:$\mathbf m = a \mathbf 1 + b \mathbf i + c \mathbf j + d \mathbf k$
:$\mathbf n = -w \mathbf 1 + x \mathbf i + y \mathbf j + z \mathbf k$
be two quaternions.
Then:
{{begin-eqn}}
{{eqn | l = \size {\mathbf m} \size {\mathbf n}
      | r = \size {\mathbf m \mathbf n}
      | c = Quaternion Modulus of Product of Quaternions
}}
{{eqn | ll= \leadsto
      | l = \paren {a^2 + b^2 + c^2 + d^2} \paren {\paren {-w}^2 + x^2 + y^2 + z^2}
      | r = \size {\mathbf m \mathbf n}
      | c = {{Defof|Quaternion Modulus}}
}}
{{eqn | r = \paren {-a w - b x - c y - d z}^2
      | c = {{Defof|Quaternion Multiplication}}
}}
{{eqn | o = 
      | ro= +
      | r = \paren {a x - b w + c z - d y}^2
      | c = 
}}
{{eqn | o = 
      | ro= +
      | r = \paren {a y - b z - c w + d x}^2
      | c = 
}}
{{eqn | o = 
      | ro= +
      | r = \paren {a z + b y - c x - d w}^2
      | c = 
}}
{{eqn | ll= \leadsto
      | l = \paren {a^2 + b^2 + c^2 + d^2} \paren {w^2 + x^2 + y^2 + z^2}
      | r = \paren {a w + b x + c y + d z}^2
      | c = simplifying
}}
{{eqn | o = 
      | ro= +
      | r = \paren {a x - b w + c z - d y}^2
      | c = 
}}
{{eqn | o = 
      | ro= +
      | r = \paren {a y - b z - c w + d x}^2
      | c = 
}}
{{eqn | o = 
      | ro= +
      | r = \paren {a z + b y - c x - d w}^2
      | c = 
}}
{{end-eqn}}
{{qed}}
Theorem: Green's Theorem

Let $\Gamma$ be a positively oriented piecewise smooth Jordan curve in $\R^2$.
Let $U = \Int \Gamma$, that is, the interior of $\Gamma$.
Let $A$ and $B$ be functions of $\tuple {x, y}$ defined on an open region containing $U$ and have continuous partial derivatives in such a set.
Then:
:$\ds \oint_\Gamma \paren {A \rd x + B \rd y} = \iint_U \paren {\frac {\partial B} {\partial x} - \frac {\partial A} {\partial y} } \rd x \rd y$
where the {{LHS}} is a contour integral.

Proof:
It suffices to demonstrate the theorem for rectangular regions in the $x y$-plane.
The Riemann-sum nature of the double integral will then guarantee the proof of the theorem for arbitrary regions, because a Riemann-sum is technically a summation of the areas of arbitrarily small rectangles. 
As the proof is for a rectangle, the proof will work for arbitrary regions, which can be approximated by collections of ever smaller rectangles.
{{handwaving|To do this formally involves the Monotone Convergence Theorem (Measure Theory) for $U$ and approximating $\Gamma$ by means of a "piecewise rectification". Its arduous, yet necessary.}}
Let $R = \set {\tuple {x, y}: a \le x \le b, c \le y \le d}$ be a rectangular region.
Let the boundary $C$ of $R$ be oriented counterclockwise.
We break the boundary into $4$ pieces:
:$C_1$, which runs from $\tuple {a, c}$ to $\tuple {b, c}$
:$C_2$, which runs from $\tuple {b, c}$ to $\tuple {b, d}$
:$C_3$, which runs from $\tuple {b, d}$ to $\tuple {a, d}$
:$C_4$, which runs from $\tuple {a, d}$ to $\tuple {a, c}$
Then:
{{begin-eqn}}
{{eqn | l = \iint_R \frac {\partial B} {\partial x} \rd x \rd y
      | r = \int_c^d \int_a^b \frac {\partial B} {\partial x} \rd x \rd y
      | c = 
}}
{{eqn | r = \int_c^d \paren {\map B {b, y} - \map B {a, y} } \rd y
      | c = 
}}
{{eqn | r = \int_c^d  \map B {b, y} \rd y + \int_d^c \map B {a, y} \rd y
      | c = 
}}
{{eqn | r = \int_{C_2} B \rd y + \int_{C_4} B \rd y
      | c = 
}}
{{end-eqn}}
We note that $y$ is constant along $C_1$ and $C_3$.
So:
:$\ds \int_{C_1} B \rd y = \int_{C_3} B \rd y = 0$
Hence:
{{begin-eqn}}
{{eqn | l = \int_{C_2} B \rd y + \int_{C_4} B \rd y
      | r = \int_{C_1} B \rd y + \int_{C_2} B \rd y + \int_{C_3} B \rd y + \int_{C_4} B \rd y
      | c = 
}}
{{eqn | r = \oint_C B \rd y
      | c = 
}}
{{end-eqn}}
A similar argument demonstrates that:
:$\ds \iint_R \frac {\partial A} {\partial y} \rd x \rd y = -\oint_C A \d x$
and hence:
:$\ds \oint_C \paren {A \rd x + B \rd y} = \iint_R \paren {\frac {\partial B} {\partial x} - \frac {\partial A} {\partial y} } \rd x \rd y$
{{qed}}
Theorem: Gauss-Ostrogradsky Theorem

Let $U$ be a subset of $\R^3$ which is compact and has a piecewise smooth boundary $\partial U$.
Let $\mathbf V: \R^3 \to \R^3$ be a smooth vector field defined on a neighborhood of $U$.
Then:
:$\ds \iiint \limits_U \paren {\nabla \cdot \mathbf V} \rd v = \iint \limits_{\partial U} \mathbf V \cdot \mathbf n \rd S$
where $\mathbf n$ is the unit normal to $\partial U$, directed outwards.
That is, the sum of all expansions of $\mathbf V$ within $U$ equals the flux of $\mathbf V$ across the boundary of $U$.

Proof:
Let $S$ be the surface of $U$.
By definition, the surface integral of $\mathbf V$ over $S$ is therefore defined as:
:$\ds \iint_S \mathbf V \cdot \mathbf n \rd S = \iint_S \mathbf V \cdot \rd \mathbf S$
where $\rd \mathbf S$ is the differential of vector area.
:700px
Let $\d v$ be an element of volume within $U$.
In the above diagram, a small cuboid has been used for convenience.
The total flux emerging from $\d v$ is:
:$\operatorname {div} \mathbf V \rd v$
where in this context $\mathbf V$ is the vector quantity at the center of $\d v$.
For the face $abcd$, the positive direction of the $x$ component of $\mathbf V$ and the outward normal are in the same sense, and so the flux is positive.
Consider another small cuboid contiguous to $\d v$, shown dotted in the above diagram.
This has the same component of $\mathbf V$, but this is acting inward, and so is negative.
Hence the flux through the face of one cuboid cancels out the flux through the adjacent face of the next cuboid.
This argument can be applied by imagining a sequence of such cuboids until you reach the surface of $U$.
This is the only part of the construction we have just built which contributes to the flux through $S$.
By applying the same treatment to the surfaces of all such small elements of volume, we arrive at a total flux:
:$\ds \iint_S \mathbf V \cdot \rd \mathbf S$
But at the same time we have integrated $\operatorname {div} \mathbf V \rd v$ throughout the whole of the volume of $U$.
This also measures the total flux through $S$.
Hence:
{{begin-eqn}}
{{eqn | l = \iint_S \mathbf V \cdot \rd \mathbf S
      | r = \iint_S \mathbf V \cdot \mathbf n \rd S
      | c = 
}}
{{eqn | r = \iiint \operatorname {div} \mathbf V \rd v
      | c = 
}}
{{eqn | r = \iiint \nabla \cdot \mathbf V \rd v
      | c = 
}}
{{end-eqn}}
Theorem: Implications of Stokes' Theorem

Stokes' Theorem implies all of the following results:
* Classical Stokes' Theorem
* Green's Theorem
* Gauss's Theorem
* Fundamental Theorem of Calculus
* Cauchy's Residue Theorem

Proof:
=== Classical Stokes' Theorem ===
We note that given $f_1, f_2, f_3: \R^3 \to \R$, $f_1 \rd x + f_2 \rd y + f_3 \rd z$ is a $1$-form defined on a surface.
Since we are integrating over the boundary of a surface, which is a $1$-manifold, we may use the general Stokes' theorem:
{{begin-eqn}}
{{eqn | l = \oint_{\partial S} f_1 \rd x + f_2 \rd y + f_3 \rd z
      | r = \iint_S \map \rd {f_1 \rd x + f_2 \rd y + f_3 \rd z}
      | c = 
}}
{{eqn | r = \iint_S \frac {\partial f_1} {\partial x} \rd x \wedge \d x + \frac {\partial f_1} {\partial y} \rd y \wedge \d x + \frac {\partial f_1} {\partial z} \rd z \wedge \d x
      | c = 
}}
{{eqn | o =
      | ro= +
      | r = \frac {\partial f_2} {\partial x} \rd x \wedge \d y + \frac {\partial f_2} {\partial y} \rd y \wedge \d y + \frac {\partial f_2} {\partial z} \rd z \wedge \d y
      | c = 
}}
{{eqn | o =
      | ro= +
      | r = \frac {\partial f_3} {\partial x} \rd x \wedge \d z + \frac {\partial f_3} {\partial y} \rd z \wedge \d y + \frac {\partial f_3} {\partial z} \rd z \wedge \d z
      | c = 
}}
{{eqn | r = \iint \paren {\frac {\partial f_2} {\partial x} - \frac {\partial f_1} {\partial y} } \rd x \wedge \d y + \paren {\frac {\partial f_3} {\partial y} - \frac {\partial f_2} {\partial z} } \rd y \wedge \d z + \paren {\frac {\partial f_1} {\partial z} - \frac{\partial f_3}{\partial x} } \rd z \wedge \d x
      | c = 
}}
{{end-eqn}}
If we define $\mathbf F: \R^3 \to \R^3$ as:
:$\mathbf F = f_1 \mathbf e_1 + f_2 \mathbf e_2 + \mathbf e_3$
we recognize this expression as:
:$\nabla \times \mathbf F$
with:
:$\d y \wedge \d z$ replacing $\mathbf e_1$
:$\d z \wedge \d x$ replacing $\mathbf e_2$
and:
:$\d x \wedge \d y$ replacing $\mathbf e_3$.
{{stub|It must be shown that the wedges give area on the surface.}}
Theorem: Divergence Test

Let $\sequence {a_n}$ be a sequence in $\R$ which does not converge to $0$.
Then $\ds \sum_{i \mathop = 1}^\infty a_n$ diverges.

Proof:
We know that Terms in Convergent Series Converge to Zero.
This is the contrapositive statement of this theorem.
Thus, the theorem holds by Rule of Transposition.
{{qed}}
Theorem: Eisenstein's Lemma

Let $p$ be an odd prime.
Let $a \in \Z$ be an odd integer such that $p \nmid a$.
Let $\paren {\dfrac a p}$ be defined as the Legendre symbol.
Then:
:$\paren {\dfrac a p} = \paren {-1}^{\map \alpha {a, p} }$
where:
:$\ds \map \alpha {a, p} = \sum_{k \mathop = 1}^{\frac {p - 1} 2} \floor {\frac {k a} p}$
:$\floor {\dfrac {k a} p}$ is the floor function of $\dfrac {k a} p$.

Proof:
We will borrow some ideas and techniques from the proof of Gauss's Lemma.
Let $S = \set {a, 2 a, 3 a, \ldots, \dfrac {p - 1} 2 a}$.
Let us create $S'$ from $S$ by replacing each element of $S$ by its least positive residue modulo $p$.
Arranging $S'$ into increasing order, we get:
:$S' = \set {b_1, b_2, \ldots, b_m, c_1, c_2, \ldots, c_n}$
where $b_m < \dfrac p 2 < c_1$ and $m + n = \dfrac {p - 1} 2$.
According to the Division Theorem, we can divide any $k a \in S$ by $p$ and get:
:$k a = p \times \floor {\dfrac {k a} p} + r$
where $r \in S'$.
Let $k$ run from $1$ to $\dfrac {p - 1} 2$ and add up the resulting equations.
We get:
:$(1): \quad \ds \sum_{k \mathop = 1}^{\frac {p - 1} 2} k a = \sum_{k \mathop = 1}^{\frac {p - 1} 2} p \times \floor {\frac {k a} p} + \sum_{i \mathop = 1}^{m} b_i + \sum_{j \mathop = 1}^n c_j$
As in the proof of Gauss's Lemma, all the remainders are in fact distinct elements of $S'$.
From the proof of Gauss's Lemma again, we have that:
:$S'' = \set {b_1, b_2, \ldots, b_m, p - c_1, p - c_2, \ldots, p - c_n} = \set {1, 2, 3, \ldots, \dfrac {p - 1} 2}$
Adding up all the elements of $S''$ we have:
:$\ds \sum_{k \mathop = 1}^{\frac {p - 1} 2} k = \sum_{i \mathop = 1}^m b_i + \sum_{j \mathop = 1}^n \paren {p - c_j} = \sum_{i \mathop = 1}^m b_i + p n - \sum_{j \mathop = 1}^n c_j$
Subtracting this equation from $(1)$ above gives us:
:$\ds \sum_{k \mathop = 1}^{\frac {p - 1} 2} k a - \sum_{k \mathop = 1}^{\frac {p - 1} 2} k = \sum_{k \mathop = 1}^{\frac {p - 1} 2} p \times \floor {\frac {k a} p} + 2 \sum_{j \mathop = 1}^n c_j - p n$
That is:
:$\ds \paren {a - 1} \sum_{k \mathop = 1}^{\frac {p - 1} 2} k = p \times \map \alpha {a, p} + 2 \sum_{j \mathop = 1}^n c_j - p n$
Now we are going to write this equation modulo $2$.
Note that:
:$\ds \paren {a - 1} \sum_{k \mathop = 1}^{\frac {p - 1} 2} k$ is even, as $a$ is odd
:$\ds 2 \sum_{j \mathop = 1}^n c_j$ is also of course even.
So:
:$p \map \alpha {a, p} - p n \equiv 0 \pmod 2$
Since $p$ is odd, this amounts to saying:
:$\map \alpha {a, p} \equiv n \pmod 2$
From the conclusion of Gauss's Lemma:
:$\paren {\dfrac a p} = \paren {-1}^n$
Hence the result.
{{qed}}
{{proofread}}
{{Namedfor|Ferdinand Gotthold Max Eisenstein|cat = Eisenstein}}
Category:Number Theory
8c9mav8lqs4t403yabgo5qzs6cuf7el
Theorem: Law of Quadratic Reciprocity

Let $p$ and $q$ be distinct odd primes.
Then:
:$\paren {\dfrac p q} \paren {\dfrac q p} = \paren {-1}^{\dfrac {\paren {p - 1} \paren {q - 1} } 4}$
where $\paren {\dfrac p q}$ and $\paren {\dfrac q p}$ are defined as the Legendre symbol.
An alternative formulation is:
$\paren {\dfrac p q} = \begin{cases}
\quad \paren {\dfrac q p} & : p \equiv 1 \lor q \equiv 1 \pmod 4 \\
-\paren {\dfrac q p} & : p \equiv q \equiv 3 \pmod 4
\end{cases}$
The fact that these formulations are equivalent is immediate.
This fact is known as the '''Law of Quadratic Reciprocity''', or '''LQR''' for short.

Proof:
Let $p$ and $q$ be distinct odd primes.
Consider the rectangle in the Cartesian $x y$ plane with vertices at $\tuple {0, 0}, \tuple {\dfrac p 2, 0}, \tuple {\dfrac p 2, \dfrac q 2}, \tuple {0, \dfrac q 2}$.
The number of lattice points inside this rectangle is $\dfrac {p - 1} 2 \times \dfrac {q - 1} 2$.
:700px
Consider the diagonal from $\tuple {0, 0}$ to $\tuple {\dfrac p 2, \dfrac q 2}$.
This has the equation:
:$y = \dfrac q p x$
{{AimForCont}} there were a lattice point $\tuple {a, b}$ on the diagonal.
Then:
:$b = \dfrac q p a$
But as $p$ and $q$ are distinct primes, this would mean:
:$p$ divides $a$
and:
:$q$ divides $b$
This means that $\tuple {a, b}$ has to be outside the rectangle.
So there are no lattice points on the diagonal inside the rectangle.
Let $A$ and $B$ be the triangular regions inside the rectangle lying respectively above and below the diagonal.
Let $k \in \Z: 0 < k < \dfrac p 2$.
The number of lattice points in $B$ which lie directly above the point $\tuple {k, 0}$ is:
:$\floor {\dfrac q p k}$
where $\floor {\dfrac q p k}$ is the floor of $\dfrac q p k$.
So the total number of lattice points in $B$ is given by:
:$\ds N_B = \sum_{k \mathop = 1}^{\frac{p - 1} 2} \floor {\dfrac q p k}$
Let $\map \alpha {q, p}$ be defined as:
:$\ds \map \alpha {q, p} = \sum_{k \mathop = 1}^{\frac {p - 1} 2} \floor {\dfrac q p k}$
In the same way, by counting the lattice points to the right of $\tuple {0, k}$, the total number of lattice points in $A$ is given by:
:$N_A = \map \alpha {p, q}$
Now the total number of lattice points in the rectangle is $N_A + N_B$.
But this is also equal to:
:$\dfrac {p - 1} 2 \times \dfrac {q - 1} 2 = \dfrac {\paren {p - 1} \paren {q - 1} } 4$
So we have that:
:$\map \alpha {p, q} + \map \alpha {q, p} = \dfrac {\paren {p - 1} \paren {q - 1} } 4$
The fact that these counts are equal relies upon the fact that there are no lattice points on the diagonal, as demonstrated above.
Now we invoke Eisenstein's Lemma:
:$\paren {\dfrac a p} = \paren {-1}^{\map \alpha {a, p} }$
So this gives us:
{{begin-eqn}}     
{{eqn | l = \paren {\frac p q} \paren {\frac q p}
      | r = \paren {-1}^{\map \alpha {p, q} } \times \paren {-1}^{\map \alpha {q, p} }
      | c = 
}}
{{eqn | r = \paren {-1}^{\map \alpha {p, q} + \map \alpha {q, p} }
      | c = 
}}
{{eqn | r = \paren {-1}^{\frac {\paren {p - 1} \paren {q - 1} } 4}
      | c = 
}}
{{end-eqn}}
which is LQR.
{{qed}}
Theorem: Manipulation of Exterior Derivative

For the exterior derivative, the following statements are true:
:$\map \d {a b} = a \rd b + \paren {\d a} b$
:$\map \d {a \wedge b} = \d a \wedge b - a \wedge \d b$
where $a \wedge b$ is the wedge product.

Proof:
{{ProofWanted}}
Category:Topology
55sgp4p9ccfvrwj3rtqotazriywkwoc
Theorem: General Stokes' Theorem

Let $\omega$ be a smooth $\paren {n - 1}$-form with compact support on a smooth $n$-dimensional oriented manifold $X$.
Let the boundary of $X$ be $\partial X$.
Then:
:$\ds \int_{\partial X} \omega = \int_X \rd \omega$
where $\d \omega$ is the exterior derivative of $\omega$.

Proof:
=== Special Case ===
Let there exist a chart:
:$x = \tuple {x_1, \ldots, x_n}: V \subseteq X \to \R^n$
such that:
:$\map \supp \omega \subseteq V$
where:
:$\map \supp \omega = \overline {\set {p \in M : \map \omega p \ne 0} }$
{{explain|Yes very good but what *is* it?}}
We may suppose that $V$ is relatively compact.
Thus, by composing $x$ with a translation, we may suppose that:
:$\ds \map x V \subseteq \mathbb H^n = \set {\tuple {x_1, \ldots, x_n} \in \R^n : x_1 < 0}$
We have, in the coordinates $x$:
:$\ds \omega = \sum_{i \mathop = 1}^n f_i \rd x_1 \wedge \cdots \wedge \hat {\d x}_i \wedge \cdots \wedge \d x_n$
The forms:
:$\hat {\d x}_i := \d x_1 \wedge \cdots \wedge \hat {\d x}_i \wedge \cdots \wedge \d x_n$ 
vanish on the tangent space to $\mathbb H^n$ for $i > 1$
Hence we have:
:$(1): \quad \ds \int_{\partial \mathbb H^n} \omega = \int_{\partial \mathbb H^n} f_1 \hat {\d x}_1$
{{explain|Clarification needed of what $\hat {\d x}_i$ means.  The inclusion of $\hat {\d x}_i$ within that long $\wedge$ expression looks wrong to me.}}
{{explain|What does $\wedge$ mean in this context?}}
Moreover:
{{begin-eqn}}
{{eqn | l = \d \omega
      | r = \sum_{i \mathop = 1}^n \rd f_i \wedge \hat {\d x}_i
      | c = 
}}
{{eqn | r = \sum_{i \mathop = 1}^n \frac {\partial f} {\partial x_i} \d x_i \wedge \hat {\d x}_i
      | c =
}}
{{eqn | r = \paren {\sum_{i \mathop = 1}^n \frac {\partial f} {\partial x_i} } \rd x_1 \wedge \cdots \wedge \d x_n
      | c = 
}}
{{end-eqn}}
so that:
:$\ds \int_{\mathbb H^n} \rd \omega = \sum_{i \mathop = 1}^n \int_{\mathbb H^n} \frac {\partial f_i} {\partial x_i} \rd x_1 \wedge \cdots \wedge \d x_n$
Let $i > 1$.
Then:
{{begin-eqn}}
{{eqn | l = \int_{\mathbb H^n} \frac {\partial f_i} {\partial x_i} \rd x_1 \wedge \cdots \wedge \d x_n
      | r = \int \cdots \int \paren {\int_{-\infty}^\infty \frac {\partial f_i} {\partial x_i} } \hat {\d x}_i
      | c = Fubini's Theorem
}}
{{eqn | r = 0
      | c = 
}}
{{end-eqn}}
Let $i = 1$.
Then:
{{begin-eqn}}
{{eqn | l = \int_{\mathbb H^n} \frac {\partial f_1} {\partial x_1} \d x_1 \wedge \cdots \wedge \d x_n
      | r = \int \cdots \int \paren {\int_{-\infty}^0 \frac {\partial f_i} {\partial x_i} } \hat {\d x}_i
      | c = Fubini's Theorem
}}
{{eqn | r = \int_{\partial \mathbb H^n} f_1 \hat {\d x}_1
      | c = 
}}
{{end-eqn}}
So:
:$\ds \int_{\mathbb H^n} \rd \omega = \int_{\partial \mathbb H^n} f_1 \, \hat {\d x}_1$
Together with $(1)$, this establishes the result.
{{Qed|lemma}}
Theorem: Even Convergent of Simple Continued Fraction is Strictly Smaller than Odd Convergent

Every even convergent is strictly smaller than every odd convergent.

Proof:
Let $k \ge 1$.
From Denominators of Simple Continued Fraction are Strictly Positive, $q_k q_{k - 1}>0$.
From Difference between Adjacent Convergents of Simple Continued Fraction:
:$C_k - C_{k - 1} = \dfrac {\paren {-1}^{k + 1} } {q_k q_{k - 1} }$
Let $k$ be even.
Then:
:$\exists s \in \N: k = 2 s$
and:
:$C_{2 s} < C_{2 s - 1}$
Let $k$ be odd.
Then:
:$\exists t \in \N: k = 2 t + 1$
and:
:$C_{2 t + 1 } > C_{2 t}$
Now, consider any even convergent $C_{2 s}$ and any odd convergent $C_{2 t + 1}$, with $s, t \ge 0$.
If $2 s < 2 t + 1$, then $2 s \le 2 t$, so:
{{begin-eqn}}
{{eqn | l = C_{2 s}
      | o = \le
      | r = C_{2 t}
      | c = Even Convergents of Simple Continued Fraction are Strictly Increasing
}}
{{eqn | o = <
      | r = C_{2 t + 1}
      | c = by the above
}}
{{end-eqn}}
If $2 s > 2 t + 1$, then $2 s - 1 \ge 2 t + 1$, so:
{{begin-eqn}}
{{eqn | l = C_{2 s}
      | o = <
      | r = C_{2 s - 1}
      | c = by the above
}}
{{eqn | o = \le
      | r = C_{2 t + 1}
      | c = Odd Convergents of Simple Continued Fraction are Strictly Decreasing
}}
{{end-eqn}}
In either case:
:$C_{2 s} < C_{2 t + 1}$
Hence the result.
{{qed}}
Theorem: Denominators of Simple Continued Fraction are Strictly Increasing

Let $N \in \N \cup \set \infty$ be an extended natural number.
Let $\sqbrk {a_0, a_1, a_2, \ldots}$ be a simple continued fraction in $\R$ of length $N$.
Let $q_0, q_1, q_2, \ldots$ be its denominators.
Then with the possible exception of $q_0 = q_1$, the sequence $\sequence {q_n}$ is strictly increasing.

Proof:
By definition of simple continued fraction, all partial denominators of $\sqbrk {a_0, a_1, a_2, \ldots}$ are strictly positive integers, with the possible exception of $a_0$.
So:
:$q_1 = a_1 \ge 1 = q_0$
:$q_2 = a_2 a_1 + 1 \ge a_1 + 1 > a_1 = q_1$
hence:
:$1 = q_0 \le q_1 < q_2$
Suppose $q_k > q_{k - 1} \ge 1$ for some $k \ge 2$.
Then:
:$q_{k + 1} = a_{k + 1} q_k + q_{k - 1} \ge q_k + q_{k - 1} \ge q_k + 1 > q_k$
So, by induction, $\sequence {q_n}$ is strictly increasing except when possibly $q_0 = q_1 = 1$.
{{qed}}