nyuuzyou/wikis huggingface dataset:
A.E. Equal Positive Measurable Functions have Equal Integrals/Corollary 1

== Theorem ==
Let $\struct {X, \Sigma, \mu}$ be a measure space.
Let $f: X \to \overline \R$ be a $\mu$-integrable function.
Let $g: X \to \overline \R$ be $\Sigma$-measurable.
Suppose that $f = g$ almost everywhere.
Then $g$ is also $\mu$-integrable, and:
:$\ds \int f \rd \mu = \int g \rd \mu$
== Proof ==
From Function Measurable iff Positive and Negative Parts Measurable, we have that: 
:$g^+$, $f^+$, $g^-$ and $f^-$ are all $\Sigma$-measurable.
From Functions A.E. Equal iff Positive and Negative Parts A.E. Equal, we have that: 
:$f^+ = g^+$ and $f^- = g^-$ $\mu$-almost everywhere. 
Since $f^+$ and $g^+$ are positive $\Sigma$-measurable functions that are equal $\mu$-almost everywhere, we have: 
:$\ds \int f^+ \rd \mu = \int g^+ \rd \mu$ 
from A.E. Equal Positive Measurable Functions have Equal Integrals.
Similarly, we have that $f^-$ and $g^-$ are positive $\Sigma$-measurable functions that are equal $\mu$-almost everywhere, and so:
:$\ds \int f^- \rd \mu = \int g^- \rd \mu$
Since $f$ is $\mu$-integrable, we have that: 
:$\ds \int f^+ \rd \mu < \infty$
and:
:$\ds \int f^- \rd \mu < \infty$
Hence: 
:$\ds \int g^+ \rd \mu < \infty$
and:
:$\ds \int g^- \rd \mu < \infty$
So $g$ is $\mu$-integrable.
We also have: 
Category:A.E. Equal Positive Measurable Functions have Equal Integrals
A.E. Equal Positive Measurable Functions have Equal Integrals/Corollary 2

== Corollary to A.E. Equal Positive Measurable Functions have Equal Integrals ==
Let $\struct {X, \Sigma, \mu}$ be a measure space.
Let $f, g : X \to \overline \R$ be positive $\Sigma$-measurable functions. 
Suppose that $f = g$ $\mu$-almost everywhere. 
Let $A \in \Sigma$. 
Then: 
:$\ds \int_A f \rd \mu = \int_A g \rd \mu$
== Proof ==
From the definition of a $\mu$-integral over $A$, we have: 
:$\ds \int_A f \rd \mu = \int \paren {f \times \chi_A} \rd \mu$
and:
:$\ds \int_A f \rd \mu = \int \paren {g \times \chi_A} \rd \mu$
We show that: 
:$f \times \chi_A = g \times \chi_A$ $\mu$-almost everywhere.
Since: 
:$f = g$ $\mu$-almost everywhere
there exists a $\mu$-null set $N \subseteq X$ such that:
:whenever $x \in X$ has $\map f x \ne \map g x$, we have $x \in N$. 
Suppose that $x \in X$ is such that: 
:$\map {\paren {f \times \chi_A} } x \ne \map {\paren {g \times \chi_A} } x$
From the definition of pointwise multiplication, we have: 
:$\map f x \map {\chi_A} x \ne \map g x \map {\chi_A} x$ 
So:
:$\map f x \ne \map g x$
Hence: 
:$x \in N$ 
So:
:$f \times \chi_A = g \times \chi_A$ $\mu$-almost everywhere.
From A.E. Equal Positive Measurable Functions have Equal Integrals, we therefore have: 
:$\ds \int \paren {f \times \chi_A} \rd \mu = \int \paren {g \times \chi_A} \rd \mu$
and so:
:$\ds \int_A f \rd \mu = \int_A g \rd \mu$
Category:A.E. Equal Positive Measurable Functions have Equal Integrals
A.E. Equal Positive Measurable Functions have Equal Integrals/Proof 1

== Theorem ==
== Proof ==
Let $N$ be the set defined by:
:$N = \set {x \in X: \map f x \ne \map g x}$
By hypothesis, $N$ is a $\mu$-null set.
If $N = \O$, then $f = g$, trivially implying the result.
If $N \ne \O$, then by Set with Relative Complement forms Partition:
:$X = N \cup \paren {X \setminus N}$
Now:
which establishes the result.
Category:A.E. Equal Positive Measurable Functions have Equal Integrals
A.E. Equal Positive Measurable Functions have Equal Integrals/Proof 2

== Theorem ==
== Proof ==
Let: 
:$A = \set {x \in X : \map f x \ne \map g x}$
From Measurable Functions Determine Measurable Sets, we have that: 
:$A$ is $\Sigma$-measurable.
Define $h : X \to \overline \R$ by: 
:$\map h x = \begin{cases}+\infty & x \in A \\ 0 & x \not \in A\end{cases}$
We can show that $h$ is $\Sigma$-measurable.
If $t < 0$, we have that: 
:$\set {x \in X : \map h x \le t} = \O$
From Sigma-Algebra Contains Empty Set, we therefore have that: 
:$\set {x \in X : \map h x \le t}$ is $\Sigma$-measurable if $t < 0$. 
If $t \ge 0$, we have that: 
:$\set {x \in X : \map h x \le t} = X \setminus A$
This is also $\Sigma$-measurable, so $h$ is a $\Sigma$-measurable function.
Clearly also $h \ge 0$. 
We can show that: 
:$f \le g + h$ 
If $x \in A$, then $\map h x = +\infty$, so $\map g x + \map h x = +\infty$, and the inequality is satisfied trivially.
If $x \not \in A$, we have $\map f x = \map g x = \map g x + \map h x$, and so $\map f x \le \map g x + \map h x$ again.
From Integral of Positive Measurable Function is Monotone, we then have: 
:$\ds \int f \rd \mu \le \int \paren {g + h} \rd \mu$
Applying Integral of Integrable Function is Additive, we then have: 
:$\ds \int f \rd \mu \le \int g \rd \mu + \int h \rd \mu$
To compute the latter integral, note that: 
:$x \in X$ is such that $\map h x \ne 0$ $x \in A$.
Since by hypothesis, $f = g$ $\mu$-almost everywhere, we have that: 
:$\map \mu A = 0$ 
So:
:$h = 0$ $\mu$-almost everywhere.
Then, by Measurable Function Zero A.E. iff Absolute Value has Zero Integral, we have: 
:$\ds \int \size h \rd \mu = 0$
Since $h \ge 0$, this gives: 
:$\ds \int h \rd \mu = 0$
and so: 
:$\ds \int f \rd \mu \le \int g \rd \mu$
Similarly, we have: 
:$g \le f + h$
and obtain: 
:$\ds \int g \rd \mu \le \int f \rd \mu$
simply swapping $g$ for $f$ in the prior computation. 
We therefore have: 
:$\ds \int f \rd \mu = \int g \rd \mu$
Category:A.E. Equal Positive Measurable Functions have Equal Integrals
AM-HM Inequality

== Theorem ==
Let $x_1, x_2, \ldots, x_n \in \R_{> 0}$ be strictly positive real numbers.
Let $A_n $ be the arithmetic mean of $x_1, x_2, \ldots, x_n$.
Let $H_n$ be the harmonic mean of $x_1, x_2, \ldots, x_n$.
Then $A_n \ge H_n$.
== Proof ==
$A_n$ is defined as:
:$\ds A_n = \frac 1 n \paren {\sum_{k \mathop = 1}^n x_k}$
$H_n$ is defined as:
:$\ds \frac 1 H_n = \frac 1 n \paren {\sum_{k \mathop = 1}^n \frac 1 {x_k} }$
We have that:
:$\forall k \in \closedint 1 n: x_k > 0$
From Positive Real has Real Square Root, we can express each $x_k$ as a square:
:$\forall k \in \closedint 1 n: x_k = y_k^2$
without affecting the result.
Thus we have:
:$\ds A_n = \frac 1 n \paren {\sum_{k \mathop = 1}^n y_k^2}$
:$\ds \frac 1 {H_n} = \frac 1 n \paren {\sum_{k \mathop = 1}^n \frac 1 {y_k^2} }$
Multiplying $A_n$ by $\dfrac 1 {H_n}$:
So:
:$\dfrac {A_n} {H_n} \ge 1$
and so from :
:$A_n \ge H_n$
== Also see ==
* Cauchy's Mean Theorem
* GM-HM Inequality
== Sources ==
* : $\S 1$: Real Numbers: Exercise $\S 1.12 \ (5)$
* 
* 
Category:Inequalities
Category:Arithmetic Mean
Category:Harmonic Mean
ARIMA Model subsumes ARMA Model

== Theorem ==
Let $S$ be a stochastic process based on an equispaced time series.
Let $M$ be an ARMA model for $S$.
Then $M$ is also an implementation of an ARIMA model.
== Proof ==
Let the values of $S$ at timestamps $t, t - 1, t - 2, \dotsc$ be $z_t, z_{t - 1}, z_{t - 2}, \dotsc$
Let $\tilde z_t, \tilde z_{t - 1}, \tilde z_{t - 2}, \dotsc$ be deviations from a constant mean level $\mu$:
:$\tilde z_t = z_t - \mu$
Let $a_t, a_{t - 1}, a_{t - 2}, \dotsc$ be a sequence of independent shocks at timestamps $t, t - 1, t - 2, \dotsc$
By definition of ARMA model, $M$ is implemented as:
:$(1): \quad \tilde z_t = \phi_1 \tilde z_{t - 1} + \phi_2 \tilde z_{t - 2} + \dotsb + \phi_p \tilde z_{t - p} + a_t - \theta_1 a_{t - 1} - \theta_2 a_{t - 2} - \dotsb - \theta_q a_{t - q}$
The general ARIMA model is implemented as:
:$w_t = \phi_1 w_{t - 1} + \phi_2 w_{t - 2} + \dotsb + \phi_p w_{t - p} + a_t - \theta_1 a_{t - 1} - \theta_2 a_{t - 2} - \dotsb - \theta_q a_{t - q}$
where:
:$w_t = \nabla^d z_t$
Letting $d = 0$ we have:
:$w_t = z_t$
Setting $w_t = z_t - \mu = \tilde z$, we recover $(1)$.
== Sources ==
* :
::$1$: Introduction:
:::$1.2$ Stochastic and Deterministic Dynamic Mathematical Models
::::$1.2.1$ Stationary and Nonstationary Stochastic Models for Forecasting and Control: Nonstationary models
Category:ARIMA Models
Category:ARMA Models
ARIMA Model subsumes Autoregressive Model

== Theorem ==
Let $S$ be a stochastic process based on an equispaced time series.
Let $M$ be an autoregressive model for $S$.
Then $M$ is also an implementation of an ARIMA model.
== Proof ==
Let the values of $S$ at timestamps $t, t - 1, t - 2, \dotsc$ be $z_t, z_{t - 1}, z_{t - 2}, \dotsc$
Let $\tilde z_t, \tilde z_{t - 1}, \tilde z_{t - 2}, \dotsc$ be deviations from a constant mean level $\mu$:
:$\tilde z_t = z_t - \mu$
Let $a_t, a_{t - 1}, a_{t - 2}, \dotsc$ be a sequence of independent shocks at timestamps $t, t - 1, t - 2, \dotsc$
By definition of autoregressive model, $M$ is implemented as:
:$(1): \quad \tilde z_t = \phi_1 \tilde z_{t - 1} + \phi_2 \tilde z_{t - 2} + \dotsb + \phi_p \tilde z_{t - p} + a_t$
The general ARIMA model is implemented as:
:$w_t = \phi_1 w_{t - 1} + \phi_2 w_{t - 2} + \dotsb + \phi_p w_{t - p} + a_t - \theta_1 a_{t - 1} - \theta_2 a_{t - 2} - \dotsb - \theta_q a_{t - q}$
where:
:$w_t = \nabla^d z_t$
Let $\theta_j = 0$ for all $j$.
Letting $d = 0$ we have:
:$w_t = z_t$
Setting $w_t = z_t - \mu = \tilde z$, we recover $(1)$.
== Sources ==
* :
::$1$: Introduction:
:::$1.2$ Stochastic and Deterministic Dynamic Mathematical Models
::::$1.2.1$ Stationary and Nonstationary Stochastic Models for Forecasting and Control: Nonstationary models
Category:ARIMA Models
Category:Autoregressive Models
ARIMA Model subsumes Moving Average Model

== Theorem ==
Let $S$ be a stochastic process based on an equispaced time series.
Let $M$ be a moving average model for $S$.
Then $M$ is also an implementation of an ARIMA model.
== Proof ==
Let the values of $S$ at timestamps $t, t - 1, t - 2, \dotsc$ be $z_t, z_{t - 1}, z_{t - 2}, \dotsc$
Let $\tilde z_t, \tilde z_{t - 1}, \tilde z_{t - 2}, \dotsc$ be deviations from a constant mean level $\mu$:
:$\tilde z_t = z_t - \mu$
Let $a_t, a_{t - 1}, a_{t - 2}, \dotsc$ be a sequence of independent shocks at timestamps $t, t - 1, t - 2, \dotsc$
By definition of moving average model, $M$ is implemented as:
:$(1): \quad \tilde z_t = a_t - \theta_1 a_{t - 1} - \theta_2 a_{t - 2} - \dotsb - \theta_q a_{t - q}$
The general ARIMA model is implemented as:
:$w_t = \phi_1 w_{t - 1} + \phi_2 w_{t - 2} + \dotsb + \phi_p w_{t - p} + a_t - \theta_1 a_{t - 1} - \theta_2 a_{t - 2} - \dotsb - \theta_q a_{t - q}$
where:
:$w_t = \nabla^d z_t$
Let $\phi_i = 0$ for all $i$.
Letting $d = 0$ we have:
:$w_t = z_t$
Setting $w_t = z_t - \mu = \tilde z$, we recover $(1)$.
== Sources ==
* :
::$1$: Introduction:
:::$1.2$ Stochastic and Deterministic Dynamic Mathematical Models
::::$1.2.1$ Stationary and Nonstationary Stochastic Models for Forecasting and Control: Nonstationary models
Category:ARIMA Models
Category:Moving Average Models
A Priori (Logic)/Examples/Legal

== Example of Use of A Priori ==
The following is an example of reasoning:
:If Fred Bloggs has committed a crime, then he is guilty.
The truth of the above statement is dependent upon the definition of the term guilty, which, in natural language simply means having committed a crime.
Contrast this with:
:Fred Bloggs has committed the crime of usury
the truth value of which is dependent upon knowing whether or not he actually did it.
Category:Examples of Logic
Abel's Lemma

== Lemma ==
Let $\sequence a$ and $\sequence b$ be sequences in an arbitrary ring $R$.
=== Formulation 1 ===
=== Formulation 2 ===
== Also known as ==
== Also see ==
* Integration by Parts
Category:Abel's Lemma
Category:Algebra
Category:Finite Calculus
Category:Proof Techniques
Category:Lemmata
Abel's Lemma/Formulation 1

== Lemma ==
Let $\sequence a$ and $\sequence b$ be sequences in an arbitrary ring $R$.
Then:
:$\ds \sum_{k \mathop = m}^n a_k \paren {b_{k + 1} - b_k} = a_{n + 1} b_{n + 1} - a_m b_m - \sum_{k \mathop = m}^n \paren {a_{k + 1} - a_k} b_{k + 1}$
Note that although proved for the general ring, this result is usually applied to one of the conventional number fields $\Z, \Q, \R$ and $\C$.
=== Corollary ===
== Proof ==
== Also reported as ==
Some sources give this as:
:$\ds \sum_{k \mathop = m}^n \paren {a_{k + 1} - a_k} b_k = a_{n + 1} b_{n + 1} - a_m b_m - \sum_{k \mathop = m}^n a_{k + 1} \paren {b_{k + 1} - b_k}$
which is obtained from the main result by interchanging $a$ and $b$.
Others take the upper index to $n - 1$:
:$\ds \sum_{k \mathop = m}^{n - 1} \paren {a_{k + 1} - a_k} b_k = a_n b_n - a_m b_m - \sum_{k \mathop = m}^{n - 1} a_{k + 1} \paren {b_{k + 1} - b_k}$
== Also known as ==
== Also see ==
* Integration by Parts
Category:Abel's Lemma
Abel's Lemma/Formulation 1/Corollary

== Corollary to Abel's Lemma: Formulation 1 ==
Let $\sequence a$ and $\sequence b$ be sequences in an arbitrary ring $R$.
Then:
:$\ds \sum_{k \mathop = 1}^n a_k \paren {b_{k + 1} - b_k} = a_{n + 1} b_{n + 1} - a_1 b_1 - \sum_{k \mathop = 1}^n \paren {a_{k + 1} - a_k} b_{k + 1}$
Note that although proved for the general ring, this result is usually applied to one of the conventional number fields $\Z, \Q, \R$ and $\C$.
== Proof ==
From Abel's Lemma: Formulation 1, we have:
:$\ds \sum_{k \mathop = m}^n a_k \paren {b_{k + 1} - b_k} = a_{n + 1} b_{n + 1} - a_m b_m - \sum_{k \mathop = m}^n \paren {a_{k + 1} - a_k} b_{k + 1}$
The result follows by setting $m = 1$.
== Also reported as ==
Some sources give this as:
:$\ds \sum_{k \mathop = 1}^n \paren {a_{k + 1} - a_k} b_k = a_{n + 1} b_{n + 1} - a_1 b_1 - \sum_{k \mathop = 1}^n a_{k + 1} \paren {b_{k + 1} - b_k}$
which is obtained from the main result by interchanging $a$ and $b$.
Others take the upper index to $n - 1$:
:$\ds \sum_{k \mathop = 1}^{n - 1} \paren {a_{k + 1} - a_k} b_k = a_n b_n - a_1 b_1 - \sum_{k \mathop = 1}^{n - 1} a_{k + 1} \paren {b_{k + 1} - b_k}$
== Sources ==
* : $\S 1.2.7$: Harmonic Numbers: Exercise $10$
Category:Abel's Lemma
Abel's Lemma/Formulation 2

== Lemma ==
Let $\sequence a$ and $\sequence b$ be sequences in an arbitrary ring $R$.
Let $\ds A_n = \sum_{i \mathop = m}^n {a_i}$ be the partial sum of $\sequence a$ from $m$ to $n$.
Then:
:$\ds \sum_{k \mathop = m}^n a_k b_k = \sum_{k \mathop = m}^{n - 1} A_k \paren {b_k - b_{k + 1} } + A_n b_n$
Note that although proved for the general ring, this result is usually applied to one of the conventional number fields $\Z, \Q, \R$ and $\C$.
=== Corollary ===
== Proof 1 ==
== Proof 2 ==
== Also known as ==
== Also see ==
* Integration by Parts
== Sources ==
* 
* 
* 
Category:Abel's Lemma
Abel's Lemma/Formulation 2/Corollary

== Lemma ==
Let $\sequence a$ and $\sequence b$ be sequences in an arbitrary ring $R$.
Let $\ds A_n = \sum_{i \mathop = m}^n a_i$ be the partial sum of $\sequence a$ from $m$ to $n$.
Then:
:$\ds \sum_{k \mathop = 0}^n a_k b_k = \sum_{k \mathop = 0}^{n - 1} A_k \paren {b_k - b_{k + 1} } + A_n b_n$
== Proof ==
From Abel's Lemma: Formulation 2, we have:
:$\ds \sum_{k \mathop = m}^n a_k b_k = \sum_{k \mathop = m}^{n - 1} A_k \paren {b_k - b_{k + 1} } + A_n b_n$
The result follows by setting $m = 0$.
Category:Abel's Lemma
Abel's Lemma/Formulation 2/Proof 1

== Lemma ==
== Proof ==
Proof by induction:
For all $n \in \N$ where $n \ge m$, let $\map P n$ be the proposition:
:$\ds \sum_{k \mathop = m}^n a_k b_k = \sum_{k \mathop = m}^{n - 1} A_k \paren {b_k - b_{k + 1} } + A_n b_n$
=== Basis for the Induction ===
First consider $\map P m$.
When $n = m$, we have that:
:$\ds \sum_{k \mathop = m}^{n - 1} A_k \paren {b_k - b_{k + 1} } = 0$
is a vacuous summation, as the upper index is smaller than the lower index.
We also have that:
:$\ds A_m = \sum_{i \mathop = m}^m {a_i} = a_m$
Thus we see that $\map P m$ is true, as this just says:
:$a_m b_m = 0 + A_m b_m = a_m b_m$
which is clearly true.
This is our basis for the induction.
=== Induction Hypothesis ===
Now we need to show that, if $\map P r$ is true, where $r \ge m$, then it logically follows that $\map P {r + 1}$ is true.
So this is our induction hypothesis:
:$\ds \sum_{i \mathop = m}^r a_k b_k = \sum_{k \mathop = m}^{r - 1} A_k \paren {b_k - b_{k + 1} } + A_r b_r$
Then we need to show:
:$\ds \sum_{k \mathop = m}^{r + 1} a_k b_k = \sum_{k \mathop = m}^r A_k \paren {b_k - b_{k + 1} } + A_{r + 1} b_{r + 1}$
where:
:$\ds A_{r + 1} = \sum_{i \mathop = m}^{r + 1} {a_i}$
=== Induction Step ===
This is our induction step:
So $\map P n \implies \map P {r + 1}$ and the result follows by the Principle of Mathematical Induction.
Therefore:
:$\ds \forall n \ge m: \sum_{k \mathop = m}^n a_k b_k = \sum_{k \mathop = m}^{n - 1} A_k \paren {b_k - b_{k + 1} } + A_n b_n$
Category:Abel's Lemma
Abel's Lemma/Formulation 2/Proof 2

== Lemma ==
== Proof ==
First note that:
:$\ds A_{m - 1} = \sum_{i \mathop = m}^{m - 1} a_i = 0$
is a vacuous summation, as the upper index is smaller than the lower index.
Then we have:
Therefore:
:$\ds \forall n \ge m: \sum_{k \mathop = m}^n a_k b_k = \sum_{k \mathop = m}^{n - 1} A_k \paren {b_k - b_{k + 1} } + A_n b_n$
Category:Abel's Lemma
Abel's Limit Theorem

== Theorem ==
Let $\ds \sum_{k \mathop = 0}^\infty a_k$ be a convergent series in $\R$.
Then:
:$\ds \lim_{x \mathop \to 1^-} \paren {\sum_{k \mathop = 0}^\infty a_k x^k} = \sum_{k \mathop = 0}^\infty a_k$
where $\ds \lim_{x \mathop \to 1^-}$ denotes the limit from the left.
== Proof 1 ==
== Proof 2 ==
== Examples ==
== Also known as ==
== Also see ==
* Definition:Abel Summation Method
* Definition:Cesàro Summation
== Sources ==
* : Appendix: $\S 18.9$: Continuity and differentiation of power series
* 
* 
* 
Category:Abel's Limit Theorem
Category:Series
Category:Analysis
Category:Abel's Theorem
Abel's Limit Theorem/Also known as

== Abel's Limit Theorem: Also known as == 
Abel's Limit Theorem is also known just as Abel's Theorem.
However, the latter name has more than one theorem attached to it, so the full name is preferred.
Again, Abel's Limit Theorem can also be found as Abel's Lemma.
However, the latter name is also found attached to a completely different result, so again, it is preferred that it not be used in this context.
== Sources ==
* : Appendix: $\S 18.9$: Continuity and differentiation of power series
Category:Abel's Limit Theorem
Abel's Limit Theorem/Examples/Arbitrary Example 1

== Examples of Use of Abel's Limit Theorem ==
Let $\ds \map g x = \sum_{n \mathop \ge 1} \paren {-1}^{n - 1} \dfrac {x^n} n$ for $\size x < 1$.
Then:
:$\map g x = \map \ln {1 + x}$
for $\size x < 1$.
The series $\map g 1$ converges by Alternating Series Test,
so by Abel's Limit Theorem:
:$\map g 1 = \ds \lim _{x \mathop \to 1^{-} } \map g x = \lim_{x \mathop \to 1^{-} } \map \ln {1 + x} = \ln 2$
since the logarithm is a continuous function.
Category:Examples of Use of Abel's Limit Theorem
Abel's Limit Theorem/Examples/Arbitrary Example 2

== Examples of Use of Abel's Limit Theorem ==
Let:
:$\ds \map g x = \sum_{n \mathop \ge 0} \frac {\paren {-1}^{n - 1} \paren {2 } !} {2^{2 n} n!^2 \paren {2 n - 1} } x^n$
for $\size x < 1$.
Then:
:$\map g x = \sqrt {1 + x}$
for $\size x < 1$.
The series $\map g 1$ is absolutely convergent
so by Abel's Limit Theorem and the continuity of $\sqrt {1 + x}$:
:$\map g 1 = \ds \lim_{x \mathop \to 1^{-} } \map g x = \lim_{x mathop \to 1^{-} } \sqrt {1 + x} = \sqrt 2$
Category:Examples of Use of Abel's Limit Theorem
Abel's Limit Theorem/Examples/Arbitrary Example 3

== Examples of Use of Abel's Limit Theorem ==
Let:
:$\map g x = \dfrac 1 {1 + x^2}$
which is differentiable for all real $x$.
When $\size x < 1$, $\map g x = \ds \sum_{n \mathop \ge 0} \paren {-1}^n x^{2 n}$ by expanding a geometric series.
While $\map g x$ has a limit as $x \to 1^{-}$ (namely $1/2$), the power series does not converge at $x = 1$.
Category:Examples of Use of Abel's Limit Theorem
Abel's Limit Theorem/Proof 1

== Theorem ==
== Proof ==
Let $\epsilon > 0$.
Let $\ds \sum_{k \mathop = 0}^\infty a_k$ converge to $s$.
Then its sequence of partial sums $\sequence {s_N}$, where $\ds s_N = \sum_{n \mathop = 1}^N a_n$, is a Cauchy sequence.
So:
:$\ds \exists N: \forall k, m: k \ge m \ge N: \size {\sum_{l \mathop = m}^k a_l} < \frac \epsilon 3$
From Abel's Lemma: Formulation 2, we have:
:$\ds \sum_{k \mathop = m}^n u_k v_k = \sum_{k \mathop = m}^{n - 1} \paren {\paren {\sum_{l \mathop = m}^k u_l} \paren {v_k - v_{k + 1} } } + v_n \sum_{k \mathop = m}^n u_k$
Let $u_k = a_k$ and $v_k = x^k$.
We have:
:$\ds \sum_{k \mathop = m}^n a_k x^k = \sum_{k \mathop = m}^{n - 1} \paren {\paren {\sum_{l \mathop = m}^k a_l} \paren {x^k - x^{k + 1} } } + x^n \sum_{k \mathop = m}^n a_k$
So $\forall n \in \N: n \ge m \ge N$ and $\forall x \in \R: 0 < x < 1$, we have:
We conclude that:
:$\ds \size {\sum_{k \mathop = N}^\infty a_k x^k} \le \frac \epsilon 3$
From the above, we have $\forall x \in \R: 0 < x < 1$:
:$\ds \size {\sum_{k \mathop = 0}^\infty a_k x^k - \sum_{k \mathop = 0}^\infty a_k} \le \sum_{k \mathop = 0}^{N - 1} \size {a_n} \paren {1 - x^n} + \frac \epsilon 3 + \frac \epsilon 3$
But for finite $n$, we have:
:$1 - x^n \to 0$ as $x \to 1^-$.
Thus:
:$\ds \sum_{k \mathop = 0}^{N-1} \size {a_n} \paren {1 - x^n} \to 0$ as $x \to 1^-$
So:
:$\ds \exists \delta > 0: \forall x: 1 - \delta < x < 1: \sum_{k \mathop = 0}^{N - 1} \size {a_n} \paren {1 - x^n} < \frac \epsilon 3$
So, for any given $\epsilon > 0$, there exists $\delta > 0$, such that:
:$\forall x \in \R: 1 - \delta < x < 1 \implies \ds \size {\sum_{k \mathop = 0}^\infty a_k x^k - \sum_{k \mathop = 0}^\infty a_k} < \epsilon$
The result follows from the definition for the limit from the left.
== Also known as ==
== Also see ==
* Definition:Abel Summation Method
* Definition:Cesàro Summation
== Sources ==
* : Appendix: $\S 18.9$: Continuity and differentiation of power series
Category:Abel's Limit Theorem
Abel's Limit Theorem/Proof 2

== Theorem ==
== Proof ==
Since $\ds \sum_{k \mathop = 0}^\infty a_k$ converges and $\cmod{x^k}\le1$ and $\sequence{x^k}$ is decreasing,
by Abel's Test for Uniform Convergence $\ds \sum_{k \mathop = 0}^\infty a_kx^k$ converges uniformly on $[0,1]$ and hence to a continuous function by Uniform Limit Theorem.
== Also known as ==
== Also see ==
* Definition:Abel Summation Method
* Definition:Cesàro Summation
Category:Abel's Limit Theorem
Abel's Summation Formula

== Theorem ==
Let $\sequence {a_n}_{n \in \N_{>0} }$ be a sequence in $\R$.
Let $f : \R_{\ge 1} \to \R$ be a continuously differentiable function.
Let $A : \R_{\ge 1} \to \R$ be defined as:
:$\ds \map A x := \sum_{1 \mathop \le n \mathop \le x} a_n$
Then for all $x \in \R_{\ge 1}$:
:$\ds \sum_{1 \mathop \le n \mathop \le x} a_n \map f n = \map A x \map f x - \int _1 ^x \map A u \map {f'} u \rd u$
== Proof ==
Category:Integral Calculus
Category:Analytic Number Theory
Abel's Test

== Theorem ==
Let $\ds \sum b_n$ be a convergent real series.
Let $\sequence {a_n}$ be a decreasing sequence of positive real numbers.
Then the series $\ds \sum a_n b_n$ is also convergent.
=== Abel's Test for Uniform Convergence ===
== Proof ==
Let $b_0 = 0$.
Let $B_N = \ds \sum_{k \mathop = 0}^N b_k$.
Then: 
:$\forall n \ge 1: b_n = B_n − B_{n − 1}$
From Abel's Lemma:
$\ds \sum_{k \mathop = 1}^N a_k b_k = \sum_{k \mathop = 1}^{N - 1} B_k \paren {a_k - a_{k + 1} } + a_N B_N$
By the Monotone Convergence Theorem:
:$\sequence {a_n}$ converges
:$\sequence {B_N}$ converges since $\ds \sum b_n$ converges.
Hence the second addend $a_N B_N$ converges.
It remains to prove the first addend $\ds \sum B_k \paren {a_k − a_{k + 1} }$ converges.
Since $\sequence {B_N}$ converges, $\size {B_N} \le M$ for some $M \in \N$.
Hence $\ds \sum B_k \paren {a_k − a_{k + 1} }$ converges absolutely.
Hence $\ds \sum a_n b_n$ converges.
== Sources ==
* 
* 
* 
* 
* 
Category:Abel's Test
Category:Convergence