# Getting started on Apache Ignite

Setup Apache Ignite

[2] Install Ignite

unzip apache-ignite-2.9.1-bin.zip

[3] Add Ignite to the PATH environment variable

export PATH=$PATH:$JAVA_HOME/bin:$IGNITE_HOME/bin [3] Start Ignite by executing the following command ignite.sh You should see: (^c to stop Ignite) [4] Enable Ignite restful interface copy the folder$IGNITE_HOME/libs/optional/ignite-rest-http to $IGNITE_HOME/libs Setup Apache Maven [1] Download Maven from https://maven.apache.org/download.cgi : [2] install Maven unzip apache-maven-3.8.1-bin.zip [3] Add Maven to the PATH environment variable export PATH=$PATH:$JAVA_HOME/bin:$IGNITE_HOME/bin:\$MAVEN_HOME/bin

[4] Test Maven, run

mvn -v

You should see:

Rest Client communicates with Ignite

[1] Create Ignite cache

Open your web-browser and enter the following URL:

http://localhost:8080/ignite?cmd=getorcreate&cacheName=testCache

It should return :

Cache is created.

[2] Put data into Ignite cache:

http://localhost:8080/ignite?cmd=put&key=pi&val=3.1415926&cacheName=testCache

[3] Get data from Ignite cache:

http://localhost:8080/ignite?cmd=get&key=pi&cacheName=testCache

You should see:

You can also use ignitevisorcmd to view your cache:

ignitevisorcmd.sh

You should see:

First Ignite Application

[1] Write your first Ignite application

[2] Build your first Ignite application

mvn clean install

You should see:

[3] Run your first Ignite application

java -jar target/HelloIgnite-runnable.jar

[4] Verify cache entries using ignitevisorcmd:

cache -scan

# Deriving Generalized Leibniz’s Integral Rule

The general form of Leibniz’s Integral Rule with variable limits states:

Suppose $f(x, t)$ satisfies the condition stated previously for the basic form of Leibniz’s Rule (LR-1, see “A Semi-Rigorous Derivation of Leibniz’s Rule“) . In addition, $a(t), b(t)$ are defined and have continuous derivatives for $t_1\le t\le t_2.$ Then for $t_1\le t \le t_2,$

$\frac{d}{d t}\int\limits_{a(t)}^{b(t)}f(x, t)\;dx = f(b(t),t)\cdot b'(t) -f(a(t),t)\cdot a'(t)+ \int\limits_{a(t)}^{b(t)}\frac{\partial}{\partial t}f(x, t)\;dx.\quad\quad\quad(1)$

(1) can be derived as a consequence of LR-1, the Multivariable Chain Rule, and the Fundamental Theorem of Calculus (FTC):

Clearly,

$\int\limits_{a(t)}^{b(t)}f(x, t)\;dx\quad\quad\quad(2)$

on the left side of (1) is a function of $t$.

Let

$u = a(t),\quad\quad\quad(3)$

$v = b(t),\quad\quad\quad(4)$

$w = t,\quad\quad\quad(5)$

(2) can be expressed as

$G(u,v,w) = \int\limits_{u}^{v}f(x,w)\;dx.$

Hence, by the chain rule,

$\frac{d}{d t}\int\limits_{u}^{v}f(x, w)\;dx = \frac{\partial}{\partial u}G(u,v,w)\cdot \frac{du}{dt} + \frac{\partial}{\partial v} G(u,v,w)\cdot \frac{dv}{dt} + \frac{\partial}{\partial w}G(u,v,w)\cdot \frac{dw}{dt}$

where

$\frac{\partial}{\partial u}G(u,v,w)=\frac{\partial}{\partial u}\int\limits_{u}^{v}f(x, w)\;dx$

$= \frac{\partial}{\partial u}\left(-\int\limits_{v}^{u}f(x, w)\;dx\right)$

$= -\frac{\partial}{\partial u}\int\limits_{v}^{u}f(x, w) \;dx$

$\overset{\textbf{FTC}}{=} -f(u, w)$

$\overset{(3), (5)}{=} -f(a(t), t),$

$\frac{\partial}{\partial v}G(u,v,w)=\frac{\partial}{\partial v}\int\limits_{u}^{v}f(x, w)\;dx \overset{\textbf{FTC}}{=}f(v, w) \overset{(4), (5)}{=} f(b(t), t)$

and,

$\frac{\partial}{\partial w}G(u,v,w)=\frac{\partial}{\partial w}\int\limits_{u}^{v}f(x, w)\;dx \overset{\textbf{LR-1}}{=} \int\limits_{u}^{v}\frac{\partial}{\partial w}f(x, w)\;dx\overset{(3), (4), (5)}{=} \int\limits_{a(t)}^{b(t)}\frac{\partial}{\partial t}f(x, t)\;dx.$

It follows that

$\frac{d}{d t}\int\limits_{a(t)}^{b(t)}f(x, t)\;dx =-f(a(t),t)\cdot a'(t) + f(b(t),t)\cdot b'(t) + \int\limits_{a(t)}^{b(t)}\frac{\partial}{\partial t}f(x, t)\;dx,$

i.e.,

$\frac{d}{d t}\int\limits_{a(t)}^{b(t)}f(x, t)\;dx = f(b(t),t)\cdot b'(t) -f(a(t),t)\cdot a'(t)+ \int\limits_{a(t)}^{b(t)}\frac{\partial}{\partial t}f(x, t)\;dx.$

# A Semi-Rigorous Derivation of Leibniz’s Rule

Leibniz’s rule (LR-1) states:

Let $f(x, \beta)$ be continuous and have a continuous derivative $\frac{\partial}{\partial \beta}$ in a domain of $x\beta-$plane that includes the rectangle $a \le x \le b, \beta_1 \le \beta \le \beta_2,$

$\frac{d}{d\beta}\int\limits_{a}^{b}f(x, \beta)\;dx =\int\limits_{a}^{b}\frac{\partial}{\partial \beta}f(x, \beta)\;dx.$

I will derive LR-1 semi-rigorously as follows:

Let

$g(t) = \int\limits_{a}^{b} \frac{\partial}{\partial t}f(x, t)\;dx.\quad\quad\quad(1-1)$

Integrate (1-1) with respect to $t$ from a constant $\alpha$ to a variable $\beta$, we have

$\int\limits_{\alpha}^{\beta} g(t)\;dt = \int\limits_{\alpha}^{\beta}\left(\int\limits_{a}^{b} \frac{\partial}{\partial t}f(x, t)\;dx\right)\;dt$

$\overset{(\star)}{=}\int\limits_{a}^{b}\left(\int\limits_{\alpha}^{\beta} \frac{\partial}{\partial t}f(x, t)\;dt\right)\;dx$

$=\int\limits_{a}^{b}f(x, \beta) - f(x, \alpha)\; dx$

$=\int\limits_{a}^{b}f(x, \beta)\;dx - \int\limits_{a}^{b}f(x,\alpha)\; dx.$

That is,

$\int\limits_{\alpha}^{\beta} g(t)\;dt = \int\limits_{a}^{b}f(x, \beta)\;dx - \int\limits_{a}^{b}f(x,\alpha)\; dx.\quad\quad\quad(1-2)$

While $\int\limits_{\alpha}^{\beta}g(t)\;dt$ and $\int\limits_{a}^{b}f(x,\beta)\;dx$ are functions of $\beta$, $\int\limits_{a}^{b}f(x,\alpha)\;dx$ is a constant.

Since

$\frac{d}{d\beta}\int\limits_{\alpha}^{\beta} g(t)\;dt \overset{\textbf{FTC}}{= }g(\beta), \quad\frac{d}{d\beta}\left(\int\limits_{a}^{b}f(x, \beta)\;dx - \int\limits_{a}^{b}f(x,\alpha)\; dx\right)=\frac{d}{d\beta}\int\limits_{a}^{b}f(x, \beta)\;dx,$

differentiate (1-2) with respect to $\beta$ gives

$g(\beta) = \frac{d}{d\beta}\int\limits_{a}^{b}f(x, \beta)\;dx\overset{(1-1)}{\implies} \int\limits_{a}^{b} \frac{\partial}{\partial \beta}f(x, \beta)\;dx= \frac{d}{d\beta}\int\limits_{a}^{b}f(x, \beta)\;dx;$

i.e.,

$\frac{d}{d\beta}\int\limits_{a}^{b}f(x, \beta)\;dx = \int\limits_{a}^{b} \frac{\partial}{\partial \beta}f(x, \beta)\;dx.$

In the three-dimensional $x, y, z$-space, the double integral of a continuous function with two independent variables, $V=\iint_{R} f(x, y) dx dy$, may be interpreted as a volume between the surface $z=f(x, y)$ and the $x, y$-plane:

Fig. 1 $V = \iint_R f(x,y) dx dy$

We see from Fig. 1 that on one hand,

$V=\int\limits_{c}^{d}\int\limits_{a}^{b}f(x,y)\;dx dy,\quad\quad\quad(2-1)$

but on the other hand,

$V=\int\limits_{a}^{b}\int\limits_{c}^{d}f(x,y)\;dy dx.\quad\quad\quad(2-2)$

Since (2-1) and (2-2) amounts to the same thing, it must be true that

$\int\limits_{c}^{d}\int\limits_{a}^{b}f(x,y)\;dx dy=\int\limits_{a}^{b}\int\limits_{c}^{d}f(x,y)\;dy dx.\quad\quad\quad(\star)$

In other words, the order of integration can be interchanged.