OpenLLM-France/wikimedia huggingface dataset:
# Novial/AIL Participles
## Participles

It seems really better to have only one active and one passive participle, and then the endings are offered quite naturally by the existing languages, namely -nt and -t. The former is known from all the Romanic languages, from which a certain number are taken over into English: fluent, different, latent, defendant, etc. Compare also D -end (pronounced -ent): sehend, etc., and Sc -nde: seende, kallande, etc. The passive participle in -t is found in I S, in E in verbs like kept, in Sc regularly: set, kallat, in D in the weak verbs: gesegnet, gebracht, in R in some verbs.
Thus from the verbal stems ama love, protekte protect, mari marry, konstitu constitute, es be, we form the participles amant amat, protektent protektet, marient marit, konstituent konstitut, esent (no passive). It will be seen that while -nt can be added immediately after a, and -t after any vowel, the vowel -e- is put in between -i, -u or a consonant on the one hand, and -nt on the other. To these forms the endings -i, -e, -o, -a can of course be added: amanti fema, amanto lover, men amata by beloved one; marito husband, etc. Stress amánt(i), amát(i), etc.
# Novial/AIL Present and Past
## Present and Past

The simple stem is used further in the present tense: me ama vu; lo protekte nus, etc.
If we look around us in the national languages to find an expression for the past tense (the tense most often used in ordinary conversation, in narratives, etc., and for which it is therefore urgent to find an easily formed and easily used expression), we discover that the Romanic languages are of no use for our purpose, as the formation of their two preterit tenses is not uniform enough to serve as basis for an interlingual formation; the Slavic -l is not sufficiently known in the rest of Europe and in America. But fortunately we find in English, Scandinavian and Dutch an ending which is excellent for our purpose, namely -d in E lived, Dan levede, Sw levade, Dutch leevde, etc. Nor is this far from D -te. Thus from the verbs already mentioned we form the past: amad, protekted, marid, konstitud, esed, an added -e being only necessary between a consonant and the ending -d. Stress as in the infinitive.
By the side of this it is advantageous to have an analytic form for the past, and E did naturally offers itself for this purpose: me did ama, protekte etc. It should not, however, be used with the same emphatic meaning as in present-day English, but rather as often in Elizabethan English, where he did say meant nothing more than he said. Did may be added to two verbs: lo did nek ama ni estima la, and be used by itself: lo ama la, kom lo did sen unesmi marita he loves her as he did his first wife.
# LMIs in Control/pages/Inverse Problem of Optimal Control

LMIs in Control/pages/Inverse Problem of Optimal Control
In some cases, it is needed to solve the inverse problem of optimal control within an LQR framework. In this inverse problem, a given controller matrix needs to be verified for the system by assuring that it is the optimal solution to some LQR optimization problem that is controllable and detectable. In other words: in this inverse problem, the controller is known and the LQR gain matrices are to be calculated such that the controller is the optimal solution.

## The System

The system is a linear time-invariant system, that can be represented in state space as shown below:
    ${\begin{aligned}{\dot {x}}&=Ax+Bu,\\z&={\begin{bmatrix}Q^{1/2}&0\\0&R^{1/2}\end{bmatrix}}{\begin{bmatrix}x\\u\end{bmatrix}}\end{aligned}}$
where $x\in R^{n},y\in R^{l},z\in R^{m}$ represent the state vector, the measured output vector, and the output vector of interest, respectively, $w\in R^{p}$ is the disturbance vector, and $A,B,C,Q,R$ are the system matrices of appropriate dimension. To further define: $x$ is $\in R^{n}$ and is the state vector, $A$ is $\in R^{n*n}$ and is the state matrix, $B$ is $\in R^{n*r}$ and is the input matrix, $w$ is $\in R^{r}$ and is the exogenous input, $C,Q,R$ is $\in R^{m*n}$ and are the output matrices, and $y$ and $z$ are $\in R^{m}$ and are the output and the output of interest, respectively.

## The Data

The matrices $A,B,C$ that define the system, and a given controller $K$ for which the inverse problem is to be solved.

## The Optimization Problem

In this LMI, the following cost function is to be minimized for a given controller K by finding an optimal input:
    ${\begin{aligned}\int _{0}^{\infty }z^{T}zdt\end{aligned}}$
the solution of the problem can be formulated as a state feedback controller given as:
    ${\begin{aligned}K&=-R^{-1}B^{T}P,\\A^{T}P+PA-PBR^{-1}B^{T}P+Q&=0\end{aligned}}$

## The LMI: Inverse Problem of Optimal Control

the inverse problem of optimal control is the following: Given a matrix $K$, determine if there exist $Q\geq 0$ and $R>0$, such that $(Q,A)$ is detectable and $u=Kx$ is the optimal control for the corresponding LQR problem. Equivalently, we seek $R>0$ and $Q\geq 0$ such that there exist $P$ nonnegative and $P_{1}$ positive-definite satisfying
    ${\begin{aligned}(A+BK)^{T}P+P(A+BK)+K^{T}RK+Q&=0\\B^{T}P+RK&=0\\A^{T}P_{1}+P_{1}A&<Q\end{aligned}}$

## Conclusion:

If the solution exists, then $K$ is the optimal controller for the LQR optimization on the matrices $Q$ and $R$

## Implementation

This implementation requires Yalmip and Sedumi.
https://github.com/rezajamesahmed/LMImatlabcode/blob/master/inverseprob.m

## Related LMIs

      * Multi-Criterion LQG]

## External Links

* LMI Methods in Optimal and Robust Control - A course on LMIs in Control by Matthew Peet.
* LMI Properties and Applications in Systems, Stability, and Control Theory - A List of LMIs by Ryan Caverly and James Forbes.
* LMIs in Systems and Control Theory - A downloadable book on LMIs by Stephen Boyd.
# LMIs in Control/pages/LMI for Attitude Control of BBT Missiles

LMI for Attitude Control of BTT Missles, Roll Channel
For this dynamic model of a Bank-to-Turn (BTT) missile, we will consider a simplified attitude system model for the roll channel in the system. We aim to achieve a non-rotating motion of missiles. It is worthwhile to note that the attitude control design for the pitch channel and the yaw/roll channel can be solved exactly in the same way while representing matrices of the system are different.
# LMIs in Control/pages/KBH theorem

WIP, Description in progress

## Kharitonov-Bernstein-Haddad (KBH) Theorem

Consider the set of matrices
${\mathcal {A}}=\{{\textbf {A}}={\begin{bmatrix}{\textbf {0}}_{(n-1)\times 1}&&{\textbf {1}}_{(n-1)\times (n-1)}\\-a_{0}&\cdots &-a_{n-1}\end{bmatrix}}|{\underline {a_{j}}}\leq a_{j}\leq {\overline {a_{j}}},j=0,1,2,\cdots ,n-1\}$
Every matrix in the set ${\mathcal {A}}$ is Hurwitz if and only if there exist ${\textbf {P}}_{i}\in mathbb{S}^{n},i=1,2,3,4$ where ${\textbf {P}}_{i}>0,i=1,2,3,4,$ such that
${\textbf {P}}_{i}A_{i}+{\textbf {A}}_{i}^{T}{\textbf {P}}_{i}<0,\quad i=1,2,3,4,$
where
${\textbf {A}}_{i}={\begin{bmatrix}{\begin{bmatrix}{\textbf {0}}_{(n-1)\times 1}&{\textbf {1}}_{(n-1)\times (n-1)}\end{bmatrix}}\\{\textbf {a}}_{i}\end{bmatrix}},\quad i=1,2,3,4,$
${\textbf {a}}_{1}=-[{\underline {a}}_{0}\quad {\underline {a}}_{1}\quad {\overline {a}}_{2}\quad {\overline {a}}_{3}\quad \cdots \quad {\underline {a}}_{n-4}\quad {\underline {a}}_{n-3}\quad {\overline {a}}_{n-2}\quad {\overline {a}}_{n-1}],$
${\textbf {a}}_{2}=-[{\underline {a}}_{0}\quad {\overline {a}}_{1}\quad {\overline {a}}_{2}\quad {\underline {a}}_{3}\quad \cdots \quad {\underline {a}}_{n-4}\quad {\overline {a}}_{n-3}\quad {\overline {a}}_{n-2}\quad {\underline {a}}_{n-1}],$
${\textbf {a}}_{3}=-[{\overline {a}}_{0}\quad {\underline {a}}_{1}\quad {\underline {a}}_{2}\quad {\overline {a}}_{3}\quad \cdots \quad {\overline {a}}_{n-4}\quad {\underline {a}}_{n-3}\quad {\underline {a}}_{n-2}\quad {\overline {a}}_{n-1}],$
${\textbf {a}}_{3}=-[{\overline {a}}_{0}\quad {\underline {a}}_{1}\quad {\underline {a}}_{2}\quad {\overline {a}}_{3}\quad \cdots \quad {\overline {a}}_{n-4}\quad {\underline {a}}_{n-3}\quad {\underline {a}}_{n-2}\quad {\overline {a}}_{n-1}],$
${\textbf {a}}_{4}=-[{\overline {a}}_{0}\quad {\overline {a}}_{1}\quad {\underline {a}}_{2}\quad {\underline {a}}_{3}\quad \cdots \quad {\overline {a}}_{n-4}\quad {\overline {a}}_{n-3}\quad {\underline {a}}_{n-2}\quad {\underline {a}}_{n-1}].$
WIP, additional references to be added

## External Links

A list of references documenting and validating the LMI.
# LMIs in Control/pages/L2 gain of Lure systems
## The System

    ${\begin{aligned}{\dot {x}}(t)&=Ax(t)+B_{p}p(t)+B_{w}w(t),\\z(t)&=C_{z}x(t)\\p_{i}(t)&=\phi _{i}(q_{i}(t)),i=1,\dots ,n_{p}\\q&=C_{q}x,\\0&\leq \sigma \phi _{i}(\sigma )\leq \sigma ^{2}\ \forall \sigma \in \mathbb {R} \end{aligned}}$

## The Data

The matrices $A,B_{p},B_{w},C_{q},C_{z}$.

## The Optimization Problem:

The following semi-definite problem should be solved.
       ${\begin{aligned}&\min _{\{P\succ 0,\Lambda =diag(\lambda _{1},\dots ,\lambda _{n_{p}})\succeq 0,T=diag(\tau _{1},\dots ,\tau _{n_{p}})\succeq 0\}}\gamma ^{2}\;\\&\quad \quad \quad \quad \quad \quad \quad \quad s.t.\quad {\begin{bmatrix}A^{\top }P+PA+C_{z}^{\top }C_{z}&PB_{p}+A^{\top }C_{q}^{\top }\Lambda +C_{q}^{\top }T&PB_{w}\\B_{p}^{\top }P+\Lambda C_{q}A+TC_{q}&\Lambda C_{q}B_{p}+B_{p}^{\top }C_{q}^{\top }\Lambda -2T&\Lambda C_{q}B_{w}\\B_{w}^{\top }P&B_{w}^{\top }C_{q}^{\top }\Lambda &-\gamma ^{2}I\end{bmatrix}}\preceq 0\\\end{aligned}}$

## Implementation

https://github.com/mkhajenejad/Mohammad-Khajenejad/commit/12a7039f9e3d966e24b43fd58a3cce3725282ed2

## Conclusion

The value function returns the square of the smallest provable upper bound on the ${\mathcal {L}}_{2}$ gain of the Lure's system.

## Remark

The Lyapunov function which is used to proof is similar to the one for the systems with unknown parameters.

## External Links

* LMI Methods in Optimal and Robust Control - A course on LMIs in Control by Matthew Peet.
* LMI Properties and Applications in Systems, Stability, and Control Theory - A List of LMIs by Ryan Caverly and James Forbes.
* LMIs in Systems and Control Theory - A downloadable book on LMIs by Stephen Boyd.
# LMIs in Control/pages/KYP Lemma (Bounded Real Lemma)

KYP Lemma (Bounded Real Lemma)
The Kalman–Popov–Yakubovich (KYP) Lemma is a widely used lemma in control theory. It is sometimes also referred to as the Bounded Real Lemma. The KYP lemma can be used to determine the $H_{\infty }$ norm of a system and is also useful for proving many LMI results.

## The System

       ${\begin{aligned}{\dot {x}}(t)&=Ax(t)+Bu(t)\\y(t)&=Cx(t)+Du(t)\\x(0)&=x_{0}\end{aligned}}$
where $x(t)\in \mathbb {R} ^{n}$, $y(t)\in \mathbb {R} ^{m}$, $u(t)\in \mathbb {R} ^{q}$, at any $t\in \mathbb {R}$.

## The Data

The matrices $A,B,C,D$ are known.

## The Optimization Problem

The following optimization problem must be solved.
    ${\begin{aligned}&{\underset {\gamma ,\;X}{\operatorname {minimize} }}\quad \gamma \\&\operatorname {subject\;to} &X>0\\&&{\begin{bmatrix}A^{T}X+XA&XB\\B^{T}X&-\gamma I\end{bmatrix}}+{\frac {1}{\gamma }}{\begin{bmatrix}C^{T}\\D^{T}\end{bmatrix}}{\begin{bmatrix}C&D\end{bmatrix}}<0\\\end{aligned}}$

## The LMI: The KYP or Bounded Real Lemma

Suppose ${\hat {G}}(s)(A,B,C,D)$ is the system. Then the following are equivalent.
       $1)\quad \left\|G\right\|_{H_{\infty }}\leq \gamma$
       $2)\quad {\text{There exists a}}\;X>0\;{\text{such that}}$
             ${\begin{bmatrix}A^{T}X+XA&XB\\B^{T}X&-\gamma I\end{bmatrix}}+{\frac {1}{\gamma }}{\begin{bmatrix}C^{T}\\D^{T}\end{bmatrix}}{\begin{bmatrix}C&D\end{bmatrix}}<0$

## Conclusion:

The KYP Lemma can be used to find the bound $\gamma$ on the $H_{\infty }$ norm of a system. Note from the (1,1) block of the LMI we know that $A$ is Hurwitz.

## Implementation

Since the KYP lemma shown above is nonlinear in gamma, in order to implement it in MATLAB we must first linearize it by using the Schur Complement to arrive at the dual formulation below:
       ${\begin{bmatrix}A^{T}X+XA&XB&C^{T}\\B^{T}X&-\gamma I&D^{T}\\C&D&-\gamma I\end{bmatrix}}<0$.
This dual KYP LMI is now linear in both $X$ and $\gamma$.
This implementation requires the use of Yalmip and Sedumi.
https://github.com/eoskowro/LMI/blob/master/KYP_Lemma_LMI.m

## Related LMIs

Positive Real Lemma

## External Links

A list of references documenting and validating the LMI.
* LMI Methods in Optimal and Robust Control - A course on LMIs in Control by Matthew Peet.
* LMI Properties and Applications in Systems, Stability, and Control Theory - A List of LMIs by Ryan Caverly and James Forbes.
* LMIs in Systems and Control Theory - A downloadable book on LMIs by Stephen Boyd.
# LMIs in Control/pages/KYP Lemma without Feedthrough
## The Concept

It is assumed in the Lemma that the state-space representation (A, B, C, D) is minimal. Then Positive Realness (PR) of the transfer function C(SI − A)$^{-1}$B + D is equivalent to the solvability of the set of LMIs given in this page. Consider now the following scalar example, where (A, B, C, D)=(−α, 0, 0, 1), with α > 0. The transfer function is H(s) = 0 that is PR

## The System

Consider a contiuous-time LTI system, ${\mathcal {G}}:{\mathcal {L}}_{2e}\rightarrow {\mathcal {L}}_{2e}$, with minimal state-space relization (A, B, C, 0), where ${\mathcal {A}}\in {\mathcal {R}}^{n\times n},{\mathcal {B}}\in {\mathcal {R}}^{n\times m},$ and ${\mathcal {C}}\in {\mathcal {R}}^{m\times n},$.
    ${\begin{aligned}{\dot {x}}(t)&=Ax(t)+Bu(t),\\y(t)&=Cx(t)\\\end{aligned}}$

## The Data

The matrices The matrices $A,B$ and $C$

## LMI : KYP Lemma without Feedthrough

The system ${\mathcal {G}}$ is positive real (PR) under either of the following equivalet necessary and sufficient conditions.
    1. There exists $P\in {\mathcal {S}}^{n},$ where $p>0$ such that
       ${\begin{aligned}PA+A^{T}P\geq 0\\PB=C^{T}\end{aligned}}$
    2. There exists $Q\in {\mathcal {S}}^{n},$ where $Q>0$ such that
       ${\begin{aligned}AQ+QA^{T}\geq 0\\B=QC^{T}\end{aligned}}$
This is a special case of the KYP Lemma for QSR dissipative systems with Q = 0, Q = 0.5 and R = 0.
The system ${\mathcal {G}}$ is strictly positive real (SPR) under either of the following equivalet necessary and sufficient conditions.
    1. There exists $P\in {\mathcal {S}}^{n},$ where $p>0$ such that
       ${\begin{aligned}PA+A^{T}P<0\\PB=C^{T}\end{aligned}}$
    2. There exists $Q\in {\mathcal {S}}^{n},$ where $Q>0$ such that
       ${\begin{aligned}AQ+QA^{T}<0\\B=QC^{T}\end{aligned}}$
This is a special case of the KYP Lemma for QSR dissipative systems with Q = ε1, Q = 0.5 and R = 0. where ε $\in {\mathcal {R}}_{>0}.$

## Conclusion:

If there exist a positive definite $P$ for the the selected Q,S and R matrices then the system ${\mathcal {G}}$ is Positive Real.

## Implementation

Code for implementation of this LMI using MATLAB. https://github.com/VJanand25/LMI

## Related LMIs

KYP Lemma
State Space Stability Discrete Time KYP Lemma with Feedthrough

## References

1. J. C. Willems, “Dissipative dynamical systems - part I: General theory,” Archive Rational Mechanics and Analysis, vol. 45, no. 5, pp. 321–351, 1972.
2. D. J. Hill and P. J. Moylan, “The stability of nonlinear dissipative systems,” IEEE Transac- tions on Automatic Control, vol. 21, no. 5, pp. 708–711, 1976.
3. LMI Properties and Applications in Systems, Stability, and Control Theory, by Ryan James Caverly1 and James Richard Forbes2
4. Brogliato B., Maschke B., Lozano R., Egeland O. (2007) Kalman-Yakubovich-Popov Lemma. In: Dissipative Systems Analysis and Control. Communications and Control Engineering. Springer, London
# LMIs in Control/pages/L2 gain of systems with multiplicative noise
## The System

    ${\begin{aligned}x(k+1)&=Ax(k)+B_{w}w(k)+\sum _{i=1}^{L}(A_{i}x(k)+B_{w,i}w(k))p_{i}(k),\quad x(0)=0,\\z(k)&=C_{z}x(k)+D_{zw}w(k)+\sum _{i=1}^{L}(C_{z,i}x(k)+D_{zw,i}w(k))p_{i}(k),\end{aligned}}$
where $p(0),p(1),\dots$, are independent, identically distributed random variables with $Ep(k)=0,Ep(k)p^{\top }(k)=\Sigma =diag(\sigma _{1},\dots ,\sigma _{L})$ and $x(0)$ is independent of the process $p$.

## The Data

The matrices $A,B_{w},\{A_{i}.B_{w,i}\}_{i=1}^{L},C_{z},D_{zw},\{C_{z,i},D_{zw,i}\}_{i=1}^{L},\{\sigma _{i}\}_{i=1}^{L}$.

## The LMI:

       ${\begin{aligned}&\min _{\{P\succ 0,\gamma ^{2}\}}\gamma ^{2}\\&\quad s.t.{\begin{bmatrix}A&B_{w}\\C_{z}&D_{zw}\end{bmatrix}}^{\top }{\begin{bmatrix}P&0\\0&I\end{bmatrix}}{\begin{bmatrix}A&B_{w}\\C_{z}&D_{zw}\end{bmatrix}}-{\begin{bmatrix}P&0\\0&\gamma ^{2}I\end{bmatrix}}+\sum _{i=1}^{L}\sigma _{i}^{2}{\begin{bmatrix}A_{i}&B_{w,i}\\C_{z,i}&D_{zw,i}\end{bmatrix}}^{\top }{\begin{bmatrix}P&0\\0&I\end{bmatrix}}{\begin{bmatrix}A_{i}&B_{w,i}\\C_{z,i}&D_{zw,i}\end{bmatrix}}^{\top }\preceq 0\end{aligned}}$

## Implementation

https://github.com/mkhajenejad/Mohammad-Khajenejad/commit/a34713575cd8ae9831cb5b7eb759d0fd45a8c37f

## Conclusion

The optimal $\gamma$ returns an upper bound on the ${\mathcal {L}}_{2}$ gain of the system. .

## Remark

It is straightforward to apply scaling method [Boyd, sec.6.3.4] to obtain component-wise results.

## External Links

* LMI Methods in Optimal and Robust Control - A course on LMIs in Control by Matthew Peet.
* LMI Properties and Applications in Systems, Stability, and Control Theory - A List of LMIs by Ryan Caverly and James Forbes.
* LMIs in Systems and Control Theory - A downloadable book on LMIs by Stephen Boyd.
# Pinyin/Prose

Proses is a literary form, the opposite are verses and poems. Proses can be broad and narrow in explanations, the method of which can easily distinguish and define broad sense prose, is that it has a loose structure. That is to say, to delete the overlapping features of other literary forms, use common grammar structure, do not fuzz over rhymes, do not fuzz over forms. There is no any restraints and limitations to the way words are organized. This can be call proses. Furthermore, narrow sense proses are singly within the area of literature, the compositions are loose in structures and do not rhyme. Proses in literature, have different historical compositions, that have different changing sequences of thoughts of different decades, and because of the freedom bring about by loose structure, the thoughts expressed by prose compositions usually have a rich and satisfactory characteristics.
Sǎnwén shì yīzhǒng wénxué xíngshì, yǔzhī xiāngduì de chēngwéi yùnwén huò shīwén. Sǎnwén kě fēnwéi guǎngyì hé xiáyì de jiěshì, qízhōng, guǎngyì sǎnwén zuì róngyì biànshì hé dìngyì de fāngshì, shì “sōngsǎn” de jiégòu. Yějiùshì shuō, kòuchú qítā wénxué xíngshì chóngdié bùfèn, yùnyòng pǔtōng yǔfǎ jiégòu, bù jiǎngjiū yīnyùn, bù jiǎngjiū páibǐ, méiyǒu rènhé shùfù hé xiànzhì de wénzì shūlǐ fāngshì, dōu kě chēngwéi sǎnwén. Chúcǐ, xiáyì sǎnwén shì dānzhǐ wénxué fànchóu nèi, jiégòu sōngsǎn de fēi-yùnwén zuòpǐn. Wénxué zhuānzhǐ de sǎnwén, lìdài zuòpǐn yǒuzhe gè shídài bùtóng liúbiàn de màiluò, ér zhèng yīnwèi sōngsǎn dàilái de zìyóu, sǎnwén zuòpǐn biǎodá chū de sīxiǎng tōngcháng yǒuzhe fēngfù hé yuánmǎn de tèsè.
# Pinyin/Properly

See also
# Pinyin/Protein

Search Pinyin protein
Proteins are essential nutrients for the human body. They are one of the building blocks of body tissue, and can also serve as a fuel source. As a fuel, proteins contain 4 kcal/g, just like carbohydrates and unlike lipids, which contain 9 kcal/g. The most important aspect and defining characteristic of protein from a nutritional standpoint is its amino acid composition.
Dànbáizhì shì réntǐ bìxū d yíngyǎngsù, shì shēntǐ zǔzhī d jīchǔ zhīyī, kěyǐ zuòwéi ránliào. Zuòwéi ránliào, dànbáizhì hányǒu 4 kcal/g, jiù xiàng tahnshuiiwuh, ér bùtóng yú zhīfáng hán 9 kcal/g. Cóng yíngyǎngxué guāndiǎn láikàn, dànbáizhì zuì zhòngyào fāngmiàn hé jièdìng tèzhēng shì ta d ānjīsuān chéngfèn.
# Pinyin/Proposal

Search proposal
# Pinyin/Protection

See also
# Pinyin/Protect

See also
# Muggles' Guide to Harry Potter/Magic/Expelliarmus
## Overview

The Expelliarmus spell is the Disarming spell; it knocks the victim's weapon (usually a wand) out of his/her hands. It can also be used to knock other items out of the target's hands, but we seldom see this use. When it is performed with more power, one can blast the opponent off their feet.

## Extended Description

We see this spell first used by Professor Snape in the Dueling Club; Snape casts it on Professor Lockhart, disarming him and throwing him to the end of the platform on which they are demonstrating. Later in the same book, Harry casts the same spell on Draco Malfoy, causing the book that he is holding, Riddle's diary, to fly out of his hand. From this point on, it is used quite often when there is a duel between wizards; if you can get your opponent to lose his wand, you do have a great advantage over him. Later on, Draco Malfoy would use this spell on Dumbledore, unintentionally earning ownership of the Elder Wand.
In Harry Potter and the Deathly Hallows, Harry uses the spell against a Dark Eater during the chase, which causes them to recognize him. Because of this, Lupin warns Harry against the further use of this spell, suggesting that he use more powerful defensive magic.
It is, interestingly, the spell Harry uses to defeat Voldemort in the final book. Harry was, by this point, aware of his own mastery of the Elder Wand and was effectively immune to Voldemort's magic by this, but he needed to cast at least one spell, and it was the Disarming Charm which Harry chose to use. This contrasts with Lupin's earlier warning against using it too often.

## Analysis

As mentioned, at one point Harry does use this spell to get someone to drop a book; as the diary is obviously not a weapon, we can see that this spell will act on anything that is held in the hand, not just a weapon or wand. It's possible that there is a mental part of this spell as well, envisioning the specific object that you want to have it act on; alternately, it could work on whatever you are pointing your wand at. In Harry Potter and the Prisoner of Azkaban, in the Shrieking Shack, Harry, Ron, and Hermione all cast this spell at the same time, and it not only pulls Snape's wand out of his hand, but also throws him against the wall. This leads us to believe that the power of the spell is additive when cast by multiple wizards on the same object.
Despite the fact that the charm can do things other than disarm, its main function is to get an attacker to drop his wand; as such, it is referred to as the Disarming spell. This is confirmed by the word's origin, "expelliar" meaning to "be expelled" and "armus" meaning "weapon".

## Questions

Study questions are meant to be left for each student to answer; please don't answer them here.
* In the seventh book, it appears that this is taken to be Harry's "signature spell". Why would that be?
* In the seventh book, we learn that there can be other effects from the use of this spell. What would those be?