text:
By Vikram K. Mulligan, Baker laboratory. Documentation written 4 April 2014, and last updated on 11 March 2016.

## Short summary

GeneralizedKIC (short for Generalized Kinematic Closure) is a generalization of the existing kinematic closure machinery. The generalized version works with arbitrary backbones, and with loops that go through side-chains, ligands, etc. It has also been implemented with full RosettaScripts support, making it easy to incorporate into more complicated protocols. The RosettaScripts interface gives full control over the inner workings of the algorithm, making it quite a powerful and versatile tool.

## Usage cases

Generalized KIC is useful for the following situations:

• Given an arbitrary unbranched, covalently-connected part of a structure, with well-defined, fixed starting and ending points, one wishes to sample alternative conformations that preserve ideal geometry. An example would be a case in which the N- and C-termini of a protein were linked by a disulfide bond, and one wished to sample alternative conformations for a short stretch of the N-terminal backbone, the disulfide bond, and a short stretch of the C-terminal backbone, subject to the condition that the disulfide remains closed.
• Given a part of a structure that is open but which ought to be covalently connected, with well-defined, fixed starting and ending points, one wishes to sample conformations that close the segment. For example, given unjoined antiparallel beta strands forming a two-strand sheet, one might wish to sample hairpin conformations for a loop connecting the strands. This closure can be through side-chains or ligands, too: given two segments containing metal-binding residues, one might want to sample conformations that allow a metal to be bound with ideal geometry.
• Given a covalent linkage between distant parts of a FoldTree and a structural perturbation that pulls the atoms involved in the covalent linkage apart, one wishes to find compensatory structural perturbations that maintain the closed covalent geometry while minimally altering the structure.

The above scenarios tend to be sub-problems of more complicated problems, particularly involving heavily cross-linked molecules for which one might wish to sample many conformations.

## Input and output

GeneralizedKIC takes as input a pose that contains a covalently-connected chain of residues (a "loop", where the loop need not be connected solely by conventional backbone connections) to which the algorithm will be applied. The chain of residues need not start out in a sensible conformation, nor need it consist of canonical alpha-amino acids.

GeneralizedKIC will return a pose in which the loop in question has been put in a new, closed conformation (subject to user-specified sampling methods). Only the loop residues (and any "tail" residues specified) will be moved by this mover; all other residues will remain in their starting positions. The FoldTree is disregarded by the GeneralizedKIC mover, and will not be altered by it. (Note: if a pre-selection mover is defined as described in the GeneralizedKIC selector documentation, this mover can alter geometry outside of the loop to be closed, or could alter the FoldTree.)

If no closed solution could be found, GeneralizedKIC will return the original input pose. If a ContingentFilter has been supplied, its value will be set to "false" in this case (and "true" otherwise). By default, GeneralizedKIC will set the mover status to failure if no solution is found, but the user can override this with dont_fail_if_no_solution_found=true.

## General workflow

In general, one must:

1. Build or import a structure, which must include the loop residues to be closed (though these need not be in a closed conformation to start).
2. Ensure that covalent linkages have been declared with the DeclareBond mover.
3. Set the GeneralizedKIC options.
4. Define the loop to be closed, and set pivots.
5. Set up one or more GeneralizedKIC perturbers, which determine how loop conformations will be sampled.
6. Define one or more GeneralizedKIC filters.
7. Set the GeneralizedKIC selector used to choose a solution from among those found (and an optional pre-selection mover).

These steps are discussed in detail in the next section.

## Detailed workflow

1. Build or import a structure. GeneralizedKIC cannot add residues or geometry. If one wants to build a new loop, for example, one must add the new loop residues before calling GeneralizedKIC. (In RosettaScripts, the PeptideStubMover is useful for building geometry from scratch or for adding residues to existing geometry.)

2. Ensure that covalent linkages have been declared. GeneralizedKIC will move atoms about to ensure ideal geometry, but cannot declare new chemical bonds. (In RosettaScripts, the DeclareBond mover lets Rosetta know that certain residues are covalently attached to one another.)

3. Set the GeneralizedKIC options (number of closure attempts, whether the algorithm should accept the first successful closure or choose from all successful closure attempts, etc.). In RosettaScripts, this is handled inside the block as follows:

<MOVERS>
...
<GeneralizedKIC name="&string" low_memory_mode="(false &bool)" closure_attempts="(100 &int)" stop_if_no_solution="(0 &int)" stop_when_n_solutions_found="(0 &int)" selector="&string" selector_scorefunction="&string" selector_kbt="(1.0 &Real)" contingent_filter="&string" dont_fail_if_no_solution_found="(false &bool)" correct_polymer_dependent_atoms="(false &bool)">
...
</GeneralizedKIC>
...
</MOVERS>
The low_memory_mode option can be used to limit the amount of information about each solution found that is stored, in order to reduce memory consumption. See the note below for more details about the advantages and risks of this non-default mode. The closure_attempts parameter sets the number of times the algorithm will try to close the loop. A setting of 0 means that it will keep trying indefinitely. The stop_when_n_solutions_found option allows the algorithm to stop after finding at least N successful solutions, or, if this is set to 0, to keep going until it has done as many attempts as specified by closure_attempts; in either case a solution is then chosen by the selector. (Note that, because a single attempt returns up to 16 closure solutions, the selector will be applied even if stop_when_n_solutions_found is set to 1, since more than one solution might have been found in the first successful attempt). The selector flag is mandatory, and specifies the way in which a solution is chosen from among the successful solutions. The selector_scorefunction flag allows a separate scorefunction to be used by those selectors that select based on energy or score; this is recommended since score terms based on side-chain packing may produce poor results, since the GeneralizedKIC algorithm does not call the packer. Some selectors also take a temperature value, set by the selector_kbt option. See the GeneralizedKIC selector documentation for more details.
In some cases, the GeneralizedKIC mover will find no solution. This could be because no solution exists (e.g. if the loop is too short for the endpoint separation, or if there is geometry blocking any path between the endpoints), because the sampling method used was too restrictive, or because too few attempts were made. If this happens, the pose is left unaltered. If the loop geometry is open, it is useful to have a means of aborting the trajectory in this case. By default, the mover returns failure status, aborting the trajectory. This can be overridden by setting dont_fail_if_no_solution_found=true. A ContingentFilter can also be used to record whether the mover failed. The ContingentFilter is a specialized filter that has its value set by a mover. GeneralizedKIC can set the value of a ContingentFilter, specified using the contingent_filter flag, to true or false depending on whether the closure was successful or unsuccessful. Subsequent application of the filter, possibly at a later point, can then abort trajectories involving unsuccessful loop closure. As a final note, GeneralizedKIC can correct positions of atoms that depend on polymer bonds (such as amide protons, carbonyl oxygens, or N-methyl groups in N-methylated amino acids) to ideal positions if the correct_polymer_dependent_atoms option is set to "true".
4. Define a series of residues for the GeneralizedKIC closure problem. This must be an unbranched chain of residues with continuous covalent linkages, listed in order from one end of the chain to the other. When the GeneralizedKIC::apply() function is called, a continuous chain of atoms running through the selected residues is automatically chosen. Residues are specified with AddResidue tags within a GeneralizedKIC block. Pivot points must also be indicated explicitly, using the SetPivots tag. Pivots are atoms in the chain of atoms to be closed that are flanked by bonds whose dihedral values will be solved for analytically by the closure algorithm in order to close the loop. Currently, due to hard-coded assumptions in the kinematic closure numerical library, the first pivot must be the second atom in the chain to be closed, and the last pivot must be the second-to-last atom in the chain to be closed. This restriction will be eliminated in a future version of GeneralizedKIC.
Optionally, additional "tail" residues can also be listed. These are residues that are either connected directly to the loop to be closed, or connected indirectly to this loop through other tail residues, and which move with the loop to be closed. The AddTailResidue tag can be used to specify these. Order is not important for the AddTailResidue flag.

<MOVERS>
...
<GeneralizedKIC ...>
...
...
<SetPivots res1="(&int)" atom1="&string" res2="(&int)" atom2="&string" res3="(&int)" atom3="&string" />
...
</GeneralizedKIC>
...
</MOVERS>
For example, if one were closing a loop consisting of residues ALA44, CYS45, LYS46, CYS47, CYS23, ASP22, and PHE21, where CYS47 and CYS23 were linked by a disulfide bond, and where CYS45 were linked to CYS12 by another disulfide such that CYS12 was expected to move with CYS45, one would write:
<MOVERS>
...
<GeneralizedKIC ...>
<SetPivots res1="4" atom1="CA" res2="23" atom2="SG" res3="21" atom3="CA" />
...
</GeneralizedKIC>
...
</MOVERS>
From the above example, we can see that loop segments may run backwards or forwards, or may involve residues that are far apart in linear sequence provided they are covalently linked. Note that while the sequence of residues matter, the overall direction of the loop does not: we could just as happily have added residues in the reverse order (21->22->23->47->46->45->44). In this example, we have arbitrarily chosen CYS23's SG atom as the middle pivot point, though any atom in the chain that is flanked by bonds that can rotate freely could have been chosen.
5. Define one or more GeneralizedKIC perturbers. Each perturber samples conformation space for each closure attempt. See GeneralizedKIC perturbers for details.

6. Define one or more GeneralizedKIC filters. Filters are applied after each closure attempt, and eliminate solutions that don't meet some criterion. See GeneralizedKIC filters for details.

7. Set the GeneralizedKICselector. The selector chooses a single solution from the set of solutions found that pass all of the filters. See the documentation on GeneralizedKIC selectors for details, including details on the optional pre-selection mover (a mover applied to all solutions passing filters prior to application of the selector).

That's it! You should be happily closing loops, now.

## Algorithm details

The kinematic closure algorithm is described in detail in Coutsias et al. (2004). J. Comput. Chem. 25(4):510-28. and in Mandell et al. (2009). Nat. Methods. 6(8)551-2.. Briefly, given a rigid-body transform describing the position and orientation of the residue immediately following a loop relative to a residue immediately preceding a loop, a loop with N degrees of freedom now has N-6 effective degrees of freedom, the remaining 6 degrees of freedom being determined by the condition that the loop be closed. This means that all but 6 degrees of freedom of the loop (e.g. all but 6 chain dihedral angles) may be sampled, perturbed, or set as one sees fit, and the remaining 6 degrees of freedom may be solved for analytically to impose the closure condition. The GeneralizedKIC algorithm does just this, allowing the user to sample all but 6 degrees of freedom of the loop, then solve analytically for the remainining 6 (the dihedral angles flanking 3 pivot atoms) to ensure that the loop is closed. For any given set of values for the other N-6 degrees of freedom, there may be anywhere from 0 to 16 solutions to the system of equations solved by the kinematic closure algorithm.

## Low-memory mode

By default, all solutions found by GenKIC are stored as full poses until one is selected by the selector. Unfortunately, this means that if instructed to store many solutions, GenKIC can exhaust available memory. As of 13 March 2016, a low-memory mode has been added ("low_memory_mode=true"). In this mode, only loop degree-of-freedom values are stored prior to selection. The drawback, however, is it can be risk to use low-memory mode with a preselection mover. Any changes to the pose made by the preselection mover will not be stored, so the preselection mover will have to be re-applied to the selected pose after selection. This costs additional time. In addition, if the preselection mover has any stochastic component to its behaviour, then the second application may not produce identical results to the first. This means that what might have been the lowest-energy pose when the preselection mover was first applied could now be a relatively high-energy pose, for example. For this reason, be judicious in the use of low-memory mode. The preferred course is to stop GenKIC when N solutions have been found to limit memory usage.

## Example RosettaScripts

This example creates a 10-residue cyclic peptide with a disulfide bond between the N- and C-termini. It then defines a loop starting at residue 3, going backwards to residue 1, through the disulfide to residue 10, and back to residue 5. This loop is closed by kinematic closure, with loop conformations sampled. A bump check filter is applied (AddFilter type="loop_bump_check"), and a solution chosen randomly (selector="random_selector") from the solutions found.

<ROSETTASCRIPTS>
<SCOREFXNS>
<ScoreFunction name="tala" weights="talaris2013" symmetric="0" />
</SCOREFXNS>
<FILTERS>
</FILTERS>
<MOVERS>
<PeptideStubMover name="pep_stub" reset="1">
<Append resname="CYD" />
<Append resname="ALA" />
<Append resname="ALA" />
<Append resname="ALA" />
<Append resname="ALA" />
<Append resname="ALA" />
<Append resname="ALA" />
<Append resname="ALA" />
<Append resname="ALA" />
<Append resname="CYD" />
</PeptideStubMover>

<SetTorsion name="tor1">
<Torsion residue="ALL" torsion_name="phi" angle="-64.7"/>
<Torsion residue="ALL" torsion_name="psi" angle="-41.0"/>
<Torsion residue="ALL" torsion_name="omega" angle="180"/>
</SetTorsion>

<SetTorsion name="tor2">
<Torsion residue="pick_atoms" angle="random">
<Atom1 residue="1" atom="N"/>
<Atom2 residue="1" atom="CA"/>
<Atom3 residue="1" atom="CB"/>
<Atom4 residue="1" atom="SG"/>
</Torsion>
<Torsion residue="pick_atoms" angle="random">
<Atom1 residue="10" atom="N"/>
<Atom2 residue="10" atom="CA"/>
<Atom3 residue="10" atom="CB"/>
<Atom4 residue="10" atom="SG"/>
</Torsion>
</SetTorsion>

<DeclareBond name="bond" res1="1" atom1="SG" res2="10" atom2="SG"/>

<GeneralizedKIC name="genkic" closure_attempts="200" stop_when_n_solutions_found="0" selector="random_selector">
<SetPivots res1="3" atom1="CA" res2="1" atom2="SG" res3="5" atom3="CA" />
<CloseBond prioratom_res="10" prioratom="CB" res1="10" atom1="SG" res2="1" atom2="SG" followingatom_res="1" followingatom="CB" bondlength="2.05" angle1="103" angle2="103" randomize_flanking_torsions="true" />
</GeneralizedKIC>

</MOVERS>
<APPLY_TO_POSE>
</APPLY_TO_POSE>
<PROTOCOLS>
</ROSETTASCRIPTS>


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I. F. Quinby Horatio Nelson Robinson.

# A new treatise on the elements of the differential and integral calculus online

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65, Taylor's Theorem may be enunciated as follows:
When a function F{x -\- h) of the algebraic sum of two varia-
bles can be developed into a series arranged according to the
ascending powers of either taken as the leading variable, the
first term is what the function becomes when this variable is
made equal to zero ; the second term is the first power of the
leading variable multiplied by the first diflorential co-efScient
of the first term taken with respect to the other variable ;
the third term is the second power of the leading variable
divided by 1 X 2, and this quotient multiplied by the second
differential co-efficient of the first term ; and tlie {n + 1)'*, or
general term, is the n'* power of the leading variable divided
by the product of the natural numbers from 1 to n inclusive,
and this quotient multiplied by the w'* differential co-efficient
of the first term.

66* In Taylor^s Formula, the co-efficients of the different
powers of the leading variable are functions of the other
variable. When one or more of these functions are such, that,
for a particular value of the second variable, they become in-
finite, the formula fails to give the development of the origi-
nal function for that value of the second variable ; for then
the function ceases to depend on the second variable, and is a

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DIFFERENTIAL CO-EFFICIENTS. 85

function of the first variable alone, and will not necessarily be
infinite for the assigned value of the second variable.
For example, if we have

F{x) = ^,^ZI^^
then F(x + h) = ^(x — a + A).

When rr = a, F{x) = 0, and the first and all the higher
difibrential co-efficients of F(x) become infinite for this partic-
ular value of X ; while, for this value. F{x -\- h) = \/h.

It will be observed that there is a marked difference be-
tween the failing cases for Maclaurin^s and Taylor's Theorems.
When Maclaurin's fails for one value of the variable {x =. 0),
it fails for all ; whereas Taylor's may fail for one value of the
second variable, but give the true development of the function
for all other values.

67m If a function becomes infinite for a finite value of the
variable, its difl'erential co-efficient will be infinite at the same
time. In the case of an algebraic function, this follows from
the fact that such function can become infinite for a finite
value of the variable, only when it is in the forni of a fraction
whose denominator reduces to zero. But the denominator of a
fraction never disappears in the process of differentiation :
hence, if the function has a vanishing denominator, so will its
differential co-efficient. In the case of transcendental func-
tions, it is only by the examination of the different forms that
the truth of this proposition can be established. Thus, in the
logarithmic function y = Zx, y becomes infinite for a? = ;

-J =1 - is also infinite for this value of x ; and for the expo-
ox x

1

nential function y =ia''\ which, if a>l, becomes infinite when

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86 DIFFERENTIAL CALCULUS,

a; = 0, the differential coefficient is -^^ = — - ^a^y 'which is

cue QC

infinite when cc = 0.

The circular functions tan. x, cot. x, sec. x, cosec. x, which
may become infinite for finite values of x, when expressed in
terms of sin. x, cos. x^ are fractional forms to which the reason-
ing in reference to algebraic functions applies.

If a function becomes infinite for an infinite value of the
variable, it does not follow that the differential co-efficient
becomes infinite at the same time.

Thus, in the example y = Zx, ^^ == -, and y is infinite when

dX X

a; = Qo ; but - =r for this value of x,

68. It was remarked in Art. 62, that, unless F{x) and
F{X'-\' h) are such that

i?'(0) + a;Ji"(0) + gj"'(0) + ...,

F{x) + hF'{x) + l'-^F"{x)Jr - .,

give rise to converging series, the formulas of Maclaurin and
Taylor will not serve for the expansion of these functions.

A series in general is a succession of quantities any one of
which is derived, according to a fixed law, from one or more
of those which precede it. If t^07^b^^2?^3j • • • '^n? ^re such
quantities called the terms of the series, then we have

Sn = ^0 + ^1 +^2 + ^3 + • . ''^n-X

for the sum of the first n terms. When this sum approaches
indefinitely a finite limit S, as n continually increases, the
series is said to be converging, and the limit in question is
called the sum of the series ; but, if the sum S„ does not thus

Digitized by

DIFFERENTIAL CO-EFFICIENTS. 87

approach any fixed limit as n increases indefinitely, the series
is said to be diverging, and has no sum.
The geometrical series

a, ar, ar'^, . . . ar"j

having ar** for its general term, has for its sum

1 — r" fi nr**

" ^ ^ 1— r 1 — r 1 — r

It is evident that, as n increases, this sum converges towards

a

the fixed limit if r is less than 1 ; and that, on the con-

1 — r

trary, as n increases, the sum also increases indefinitely if r is

greater than 1.

We are assured of the convergence of the series

when, as n increases, the sum

Sn = Uo + Ui+U2+ . . . U„_i

converges to a fixed limit Sf and when, at the same time, the
differences

Srt + l — Sn = Un, S„^2 — S„ = U^ + U„^l • • • J

vanish when n is made infinite.

The limits assigned this work do not permit an investiga-
tion of the rules by which, in many cases, the convergence or
divergence of a series may be ascertained.

69. Admitting that F(x) can be expanded into a series
arranged according to the ascending integral powers of Xj
Maclaurin's Theorem may be demonstrated as follows : —

Assume

F{x) = Ao + A,x<' + A,x' + . . . + A,xP
in which A^, A^, A^ ^ , » , Ao not contain x, and the exponents

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88 DIFFERENTIAL CALCULUS,

a,b,c . . . J are written in the order of their magnitude, a being
the least ; then, by successive differentiation, we have
F'{x) = aA^x"-^ + bA.,x^-^ + . . . +pA^x^-\
F''{x)=a{a — l)A^x''-^ + b{b - l)A.,x'-^ + .. .

+ p{p^l)A„xP-\
F'''{x) = a(a — 1) (a — 2)AiX^-' + b{b — 1) (6 — 2) A^x^-^
+ ...+pip-l){p-2)A,xP-\

The assumed and all the following equations, being true for
all values of x, make x = 0; then, since F{0), i^'(O), F''{0) . . . ,
i^puld in general reduce neither to nor to oo, we should have

J, = F{0), a=l, A = F'{0), 6 = 2,

1.2 ' ' ' 1.2.3

F{x)=F{0) + xF'iO) + ^F"iO) + ^F"'{0) + ...,

which is identical with the formula of Art, 62.

70. Xaylor^s Theorem also admits of the following simple
demonstration when the function F(x + h) can be expanded
into a series arranged according to the ascending integral
powers of one of the variables with co-efficients which are
functions of the other variable only.

Assume

Fix + h) =/{x) +Mx) h" +Mx) A» + . . . +/„(x) A",
and differentiate with respect to x, and also with respect to A;
then

dF{x + h) ^df{x) dMx)^^ (y,(x)^, ^/"(^)a'

dx dx dx dx ' " dx

^l^±^\ = oMx)h''-^ + bAix)h''-'+. . . -\-pA(x)b'"K

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DIFFERENTIAL CO-EFFICIENTS, 89

But F{x -\- h) involves h in precisely the same way that it
it does x; and, if we place a; -f 7i = y, we have (Art. 42)

dF{x + h) dF{y) dy _dF{y)

dx dy dx dy

dF{x + h) ^ dF{y) dy^ dFjy)
dh dy dh dy

Xl:

hence dF{x + h) _ dF{x + h) ^
hence ^ _ - ,

that is, these differential coeflScients are equal for all values
of X and h, which can only be the case when they are identi-
cally the same. This requires that

a-\ f(x'\-'ltM. 6-2 /•ra;•>-i-^iM•
also, by making A = in the assumed development, we find

f{x) = F{x);

whence .A {x) = F'{x), /, (x) = ^^ ...:

therefore

Fix + h) = Fix) + hF'ix) + ^F"ix)+...

12

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SECTION VI.

EXPANSION OF FUNCTIONS.

7i. The application of the formulas, demonstrated in tlie
preceding section for the expansion of functions, gives rise
to many important series, some of which we shall now deduce.

1. I{ F{x) = {l+x)"',t\ien

F"{x) =zm(m— 1) (1 + x)""-',

i^('»-i>(x) = m(7n — 1) . . . (m— M + 2)(l -[-a;)"*-""'-^
i^(«) (x) = m (m — 1 ) . . . (/w — 71 + 1) (1 +-a:)"»-'» ;
therefore F{0) = 1. F'{0) = m, F''{0) ^ m (m — 1) . . . ,

^,n-i)(0)=rm (m — 1) ... (m — 71 + 2);
and hence, by Art. 60,

{1 + x)"" = 1 + mx + m -j-2~^' + • • •

+ "* 1.2.3.. .(n-1) "^

When X is less than unity, the last term in this development
will diminish as n increases ; and, by making n sufficiently
great, the series

1 . . rn—l , , (m—l)(m — 2) , ,
Ij^rnx + m -j-^ x- + m ^ j-^^ -' x^ + . . .

90

1
J

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EXPANSTOy OF FUNCTIONS. 91

will approximate more and more nearly the true value of
(1 -|- a:)"* the greater the number of terms taken.

2. Let F{x) — e''; then

F\x) = 6"^ = F" {x) = F^" {x) = .. . = F^""-^^ (x) = i^<"> (x),
F'{0)=:1 = F'' (0) = F'^' (0) ^ , . . = i^(»-i>(0);

X X X

therefore ^' = -^ + T + r2 + TTS "^

^" &r

+ 1.2.3. ..(n-l)"''l.2. ..»*
Making in this x^l, we have

a series that may be used for .finding the approximate value
of e.

3. Let F(x) = sin.rc; then

F^ {x)=: cos.a;= sin. f a; + -),
rfsin. [oj + oj

^"<»^)= — W^=»'-(-+l)=""-('+¥}

n

Therefore F{0) = 0, F' (0) = 1, F" (0) = 0, F'" (0) = - 1,
and we have

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92 DIFFERENTIAL CALCULUS,

x^ x^

a;"-^ . n— 1

+ i72:3:TT(;r:riy'°- -2- "

+ 172737:7 n^'^-l^^^+T/

4. Let -f (x) = COS. X ; then

i?''(a;) = — sin. x = cos.fx + ^ j, jP" (x) = co8.f a; + — - j,

F'" (x) = COS. fx + ~\... F'"^ (x) = COS. (x + ^\
i^(0) = 1, i?" (0) = 0, i?"' (0) = - 1, F'" (0) = . • .
i?'<"-i>(0) = cos.^^ff:

k ^ X'^ X*'

hence cos.xzir 1 — j^+ ^ 0.3.4

H COS. 7t

^1.2.3. ..(/I- 1) 2

4- :r- COS. to -^ ).

^ 1.2.6. ,.n \ ^ 2/
By Art. 63, it will be observed that the last terms in Exs.
2, 3, and 4, diminish as n is increased, and finally vanish when
n becomes infioite.

5. Let i^(a;) = f(l+x); then

F(0) = 0, F' (x) = 1, F" {x) = — \, F'" (x) = 2 . . .
i?'"' (0) = ( — 1)»- > 1.2.3. . . (n - 1) ; and therefore
7/1 . \ x' x' X* , (-1)"-* „ ,

(-1)"-^ _x^

+ » (I + to)"'

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EXPANSION OF FUNCTIONS. . 93

An examination of the last term of this expansion shows,
that^ when x does not exceed unity, this term necessarily de-
creases as n increases, and vanishes wlien n becomes infinite.

And, since the factor [ -r-r-ir ) under this hypothesis cannot
exceed unity, the sum of the series, up to the n^^ term in-
clusive, cannot differ from the true sum by more than - • and

, hence, by increasing n sufficiently, this difference can be made
as small as we please.

Changing the sign of a;, we have

7/1 N ^' ^' ( - l)«-2 „ ,

n (X — Oxf
6. Let F{x)z=zi^xir^x; then

1-1- a;- 2Vl-a;V-l ^l + a;\/-l
= I ((1 - xV^)-i + (1 + xy/^Vr
F"(x)=.\ X ^ 1 X - \/=l (1 - x-oZ—iy
+ ^ X - 1 X V=l ( 1 + a; V^^ )-'

= V:

- 1 X ^((l - ^ V- ir - (- 1)^ (1 + a;^/^!)-* j

F"'{x)

1.2

= ( V- 1)' y ((1 - xV^^)-' - (- 1)' (1 + x V^I)-')

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94 DIFFERENTIAL CALCULUS.

therefore

i^c. (0) = (v/^T)"-' kM^^(!L-_L) U - (- 1)").

Whence it follows, that, if n is an even number, i^<'"(0) = 0;
but, if n is uneven, then

i?'<»> (0) = (v/=ri)»-i L?:!i_-^(?i:zl) X 2

= i^-lZ 1.2.3.. (7i-l) = ± -7-''^^-1.2.3..(n-l).

Hence we have

, x^ ^ x^ x^~^

tan.~* x^=zx rr + -r- • • • ± -i

3 5 71 — 1

^ (1— ^^V^)~"^ (1 + rV^)"" ^

^"^ 2V-T

The final term in this development is not in a convenient
form, as it stands, to decide whether the series is converging
or diverging ; but by referring to Ex. 18, p. G7, making

a = 1, and observing that there ^ = ^ — tan.~^ x, we have

F^n) (^) ^ (_ 1) n-1 1.2.3. ..(n-l) ^.^^ fn^ _ ^ ^^^ _i \ .

{i + x'f \2 y

therefore
tan "^ a:i=:iC— — +-r & + •••

o o

(-ir^^ — ^sin.f^;-ntan.-^a:

This form of the final term shows, that, if x is less than
unity, the numerical value of the term may be made as small
as we please by giving to n a value sufficiently great.

The above form for i^^"^ {x) might have been used for find-
ing all the differential co-efficients of tan.~^ x as readily as that
specially deduced for that purpose.

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EXPANSION OF FUNCTIONS. 95

The following is a more simple process for getting the
expansion of tan.~^ x : —

Assume

tan.-^ x=:A + Bx+Cx'- + Dx^ + &c. (1 ),
and differentiate both members with respect to x ; then

j^^ = B + 2Cx + 3J)x-^ + &c. (2);

but by division, or by the Binomial Theorem,

, } , =:zl-x'' + x'- X^ + X^ -.& + &C. (3).

The second members of (2) and (3) must be identical: hence,
equating the co-efficients of like powers of x, we have

o

and, since the assumed development must be true for all values
of ic, make a; = in (1), and we find A = 0: therefore

tan -1 0? ^ X — —+ - - - + &

7. If y = sin.~^x, assume

sin -^x = A + Bx+ Cx^ + Dx"" + . . . (1),
and differentiate both members ; then

^ ^ = B + 2Cx-\'ZDx''-\-4.Ex^ + . . . (2);

yf X ~-~ X

but, by the Binomial Theorem, we find

1 „ . 1.3 . . 1.3.5

The co-efficients of the like powers of x in the second
members of (2) and (3) must be equal: hence

and, by making ic = in (1), we get ^ = : therefore

. _i , li»' , 1.3 a;^ , 1.3.5 a:^ ,
Bin,~* X:=.X -+• V- f- . . .

^2 3 ^2.4 5 ^2.4.6 7 ^

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96

DIFFERENTIAL CALCULUS:

8. \jei y =z e' ""■ *, and assume
y=A, + A^xJrA^x' - \-.. . + J„x" + . . . (1).
Differentiate twice ; then

^-f =Ai-\-2A^x + ZAiX^ + ... + nA^x''-^-lr. . . (2)

= 2Jj + 2.3J,x + ... + («_ l)n^„a;— * + . . .• (3).
But

dy _ a.in-^x «

dx Vl-a;«

and hence

— e -|- , ;

Substitute in (4) the values of -^y - 4, taken from (2) and

dx dx^

(3), and we have

2A2+2.SA3X + SAA.x^-] h (^ - l)n^„a?""' + . .

'-'{2A2X^' + 2.SA,x^+SAA,a^-\ \'{n — l)nA„x'' + ..)

— ( AiX + 2A^x^ + 3A^x^ + 4:A^x* -] f-^^«^'*+ • •) .

Equating the co-efBcients of the same powers of x in the
two members of this equation, we find

T^" "^'-~2.T

and generally

-<»2 — "5" -^c -^» o"? — -^1' ""* — g~J~-^«**''

^. = «_!±(!L^A_....(5).

(n — l)n

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EXPANSION OF FUNCTIONS, 97

If then ^0 and ^1 be found, formula 5 will give all the
following co-efficients in terms of these two.

Aq is what e"**'"" "^ becomes when x=zO: hence ^q = 1.

And ^1 is what ^r^e"**^-"^^ — r=^ l^^comes whena7 = 0:
dx Vl-x'

hence Ai==^a.

In formula 5, making n equal to 2, 3, 4, &c., successively,
we get

^ _ «' ^ -i^!+A)^ A - («' + ^ ')«'

^""O' '~ 1.2.3 ' ^*- 1.2.3.4 •••

Substituting these values ofu4«, AifA^. . . in the assumed
development, it becomes

«.,«-u 1 , , a^x'- , a(a2+l) . , aVa^ + 2') ,

+ 172^75 ^■+-"

By Ex. 2 of this article, avo also have
c'""'"''=l + asin.-ix-h~(sin.-ia;)^ + ,-^(sin.-'a;)3 + ..

Equating the co-efficient of the first power of a in this
series with that of the same power of a in the preceding
series, we have

. , , 1 a;3 1.3 a;^ , 1.3.5 X' , ,

^2 S^2A 5 ^2.4.6 7 ^ '

as in Ex. 7. By equating the co-efficients of a^, we should

also find

/ • -1 v> , 22 ^ , 2^4'-^ 6 , 2"^. 41 62. 3
(sm. ^-)- = -+o" +3.47570" +3.4X(>:T:8^-"

9. V = Hl+en^ y=^^+-2+h-2^ik3A+^

13

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98 DIFFERENTIAL CALCULUS.

11. 2^ = Z(l + sin.a;), y nz a; — ^4-^ . . .

12. y = e'"- ', y = i + x + ^_^-^|...

13. If y = ( — J __-w — ^ 1 , sliow for what values of x

Taylor's Theorem fails to give the development.
It fails for a; = c; 1st term is then infinite.
It fails for x=:a; 2d differential co-efficient is then infinite.

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i

SECTION YIT.

APPLICATI05J OP SOME OP THE PRECEDING SERIES TO TRIGO-
NOMETRICAL AND LOGARITHMIC EXPRESSIONS.

72. Let a and b represent any two real quantities what-
ever; then a-^bs^— 1 will bo the most general symbol for
quantity, since, by giving to a and b suitable values, it may be
made to embrace every conceivable quantity, real or imaginary.

The two expressions, a + b V— I, a — 6 V— 1, which dif-
fer only in the signs of their second terms, are said to be conju-
gate ; and their product, (a + 6 V — 1 ) (a — 6 V — f ) = a^ -[- 6^,
is always real and positive. The numerical value of the square
root of a^ 4" ^^ IS the modulus of either of the conjugate ex-
pressions. Denote this modulus by r; then it may be shown
that the expression a-\-b\^— 1 can be put under the form

r{cos.O + V- Isin. 6).
For let a 1= r cos. Ojb^=zr sin. :

tan./? = -, r2(cos.2/? + sin.2/9)iz=r2=:a2-fi2,

r = \/a''-\-b\

7t

^ Now, if we suppose the arc of a circle to start from— -, and

^, to increase by continuous degrees to -f- o? passing through

t^ zero, the tangent will at the same time increase by continuous
degrees, and pass through all possible values between — oo
and -|- 30 . Among tliese values of the tangent, there must be

one that will satisfy the eauatky^an. ^ = - ; and the arc an-

478155. ^ ,

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100 DIFFERENTIAL CALCULUS,

Bwering to this tangent Avill be that whose sine and cosine will
satisfy the equations a = r cos. Oj b=zr sin. 6, and therefore
render r{co&.0 + V— I sin. 6?) the equivalent of a-|-6 V — 1.
73. Let us resume the series (Art. 71, Exs. 2, 3, 4).

cos.x = l - +^^^^-...(3)
and in (1) write x V— 1, — x V— 1, for x successively ; then

, ; x^ x^\/—\ ^*

-i-t-a;v i ^^ ^2.3 ^^ 1.2.3.4

"^1~2"3"4~5'' == cos.a; + V-lsin.a;,

as is seen by comparing this result with the second members
of(2)and(3).

x'^ x^

Also e-^^-^ = 1- xV- 1 - Ti; +

V^

1.2 ' 1.2.3. • 1.2.3.4

COS. X — V — 1 sm. 03 :

1.2.3.4.5 '
therefore cos. x + V— 1 sin. x r= e^^^-^ • • • (4)

cos.ic — V— 1 sin.a;=: g-^^-i . . . (5),
also COS. y + V— 1 sin. y = e^^-^ • • . (6);

multiplying (4) by (6)
(cos. 0? + V— 1 sin. x) (cos. y + V — 1 sin. y) = e(^+y)^-i

= cos. (iz^ + y) + V— 1 sin. (x + 2/)-

Eflfecting the multiplication in the first member, and then

equating the real part in one member with the real part in the

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TRIGONOMETRICAL EXPRESSIONS, 101

other, and the imaginary part in the one with the imaginary

part in the other, wo find

COS. [x-\-y) =■ COS. X COS. y — sin. x sin. y

sin. (x -\-y) =z sin. x cos. y -[- sin. y cos. x.
Again :

(cos.a:-f-V — Isin.ir) ^cos.?/4-V — Isin-y) (cos.z + V— Isin.^) • •
= e^^+^+'- • >"^-i= COS. (x+y-\-z + ' •)+ V^sin.(aj+y+2+. ^),
from Avhich, by making x =^y =iz= • • • , wo have

(cos. X + V — 1 sin. ir) "* zzr cos. twj; -{- V — i sin. mXj
and generally

(cos. a; ± V — 1 sin. x) "* = cos. mx i V — 1 sin. mXj
which is known as De Moivro'.s Formula.

Hence the multiplication of expressions of the form of
COS. X -\- \^ — i sin. Xj and therefore of all imaginary expres-
sions, is thus reduced to an addition, and the raising to
powers to a multiplication.

7^:. Dividing formula (4) of the preceding article by (5)
of the same, member by member, we have
e"^^"^ _ 2^v— I __ ^'^^'^ + V— 1 sin X __1 -\- V— 1 tan. a?.

€~^^~^ cos. a; — \/ — Isln. a; 1 — V— 1 tan. a:

whence, by taking the Napierian logarithms of both members,

2xV^^ = I (1 + V^i tan. x) —I (I— V^^l tan. x) .

Expanding the terms in the second member by Ex. 5, Art. 71,

/ — - / — - , tan.^x / — tan.^ic tan.'* a;

2xV— 1 = V— Itan.irH ry V — 1

/ — ^-tan.^ic

2 ^ " 3 4

x
5~

-f- y_-I tan. .V + -'— + V^n.
tan."*.:; , tui.'^.j

tan.' X

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102 DIFFERENTIAL CALCULUS.

Equating tlio imaginary parts in the two members of this
equation, and then dividing through by 2 V— 1, wo have

tan.^ir , tan.^a; i\\n? x , . .
X = tan. X — H > — _ f- & — <fcc.,

o O I

a series that may be used for the calculation of tZj and which*
agrees with the formula in Ex. 6, Art. 71.

7 J. To find the expansion of cos.'*aj in terms of the cosines
of multiples of re.

Make e'^^-'^^y; tlien e''^^~~^=y'^j e-'"^^ = ",

^_w-^~i___

y
1

From formulas 4, 5, Art. 73, Ave find

2 cos. X = 6^""^! + c-'^-^ = 2/ + ~

2 V — 1 sin. X = e-^^-^ — - '""

y

1

^ y

also, from De Moivre's Theorem, we deduce

2 COS. mx = i/"' -}- - ^7 2 V — 1 sin. mx = y''* — -^.
»/ y

1 / l\n

Because 2 cos. a; = y + - , 2" cos." a; = ( y + - ) 5
but

. n-1 1 , 1,1

•••+'' i.2~r^^ + ''p^+r'

by combining terms at equal distances from the extremes;
hence

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TRIGONOMETRICAL EXPRESSIONS, 103

COS. "a; = -K^zi ( ^^s. 7ix-\-n cos. {n — 2)x

Since there are n-}-l terms in series (a), when n is even, the
number of terms is odd, and the middle term, that is,

n(n-l)(n-2)...Q+2)0+l)

,2 / 2

will be independent of y, and consequently of x ; but, when n
is odd, n + 1 is even, and there is no middle term in series (a),
and therefore no term independent of x. In the first case,
there will be within the ( ) in formula (6), besides the term

that does not depend on x, ~ terms, containing as factors the

first COS. no:, the second cos. (n — 2) a:; and so on to the last,
which will have cos. 2x for a factor. In the second case, that
is, when n is odd, there is no term within the ( ) in formula

(6) that does not involve x; but the — ' — terms will then have

Ex. 1. cos.'*a; = -^ ( cos. 4a; + 4 cos. 2x -[- 6

for factors, severally, cos. nx, cos. (n — 2)x, . . , cos. Zx, cos. x.
23

Ex. 2. cos.^a; = — f cos. 5a; + 5 cos. 3x + 10 cos.x ].

2^\ /

76. To find the expansion of sin.^cc in terms of the sines
of multiples of x.

By formulas 4 and 5 of Art. 73, we have, employing the
notation of the last* article,

y

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(«)•

104 DIFFERENTIAL CALCULUS.

therefore 2''(V— 1)' 9in.»a; = 2''(— 1)^ 8in."a; = (y — -]

^ \

= y» - «y»-« + n -y-^- y"-* - &-\

An examination of tliis series shows, that, when n is even,
the second terms within the ( ) are all plus ; and, when n is
odd, they are all minus. In the first case, the expansion of
^ui.^x will involve only the cosines of multiples of x ; and, in
the second case, it Avill involve only the sines of these multi-
ples.

n

The factor (—1)^ in the first member will be positive and
real when n is any one of the alternate even numbers begin-
ning with 0; that is, when n is or 4 or 8 or 12, &c. ; and
negative and real when n is one of the alternate even num-

n

bers beginning with 2. In like manner, (—1)'^ will be imagi-
nary and positive when n is any one of the alternate odd
numbers beginning with 1 ; and it will be imaginary and nega-
tive when n is any other odd number.

Let h represent any positive whole number, zero included ]
then the different series of values above indicated for n will be
embraced in the four fonns, 44, 44 + 2 ; 44 + 1, 4fc + 3.

It would be of no advantage to make formula (a) conform
to each of these cases by special notation, as it can be easily
applied, as it now stands, to the examples falling under it.

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TRIGONOMETRICAL EXPRESSIONS. 105

Ex. 1. Expand sin.' a? in terms of the sines of the mul-
tiples of X,

2'(V^)'sin.»x = ^y» _ 1) _ 3 ^y - ^)

= 2V-^ 1 sin. Sx — 6V— 1 sin. a?;

1

sin.'ic = — — (sin. 3a; — 3 sin. x).

Ex. 2. Expand sin.*aj in terms of the cosines of the mul-
tiples of x.

= 2 cos. 4a; — 8 cos. 2a; -|- 12 :

sin.^a; = — f COS. 4a; — 4 COS. 2x + 6 ].

Ex. 3. Expand sin.^o; iu terms of the sines of the mul-
tiples of a;.

= 2 V^^ sin. 5a; — 10 V^^ sin. 3a; -f 20 V^sin. a?:

sin.^ic ^^ o4 ( ®'^* ^^ — ^ ^^^* ^^ "f" ^^ ^^^' ^ )•

Ex. 4. Expand sin.^o? in terms of the cosines of the mul-
tiples of X.

2e( V:ri)esin.ea;z= ^ye+ 1) - 6 {y^^ 1) + 15 (y^+l) - 20

= 2 cos. QiX — 12 COS. 4a; + 30 cos. 2a; — 20 :

8in.*a;=: — — (cos. 6a;— 6 cos. 4a; + 15 cos. 2a; — 10).
2'\ /

77. To find the different n*^ roots of unity.
Let X represent the general value of the n*^ root of unity;
then, by the definition of the root of a number, a;" = l, or

14

(

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106 DIFFERENTIAL CALCULUS.

a;" — 1 = 0; and the object of the investigation is to find all of
the values of a: that will satisfy the equation x^ — 1 = 0.
By Do Moivre's Theorem, Art. 73, we have

(cos. y dt= V— i sin. ?/)'" = cos. my dt= V— 1 sin. my;
an equation which holds, whether m is entire or fractional,
positive or negative. Now, if h be any whole number, 2k7t

Online LibraryI. F. Quinby Horatio Nelson RobinsonA new treatise on the elements of the differential and integral calculus → online text (page 5 of 22)
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text:
# Control Theory for Linear Systems

Harry L. Trentelman, Antonie Arij Stoorvogel, Malo Hautus

4 Citations (Scopus)

### Abstract

Control Theory for Linear Systems deals with the mathematical theory of feedback control of linear systems. It treats a wide range of control synthesis problems for linear state space systems with inputs and outputs. The book provides a treatment of these problems using state space methods, often with a geometric flavour. Its subject matter ranges from controllability and observability, stabilization, disturbance decoupling, and tracking and regulation, to linear quadratic regulation, $H_2$ and $H_\infty$ control, and robust stabilization. Each chapter of the book contains a series of exercises, intended to increase the reader's understanding of the material. Often, these exercises generalize and extend the material treated in the regular text.
Original language Undefined London, UK Springer 389 978-1-85233-316-4 Published - 2001

### Publication series

Name Communications and Control Engineering Springer Verlag Ltd

### Keywords

• H-control theory
• Linear quadratic regulator problem
• IR-68841
• System Theory
• EWI-16619
• Geometric control

## Cite this

Trentelman, H. L., Stoorvogel, A. A., & Hautus, M. (2001). Control Theory for Linear Systems. (Communications and Control Engineering). London, UK: Springer.


text:
# Weigh a scale with itself

Using nothing but a pocket scale and a flat surface such as a countertop, find the total mass of the scale, including the attached metal plate and the plastic tray/cover (left).

• If you press the TARE button, the scale's zero point will be set to whatever weight is currently on the scale.
• After using the TARE function, any weight placed on the scale will be displayed as a positive number and any weight removed from the scale will be displayed as a negative number.
• When you power on the scale with the ON/OFF button, it will wait for 3 seconds, then automatically trigger the TARE function.
• Values are displayed in grams.
• The scale's strain gauge is located between the metal plate and the body of the scale. The bar that the strain gauge is attached to has negligible mass.
• You may not use the UNIT or PCS functions.

• When sitting upright, the metal plate is the highest point on the scale. The metal plate does not come in contact with the plastic cover when the cover is on.
• The scale weighs less than the weight limit (300 grams).
• The center of mass in the X-Y plane is within the metal plate, as determined by this method — the scale will not tip if set upside down.

• Does the scale weigh less than 300 g? – msh210 May 15 at 8:55
• @msh210 Good question! It does. – Snowball May 15 at 11:48

You can:

1. Place the pocket scale upside down on the flat surface, turn it on, wait a bit, then turn it around. That should give you the weight of the body 1 (negative value), minus the weight of metal plate.
2. Place the pocket scale upside down sideways over the cover you showed, so that the metal plate hangs freely, but the body of scale is supported by the cover. (The metal plate does not come in contact with the plastic cover when the cover is on.) Turn it on, wait a bit, then turn it around. That should give you twice the weight of the metal plate (positive value). It's double since it first measures the plate hanging (and calls it zero), then it measures the plate on top. The difference will be twice the weight of the plate, which is what we want, since the plate was subtracted twice in step 1.
3. Press the "Tare" button, and put the cover on the weight. That should give you the weight of the cover (positive value).
4. Add the three numbers together to get the total weight of the scale.

Note: Corrected after comment by Jaap Scherphuis.

• Great solution. The scale is 104 g with the included batteries, or 90 g without them. – Snowball May 15 at 12:27
• Doesn't step 1 give you the the weight of the body minus the weight of the plate? (i.e. not the weight of the body alone, which is the weight of the whole scale minus the plate). So later in step 4 you should add twice the weight of the plate back on. – Jaap Scherphuis May 15 at 12:31
• @JaapScherphuis of course, you're right... Edited and credited :) – CG. May 15 at 12:52
• @TCooper With this scale, it doesn't fail because the center of mass in the X-Y plane is within the metal plate. I'll add that information to the question. Thanks for pointing out the omission! – Snowball May 15 at 20:05
• Gotta say, this was quite a fun little puzzle! – tgm1024--Monica was mistreated May 15 at 23:37

Hold the scale in the air by the metal plate, so that the base of the scale is pulling down and creating tension on the strain gauge equal to the base's weight. Turn the scale on and let it tare. Now place the scale on the table. Relative to the previous tare, this new measurement no longer has the base pulling down (so +weight of base), and it also now has the weight of the plate on top (so +weight of plate). This means the displayed measurement is weight of base + plate, i.e. weight of the scale.

• That's a clever solution, but I'm not sure it would work in practice. I believe you would have to keep incredibly steady. D=0.01 grams, which is very very little... Note: I'm not sure, it could work. – CG. May 16 at 14:30
• @Snowball, can you try this out and let us know how it compares to the accepted answer? – knrumsey May 16 at 14:59
• @knrumsey: This was a bit tricky to pull off, but it works! CG.'s solution gave 104.165±0.004 g (SEM, N=10) and hpp3's solution gave 104.163±0.027 g (SEM, N=10). – Snowball May 16 at 19:21

Can you

power it on, put it upside down (with the plastic cover on top), wait for 3 seconds, and then put it upright again?

If

it says, for instance, -500 grams, the scale weighs 500 grams?

• this is what I thought too, but I think it should not be that obvious :D when putting upside down might be making scale touches on the ground from other points? that information is not given though. – Oray May 15 at 8:09
• @Glorfindel exactly what I thought! – William Pennanti May 15 at 8:14
• Not quite, that wouldn't include the mass of the metal plate. – Snowball May 15 at 8:21
• @Oray: The metal plate is the highest point on the scale. Thanks for pointing out that omission; I'll add it to the details in the question. – Snowball May 15 at 8:28

I'd like to examine the pieces of the puzzle so that we can build to a solution. The assumption going forward is that each "setup" is stable such that nothing is moving or being held in place by anything other than the weight of the pieces. Also, that the scale isn't sideways, as it's probably not designed to handle that.

What's the meaning of the number that appears on the screen? It's not necessarily a reak weight, but the difference in weight between the current setup and a reference setup - thus the need to use the TARE function.

So then, how do different setups affect the measurement? Each relative number measures the compression on the contact point (i.e., strain gauge) between the plate and the base. If the plate and base are being pushed together, the pieces sitting on top of the contact point are measured. But if the plate and base are being pulled apart, the pieces hanging from the contact point are measured as negative values.

How can we use this? There are a handful of meaningful setups we can use as our reference points. From there, we can place the scale in setup1, use TARE, move it to setup2, and record setup2 - setup1 = number_on_screen. With a couple of those in hand, it's simple to solve for each unknown.

Here are some relevant setups (the list may not be exhaustive, but is sufficient):

1. Scale right-side-up, lid to the side: measures plate
2. Scale right-side-up, lid on plate: measures plate + lid
3. Scale up-side-down, lid to the side: measures base
4. Scale up-side-down, lid on base: measures base + lid
5. Scale, with lid attached, up-side-down (the plate hangs): measures plate, but negative

Finally, we can combine each of those in pairs, record the value displayed, and form equations. Ideally we can isolate a piece so that we can get its weight directly. If/once we have one weight, we can discount it as a "variable" and solve for another weight in a shared equation. Fortunately, we can isolate the lid and the plate as follows: 1. TARE for setup 2, then setup 1 => X = (plate + lid) - (plate) = lid 2. TARE for setup 5, then setup 1 => Y = (plate) - (-plate) = 2*plate

So we have the weight of the lid and the plate, but need to get the base. We need any setup that has the base as well as any setup that does not have the base. The simplest is: 3. Tare for setup 1, then setup 3 => Z = (base) - (plate)

Finally, with some algebra, total = (lid) + (plate) + (base) = (X) + (Y/2) + (Z + Y/2).


text:
Propagation of singularities for the wave equation on manifolds with corners
Séminaire Équations aux dérivées partielles (Polytechnique) dit aussi "Séminaire Goulaouic-Schwartz" (2004-2005), Exposé no. 20, 16 p.

In this talk we describe the propagation of ${𝒞}^{\infty }$ and Sobolev singularities for the wave equation on ${𝒞}^{\infty }$ manifolds with corners $M$ equipped with a Riemannian metric $g$. That is, for $X=M×{ℝ}_{t}$, $P={D}_{t}^{2}-{\Delta }_{M}$, and $u\in {H}_{\text{loc}}^{1}\left(X\right)$ solving $Pu=0$ with homogeneous Dirichlet or Neumann boundary conditions, we show that ${WF}_{\text{b}}\left(u\right)$ is a union of maximally extended generalized broken bicharacteristics. This result is a ${𝒞}^{\infty }$ counterpart of Lebeau’s results for the propagation of analytic singularities on real analytic manifolds with appropriately stratified boundary, [7]. Our methods rely on b-microlocal positive commutator estimates, thus providing a new proof for the propagation of singularities at hyperbolic points even if $M$ has a smooth boundary (and no corners).

These notes are a summary of [17], where the detailed proofs appear.

Classification : 58J47,  35L20
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title = {Propagation of singularities for the wave equation on manifolds with corners},
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publisher = {Centre de math\'ematiques Laurent Schwartz, \'Ecole polytechnique},
year = {2004-2005},
mrnumber = {2182064},
language = {fr},
url = {http://www.numdam.org/item/SEDP_2004-2005____A20_0/}
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Vasy, András. Propagation of singularities for the wave equation on manifolds with corners. Séminaire Équations aux dérivées partielles (Polytechnique) dit aussi "Séminaire Goulaouic-Schwartz" (2004-2005), Exposé no. 20, 16 p. http://www.numdam.org/item/SEDP_2004-2005____A20_0/

[1] J. Cheeger and M. Taylor. Diffraction by conical singularities, I, II. Comm. Pure Applied Math., 35 :275–331, 487–529, 1982. | Zbl 0536.58032

[2] J. J. Duistermaat and L. Hörmander. Fourier integral operators, II. Acta Mathematica, 128 :183–269, 1972. | MR 388464 | Zbl 0232.47055

[3] A. Hassell, R. B. Melrose, and A. Vasy. Scattering for symbolic potentials of order zero and microlocal propagation near radial points. Preprint, 2005.

[4] A. Hassell, R. B. Melrose, and A. Vasy. Spectral and scattering theory for symbolic potentials of order zero. In Séminaire : Équations aux Dérivées Partielles, 2000–2001, Sémin. Équ. Dériv. Partielles, pages Exp. No. XIII, 21. École Polytech., Palaiseau, 2001. | Numdam | MR 1860685 | Zbl 1063.35126

[5] L. Hörmander. Fourier integral operators, I. Acta Mathematica, 127 :79–183, 1971. | MR 388463 | Zbl 0212.46601

[6] L. Hörmander. The analysis of linear partial differential operators, vol. 1-4. Springer-Verlag, 1983.

[7] G. Lebeau. Propagation des ondes dans les variétés à coins. Ann. Scient. Éc. Norm. Sup., 30 :429–497, 1997. | Numdam | MR 1456242 | Zbl 0891.35072

[8] R. B. Melrose and J. Sjöstrand. Singularities of boundary value problems. I. Comm. Pure Appl. Math, 31 :593–617, 1978. | MR 492794 | Zbl 0368.35020

[9] R. B. Melrose and J. Sjöstrand. Singularities of boundary value problems. II. Comm. Pure Appl. Math, 35 :129–168, 1982. | MR 644020 | Zbl 0546.35083

[10] R. B. Melrose, A. Vasy, and J. Wunsch. Propagation of singularities for the wave equation on manifolds with edges. In preparation.

[11] R. B. Melrose and J. Wunsch. Propagation of singularities for the wave equation on conic manifolds. Invent. Math., 156(2) :235–299, 2004. | MR 2052609 | Zbl 1088.58011

[12] R. B. Melrose. Transformation of boundary problems. Acta Math., 147(3-4) :149–236, 1981. | MR 639039 | Zbl 0492.58023

[13] R. B. Melrose and P. Piazza. Analytic $K$-theory on manifolds with corners. Adv. Math., 92(1) :1–26, 1992. | MR 1153932 | Zbl 0761.55002

[14] M. Mitrea, M. Taylor, and A. Vasy. Lipschitz domains, domains with corners and the Hodge Laplacian. Preprint, 2004. | MR 2182300

[15] A. Vasy. Propagation of singularities in many-body scattering. Ann. Sci. École Norm. Sup. (4), 34 :313–402, 2001. | Numdam | MR 1839579 | Zbl 0987.81114

[16] A. Vasy. Propagation of singularities in many-body scattering in the presence of bound states. J. Func. Anal., 184 :177–272, 2001. | MR 1847427 | Zbl 1085.35010

[17] A. Vasy. Propagation of singularities for the wave equation on manifolds with corners. Preprint, 2004.


text:
# Creating universal Emacs Lisp packages and scripts

A few weeks ago, I decided to create a dedicated Emacs package for work, which contains helpful commands specific to our team’s workflow. Of course, not everyone in the company uses Emacs as their editor (and you cannot expect them to). This problem got me thinking - how can I create a fantastic DevX for the Emacs users, but at the same time without leaving out non-Emacs users?

## Emacs as a Lisp interpreter

Modern Emacs allows to not only byte compile (and even natively compile) packages, but also still interpret Emacs Lisp scripts in a completely headless way. This means that you can use the full power of a high level, homo-iconic language like Elisp to automate common chores, and also use all of the benefits of the highly productive Emacs development workflow. However, we also want to use this script as a normal Emacs package. This leads us into the next section.

## How to make a package behave like a script

Unfortunately there are barely any guides on how to implement this polymorphic behavior, which is why I am writing this guide.

First, let’s create a minimal Emacs package and call it foo.el:

;;; foo.el --- Foo development and productivity utils  -*- lexical-binding: t; -*-

(defun bar ()
"Print foo bar."
(interactive)
(message "Foo Bar!"))

(provide 'foo)

Next, to allow us to simply call the script as if it was a regular shell script, we need to add this shebang to the first line of the file:

#!/usr/bin/env emacs --script

This should locate emacs on almost any system, as well as switch on batch mode to interpret the script.

Next, let’s make a special “CLI” section at the end of the file, and also create a small predicate function to easily tell when the file is running as a script or not:

;;;; CLI

(defun foo-running-as-script-p ()
"Return truthy if running as Elisp script."

Finally, we need to define an entrypoint (think of it like a main in mainstream languages). I found that the simplest way is to create a main function and run it when the script is being interpreted:

(defun main ()
"Entrypoint for foo"
(pprint command-line-args-left)
(message "Do stuff here"))

(when (foo-running-as-script-p)
(main))

The best part is that you can use the same message and yes-or-no-p as you would with an Emacs package, and they will behave in an interactive way, even when executed from the command line.

One final thing to note: you might have to explicitly require some Elisp modules if you are using certain functions (e.g. cl-defun from cl-lib). The easiest way to tell which ones, is to try to execute your script interactively and add any missing modules like so:

(require 'cl-lib)

The full example looks like this:

Make the script executable with chmod +x foo.el and call it like so:

\$ ./foo.el somearg
("somearg")
Do stuff here

Congratulations! You now have a universal Emacs package + script. Hopefully this guide has been helpful.

## Honorable mentions

https://leancrew.com/all-this/2008/04/emacs-lisp-as-a-scripting-language/

https://kitchingroup.cheme.cmu.edu/blog/2014/08/06/Writing-scripts-in-Emacs-lisp/


text:
Janae G.

# #11 step by step please

Games at the carnival cost $3.50 each. The prizes awarded to winners cost the owner$135. How many games must be played for the owner of the game to make at least \$45?

By:


text:
# Multi-product profit maximization

I was just thinking about the question

How does a firm choose the optimal input vector as to maximize profits when the firm produces more than one final product

More precisely if $\mathbb{R}_{\geq 0}^{\ell}=\{x \in \mathbb{R}^{\ell} : x \geq 0\}$, the production function would be $$f:\mathbb{R}_{\geq 0}^{\ell}\rightarrow \mathbb{R}_{\geq 0}^{n}, \quad n\geq 2$$ For simplicity we can assume the firm operates in a competitive market so that the price is given.

My question is, how does the firm choose the input vector $x \in \mathbb{R}_{\geq 0}^{\ell}$ as to maximize profits?

I did some research and apparently the type of problem this is a “multi-objective” optimization problem, and it appears to be an active field in engineering. Some of the methods to solve this kind of problems use pareto-dominance style arguments.

I wanted to know if anyone could clarify the specific question of multi product profit maximization, and more generally I would also appreciate any literature or books on this topic (multi objective optimization), written for economists, as I have been searching in google and every book I find appears to be written for engineers.

I suspect that a treatment of this topic from an economics point of view would involve some discussion of social choice theory.

Any help is most welcomed!

• Sounds to me like any notion of equilibrium used in economics is a “multi-objective” optimisation problem. – Theoretical Economist Oct 26 '17 at 23:08

At the end of the day, the firm seeks to maximize profits, which is a one-dimensional object. So the firm simply solves $$\max_{\mathbf x}\;\mathbf{p}\cdot\mathbf f(\mathbf x)-c(\mathbf x)$$ where $\mathbf x\in\mathbb R_+^\ell$ is a vector of $\ell$ inputs, $\mathbf p\in\mathbb R_+^n$ is a vector of $n$ prices for the final output, $\mathbf f:\mathbb R_+^\ell\to\mathbb R_+^n$ is the production function mapping $\ell$ inputs to $n$ outputs, and $c:\mathbb R_+^\ell\to \mathbb R_+$ is the cost function. This is a standard unconstrained optimization exercise that any first year grad student in economics should know how to solve. The standard textbooks --- MWG, Jehle and Reny, Simon and Blume --- have sections that cover the techniques.

• You are absolutely right. I didnt think the “economic example” through... the reason is that I just wanted to ask about economic literature of multi objective optimization, and thought that an economic example would make it suitable for the site. Would it be ok if i edit the question accordingly? I guess that the “economic example” would be if the firm wanted to maximize production instead of profits. – thekiciminister Oct 26 '17 at 17:05
• @agustromo: Yes, feel free to edit your question to reflect what you really want to ask. – Herr K. Oct 26 '17 at 18:18


text:
# 1

Jump to navigation Jump to search

 ← 0 1 2 →
Cardinalone
Ordinal1st
(first)
Numeral systemunary
Factorization
Divisors1
Greek numeralΑ´
Roman numeralI, i
Greek prefixmono-/haplo-
Latin prefixuni-
Binary12
Ternary13
Octal18
Duodecimal112
Hexadecimal116
Greek numeralα'
Arabic, Kurdish, Persian, Sindhi, Urdu١
Assamese & Bengali
Chinese numeral一/弌/壹
Devanāgarī
Ge'ez
Georgian Ⴁ/ⴁ/ბ(Bani)
Hebrewא
Japanese numeral一/壱
Kannada
Khmer
Malayalam
Thai
Tamil
Telugu
Countin' rod𝍠

1 (one, also called unit, and unity) is a number and an oul' numerical digit used to represent that number in numerals. Be the holy feck, this is a quare wan. It represents a feckin' single entity, the bleedin' unit of countin' or measurement. For example, a line segment of unit length is a line segment of length 1, enda story. In conventions of sign where zero is considered neither positive nor negative, 1 is the feckin' first and smallest positive integer.[1] It is also sometimes considered the bleedin' first of the feckin' infinite sequence of natural numbers, followed by 2, although by other definitions 1 is the bleedin' second natural number, followin' 0.

The fundamental mathematical property of 1 is to be a multiplicative identity,[2] meanin' that any number multiplied by 1 returns that number. Most if not all properties of 1 can be deduced from this. In advanced mathematics, a bleedin' multiplicative identity is often denoted 1, even if it is not an oul' number, begorrah. 1 is by convention not considered a prime number; although universal today, this was an oul' matter of some controversy until the mid-20th century.

## Etymology

The word one can be used as a bleedin' noun, an adjective and a bleedin' pronoun.[3]

It comes from the oul' English word an,[3] which comes from the oul' Proto-Germanic root *ainaz.[3] The Proto-Germanic root *ainaz comes from the bleedin' Proto-Indo-European root *oi-no-.[3]

Compare the Proto-Germanic root *ainaz to Old Frisian an, Gothic ains, Danish en, Dutch een, German eins and Old Norse einn.

Compare the oul' Proto-Indo-European root *oi-no- (which means "one, single"[3]) to Greek oinos (which means "ace" on dice[3]), Latin unus (one[3]), Old Persian aivam, Old Church Slavonic -inu and ino-, Lithuanian vienas, Old Irish oin and Breton un (one[3]).

## As a number

One, sometimes referred to as unity,[4][1] is the oul' first non-zero natural number. It is thus the bleedin' integer after zero.

Any number multiplied by one remains that number, as one is the oul' identity for multiplication, the hoor. As an oul' result, 1 is its own factorial, its own square and square root, its own cube and cube root, and so on. Soft oul' day. One is also the feckin' result of the bleedin' empty product, as any number multiplied by one is itself. It is also the feckin' only natural number that is neither composite nor prime with respect to division, but is instead considered a bleedin' unit (meanin' of rin' theory).

## As a holy digit

The glyph used today in the feckin' Western world to represent the oul' number 1, a bleedin' vertical line, often with a feckin' serif at the oul' top and sometimes a short horizontal line at the oul' bottom, traces its roots back to the Brahmic script of ancient India, where it was an oul' simple vertical line. It was transmitted to Europe via Arabic durin' the Middle Ages.

In some countries, the serif at the top is sometimes extended into a feckin' long upstroke, sometimes as long as the vertical line, which can lead to confusion with the oul' glyph for seven in other countries. Sufferin' Jaysus. Whereas the bleedin' digit 1 is written with a long upstroke, the digit 7 has a feckin' horizontal stroke through the bleedin' vertical line.

While the oul' shape of the bleedin' character for the oul' digit 1 has an ascender in most modern typefaces, in typefaces with text figures, the feckin' glyph usually is of x-height, as, for example, in .

The 24-hour tower clock in Venice, usin' J as a symbol for 1.

Many older typewriters do not have a separate symbol for 1, and use the oul' lowercase letter l instead. In fairness now. It is possible to find cases when the oul' uppercase J is used, while it may be for decorative purposes.

## Mathematics

### Definitions

Mathematically, 1 is:

Formalizations of the natural numbers have their own representations of 1. In the Peano axioms, 1 is the successor of 0, would ye swally that? In Principia Mathematica, it is defined as the feckin' set of all singletons (sets with one element), and in the feckin' Von Neumann cardinal assignment of natural numbers, it is defined as the feckin' set {0}.

In a multiplicative group or monoid, the feckin' identity element is sometimes denoted 1, but e[2] (from the feckin' German Einheit, "unity") is also traditional. However, 1 is especially common for the oul' multiplicative identity of an oul' rin', i.e., when an addition and 0 are also present. When such a rin' has characteristic n not equal to 0, the feckin' element called 1 has the bleedin' property that n1 = 1n = 0 (where this 0 is the oul' additive identity of the oul' rin'). Would ye believe this shite?Important examples are finite fields.

By definition, 1 is the magnitude, absolute value, or norm of a feckin' unit complex number, unit vector, and a unit matrix (more usually called an identity matrix), be the hokey! Note that the term unit matrix is sometimes used to mean somethin' quite different.

By definition, 1 is the probability of an event that is absolutely or almost certain to occur.

In category theory, 1 is sometimes used to denote the oul' terminal object of a feckin' category.

In number theory, 1 is the value of Legendre's constant, which was introduced in 1808 by Adrien-Marie Legendre in expressin' the oul' asymptotic behavior of the oul' prime-countin' function. Legendre's constant was originally conjectured to be approximately 1.08366, but was proven to equal exactly 1 in 1899.

### Properties

Tallyin' is often referred to as "base 1", since only one mark – the oul' tally itself – is needed, for the craic. This is more formally referred to as a unary numeral system. Arra' would ye listen to this shite? Unlike base 2 or base 10, this is not a bleedin' positional notation.

Since the oul' base 1 exponential function (1x) always equals 1, its inverse does not exist (which would be called the feckin' logarithm base 1 if it did exist).

There are two ways to write the bleedin' real number 1 as a recurrin' decimal: as 1.000..., and as 0.999.... Here's a quare one. 1 is the feckin' first figurate number of every kind, such as triangular number, pentagonal number and centered hexagonal number, to name just a feckin' few.

In many mathematical and engineerin' problems, numeric values are typically normalized to fall within the bleedin' unit interval from 0 to 1, where 1 usually represents the oul' maximum possible value in the feckin' range of parameters. Jesus, Mary and holy Saint Joseph. Likewise, vectors are often normalized into unit vectors (i.e., vectors of magnitude one), because these often have more desirable properties, you know yourself like. Functions, too, are often normalized by the condition that they have integral one, maximum value one, or square integral one, dependin' on the feckin' application.

Because of the bleedin' multiplicative identity, if f(x) is a holy multiplicative function, then f(1) must be equal to 1.

It is also the feckin' first and second number in the Fibonacci sequence (0 bein' the zeroth) and is the feckin' first number in many other mathematical sequences.

The definition of a holy field requires that 1 must not be equal to 0. Chrisht Almighty. Thus, there are no fields of characteristic 1, to be sure. Nevertheless, abstract algebra can consider the bleedin' field with one element, which is not a feckin' singleton and is not a holy set at all.

1 is the oul' most common leadin' digit in many sets of data, a holy consequence of Benford's law.

1 is the bleedin' only known Tamagawa number for a holy simply connected algebraic group over a number field.

The generatin' function that has all coefficients 1 is given by

${\displaystyle {\frac {1}{1-x}}=1+x+x^{2}+x^{3}+\ldots }$

This power series converges and has finite value if and only if ${\displaystyle |x|<1}$.

### Primality

1 is by convention neither a bleedin' prime number nor a feckin' composite number, but an oul' unit (meanin' of rin' theory) like −1 and, in the oul' Gaussian integers, i and −i. Bejaysus.

The fundamental theorem of arithmetic guarantees unique factorization over the integers only up to units. Arra' would ye listen to this shite? For example, 4 = 22, but if units are included, is also equal to, say, (−1)6 × 123 × 22, among infinitely many similar "factorizations".

1 appears to meet the naïve definition of a prime number, bein' evenly divisible only by 1 and itself (also 1). Be the hokey here's a quare wan. As such, some mathematicians considered it an oul' prime number as late as the feckin' middle of the feckin' 20th century, but mathematical consensus has generally and since then universally been to exclude it for a bleedin' variety of reasons (such as complicatin' the oul' fundamental theorem of arithmetic and other theorems related to prime numbers).

1 is the feckin' only positive integer divisible by exactly one positive integer, whereas prime numbers are divisible by exactly two positive integers, composite numbers are divisible by more than two positive integers, and zero is divisible by all positive integers.

### Table of basic calculations

Multiplication 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 50 100 1000
1 × x 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 50 100 1000
Division 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
1 ÷ x 1 0.5 0.3 0.25 0.2 0.16 0.142857 0.125 0.1 0.1 0.09 0.083 0.076923 0.0714285 0.06
x ÷ 1 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
Exponentiation 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20
1x 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1
x1 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20

## In science

• Dimensionless quantities are also known as quantities of dimension one.
• 1 is the atomic number of hydrogen.
• +1 is the bleedin' electric charge of positrons and protons.
• Group 1 of the periodic table consists of the alkali metals.
• Period 1 of the oul' periodic table consists of just two elements, hydrogen and helium.
• The dwarf planet Ceres has the minor-planet designation 1 Ceres because it was the feckin' first asteroid to be discovered.
• The Roman numeral I often stands for the oul' first-discovered satellite of a planet or minor planet (such as Neptune I, a.k.a. Triton). Here's another quare one for ye. For some earlier discoveries, the oul' Roman numerals originally reflected the oul' increasin' distance from the feckin' primary instead.

## In philosophy

In the philosophy of Plotinus (and that of other neoplatonists), The One is the oul' ultimate reality and source of all existence.[8] Philo of Alexandria (20 BC – AD 50) regarded the number one as God's number, and the oul' basis for all numbers ("De Allegoriis Legum," ii.12 [i.66]).

## References

1. ^ a b Weisstein, Eric W. "1", you know yerself. mathworld.wolfram.com. Here's another quare one. Retrieved 2020-08-10.
2. ^ a b "Compendium of Mathematical Symbols". Be the hokey here's a quare wan. Math Vault. 2020-03-01. Retrieved 2020-08-10.
3. "Online Etymology Dictionary". Here's a quare one for ye. etymonline.com, the cute hoor. Douglas Harper.
4. ^ Skoog, Douglas. Jaykers! Principles of Instrumental Analysis. Jasus. Brooks/Cole, 2007, p. Chrisht Almighty. 758.
5. ^ "Plastic Packagin' Resins" (PDF), the hoor. American Chemistry Council. Archived from the original (PDF) on 2011-07-21.
6. ^ Woodford, Chris (2006), Digital Technology, Evans Brothers, p. 9, ISBN 978-0-237-52725-9
7. ^ Godbole, Achyut S. Whisht now and listen to this wan. (1 September 2002), Data Comms & Networks, Tata McGraw-Hill Education, p. 34, ISBN 978-1-259-08223-8
8. ^ Olson, Roger (2017), for the craic. The Essentials of Christian Thought: Seein' Reality through the oul' Biblical Story. Arra' would ye listen to this shite? Zondervan Academic, be the hokey! ISBN 9780310521563.


text:
# How do you differentiate f(x)=1/(sqrt(x-3) using the chain rule?

Jan 1, 2016

Chain rule states that $\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{\mathrm{dy}}{\mathrm{du}} \frac{\mathrm{du}}{\mathrm{dx}}$.
We'll also need two power rules here.

#### Explanation:

• ${a}^{-} n = \frac{1}{a} ^ n$

• ${a}^{\frac{m}{n}} = \sqrt[n]{{a}^{m}}$

So, we can rewrite it as $f \left(x\right) = {\left(x - 3\right)}^{- \frac{1}{2}}$

Renaming $u = x - 3$, we get $f \left(x\right) = {u}^{- \frac{1}{2}}$ and can now differentiate it.

$\frac{\mathrm{dy}}{\mathrm{dx}} = - \frac{1}{2} {u}^{- \frac{3}{2}} \left(1\right) = - \frac{1}{2 \left({u}^{\frac{3}{2}}\right)} = - \frac{1}{2 \sqrt{{\left(x - 3\right)}^{3}}}$


text:
# Publications

(2017). $\operatorname{Aut}(\mathbb{F}_5)$ has property (T). accepted in Mathematische Annalen.

(2017). Certifying Numerical Estimates of Spectral Gaps. Groups Compl. Crypt. Vol. 10 No. 1 (2018).

(2015). Effective topological complexity of spaces with symmetries. Publ. Math. Vol. 62 No. 1 (2018).

(2014). Group actions on complex projective spaces. Proc. RIMS Kokyuroku, Vol 1922, 2014.

(2014). On Representation of the Reeb Graph as a Sub Complex of Manifold. Topol. Methods Nonlinear Anal., Vol 45 No. 1, 2015.

(2011). Non Symplectic Actions on Complex Projective Spaces. J. Symplectic Geom., Vol 10 No. 1, 2012.


text:
How to calculate the area of a circle ( given: origin, radius ) on a sphere ( Earth )?

I know that the Earth isn't a sphere, not even an ellipsoid, but for my measurements, its an acceptable approximation. Assuming I have a coordinate(lat,lon) and a distance( e.g.: 1000km ), what is the surface area on the earth with that distance radius?

Just a small amplification of the answer by Ross Millikan. Use the same notation as the article he linked to. I take it that your $1000$ km is the surface of the Earth distance from the center $C$ of your circle to the furthest points $P$ from the center. In the picture linked to, $C$ is the top of the sphere, and $P$ is any point on outer edge of the bottom of the cap.

Assume that this surface of the Earth distance is $d$, and that is is $\le$ $1/4$ of the circumference of the Earth (that's not necessary, but it makes visualization easier). Let the radius of the Earth be $r$.

Then the angle $\theta$ subtended by the arc $CP$ at the centre of the Earth is given by $$\theta=\frac{d}{r}.\tag{\ast}$$ The "$h$" in the linked picture is given by $h=r-r\cos\theta$. The surface area is $2\pi rh$, which is $$2\pi r^2(1-\cos\theta).\tag{\ast\ast}$$ Compute $\theta$ using $(\ast)$, and then use $(\ast\ast)$ to find the surface area.

• Thank you for the detailed answer. I was missing the (*) equation. – p1100i Apr 23 '12 at 13:53
• Actually this answer is incorrect. θ = (d / r), in radians, or θ = (d * 360 / (2 πr)), in degrees. – Hans Brende Feb 22 '16 at 3:17
• Unless, of course, you are speaking in turns ;) – Hans Brende Feb 24 '16 at 5:55
• @HansBrende: Thank you. I will look at it again, tomorrow perhaps. – André Nicolas Feb 24 '16 at 6:05
• @HansBrende: Thank you for spotting the incorrect $\theta$. Fixed. – André Nicolas Mar 1 '16 at 6:12

You could look at Spherical Cap for the formula. If you treat the earth as a sphere, the coordinates of the center do not matter, just the radius and the radius of the earth.

• Thx for the reply, if there would be a distributed solution option I would give the half for u, but since Andrés answer was more detailed... – p1100i Apr 23 '12 at 13:55


text:
Home > Mean Square > Minimum Mean Square Error Equalizer Wiki

# Minimum Mean Square Error Equalizer Wiki

## Contents

Retrieved from "https://en.wikipedia.org/w/index.php?title=Zero_forcing_equalizer&oldid=699279334" Categories: Filter theory Navigation menu Personal tools Not logged inTalkContributionsCreate accountLog in Namespaces Article Talk Variants Views Read Edit View history More Search Navigation Main pageContentsFeatured contentCurrent eventsRandom What's that? This will be useful when ISI is significant compared to noise. Lee, David G. http://codecove.net/mean-square/minimum-mean-square-error-equalizer.html

The negative sign indicates that, we need to change the weights in a direction opposite to that of the gradient slope. Estimator The MSE of an estimator θ ^ {\displaystyle {\hat {\theta }}} with respect to an unknown parameter θ {\displaystyle \theta } is defined as MSE  ( θ ^ ) But, I think the variance does not change even if we compute in frequency domain or in time domain. I have also sent an email to you. https://en.wikipedia.org/wiki/Minimum_mean_square_error

## Minimum Mean Square Error Estimation Example

Thank you. Reply kakuna October 31, 2010 at 8:01 pm Hi Krishna, How about MMSE equalization in STBC? Singer) Reply Krishna Sankar March 28, 2010 at 2:28 pm @Sanka: Well, I have not studied Turbo decoder well enough to reply. Viterbi equalizer: Finds the maximum likelihood (ML) optimal solution to the equalization problem.

Reply Krishna Sankar November 15, 2010 at 2:26 am @kakuna: For 2 Tx, 1 Rx STBC, having ZF equalization is optimal. Zero-mean white Gaussian noise v(i) with variance sigma 2 is added to the channel's output so that Y(i)=s(i)+0.5s(i-1)+v(i) The sequence ݏሺ݅ሻ is white and is independent of noise v. x ^ M M S E = g ∗ ( y ) , {\displaystyle {\hat ^ 2}_{\mathrm ^ 1 }=g^{*}(y),} if and only if E { ( x ^ M M Mmse Estimator Derivation Thanking You Regards Varun Reply Krishna Sankar April 30, 2009 at 5:05 am @ Varun Raj: you may find my email address in the page http://www.dsplog.com/contact-us/ Further, you may find many

Reply Krishna Sankar May 26, 2011 at 5:49 am @Nikes: Each symbol gets a different channel realization. could you show how did you get this result formula? Reply Krishna Sankar June 1, 2009 at 5:18 am @maya: Well, I diversity in the general sense means - the using the extra information which is available and/or transmitted to improve https://en.wikipedia.org/wiki/Recursive_least_squares_filter Why does MMSE provides better performance than Zero-Forcing in terms of system spectral efficiency?

Solving, . Minimum Mean Square Error Matlab This must be true for both the receivers… so if they have the same slope they won't merge in log scale. I'm facing some problem in simulation with matlab. effect of fluorescent lights, doppler for different indoor multipath characteristics. (2). (3).

## Least Mean Square Error Algorithm

Even in low Snr's the difference between MMSE and ZF is very less, in the range of 10^-4. https://en.wikipedia.org/wiki/Zero_forcing_equalizer Wikipedia® is a registered trademark of the Wikimedia Foundation, Inc., a non-profit organization. Minimum Mean Square Error Estimation Example Tgn Channel Models, Vinko Erceg et al. Minimum Mean Square Error Algorithm The expression for optimal b {\displaystyle b} and W {\displaystyle W} is given by b = x ¯ − W y ¯ , {\displaystyle b={\bar − 6}-W{\bar − 5},} W =

Wiener filter solutions The Wiener filter problem has solutions for three possible cases: one where a noncausal filter is acceptable (requiring an infinite amount of both past and future data), the weblink Let the fraction of votes that a candidate will receive on an election day be x ∈ [ 0 , 1 ] . {\displaystyle x\in [0,1].} Thus the fraction of votes NLRLS algorithm summary The algorithm for a NLRLS filter can be summarized as Initialization: For i = 0,1,...,N δ ¯ ( − 1 , i ) = 0 {\displaystyle {\overline Yes, a practical MMSE implementation needs to know the measure the SNR at the receiver. Minimum Mean Square Error Pdf

More succinctly put, the cross-correlation between the minimum estimation error x ^ M M S E − x {\displaystyle {\hat − 2}_{\mathrm − 1 }-x} and the estimator x ^ {\displaystyle Simplifications For most systems the expectation function E { x ( n ) e ∗ ( n ) } {\displaystyle {E}\left\{\mathbf {x} (n)\,e^{*}(n)\right\}} must be approximated. Thus, we may have C Z = 0 {\displaystyle C_ σ 4=0} , because as long as A C X A T {\displaystyle AC_ σ 2A^ σ 1} is positive definite, navigate here A more balanced linear equalizer in this case is the minimum mean-square error equalizer, which does not usually eliminate ISI completely but instead minimizes the total power of the noise and

thinks Reply chaouki March 24, 2009 at 9:32 pm cher monsieur J'essaye de simuler un système SC-FDMA qui est proche du OFDM. Mean Square Estimation Contents 1 Definition and basic properties 1.1 Predictor 1.2 Estimator 1.2.1 Proof of variance and bias relationship 2 Regression 3 Examples 3.1 Mean 3.2 Variance 3.3 Gaussian distribution 4 Interpretation 5 Hence doing ifft() is not needed.

## Estare esperando tu respuesta.

ISBN0-13-042268-1. By using this site, you agree to the Terms of Use and Privacy Policy. Examples Example 1 We shall take a linear prediction problem as an example. Minimum Mean Square Error Estimation Matlab While these numerical methods have been fruitful, a closed form expression for the MMSE estimator is nevertheless possible if we are willing to make some compromises.

If I would need to do a MIMO system 4×4, how could I change the inverse section? Also in regression analysis, "mean squared error", often referred to as mean squared prediction error or "out-of-sample mean squared error", can refer to the mean value of the squared deviations of By using this site, you agree to the Terms of Use and Privacy Policy. his comment is here Statistical Digital Signal Processing and Modeling.


text:
# Find the Maximum area for the given perimeter of a rectangle. State the length and width of the rectangle.

28 inches

Well, finally a calculus problem.
Ok, we know that the area for a rectangle is
A=l*w and the perimeter is P=2(l+w)
In this problem P= 28, so let's express one of the dimensions in terms of the other an substitute into the area formula. Thus 28=2(l+w). Let's solve for l in terms of w, thus 14=l+w, or l=14-w When we substitute this into the area formula we get
A=(14-w)*w. So A=14w-w^2
Now find dA/dw and evaluate the critical points.
dA/dw = 14-2w and dA/dw = 0 means 14-2w=0
So w = 7 is a critical point. I'll let you verify that this is the max. (the second deriv is -2, what does that mean?)
Thus the rectangle of maximum area has w=7. If you put this back into the formula for the perimeter, you'll find that l=7 too. This means that the rectangle of max. area is a square with a side=P/4.

Considering all rectangles with the same perimeter, the square encloses the greatest area.
Proof: Consider a square of dimensions x by x, the area of which is x^2. Adjusting the dimensions by adding a to one side and subtracting a from the other side results in an area of (x + a)(x - a) = x^2 - a^2. Thus, however small the dimension "a" is, the area of the modified rectangle is always less than the square of area x^2.

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text:
# Understanding minimal casting cost for area spells with high spell level?

On the specific example of the spell:

Sense Foes - Information; Area; Info

Tells the caster if the subject has hostile intent, and what the degree of hostility is. Can be cast on one person or a whole area. If cast over an area, this spell only detects that someone is hostile, without telling who.

Base cost: 1 (minimum 2).

[Source: GURPS 4th Edition Basic Set - Characters p.245 or GURPS Magic p.45]

The section on Area Spells (B239 or M11) explains this minimum as:

The cost listed for an Area spell is always its “base cost.” The actual cost to cast the spell is equal to its base cost multiplied by the radius of the area of effect in yards (minimum one yard): base cost x1 for a one-yard radius, x2 for a two-yard radius, x3 for a threeyard radius, and so on.

Area spells with a fractional base cost, such as 1/2 or 1/10, cost a minimum of one energy point. A few Area spells specify a minimum cost, which you must always pay, even if it is larger than the base cost multiplied by the desired radius.

If I have a high enough skill in Sense Foes to generally be eligible for reduction to casting cost (B237 or M8), does that apply to the minimum as well?

[...] the cost reduction for high skill only applies to the total cost of the spell.

[Source: GURPS Official FAQ 4.3.3] thanks to @Ian Pugsley

Examples:

Sense Foes cast on an area of 1 yard radius calculates at 1FP,
but is raised to 2FP because of the minimum.

Do i get those 2FP waived if I have a skill of 20 or above?

Sense Foes cast on an area of 3 yards radius calculates at 3FP,
not increased because i't above the minimum.

Will skill 20 reduce my cost to 2FP (the minimum) or to 1FP (the standard high skill bonus)?

-
i recall that information spells have their own ranges – Dennis Christian Jun 7 at 10:45
@dennis-christian What does range of information spells got to do with the question? – Yolgie Jun 7 at 10:51
if i recall correctly it is range and area that are replaced, i'll look that up when i'm back home – Dennis Christian Jun 7 at 11:23

This is from p. B236:

Calculate the entire cost for a spell (for instance, by multiplying cost for the size of the subject or the area affected) before applying energy cost reductions for high skill.

You use an area spell's minimum cost to figure what it's total cost is, then you apply your skill-based reductions to that total cost.

So, casting Sense Foes over a 1- or 2-yard area has a total cost of 2FP. That gets reduced to 0 for a Skill 20 caster. Casting it over a 3-yard area costs 3FP, reduced to 1FP.

In other words, "minimum cost" is a rule you use when figuring the total cost of a spell. It doesn't interact directly with the rule for skill-based cost reduction.

-

In general, high skill reductions of FP costs of spells are taken off the total cost of the spell. Meaning you'd evaluate all costs of the spell first, then reduce it based on high skill from the end cost - as you suspected, you'd have the 2FP waived at skill 20 for your first example, and reduced to 1FP in your second. The "minimum costs" are the minimum costs for effective area, not the minimum cost for the entire spell.

It isn't directly relevant, but the "total cost" bit is supported by FAQ 4.3.3:

Note that the cost reduction for high skill only applies to the total cost of the spell.

-
If you could give me a source for "The "minimum costs" are the minimum costs for effective area, not the minimum cost for the entire spell." I'd imediately accept this. Unfortunately I'm looking for more than 'just' opinions. – Yolgie Apr 26 '12 at 11:40


text:
# Persistently damped transport on a network of circles

3 GECO - Geometric Control Design
Inria Saclay - Ile de France, X - École polytechnique, CNRS - Centre National de la Recherche Scientifique : UMR7641
Abstract : In this paper we address the exponential stability of a system of transport equations with intermittent damping on a network of $N \geq 2$ circles intersecting at a single point $O$. The $N$ equations are coupled through a linear mixing of their values at $O$, described by a matrix $M$. The activity of the intermittent damping is determined by persistently exciting signals, all belonging to a fixed class. The main result is that, under suitable hypotheses on $M$ and on the rationality of the ratios between the lengths of the circles, such a system is exponentially stable, uniformly with respect to the persistently exciting signals. The proof relies on an explicit formula for the solutions of this system, which allows one to track down the effects of the intermittent damping.
Keywords :
Type de document :
Article dans une revue
Transactions of the American Mathematical Society, American Mathematical Society, 2017, 369 (6), pp.3841-3881. 〈10.1090/tran/6778〉
Domaine :

Littérature citée [36 références]

https://hal.inria.fr/hal-00999743
Contributeur : Guilherme Mazanti <>
Soumis le : lundi 23 novembre 2015 - 10:38:59
Dernière modification le : jeudi 10 mai 2018 - 02:05:54
Document(s) archivé(s) le : vendredi 28 avril 2017 - 18:30:56

### Fichier

Transport.pdf
Fichiers produits par l'(les) auteur(s)

### Citation

Yacine Chitour, Guilherme Mazanti, Mario Sigalotti. Persistently damped transport on a network of circles. Transactions of the American Mathematical Society, American Mathematical Society, 2017, 369 (6), pp.3841-3881. 〈10.1090/tran/6778〉. 〈hal-00999743〉

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text:
# Calculate expected voltage around a resistor [duplicate]

It's been a few years since I first studied the Ohm's law at school. Now I'm really getting into electronics, and I must admit a part of it still puzzles me.

I know this is kind of a strange question, but I have no clue how to express it otherwise, so I'll just lay out my reasoning. The following train of thoughts is wrong and/or incomplete. Please stop me whenever I'm wrong (even for the sligthest thing), and complete this reasoning.

Here it comes:

Let's make a circuit with:

• 6V battery (typically 4 AA batteries)
• 2k ohms resistor
• LED, for the sake of doing something

Circuit map:

simulate this circuit – Schematic created using CircuitLab

From Ohm's law, we know that the current in the circuit will be of (considering only resistor has any resistance)

$I = \dfrac{V}{R} = 3mA$

I do understand that the current is limited due to the resistor... crystal clear.

The book tells me that, if I measure the potential difference between both sides of the resistor, I will have something between 0 and 6 volts.

Sounds legit:

• can't be 6 volts because the two ends of the battery are connected, therefore lowering the potential
• can't be 0 volts, since the current does not freely move around, due to the resistor; resulting in one side being more negatively charged, and the other more positively charged, since the electrons can move faster to an end of the battery than through the resistor.

What I don't understand (and if my whole reasoning is right until now), is: How can I calculate the expected voltage around the resistor ?

• For the next time: we have an on-site schematic editor, you can launch it with Ctrl-M. – user17592 Apr 16 '13 at 16:52
• Brilliant ! Will do :) – Antoine_935 Apr 16 '13 at 17:01
• I'd like to clarify that the question is not an exact duplicate of the linked one. Just, it provides a complete and thorough explanation about how circuit with LEDs behave, since your question - while being about the resistor - is strongly dependent on the LED behavior. I hope you can find there the answer to your question, otherwise you can tell what's missing and ask for reopening. – clabacchio Apr 16 '13 at 20:15
• Excellent, thanks for the link and for adding the schema! – Antoine_935 Apr 17 '13 at 8:26

From the Ohm's law, we know that the current in the circuit will be of (considering only resistor has any resistance)

Na-ah, the LED has a resistance too. Common LEDs use 2.5V at 20mA. We can calculate the LED's resistance (on 2.5V) with $R=\frac{V}{I}=\frac{2.5}{0.02}=125\Omega$.

Now, when it comes to your circuit, you shouldn't just use a 2K resistor, you should think about that and calculate the needed value. Now, the voltage over the LED probably won't be 2.5V, which means the LED has a different resistance, different current is flowing, etc. You cannot calculate the current flowing in this circuit and therefore you cannot calculate the voltage over the resistor (we'll come to that later). Here's how you should design your circuit:

You want a 0.02A current through the circuit (the current is the same everywhere as it is a series circuit). That means the total resistance has to be $R_{tot}=\frac{V}{I}=\frac{6}{0.02}=300\Omega$. We already saw the resistance of the LED was $125\Omega$, that means your resistor has to be $300-125=175\Omega$.

Now, how do you calculate the voltage over the resistor? Take Ohm's law, the current through the resistor (0.02A) and the resistance ($175\Omega$):

$$V=I\cdot{}R = 0.02\cdot175 = 3.5V$$

Note that this also equals $6V-2.5V$.

• Excellent answer! Many thanks. For the sake of curiosity, what happens if the resistor was less than 175 ohms? Would that create a so-called "short circuit", because not all voltage is taken? On the opposite, what happens to the voltage if I insist in using this 2k resistor? It really cannot be calculated? – Antoine_935 Apr 16 '13 at 17:20
• @Antoine_935 "not all voltage is taken" - no, all voltage will be taken, but because of the lower resistance, the current will increase (V=IR). So it might damage your components. Your LED will be brighter though. The problem with the 2K is that it falls out of the LEDs specs. LEDs don't follow Ohm's law, so the specs can only be applied on the intentional circumstances. It really cannot be calculated thus, but you can of course just measure it with a voltmeter. – user17592 Apr 16 '13 at 17:22
• Waw, some things do not follow ohm's law? Totally ignored that. I guess I'll ask another question "only" for this. Again, many thx! – Antoine_935 Apr 16 '13 at 17:25
• @Antoine_935 it's indeed best to ask a separate question for that. And you're welcome, of course :) – user17592 Apr 16 '13 at 17:26
• It is bad practice to calculate the effective resistance of an LED as it varies wildly with current. It is better to model an LED with a voltage drop. – jippie Apr 16 '13 at 18:13

The sum of all voltage drops must equal the source voltage.

$V_{source} = V_{resistor} + V_{led}$

Depending on what wavelength the LED the operates at, it will essentially drop the same the voltage (when you are within the current restrictions of the LED).

If we assume that the LED drops 1.2V, then that means that the resistor must drop 4.8V.

The LED forward voltage will always be a constant (might deviate slightly), when you are within the current limits.

If the LED can take a max of 20mA, then at 15mA and 10mA, the voltage drop across the LED will always be the forward voltage drop (in this example, the 1.2V).

Make sense ?

• Nice answer, it's helping. I'd upvote it too if I could, thanks. – Antoine_935 Apr 16 '13 at 17:21
• The forward voltage is not constant. As LEDs heat up, the forward voltage drops and the current passing through the LED increases, which creates additional heat. This can lead to thermal runaway and even destroy the diode. This is why constant current source are preferred when driving LEDs. – Rev Apr 16 '13 at 20:41

Your LED will have about 2 volts across it for a standard LED which means you'll have about 2mA flowing in the circuit. It won't be very bright.

When you said: -

I = V/R -> 3mA

That's where you made an error. Even though the LED may be talked-about as having little resistance it still drops a voltage across it and for your average LED it will be about 2V whether you're feeding it with 2mA or 20mA.

• Nice answer, it's helping. I'd upvote if I could, thanks. – Antoine_935 Apr 16 '13 at 17:20
• Note that the actual forward voltage depends on the forward current. This is why most diode data-sheets provide a "fwd voltage VS current" graph. From the answer is sounds like this isn't the case. – Rev Apr 16 '13 at 20:24
• @Rev1.0 - it is simplistic i agree – Andy aka Apr 16 '13 at 20:54