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A question about prefibrations and lax tranformations
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Let $P: \mathcal{C}\to \mathcal{A}$ be a functor. We call a morphism $v$ of $\mathcal{C}$ vertical (over $A\in \mathcal{A}$) if $P(v)=1_A$; then we have the fibre category $\mathcal{C}_A$ of vertical
morphisms over $A$. We call a morphism $f: X\to Y$ in $\mathcal{C}$ cartesian if for $f': X'\to Y$ with $P(f)=P(f')$, there exists a unique vertical morphism $v: X'\to X$ such that $f'=f\circ v$.
If $f: A\to B$ is in $\mathcal{A}$ and $Y\in \mathcal{C}(B)$, we call a cartesian morphism of type $\theta_f(Y): f^\ast Y\to Y$ with $P(\theta_f)=f)$ a cartesian $f$-lifting of $Y$. If there exists
such for each $f$ and $Y$, we call $P$ a prefibration, and a precleavage is a choice of $\theta_f(Y)$ for each $f$ and $Y$. If we choose $\theta_{1_A}=1_A$, the preclivage is called normal.
After fixing a cleavage, we obtain a functor $\theta_f: \mathcal{C}(B)\to \mathcal{C}(A)$, and for $X\xrightarrow{f}Y\xrightarrow{g}Z$ there is a transformation $c_{g, f}: f^\ast\circ g^\ast\
Rightarrow (g\circ f)^\ast$ (considering $ \theta_g(Z)\circ \theta_f(g^\ast(Z))$). Thus, we have a lax functor $\textbf{C}: \mathcal{A}^{op}\to CAT$ defined as $\textbf{C}(A)=\mathcal{C}(A)\ A\in\
mathcal{A}$, $\textbf{C}(f)=f^{\ast}$,
$C_{g,f}=c_{g, f}: \textbf{C}(f)\circ \textbf{C}(g)\Rightarrow \textbf{C}(g\circ f)$.
Let $P: \mathcal{C}\to \mathcal{A}$ and $Q: \mathcal{D}\to \mathcal{A}$ be prefibrations with fixed normal cleavages. If $T: \mathcal{C}\to \mathcal{D}$ is a functor over $\mathcal{A}$, i.e. with $Q\
circ T= P$, then by restriction we have functors $T_A: \textbf{C}(A)\to \textbf{D}(A)$. Applying $T$ to $\theta_f(Y)$ and considering the cartesian $\theta_f(T(Y))$ (in $\mathcal{D}$), we have a
natural vertical morphism $T_f(Y): T(f^\ast(Y))\to f^\ast(T(Y))$, which defines a transformation $T_f: T_A\circ \textbf{C}(f)\Rightarrow \textbf{D}(f)\circ T_B$. The data $(T_A, T_F)_{A, f}$ is what
is called an oplax transformation. In the other direction, from an oplax transformation $(T_A, T_F)_{A, f}$ we can make a functor $T: \mathcal{C}\to \mathcal{D}$ over $\mathcal{A}$ (considering that
each morphism over $\mathcal{C}$ or in $\mathcal{D}$ has a unique factorization as a vertical morphism followed by a cartesian morphism).
Thus the oplax transformations are identified with functors $T$ over $\mathcal{A}$.
My question is:
If (instead of an oplax tranformation) we consider a lax tranformation $T$, i.e. a family
$T_A: \textbf{C}(A)\to \textbf{D}(A)$ and $T_f: \textbf{D}(f)\circ T_B \Rightarrow T_A\circ \textbf{C}(f)$
with the coherence conditions:
$T_{gf}\ast (d_{g, f})\circ T_C)=$
$(T_A\circ c_{g, f})\ast(T_f\circ \textbf{C}(g))\ast(\textbf{D}(f)\circ T_g): \textbf{D}(f)\circ\textbf{D}(g)\circ T_C\Rightarrow T_A\circ \textbf{D}(f)\circ\textbf{D}(g)$
(the unitary condition being unnecessary, by our choice of normal cleavages).
Is there some way to represent $T$ as a categorical construction?
If I understand correctly what you are asking, this is a good question. But before I edited it for spelling, grammar, and punctuation, it was almost impossible to understand. I appreciate that
English can be a difficult language for non-native speakers, but as a practical matter, you might have more luck getting answers if your questions were written in a way that was easier to
understand. – Mike Shulman Aug 13 '12 at 6:14
And as long as I'm giving advice, it's probably not necessary to rehash the definitions of cartesian arrow, cleavage, and (pre)fibration in the question. Giving background is good, but not so much
background that it overwhelms the question being asked -- conciseness is also a virtue, and you can always include a link for more background details. I suppose opinions may differ on this point,
though. – Mike Shulman Aug 13 '12 at 6:17
One more thing: if you can be more precise in the question title than "A question about ...", you'll get more people clicking on it. A title for this question might be "Can lax transformations be
represented in terms of fibrations?" – Mike Shulman Aug 13 '12 at 6:19
thank you for suggestions – Buschi Sergio Aug 13 '12 at 16:35
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1 Answer
active oldest votes
I guess that what you mean is, is there some way to represent $T$ in terms of the (pre)fibrations $P:\mathcal{C}\to \mathcal{A}$ and $Q:\mathcal{D}\to \mathcal{A}$? As far as I know, the
answer is no — at least, nothing so natural as how oplax transformations are represented by plain functors over $\mathcal{A}$.
Here's a bit more abstract context, to make this answer seem less negative. (-: There's a general construction (due originally to Hermida, I believe -- see his paper From coherent
structures to universal properties) which, given a well-behaved 2-monad $M$ on a well-behaved 2-category $\mathcal{X}$, produces a new 2-monad $\hat{M}$ on a new 2-category $\hat{\mathcal
{X}}$ such that:
• $M$-algebras can be identified with $\hat{M}$-algebras;
• pseudo $M$-morphisms can be identified with pseudo $\hat{M}$-morphisms;
• colax $M$-morphisms can be identified with colax $\hat{M}$-morphisms; and
• $\hat{M}$ is colax-idempotent, i.e. every morphism in $\hat{\mathcal{X}}$ between $\hat{M}$-algebras is a colax $\hat{M}$-morphism in a unique way.
up vote 2
down vote One property of colax-idempotent 2-monads is that if a morphism between algebras is lax, then the lax structure map automatically becomes an inverse to the unique colax structure map and
makes it pseudo. Thus every lax morphism is automatically pseudo. Hence, in the above theorem we cannot expect to identify lax $\hat{M}$-morphisms with lax $M$-morphisms, unless $M$ already
had the very special property that all lax morphisms are pseudo.
For the case in point, $\mathcal{X}=\mathrm{Cat}^{\mathrm{ob}(\mathcal{A})}$ and $M$ is the 2-monad whose algebras are normal lax functors. Then $\hat{\mathcal{X}}$ is $\mathrm{Cat}/\
mathcal{A}$ and $\hat{M}$ is the 2-monad whose algebras are prefibrations. Thus, every morphism in $\mathrm{Cat}/\mathcal{A}$ between prefibrations is a colax $\hat{M}$-morphism,
corresponding to a colax $M$-morphism, i.e. an oplax transformation — but there is no natural way to see the lax $M$-morphisms in terms of $\hat{M}$-algebras.
Another example (perhaps the other canonical example) is when $\mathcal{X}=\mathrm{Cat}$ and $M$ is the 2-monad whose algebras are monoidal categories. Then $\hat{\mathcal{X}}$ is the
2-category of multicategories, and $\hat{M}$-algebras are representable multicategories.
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Ideals generated by idempotent elements
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Let $R$ be a unitary associative ring and $J$ be an ideal of $M_{n}(R)$. We know that there is an ideal $I$ of $R$ such that $J=M_{n}(I)$. Now there is a question.
Question: If $J$ is generated by a subset of idempotent elements of $M_{n}(R)$ say $S$, is $I$ generated by a subset of idempotent elements of $R$, which related to $S$?
1 Dear all, I think if $R$ is an integral domain, in the ring $M_2(R)$ your statement is trivially true. because in this case, we could characterize all of idempotents. all of idempotents have the
form $0$, $I_2$ or a matrix whose entries are $a_{11}=a$, $a_{12}=b$ , $a_{21}=c$, $a_{22}=1-a$ and $a,b,c$ satisfy in the relation $bc=a-a^2$. so in this case the only nonzero ideal in $M_2(R)$
which contains an idempotent is the total ring. so in this case, your claim is true. – Ali Reza Nov 27 '12 at 7:34
What do you mean by "$J$ is generated by idempotents?" Wouldn't it contain one of such, and then contain a unit, thus being the whole $M_n(R)$? – Filippo Alberto Edoardo Nov 28 '12 at 1:56
Dear Filippo, $J$ is generated by a subset of idempotent elements as a two-sided ideal of $M_[n}(R)$. If one is unit, then is trivial. – Ali Nov 28 '12 at 4:44
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1 Answer
active oldest votes
I don't know if this helps. We readily obtain the following assertions:
(1) If the ideal $J=M_n(I)$ of $M_n(R)$ is generated by some idempotents, then $I$ is an idempotent ideal, i.e., $I^2=I$.
up vote 2 (2) If $J=M_n(I)$ is generated by finitely many idempotents, then $I$ is a finitely generated ideal.
down vote
Thus in the case where $R$ is commutative and $J$ is generated by finitely many idempotents, there exists an idempotent $e\in R$ such that $I=Re$.
1 I think this solves the problem. Choose a set of generators of $J$. For each generator, look at the ideal generated by just that, and apply this argument to find an idempotent. Then
the ideal generated by all those idempotents is $I$. – Will Sawin Jan 18 '13 at 18:04
Dear all, thank of yours answer, but these are not true. – Ali Jan 19 '13 at 6:35
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Various Formulae
Recruit Minion XP
LMXP = Largest Minion XP = the total XP (trained or untrained) on the largest minion
• Cheap Option Cost = aprox. LMXP*0.13(3) ; XP on new minion = Cost/5 = aprox. LMXP/37.5
• Expensive Option Cost = aprox. LMXP*0.33(3) ; XP on new minion = Cost/3 = aprox. LMXP/9
Average Rewards
THIS IS OUTDATED (Natasha')
The formula below can be attributed to the user Avoid and is commonly referred to as Avoid's method. It works best if you do your calculations on Sunday when no clan bonuses are in play.
The first step is to attack a character twice and record the cash rewards from each battle. If you get the same number twice, or if they are only a dollar or two off, you'll need a third battle.
Then Take the higher of your two values and divide it by the lower.
• If you get 2.25,2,1.75 or 1.25, multiply the lower cash value by 13/8 to get your average cash reward.
• If you get 1.8,1.6,1.4 or 1.2, multiply the lower cash value by 13/10 to get your average cash reward.
• If you get 1.333 or 1.667, multiply the lower cash value by 13/12 to get your average cash reward.
• If you get 1.286 or 1.143, multiply the lower cash value by 13/14 to get your average cash reward.
• If you get 1.125, multiply the lower cash value by 13/16 to get your average cash reward.
• If you get 1.5, there are two possibilities for what your average cash reward can be - low*13/8 or low*13/12. You can look at both and guess which is right, or you can fight another battle and
try again, or you can find a nice quiet corner and cry.
This doesn't work for xp; only cash, but you can get your average xp rewards by dividing your average cash rewards by 1.655. This method is hardly accurate anymore since it does not take account the
Challenge Bonus and how its different per character. Since it can range from +100% to -100%.
• If you get 2, 1.75, 1.5, or 1.25, multiply the lower cash value by 3/2 to get your average cash reward.
• If you get 1.6, 1.4, or 1.2 multiply the lower cash value by 6/5 to get your average reward.
• If you get 1.33 or 1.167 the lower cash value is your average reward.
• If you get 1.142 multiply the lower cash value by 5/6 to get your average reward.
BA Regen Thresholds
TopMPR in the following formula is defined as the highest MPR character in the game. This can be found from your "Home" page>Character standings>Minions PR link. The calculated value for moving to a
different BA Regen threshold is based on your VPR (MPR+experience) and the top MPR.
• 9/20=TopMPR*.025
• 8/20=TopMPR*.15
• 7/20=TopMPR*.4
• 6/20=TopMPR*.6
Character Transfer Cost
VMPR is equal to all trained & untrained XP. If all of your XP is trained, then VMPR is equal to MPR.
• 4/3*VMPR + 100k
Solving for XP given MPR & vice versa
This equation is accurate to within 1% and was provided by Miandrital & Lochnivar.
• EXP = (1.4307*(MPR^1.2501))
• MPR = (EXP/1.4307)^(1/1.2501)
Solving for MTL given MPR & vice versa
• MTL = (MPR^1.2501)*1.4307/24
• MPR = (MTL*24/1.4307)^(1/1.2501)
Tattoo growth rates
• "Lesser" tattoos grow at 1/3 the rate of MTL = one level for every 72 XP earned by the character's team
• "Regular" tattoos grow at 2/3 the rate of MTL = one level for every 36 XP earned by the character's team
Physical damage formula
• Weapons are listed as Weapon name [CxWDM] (+PTH), where C - some constant determining a rough power of the weapon ; WDM - weapon damage multiplier (a.k.a. "weapon x") ; PTH - weapon added point
to hit (a.k.a. "weapon +")
• Damage = Base damage * SQRT ( (ST/20) * Weapon_damage_multiplier )
• Base damage is the (min/average/max) damage dealt by that particular weapon at the base weapon stats from a 20 ST minion
• That base damage is NOT the "C" number listed in the weapon description, but is based on it (exact formula unknown at this time)
• SQRT is the Square Root function (or, in other words X^0.5)
What this means in layman's terms is that the base reference points of all weapons is the damage dealt by an x1 weapon of that type wielded by a minion with 20 ST.
In order to double the damage dealt (i.e. increase damage by a factor of two), you can either:
• increase ST by a factor of four
• or you can increase the weapon X by a factor of four
• or you can increase both ST and weapon X by a factor of two each.
To-Hit Calculations
Since we have a page devoted to the To-hit calculation, follow the link to see the in-depth method of calculation.
last edited by Fuhgawz at Aug 19 2010 - Edit Various Formulae
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search results
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1. CMB 2011 (vol 55 pp. 821)
New Examples of Non-Archimedean Banach Spaces and Applications
The study carried out in this paper about some new examples of Banach spaces, consisting of certain valued fields extensions, is a typical non-archimedean feature. We determine whether these
extensions are of countable type, have $t$-orthogonal bases, or are reflexive. As an application we construct, for a class of base fields, a norm $\|\cdot\|$ on $c_0$, equivalent to the canonical
supremum norm, without non-zero vectors that are $\|\cdot\|$-orthogonal and such that there is a multiplication on $c_0$ making $(c_0,\|\cdot\|)$ into a valued field.
Keywords:non-archimedean Banach spaces, valued field extensions, spaces of countable type, orthogonal bases
Categories:46S10, 12J25
2. CMB 2007 (vol 50 pp. 588)
Cohomological Dimension and Schreier's Formula in Galois Cohomology
Let $p$ be a prime and $F$ a field containing a primitive $p$-th root of unity. Then for $n\in \N$, the cohomological dimension of the maximal pro-$p$-quotient $G$ of the absolute Galois group of
$F$ is at most $n$ if and only if the corestriction maps $H^n(H,\Fp) \to H^n(G,\Fp)$ are surjective for all open subgroups $H$ of index $p$. Using this result, we generalize Schreier's formula for
$\dim_{\Fp} H^1(H,\Fp)$ to $\dim_{\Fp} H^n(H,\Fp)$.
Keywords:cohomological dimension, Schreier's formula, Galois theory, $p$-extensions, pro-$p$-groups
Categories:12G05, 12G10
3. CMB 2003 (vol 46 pp. 388)
Tracially Quasidiagonal Extensions
It is known that a unital simple $C^*$-algebra $A$ with tracial topological rank zero has real rank zero. We show in this note that, in general, there are unital $C^*$-algebras with tracial
topological rank zero that have real rank other than zero. Let $0\to J\to E\to A\to 0$ be a short exact sequence of $C^*$-algebras. Suppose that $J$ and $A$ have tracial topological rank zero. It
is known that $E$ has tracial topological rank zero as a $C^*$-algebra if and only if $E$ is tracially quasidiagonal as an extension. We present an example of a tracially quasidiagonal extension
which is not quasidiagonal.
Keywords:tracially quasidiagonal extensions, tracial rank
Categories:46L05, 46L80 | {"url":"http://cms.math.ca/cmb/kw/extensions","timestamp":"2014-04-20T23:30:10Z","content_type":null,"content_length":"29572","record_id":"<urn:uuid:d7935716-7ee2-4124-937d-dcb9ea3d87ca>","cc-path":"CC-MAIN-2014-15/segments/1397609539337.22/warc/CC-MAIN-20140416005219-00127-ip-10-147-4-33.ec2.internal.warc.gz"} |
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dc-dc converter- averaging
1. 30th December 2008, 11:23 #1
Full Member level 5
Join Date
Jul 2006
23 / 23
state space averaging
usually to analyze a switched mode dc-dc converter we need to average the equations across the 2 states(ON and OFF). now my doubt is why must we linearize it(by perturbation) after
does not averaging itself linearize the switched circuit, as the only nonlinear element was the switch??
thank you,
2. 30th December 2008, 15:32 #2
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Jun 2008
35 / 35
state-space average nonlinear waveform converter
i have not heard of this method of which you speak...
if you google "feedback compensation"
"type 3 compensation networks"
(also 1 and 2)
you'll get some ideas...but i have never heard of your methods of averaging on/off etc.
Lloyd dixon does some good notes for googling.
□ 30th December 2008, 15:32
3. 30th December 2008, 19:47 #3
Advanced Member level 4
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Dec 2006
391 / 391
dynamic model dc dc converters
Also try the patents on http://www.google.com/patents you will find the modern methods
4. 31st December 2008, 16:25 #4
Full Member level 5
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Jul 2006
23 / 23
dc-dc converter state space
i am not talking about the compensation scheme. the averaging and linearization i mentioned was that pertaining to the small-signal representation of the switching converters.
5. 4th January 2009, 07:25 #5
Member level 2
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Oct 2006
11 / 11
average model dcdc converter
The averaging only screen out the effect of switching. The average model still be a non-linear model. Linearization is necessary to apply linear control theory.
6. 4th January 2009, 08:52 #6
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Jan 2008
Bochum, Germany
7949 / 7949
averaging state space dc dc converter
I think there are different aspects involved with your question, and there is possibly no simple yes or no answer. You're talking of a small-signal equivalent ciruit, which is a usual method
for modelling of switching circuits, I think. By definition, it's linear but valid for a particular operation point only. If your analysis also implies larger variations of the operation
point, the linear model is more-or-less incorrect. You have to find out, if it can give meaningful results though.
Because of this basic limitation, I'm not sure if it's of much use, to perform an accurate averaging of equations. Personally, I prefer simple equivalent circuits, that represent the basic
properties of the circuit. For some circuits, that have a basical nonlinear behaviour, e. g. a buck converter in non-continuous mode, you may be able to define an equivalent circuit composed
of a linear term and operation point dependant gain. But I don't think, there's a general numerical method, it's rather a matter of intuitive understanding.
7. 4th January 2009, 11:51 #7
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Jul 2006
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linear state feedback boost converter
as far as switching circuits are concerned averaging is basically done to remove the switch and decompose the switched circuit into a continuous time equivalent circuit model. this decomposed
model has only resistors, capacitors and inductors. so it is a completely linear circuit. (we don't do small signal analysis for completely passive networks).
but, still all the text books carry on the step of linearization after averaging. why is that so? am i missing out on anything?
8. 4th January 2009, 13:55 #8
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Jan 2008
Bochum, Germany
7949 / 7949
reference model of dc dc converter
You may want to give an example or a literature reference to make "linearization after averaging" understandable. As you said, a linear cicrcuit can't be linearized anymore.
9. 5th January 2009, 06:46 #9
Member level 2
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Oct 2006
11 / 11
state space model of dc dc converter
To amriths04,
IT IS A NON-LINEAR MODEL! if you took the duty ratio as a state variable. You CANNOT took the duty ratio as a constant in deriving the dynamic model of the system.
10. 5th January 2009, 12:13 #10
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Jul 2006
23 / 23
averaging in power electronic
fundamentals of power electronics by robert erickson and any TI manual dealing with small-signal analysis on buck design.
after averaging we might have controlled sources dependent on duty ratio in our circuit. but how do you say it is non-linear? could you please elaborate this one out?
11. 8th January 2009, 05:47 #11
Member level 2
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Oct 2006
11 / 11
average model of dc-dc converter
fundamentals of power electronics by robert erickson and any TI manual dealing with small-signal analysis on buck design.
after averaging we might have controlled sources dependent on duty ratio in our circuit. but how do you say it is non-linear? could you please elaborate this one out?
To amriths04,
I do not like to involve too much math here, so I just give some hints and you find out the rest yourself.
I think you already have the book "Fundamental of Power Electronics" by Robert Erickson, so you have the state-space averaging model of one of the DC-DC converters. Now I tell you that the
duty ratio is a state variable as the cap voltage and inductor current. Take a look of the model, you will find that the product term of the duty ratio and the other state variables make it
non-linear. So IT IS A NON-LINEAR MODEL.
12. 8th January 2009, 21:01 #12
Newbie level 4
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Jan 2009
0 / 0
robert erickson dc dc converter
I am currently studying this book for my MSEE. My understanding is that we use inductor volt-second balance and Capacitor Charge balance andt eh small-ripple approximation to eliminate the
hassle in using long mathematical equations. Therefore the book focuses on steady-state and at the same time neglecting the ripple effect seen on the DC output. When we do this and look at
the waveform in steady state we see a linear response in a buck boost converter etc.
13. 14th January 2009, 16:33 #13
Newbie level 1
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Jan 2009
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non linear buck converter circuit
if you take the duty cycle ratio as input variable; you will see that duty cycle cannot be greater than 1(%100) as its definition. (for full bridge it swings between +1 and -1)
So there is a saturation effect..
So there comes nonlinearity..
14. 27th June 2011, 10:50 #14
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Jun 2011
0 / 0
Re: state space averaging
usually to analyze a switched mode dc-dc converter we need to average the equations across the 2 states(ON and OFF). now my doubt is why must we linearize it(by perturbation) after
does not averaging itself linearize the switched circuit, as the only nonlinear element was the switch??
thank you,
averaging does not linearize the model, rather averaging is done to cancel off the switching harmonics brought about by the switch... i think u got the logic wrong... go backto erickson and
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The intermediate value theorem
14 The intermediate value theorem
We can now prove the intermediate value theorem within ASD, in the two forms (non-singular and singular) that we discussed in Section 2. In the non-singular case on a closed bounded interval, the
approximate forms of the theorem in the previous section are enough to ensure that the solution-set S[f]=Z[f] is compact, overt and non-empty, and therefore has a maximum element. We begin by
formulating the relevant definitions from Sections 1 and 2 in ASD.
Definition 14.1 A function …, x:ℝ ⊢ f x:ℝ (Definition 5.3)
1. doesn’t hover (cf. Definition 1.4) if, for d,u:ℝ,
d< u =⇒ ∃ x.(d< x< u) ∧ (f x≠ 0);
2. doesn’t touch from below without crossing if
(d< u) ∧ (f d< 0) ∧ (f u< 0) =⇒ (∀ x:[d,u].f x< 0) ∨ (∃ x:[d,u].f x> 0);
3. and doesn’t touch from above without crossing if
(d< u) ∧ (f d> 0) ∧ (f u> 0) =⇒ (∀ x:[d,u].f x> 0) ∨ (∃ x:[d,u].f x< 0).
4. A real number a:ℝ is a stable zero of f (cf. Definition 1.8) if
(d< a< u) =⇒ ∃ e t.(d< e< t< u) ∧ (f e< 0< f t ∨ f e> 0> f t);
5. and the possibility modal operator ◊:Σ^Σ^ℝ is, for φ:Σ^ℝ,
◊φ ≡ ∃ d< u. (∀ x:[d,u].φ x) ∧ (f d< 0< f u ∨ f d> 0> f u),
so f d< 0< f u=⇒◊⊤.
6. The solution co-classifier (or non-solution classifier) is ω x ≡ (f x≠ 0), which defines an open subspace W[f] and a closed one Z[f]; and,
7. restricting to an interval I≡[d,u], the necessity modal operator ▫:Σ^Σ^I is
▫φ ≡ ∀ x:I.(f x≠ 0)∨φ x,
which makes Z[f] a compact subspace.
Proposition 14.2 The stable zeroes are exactly the members or accumulation points of ◊ in the sense of Definition 11.5.
Proof If a is a stable zero then, by Theorem 10.2,
φ a ⇒ ∃ d u.(d< a< u) ∧ ∀ x:[d,u].φ x
⇒ ∃ d e t u.(d< e< t< u) ∧ (f e< 0< f t∨ f e> 0> f t) ∧ ∀ x:[e,t].φ x
⇒ ◊φ.
Conversely, suppose that φ a⇒◊φ and let d< a< u. With φ x≡(d< x< u),
⊤ ⇔ φ a ⇒ ◊φ ⇒ ∃ e t. (d< e< t< u)∧(f e< 0< f t∨ f e> 0> f t). ▫
Recall from Definition 11.1 that ◊ and ω define a closed overt subspace iff they satisfy ◊ω⇔⊥ and relative instantiation, φ x⇒ω x∨◊φ. The first of these comes for free:
Proof With φ x≡(f x> 0) and ψ x≡(f x< 0),
◊ω ≡ ∃ d< u.(∀ x:[d,u].f x≠ 0) ∧(f d< 0< f u∨ f u< 0< f d)
⇒ ∃ d< u.(∀ x:[d,u].φ x∨ψ x) ∧(∃ y:[d,u].φ y) ∧(∃ z:[d,u].ψ z)
⇒ ∃ d< u.∃ w:[d,u].φ w∧ψ w
⇒ ∃ w:ℝ.0< f w< 0 ⇔ ⊥
by connectedness of [d,u] in the form of Proposition 13.7(b). ▫
Hence if a∈◊ then a∈{x ∣ ¬ω x}, or S[f]⊂ Z[f] in the notation of Section 2. In order to characterise the non-singular case, S[f]=Z[f], we therefore show that relative instantiation is equivalent to
the conditions in Definition 14.1.
Lemma 14.4 If f doesn’t hover or touch without crossing then ◊ and ω satisfy relative instantiation.
Proof If φ x then ∃ d u.(d< x< u) ∧ ∀ y:[d,u].φ y by Theorem 10.2, and, since f doesn’t hover in (d,x) or (x,u), there are d< e< x< t< u with f e≠ 0 and f t≠ 0, so without loss of generality f d≠
0 and f u≠ 0.
This gives four cases, according as f d< 0 or f d> 0 and as f u< 0 or f u> 0. Two of them say
∃ d< u. (f d< 0< f u∨ f d> 0> fu)∧∀ y:[d,u].φ y,
which is ◊φ. Since f doesn’t touch from below without crossing, i.e.
(f d< 0∧ f u< 0) =⇒(∀ y:[d,u].f y< 0)∨(∃ y:[d,u].f y> 0),
in the third case we deduce (f x≠ 0) from the ∀ disjunct, whilst the other one provides a straddling interval [d,y] and so ◊φ. The fourth yields the same conclusion since f doesn’t touch from above
without crossing. Hence φ x⇒ω x∨◊φ in all four cases. ▫
Lemma 14.5 If ◊ and ω satisfy relative instantiation then f doesn’t hover.
Proof Given d< u, consider φ x ≡ (d< x< u), for which
◊φ ≡ ∃ e t.(d< e< t< u)∧(f e< 0< f t∨ f e> 0> f t)
⇒ ∃ y.(d< y< u)∧ (f y≠ 0).
Then (d< x< u) =⇒ (f x≠ 0) ∨ ◊φ =⇒ ∃ y.(d< y< u)∧ (f y≠ 0). ▫
Lemma 14.6 If ◊ and ω satisfy relative instantiation then f doesn’t touch from below without crossing.
Proof We are given e< t with f e< 0 and f t< 0. By Theorem 10.2, there are d< e< t< u with ∀ x:[d,e].f x< 0 and ∀ x:[t,u].f x< 0. Relative instantiation for φ x≡(d< x< u) gives
⊤ ⇔ ∀ x:[e,t].φ x
⇒ ∀ x:[e,t].(◊φ∨ f x< 0∨ f x> 0)
⇔ ◊φ ∨ ∀ x:[e,t].(f x< 0∨ f x> 0) Proposition 8.2(c)
⇒ ◊φ ∨ (∀ x:[e,t].f x< 0) ∨ (∀ x:[e,t].f x> 0)
by compact connectedness of [e,t], since f x< 0 and f x> 0 are disjoint. But the last disjunct contradicts the hypotheses that f e< 0 and f t< 0. Finally,
◊φ =⇒ ∃ x:[d,u].f x> 0 =⇒ ∃ x:[e,t].f x> 0
since ∀ x:[d,e].f x< 0 and ∀ x:[t,u].f x< 0. ▫
Theorem 14.7 The following are equivalent for f:ℝ→ℝ:
1. all zeroes are stable;
2. f doesn’t hover or touch without crossing;
3. ◊ and ω satisfy relative instantiation, φ x =⇒ ω x∨ ◊φ, so they define a closed overt subspace; and
4. in any interval, ◊ and ▫ satisfy the mixed modal law ▫(φ∨ψ) ⇒ ▫φ∨◊ψ, so they define a compact overt subspace of that interval.
Proof [a⊣⊢c] by Lemma 14.3 and Proposition 11.6, [b⊣⊢c] by the previous three lemmas and [c⊣⊢d] by Proposition 12.1. ▫
Corollary 14.8 In this case, Bolzano’s formula for a zero in Theorem 1.1(a),
is, after all, not only valid in ASD but also computationally meaningful.
Proof The compact overt subspace is either empty or has a maximum (Theorem 12.9), but it cannot be empty by Propositions 13.4f. ▫
Now we turn to the singular case, of functions that may touch without crossing. We are still looking for stable zeroes, because finding tangential points requires other evidence that the function
does take a value that is equal to zero (Section 1). However, the impact of the following results (which were developed while this paper was already in the reviewing process) is on the case where the
stable and unstable zeroes are densely mixed, rather than simply for polynomials with double zeroes.
In the non-singular case the operator ◊ had very strong topological properties (relative instantiation and the mixed modal law) that related it to the closed or compact space of all zeroes, but this
knowledge is no longer of any help now that the two spaces are different. Note that the Brouwer degree is not defined in this situation either.
Nevertheless, we argued that Theorems 2.5 and 2.8 capture the computational ideas behind solving equations. However, we leave the translation of the former into ASD as an exercise, and instead
consider a new argument that has the generality of the classical intermediate value theorem.
Example 1.2 is typical of intermediate-value questions that arise in the real world, such as exactly when the boom of 2007 turned into the crash of 2009: the more economic indicators we investigate,
the muddier it becomes. The common-sense answer is that it happened during a certain interval, on which we may place upper and lower bounds. However, since these are extremely imprecise, this
concession to classical pure mathematicians will not appeal to numerical analysts.
We therefore propose
Definition 14.9 A solution of an equation f x=0 (where f:[d,u]→ℝ with f d< 0< f u) is not necessarily a point but an occupied compact interval, i.e. a Dedekind pseudo-cut [δ,υ] in the sense of
Definition 9.9. Of course, there may be many such intervals, just as f may have many zeroes in the usual pointwise sense.
Definition 14.10 An open subspace φ:Σ^ℝ is contour-closed if it is a union of open intervals at whose (Euclidean) endpoints the function is non-zero:
φ x ⇔ ∃ d u.(d< x< u) ∧ (f d≠ 0) ∧ (f u≠ 0) ∧ ∀ y:[d,u].φ y.
We are thinking of geographical contours here: if φ includes part of a hovering interval then it contains the whole of it. In Example 1.2, the interval (0,3) is contour-closed but (0,1⅔) and (1⅓,3)
are not. On the other hand, f doesn’t hover iff every open subspace is contour-closed.
Lemma 14.11 Any union or binary intersection of contour-closed open subspaces is contour-closed. If φ is contour-closed then so are its components, the convex open intervals in Theorem 13.15. If
(unlike Example 1.11) f is not eventually constantly zero then ⊤ is contour-closed. ▫
Then the generalisation of Theorem 2.5 is
Theorem 14.12 Restricted to contour-closed open sets, the operator ◊ preserves joins and satisfies the (easy) mixed modal law ▫φ∧◊ψ⇒◊(φ∧ψ).
Proof By Definition 14.1, ◊(∃ i.θ[i] ) says that there is a straddling interval [d,u] that is covered by the union ∃ i.θ[i] . By Definition 14.10, each member θ[i] of this union is itself a union
of intervals (e,t) with f e≠ 0≠ f t, so
◊(∃ i.θ[i] ) ≡ ∃ d u.(f d< 0< fu∨ fd > 0> f u)
∧ ∀ x:[d,u].∃ i.∃ e t.(e< x< t)
∧ (f e≠ 0∧ f t≠ 0) ∧ ∀ y:[e,t].θ[i] y.
Now we apply the Heine–Borel theorem in its traditional form to the open cover indexed by (i,e,t). This uses general Scott continuity (Axiom 9.1), the hyper-space of lists from an overt discrete
space (Remark 12.16) and a combinatorial argument like those in the next section. It yields
◊(∃ i.θ[i] ) ⇔ ∃ d u:ℚ.(f d< 0< f u∨ f d > 0> f u)
∧ ∃ℓ:List (I×ℚ×ℚ).
(∀ x:[d,u].∃(i,e,t)∈ℓ.(e< x< t))
∧ (∀(i,e,t)∈ℓ.(f e≠ 0∧ f t≠ 0) ∧ ∀ y:[e,t].θ[i] y),
in which, by Theorem 10.2, we may assume that d, u and all of the e and t in the list ℓ are distinct. Arranging them in numerical order, some successive pair (p,q) must have f p< 0< f q or f p> 0> f
q. Although p and q are not the e and t of any (i,e,t)∈ℓ because the intervals must overlap, nevertheless the interval [p,q] is part of some such [e,t], and therefore ∃ i.∀ y:[p,q].θ[i] y. Hence ∃ i
.◊θ[i] and the modal law also follows using Proposition 12.1 and Lemma 14.3. ▫
Corollary 14.13 For any function f:I≡[d,u]→ℝ with f d< 0< f u, there is a non-deterministic program ◊⊤ that finds arbitrarily close approximations to interval-valued stable zeroes.
Proof In Theorem 1.7 we found a nested sequence of open intervals for which the function takes non-zero values with opposite signs at the endpoints. The non-hovering condition ensured that we
could divide any such interval approximately in half at a new non-zero value (Theorem 2.8), although interval-Newton methods can make much better choices (Remark 2.9). Without this condition, we can
apparently do no better than prod randomly (and in parallel) in search of non-zero values. Nevertheless, if we find them then the subintervals under consideration are contour-closed, so the Theorem
applies to them.
The endpoints of the sequence that we choose generate a pseudo-cut, cf. Theorem 6.14. If the non-deterministic search for non-zero values is fair (in the sense that it will eventually consider any
possibility that exists, and distinguish any value from zero if it is non-zero), the function is constantly zero on the interval that this pseudo-cut defines. ▫
In Remark 2.6 we mentioned the classical objection that the non-hovering condition is redundant, because both hovering and non-hovering functions have zeroes, respectively for trivial and
constructive reasons. Instead of this disjunction of untestable cases, we can factorise the given map f:I→ℝ into a composite of maps that capture the purely classical and purely constructive ideas.
(This also re-admits the constantly zero function.)
Lemma 14.14 The relation x#[f] z (or just x# y) defined by
∃ y. (x< y< z ∨ x> y> z) ∧ f y≠ 0
co-classifies a closed equivalence relation: it is irreflexive, symmetric and co-transitive,
x# z=⇒ x≠ z, x# z=⇒ z# x and x# z=⇒ x# y∨ y# z,
and its equivalence classes are compact intervals (Definition 9.9).
Proof Observe that x# z ⇔ δ[z] x∨υ[z] x, where
δ[z] x ≡ ∃ y.(x< y< z)∧ f y≠ 0 and υ[z] x ≡ ∃ y.(x> y> z)∧ f y≠ 0
are rounded and disjoint but not necessarily bounded or located. For co-transitivity, suppose δ[z] x. Then, by Theorem 10.2,
∃ u w. (x< u< w< z) ∧ ∀ v:[u,w].f v≠ 0, whilst u< y∨ y< w,
of which the first disjunct gives δ[y] x and the second δ[z] y. ▫
Theorem 14.15 Any function f:I≡[d,u]→ℝ factorises as f=g· q where
1. q:I→I/(#[f]) is a proper surjection onto a compact Hausdorff space and has compact connected fibres; and
2. g:I/(#[f])→ℝ doesn’t hover, but if f d< 0< f u then g d< 0< g u.
If f doesn’t hover then #[f] is ≠ and q=id[I], whilst if f is constantly zero then # is ⊥ and I/(#[f]) is a singleton.
Proof The quotient of an overt discrete space by an open equivalence relation is constructed in Section C 10, but that paper only used the part of the calculus that is strictly lattice dual
(without assuming Scott continuity or anything about ℕ or ℝ), so the result is also directly applicable to the quotient of a compact Hausdorff space by a closed equivalence relation. However, this
uses the more general theory of Σ-split subspaces that was sketched in Remark 7.12 and does not belong to the class of simpler types that are generated by Axiom 4.1.
The condition f d< 0< f u ensures that δ[z] and υ[z] are bounded, so the fibres are compact connected (Theorem 15.10). The function f respects the closed equivalence relation #[f] because
and therefore factors through the quotient. The relation #[g] is ≠ on I/(#[f]), so g doesn’t hover and the constructive intermediate value theorem applies to it. ▫
This answers the opening discussion in Section 1:
Corollary 14.16 The difference between the general classical intermediate value theorem and its restricted constructive version lies exactly in the distinction between being occupied and inhabited
(Definitions 8.6 and 11.5). ▫
Remark 14.17 In order to generalise this argument to Brouwer’s fixed point theorem or to finding zeroes of f:ℝ^n→ℝ^n, we must replace the intervals in Definition 14.10 by polyhedra on whose faces (or
(n−1)-skeleton) f≠ 0, cf. Remark 2.7.
The argument with alternating signs that we used in Theorem 14.12 has a well known generalisation in combinatorial topology called Sperner’s Lemma [Spe28]. For example, Dirk van Dalen[vD09] used this
to generalise the approximate intermediate value theorem (Proposition 13.4). In Günter Baigger’s counterexample [Bai85, Pot07] there is a grid of zeroes (or fixed points), so its only contour-closed
open subset is the whole square. ▫ | {"url":"http://www.paultaylor.eu/ASD/lamcra/asdivt","timestamp":"2014-04-17T00:49:02Z","content_type":null,"content_length":"36328","record_id":"<urn:uuid:0ba793b3-b2e2-4fe0-b7fc-af6e884eecdd>","cc-path":"CC-MAIN-2014-15/segments/1398223201753.19/warc/CC-MAIN-20140423032001-00473-ip-10-147-4-33.ec2.internal.warc.gz"} |
WyzAnt Resources
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Topic: [HM] Kerry, B
Replies: 3 Last Post: Apr 11, 2006 9:15 AM
Messages: [ Previous | Next ]
[HM] Kerry, B
Posted: Apr 2, 2006 10:34 AM
Il giorno 02/apr/06, alle ore 08:05, HISTORIA MATEMATICA ha scritto:
> Date: Fri, 24 Mar 2006 15:17:16 +0100 (CET)
> From: Gunnar Berg <gunnar@math.uu.se>
> Subject: [HM] A certain Kerry
Might be Benno Kerry. He studied, I think, in Strasbourg, where he
wrote his dissertation: Researches on the problem of causation based
on a critic of Mill's theory (Untersuchungen über das Causalproblem
auf dem Boden einer Kritik der einschlägigen Lehren J. St. Mills,
1881), and with Brentano in Wien. He devoted his Habilitationsschrift
of 1884 to the Fundamentals of a theory of mathematical and not-
mathematical concepts of limit as a contribution to the theory of
knowledge (Grundzüge einer Theorie der mathematischen und nicht-
mathematischen Grenzbegriffe. Ein Beitrag zur Erkenntnistheorie),
which was at the time, I'd say, quite a neo-kantian theme (Cohen's
book on Infinitesimalmethode dates 1883); but K.'s works mostly show
influences from Brentano and Alois Höfler, and he was one of those
students of Brentano's who also worked much on Bolzano.
His most famous work is a series of eight articles on Intuition and
its physical reworking (Über Anschauung und ihre psychische
Verarbeitung) in the Vierteljahrsschrift fuer wissenschaftliche
Philosophie, 1885 to 1891, in response to which Frege wrote his
Concept and Object (Ueber Begriff und Gegenstand, same periodical,
Posthumous appeared a System of a theory of the concepts of limit
(System einer Theorie der Grenzbegriffe. Ein Beitrag zur
Erkenntnistheorie, ed. by Gustav Kohn, Deuticke, Leipzig und Wien,
1890, 1st part; the second, of which according to the editor only a
chapter was found among his mss., was never published).
It seems that K. also edited the paedagogist Ernst Laas' posthumous
mss (Literarischer Nachlass, 1887). He might be the same B.K. who
wrote in 1883 a True solution of anti-semitism (Die wahre Erlösung
vom Antisemitismus, Wigand, Leipzig, 61 p.) "von einem getauften
Juden", i.e. "by a baptised Jew".
Volker Peckhaus should know everything about him (just as on many
other subjects), since he wrote a Contribution to K.'s biography some
years ago.
// enrico pasini //
// enrico.pasini()unito.it //
Date Subject Author
4/2/06 Enrico Pasini
4/2/06 Re: [HM] Axiom of Infinity Jose Ferreiros
4/11/06 Re: [HM] Axiom of Infinity Irving Anellis | {"url":"http://mathforum.org/kb/message.jspa?messageID=4603823","timestamp":"2014-04-16T22:12:40Z","content_type":null,"content_length":"20503","record_id":"<urn:uuid:016895ed-a51d-49fc-8f91-31171ff700b0>","cc-path":"CC-MAIN-2014-15/segments/1398223206647.11/warc/CC-MAIN-20140423032006-00186-ip-10-147-4-33.ec2.internal.warc.gz"} |
Heatsink thermal resistance calculation
If I have a heatsink which I know the natural convection thermal resistance (TR), is there a way to calculate or estimate the TR of the same heatsink with forced convection?
I am trying to find a heatsink with a certain TR, but data sheets seem to all give natural convection TR information and I am trying to figure out what would happen to the TR if I add fans. | {"url":"http://www.physicsforums.com/showthread.php?p=3865722","timestamp":"2014-04-16T10:32:20Z","content_type":null,"content_length":"19728","record_id":"<urn:uuid:2b1263d3-d3d6-4cc9-a80b-a89f73dddb39>","cc-path":"CC-MAIN-2014-15/segments/1397609523265.25/warc/CC-MAIN-20140416005203-00203-ip-10-147-4-33.ec2.internal.warc.gz"} |
Finding Young's Modulus and Poisson's Ratio
A rod-like specimen subjected to uniaxial tension will exhibit some shrinkage in the lateral direction for most materials. The ratio of lateral strain and axial strain is defined as Poisson's ratio n
The Poisson ratio for most metals falls between 0.25 to 0.35. Rubber has a Poisson ratio close to 0.5 and is therefore almost incompressible. Theoretical materials with a Poisson ratio of exactly 0.5
are truly incompressible, since the sum of all their strains leads to a zero volume change. Cork, on the other hand, has a Poisson ratio close to zero. This makes cork function well as a bottle
stopper, since an axially-loaded cork will not swell laterally to resist bottle insertion.
The Poisson's ratio is bounded by two theoretical limits: it must be greater than -1, and less than or equal to 0.5,
The proof for this stems from the fact that E, G, and K are all positive and mutually dependent. However, it is rare to encounter engineering materials with negative Poisson ratios. Most materials
will fall in the range, | {"url":"http://www.efunda.com/formulae/solid_mechanics/mat_mechanics/elastic_constants_E_nu.cfm","timestamp":"2014-04-20T18:25:46Z","content_type":null,"content_length":"20933","record_id":"<urn:uuid:085e214b-01d9-4849-84ea-2d810a94147e>","cc-path":"CC-MAIN-2014-15/segments/1398223211700.16/warc/CC-MAIN-20140423032011-00333-ip-10-147-4-33.ec2.internal.warc.gz"} |
Mathematical intuition??
Hello kdinser, not too long ago I read
this article
(scientific american). In it, it details the possible thought proccess of a chess grandmaster. Aparently( and I agree with it) chess masters dont actually calculate all posible moves like a computer
does. Instead, for a paticular arrangement of the pieces of the board they know based on experience what are usually the best paths to follow.
Math is the same way, there are many paths that you can follow but only a few will get you to the end with relative ease. I think only experience and lots of practice will get you better at it. The
more you practice the fundamentals, the deeper you are going to "see" the problem at a subconscious level. For example instead of readind (x^2-1) you read (x+1)(x-1). This is a very basic example but
I bet it applies to much deeper processes.
I would assume that the problem you are having is due to lack of practice of the fundamentals. I encounter the same problem that you do, but since I've become a pre-calc tutor I get more than enough
practice and the paths to follow are becoming ever simpler. If I were you I would try becoming a homework helper for pre-calc level exersices. You ought to find more than enough practice there and
you will be helping others. Just my sugestion. | {"url":"http://www.physicsforums.com/showthread.php?t=134228","timestamp":"2014-04-18T21:20:22Z","content_type":null,"content_length":"60386","record_id":"<urn:uuid:b2bcb100-6446-49ed-8dd4-c4062f1491f4>","cc-path":"CC-MAIN-2014-15/segments/1398223210034.18/warc/CC-MAIN-20140423032010-00138-ip-10-147-4-33.ec2.internal.warc.gz"} |
the first resource for mathematics
Characterizations of the set of rational parametric curves with rational offsets.
(English) Zbl 0813.65032
Laurent, Pierre-Jean (ed.) et al., Curves and surfaces in geometric design. Papers from the 2nd international conference on curves and surfaces, held in Chamonix-Mont-Blanc, France, June 10-16, 1993.
Wellesley, MA: A K Peters. 153-160 (1994).
Summary: Rational curves do not generally have rational offsets. We give an analytic characterization of rational curves with rational offsets. We also propose a geometric characterization of these
curves after showing that the set of rational curves with rational or piecewise rational arc length is equal to the set of caustics of the rational curves.
65D17 Computer aided design (modeling of curves and surfaces)
41A20 Approximation by rational functions | {"url":"http://zbmath.org/?q=an:0813.65032","timestamp":"2014-04-23T22:19:11Z","content_type":null,"content_length":"20432","record_id":"<urn:uuid:d31ed617-8911-438f-be77-18b4882d383d>","cc-path":"CC-MAIN-2014-15/segments/1398223203841.5/warc/CC-MAIN-20140423032003-00118-ip-10-147-4-33.ec2.internal.warc.gz"} |
[Numpy-discussion] how to deal with large arrays
Darren Dale dd55 at cornell.edu
Fri Oct 15 14:18:41 CDT 2004
I have two 2D arrays, Q is 3-by-q and R is r-by-3. At each q, I need to sum
q(r) over R, so I take a dot product RQ and then sum along one axis to get a
1-by-q result.
I'm doing this with dot products because it is much faster than the equivalent
for or while loop. The intermediate r-by-q array can get very large though
(200MB in my case), so I was wondering if there is a better way to go about
If not, I can slice up R and deal with it one chunk at a time, then the
intermediate arrays fit within the available system resources. Would somebody
offer a suggestion of how to do this intelligently? Should the intermediate
array be about the size of the processor cache, some fraction of the
available memory, or is there something else I need to consider?
Thank you,
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week beginning
Seminar: KUWAIT LECTURE
Location and Time: Meeting Room 2 (CMS) at 5.00pm
Speaker: Professor S. Donaldson (Imperial College)
Title: Existence theorems and numerical solutions in complex differential geometry
Seminar: Number Theory Seminar
Location and Time: Meeting Room 13 (CMS) at 2.30pm
Speaker: Marie-France Vigneras (Paris VII)
Title: On the p-adic local Langlands correspondence
Seminar: Category Theory Seminar
Location and Time: Meeting Room 9 (CMS) at 2.15pm
Speaker: Peter Johnstone (Cambridge)
Title: Universal Normal Covers
Seminar: Probability Seminars
Location and Time: Meeting Room 12 (CMS) at 2.00pm
Speaker: James Norris (Cambridge)
Title: Differential equation approximations to Markov chains - with an application to a virus growth model with fast and slow variables
Seminar: Complex Analysis Seminar
Location and Time: Meeting Room 15 (CMS) at 4.00pm
Speaker: Professor Alan Beardon (University of Cambridge)
Title: Analytic maps as Euclidean contractions
Seminar: Algebra Seminar
Location and Time: Meeting Room 12 (CMS) at 4.30pm
Speaker: Rachel Camina (Cambridge)
Title: Conjugacy class sizes and the IP-graph
<>Seminar: Analysis Seminar
Location and Time: Meeting Room 12 (CMS) at 2.00pm
Speaker: András Zsák (Cambridge)
Title: The lattice of closed ideals in the Banach algebra of operators on a certain dual Banach space | {"url":"https://www.dpmms.cam.ac.uk/Seminars/Weekly/2005-2006/seminars-24-Oct-2005.html","timestamp":"2014-04-17T21:22:50Z","content_type":null,"content_length":"6985","record_id":"<urn:uuid:ae41ef1e-46bd-4760-95f1-259504f2df85>","cc-path":"CC-MAIN-2014-15/segments/1398223201753.19/warc/CC-MAIN-20140423032001-00277-ip-10-147-4-33.ec2.internal.warc.gz"} |
Squares, Congruency and Perpendicular bisectors
April 17th 2010, 09:20 AM #1
Oct 2009
Squares, Congruency and Perpendicular bisectors
I have a question from my textbook:
Given Square PINK with vertices P (-1,-1), I (3, 2), N (6, -2) and K (2, -5) prove that its diagonals are congruent and are perpendicular bisectors of each other.
I'm really not sure what I am supposed to do once I have this sketched up.
squares etc
Plot the four points. Connect the points Draw the slope diagrams for each side.Prove that each side is equal to 5.Write the slopes for any two lines meeting at one of the points and show that
they are perpendicular to each other.You go the rest of the way.
Last edited by bjhopper; April 17th 2010 at 11:27 AM. Reason: add clarification
Hello, (?)G!
Given Square PINK with vertices P (-1,-1), I (3, 2), N (6, -2) and K (2, -5)
Prove that its diagonals are congruent and are perpendicular bisectors of each other.
I'm really not sure what I am supposed to do once I have this sketched up.
How about doing the part in blue?
Determine the lengths of diagonals $PN$ and $I\!K$, and show that they are equal.
Find the slopes of $PN$ and $I\!K$ and show they are perpendicular.
Find the intersection of $PN$ and $I\!K$ and show that it is their mutual midpoint.
April 17th 2010, 11:24 AM #2
Super Member
Nov 2007
Trumbull Ct
April 17th 2010, 03:04 PM #3
Super Member
May 2006
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You did letter B for me, can you check my work?
Posted by For Reiny on Wednesday, September 5, 2012 at 11:10pm.
if f(x)=lnx, g(x)=e^3x, and h(x)=x^4 find following
A)(f of g)(x)and the domain of f of g
B) (g of f)(x) and the domain of g of f
C) (f of h)(x) and the domain of f of h
A) f(g(x))
domain x>0
• You did letter B for me, can you check my work? - Reiny, Wednesday, September 5, 2012 at 11:17pm
A) close , but not quite
= f(e^3x)
= ln (e^(3x)) , you have an extra x in there
f(x) was defined as lnx, so whatever x is replaced by on the left side, must be replaced by on the right side ....e.g.
f(2) = ln2
f(happy face) = ln(happy face)
f(e^3x) = ln(e^3x)
C) you made the same error
f(h(x)) = f(x^4) = ln x^4
the domain would be x > 0
• You did letter B for me, can you check my work? - For Reiny, Wednesday, September 5, 2012 at 11:22pm
i understand now thank you
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FOM: LICS (Logic in Computer Science) Newsletter 79
Stephen G Simpson simpson at math.psu.edu
Fri Jun 14 15:47:56 EDT 2002
From: Martin Grohe <grmail at dcs.ed.ac.uk>
To: LICS List <grmail at dcs.ed.ac.uk>
Subject: LICS Newsletter 79
Date: Fri, 14 Jun 2002 19:30:09 +0100
* Past issues of the newsletter are available at
* Instructions for submitting an announcement to the newsletter
can be found at
Workshop on Term Graph Rewriting
Workshop on Domains
Workshop on Logical Frameworks and Meta-Languages
Workshop on Hybrid Logics
Isabelle/HOL: A Proof Assistant for Higher-Order Logic
by Tobias Nipkow, Lawrence C. Paulson and Markus Wenzel
* JOURNALS
Special Issue of the Jornal of Automated Reasoning
INTERNATIONAL WORKSHOP ON TERM GRAPH REWRITING (TERMGRAPH 2002)
(A satellite event of The First International Conference on Graph
Transformation, ICGT 2002)
Second Call for Papers
Barcelona, Spain, October 7, 2002
* Topics of interest. All aspects of term graphs and sharing of common
subexpressions in rewriting, programming, automated reasoning and symbolic
computation. This includes (but is not limited to): Theory of first-order
and higher-order term graph rewriting; graph rewriting in lambda calculus
(sharing graphs, optimality); applications in functional, logic and
functional-logic programming; applications in automated reasoning and
symbolic computation; implementation issues; system descriptions
* Submissions. Authors are invited to submit an extended abstract of 5 to
10 pages by e-mail to the program chair (det at cs.york.ac.uk). Submissions
should be in PostScript format. It is strongly recommended to use LaTeX
and ENTCS style files (http://math.tulane.edu/~entcs/).
* Publication. Accepted contributions will appear in an issue of Elsevier's
Electronic Notes in Theoretical Computer Science.
* Important dates.
Submission deadline: June 15, 2002
Notification: July 15, 2002
Final version due: September 6, 2002
* Program committee. Zena M. Ariola (University of Oregon, Eugene),
Richard Banach (University of Manchester), Rachid Echahed (IMAG, Grenoble),
Richard Kennaway (University of East Anglia, Norwich), Jan Willem Klop
(Free University of Amsterdam), Rinus Plasmeijer (University of Nijmegen),
Detlef Plump (University of York, chair)
Call for Abstracts
Birmingham, UK, 16-19 September 2002
* The Workshop on Domains is aimed at computer scientists and
mathematicians alike who share an interest in the mathematical
foundations of computation. The workshop will focus on domains,
their applications and related topics.
* Invited Speakers: Ulrich Berger (University of Wales Swansea),
Thierry Coquand (Goeteborg University), Jimmie Lawson (Louisiana
State University), John Longley (University of Edinburgh), Dag
Normann (University of Oslo), Prakash Panangaden (McGill
University), Uday Reddy (University of Birmingham), Thomas Streicher
(Darmstadt University)
* As such, domain theory is highly interdisciplinary. Topics of
interaction with domain theory for this workshop include, but are
not limited to: program semantics, program logics, probabilistic
computation, exact computation over the real numbers, lambda
calculus, games, models of sequential computation, constructive
mathematics, recursion theory, realizability, real analysis,
topology, locale theory, metric spaces, category theory, topos
theory, type theory
* Submission of Abstracts: One-page abstracts should be submitted to
domainsvi at cs.bham.ac.uk. Shortly after an abstract is submitted
(usually one or two weeks), the authors will be notified by the
programme committee. The criterion for acceptance is relevance to
the meeting. In particular, talks on subjects presented at other
conferences and workshops are acceptable. Abstracts will be dealt
with on a first-come/first-served basis.
* Programme Committee: Martin Escardo (University of Birmingham),
Achim Jung (University of Birmingham), Klaus Keimel (Darmstadt
University), Alex Simpson (University of Edinburgh)
WORKSHOP ON LOGICAL FRAMEWORKS AND META-LANGUAGES (LFM'02)
(Affiliated with FLoC'02)
Call for Participation
Copenhagen, Denmark, July 26, 2002
* Logical frameworks and meta-languages form a common substrate for
representing, implementing, and reasoning about a wide variety of
deductive systems of interest in logic and computer science. Their
design and implementation has been the focus of considerable
research over the last two decades, using competing and sometimes
incompatible basic principles. This workshop will bring together
designers, implementors, and practitioners to discuss all aspects of
logical frameworks.
* Informal proceedings will be published as a volume in the Electronic
Notes in Theoretical Computer Science (ENTCS) and will be available
at the workshop.
* The workshop is open for any interested party. If you are planning
to attend you are welcome to participate in the informal system
demonstration session. Please send mail to the workshop chair
(Frank Pfenning <fp at cs.cmu.edu>) if you are interested in showing
your implementation.
* The program can be found at http://www-2.cs.cmu.edu/~lfm02/program.html.
WORKSHOP ON HYBRID LOGICS (HyLo at LICS)
(Affiliated with FLoC'02)
Call for Participation
Copenhagen, Denmark, July 25, 2002
* Hybrid logic is a branch of modal logic in which it is possible to
directly refer to worlds/times/states or whatever the elements of
the (Kripke) model are meant to represent. Although they date back
to the late 1960s, and have been sporadically investigated ever
since, it is only in the 1990s that work on them really got into its
* It is easy to justify interest in hybrid logic on applied grounds,
with the usefulness of the additional expressive power. For example,
when reasoning about time one often wants to build up a series of
assertions about what happens at a particular instant, and standard
modal formalisms do not allow this. What is less obvious is that the
route hybrid logic takes to overcome this problem (the basic
mechanism being to add nominals --- atomic symbols true at a unique
point --- together with extra modalities to exploit them) often
actually improves the behavior of the underlying modal
formalism. For example, it becomes far simpler to formulate modal
tableau and resolution in hybrid logic, and completeness and
interpolation results can be proved of a generality that is simply
not available in modal logic. That is, hybridization --- adding
nominals and related apparatus --- seems a fairly reliable way of
curing many known weaknesses in modal logic.
* HyLo at LICS is likely to be relevant to a wide range of people,
including those interested in description logic, feature logic,
applied modal logics, temporal logic, and labelled
deduction. Moreover, if you have an interest in the work of the late
Arthur Prior, note that this workshop is devoted to exploring ideas
he first introduced 30 years ago --- it will be an ideal opportunity
to see how his ideas have been developed in the intervening period.
* In this workshop we hope to bring together researchers from all the
different fields just mentioned (and hopefully some others) in an
attempt to explore what they all have (and do not have) in
common. If you're unsure whether the workshop is of relevance to
your work , please check out the Hybrid Logics homepage. And do not
hesitate to contact the workshop organisers for more
information. We'd be delighted to tell you more. Contact details are
given below.
* The program includes invited talks by Melvin Fitting and Moshe
Vardi. The full program is available at the workshop webpages:
Isabelle/HOL: A Proof Assistant for Higher-Order Logic
by Tobias Nipkow, Lawrence C. Paulson and Markus Wenzel
Springer LNCS 2283, ISBN 3-540-43376-7
* This book is a self-contained introduction to interactive proof in
higher-order logic (HOL), using the proof assistant Isabelle2002. It is a
tutorial for potential users rather than a monograph for researchers.
* The book has three parts.
1. Elementary Techniques shows how to model functional programs in
higher-order logic. Early examples involve lists and the natural
numbers. Most proofs are two steps long, consisting of induction on a
chosen variable followed by the auto tactic. But even this elementary part
covers such advanced topics as nested and mutual recursion.
2. Logic and Sets presents a collection of lower-level tactics that you can
use to apply rules selectively. It also describes Isabelle/HOL's treatment
of sets, functions and relations and explains how to define sets
inductively. One of the examples concerns the theory of model checking, and
another is drawn from a classic textbook on formal languages.
3. Advanced Material describes a variety of other topics. Among these are
the real numbers, records and overloading. Advanced techniques are
described involving induction and recursion. A whole chapter is devoted
to an extended example: the verification of a security protocol.
* The book is available online.
Bytecode Verification
Call for Papers
* Topic: Bytecode verification for the Java Virtual Machine and related
approaches to program analysis of low level code for the enforcement of
safety properties. Particularly welcome are contributions emphasizing the
correctness of the analysis and the application of automated or interactive
verification techniques.
* Submission: Manuscripts should be unpublished works and not
submitted elsewhere. Revised and enhanced versions of papers
published in conference proceedings that have not appeared in
archival journals are eligible for submission. All submissions will
be reviewed according to the usual standards of scholarship and
originality. *The deadline for submissions is October 13, 2002.*
* Guest editor: Tobias Nipkow, nipkow at in.tum.de.
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Cadlag modification
I was reading Stochastic Integration and Differential Equations by Protter and it had a nice theorem that every Levy process in law which has continuity in probability, admits a cadlag modification.
The proof is very confusing and I was wondering if anyone could help me clear it up a bit.
I wish to prove a slight modification, which is that the set
[tex]\{\omega : \not \exists \lim_{\mathbb{Q}\ni s\downarrow t}X_t(\omega) \quad t \geq 0 \}[/tex]
is measurable and has a measure zero. I have that X has stationary independent increments and is continuous in probability (also X starts at 0 a.s.). Any help would be much appreciated. | {"url":"http://www.physicsforums.com/showthread.php?t=296322","timestamp":"2014-04-18T21:25:36Z","content_type":null,"content_length":"19519","record_id":"<urn:uuid:129c9a62-9016-4368-b23f-05bd17112b00>","cc-path":"CC-MAIN-2014-15/segments/1398223202774.3/warc/CC-MAIN-20140423032002-00424-ip-10-147-4-33.ec2.internal.warc.gz"} |
Re: st:Confidence interval of difference between two proportions and-csi
[Date Prev][Date Next][Thread Prev][Thread Next][Date index][Thread index]
Re: st:Confidence interval of difference between two proportions and-csi-
From Garry Anderson <g.anderson@unimelb.edu.au>
To statalist@hsphsun2.harvard.edu
Subject Re: st:Confidence interval of difference between two proportions and-csi-
Date Fri, 19 Mar 2004 19:11:05 +1100
Thank you Roger for your suggestion of using -exactcci-
However, this does not calculate a risk difference and the 95% CI.
I suppose I am suggesting to the folks at Stata that a revised formula be used in -csi- to calculate the 95% CI of the risk difference when there is a small number of observations, and one or both
have a 100% risk. Currently it is possible for the upper bound to go beyond the theoretical maximum of 100% difference.
I would appreciate opinions on this by others on the list.
Newcombe RG (1998) Interval estimation for the difference between independent proportions: comparison of eleven methods. Statistics in Medicine 17: 873-890
Kind regards, Garry
At 05:17 PM 18/03/2004 +0000, you wrote:
At 16:37 18/03/04 +1100, Garry Anderson wrote:
I am enquiring if a more appropriate method could be used please to calculate the 95% CI of the difference between two proportions in the -csi- command?
At the moment it is possible for the upper bound of the confidence interval of the difference between two proportions to be greater than 1.0. I realize that the approximation that is used is
not appropriate for small sample sizes, however I think that reporting of results that are impossible should be avoided.
One possibility is to use the -somersd- package, downloadable (complete with a .pdf manual) from SSC, using its -transform()- option. The difference between 2 proportions is a special case of
Somers' D, and the -somersd- package offers a choice of transformations appropriate for Somers' D, notably the hyperbolic arctangent (or z) transformation or the arcsine transformation. If
-diseased- and -exposed- are 2 binary (0,1) variables indicating disease and exposure, respectively, then Garry might type
somersd exposed diseased, tr(z)
or, alternatively,
somersd exposed diseased, tr(asin)
and get a confidence interval for the difference between the proportion of exposed individuals with the disease and the proportion of unexposed individuals with the disease, using a normalizing
and variance-stabilizing transformation.
However, it should be stressed that, with Garry's example, there is a zero cell (for exposed noncases), so one of the proportions is either zero or one, so a normalizing or variance-stabilizing
transformation might be inappropriate because the sample size is so low. In such circumstances, it might be better to use the -exactcci- package to define a conservative confidence interval for
the odds ratio, which may have an infinite upper limit or a zero lower limit. If Garry uses -findit- to find and install the -exactcci- fackage and types
exactcci 5 1 0 4, exact
then the so-called "exact" confidence interval is generated. (Note, however, that this confidence interval is conservative, not exact. It is called "exact" because it uses the exact
hypergeometric distribution to calculate conservative confidence limits.)
I hope this helps.
Roger Newson
Lecturer in Medical Statistics
Department of Public Health Sciences
King's College London
5th Floor, Capital House
42 Weston Street
London SE1 3QD
United Kingdom
Tel: 020 7848 6648 International +44 20 7848 6648
Fax: 020 7848 6620 International +44 20 7848 6620
or 020 7848 6605 International +44 20 7848 6605
Email: roger.newson@kcl.ac.uk
Website: http://www.kcl-phs.org.uk/rogernewson
Opinions expressed are those of the author, not the institution.
* For searches and help try:
* http://www.stata.com/support/faqs/res/findit.html
* http://www.stata.com/support/statalist/faq
* http://www.ats.ucla.edu/stat/stata/
* For searches and help try:
* http://www.stata.com/support/faqs/res/findit.html
* http://www.stata.com/support/statalist/faq
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Posts by
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Total # Posts: 618
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nag_ode_ivp_rk_onestep (d02pdc)
NAG Library Function Document
nag_ode_ivp_rk_onestep (d02pdc)
1 Purpose
nag_ode_ivp_rk_onestep (d02pdc) is a one-step function for solving the initial value problem for a first order system of ordinary differential equations using Runge–Kutta methods.
2 Specification
#include <nag.h>
#include <nagd02.h>
nag_ode_ivp_rk_onestep (Integer neq,
void void (*f)(Integer neq, double t, const double y[], double yp[], Nag_User *comm),
double *tnow, double ynow[], double ypnow[], Nag_ODE_RK *opt, Nag_User *comm, NagError *fail)
3 Description
nag_ode_ivp_rk_onestep (d02pdc) and its associated functions (
nag_ode_ivp_rk_setup (d02pvc)
nag_ode_ivp_rk_reset_tend (d02pwc)
nag_ode_ivp_rk_interp (d02pxc)
nag_ode_ivp_rk_errass (d02pzc)
) solve the initial value problem for a first order system of ordinary differential equations. The functions, based on Runge–Kutta methods and derived from RKSUITE (
Brankin et al. (1991)
), integrate
$y ′ = f t,y given y t 0 = y 0$
is the vector of
solution components and
is the independent variable.
This function is designed to be used in complicated tasks when solving systems of ordinary differential equations. You must first call
nag_ode_ivp_rk_setup (d02pvc)
to specify the problem and how it is to be solved. Thereafter you (repeatedly) call nag_ode_ivp_rk_onestep (d02pdc) to take one integration step at a time from
in the direction of
(as specified in
nag_ode_ivp_rk_setup (d02pvc)
). In this manner nag_ode_ivp_rk_onestep (d02pdc) returns an approximation to the solution
and its derivative
at successive points
. If nag_ode_ivp_rk_onestep (d02pdc) encounters some difficulty in taking a step, the integration is not advanced and the function returns with the same values of
as returned on the previous successful step. nag_ode_ivp_rk_onestep (d02pdc) tries to advance the integration as far as possible subject to passing the test on the local error and not going past
. In the call to
nag_ode_ivp_rk_setup (d02pvc)
you can specify either the first step size for nag_ode_ivp_rk_onestep (d02pdc) to attempt or that it compute automatically an appropriate value. Thereafter nag_ode_ivp_rk_onestep (d02pdc) estimates
an appropriate step size for its next step. This value and other details of the integration can be obtained after any call to nag_ode_ivp_rk_onestep (d02pdc) by examining the contents of the
, see
Section 5
. The local error is controlled at every step as specified in
nag_ode_ivp_rk_setup (d02pvc)
. If you wish to assess the true error, you must set
in the call to
nag_ode_ivp_rk_setup (d02pvc)
. This assessment can be obtained after any call to nag_ode_ivp_rk_onestep (d02pdc) by a call to the subroutine
nag_ode_ivp_rk_errass (d02pzc)
If you want answers at specific points there are two ways to proceed:
The more efficient way is to step past the point where a solution is desired, and then call
nag_ode_ivp_rk_interp (d02pxc)
to get an answer there. Within the span of the current step, you can get all the answers you want at very little cost by repeated calls to
nag_ode_ivp_rk_interp (d02pxc)
. This is very valuable when you want to find where something happens, e.g., where a particular solution component vanishes. You cannot proceed in this way with
The other way to get an answer at a specific point is to set
to this value and integrate to
. nag_ode_ivp_rk_onestep (d02pdc) will not step past
, so when a step would carry it past, it will reduce the step size so as to produce an answer at
exactly. After getting an answer there (
), you can reset
to the next point where you want an answer, and repeat.
could be reset by a call to
nag_ode_ivp_rk_setup (d02pvc)
, but you should not do this. You should use
nag_ode_ivp_rk_reset_tend (d02pwc)
because it is both easier to use and much more efficient. This way of getting answers at specific points can be used with any of the available methods, but it is the only way with
. It can be inefficient. Should this be the case, the code will bring the matter to your attention.
4 References
Brankin R W, Gladwell I and Shampine L F (1991) RKSUITE: A suite of Runge–Kutta codes for the initial value problems for ODEs SoftReport 91-S1 Southern Methodist University
5 Arguments
1: neq – IntegerInput
On entry: the number of ordinary differential equations in the system to be solved.
Constraint: $neq≥1$.
2: f – function, supplied by the userExternal Function
must evaluate the first derivatives
$y i ′$
(that is the functions
$f i$
) for given values of the arguments
$t , y i$
The specification of
void f (Integer neq, double t, const double y[], double yp[], Nag_User *comm)
1: neq – IntegerInput
On entry: the number of differential equations.
2: t – doubleInput
On entry: the current value of the independent variable, $t$.
3: y[neq] – const doubleInput
On entry: the current values of the dependent variables, $y i$ for $i = 1 , 2 , … , neq$.
4: yp[neq] – doubleOutput
On exit: the values of $f i$ for $i = 1 , 2 , … , neq$.
5: comm – Nag_User *
Pointer to a structure of type Nag_User with the following member:
p – Pointer
On entry/exit
: the pointer
should be cast to the required type, e.g.,
struct user *s = (struct user *)comm → p
, to obtain the original object's address with appropriate type. (See the argument
3: tnow – double *Output
On exit: the value of the independent variable $t$ at which a solution has been computed.
4: ynow[neq] – doubleOutput
On exit
: an approximation to the solution at
. The local error of the step to
was no greater than permitted by the specified tolerances (see
nag_ode_ivp_rk_setup (d02pvc)
5: ypnow[neq] – doubleOutput
On exit
: an approximation to the derivative of the solution at
6: opt – Nag_ODE_RK *
Pointer to a structure of type Nag_ODE_RK as initialized by the setup function
nag_ode_ivp_rk_setup (d02pvc)
with the following members:
totfcn – IntegerOutput
On exit
: the total number of evaluations of
used in the primary integration so far; this does not include evaluations of
for the secondary integration specified by a prior call to
nag_ode_ivp_rk_setup (d02pvc)
stpcst – IntegerOutput
On exit
: the cost in terms of number of evaluations of
of a typical step with the method being used for the integration. The method is specified by the argument
in a prior call to
nag_ode_ivp_rk_setup (d02pvc)
waste – doubleOutput
On exit: the number of attempted steps that failed to meet the local error requirement divided by the total number of steps attempted so far in the integration. A ‘large’ fraction indicates
that the integrator is having trouble with the problem being solved. This can happen when the problem is ‘stiff’ and also when the solution has discontinuities in a low order derivative.
stpsok – IntegerOutput
On exit: the number of accepted steps.
hnext – doubleOutput
On exit: the step size the integrator plans to use for the next step.
7: comm – Nag_User *
Pointer to a structure of type Nag_User with the following member:
p – Pointer
On entry/exit
: the pointer
, of type Pointer, allows you to communicate information to and from
. An object of the required type should be declared, e.g., a structure, and its address assigned to the pointer
by means of a cast to Pointer in the calling program, e.g.,
comm.p = (Pointer)&s
. The type pointer will be
void *
with a C compiler that defines
void *
char *
8: fail – NagError *Input/Output
The NAG error argument (see
Section 3.6
in the Essential Introduction).
6 Error Indicators and Warnings
Internally allocated memory has been freed by a call to
nag_ode_ivp_rk_free (d02ppc)
without a subsequent call to the setup function
nag_ode_ivp_rk_setup (d02pvc)
The value of neq supplied is not the same as that given to the setup function
nag_ode_ivp_rk_setup (d02pvc)
but the value given to
nag_ode_ivp_rk_setup (d02pvc)
The setup function
nag_ode_ivp_rk_setup (d02pvc)
has not been called.
The previous call to a function had resulted in a severe error. You must call
nag_ode_ivp_rk_setup (d02pvc)
to start another problem.
The previous call to the function nag_ode_ivp_rk_onestep (d02pdc) had resulted in a severe error. You must call
nag_ode_ivp_rk_setup (d02pvc)
to start another problem.
The function to be called as specified in the setup function
nag_ode_ivp_rk_setup (d02pvc)
nag_ode_ivp_rk_range (d02pcc)
. However the actual call was made to nag_ode_ivp_rk_onestep (d02pdc). This is not permitted.
The global error assessment algorithm failed at the start of the integration.
The global error assessment may not be reliable for t past
More than 100 output points have been obtained by integrating to
. They have been sufficiently close to one another that the efficiency of the integration has been degraded. It would probably be (much) more efficient to obtain output by interpolating with
nag_ode_ivp_rk_interp (d02pxc)
(after changing to
if you are using
In order to satisfy the error requirements nag_ode_ivp_rk_onestep (d02pdc) would have to use a step size of $value$ at current $t=value$. This is too small for the machine precision.
has been reached already. To integrate further with same problem the function
nag_ode_ivp_rk_reset_tend (d02pwc)
must be called with a new value of
The problem appears to be stiff.
Approximately $value$ function evaluations have been used to compute the solution since the integration started or since this message was last printed.
7 Accuracy
The accuracy of integration is determined by the arguments
in a prior call to
nag_ode_ivp_rk_setup (d02pvc)
. Note that only the local error at each step is controlled by these arguments. The error estimates obtained are not strict bounds but are usually reliable over one step. Over a number of steps the
overall error may accumulate in various ways, depending on the properties of the differential system.
8 Further Comments
If nag_ode_ivp_rk_onestep (d02pdc) returns with
and the accuracy specified by
is really required then you should consider whether there is a more fundamental difficulty. For example, the solution may contain a singularity. In such a region the solution components will usually
be of a large magnitude. Successive output values of
should be monitored with the aim of trapping the solution before the singularity. In any case numerical integration cannot be continued through a singularity, and analytical treatment may be
If nag_ode_ivp_rk_onestep (d02pdc) returns with a non-trivial value of
(i.e., those not related to an invalid call) then performance statistics are available by examining the structure
Section 5
). Furthermore if
then global error assessment is available by a call to the function
nag_ode_ivp_rk_errass (d02pzc)
. The approximate extra number of evaluations of
used is given by
$2 × stpsok × stpcst$
$3 × stpsok × stpcst$
After a failure with
the diagnostic function
nag_ode_ivp_rk_errass (d02pzc)
may be called only once.
If nag_ode_ivp_rk_onestep (d02pdc) returns with $fail.code=NE_STIFF_PROBLEM$ then it is advisable to change to another code more suited to the solution of stiff problems. nag_ode_ivp_rk_onestep
(d02pdc) will not return with $fail.code=NE_STIFF_PROBLEM$ if the problem is actually stiff but it is estimated that integration can be completed using less function evaluations than already
9 Example
We solve the equation
$y ′′ = -y , y 0 = 0 , y ′ 0 = 1$
reposed as
$y 1 ′ = y 2 y 2 ′ = - y 1$
over the range
with initial conditions
$y 1 = 0.0$
$y 2 = 1.0$
. We use relative error control with threshold values of
for each solution component and print the solution at each integration step across the range. We use a medium order Runge–Kutta method (
) with tolerances
$tol = 1.0e−4$
$tol = 1.0e−5$
in turn so that we may compare the solutions. The value of
is obtained by using
nag_pi (X01AAC)
9.1 Program Text
9.2 Program Data
9.3 Program Results | {"url":"http://www.nag.com/numeric/cl/nagdoc_cl23/html/D02/d02pdc.html","timestamp":"2014-04-19T00:12:01Z","content_type":null,"content_length":"36746","record_id":"<urn:uuid:994d24eb-fa41-4726-a118-b75eabf59b85>","cc-path":"CC-MAIN-2014-15/segments/1397609535535.6/warc/CC-MAIN-20140416005215-00281-ip-10-147-4-33.ec2.internal.warc.gz"} |
[Numpy-discussion] numarray.where confusion
Francesc Alted falted at pytables.org
Thu May 27 00:47:04 CDT 2004
A Dimecres 26 Maig 2004 21:01, Perry Greenfield va escriure:
> correct. You'd have to break apart the m1 tuple and
> index all the components, e.g.,
> m11, m12 = m1
> x[m11[m2],m12[m2]] = ...
> This gets clumsier with the more dimensions that must
> be handled, but you still can do it. It would be most
> useful if the indexed array is very large, the number
> of items selected is relatively small and one
> doesn't want to incur the memory overhead of all the
> mask arrays of the admittedly much nicer notational
> approach that Francesc illustrated.
Well, boolean arrays have the property that they use very little memory
(only 1 byte / element), and normally perform quite well doing indexing.
Some timings:
>>> import timeit
>>> t1 = timeit.Timer("m1=where(x>4);m2=where(x[m1]<7);m11,m12=m1;x[m11[m2],m12[m2]]","from numarray import arange,where;dim=3;x=arange(dim*dim);x.shape=(dim,dim)")
>>> t2 = timeit.Timer("x[(x>4) & (x<7)]","from numarray import arange,where;dim=3;x=arange(dim*dim);x.shape=(dim,dim)")
>>> t1.repeat(3,1000)
[3.1320240497589111, 3.1235389709472656, 3.1198310852050781]
>>> t2.repeat(3,1000)
[1.1218469142913818, 1.117638111114502, 1.1156759262084961]
i.e. using boolean arrays for indexing is roughly 3 times faster.
For larger arrays this difference is even more noticeable:
>>> t3 = timeit.Timer("m1=where(x>4);m2=where(x[m1]<7);m11,m12=m1;x[m11[m2],m12[m2]]","from numarray import arange,where;dim=1000;x=arange(dim*dim);x.shape=(dim,dim)")
>>> t4 = timeit.Timer("x[(x>4) & (x<7)]","from numarray import arange,where;dim=1000;x=arange(dim*dim);x.shape=(dim,dim)")
>>> t3.repeat(3,10)
[3.1818649768829346, 3.20477294921875, 3.190640926361084]
>>> t4.repeat(3,10)
[0.42328095436096191, 0.42140507698059082, 0.41979002952575684]
as you see, now the difference is almost an order of magnitude (!).
So, perhaps assuming the small memory overhead, in most of cases it is
better to use boolean selections. However, it would be nice to know the
ultimate reason of why this happens, because the Perry approach seems
intuitively faster.
Francesc Alted
More information about the Numpy-discussion mailing list | {"url":"http://mail.scipy.org/pipermail/numpy-discussion/2004-May/003028.html","timestamp":"2014-04-16T05:04:15Z","content_type":null,"content_length":"5000","record_id":"<urn:uuid:048b0750-c726-4033-ba4d-78f3763c39f8>","cc-path":"CC-MAIN-2014-15/segments/1397609525991.2/warc/CC-MAIN-20140416005205-00577-ip-10-147-4-33.ec2.internal.warc.gz"} |
Figure 1: Analysis of AP clusters and estimation of AP time precision in the CA3 neuron. The effect of the channel noise. (a) Overlapped output spike trains generated by identical input presented to
neuron times. (b) Extracted and overlapped shapes of APs-forming AP clusters. Values depicted on the top of each cluster show the —the measure of AP time precision calculated as a standard deviation
of AP times found in the AP cluster. (c) Shapes of -AP cluster’s mean voltage course ending with triggering an AP. (d) of consecutive AP clusters for 3s duration stimulation. (e) Visualization of
spike train groups. Based on the Hamming distance between spike trains elicited by repeating the same input, the trains were sorted according to their firing pattern into different spike train groups
( is the number of spike trains in each group). In this panel three groups are depicted by three lines differing in color. The ordinate shows the order of APs and the abscissa the firing time of AP.
Despite the fact that each group contains different numbers of fired APs (some APs failed to be fired in individual trials), firing times in-between groups are well aligned. This indicates an
interesting preference of the spiking mechanism to follow some pattern more frequently than others. (f) Aligned courses of of different AP clusters. (g) Correlations between voltages of all AP
clusters at given times , and all clusters’ were calculated under the channel noise effect. Correlation curves are delimited with their 0.95 confidence intervals (indicated in red), obtained from 40
correlation curves obtained over 40 various discharge patterns of presynaptic neurons repeated times. (h) The histogram demonstrates the common distribution of ISIs in our model. | {"url":"http://www.hindawi.com/journals/cmmm/2012/595398/fig1/","timestamp":"2014-04-20T11:58:56Z","content_type":null,"content_length":"22584","record_id":"<urn:uuid:611bc3d7-aa24-42cc-b625-d835e57ee107>","cc-path":"CC-MAIN-2014-15/segments/1397609538423.10/warc/CC-MAIN-20140416005218-00362-ip-10-147-4-33.ec2.internal.warc.gz"} |
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You are currently browsing the tag archive for the ‘tridiagonal matrix’ tag.
Van Vu and I have just uploaded to the arXiv our paper A central limit theorem for the determinant of a Wigner matrix, submitted to Adv. Math.. It studies the asymptotic distribution of the
determinant ${\det M_n}$ of a random Wigner matrix (such as a matrix drawn from the Gaussian Unitary Ensemble (GUE) or Gaussian Orthogonal Ensemble (GOE)).
Before we get to these results, let us first discuss the simpler problem of studying the determinant ${\det A_n}$ of a random iid matrix ${A_n = (\zeta_{ij})_{1 \leq i,j \leq n}}$, such as a real
gaussian matrix (where all entries are independently and identically distributed using the standard real normal distribution ${\zeta_{ij} \equiv N(0,1)_{\bf R}}$), a complex gaussian matrix (where
all entries are independently and identically distributed using the standard complex normal distribution ${\zeta_{ij} \equiv N(0,1)_{\bf C}}$, thus the real and imaginary parts are independent with
law ${N(0,1/2)_{\bf R}}$), or the random sign matrix (in which all entries are independently and identically distributed according to the Bernoulli distribution ${\zeta_{ij} \equiv \pm 1}$ (with a $
{1/2}$ chance of either sign). More generally, one can consider a matrix ${A_n}$ in which all the entries ${\zeta_{ij}}$ are independently and identically distributed with mean zero and variance ${1}
We can expand ${\det A_n}$ using the Leibniz expansion
$\displaystyle \det A_n = \sum_{\sigma \in S_n} I_\sigma, \ \ \ \ \ (1)$
where ${\sigma: \{1,\ldots,n\} \rightarrow \{1,\ldots,n\}}$ ranges over the permutations of ${\{1,\ldots,n\}}$, and ${I_\sigma}$ is the product
$\displaystyle I_\sigma := \hbox{sgn}(\sigma) \prod_{i=1}^n \zeta_{i\sigma(i)}.$
From the iid nature of the ${\zeta_{ij}}$, we easily see that each ${I_\sigma}$ has mean zero and variance one, and are pairwise uncorrelated as ${\sigma}$ varies. We conclude that ${\det A_n}$ has
mean zero and variance ${n!}$ (an observation first made by Turán). In particular, from Chebyshev’s inequality we see that ${\det A_n}$ is typically of size ${O(\sqrt{n!})}$.
It turns out, though, that this is not quite best possible. This is easiest to explain in the real gaussian case, by performing a computation first made by Goodman. In this case, ${\det A_n}$ is
clearly symmetrical, so we can focus attention on the magnitude ${|\det A_n|}$. We can interpret this quantity geometrically as the volume of an ${n}$-dimensional parallelopiped whose generating
vectors ${X_1,\ldots,X_n}$ are independent real gaussian vectors in ${{\bf R}^n}$ (i.e. their coefficients are iid with law ${N(0,1)_{\bf R}}$). Using the classical base-times-height formula, we thus
$\displaystyle |\det A_n| = \prod_{i=1}^n \hbox{dist}(X_i, V_i) \ \ \ \ \ (2)$
where ${V_i}$ is the ${i-1}$-dimensional linear subspace of ${{\bf R}^n}$ spanned by ${X_1,\ldots,X_{i-1}}$ (note that ${X_1,\ldots,X_n}$, having an absolutely continuous joint distribution, are
almost surely linearly independent). Taking logarithms, we conclude
$\displaystyle \log |\det A_n| = \sum_{i=1}^n \log \hbox{dist}(X_i, V_i).$
Now, we take advantage of a fundamental symmetry property of the Gaussian vector distribution, namely its invariance with respect to the orthogonal group ${O(n)}$. Because of this, we see that if we
fix ${X_1,\ldots,X_{i-1}}$ (and thus ${V_i}$, the random variable ${\hbox{dist}(X_i,V_i)}$ has the same distribution as ${\hbox{dist}(X_i,{\bf R}^{i-1})}$, or equivalently the ${\chi}$ distribution
$\displaystyle \chi_{n-i+1} := (\sum_{j=1}^{n-i+1} \xi_{n-i+1,j}^2)^{1/2}$
where ${\xi_{n-i+1,1},\ldots,\xi_{n-i+1,n-i+1}}$ are iid copies of ${N(0,1)_{\bf R}}$. As this distribution does not depend on the ${X_1,\ldots,X_{i-1}}$, we conclude that the law of ${\log |\det A_n
|}$ is given by the sum of ${n}$ independent ${\chi}$-variables:
$\displaystyle \log |\det A_n| \equiv \sum_{j=1}^{n} \log \chi_j.$
A standard computation shows that each ${\chi_j^2}$ has mean ${j}$ and variance ${2j}$, and then a Taylor series (or Ito calculus) computation (using concentration of measure tools to control tails)
shows that ${\log \chi_j}$ has mean ${\frac{1}{2} \log j - \frac{1}{2j} + O(1/j^{3/2})}$ and variance ${\frac{1}{2j}+O(1/j^{3/2})}$. As such, ${\log |\det A_n|}$ has mean ${\frac{1}{2} \log n! - \
frac{1}{2} \log n + O(1)}$ and variance ${\frac{1}{2} \log n + O(1)}$. Applying a suitable version of the central limit theorem, one obtains the asymptotic law
$\displaystyle \frac{\log |\det A_n| - \frac{1}{2} \log n! + \frac{1}{2} \log n}{\sqrt{\frac{1}{2}\log n}} \rightarrow N(0,1)_{\bf R}, \ \ \ \ \ (3)$
where ${\rightarrow}$ denotes convergence in distribution. A bit more informally, we have
$\displaystyle |\det A_n| \approx n^{-1/2} \sqrt{n!} \exp( N( 0, \log n / 2 )_{\bf R} ) \ \ \ \ \ (4)$
when ${A_n}$ is a real gaussian matrix; thus, for instance, the median value of ${|\det A_n|}$ is ${n^{-1/2+o(1)} \sqrt{n!}}$. At first glance, this appears to conflict with the second moment bound $
{\mathop{\bf E} |\det A_n|^2 = n!}$ of Turán mentioned earlier, but once one recalls that ${\exp(N(0,t)_{\bf R})}$ has a second moment of ${\exp(2t)}$, we see that the two facts are in fact perfectly
consistent; the upper tail of the normal distribution in the exponent in (4) ends up dominating the second moment.
It turns out that the central limit theorem (3) is valid for any real iid matrix with mean zero, variance one, and an exponential decay condition on the entries; this was first claimed by Girko,
though the arguments in that paper appear to be incomplete. Another proof of this result, with more quantitative bounds on the convergence rate has been recently obtained by Hoi Nguyen and Van Vu.
The basic idea in these arguments is to express the sum in (2) in terms of a martingale and apply the martingale central limit theorem.
If one works with complex gaussian random matrices instead of real gaussian random matrices, the above computations change slightly (one has to replace the real ${\chi}$ distribution with the complex
${\chi}$ distribution, in which the ${\xi_{i,j}}$ are distributed according to the complex gaussian ${N(0,1)_{\bf C}}$ instead of the real one). At the end of the day, one ends up with the law
$\displaystyle \frac{\log |\det A_n| - \frac{1}{2} \log n! + \frac{1}{4} \log n}{\sqrt{\frac{1}{4}\log n}} \rightarrow N(0,1)_{\bf R}, \ \ \ \ \ (5)$
or more informally
$\displaystyle |\det A_n| \approx n^{-1/4} \sqrt{n!} \exp( N( 0, \log n / 4 )_{\bf R} ) \ \ \ \ \ (6)$
(but note that this new asymptotic is still consistent with Turán’s second moment calculation).
We can now turn to the results of our paper. Here, we replace the iid matrices ${A_n}$ by Wigner matrices ${M_n = (\zeta_{ij})_{1 \leq i,j \leq n}}$, which are defined similarly but are constrained
to be Hermitian (or real symmetric), thus ${\zeta_{ij} = \overline{\zeta_{ji}}}$ for all ${i,j}$. Model examples here include the Gaussian Unitary Ensemble (GUE), in which ${\zeta_{ij} \equiv N(0,1)_
{\bf C}}$ for ${1 \leq i < j \leq n}$ and ${\zeta_{ij} \equiv N(0,1)_{\bf R}}$ for ${1 \leq i=j \leq n}$, the Gaussian Orthogonal Ensemble (GOE), in which ${\zeta_{ij} \equiv N(0,1)_{\bf R}}$ for ${1
\leq i < j \leq n}$ and ${\zeta_{ij} \equiv N(0,2)_{\bf R}}$ for ${1 \leq i=j \leq n}$, and the symmetric Bernoulli ensemble, in which ${\zeta_{ij} \equiv \pm 1}$ for ${1 \leq i \leq j \leq n}$ (with
probability ${1/2}$ of either sign). In all cases, the upper triangular entries of the matrix are assumed to be jointly independent. For a more precise definition of the Wigner matrix ensembles we
are considering, see the introduction to our paper.
The determinants ${\det M_n}$ of these matrices still have a Leibniz expansion. However, in the Wigner case, the mean and variance of the ${I_\sigma}$ are slightly different, and what is worse, they
are not all pairwise uncorrelated any more. For instance, the mean of ${I_\sigma}$ is still usually zero, but equals ${(-1)^{n/2}}$ in the exceptional case when ${\sigma}$ is a perfect matching (i.e.
the union of exactly ${n/2}$${2}$-cycles, a possibility that can of course only happen when ${n}$ is even). As such, the mean ${\mathop{\bf E} \det M_n}$ still vanishes when ${n}$ is odd, but for
even ${n}$ it is equal to
$\displaystyle (-1)^{n/2} \frac{n!}{(n/2)!2^{n/2}}$
(the fraction here simply being the number of perfect matchings on ${n}$ vertices). Using Stirling’s formula, one then computes that ${|\mathop{\bf E} \det A_n|}$ is comparable to ${n^{-1/4} \sqrt
{n!}}$ when ${n}$ is large and even. The second moment calculation is more complicated (and uses facts about the distribution of cycles in random permutations, mentioned in this previous post), but
one can compute that ${\mathop{\bf E} |\det A_n|^2}$ is comparable to ${n^{1/2} n!}$ for GUE and ${n^{3/2} n!}$ for GOE. (The discrepancy here comes from the fact that in the GOE case, ${I_\sigma}$
and ${I_\rho}$ can correlate when ${\rho}$ contains reversals of ${k}$-cycles of ${\sigma}$ for ${k \geq 3}$, but this does not happen in the GUE case.) For GUE, much more precise asymptotics for the
moments of the determinant are known, starting from the work of Brezin and Hikami, though we do not need these more sophisticated computations here.
Our main results are then as follows.
Theorem 1 Let ${M_n}$ be a Wigner matrix.
□ If ${M_n}$ is drawn from GUE, then
$\displaystyle \frac{\log |\det M_n| - \frac{1}{2} \log n! + \frac{1}{4} \log n}{\sqrt{\frac{1}{2}\log n}} \rightarrow N(0,1)_{\bf R}.$
□ If ${M_n}$ is drawn from GOE, then
$\displaystyle \frac{\log |\det M_n| - \frac{1}{2} \log n! + \frac{1}{4} \log n}{\sqrt{\log n}} \rightarrow N(0,1)_{\bf R}.$
□ The previous two results also hold for more general Wigner matrices, assuming that the real and imaginary parts are independent, a finite moment condition is satisfied, and the entries match
moments with those of GOE or GUE to fourth order. (See the paper for a more precise formulation of the result.)
Thus, we informally have
$\displaystyle |\det M_n| \approx n^{-1/4} \sqrt{n!} \exp( N( 0, \log n / 2 )_{\bf R} )$
when ${M_n}$ is drawn from GUE, or from another Wigner ensemble matching GUE to fourth order (and obeying some additional minor technical hypotheses); and
$\displaystyle |\det M_n| \approx n^{-1/4} \sqrt{n!} \exp( N( 0, \log n )_{\bf R} )$
when ${M_n}$ is drawn from GOE, or from another Wigner ensemble matching GOE to fourth order. Again, these asymptotic limiting distributions are consistent with the asymptotic behaviour for the
second moments.
The extension from the GUE or GOE case to more general Wigner ensembles is a fairly routine application of the four moment theorem for Wigner matrices, although for various technical reasons we do
not quite use the existing four moment theorems in the literature, but adapt them to the log determinant. The main idea is to express the log-determinant as an integral
$\displaystyle \log|\det M_n| = \frac{1}{2} n \log n - n \hbox{Im} \int_0^\infty s(\sqrt{-1}\eta)\ d\eta \ \ \ \ \ (7)$
of the Stieltjes transform
$\displaystyle s(z) := \frac{1}{n} \hbox{tr}( \frac{1}{\sqrt{n}} M_n - z )^{-1}$
of ${M_n}$. Strictly speaking, the integral in (7) is divergent at infinity (and also can be ill-behaved near zero), but this can be addressed by standard truncation and renormalisation arguments
(combined with known facts about the least singular value of Wigner matrices), which we omit here. We then use a variant of the four moment theorem for the Stieltjes transform, as used by Erdos, Yau,
and Yin (based on a previous four moment theorem for individual eigenvalues introduced by Van Vu and myself). The four moment theorem is proven by the now-standard Lindeberg exchange method, combined
with the usual resolvent identities to control the behaviour of the resolvent (and hence the Stieltjes transform) with respect to modifying one or two entries, together with the delocalisation of
eigenvector property (which in turn arises from local semicircle laws) to control the error terms.
Somewhat surprisingly (to us, at least), it turned out that it was the first part of the theorem (namely, the verification of the limiting law for the invariant ensembles GUE and GOE) that was more
difficult than the extension to the Wigner case. Even in an ensemble as highly symmetric as GUE, the rows are no longer independent, and the formula (2) is basically useless for getting any
non-trivial control on the log determinant. There is an explicit formula for the joint distribution of the eigenvalues of GUE (or GOE), which does eventually give the distribution of the cumulants of
the log determinant, which then gives the required central limit theorem; but this is a lengthy computation, first performed by Delannay and Le Caer.
Following a suggestion of my colleague, Rowan Killip, we give an alternate proof of this central limit theorem in the GUE and GOE cases, by using a beautiful observation of Trotter, namely that the
GUE or GOE ensemble can be conjugated into a tractable tridiagonal form. Let me state it just for GUE:
Proposition 2 (Tridiagonal form of GUE) \cite{trotter} Let ${M'_n}$ be the random tridiagonal real symmetric matrix
$\displaystyle M'_n = \begin{pmatrix} a_1 & b_1 & 0 & \ldots & 0 & 0 \\ b_1 & a_2 & b_2 & \ldots & 0 & 0 \\ 0 & b_2 & a_3 & \ldots & 0 & 0 \\ \vdots & \vdots & \vdots & \ddots & \vdots & \vdots \
\ 0 & 0 & 0 & \ldots & a_{n-1} & b_{n-1} \\ 0 & 0 & 0 & \ldots & b_{n-1} & a_n \end{pmatrix}$
where the ${a_1,\ldots,a_n, b_1,\ldots,b_{n-1}}$ are jointly independent real random variables, with ${a_1,\ldots,a_n \equiv N(0,1)_{\bf R}}$ being standard real Gaussians, and each ${b_i}$
having a ${\chi}$-distribution:
$\displaystyle b_i = (\sum_{j=1}^i |z_{i,j}|^2)^{1/2}$
where ${z_{i,j} \equiv N(0,1)_{\bf C}}$ are iid complex gaussians. Let ${M_n}$ be drawn from GUE. Then the joint eigenvalue distribution of ${M_n}$ is identical to the joint eigenvalue
distribution of ${M'_n}$.
Proof: Let ${M_n}$ be drawn from GUE. We can write
$\displaystyle M_n = \begin{pmatrix} M_{n-1} & X_n \\ X_n^* & a_n \end{pmatrix}$
where ${M_{n-1}}$ is drawn from the ${n-1\times n-1}$ GUE, ${a_n \equiv N(0,1)_{\bf R}}$, and ${X_n \in {\bf C}^{n-1}}$ is a random gaussian vector with all entries iid with distribution ${N(0,1)_{\
bf C}}$. Furthermore, ${M_{n-1}, X_n, a_n}$ are jointly independent.
We now apply the tridiagonal matrix algorithm. Let ${b_{n-1} := |X_n|}$, then ${b_n}$ has the ${\chi}$-distribution indicated in the proposition. We then conjugate ${M_n}$ by a unitary matrix ${U}$
that preserves the final basis vector ${e_n}$, and maps ${X}$ to ${b_{n-1} e_{n-1}}$. Then we have
$\displaystyle U M_n U^* = \begin{pmatrix} \tilde M_{n-1} & b_{n-1} e_{n-1} \\ b_{n-1} e_{n-1}^* & a_n \end{pmatrix}$
where ${\tilde M_{n-1}}$ is conjugate to ${M_{n-1}}$. Now we make the crucial observation: because ${M_{n-1}}$ is distributed according to GUE (which is a unitarily invariant ensemble), and ${U}$ is
a unitary matrix independent of ${M_{n-1}}$, ${\tilde M_{n-1}}$ is also distributed according to GUE, and remains independent of both ${b_{n-1}}$ and ${a_n}$.
We continue this process, expanding ${U M_n U^*}$ as
$\displaystyle \begin{pmatrix} M_{n-2} & X_{n-1} & 0 \\ X_{n-1}^* & a_{n-1} & b_{n-1} \\ 0 & b_{n-1} & a_n. \end{pmatrix}$
Applying a further unitary conjugation that fixes ${e_{n-1}, e_n}$ but maps ${X_{n-1}}$ to ${b_{n-2} e_{n-2}}$, we may replace ${X_{n-1}}$ by ${b_{n-2} e_{n-2}}$ while transforming ${M_{n-2}}$ to
another GUE matrix ${\tilde M_{n-2}}$ independent of ${a_n, b_{n-1}, a_{n-1}, b_{n-2}}$. Iterating this process, we eventually obtain a coupling of ${M_n}$ to ${M'_n}$ by unitary conjugations, and
the claim follows. $\Box$
The determinant of a tridiagonal matrix is not quite as simple as the determinant of a triangular matrix (in which it is simply the product of the diagonal entries), but it is pretty close: the
determinant ${D_n}$ of the above matrix is given by solving the recursion
$\displaystyle D_i = a_i D_{i-1} + b_{i-1}^2 D_{i-2}$
with ${D_0=1}$ and ${D_{-1} = 0}$. Thus, instead of the product of a sequence of independent scalar ${\chi}$ distributions as in the gaussian matrix case, the determinant of GUE ends up being
controlled by the product of a sequence of independent ${2\times 2}$ matrices whose entries are given by gaussians and ${\chi}$ distributions. In this case, one cannot immediately take logarithms and
hope to get something for which the martingale central limit theorem can be applied, but some ad hoc manipulation of these ${2 \times 2}$ matrix products eventually does make this strategy work.
(Roughly speaking, one has to work with the logarithm of the Frobenius norm of the matrix first.)
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formula remains true Fibonacci sequence
December 4th 2011, 02:47 PM #1
Dec 2011
formula remains true Fibonacci sequence
Fn+4 + Fn = 3Fn+2
for n = 0,1,2, . . .
Would the formula, necessarily remain true if the
sequence Fn is replaced by a sequence with the same recurrence
relation as the Fibonacci sequence but different initial terms?
Justify your answer briefly.
any ideas? thanks
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ISBN: 978-1-118-09570-6
304 pages
September 2012, Jossey-Bass
Read an Excerpt
Teaching K-12 math becomes an easier task when everyone understands the language, symbolism, and representation of math concepts
Published in partnership with SEDL, The Problem with Math Is English illustrates how students often understand fundamental mathematical concepts at a superficial level. Written to inspire aha
moments, this book enables teachers to help students identify and comprehend the nuances and true meaning of math concepts by exploring them through the lenses of language and symbolism, delving into
such essential topics as multiplication, division, fractions, place value, proportional reasoning, graphs, slope, order of operations, and the distributive property.
• Offers a new way to approach teaching math content in a way that will improve how all students, and especially English language learners, understand math
• Emphasizes major attributes of conceptual understanding in mathematics, including simple yet deep definitions of key terms, connections among key topics, and insightful interpretation
This important new book fills a gap in math education by illustrating how a deeper knowledge of math concepts can be developed in all students through a focus on language and symbolism.
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The Author xiii
About SEDL xv
About This Book xvii
Introduction xix
Julian’s Story xix
Rationale and Purpose xx
Who Benefits from This Book? xxii
ONE The Problem with Math Is English (and a Few Other Things) 1
Why Language and Symbolism? 1
What We Are Teaching 4
Turning the tide: A Sampling of Approaches 6
Mathematics Is About Relationships 8
Connecting the Pieces and Looking Ahead 9
TWO Why a Language Focus in Mathematics? 11
The Convergence of Mathematics and English: More Than Just Vocabulary 11
Problems Based on the English Language 13
A Number of Problems with Number 16
THREE Language and Symbolism in Traditional Instruction 21
Shortcomings of Traditional Instruction 22
More Language and Symbolism Issues: Adding Fuel to the Fire 32
Tell Me Again Why the Language Focus in Math? 38
FOUR So What Does Conceptual Understanding Look Like? 41
It Starts with Definitions 42
Making Connections in Math: Beyond Connecting Dots 51
The Interpretation and Translation Of Math 55
Conclusion 61
F I V E The Order of Operations: A Convention or a Symptom of What
Ails Us? 63
The Roots of the Rules 64
The Natural Order: A Mathematical Perspective 65
Conclusion: A Conceptual Understanding of the Order of Operations 78
S I X Using Multiplication as a Critical Knowledge Base 81
Understanding Key Definitions and Connections 81
Interpreting Multiplication 86
Using The Power of the Distributive Property 88
Feeling Neglected: The Units in Multiplication 100
Conclusion: Small Details, Huge Impact 103
SEVEN Fractions: The ‘‘F Word’’ in Mathematics 105
Defining Fractions: Like Herding Cats 105
The Fraction Kingdom 107
Interpreting Fractions 116
Conclusion 124
E I G H T Operations with Fractions 127
Adding and Subtracting Fractions 127
Multiplying Fractions 131
Dividing Fractions 150
Conclusion 160
NINE Unlocking the Power of Symbolism and Visual Representation 161
Symbolism 161
Visual Representation 168
The Power of Interpretation: Three Perspectives of Trapezoids 178
Conclusion 187
TEN Language-Focused Conceptual Instruction 189
Language Focus: Beyond the Definitions 190
The Secrets to Solving Word Problems 192
Suggested Instructional Strategies 197
Conclusion 216
ELEVEN Mathematics: It’s All About Relationships! 219
Language and Symbolism: Vehicles for Relationship Recognition 220
Relationships and Fractions 224
Proportional Reasoning 227
Relationships: Important Considerations 230
Relationships: Making Powerful Connections 234
Conclusion 249
TWELVE The Perfect Non-Storm: Understanding the Problem and Changing the System 251
A Systemic Issue 251
Math Makeover 257
Conclusion 264
Bibliography 267
Index 269
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Concepcion Molina, Ed.D., is a program associate with SEDL, a private, nonprofit education research, development, and dissemination corporation based in Austin, Texas. Dr. Molina supports systemic
reform efforts in mathematics and works to assist state and intermediate education agencies in their efforts to improve instruction and student achievement.
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October 22, 2012
The Problem with Math is English: A Language-Focused Approach to Giving All Students a Deeper Understanding of Mathematics
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All teachers know that teaching math isn’t easy. It’s a challenge just to get students to pay attention to the numbers and many students struggle with the subject. What if teachers took a different
approach? What if they looked at math from a different perspective, and didn’t see the subject as just numbers and procedures?
That’s what author Concepcion Molina suggests in his new book THE PROBLEM WITH MATH IS ENGLISH: A Language-Focused Approach to Helping All Students Develop a Deeper Understanding of Mathematics
(September 2012, Jossey-Bass).
Molina shows teachers that there are inherent problems with the way language, representation and symbols are used in teaching and learning mathematics and offers a new, language focused approach. He
does this by providing the tools needed to help students understand the underlying concepts of math.
Published in partnership with SEDL, the book illustrates conceptual-level mathematics by using a multitude of perspectives. It illustrates the importance of language, symbolism, and representation in
a way that enables teachers to use this knowledge to deepen the level of their students' understanding.
Through purposeful problems and activities, THE PROBLEM WITH MATH emphasizes major attributes of conceptual understanding, offers simple yet deep definitions of key terms, and provides an insightful
interpretation of mathematics.
“I’ve seen math frustrate so many students, especially English Language Learners,” says Molina. “I wanted to give teachers a new approach because traditional instruction does not focus on math. As a
unique language that any student can learn – students just need the proper guidance.”
This important book fills a gap in math education by illustrating how a deeper understanding of math concepts can be developed in all students.
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SSAC Geology of National Parks Modules
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What is the Discharge of the Congaree River at Congaree National Park? part of Pedagogy in Action:Partners::Examples
Denise Davis
Spreadsheets Across the Curriculum module/Geology of National Parks course. Students use a rating curve to determine discharge at various stage heights.
Quantitative Concepts: Measurement; Data presentation and analysis; Probability:Descriptive statistics; trend lines , Measurement; Data presentation and analysis; Probability, :Descriptive
statistics; trend lines :Goodness of fit (R2), Geometry; Trigonometry :Squares and rectangles, Algebra; Modeling; Functions:Modeling:Forward modeling, Basic arithmetic; Number sense:Logarithms;
orders of magnitude; scientific notation, Measurement; Data presentation and analysis; Probability:Gathering data , Visual display of data :XY scatter plots, Measurement; Data presentation and
analysis; Probability:Visual display of data , Visual display of data :Logarithmic scale, Measurement; Data presentation and analysis; Probability:Descriptive statistics; trend lines :Line- and
Excel Skills: Graphs and Charts:XY Scatterplot:Trendlines, Log Scale, Other Elementary Math Functions:LOG, LOG10, Graphs and Charts:XY Scatterplot, Basic Arithmetic
Subject: Environmental Science:Natural Hazards, Earth & Space Science:Geology, Ground & Surface Water, Environmental Science:Resource Use & Consequences:Land Use & Planning, Environmental Science:
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Tribulation Network
ISRAEL’S ALTARS
Moses, Solomon and Ezekiel
Section IV, Part 5 – The Wonderful Numbers of Sacred Canon
Doug Krieger
Page Flip version
Altar of Burnt Offering & Incense Altar
“And let them make Me a sanctuary (lit. “Sacred place”) that I may dwell among them. According to all that I show you, that is, the pattern of the tabernacle and the pattern of all its furnishings,
just so you shall make it” (Exodus 25:1).
It’s as if we’re dealing with the “God of Dimension” Who is meticulous in “pattern” – for “…see to it that you make them according to the pattern which was shown you on the mountain” (Exodus 25:40).
This “pattern” is inextricably adumbrative in its configurations – for the writer of Hebrews declares:
“Who serve the copy and shadow of the heavenly things, as Moses was divinely instructed when he was about to make the tabernacle. For He said, ‘See that you make all things according to the pattern
shown you on the mountain’…. Therefore it was necessary that the copies of the things in the heavens should be cleansed with these, but the heavenly things themselves with better sacrifices than
these” (Hebrews 8:5; 9:23). Therefore, it is recorded: “So Christ was offered once to bear the sins of many. To those who eagerly wait for Him He will appear a second time, apart from sin, for
salvation” (Hebrews 9:28) . . . . Yes, we are dealing here with “a shadow of the good things to come, and not the very image of the things” (Hebrews 10:1).
Please allow me, at this time, to preface this document as both a devotional, as well as a tedious research paper – a rare combination of hopeful inspiration and amazing calculations validating the
authenticity of these Sacred Altar measurements – measurements whose dimensions clearly detail their ultimate destination: The New Jerusalem.
The three Altars of Israel under consideration are those of Moses – the Tabernacle in the Wilderness (a.k.a. the Altar of Burnt Offering); Solomon – The Altar of Sacrifice in Israel’s First Temple
(a.k.a. the Bronze Altar); and, the futuristic Millenarian Altar of Sacrifice found in Ezekiel. Truly, these three epic Altars of Israel reveal the “Triple Crown” of the Almighty’s intentions for
humankind in that Moses, through whom the priesthood was initiated, encapsulates the “Priestly” aspect of the altar – in bringing God to man, and man to God; the Solomonic Temple Altar depicts the
kingly ministry – the Lion of the Tribe of Judah; and the futuristic Millenarian Altar of Ezekiel is representative of Messiah’s prophetic ministry…our Priest, our King and our Prophet.
Moses’ Altar initiates a most mandatory work on behalf of the New Jerusalem and of Messiah’s coming as the “Suffering Servant” Whose first coming is likewise depicted in Zechariah as Joshua, the High
Priest, who took our “filthy garments” and bore our guilt and shame, and was given “rich robes” for it was said of Joshua: “‘Take away the filthy garments from him,’ And to him He said, ‘See, I
have removed your iniquity from you, and I will clothe you with rich robes.’”
Thus, was this high priestly ministry revealed – revealed to all who have eyes to see “Joshua, the high priest” for who he is: “You and your companions who sit before you, for they are men of a sign
of wonder; for behold, I am bringing forth My servant the BRANCH. For behold, the stone that I have laid before Joshua: Upon the stone are seven eyes. Behold, I will engrave its inscription,’ Says
the LORD of hosts, ‘And I will remove the iniquity of that land in one day’” (Zechariah 3 – excerpts).
This “priestly altar” (for through Moses was the priesthood initiated) is followed by the “Altar of Kingship” as the apex of Israel’s Kingdom, the Solomonic Altar – the Bronze Altar – depicts the
Kingship – which also is seen in Zerubbabel the Governor juxtaposed to Joshua, the High Priest. It is this Altar whose immediate worth bears prophetic implications for the Kingship’s building of the
Temple – for its extensions are amplified in the rededication under Zerubbabel to whom it was said:
“‘This is the word of the LORD to Zerubbabel: ‘Not by might nor by power, but by My Spirit,’ says the LORD of hosts, ‘Who are you, O great mountain? Before Zerubbabel you shall become a plain! And
he shall bring forth the capstone with shouts of “Grace, grace to it!””” (Zechariah 4:6-7).
The “Lion of the Tribe of Judah” bears the crown of the Kingship – for though He at once is Joshua, our High Priest, He is simultaneously worthy to bear the crown of Kingship the second time – for it
is Joshua who is crowned, not Zerubbabel: “Behold, the Man whose name is the BRANCH! From His place He shall branch out, and He shall build the temple of the LORD; yes, He shall build the temple
of the LORD. He shall bear the glory, and shall sit and rule on His throne; so He shall be a priest on His throne, and the counsel of peace shall be between them both.”’ (Zechariah 6:12-13).
But it is not until we arrive at the futuristic, Millenarian Altar found in Ezekiel’s accounting that the Prophet is revealed – for, indeed, Ezekiel is among the most prophetic of texts. The Altar
seen in Ezekiel is by far the most majestic and awesome in its centricity. In it the ministry of the Prophet is wholly discovered. Even so, this part of the “Triple Crown” is seen in Zechariah’s
revelation as the glorious and universal Lampstand, taking central position betwixt the “Two Sons of Oil” – Joshua and Zerubbabel, the High Priest and the King, the Priesthood and the Kingship.
Ultimately, it is this glorious Menorah, Lampstand, that shines forth in the Apocalypse wherein, “One like the Son of Man” (Revelation 1:13) shines brilliant in the midst of the Seven Golden
Lampstands (the Churches). Here is the Prophet with “His voice as the sound of many waters” and from whose mouth “…went a sharp two-edged sword” (Rev. 1:16) … for the Prophet speaks and illuminates,
and the Truth is revealed…for “the Testimony of Jesus is the spirit of prophecy” (Revelation 19:10).
Moses, Solomon and Ezekiel – High Priest, King and Prophet – three Altars – three epic revelations and ministries which build the Temple, which sanctify His people and make them a “kingdom of
priests” and, finally, call them to be His prophets: “You must prophesy again about many peoples, nations, tongues, and kings . . . they will prophesy” (Revelation 10:11; 11:3) – for they are the
“measured ones” found at the Apocalyptic Altar fulfilling their mission in following Him, their Priest, their King and their Prophet (Revelation 11:1-3).
MOSES – the Priest - Measurements of the Tabernacle’s Altars – Burnt Offering and Incense
In that the Altar of Burnt Offering and the Golden Incense Altar – whose patterns were given to Moses in the first “sacred place” in a barren wilderness – appear in stark contrast to the bleak scene
wherein they made their first appearance. God’s ways are entirely different than ours and often we find that the “wilderness” – a place of endurance, patience, testing and trial – is precisely where
He desires to initiate His glory and commence His “pattern.”
We have fused these two furnishings – the Altar of Burnt Offering and the Golden Altar of Incense – together in that their ultimate interchange in the Apocalypse does as well, i.e., later we shall
discover that the Golden Altar before the Throne is readily in view in John’s Revelation, but the Altar of Sacrifice as seen clearly in Revelation 11:1 is as well. These two are, in essence, but one
heavenly Altar – and indeed, for the “Sacrifice of Praise” and our “ascended prayers” before the Holiest of All…before the Throne of God…are merged in their consecration and ultimate sacrifice of
praise before Him.
Therefore, let us commence these “measurement proceedings” and determine the extent of the pattern shown to Moses on the Mount:
We read:
“(1) You shall make an altar of acacia wood, five cubits long and five cubits wide—the altar shall be square—and its height shall be three cubits. (2) You shall make its horns on its four corners;
its horns shall be of one piece with it. And you shall overlay it with bronze. (3) Also you shall make its pans to receive its ashes, and its shovels and its basins and its forks and its firepans;
you shall make all its utensils of bronze. (4) You shall make a grate for it, a network of bronze, and on the network you shall make four bronze rings at its four corners. (5) You shall put it
under the rim of the altar beneath, that the network may be midway up the altar. (6) And you shall make poles for the altar, poles of acacia wood, and overlay them with bronze. (7) The poles shall be
put in the rings, and the poles shall be on the two sides of the altar to bear it. (8) You shall make it hollow with boards; as it was shown you on the mountain, so shall they make it” (Exodus 27:1-8
The simplicity of the Altar of Burnt Offering – bronze or copper being its overlay – should not be held in contempt because its size was less than magnanimous than, let’s say, either Solomon’s Altar
and Ezekiel’s designed for the same purpose…for within its dimensions are wondrous truths to explore.
· Length and Width is square – i.e., 5 cubits by 5 cubits – in that we have discovered the precise length of the Hebrew Sacred Cubit – thanks to astounding and persistent research by metrologists
throughout the past two hundred years, as well as their counterparts within archaeology and other sciences (e.g., Sir Isaac Newton of physics fame); therefore….
· 5 cubits = 5 x 25.20” = 126” and this is equal to “9” (as in 1 + 2 + 6 = 9)
· 126” x 4 equal edges of this square = 504” which also is “9” for the top of the Altar and “9” for the bottom of the Altar of 504 + 504 = 9 + 9 = “18” (or precisely 1008 linear inches top and
bottom) which is the Standard Measurements of the New Jerusalem and, as well, 5040 is Plato’s Optimal Number of social harmony, as well as governance; therefore, when divided by 2 as in 5040/2 =
25.20 which is the measurement of the Sacred Cubit.
· 3 cubits is the Altar’s height – therefore: 3 x 25.20 = 75.6” (which equals “18” New Jerusalem Standard of Measurement as in NJS*) and this one edge is a fractal of the 756’ of the base
measurement on one edge of the Great Pyramid of Giza – having, therefore, all the potential to extrapolate into the measurements of the New Jerusalem in that 756 x 12 = 9072 (a cubed GPG) and this
“18” is equal in inches to 9072 x 12 = 108,864 (or two number sets of “108” – which is a fractal of the radius of Earth’s Moon @ 1,080 Statute Miles and the “864” number set which is a fractal of the
Sun’s diameter of 864,000 miles and both number sets are equal to “18” (1080) and “864” (8 + 6 + 4 = 18) and these two sets of “18” when added together produce the sum of “36” (as in 18 + 18 = 36)
and “36” is the fractal of a circle’s 360° which connotes the Eternal God without Beginning or Ending (Please read below only if you can tolerate “Sacred Geometry” math!).
· The total linear cubits of Moses’ Altar in the Wilderness: 20 cubits (top) + 20 cubits (bottom) + 3x4 cubits or four vertical edges or 12 = 52 cubits x 25.20” = 1,310.4”
· *The New Jerusalem Standard of Measurement is “18” derived from Revelation 21:16 wherein it is said that “The city is laid out as a square; its length is as great as its breadth.…And he measured
the city with the reed: twelve thousand furlongs…Its length, breadth, and height are equal.” A furlong is equal to 660 Imperial Feet (12” in one foot); therefore, 12,000 furlongs x 660’ equal
7,920,000 Linear Feet divided by 5,280 Linear Feet in one Imperial Mile equals 1,500 miles times 12 edges of a cubed New Jerusalem as so described is equal to 18,000 miles (as in 12 x 1,500 miles =
18,000 miles) – henceforth, this “18” is the Standard Bearer of the total six-sided cubed New Jerusalem and its 12 edges.
In conjunction with the immediate configurations above, we would highlight the fact that in John Michell’s Dimensions of Paradise (pp. 43-47) he discovered this set of numbers (108,864) in
configuring the number of NJ units (i.e., NJ = New Jerusalem Units of 576 square feet per NJ Unit when using 22/7 as π vs. 576.19 square feet per NJ Unit when using 864/275 as π); however, it does
not appear he connected the “dots” to the New Jerusalem nor to its most immediate planetary associations: Earth’s Moon and Sun. In both cases, Michell uses the Earth’s radius as a fractal of 3,960’
(Earth’s radius is 3,960 miles), then calculates the area of its artificial square (replicating the square of the New Jerusalem found in the Revelation) and then calculates the area of its circle but
using two different calculations for π as so stated (Note: Area of a circle = π * r * r) and coming up with two separate number sets for NJ Units – by using 22/7 he comes up with 108,900 NJ units
(i.e., 576 sq. feet per NJ Unit) and by using 864/275 (which is 3.1418181818+) he comes up with 108,864 NJ Units (i.e., 576.19 per NJ Unit) which is more accurate in his “New Jerusalem Diagram.” His
“square area” – using 7,920’ as the diameter/edge of the squared side remains the same (i.e., 7920’ x 7920’ = 62,726,400 / 43,560 sq. feet (1 acre) = 1,440 acres … naturally, in both illustrations
this number is the same since we’re dealing with a squared area).
However, when calculating the area of the miniature Earth (i.e., 7920 miles vs. 7920’ or its radius of 3960 miles vs. 3960’) and using π at 22/7 (3.14285714) vs. using π at 864/275 (3.14181818) to
determine the area of its circle in NJ Units, the product of the NJ Units is altered. The NJ Units go from 576 (22/7) to 576.19 (864/275) or the total number of NJ Units of the circle’ area itself
is diminished (which was Michell’s preference). The resulting number of New Jerusalem Units were determined under 22/7 to equal 108,900 NJ Units and when using 864/275 resulted in 108,864, a
discrepancy of 36 NJ Units which Michell claims “is the area of a circle with a radius of 72 feet”; therefore, he suggests that this circle/radius be placed at the center of his New Jerusalem Diagram
thereby making the area of both “Pi’s” equal
to 108,864 NJ units. Michell wants to get to the 3,024 NJ units by using 864/275 for π juxtaposed to using 22/7 as π and arriving at 3025 NJ units in his 660’ radius example and doing a similar fete
by using a radius of 3960’ and arriving at 108,900 NJ units using 22/7 as π juxtaposed to using 864/275 for π and arriving at 108,864 NJ units. Oddly enough – and Michell doesn’t see it (at least
doesn’t mention it) – by using the more accurate 864/275 as π he can achieve 3,024 NJ units in his 660’ radius/circle and using 864/275 as π he can achieve 108,864 NJ units in his 3,960 radius
Why are 108,864 NJ units preferred? Because, says Michell, 108,864 is divisible by 504 and chimes in with Plato’s Optimal Number for social governance (likewise, 108,864/504 = 216 = The Name of God
or is the fractal of the Moon’s diameter and a fractal of Earth’s Nautical Mile circumference of 21,600 and is also a fractal of six “Earth rotation days” @ 360° x 6 days = 2160° wherein man was
created on the sixth day). However, Mitchell’s point, though well taken, misses the cardinal observation that 3,024 is the perimeter of the four-edged-base of the Great Pyramid of Giza (GPG) * 3
(cubed) = 9,072’ x 12” = 108,864” of a cubed 12-edged GPG. Both “number sets” found in “108,864” are equal to “18” (as in 108 = 18 and 864 = 18) which “18” is the Standard Bearer/Measurement of the
New Jerusalem’s 18,000 Mile 12-edged perimeter. Thus, the New Jerusalem’s 7,920,000 Linear Feet (i.e., 12,000 furlongs x 660’ = 7,920,000 Linear Feet) or 7,920’ is inter-related with the Earth, the
Moon and the Sun, as well as the GPG – a most amazing correlation and, safe to say, hardly a happenstance association!
Dimensions of Paradise vs. Wonderful Numbers of Sacred Canon
I would at this juncture insert a casual but striking observation throughout our comparing and contrasting with Michell’s Dimensions of Paradise and our less than renowned Wonderful Numbers of Sacred
Canon (now in production, 2012) – or for that matter those of us who claim to be the object of Sacred Geometry and those that heretofore have written of our destination and its amazing integration
with Sacred objects and mysterious metrologies the world over – that our thesis of the same is prejudiced due to the angelic intervention by none other than the Angel Palmoni spoken of in Daniel 8:13
(Holman Literal Translation) – a.k.a. “that certain saint…the numberer of secrets…THE WONDERFUL NUMBERER” – therefore, our undeniable emphasis upon the “18” of the New Jerusalem dominates our
conversation – a number and those “number sets” which reflect its ostentatious and somewhat ubiquitous demonstration throughout Holy Writ, astronomy, music, physics, mathematics and a host of
sciences…so numerous are its sums, products, quotients and remainders that its recurrence appears more so than virtually any other ultimate combination and this having been attested to in what is
known as the *Gordonian Histogram which validates our thesis that “18” ubiquitous compatibilities insofar as the aforesaid is a proven mathematical reality – why, well, we just need a bit more
algebra and we’ll nail it! (*Gordonian Histogram was designed in 2011 to determine the recurrence in number of “18”.)
That said, not only is “18” the announced New Jerusalem Standard of Measurement, but 25.20 or 252 take on far more significance in our writings than in the world of Sacred Geometry – because it is
the key which unlocks countless dimensions given in Sacred Cubit fashion throughout Sacred Text – in particular, the 51 or 66 books of the Book of books. Furthermore, connection to the Great Pyramid
of Giza (GPG) is seen as that Altar to the LORD in the midst of Egypt and, therefore, more readily are its measurements taken into account in our calculations. Moreover, the persistence of Messiah
occurs throughout our accounting and His relationship to the New Jerusalem, as well as His ministry past and future – as in the numbers: 27, 30, 33, 45 and the name of God as in 36, 72 and 216 – we
take umbrage regarding the “Christian myth” and the “legends of the Christ” – in that our affirmations relative to Sacred Judeo-Christian text demand, not only “deliverance” but view reconciliation
as “God wrought and God bought” not affected by human manipulation but by divine intervention.
And, of course, further clarification as to the numbers which reflect the New Jerusalem as in 792 and 144 and their association with the GPG, as well as the metrologies of Moon and Sun, let alone our
entire solar system as is described in other sections in our text. On the one hand it simply is not academically fair to analyze John Michell’s Dimensions of Paradise without his retort (now that
he’s passed from this earthly clod); however, I am sure a hefty response from others within the Sacred Geometry community will be forthcoming and then an abundant synthesis will hopefully result to
the benefit of all.
The Altar in the Wilderness
Now…back to Moses’ Altar in the Wilderness….
· 75.6” x 4 height edges of this cubical (three-dimensional) Altar = 302.4” (for it is 3 cubits in height or 3 x 25.20” = 75.6”) which, of course, is the fractal of the perimeter of the Great Pyramid
of Giza at 3024 Linear Feet (4 x 756’ = 3024’) and 3024’ x 12” = 36,288 Linear Feet of a flat 4-edged base of the GPG – likewise, as we shall see it is precisely the measurement of the Altar Hearth
of Ezekiel’s Millenarian Temple which hearth is 12 Cubits in Length and 12 Cubits in Width or 12 x 25.20” = 302.4” (Ezekiel 43:13-17).
· Sum of all linear inch measurements of the Altar of Burnt Offering: 1008” (top and bottom measurements) + 302.4” = 1,310.4 Linear Inches = “9” or 1 x 3 x 1 x 4 = “12” (in both number sets – sum
and product – God (3) seeks to redeem humanity (Man= 6 as in 3 + 6 = 9 or Creation = 4 as in 3 (Triune God) x 4 = 12)
· Surface Area of the Altar: 75.6” x 126” = 9,525.6” x 4 (sides of Altar) = 38,102.4 sq. inches= “18” NJS
· Surface Area of the Altar: 126” x 126” = 15,876 x 2 (top and bottom of Altar) = 31,752 sq. in. = “18” NJS
· Combined Surface Area of Altar: 38,102.4 sq. in. + 31,752 sq. in. = 69,854.4 sq. inches = “36” (as in 6 + 9 + 8 + 4 + 5 + 4 = 36); therefore the entire Altar reflects God’s ultimate satisfaction
and design for the Altar.
· 69,854.4 sq. inches / 144 (the wall of the New Jerusalem) = 485.1 = “18” NJS OR 4 + 8 + 5 + 1 = 18.
· Volume or Capacity of the Altar of Burnt Offering: 126” x 75.6” x 126” = 1,200,225.6 = “18” NJS
· When 1,200,225.6 is divided by the volume of one square foot 12 x 12 x 12 or 1,728 cubic inches the result is: 1,200,225.6 / 1,728 = 694.575 = “36” (as in 6 + 9 + 4 + 5 + 7 + 5 = 36) which
connotes the 360° of a circle which no beginning or ending as presents the Eternal God to us.
· The “cubit number sets” themselves result in a sum of “52” as in 8 x 5 (top and bottom measurements) = 40 cubits and the height of 3 cubits x 4 = 12 cubits and 40 + 12 = 52 cubits and “52” is the
number of weeks in a Solar Year; therefore, the Altar of Burnt Offering is a very clear pattern wherein the Eternal God has purposed the Altar of burnt Offering to be a place of sacrifice and
cleansing preparing God’s People for His ultimate plan and purpose of the ages: The New Jerusalem – the evidence is overwhelming and the measurements repeatedly attest thereto. This article’s focus
is not upon the types of sacrifice and the richness of their meaning but is cognizant that God, the priests and the people are intertwined in their engagement in the Altar’s sacrifices – benefitting
all three – and that the “once for all” offering of the Lamb of God, our whole burnt offering, suffices for eternity.
The Golden Altar of Incense
The Golden Altar of Incense within the Holy Place of the Tabernacle in the wilderness finds itself in the Christian Sacred Canon “inside the veil” which was rent from top to bottom on behalf of the
“blood of the Everlasting Covenant” – Jesus, our High Priest, entered in once for all sprinkling His Own blood upon the mercy seat obtaining eternal redemption. Now, with boldness, we may enter into
His presence “through the blood of the Everlasting Covenant” that we may obtain mercy and find grace to help in time of need. Thus, now is the Golden Altar of Incense within the veil, before the
very presence of God on the Throne (Revelation 9:13). In that the Altar of Burnt Offering (Moses) and the Altar of Incense (Moses) is initially separated due to the veil – that veil in Christian
theology is removed; therefore, not only in Pauline epistles is there this removal but in John’s Revelation these two altars (the Altar of Burnt Offering and the Incense Altar) are viewed as but one.
Likewise, the composition of the incense, that sweet-smelling savor, is of inestimable worth and full of glorious meaning – but we must move on to the measurements of this Altar of Incense and review
its simplicity and brilliance, for it is overlaid thereon with gold, hence: The Golden Altar of Incense – and it keeps this title into the Revelation – for only those prayers offered by our Lord
Jesus on our behalf are of such quality and eternal value – and only those prayers uttered by the Spirit from within His people with “groaning which cannot be uttered” by our High Priest are
acceptable to Him Who sits on the Throne.
“(1) You shall make an altar to burn incense on; you shall make it of acacia wood. (2) A cubit shall be its length and a cubit its width—it shall be square—and two cubits shall be its height. Its
horns shall be of one piece with it. (3) And you shall overlay its top, its sides all around, and its horns with pure gold; and you shall make for it a molding of gold all around. (4) Two gold rings
you shall make for it, under the molding on both its sides. You shall place them on its two sides, and they will be holders for the poles with which to bear it. (5) You shall make the poles of
acacia wood, and overlay them with gold. (6) And you shall put it before the veil that is before the ark of the Testimony, before the mercy seat that is over the Testimony, where I will meet with
you. (7) Aaron shall burn on it sweet incense every morning; when he tends the lamps, he shall burn incense on it. (8) And when Aaron lights the lamps at twilight, he shall burn incense on it, a
perpetual incense before the LORD throughout your generations. (9) You shall not offer strange incense on it, or a burnt offering, or a grain offering; nor shall you pour a drink offering on it. (10)
And Aaron shall make atonement upon its horns once a year with the blood of the sin offering of atonement; once a year he shall make atonement upon it throughout your generations. It is most holy to
the LORD” (Exodus 30:1-10).
The Golden Altar of Incense Dimensions
· 1 Cubit Length and 1 Cubit Width and 2 Cubits Height
· 1 Cubit (width) x 25.20” = 25.20” x 4 (4 square top) = 100.8” (i.e., 4 linear widths top) and 100.8” (4 liner widths bottom) or “18” and “18” NJS
· 18 + 18 = “36” and 36 connotes the 360° of the Eternal God without Beginning or Ending as in a circle
· 2 Cubits the Height x 25.20 = 50.4” per edge x 2 on one side = 100.8” or “18” NJS and 100.8” (2 on the other side for an addition “18” NJS)
2 Sets of Side Edges = “18” NJS ea. and 18 + 18 = “36” which is a fractal of 360° of a circle which is the Eternal God without Beginning or Ending….and 100.8” + 100.8” = 201.6”
· 201.6” = “9” (God-3 and Man-6) and 201.6” is the fractal of the diameter of the Moon @ 2016 Miles and of the circumference of the Earth in Nautical Miles @ 21,600 and is equal to 3° of the Great
Precessional of the Mazzaroth or 216 years; and 216 is the Name of God.
· 100.8” is also a fractal of the Moon’s radius of 1080 Miles
· Sum of all 12 Edges of the Golden Altar of Incense: 100.8” (four top edges) + 100.8” (four bottom edges) + 201.6” (four side edges) = 403.2” or 4 x 3 x 2 = “24” and 24 is the constant expression
of Governmental Authority within the New Jerusalem represented by the 24 Elders which we affirm to be the same 24 given in the New Jerusalem as the 12 Patriarchs of Israel and the 12 Apostles of the
Lamb (the Church) – the Gates and Foundations of the Holy City; likewise, “24” is the Jewish Canon of Sacred Text – the revelation of the One True God in the Hebrew Canon and the “24 courses of the
priesthood (I Chronicles 24-26).”
· Square Area of the Golden Altar of Incense (top and bottom): 25.20” x 25.20” = 635.04 sq. inches. (top) = “18” NJS and 635.04 sq. inches (bottom) = “18” NJS ea. and 18 + 18 = “36” = the Eternal
God (circle, etc.) or 635.04 + 635.04 = 1,270.08 = 18 NJS
· Square Area of the Golden Altar of Incense (sides only): 50.40” x 25.20” (one side of four vertical sides of this six-sided altar – four vertical sides and its top and bottom “sides”) = 1,270.08 =
18 NJS (1 + 2 + 7 + 8 = 18) and 1,270.08 x 4 vertical sides = 5,080.32 sq. inches = 18 NJS (5 + 8 + 3 + 2 = 18)
· Combined Square Area of the Golden Altar of Incense: 5,080.32 Sq. In. (four vertical sides) + 1,270.08 (top and bottom areas) 6,350.4 (all six sides) = 18 NJS (as in 6 + 3 + 5 + 4 = 18)
· Capacity in Cubic Inches of the Golden Altar of Incense: 25.20” x 50.40” x 25.20” = 32,006.016 cubic inches = 18 NJS (3 + 2 + 6 + 1 + 6 = 18) and 32,006.016 / 144 = 222.264 = “18” New Jerusalem
Standard and 32,006.016 cu. in. / 1728 cu. in. (number of cubic inches in 1 sq. foot or 12” x 12” x 12” = 1728 cu. in.) = 18.522 sq. feet = 18 NJS (as in 1 + 8 + 5 + 2 + 2 = 18). (Note: “1728”
The ubiquity of the 18 New Jerusalem Standard upon the Golden Altar of Incense convinces this author that the supreme purpose of the Golden Altar of Incense has everything to do with the preparation
of the Bride of Messiah, who is called the Holy City, New Jerusalem. Likewise, the Eternal God (36) as seen in His Name (216) has destined His people (Israel and the Church) as seen in the (24) to
accomplish His eternal purpose for the ages. The dimensions of the Golden Altar of Incense give absolute confirmation that this is the destiny of the people of God. The prayers of the saints (24)
are offered up to the Eternal God (36) in His Name (216) and all have but one ultimate recourse: The New Jerusalem (18).
SOLOMON – the King - Altar of Sacrifice
Solomon’s Altar of Sacrifice is likewise known as the “brazen altar.” Its location was east of the Temple proper. Its form resembled the altar of the Tabernacle in the Wilderness; however, it was
much larger: twenty cubits in length (not 5 cubits like the Altar of Burnt Offering in the Wilderness), twenty cubits in breath (not 5 cubits as the Wilderness Altar) and ten cubits in height (II
Chronicles 4:1) (Wilderness Altar = 3 cubits in height). Thus, Solomon’s Altar was designed in ratio to the Altar in the Wilderness: 20x20x10 = 4,000 vs. 5x5x3 = 75; therefore, 4000:75 or 160:3.
Solomon’s linear cubit measurements on all 12 edges (i.e., 200 cubits x 25.20”) of the Altar were 5040.00” (5040 Plato’s Optimal Number) and Moses’ Altar in the Wilderness were 1,310.4” (52 cubits x
25.20”) – the difference between the two linear cubit measurements is most significant: 5040.00” (Solomon’s Altar) minus 1310.40” (Moses’ Altar) = 3729.6” or 3 + 7 + 2 + 9 + 6 = “27” and this “27”
is the full revelation of Messiah, as the Suffering Servant, in Christian Sacred Canon of 27 texts. Also, please note the tally of both altars: 5040 = “9” and 1310.4 = “9” and 9 + 9 = 18 which is
the New Jerusalem Standard of Measurement – the “Altar” paves the way to the New Jerusalem and the two altars are linked via the Messiah (27) who came as the Suffering Servant to bear the sins of the
world as per Christian theology.
The altar built by King Solomon (I Kings 8:64), became the new object of consecration during the reign of King Asa (II Chronicles 15:8); therefore, and no doubt, some restorative effort was made.
Asa removed the altar towards the north, and in its place he erected a similar altar to the one he had seen in Damascus (II Kings 16:10-15) – likewise, he removed the bulls from beneath the Cast
Bronze Basin and he also “cut off the panels of the carts, and removed the lavers from them.”
There is some evidence that indicates King Hezekiah may have restored some of the grandeur to the Temple and its furnishings, though the text does not indicate such; however, it does indicate that
Hezekiah’s son, Manasseh, polluted the Temple (II Kings 21:4-5; II Chronicles 33:4-5) which was again restored and re-dedicated to Yahweh by this same King Manasseh after Manasseh was taken by the
King of Assyria with “nose hocks, bound him with bronze fetters, and carried him off to Babylon…Now when he was in affliction, he implored the LORD his God, and humbled himself greatly before the God
of his fathers, and prayed to Him; and He received his entreaty, heard his supplication, and brought him back to Jerusalem into his kingdom. Then Manasseh knew that the LORD was God” (II Chronicles
33:11-13); then, it was at this time King Manasseh “repaired the altar of the LORD, sacrificed peace offerings and thank offerings on it, and commanded Judah to serve the LORD God of Israel” (II
Chronicles 33:16).
King Solomon Built the Altar and the First Temple to Plan
“‘Consider now, for the LORD has chosen you to build a house for the sanctuary; be strong, and do it.’ Then David gave his son Solomon the plans for the vestibule, its houses, its treasuries, its
upper chambers, its inner chambers, and the place of the mercy seat; and the plans for all that he had by the Spirit, of the courts of the house of the LORD, of all the chambers all around, of the
treasuries … ‘All this,’ said David, ‘the LORD made me understand in writing, by His hand upon me, all the details of these plans’… Furthermore King David said to all the assembly: ‘My son Solomon,
whom alone God has chosen, is young and inexperienced; and the work is great, because the temple is not for man but for the LORD God’” (I Chronicles 28:10-12) (I Chronicles 28:19; 29:1).
“Moreover he made a bronze altar: twenty cubits was its length, twenty cubits its width, and ten cubits its height” (II Chronicles 4:1).
Again, it is imperative for us to understand that all of “these plans” given to David and communicated to the future King Solomon – yes, “all the details of these plans” – had to be followed with
great precision for the Temple so designed was “not for man but for the LORD God”…and that is precisely what is going on here in all these dimensional deliberations. The question is forthright:
This house is for the LORD God, not for man – this is what God is after and has always been after – to “dwell among His people and be their God and they His people” – that’s what all these plans are
Top and Bottom of Altar Square Area and Linear Edges
Square Area or Altar’s Top and Bottom
· Top: 20 cubits x 25.20” = 504” x 504” = 254,016 sq. inches = 18 NJS (2 + 5 + 4 + 1 + 6 = 18)
(i.e., 20 cubits x 20 cubits = 400 sq. cubits or 504” x 504” = 254,016 sq. inches)
· Bottom: 20 cubits x 25.20” = 504” x 504” = 254,016 sq. inches = 18 NJS (2 + 5 + 4 + 1 + 6 = 18)
(Total 400 sq. cubits + 400 sq. cubits (top and bottom) = 800 sq. cubits)
· Top + Bottom Combined: 254,016 + 254,016 = 508,032 sq. inches = 18 NJS (5 + 8 + 3 + 2 = 18)
Linear Edges of Top and Bottom Edges and Height Edges
· Linear Inches 504” x 4 (four top edges) = 2016” x 2 (top and bottom) = 4,032” Linear Inches (8 edges) –
· NOTE: 2016” is a fractal of 2160 the diameter of the Moon and the 21,600 Nautical Mile Circumference of the Earth and the 360^ₒ Earth’s rotation x 6 days = 2160 degrees; and, of course, 216 is the
Name of God; however, to complete the Linear Edges of the given altar measurements we must add the 4 height measurements below…
· 10 Cubit Height (one vertical edge) x 25.20” = 252” (Linear) x 4 (4 height edges) = 1008 = 18 NJS and 1008 is the fractal of the 1080 mile radius of the Moon.
· Combined Linear Measurements: 4,032” Top/Bottom Edges (8 edges) + 1008” (4 height edges) = 5,040 = Plato’s Optimal Number of Society and 1080 (Radius of the Moon) + 3960 (Radius of the Earth) =
Square Area of the Four Sides of the Altar’s Measurements
· 10 Cubits Height and 12 Cubit Length or 252” x 504” = 127,008 sq. inches = 18 NJS (1 + 2 + 7 + 8 = 18)
· 127,008 x 4 (4 separate sides of the cubed height) = 508,032 sq. inches = 18 NJS (5 + 8 + 3 + 2 = 18)
· 508,032 sq. inches / 144 (sq. inches in 1 sq. foot) = 3,528 sq. feet = 18 NJS (3 + 5 + 2 + 8 = 18)
Combining the Top, Bottom and Sides of the Altar sq. inch areas:
· Top/Bottom Sq. Area in Inches: 508,032 sq. inches, added to…
· Four Area Sides of the Altar: 508,032 or 5832 + 5832 = 11,664 or 1,016,064 sq. inches = 18 NJS = (1 + 1 + 6 + 6 + 4 = 18)
Volume of the Altar:
· 20 cubits x 20 cubits x 10 cubits) or 504” x 504” x 252” = 63,631,008 sq. inches / 1,728 cubic inches (or 12” x 12” x 12” = 1,728 cubic inches) = 36,823.5 sq. feet = “27” (3 + 6 + 8 + 2 + 3 + 5 =
· “27” = The Full Revelation of Messiah in Sacred Christian Canon (27 volumes).
· Therefore, the “fullness” of the Altar is wholly representative of the “Altar of Sacrifice” wherein the “Suffering Servant” came as the offering for sin once and for all and it was “not for
himself” (Daniel 9:26: “And after the sixty-two weeks Messiah shall be cut off (lit. “suffer the death penalty”) but not for Himself.”) The entire measurement of the Solomonic Altar from top to
bottom in both its linear edge measurements and its “expression” as the “18” in its flat surface areas is wholly representative of the New Jerusalem but the “essence” – the “fullness” – of the Altar
is wholly given over to the Lamb of God Who takes away the sin of the world – He is our “Burnt Offering (Leviticus 1:1-17; Grain Offering (Leviticus 2:1-16); Peace Offering (Leviticus 3:1-17); Sin
Offering (Leviticus 4:1-35); Trespass Offering (Leviticus 5:1-13) – our fivefold offering.
The “centrality of the Solomonic Altar” – in front of the entrance, not on the side of the entrance into the temple – equates to the centrality of the cross of Christ. All that He is in the
offerings is not for Himself but for the New Jerusalem – so is the Altar’s measurements and so is the fullness of the Altar.
As we read these words we must read them with a profound understanding that as the fire of His acceptance fell, it fell upon an altar wholly committed to the Divine Purpose of that city which father
Abraham sought:
“For he (Abraham) waited for a city which has foundations, whose builder and maker is God” (Hebrews 11:10).
“1 Now when Solomon had finished praying, fire came down from heaven and consumed the burnt offering and the sacrifices, and the glory of the LORD filled the house. 2 The priests could not enter into
the house of the LORD because the glory of the LORD filled the LORD’S house. 3 All the sons of Israel, seeing the fire come down and the glory of the LORD upon the house, bowed down on the pavement
with their faces to the ground, and they worshiped and gave praise to the LORD, saying, ‘Truly He is good, truly His loving-kindness is everlasting.’”
Sacrifices Offered
“4 Then the king and all the people offered sacrifice before the LORD. 5 King Solomon offered a sacrifice of 22,000 oxen and 120,000 sheep. Thus the king and all the people dedicated the house of
God. 6 The priests stood at their posts, and the Levites also, with the instruments of music to the LORD, which King David had made for giving praise to the LORD—‘for His loving-kindness is
everlasting’—whenever he gave praise by their means, while the priests on the other side blew trumpets; and all Israel was standing. 7 Then Solomon consecrated the middle of the court that was before
the house of the LORD, for there he offered the burnt offerings and the fat of the peace offerings because the bronze altar which Solomon had made was not able to contain the burnt offering, the
grain offering and the fat.”
The Feast of Dedication
“8 So Solomon observed the feast at that time for seven days, and all Israel with him, a very great assembly who came from the entrance of Hamath to the brook of Egypt. 9 On the eighth day they held
a solemn assembly, for the dedication of the altar they observed seven days and the feast seven days. 10 Then on the twenty-third day of the seventh month he sent the people to their tents, rejoicing
and happy of heart because of the goodness that the LORD had shown to David and to Solomon and to His people Israel.”
God’s Promise and Warning
“11 Thus Solomon finished the house of the LORD and the king’s palace, and successfully completed all that he had planned on doing in the house of the LORD and in his palace. 12 Then the LORD
appeared to Solomon at night and said to him, ‘I have heard your prayer and have chosen this place for Myself as a house of sacrifice. 13 If I shut up the heavens so that there is no rain, or if I
command the locust to devour the land, or if I send pestilence among My people, 14 and My people who are called by My name humble themselves and pray and seek My face and turn from their wicked ways,
then I will hear from heaven, will forgive their sin and will heal their land. 15 Now My eyes will be open and My ears attentive to the prayer offered in this place. 16 For now I have chosen and
consecrated this house that My name may be there forever, and My eyes and My heart will be there perpetually. 17 As for you, if you walk before Me as your father David walked, even to do according to
all that I have commanded you, and will keep My statutes and My ordinances, 18 then I will establish your royal throne as I covenanted with your father David, saying, ‘You shall not lack a man to be
ruler in Israel.’ 19 But if you turn away and forsake My statutes and My commandments which I have set before you, and go and serve other gods and worship them, 20 then I will uproot you from My land
which I have given you, and this house which I have consecrated for My name I will cast out of My sight and I will make it a proverb and a byword among all peoples. 21 As for this house, which was
exalted, everyone who passes by it will be astonished and say, ‘Why has the LORD done thus to this land and to this house?’ 22 And they will say, ‘Because they forsook the LORD, the God of their
fathers who brought them from the land of Egypt, and they adopted other gods and worshiped them and served them; therefore He has brought all this adversity on them.’” (II Chronicles 4:4-22).
Perspective of the Temple Institute in Jerusalem Regarding Solomon’s Altar...
“The altar is the very heart of the Holy Temple, for all of the Divine service is centered around it: all of the daily and additional offerings, as well as the individual and congregational
sacrifices. All of the major ceremonies in the Temple also take place in the vicinity of the altar: the Passover sacrifice, the bringing of the firstfruits on Shavuot, and even the rejoicing with the
lulav branches on Sukkot all take place around the altar.
A Precise Location, From the Beginning of Time
“The exact location of the altar is extremely precise, and has been established since time immemorial. The altar built by King David and King Solomon in the days of the First Temple, as well as the
one built later in the era of the Second Temple, were both erected on the very same place: for this was the very place from which Adam, the first man, was created. The sages stated: ‘Man was created
from the very spot which atones for him’ (B'reishith Rabbah 14:6). Later, it was at this spot on Mount Moriah that Abraham bound Isaac upon the altar that he had built. Through that action, Abraham
declared that this would be the place of God's Temple for all time.
The Horns and the Ramp
“The altar was built as a perfect square and was quite large . . . It was constructed of two main parts: the altar itself, and the ascent ramp. Both were constructed of stones and earth. On top of
the altar at its four corners, there were hollow boxes which made small protrusions or ‘horns.’ These horns measured one amah [or ammah meaning “mother of the arm”] square and 5 handbreadths high,
each (or, app. 18" x 18" x 15")….” [Please note: 5 handbreadths high is considered 18” or 18”/5 = 3.6”; therefore, 7 handbreadths would be: 3.6” x 7 = 25.20”.]
Three Fires Atop the Altar
“Three separate piles of wood burned atop the altar. The largest of these arrangements was designated to receive all the sacrifices; the second provided the coals for the incense altar within the
sanctuary, and the third was the ‘perpetual fire’ which constantly burned on the altar, as the verse states (Lev. 6:5) ‘And a fire shall burn there on the altar constantly; it shall not be
“A large pile of ashes formed in the center of the altar from the remnants of these fires. God commanded that the coals be removed from here.” (Temple Institute, Jerusalem)
EZEKIEL – the Prophet - the Millenarian Altar
The Sacred Cubit is 25.20”… Ezekiel Clarifies the Measurement
Ezekiel 43:13-17: “(13) These are the measurements of the altar in cubits (the cubit is one cubit and a handbreadth) the base one cubit high and one cubit wide, with a rim all around its edge of one
span (span = 9”). This is the height of the altar: (14) from the base on the ground to the lower ledge, two cubits; the width of the ledge, one cubit, from the smaller ledge to the larger ledge, four
cubits; and the width of the ledge, one cubit. (15) The altar hearth is four cubits high, with four horns extending upward from the hearth. (16) The altar hearth is twelve cubits long, twelve
wide, square at its four corners; (17) the ledge, fourteen cubits long and fourteen wide on its four sides, with a rim of half a cubit around it, its base, one cubit all around; and its steps face
toward the east.” – Therefore its measurements are based upon the Sacred Cubit….
· “These are the measurements of the altar in cubits (the cubit is one cubit and a handbreadth)” or 21.6 (long cubit + 3.6” = 25.20”)…
· The Assyria Cubit of 21.6” or 1.8’ is the “base cubit” or equal to the fractal of 2160° or six rotations of the earth at 360° per day = 2160° +
· “a handbreadth” = 3.6” – therefore, “a cubit (Assyrian 21.6”) and a handbreadth (3.6”)” = 21.6” + 3.6” = 25.20” or the measurement of the Sacred Cubit. The 21.6” = “6” as standing for “MAN” (for
Man was created on the sixth day) and, therefore, the Altar’s “sacrifice” is on behalf of man; however, the 3.6” handbreadth is equal to “36” which is the Eternal God. The 6 needs the 36 = for Man’s
redemption before Almighty God cannot be without Sacrifice – but only when the Eternal God is added to Man’s need for redemption does it reach the “7” days (360° x 7 days = 2520° or 7 rotations of a
full circle of the Earth in 7 days) – therefore, is the 3.6” handbreadth added to reach the measurement of the Sacred Cubit of 25.20”.
· Persia Royal Cubit was equal to “seven Assyrian hand palms” – therefore, the Persian Royal Cubit approximates 25.2” / 7 (Assyrian Palms) = 3.6” per palm x 7 = 25.2” = Persian Royal Cubit = Hebrew
Sacred Cubit) (Note: “All these are multiples of the basic palm breadth of 3.6 inches (91.44 mm) which was used by the Babylonians and Hebrews [and Assyrians].” – Source: Gordon J. Row, The Cubit).
The base of the altar in Ezekiel is one cubit high and one cubit wide, with a rim all around its edge of one span (Note: The “span” was considered the distance from the thumb to the little finger
stretched out and equaled 9”). The “handbreadth” is acknowledged by the Jerusalem Temple Institute as 3.6” for when speaking of the Altar of Solomon’s Temple and the protrusion of the horns upon the
altar they state: “These horns measured one amah square and 5 handbreadths high, each (or, app. 18” x 18” x 15”) (Note: “5 handbreadths” = 3.6” x 5 = 18” …therefore, “7 handbreadths” would be:
3.6” x 7 = 25.20” = 1 Sacred Cubit measurement.)
· Here we may note that the 3.6” is discovered by dividing 18” / 3.6” = 5 or 5 x 3.6” = 18”.
· Furthermore, and of great interest to us is the fact that the “horns of the altar” have the following capacity: 18” x 18” x 15” = 4,860” = 4 + 8 + 6 = 18 New Jerusalem Standard. Thus, the “horns
of the altar” have the ultimate authority of the New Jerusalem! Likewise, the “18” is the New Jerusalem Standard and the “15” is one edge of the New Jerusalem’s Cubed shape – i.e., as a fractal the
“15” equates to the 1,500 miles of the New Jerusalem.
· “Some indeed are of the opinion that the handbreadth is to be added only to the six cubits (total – i.e., six cubits and then only 1 handbreadth added to the 6) and not to each of them; but the
text is clear and expresses that these cubits were by or according to a cubit and a hand's breadth. So the Targum paraphrases it, ‘and in the man's hand measuring reeds, one of which was six cubits
by a cubit, which is a cubit and a hand's breadth;’” (Gill’s Exposition of the Entire Bible; Ezekiel 40:5)
· “The description of the temple (for, according to what follows, הבּית is the house of Jehovah) (cf. Ezekiel 43:7) commences with the surrounding wall of the outer court, whose breadth (i.e.,
thickness) and height are measured (see the illustration, Plate I a a a a), the length of the measuring rod having first been given by way of parenthesis. This was six cubits (sc., measured) by the
cubit and handbreadth - that is to say, six cubits, each of which was of the length of a (common) cubit and a handbreadth (cf. Ezekiel 43:13); in all, therefore, six cubits and six handbreadths. The
ordinary or common cubit, judging from the statement in 2 Chronicles 3:3, that the measure of Solomon's temple was regulated according to the earlier measure, had become shorter in the course of time
than the old Mosaic or sacred cubit. From the new temple, therefore, the measure is regulated according to a longer cubit, in all probability according to the old sacred cubit of the Mosaic law,
which was a handbreadth longer than the common cubit according to the passage before us, or seven handbreadths of the ordinary cubit. הבּנין, the masonry, is the building of the wall, which was one
rod broad, i.e., thick, and the same in height. The length of this wall is not given, and can only be learned from the further description of the whole wall (see the comm. on Ezekiel 40:27).” (Keil
and Delitzsch Commentary on the Old Testament, Ezekiel 40)
· “The CUBIT was a measure of length used by the Hebrews, Egyptians and Babylonians, being the distance from the elbow to the tip of the middle finger. So the CUBIT would vary by the physical size of
the workman or overseer. This needed to be “standardized” if a project like the construction of the Temple was to proceed to a satisfactory conclusion. Archaeological research has revealed that in
Solomon’s day there were at least three different cubits in use:
(1) LAND CUBIT which was used for plotting the layout of the Temple’s courts and the surrounding terraces, it had a length of about 17.6 inches (447 mm)
(2) BUILDING CUBIT used in the construction of buildings was about 14.4 inches (366 mm) long
(3) GOLD CUBIT used in the making of the gold and silver vessels and the decorative work was about 10.8 inches (274 mm).
All these are multiples of the basic palm breadth of 3.6 inches (91.44 mm) which was used by the Babylonians and Hebrews. [Therefore, the 25.20” Sacred Cubit is comprised of 7 ea. 3.6 palm breadths
or 7 x 3.6” = 25.20” (Royal Cubit)
The Egyptian Cubit was devised about 3000BC and was standardized by a Royal Master Cubit of black granite, against which all the Cubic Sticks were checked at regular intervals. The [Egyptian] Royal
Cubit (20.62 inches, 524 mm) was subdivided in a rather complicated way. The basic unit was the DIGIT, a fingers breadth, of which there are 28 in the Royal Cubit. The digit in turn was subdivided.
The 14th digit on the Cubit stick was divided into 16 equal parts. The next digit was divided into 15 equal parts, and so on, to the 28th digit which was divided into two equal parts. Thus the
smallest division, 1/16th of a digit, was equal to 1/148th part of a Royal Cubit, (0.046 inches or 1.17 mm). The accuracy of the Cubit Stick is evidenced by the Great Pyramid of Giza; thousands were
employed on its construction, yet its sides vary no more than 0.05% from the mean length of 755.8 feet.” (The Cubit, Gordon J. Row)
· The original measures of length were derived from the human body: the finger, hand, arm, span, foot, and pace. As these measures differ with each individual, they must be reduced to a certain
definite standard for general use. The Hebrew system, therefore, had such a standard; the ell (“amah”) contained 2 spans (“zeret”), while each span was made up of 3 handbreadths (“efa”) of 4 fingers
(“eba’”) each. This division of the ell into 6 handbreadths (i.e., if 1 handbreadth equals 3.0" then 6 handbreadths equal to 6 x 3" = 18" + 1 additional handbreadth = 21" but if 1 handbreadth equals
to 3.6" then 6 x 3.6" = 21.6" and 1 additional handbreadth equals to 21.6" + 3.6" = 25.20") was the one customarily employed in antiquity, but it was supplanted in Babylonia by the sexagesimal
system. The Old Testament mentions two ells of different size. Ezekiel implies that in his measurement of the Temple the ell was equal to a "cubit and a handbreadth" (xl. 5, xliii. 13)—that is, one
handbreadth larger than the ell commonly used in his time (i.e., ell = cubit + 1 handbreadth = SPECIAL ELL). Since among all peoples the ell measured 6 handbreadths, the proportion of Ezekiel's ell
to the others was as 7 to 6. The fact that Ezekiel measured the Temple by a special ell is comprehensible and significant only on the assumption that this ell was the standard of measurement of the
old Temple of Solomon as well. This is confirmed by the statement of the Chronicler that the Temple of Solomon was built according to "cubits after the first measure" (II Chron. iii. 3), implying
that a larger ell was used at first, and that this was supplanted in the course of time by a smaller one. (Jewish Encyclopedia, Weights and Measures)
· “Cubit is a measure of length used by several early civilizations. It was based on the length of the forearm from the tip of the middle finger to the elbow. No one knows when this measure was
established. The length of the arm, or cubit, was commonly used by many early peoples, including the Babylonians, Egyptians, and Israelites. The royal cubit of the ancient Egyptians was about 20.6
inches (52.3 centimeters) long. That of the ancient Romans was 17.5 inches (44.5 centimeters). The Israelites' cubit at the time of Solomon was 25.2 inches (64 centimeters).” (Richard S. Davis, World
· This cubit was also much used by the Jews (33), and is so often referred to that it has eclipsed the 25.1 cubit in most writers. The Gemara names 3 Jewish cubits (2) of 5, 6 and 7 palms; and, as
Oppert (24) shows that 25.2 was reckoned 7 palms, 21.6 being 5 palms, we may reasonably apply this scale to the Gemara list, and read it as 18, 21.6 and 25.2 in. There is also a great amount of
medieval and other data showing this cubit of 21.6 to have been familiar to the Jews after their captivity; but there is no evidence for its earlier date, as there is for the 25 in. cubit (from the
brazen sea) and for the 18 in. cubit from the Siloam inscription. (24) J. Oppert, Etalon des mesures assyriennes (1875);
· “25.1. - The earliest sign of this cubit is in a chamber at Abydos (44) about 1400 B.C.; there, below the sculptures, the plain wall is marked out by red designing lines in spaces of 25.13 ± 0.03
in., which have no relation to the size of the chamber or to the sculpture. They must therefore have been marked by a workman using a cubit of 25.13. Apart from medieval and other very uncertain
data, such as the Sabbath day's journey being 2000 middling paces for 2000 cubits, it appears that Josephus, using the Greek or Roman cubit, gives half as many more to each dimension of the temple
than does the Talmud; this shows the cubit used in the Talmud for temple measures to be certainly not under 25 in. Evidence of the early period is given, moreover, by the statement in I Kings (vii.
26) that the brazen sea held 2000 baths; the bath being about 2300 cub. in., this would show a cubic of 25 in. The corrupt text in Chronicles of 3000 baths would need a still longer cubit; and, if a
lesser cubit of 21.6 or 18 in, be taken, the result for the size of the bath would be impossibly small. For other Jewish cubits see 18.2 and 21.6. Oppert (24) concludes from inscriptions that there
was in Assyria a royal cubit (7/6)ths of the U cubit, or 25.20; and four monuments show (25) a cubit averaging 25.28. For Persia Queipo (33) relies on, and develops, an Arab statement that the Arab
cubit was the royal Persian, thus fixing it at about 25 in.; and the Persian guerze at present is 25, the royal guerze being 1+(1/2) times this, or 371 in. As a unit of 1.013, decimally multiplied,
is most commonly to be deduced from the ancient Persian buildings, we may take 25.34 as the nearest approach to the ancient Persian.” (Weights and Measures, Encyclopedia Britannica)
· Final Comment: The notion that the 25” Cubit or 25.20” Cubit (or the 2.1’ Sacred Cubit or 25.02” Cubit) is obscured amongst the ancient measurements is without academic merit. Flinders Petrie’s
20.63” Egyptian Royal Cubit and the so-called Assyrian 21.6” Cubit are well known in the metrology community; however, the ubiquity of the 25.20” true SC or Sacred Cubit of the Hebrews (found even
among the Assyrians to be 25.20” – Oppert) is altogether a metrological phenomena which, unfortunately, has been touted about as inconsequential – a vain attempt to bury the Sacred Canon’s most
obvious standard of measurement…fortunately, the British were masters at its sly incorporation into the Imperial Standard of Measurement whereby their entire system – which prevails amongst the
Americans to this day – from inches to miles, from pounds to tons, from pints to gallons and on and on – how blessed are we to have their brilliance preserved to this day within the King James
Version of Sacred Canon these glorious measurements for us and our posterity – all hail the Sacred Cubit’s preservation – long live the British for their genius and the Americans for their stubborn
pride in keeping to this “anointed system of metrology” juxtaposed to a world absorbed by an alternative metric system with no basis in either Sacred Geometry nor the Dimensions of Paradise! What
joy to announce I am 5’10 1/2” in height and having a girth of 38” and a weight of 190 pounds, as I fill up with 15 gallons of gasoline and buy a half-gallon carton of milk at the store on a day
where the temperature is measured in Fahrenheit and not Celsius – only in America…and where she has had her influence…I could go on mile after mile (not km after km), but the joy – so casually taken
by we Americans (so oblivious are we to its merits) – leaves me exhilarated each time I read the word “cubit” in the Sacred Canon! Yet, the LIE that the metric system is more accurate continues its
perpetuation by the numbskulls – i.e., the progeny of several French savants who got it wrong, and then hid their subterfuge from the deceived public until this day! Shame upon the so-called
scientific community and their blind forbearance to this most unscientific account! (Ref. John Michell, New View Over Atlantis, 1983, pp. 125-6)
Ezekiel’s Millenarian Altar’s Measurements
Contention as to Ezekiel’s forecast persists – this tome does not enter into those contentions – but examines the measurement of those Altars which foretell, past and future, “altar dimensions” – and
affirm, that those calculations most definitely, under the sway of the Sacred Cubit’s measurement of 25.20” – abundantly proclaim the Divine Intention inexorably, irresistibly and persistently
preparing the People of God toward their ultimate destiny: The New Jerusalem. Ezekiel’s Altar was tiered: Foundation/Base, Bottom Tier, Middle and Top Tiers where the actual Hearth was located.
Therefore its measurements are based upon the above evidence and we must, therefore, conclude that the length of the cubit so mentioned by Ezekiel in Ezekiel 40:5 included both the so-called long
Assyrian Cubit of 21.6” and an additional handbreadth of 3.6” (or palm). Likewise, when mentioning “cubit” in most places in Ezekiel’s measurements, he had in mind the cubit length of Moses and
Solomon, i.e., 25.20” which was an extension of the Assyrian Cubit (short) of 21.6” by the addition of a handbreadth of 3.6” to equal the old Sacred Cubit of Moses and Solomon – a rather peculiar way
of announcing its length, however, for the day it communicated nicely.
· Foundation/Base: 1 cubit high and 1 cubit wide (base ledge) with a 9” rim around it.
· Bottom Tier: 2 cubits high and 1 cubit wide (for its ledge)
· Middle Tier: 4 cubits high and 1 cubit wide (for its ledge)
· The Altar Hearth: 4 cubits high – the “14 cubits all around the Altar Hearth” at its base is actually the top of the Middle Tier’s ledge and is not counted in the width calculations
· Beneath the Altar Hearth is a 1/2 cubit rim extending out from the Middle Tier’s ledge; it is therefore considered 1/2 of a cubit all around the base of the Altar Hearth but does not constitute an
extension of the Middle Tier Ledge itself but is considered “one cubit all around” in that .5 + .5 (on either side of the Middle Tier/Altar Hearth Rim would equal one full cubit but not a part of the
Altar’s width – please see graphics).
· The Altar Hearth itself is 12 cubits by 12 cubits in length and width – a perfect square.
· Base Dimensions: Calculating its Foundation/Base length and width we must examine all “horizontal measurements” or “width” measurements; therefore: Base = 1 cubit + Bottom Tier = 1 cubit + Middle
Tier = 1 cubit = 1 + 1 + 1 x 2 (accounting for ‘front and back’) = 3 x 2 = 6 cubits + 12 cubits for the Altar Hearth measurement = 6 + 12 = 18 cubits.
· Bottom Tier Dimensions: Deduct 1 cubit from either side…the ledge of the base…therefore: 18 cubits – 2 cubits = 16 cubits
· Middle Tier Dimensions: Deduct an additional 1 cubit from either side the ledge atop the bottom tier…16 cubits – 2 cubits = 14 cubits
· Altar Hearth dimensions are clearly given as 12 cubits
· Summary of underside measurements of each tier:
Base Dimensions: 18 cubits x 18 cubits
Bottom Tier: 16 cubits x 16 cubits
Middle Tier: 14 cubits x 14 cubits
Altar Hearth: 12 cubits x 12 cubits
· Square Measurements of the tiers’ bases only, using 25.20” as the cubit length:
(1) Base: 18 x 18 = 453.6” x 453.6” = 205,752.96 sq. inches = “36” – The Altar is based upon the Eternal God
and when the 144 Wall of the New Jerusalem integrates with the intentions of the Eternal God this is the result: 205,752.96 / 144 = 1,428.84 = “27” Messiah’s full revelation in Christian Sacred Text
(27 volumes); therefore, the Eternal God, through the offering up of the Suffering Servant on behalf of the New Jerusalem is immediately seen in the Base of Ezekiel’s Altar.
(2) Bottom Tier: 16 x 16 = 403.2” x 403.2” = 162,570.24 sq. inches = “27” – Again, Messiah Revealed…and by dividing 162,570.24 / 144 = 1,128.96 = “27” confirming that the very basis – as a “double
witness” for the Altar’s eternal purpose reveals Him as the Lamb of God, the Suffering Servant upon the Throne on that Final Day = though He is the Lion of the Tribe of Judah, He is forever the Lamb
upon the Throne!
(3) Middle Tier: 14 x 14 = 352.8” x 352.8” = 124,467.84 sq. inches = “36” – The Eternal God…and when 124,467.84 / 144 (New Jerusalem Wall) = 864.36 = “27” Messiah and, as well draws attention to the
“864” or the Diameter of the Sun connoting the “Sun of Righteousness” in fulfillment of Malachi 4:2 – “But to you who fear My name (36 – the Eternal God) the Sun of Righteousness (27 – Messiah) shall
arise with healing in His wings (for the “144” the New Jerusalem).
(4) Altar Hearth: 12 x 12 = 302.4” x 302.4” = 91,445.76 sq. inches = “36” – The Eternal God…when divided by 144 as in 91,445.76 / 144 = 635.04 = “18” New Jerusalem Standard = The Eternal God has
revealed Messiah and accepted His offering on our behalf so that we might be His forever: The New Jerusalem.
(5) Combining all “tier base” measurements: 205,752.96 + 162,570.24 + 124,467.84 + 91,445.76 = 584,236.8 = “36” as in 5 + 8 + 4 + 2 + 3 + + 8 = 36 = The Eternal God without Beginning or Ending is
wholly committed from beginning to the consummation – from the base at “36” to the height at “36” and each time the Wall of the New Jerusalem (144) “integrates” with the dimensions of the lower
tiers, the Messiah is there for them – there, until eternity and the preparation of His Bride is complete – for the Altar’s Hearth when integrated with the Eternal God’s Purpose produces the New
Jerusalem as the Bride of Messiah: 18.
· Linear Measurements by Cubit:
(1) 18 x 4 (edges) = 72 cubits (Base)
(2) 16 x 4 (edges) = 64 cubits (Bottom Tier)
(3) 14 x 4 (edges) = 56 cubits (Middle Tier)
(4) 12 x 4 (edges) = 48 cubits (Altar Hearth)
Total = 240 Linear Cubits +
Linear Rim at Base = 18 x 4 = 72 cubits (linear)
Linear Rim atop the Middle Tier = 14 x 4 = 56 cubits (linear)
Total Linear Rim Cubits = 72 + 56 = 128 Linear Cubits of Rim
Grand Total of All Linear Cubits given in the Altar’s Linear Measurements: 240 + 128 = 368 Cubits
368 Cubits x 25.20” (per cubit) = 9,273.6 Linear Inches = “27” = 9 + 2 + 7 + 3 + 6 = 27 = The full revelation of Messiah in Sacred Christian Canon – “Behold, the Lamb of God Who takes away the sin of
the world!” And for whom did He die? For Messiah was “cut off, but not for Himself” – therefore: The total 368 Cubits tell us clearly: 3 x 6 x 8 = 144 the Wall of the New Jerusalem – He came for
you, for me, to be His Bride – is it any wonder we are called “THE WOMAN, THE LAMB’S WIFE…and showed me the Holy City, Jerusalem, descending out of heaven from God” (Revelation 21:9-10).
· Surface Measurements of each tier – vertical and horizontal ledges atop each:
Base Surfaces Exposed Side: Vertical Sides – 1 cubit x 18 cubits x 4 surface/base sides = 25.20” x 453.6” = 11,430.72 sq. inches x 4 exposed sides = 45,722.88 sq. inches (all base vertical sides) +
Horizontal ledge measurements of 1 cubit wide ledge (without overlap) = 453.6” x 25.20 x 2 (full ledge sides matching vertical side panels) = 22,861.44 sq. inches + 2 “shortened horizontal ledge”
measurements or 25.20” x 16 cubits or 403.2” = 10,160.64 sq. inches x 2 shortened horizontal ledges = 20,321.28 sq. in. for a total horizontal surface area: 43,182.72 sq. inches (horizontal area).
Combined vertical and horizontal = 45,722.88 sq. in. + 43,182.72 sq. inches = 88,905.6 = “36”… But to this figure (“36”) we must add the attached 9” Rim below…for it is so attached to the surface
area at the base….(see immediately below for final calculation)….
· 9” Rim at Base (all around) or 9” x 18 cubits or 9” x 693.6” x 4 (ledge sides) = 24,969.6 sq. inches
(Note: We are aware of the “overlap” at the ends where the edges connect and have simply taken the plain surface calculation – i.e., end to end, rather than going into an elaborate removal of the
“tailings” – a straight 693.6 x 9” on each side is calculated.); therefore the Grand Total Sum: 88,905.6 sq. inches (vertical and horizontal surface areas at the base of the Altar + 24,969.6 sq.
inches (9” Perimeter Rim attached to base) = 113,875.2 = “27” (1 + 1 + 3 + 8 + 7 + 5 + 2 = 27) in that the full revelation of Messiah in the Christian Sacred Canon discloses that the base of the
Altar is the sacrifice of the Son for our full redemption and that the Father (36) sent the Son (27) to be the Redeemer and Savior – throughout all eternity we will behold Him as the “Lamb slain from
the foundation of the world” (Revelation 13:8).
· (Note: 9” on one side and 9” on the other side at the base (north and south sides) and east and west (omitting the steps would be similar at 9” + 9” = 18 therefore 2 sets of “18” = 36 (18 + 18 =
36) validating that the Eternal God (360° circle) has established the Millenarian Altar for the New Jerusalem (18)).
· Bottom Tier (2 cubits vertical x 16 cubit length/width) and 1 cubit horizontal ledge = therefore: 4 full vertical sides = 50.4” x 403.2” = 20,321.28 sq. inches x 4 = 81,285.12 sq. inches + 1 cubit
ledge atop bottom tier = 25.20” x 16 cubits or 25.20” x 403.2” = 10,160.64 sq. inches x 2 (north and south sides) = 20,321.28 sq. inches + the two overlapped sides (east and west) = 25.20” x 14
cubits or 25.20” x 352.8” = 8,890.56 sq. inches x 2 sides = 17,781.12 sq. inches. Combining the two opposite ledges atop the same tier is: 20,321.28 sq. inches + 17,781.12 sq. inches = 38,102.4
(bottom horizontal base). Grand Total of surface area of bottom tier vertical four sides and ledge atop = 81,285.12 sq. ins. + 38,102.4 sq. ins. = 119,387.52 sq. inches = “36” affording us ample
evidence that the Son (27) submitted His will to the Eternal Father-God (36).
· Middle Tier, 1 Cubit Ledge & ½ Cubit Rim:
4 full vertical sides @ 4 cubit height x 14 cubit length and width or 100.8” (4 cubit height) x 352.8” (length of middle tier) = 35,562.24 sq. inches x 4 sides = 142,248.96 sq. inches (vertical
middle tier sides) + horizontal ledge atop the middle tier of 1 cubit x opposite sides of the full length/width of the middle tier of 14 cubits or 25.20” x 352.8” = 8,890.56 sq. inches x 2 full sides
= 17,781.12 sq. inches + 2 horizontal ledges on opposites sides minus the 2 cubit overlaps or 25.20 x 302.4” (i.e., 12 cubits, not 14) = 7,620.48 x 2 (opposite ledge sides of lesser area) = 15,240.96
sq. inches. Combining the two ledge “sets” we arrive at: 17,781.12 + 15,240.96 = 33,022.08 sq. inches (ledge area) on the 1 cubit ledge atop the Middle Tier. Then there is the 1/2 cubit wide rim
surrounding the base of the Altar Hearth attached to the ledge atop the Middle Tier. Just as we did with the ledge measurements we must do with rim extensions – i.e., two full opposite side measures
of 1/2 cubit @ 12.6” x 352.8” (full 14 cubit length) = 4,445.28 x 2 full lengths = 8,890.56 sq. inches + 2 lesser/overlapped rim sides or 1/2 cubit @ 12.6” x 302.4” (12 cubits) = 3810.24 sq. inches x
2 opposite lesser sides = 7,620.48 sq. inches – these 2 figures/rim areas are then added together: 8,890.56 sq. inches + 7,620.48 sq. inches = 16,511.04 sq. inches (Rim area)
· Total Surface Areas of Middle Tier:
(1) 142,248.96 sq. inches = Middle Tier 4 vertical sides
(2) 33,022.08 sq. inches (4 ledge surface areas)
(3) 16,511.04 sq. inches (1/2 cubit rim)
(4) Grand Total of Middle Tier surface areas: 191,782.08 sq. inches = “36” indicative of the fact that the Father’s love for the Son (base tier) in eternity past (bottom tier) and in the time present
(middle tier) is unchanging and altogether affirming of His love.
· Altar Hearth
We now arrive at the Altar Hearth with its simplicity of measurement, in that the more difficult measurements have been calculated. Here we find that the Altar Hearth’s sides match those of the
Middle Tier, 4 cubits vertical height and the Altar Hearth top surface only area is 12 cubits x 12 cubits (width and length are the same as a perfect square); therefore, determining its sides = 4
cubits x 12 cubits = 1 side x 4 sides = Total Sq. Inches…
100.8” (4 cubit height for each of the four sides) x 302.4” (12 cubits) = 30,481.92 sq. inches (on one vertical side of the Altar Hearth) x 4 sides (vertical four sides of Altar Hearth) = 121,927.68
sq. inches = “36” in that the Eternal God is satisfied with the offering of the Son. If we were to add in the surface area of the Altar Hearth as in 302.4” x 302.4” = 91,445.76 sq. inches and add
this figure of “36” to the “36” on the sides of the Altar Hearth we would have this calculation (minus its immediate base):
121,927.68 sq. inches (sides of Altar Hearth) + 91,445.76 sq. inches (Altar Hearth surface area) = 213,373.44 sq. inches = “27” which is the Messiah, the Son Whose perfect sacrifice is well-pleasing
to the Father … only when we add in this measurement do we see the Son and declare: Worthy is the Lamb! The Son is the basis of the Millenarian Altar and He is its offering – sacrifice – its
· Total Surface Summations:
If we were to add all the surface measurements on the exterior of the Millenarian Altar our calculations would look like this:
(1) 113,875.2 = “27” = Base Tier surface area, ledge and 9” rim
(2) 119,387.52 sq. inches = “36” = Bottom Tier surface area and ledge
(3) 191,782.08 sq. inches = “36” = Middle Tier surface area, ledge and 1/2 cubit rim
(4) 213,373.44 sq. inches = “27” = Altar Hearth surface area – sides and hearth itself
(5) GRAND TOTAL: 638,418.24 = “36” The Eternal God has accepted the offering of the Son!
(6) 638,418.24 / 144 (New Jerusalem Wall) = 4,433.46 = “24” for all that the Father gives to the Son will come to Him = it is for these “24” – the 12 Gates and the 12 Foundations of the Holy City,
New Jerusalem that the entire Millenarian Altar bears witness!
Elevation of Ezekiel’s Millenarian Temple – and its Centrality in the Sanctuary:
A summary of the various elevations and their summation is of great interest to us.
(1) Base/Foundation Tier: 1 cubit (Ezekiel 43:13) 25.20”
(2) Bottom Tier: 2 cubits (Ezekiel 43:14) 50.40”
(3) Middle Tier: 4 cubits (Ezekiel 43:14, 17) 100.80”
(4) Top Tier – Altar Hearth: 4 cubits (Ezekiel 43:15-16) 100.80”
(Note: The “four horns extending upwards” are not given measurements – Ezk. 43:15; however, the Temple Institute in Jerusalem affirms that their measurement is 18” x 18” x 15” each.)
(5) Grand Total of “Cubit Elevations” = 277.2” = 2 + 7 + 7 + 2 = “18” New Jerusalem Standard
If the elevation of 277.2” is matched to the 18 cubit base or 277.2” x 453.6” (both of which are “18” NJS) we have a product of 125,737.92 = “36” The Eternal God has His stamp upon this majestic
Millenarian Altar and it is “eternal” in nature, for the Lamb is upon the Throne.
Area and Circumference of a Circle using the Altar’s height of 277.2” as its radius:
If the 277.2” height is used as the radius of a circle, then the circumference of that circle would be calculated as follows: C = π * d and its area as A = π * r^2 – therefore: C = 3.14 x 554.4”
(radius 277.2” x 2 = 554.4”) = C = 1,740.816 = “27” of the full revelation of Messiah in Sacred Christian Canon and the area of that circle would be: A = 3.14 x 277.2” x 277.2” – therefore, A = 3.14
x 76,839.84” = A = 241,277.098 sq. inches = “40” which connotes “judgment” or “testing/trial” in Holy Writ. This would be in keeping that the Altar signifies that judgment has been passed upon the
sacrifice and offering on our behalf.
Area and Circumference of a Circle using the Altar’s base edge measurement of 453.6” (18 cubits) as its diameter:
C = π * d = C = 3.14 x 453.6” = 1,424.304” = “18” New Jerusalem Standard
A = π * r^2 = A = 3.14 x 226.8” x 226.8” (226.8” is radius or 1/2 of 453.6” diameter) = A = 3.14 x 51,438.24 = 161,516.074 sq. inches of the circle’s area or 161,516.074 sq. inches / 144 = 1,121.6394
= “27” Messiah – therefore we conclude, using the 453.6” or 18 cubit base of the Altar and using it as the diameter of a circle that we have the Messiah as the “content” of that circle and the
circumference of that circle or its expression is the New Jerusalem.
Cubing Ezekiel’s Millenarian Altar using its Base as One Primary Edge (artificial)
If we “cube” Ezekiel’s Millenarian Altar using either 277.2” as its radius/diameter/edge or the 18 cubit base as its diameter/edge (i.e., 453.6 Linear Inches) we arrive at this calculations
multiplying either number set by 12 (12 edges in a cube to parallel the cube appearance of the New Jerusalem):
Cubing the Altar’s Height or Base to arrive at Linear Inch sums (12 edges of a cube):
277.2” x 12 = 3,326.4” = “18” New Jerusalem Standard and is a replica of the New Jerusalem in miniature.
453.6” x 12 = 5,443.2” = “18” New Jerusalem Standard and is a replica of the New Jerusalem in miniature.
Surface Area using above calculations:
277.2” x 277.2” = 76,839.84 sq. inches x 6 (sides of a cube) = 461,039.04 sq. inches = 27 Messiah or
76,839.84 / 144 = 3,201.66 = “18” New Jerusalem Standard of Measurement
453.6” x 453.6” = 205,752.96 sq. inches x 6 (sides of a cube) = 1,234,517.76 sq. inches = 36 Eternal God or 205,752.96 sq. inches / 144 = 8,573.04 = “27” Messiah…therefore, by using the elevation of
Ezekiel’s Millenarian Altar we see Messiah’s intention in achieving through redemption (i.e., the Altar – the shedding of blood) the New Jerusalem and that the Altar’s base measurements when cubed
tell us that the Eternal God (36) when “integrated” with the Wall of the New Jerusalem will bring us to see that only through Messiah (27) can that goal be achieved.
Cubic Inch Fullness/Capacity/Volume of the 277.2” (elevation of the Altar) or the 453.6” base of the Altar and determining the interior or the capacity of a cubed Altar using either one of these
calculations; therefore:
277.2” x 277.2” x 277.2” = 21,300,003,648 cubic inches = “27” Messiah is the fullness of the cubed Altar using it’s elevation as the diameter of a its cube (not its radius) and 21,300,003,648 / 1728
(12” x 12” x 12” = 1728 cubic inches in one cubic foot) = 12,326,391 cubic feet = “27” Messiah, again as its fullness.
453.6” x 453.6” x 453.6” = 93,329,542.656 cubic inches = “54” or 93,329,542.656 / 1728 cubic inches in one sq. foot = 54,010.152 = “18” New Jerusalem Standard of Measurement and “5040” is Plato’s
Optimal Number for Society – in other words only the New Jerusalem prepared by Messiah can secure the Optimal Number for society – only then can true and lasting peace, harmony and true unity and
governance be achieved for humans. This conclusion is staggeringly important for humankind: Therefore, is the Millenarian Altar foreseen by Ezekiel an altogether significant memorial reminding us
all how the Eternal God, through offering the Son as the judgment for our sin and the forgiveness thereof, obtaining redemption for humankind, was eternally accomplished for us – and we shall never
forget – in order that the Eternal God would have a Bride for His Son. This is the glorious portrait of the future Millenarian Altar of Ezekiel – do we see it?
A brief word regarding the CENTRALITY of the Altar’s position within the immediate Temple Complex of Ezekiel’s Millenarian vision of the Temple and the Altar – its position is altogether obvious and
connotes the supreme importance of that future City wherein the Builder and Maker thereof is God…for the Lamb sits upon its throne as testament, witness to His everlasting love, mercy and justice. As
the graphic above so indicates: The Altar occupies the central focus of the Millenarian Temple. It stands out with classic distinction. To the Christian it forever hearkens to the Centrality of
the Cross and bespeaks of One Who gave His All for the Bride, the Holy City, New Jerusalem.
A Final Measurement taken from the Altar’s Hearth:
The Altar’s Hearth presents the apex of the Father’s love for the Son and of the Son’s Whole Burnt Offering for the “24” – however, we must be absolutely positive about this proposition. The 12 x 12
cubit surface area of the Altar Hearth is equal to 302.4 x 302.4 = 91,445.76 sq. inches / 144 sq. inches = 635.04 sq. feet and is equal to the 18 New Jerusalem Standard – once again attesting that
“Messiah was cut off, but not for Himself” (Daniel 9:26).
If, again, we take the entire Altar Hearth as a cube having, again, 12 edges and 6 sides, we discover the following:
Top and Bottom = 91,445.76 x 2 = 182,891.52 sq. inches
4 Sides of the Altar Hearth = 100.8” (4 cubits) x 302.4” (12 cubits) = 30,481.92 x 4 sides = 121,927.68 sq. inches –
Combining all the “cubed Altar hearth surface and bottom area” to mirror the cube of the New Jerusalem, we discover:
182,891.52 sq. inches + 121,927.68 sq. inches = 304,819.2 sq. inches = “27” which again is the full revelation of Messiah as the perfect sacrifice – for the Altar is the offering of the Lamb of God
and abides the memorial for all to witness. And, just why did the Son yield Himself to the Father’s will – it was not for Himself, it was for us He suffered, bled and died – for our sin and our
shame; therefore cannot we see this? Absolutely: For 304,819.2 sq. inches (27) / 144 (the Wall of the New Jerusalem) produces: 2,116.8 = 18 The Standard of the New Jerusalem, the Bride of the
Lamb. Is not this Millenarian Altar a marvelous display of His Eternal Purpose for the Ages? I believe so…and if we were to find the Altar’s uppermost capacity we would find this: 12 cubits
(length) x 12 cubits (width) x 4 cubits (height) = 576 sq. cubits = 18 New Jerusalem and furthermore, 302.4” x 302.4” x 100.8” = 9,217,732.608 / 144 = 64,012.032 = 18 Standard of the New Jerusalem.
There is little doubt that through the offering up of the Son by the Eternal Father that He did not do this for Himself but to prepare the Bride of Messiah, the New Jerusalem.
The Horns of the Altar allow the Millenarian Altar of Ezekiel to Equate to the Great Pyramid of Giza and thence to the New Jerusalem…
Although the measurements of the authoritative horns of Ezekiel’s Millenarian Altar are not given, we do have evidence contained within the Septuagint Version of the Hebrew Sacred Text (translated
from the Hebrew into the Greek) which indicates that the horns were 1 cubit in height (Ref. Milgrom, article – “Altar” – Encyclopedia Judaica 2, col. 763: 12 cubits [height of Ezekiel’s Altar]
including the horns [1 cubit inclusive]); therefore, the overall height/elevation of Ezekiel’s Millenarian Altar would be a fractal of the perimeter of the Great Pyramid of Giza, as follows: 12
Cubits (11 cubits + 1 Cubit of the horns) x 25.20” = 302.4” (cf. GPG @ 756’ base x 4 (perimeter of its base) = 3024’.
This calculation aligns the Altar in Ezekiel with measurements of the GPG, for when we multiply the 302.4” and convert the Altar’s elevation into a cube, using its elevation of 302.4” as the primary
edge of such a “cubed altar” - then we discover the following: 302.4” x 12 (12 edges in a cube) = 3,628.8” which is a fractal of the GPG’s base measurement in inches @ 36,288 Linear Inches (i.e.,
756’ x 4 = 3024’ x 12” = 36,288 inches) and, of course, both the GPG and, therefore, the Altar, translate quickly into the measurements of the New Jerusalem. This is truly remarkable.
However, we can demonstrate the esoteric 18 x 18 x 18 Cubit Structure (see below) – but by taking the 12 cubits as its “newly discovered height” (adding in the horns upon the altar) and its length
and width (base) as 18 x 18 we would/could see a structure that is 18 (length) x 18 (width) x 12 (height) (cubed); therefore, the total number of cubits on this similar esoteric structure would be:
18 cubits (width and length or top and bottom) x 8 = 144 cubits + 4 ea. 12 cubit edges for its height or 4 x 12 = 48 cubits or 144 cubits + 48 cubits = 192 Linear Cubits (a cubed altar – given its
actual base in cubits and its actual height, including in the horns of the altar to equal 12 cubits in elevation).
Therefore: 192 cubits x 25.20” = 4,838.4” = “27” Messiah Revealed = thus, the Memorial to Messiah as the Suffering Servant is wholly revealed in these dimensions and this is confirmed by its
newly-found volume/capacity wherein 18 x 18 x 10 (all cubits) = 3,888 sq. cubits = “27” Messiah. Also, 453.6” (18 cubits) x 453.6” (18 cubits) x 302.4” (12 cubits) = 62,219,695.104 cubic inches =
“45” Messiah time of Blessing and Second Coming in Christian theology (Daniel 12:11-12) and the 45 Days of Blessing (i.e., 1335 – 1290 = 45). Likewise, 62,219,695,104 cubic inches / 1728 cubic
inches (a cubit foot as in 12” x 12” x 12” = 1728 cubic inches or 1 cubic foot) = 36,006.768 cubic feet = “36” The Eternal God without beginning or ending. Messiah First Coming as the Suffering
Servant (27) and His Second Coming in Blessing (45) is altogether acceptable by the Eternal God (36).
The Cube of Ezekiel’s Altar
In that the Millenarian Altar of Ezekiel is tiered – and that up to 6 cubits (i.e., 18 cubits x 18 cubits perimeter and the top Altar Hearth 12 cubits x 12 cubits – it is, therefore, geometrically
impossible to form as an overall cubical rectangle. Its height, minus the 18” horns of the altar upon its hearth, measures 11 cubits (1 cubit at its base; 2 cubits at the lower settle; four cubits
at the upper settle; and, finally 4 cubits at the Altar hearth).
Notwithstanding, we can construct a geometric expression of the complete Altar by ascending vertical lines from each of the four base 90° corners upwards and create a “top platform” that would equal
its base of 18x18 cubits; thus, we would have four height edges of 11 cubits each. Therefore, the total linear number of cubits would mean that the base 18x18 cubits and the artificial top at 18x18
+ the 4ea. 11 cubits forming its height.
18 x 8 (Eight edges of this artificial rectangular cube – top and bottom measurements) = 144 (notice something here?) + 11 cubits x 4 (Four edges of this artificial rectangular cube – height;
excluding the height of the horns) = 44 cubits. Therefore, 144 + 44 = 188 cubits or a total of 188 x 25.20” = 4,737.6 = “27” (Messiah fully revealed – as it should be!). Also, 18 x 18 x 11 would
give us its cubic capacity in cubits = 3,564 = “18” New Jerusalem Standard of Measurement; thus, “Messiah was cut off, but not for himself” (Daniel 9:26) = the entire Altar is for the New Jerusalem!
Likewise, this artificial capacity/volume can be measured in cubic inches: 453.6” (18 cubits) x 453.6” (18 cubits) x 277.2” (11 cubits) = 57,034,720.512 sq. inches = “36” The Eternal God. This
figure is then divided by 144 (12” x 12” = 144 sq. inches = 1 sq. foot); therefore, 57,034,720.512 sq. inches / 144 sq. inches = 396,074.448 sq. feet = “45” and this “45” is directly connected to the
“45” Days of Blessing between the terminus of Desolations (1,290 days) and those who wait until the 1,335th day or 45 days beyond Desolations (i.e., 1,335th days less 1,290th day = 45 days), the time
of the Coming of Messiah to Earth (Daniel 12:11-12). However, 12”x12”x12” = 1,728 sq. inches or 1 cubic foot (capacity); therefore, 57,034,720.512 sq. inches / 1,728 sq. inches = 33,006.204 cubic
feet = “18” The New Jerusalem Standard of Measurement. Messiah’s coming as the Lion of the Tribe of Judah (“45”) is for the express purpose that He might have His Bride, the Holy City, New Jerusalem
Now, using the base measurement of 18 cubits as the prime measurement and cubing Ezekiel’s Altar thereby we would arrive at: 18 cubits x 12 edges (perfectly square cube akin to that of the New
Jerusalem herself which “18” is her overall expression as in 15 x 12 edges = 180 or “18” – but, 18 cubits x 12 edges = 216 linear cubits (an immediately recognizable figure for it is the Name of God
(216) and the diameter of the Moon (2160) and the circumference of the nautical Earth (21,600 miles).
Furthermore, this “cubed Altar” at the center of Ezekiel’s Millenarian Temple at 18x18x18 Cubits (esoteric) = 5,832 cubic cubits or 5832 = “18” New Jerusalem Standard of Measurement. And, in
inches: 216 Cubits x 25.20” = 5,443.2” which, again, is the measurement of the New Jerusalem Standard as in 5 + 4 + 4 + 3 + 2 = 18. Likewise, 453.6” (18 cubits) x 453.6” (18 cubits) x 453.6” (18
cubits) = 93,329,542.656 cubic inches = “54” which is Plato’s Optimal Number but to determine the number of square feet we would divide 12x12x12 = 1728 sq. inches (equal to one square foot) into
93,329,542.656 = 54,010.152 = “18” New Jerusalem Standard of Measurement. Therefore, in our esoteric model of Ezekiel’s Millenarian Altar we have produced a replica of the New Jerusalem herself –
and rightly so, for the Optimal Numbered Society is realized (54) is naught but the New Jerusalem as an ALTAR in that herein is the “throne of God and of the Lamb” (Revelation 22:1) and where there
is “…no temple in it, for the Lord God Almighty and the Lamb are its temple….the city had no need of the sun or of the moon to shine in it, for the glory of God illuminated it. The Lamb is its
light…and the nations of those who are saved shall walk in its light, and the kings of the earth bring their glory and honor into it….its gates shall not be shut at all by day (there shall be no
night there)….and they shall bring the glory and the honor of the nations into it” (Revelation 21:22-26)…the LAMB upon the throne is the central figure of the New Jerusalem…thus is the Altar in the
midst thereof!
Moses, Solomon and Ezekiel’s Altars – A Comparison
Now that we have secured an artificial six-sided rectangular cube of Ezekiel’s Millenarian Altar (using its 11 cubit elevation as its height and 18 cubits x 18 cubits as its top/base), we can have a
degree of comparison with the Altars of Moses and Solomon:
Moses’ Altar for the Tabernacle in the Wilderness: Length 5 Cubits; Width/Breadth 5 Cubits; Height 3 Cubits
Solomon’s Temple Altar: Length 20 Cubits; Width/Breadth 20 Cubits; Height/Depth 10 Cubits
Ezekiel’s Millenarian Altar: Length 18 Cubits; Width/Breadth 18 Cubits; Height/Depth 11 Cubits
Moses’ Total Linear Cubits as a “cubed altar” = 20 + 20 + 12 = 52 Linear Cubits
Solomon’s Total Linear Cubits as a “cubed altar” = 80 + 80 + 40 = 200 Linear Cubits
Ezekiel’s Total Linear Cubits as an artificial “cubed altar” 72 + 72 + 44 = 188 Cubits
Firstly, if we were to add all three altars together in linear cubits: 52 + 200 + 188 = 440 Linear Cubits x 25.20” =11,088” or 1 + 1 + 8 + 8 = 18 New Jerusalem Standard of Measurement; again, all
three altars – Prophet, Priest and King was “not for Himself” but “for” the New Jerusalem – this is brought out more clearly as we divide by 12” to determine the number of feet in all these linear
measurements; thus: 11,088” / 12” = 924 = “15” which is a fractal of the one edge of the New Jerusalem’s 1,500 miles or 11,088” / 144 = “77” which connotes “completion upon completion” and the
Eternal Sabbath Rest for on the Seventh Day He rested and this is the finality of all His work past and future in time – 7 after creation and 7 at the close of time or “77.”
The height of all the altars together totals: 12 + 40 + 44 = 96 = 15 = the Edge of the New Jerusalem’s 1,500 miles or 15 is a fractal of 1,500. Also: 96 x 25.20” = 2,419.2 = “18” New Jerusalem
Standard of Measurement.
All of the base configurations of the three altars: 20 + 80 + 72 = 172 Linear Cubits x 25.20” = 4,334.4 = “18” = New Jerusalem Standard of Measurement + 172 Linear Cubits (top measurements which
would also equal “18”). When these two sets of “18” are added together we arrive at “36” The Eternal God without beginning or ending. Or, we add them together as 172 + 172 = 345 Linear Cubits x
25.20” = 8,694” = “27” Messiah Revealed. Therefore, the Eternal God (36 = the top and bottom measurements of all three altars in linear cubic edges equaling to 24 separate edges for there are 8
edges (top and bottom per altar) views these altars as the “means” (through Messiah “27”) to effect the New Jerusalem (“18”) – thus, is His hand displayed.
In capacity/volume if we were to add up all the heights: 12 + 40 + 44 = 96 Cubits; all the lengths: 20 + 80 + 72 = 172; and all the widths: 20 + 80 + 72 = 172 and multiply their sums to arrive at
cubic cubits we would have:
172 x 172 x 96 = 2,840,064 Cubic Cubits = “24” demonstrating His Divine governance in that the “24” Elders of the New Jerusalem (12 Patriarchs of Israel and 12 Apostles of the Lamb) display how “they
will rule and reign with Him a thousand years.” The measurement in inches displays the following: 172 cubits = 4,334.4” x 4,334.4” x 2,419.2” (96 cubits) (and please note all three of these
measurements of cubits converted to inches equal 18 or 18x18x18 giving us 5,832 which is “18” NJS) = 45,449,566,912.512 cubic inches = “63” and if we divide 45,449,566,912.512 cubic inches by 144 we
arrive at 315,621,992.448 = “54” with both of these together, of course, are 63 = 9 and 54 = 9 … 9 + 9 = “18” New Jerusalem Standard of Measurement; however, to arrive at the number of cubic feet in
45,449,566,912.512 cubic inches we must divide by 12x12x12 = 1,728 cubic inches (or the number of cubic inches in one cubic foot) – therefore: 45,449,566,912.512 cubic inches / 1,728 cubic inches
(or one cubic foot) = 26,301,832.704 = “36” which connotes the Eternal God without beginning or ending.
The difference between the 252 linear cubit measurements of the Altars of Moses and Solomon from that of Ezekiel’s 188 cubits is as follows: 252 – 188 = 64 cubits x 25.20” = 1,612.8” = “18” New
Jerusalem Standard of Measurement and the 12 cubit difference between Solomon’s Altar and Ezekiel’s Altar: 200 cubits – 188 cubits = 12 cubits x 25.20” = 302.4” which is a fractal of the base of the
Great Pyramid of Giza in feet or 3024’ x 12” = 3,628.8” x 3 (12 cubed edges) = 10,886.4 or 108,864 = “36” the Eternal God or “108” = “18” and “864” = “18” or 108 is a fractal of the Moon’s radius of
1080 miles and its “864” is a fractal of the Sun’s 864,000 mile diameter. The difference between Moses’ Altar and that of Solomon’s is: 200 – 52 = 148 cubits x 25.20” = 3,729.6 = “27” for Messiah
is in view as the Suffering Servant. The difference between Moses’ Altar and that of Ezekiel’s: 188 – 52 = 136 Cubits x 25.20” = 3,427.2” = “18” New Jerusalem Standard of Measurement.
The actual linear measurements of all three altar “hearth zones” – their tops – Moses: 20 (5x5 or 4x5 = 20 linear cubits); Solomon 80 (20x20 or 4x20 = 80 linear cubits); Ezekiel 48 (taking the
literal measurement of the Altar Hearth at 12 cubits x 12 cubits or 4x12 = 48 linear cubits) – or 20 + 80 + 48 = 148 x 25.20” = 3,729.6 = “27” perfectly attesting to the Suffering Messiah who came as
a offering for sin. These three top altar measurements are absolute; therefore, taking their “area” calculations: Moses: 126 in. x 126 in. = 15,876 sq. in.; Solomon: 504 in. x 504 in. = 254,016
sq. inches; Ezekiel: 302.4 in. x 302.4 in. = 91,445.76 sq. in. Now, their totals: 15,876 + 254,016 + 91,445.76 = 361,337.76 sq. inches = “36” the Eternal God without beginning or ending for “this
is My Beloved Son in Whom I am well pleased” – i.e., the price that the Suffering Servant, the Messiah (27), paid upon the Altars as our Priest, our King and our Prophet (Moses, Solomon and Ezekiel)
is well pleasing to the Father (36). For when the Father views these altars, He sees as Abraham saw in offering up Isaac on this same Altar – prefiguring that day – this is My Beloved Son:
361,337.76 sq. inches / 144 (Wall of the New Jerusalem) = 2,509.29 = “27” and 361,337.76 sq. inches / 25.20” the Sacred Cubit = 14,338.8 = “27.”
The Dimensions of the Millenarian Sanctuary Where the Ezekiel’s Altar is Located – Ezekiel 45:1-5
There are significant dimensions given in Ezekiel 40-48 concerning the future Kingdom, the Holy District, Temple, Sanctuary, the Prince’s Portion, the Levites Portion and plots for Urban
Homesteading, etc. – however, we have chosen to focus on just one of them here, the Holy District’s “Sanctuary within the land allotted the ‘Priests’ which is a plot that is 25,000 cubits by
10,000 cubits or 25,000 + 25,000 + 10,000 + 10,000 = 70,000 cubits x 25.20” = 1,764,000 Linear Inches = “18” New Jerusalem Standard.
Also, 25,000 cubits x 25.20” = 630,000 Linear Inches and 10,000 cubits x 25.20” = 252,000 Linear Inches; therefore, 630,000 LI x 252,000 LI = 158,760,000,000 sq. inches or “27” (Messiah) / 144 (Wall
of the New Jerusalem) = 1,102,500,000 = “9” and this “9” when added to the Levite portion, which is identical to that of the Priests portion = “18” New Jerusalem Standard; therefore, all “priestly”
portions are so dedicated to the New Jerusalem Standard. Now, the Sanctuary’s measurements within the Priest’s portion:
· Ezekiel 45:2 – 500x500 rods with fifty cubits around it for an open space
· 1 Rod = 16.5’ or 5 ½ yards therefore 500 x 16.5’= 8,250’ Linear Feet x 4 (4 edged perimeter of the “plot”) = 33,000 sq. ft. x 3 (arriving at a 12-edged cube to match the object of the New
Jerusalem) = 99,000 Linear Feet = 9 + 9 = “18” New Jerusalem Standard; therefore, the layout of the Holy District (cubed) matches the New Jerusalem …Likewise: There are 12 inches per foot, 33/2 feet
per rod, 4 rods per chain, and 80 chains per mile – multiplying these numbers we get 12” * 33/2 * 4 * 80 = 63,360 inches; therefore, 1 rod = 16.5’ and 16.5’ = 198” and 63,360” / 198” = 320 rods in
one mile x 16.5 feet = 5,280’ x 12” = 63,360” = “18” (6 + 3 + 3 + 6 = 18) = The Standard of the New Jerusalem, 18.
· ROD = 5.5 yards x 3 = 16.5' x 12" = 198 inches x 500 Rods = 99,000 inches x 4 = 396,000 linear inches (perimeter of the 500x500) x 3 (cubed) = 1,188,000 Linear Inches / 12" = 99,000 Linear Feet all
12 edges and 99,00 inches x 4 = 396,000 Linear Inches / 12" = 33,000 Linear Feet x 12 (edges) = 396,000 Linear Feet or 33,000 x 33,000 = 1,089,000,000 sq. feet = 18 and 1,089,000,000 / 43,560 =
25,000 acres x 6 sides = 150,000 acres. Or 1 edge in sq. feet of 1,089,000,000 sq. feet x 6 sides = 6,534,000,000 sq. feet = "18"
· The rod is a unit of length equal to 5.5 yards, 5.0292 metres, 16.5 feet, or ⅟320 of a statute mile. A rod is the same length as a perch or a pole. In old English, the term lug is also used.^[1]^
[2]^ (Wikipedia)
· Again: 500 rods x 16.5’ = 8,250’ x 8,250’ = 68,062,500 sq. feet / 43,560 sq. feet (1 acre) = 1,562.5 sq. acres =18 NJS x 6 (cubed sides if comparable to the New Jerusalem object found in the
Apocalypse as a cube) = 9,375 sq. acres = 24 and “24” is the number of “governance” for we have the 24 Elders of the New Jerusalem
· Now, if we take the surface “area” of the Holy District 9,375 sq. acres / 12 = 781.2 = “18” New Jerusalem Standard of Measurement
· “Open spaces” around the Holy District: 50 cubits surrounding the Holy District on all sides = 50 x 25.20” = 1,260” / 12” to arrive at Linear Feet = 105 Linear Feet – Note: In some respects this
is an “esoteric” measurement juxtaposed to the 500x500 Rods – i.e., it is included within the 500x500 or outside of it (“around it”); however, taking the measurement by itself as 105 Linear Feet and
extrapolating it as a cube itself…105 is in fact “15” which is the very fractal of the edge of the New Jerusalem’s Edge; therefore, the extrapolation is quite simple: 15 x 12 cubed edges = 180 =
“18” New Jerusalem Standard of Measurement.
· Please also see Ezekiel’s Vision of the Future by the United Church of God.
The Altar, the Tabernacle, the Temples – but “we seek a city yet to come…Whose Builder and Maker is God” . . . how sufficient is our High Priest, our King, our Prophet! May His Altar forever be the
center of our universe . . . .
“And I heard a loud voice from heaven saying, ‘Behold, the tabernacle of God is with men, and He will dwell with them, and they shall be His people. God Himself will be with them and be their God’ .
. . But I saw no temple in it, for the Lord God Almighty and the Lamb are its temple” (Revelation 21:3, 22).
See this article in Page-Flip | {"url":"http://www.the-tribulation-network.com/new_tribnet/ebooks/wonderful_numbers_of_sacred_canon/israels_alters_sec4_pt5.html","timestamp":"2014-04-17T07:23:27Z","content_type":null,"content_length":"238527","record_id":"<urn:uuid:02a49ac1-2557-41b8-9ed3-fd56d9f85420>","cc-path":"CC-MAIN-2014-15/segments/1397609526311.33/warc/CC-MAIN-20140416005206-00405-ip-10-147-4-33.ec2.internal.warc.gz"} |
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MULTIPLE REGRESSION. 1. The Multiple Regression Model . 2. The OLS Method of Estimation. 3. The R2 and the Adjusted R2. 4. Hypothesis Testing. 5. How to Estimate a Simple Regression in EViews
Quantitative Methods Heteroskedasticity Heterskedasticity OLS assumes homoskedastic error terms. In OLS, the data are homoskedastic if the error term does not have constant variance.
Eviews 操作 2011/12 I. Getting data into Eviews The first step is to read your data into an EViews workfile. 1.現有資料檔: File/Open/EViews workfile,選擇想要的工作檔. 2.新的資料檔(create a new
workfile) 以TEST.xls為例,有兩種方式叫入檔案: (1) File/New/ Workfile ...
EViewsLink Could we let EViews do the work? C(1) C(2) C(3) C(4) EViews Numbering Convention: Equivalence of Two-Tailed t-Tests and Wald Tests Claim: A two-tailed t-test is a special case of a Wald
Test log(Qt) = log( Const) + Plog(Pt) + Ilog ...
If you try to estimate an equation in Eviews and your equation specifications suffers from perfect multicollinearityEviews will not give you results but will give you an error message mentioning
multicollinearity in it. Imperfect . Multicollinearity.
Title: EViews 4 Guide Author: WASIN_EC0NOMICS_THAMMASAT Last modified by: WASIN_EC0NOMICS_THAMMASAT Created Date: 10/23/2005 2:27:01 PM Document presentation format
Finite Impulse Response Filters 5 - * Outline Many Roles for Filters Convolution Z-transforms Linear time-invariant systems Transfer functions Frequency responses Finite impulse response (FIR)
filters Filter design with demonstration Cascading FIR filters demonstration Linear phase 5 - * Many ...
PENGENALAN EVIEWS. Angelina Ika Rahutami. Unika Soegijapranata. Gasal 2011/2012
Chapter 15 Maximum Likelihood Estimation, Likelihood Ratio Test, Bayes Estimation, and Decision Theory Chapter 15 Bei Ye, Yajing Zhao, Lin Qian, Lin Sun, Ralph Hurtado, Gao Chen, Yuanchi Xue, Tim
Knapik, Yunan Min, Rui Li
Step 5: Estimate and Evaluate the Equation Once steps 1–4 have been completed, the estimation part is quick using Eviews or Stata to estimate an OLS regression takes less than a second! The
evaluation part is more tricky, however, ...
... ’ Stata, SAS, EViews Programming environments: R, Matlab, Gauss, Mathematica We will not use class time for software instruction ‘Lab’ work Problem sets Questions and review as requested Part 1:
Introduction [*/43] ...
Basic Analysis of Variance and the General Linear Model Psy 420 Andrew Ainsworth Assumptions of the analysis Normality of the sampling distribution of means This assumes that the sampling
distribution of each level of the IV is relatively normal.
The Eviews Program. Conclusion. Outline of Presentation. First proposed in Hodrick and Prescott (1997), "Postwar U.S. Business Cycles: An Empirical Investigation," Journal of Money, Credit, and
Banking. Statistical technique used to separate cyclical component from raw data series.
R eviews HIGH ALERT medications U nusually easy format G ives memory tools S implifies key points for testing and clinical practice This book transforms the study of pharmacology to a fun and easy
way to learn all those drugs! Transforming ...
Note: EViews sets the forecast variable equal to the observed Value for 1949.01-1960.12. Qick menu, show command window Correlogram of Seasonal Difference in log of passengers. Note there is still
structure, decay in the ACF, ...
Most used commands in EViews are described in Manual. Workshop 28 July 2010. Getting model database into EViews. Make sure datab.xls is not open. Run model.prg up to loaddata.prg.
... 1982:10 1996:12 Estimation in Eviews Consumer’s problem Monetary policy rules References: Favero (2000), Clarida, Gali and Gertler (1998). Variables: USFF ( US average Federal Funds rate), USINFL
(US inflation rate), USGAP1 ...
Cochrane Diagnostic test accuracy reviews. Introduction to meta-analysis. Jon Deeks and Yemisi Takwoingi. Public Health, Epidemiology and Biostatistics. University of Birmingham, UK. Outline.
Analysis of a single study. Approach to data synthesis. Investigating heterogeneity. | {"url":"http://ebookily.org/ppt/eviews","timestamp":"2014-04-24T01:17:54Z","content_type":null,"content_length":"40378","record_id":"<urn:uuid:c0186944-b68c-4f1c-b41d-dce9452b17d9>","cc-path":"CC-MAIN-2014-15/segments/1398223205375.6/warc/CC-MAIN-20140423032005-00492-ip-10-147-4-33.ec2.internal.warc.gz"} |
Cadlag modification
I was reading Stochastic Integration and Differential Equations by Protter and it had a nice theorem that every Levy process in law which has continuity in probability, admits a cadlag modification.
The proof is very confusing and I was wondering if anyone could help me clear it up a bit.
I wish to prove a slight modification, which is that the set
[tex]\{\omega : \not \exists \lim_{\mathbb{Q}\ni s\downarrow t}X_t(\omega) \quad t \geq 0 \}[/tex]
is measurable and has a measure zero. I have that X has stationary independent increments and is continuous in probability (also X starts at 0 a.s.). Any help would be much appreciated. | {"url":"http://www.physicsforums.com/showthread.php?t=296322","timestamp":"2014-04-18T21:25:36Z","content_type":null,"content_length":"19519","record_id":"<urn:uuid:129c9a62-9016-4368-b23f-05bd17112b00>","cc-path":"CC-MAIN-2014-15/segments/1397609535095.9/warc/CC-MAIN-20140416005215-00424-ip-10-147-4-33.ec2.internal.warc.gz"} |
Calculating Torque of a Crank Mechanism
Simon Bridge
Yes - you have drawn one of them. But I think you have your thinking backwards - you are driving the block from the wheel not the other way around - the block moves along a line so torque may not
help much here.
yup, my application requires the wheel to drive the block. but then, i need to select a suitable motor to drive the crank/crankwheel, then the linkage and then the block. Therefore, I need the torque
require to calculate my power needed right? if no, then do you have suggestion how much power/torque to drink the crankwheel in order to make the block move?
The torque you need to apply to the shaft to drive the block will vary with time - depending on how you want the block (or the wheel) to move.
eg. You should be able to plot the position of the black against the angular position of the wheel. If you intend to crank the wheel at a constant angular speed, you can use that to find a
position-time graph for the block.
yes, it will vary in time. just to let you know, the crankwheel is going to continuously run/rotate 360deg. I already set/came out with the position of the block i need. also, i want to plot a torque
of the crankwheel against time/angular position of the block.
Who knows - how did you calculate this figure? What were the conditions?
if you look at my 3rd picture, the 12.5mm is the distance and F is the force required to move the block. this is just my theory. it might be wrong or right. im not too sure. hence, i need more
suggestion & guideline :)
do you have any suggestion/guideline what formula to use or how to resolve the force to come out with the torque require to drive the crankwheel?
Your last diagram appears to show an applied force on the block, but your description says the wheel is driving the block - which is it? If you push on the block like that diagram shows, the force is
transmitted along the shaft. So the direction shown for force at the pivot between the wheel and the shaft is incorrect.
There is not enough information: for instance, the torque will be different if you want the block to have a constant speed for part of it's motion, a constant acceleration, or perhaps you want to
drive the wheel at a constant speed - or with a constant torque (and you want the minimum torque for this).
just to clarify, my wheel is driving the block. the reason why i calculate backward from the block to the wheel is because, the only information i have is the weight of the block. therefore, i
calculate the force acting from the block and resolve it to the crankwheel in order to find the torque required to drive the block.
also, i understand that the torque will be different during the motion of the crankwheel. assume my crankwheel will rotate at constant angular velocity, how do i calculate or plot the torque-time
graph or torque-angular graph.
if i can come out with the torque graph, i could know the maximum torque in the crank cycle. hence, i am able to choose the right motor :)
lastly, thanks alot for the reply. appreciate it :) | {"url":"http://www.physicsforums.com/showthread.php?t=555661","timestamp":"2014-04-17T01:01:07Z","content_type":null,"content_length":"42971","record_id":"<urn:uuid:46f3e71e-bac1-4fb3-991e-5b7a3c8a6a0e>","cc-path":"CC-MAIN-2014-15/segments/1398223202548.14/warc/CC-MAIN-20140423032002-00328-ip-10-147-4-33.ec2.internal.warc.gz"} |
Event Number 3147
Chaos & Complex Systems Seminar
A historical introduction to quantum computing
Time: 12:05 pm
Place: 4274 Chamberlin (refreshments will be served)
Speaker: Marty Lichtman, UW Department of Physics
Abstract: The quantum computer is on the horizon. If a system is small enough and isolated enough, it behaves according to the weird laws of quantum mechanics. One of the beautiful behaviors of a
quantum system is that it can exist in a "superposition" of multiple states, at the same time. In the last two decades, physicists, including the 2012 Nobel laureates, have learned to control
these systems. If we think of the state of these systems as a piece of information, we can store a superposition of data. Then by manipulating the system, a calculation is performed. The power of
this quantum computation is that many calculations may effectively be performed at the same time. The potential speedup is immense. A functional quantum computer will certainly bring advances in
cryptography, search, and physical simulation, and likely in all areas of science that have hard computational problems. This talk will present the development of the quantum computer in the
historical context of classical computing. We will discuss the basics of how a quantum computer can speed up certain calculations, and also look at one experimental attempt to build a quantum
computer using trapped neutral atoms here at UW-Madison.
Host: Clint Sprott | {"url":"https://www.physics.wisc.edu/twap/index.php?id=3147","timestamp":"2014-04-20T21:13:33Z","content_type":null,"content_length":"18483","record_id":"<urn:uuid:cd284b98-fa0f-4359-bb47-dfe9773270fc>","cc-path":"CC-MAIN-2014-15/segments/1398223210034.18/warc/CC-MAIN-20140423032010-00218-ip-10-147-4-33.ec2.internal.warc.gz"} |
From Atoms to Bits to Atoms: A Cat Toy's Journey
From Atoms to Bits to Atoms: A Cat Toy’s Journey
I thought I would follow up the previous cat toy post with another one. I’m not obsessed with cat toys! Really! They just make good geometric models. The main content of this post will soon
go into an Instructables page as an entry to win a 3D printer in the Make It Real Challenge, but I wanted to start with a copy here on my own blog.
There are some cat toys at the house that are particularly popular with the local residents. They are simple hollow spheres with cut-out slots. They are great for kitties to kick around —
small and light. After staring at one for a few minutes, piecing out the geometry, I realized that this would make a great instructional design for an introduction to 3D CAD modeling
using simple geometric shapes and boolean transforms.
Before we begin, you will need two things:
• A copy of OpenSCAD. This is “free” in every sense of the word — it costs nothing AND is an Open Source project that people from around the globe contribute to and use. You can get it for Mac,
Windows, or Linux.
• A web browser tab open to the OpenSCAD User Manual. OpenSCAD works like a programming language and although I’ll step you through the process, it is always good to have a language reference
If you look at the original cat toy, it is a ball. That is to say, it is a hollow sphere (or spherical shell if you want to get fancy). That ball then has slots cut away at regular
intervals. I did a little bit of measuring beforehand.
Let’s start by making the sphere. This is a simple one line “program.” There is an extra line up top that you can modify to produce different detail levels. I find that a value of 30 is good
to do most work in, though I bump that up to about 100 when I am ready to produce a final render. This sphere has a radius of 20mm. Remember that the radius (20mm) is half of the diameter (40mm).
DETAIL = 30;
sphere(r = 20, $fn = DETAIL);
Type this code into the text area on the left of the OpenSCAD window. Go to the “Design” pulldown menu and select “Compile.” In a few moments, you should see a sphere in the right half of the
OpenSCAD window. You can click and drag with your mouse to rotate the 3D shape and use the mouse wheel to zoom in and out.
If we were to print this sphere on a 3D printer, it would be solid. We don’t want a solid ball, but a hollow shell. Let’s subtract enough away to give us a 2mm shell. We do this with the
difference() function. This function takes a list of two or more objects, draws the first one, and then subtracts each of the other ones. In this case, we will make a slightly smaller sphere
inside and subtract it.
DETAIL = 30;
sphere(r = 20, $fn = DETAIL);
sphere(r = 18, $fn = DETAIL);
Now, select “Compile” or use its hotkey (F5). Note that on my keyboard (a Mac), the F5 key is actually keyboard brightness and I have to use the “Fn” key to use it as an F5.
In the graphic window you will see.... something that looks exactly like what you had before. Except it is hollow now. How do you know? Let’s take a quick bite out of it with a cube.
DETAIL = 30;
sphere(r = 20, $fn = DETAIL);
sphere(r = 18, $fn = DETAIL);
cube(size = [20, 20, 20]); // temporarily use this to peek at the thickness
This program starts with the big sphere, subtracts out the smaller one, then subtracts out a cube. You may have to use your mouse to rotate around, but it should look something like this:
Now remove that cube line. We will do something very much like that to cut out the slots, except instead of cubes, we will use cuboids, which are the rectangles of cubes. Let’s just start with
one to see how to line things up. We’ll do this:
DETAIL = 30;
sphere(r = 20, $fn = DETAIL);
sphere(r = 18, $fn = DETAIL);
cube(size=[80, 40, 5]);
You’ll end up with a sphere with a slit similar to a Cylon eye strip. What’s going on here? Shouldn’t that have cut out a rectangle? It did, but the rectangle was placed directly on the
origin. If you select “View -> Thrown Together” from the pulldown menus, then “Design -> Compile” you’ll see the combination of shapes. This is an extremely useful view of your virtual world
when you are trying to subtract shapes and don’t understand what is going on.
We want it to completely intersect from one edge of the sphere to the other. To do that, we’ll need to slide it over a bit. We can do that with the translate() function, which takes the object
following it and shifts it around on the x, y, and z axis. Let’s try it.
DETAIL = 30;
sphere(r = 20, $fn = DETAIL);
sphere(r = 18, $fn = DETAIL);
translate(v = [-40, 0, 0])
cube(size=[80, 40, 5]);
That’s the strip we’re looking for, but it needs to be offset just a bit from the center to give us that central support column. Let’s tweak that y value over a little bit more.
DETAIL = 30;
sphere(r = 20, $fn = DETAIL);
sphere(r = 18, $fn = DETAIL);
translate(v = [-40, 4, 0])
cube(size=[80, 40, 5]);
That’s great! Now we just need to add a few more! I’m going to adjust the height from 5mm to 4.45mm to get them to divide a little more evenly across the sphere. Let’s add some more translated
cubes. It takes a little bit of math to get the positioning right, but you can add one at a time, experiment with positioning, and play with the results as you go.
DETAIL = 30;
sphere(r = 20, $fn = DETAIL);
sphere(r = 18, $fn = DETAIL);
translate(v = [-40, 4, -20 + 2 * 4.45])
cube(size=[80, 40, 4.45]);
translate(v = [-40, 4, -20 + 4 * 4.45])
cube(size=[80, 40, 4.45]);
translate(v = [-40, 4, -20 + 6 * 4.45])
cube(size=[80, 40, 4.45]);
Finally, the other half of cutouts goes in, offset from the first set. The code looks almost exactly like the first half:
DETAIL = 30;
sphere(r = 20, $fn = DETAIL);
sphere(r = 18, $fn = DETAIL);
// right side
translate(v = [-40, 4, -20 + 2 * 4.45])
cube(size=[80, 40, 4.45]);
translate(v = [-40, 4, -20 + 4 * 4.45])
cube(size=[80, 40, 4.45]);
translate(v = [-40, 4, -20 + 6 * 4.45])
cube(size=[80, 40, 4.45]);
// left side
translate(v = [-40, -44, -20 + 1 * 4.45])
cube(size=[80, 40, 4.45]);
translate(v = [-40, -44, -20 + 3 * 4.45])
cube(size=[80, 40, 4.45]);
translate(v = [-40, -44, -20 + 5 * 4.45])
cube(size=[80, 40, 4.45]);
translate(v = [-40, -44, -20 + 7 * 4.45])
cube(size=[80, 40, 4.45]);
Finally, you can crank up that DETAIL value when everything looks good. This will give you a higher quality 3D model with more surfaces, but it takes a little longer to generate. Go to
“Design -> Compile and Render” on the pulldown menu for your final render. Assuming that looks good, go to “Design -> Export as STL” to save it as a 3D file. Most 3D printers and 3D printing
services use this file format.
Congratulations! You created a cat toy using geometry! Time to print!
On a home 3D printer, even if you build using support structures, the upper slats will have gravity working against them. They’ll end up sagging a bit and kind of stringy on the bottom. You
will get a ball that is structurally sound, but looks a little warped. If you have a small metal file or X-Acto knife, you can manually clean it up a little. Industrial printers typically
do not suffer from this sort of problem because they use a different process for printing and support.
I have a zip file (cat_ball.zip) containing source for each of these steps. It also contains a mondo-mega final OpenSCAD file called a parametric model. This is a model where all of the
important numbers are distilled and extracted to the top (much like DETAIL is in the above examples). This then lets you easily change the size of the ball, the wall thickness, the number of
slots, and the width of the support structure (i.e. the depth of the cuts). You can then make all sorts of crazy variants just by changing a few numbers. You want one that is 50mm in
diameter with a 5mm thick shell and 19 cutouts? There you go! But good luck printing this variant on a home 3D printer.
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How many seeds in a sunflower seed head?
I planted giant sunflowers in the back of the vegetable garden this year, just on a whim to see if I could provide some “natural” food for the birds. The plants did indeed become gigantic — more
than 12 feet tall (see Sept 2 post). The flowers were about 8 inches across, and attracted a variety of pollinators, but were most attractive to the bumblebees (one is working on the giant
sunflower below).
Unfortunately, it wasn’t the birds who discovered the maturing seed heads, it was the squirrels, who were climbing the stalks and pulling the flowers over and eating them. So, I had to harvest the
flower heads a little early, before the seeds in the center had a chance to mature. Even though slightly immature, these seed heads were enormous (compare with camera lens cover). You can see the
squirrel damage on the flower on the left.
Some of the disk flowers had not dropped off yet, but there were maturing seeds underneath. These flowers are a testament to the pollination efficiency of those bees, as illustrated below (with disk
flowers removed). If there ever were a demonstration of the eco-service provided by bees, this has to be it. It looks like almost every single one of those disk flowers got pollinated and
fertilized and has at least tried (less successfully in the center) to produce a seed.
You can see the beautiful geometric pattern of spirals that promotes the most efficient seed packing into limited space (read about the “golden angle” in the post on Sept 2). But just to drive the
point home, here’s a close-up.
So you can better appreciate the geometric spiralling pattern, I drew on the seed head to help me keep track while counting seeds. Yes, counting the seeds, because I wanted to know how many seeds
were in a 6.25 inch diameter sunflower seed head.
Now, here’s the interactive part of this post.
How many seeds do you think there were in this seed head?
12 thoughts on “How many seeds in a sunflower seed head?”
1. Using no scientific method at all, I’ll guess 5,000 seeds.
2. Good try, Mike. You are closer than anyone else so far.
3. I would go lower at 1,500 but it is only a guess. My excuse is I’ve just finished varnishing the patio windows and I’m tired. (I expect you’ve heard every excuse in the book.)
□ Very close — good guess.
4. I’m guessing 2000, but really haven’t a clue.
5. Pingback: 1115 | Back Yard Biology
6. i think its 1440
□ Hi Amy, thanks for visiting and making a guess on the seed head. Apparently I never gave the right answer to the question i posed. There were only 1080! Much lower than expected because of
all the undeveloped seeds in the middle of the flower.
7. Were you able to determine a model or any link between spiral count, head diameter and seed count? EG; number of spirals by spiral length, or seeds per diameter inch.
□ I didn’t do it, but someone else did, and this is an excellent website to show you the mathematical relationships of sunflower heads and seeds in their spiral rows. http://momath.org/home/
8. i real like this mathematical relationship because it relies on truth | {"url":"http://bybio.wordpress.com/2012/09/30/how-many-seeds-in-a-sunflower-seed-head/","timestamp":"2014-04-18T05:29:56Z","content_type":null,"content_length":"69366","record_id":"<urn:uuid:ff60ead2-f9ed-4427-9d71-fc8d1c6eab50>","cc-path":"CC-MAIN-2014-15/segments/1397609532573.41/warc/CC-MAIN-20140416005212-00364-ip-10-147-4-33.ec2.internal.warc.gz"} |
What would you want on a Lie theory cheat poster?
up vote 56 down vote favorite
For some long time now I've thought about making a poster-sized "cheat sheet" with all the data about Lie groups and their representations that I occasionally need to reference. It's a moving target,
of course -- the more I learn, the more stuff I'd want to see on such a poster!
There are many obvious things I think everybody would agree should be there. The finite and affine Dynkin diagrams, with the Bourbaki numbering and also the coefficients of the simple roots in the
highest root. The dimensions of the fundamental representations. Coordinate descriptions of each of the root systems, and of their Weyl groups. The exponents of each group, the Coxeter number, and
the structure of the center. The exceptional isomorphisms of low-rank groups.
Only slightly less obvious: Satake diagrams. Dynkin's characterization of the nilpotents. Geometric descriptions of the partial flag manifolds $G/P$. The classification of real symmetric spaces.
One suggestion per answer, please, but otherwise, go wild! I'm not promising to actually make this thing in any timely manner, but would look to the votes on answers to prioritize what actually makes
it onto the poster.
1 Funny - I was having ideas of making the same thing, to decorate my office (and maybe our new tea-room after our department migrates to a new building this summer). Let me know if you make it, and
I'll let you know as well. – Marty Apr 26 '11 at 14:54
add comment
26 Answers
active oldest votes
Homology and homotopy groups
up vote 25
down vote
1 The homotopy groups are hopeless (think of $SU(2)$). The homology should be recorded mod $p$ for all the relevant primes, along with the action of the corresponding Steenrod
algebra. – André Henriques Apr 26 '11 at 15:08
The unstable homology is also hopeless. – Jim Conant Apr 26 '11 at 21:09
16 If you don't know everything, it does not mean you know nothing. The homotopy groups of Lie groups are known up to dimension higher than most people are likely to need. And in the
stable range they are all known by Bott periodicity. – Richard Borcherds Apr 26 '11 at 22:59
+1 for homology, I often find myself searching for what are the exponents of a given group. – Reimundo Heluani Oct 19 '11 at 11:40
add comment
Maximal compact subgroups
up vote 19 down vote
add comment
Real forms. And now I complain that Mathoverflow will not let me enter an answer with less than 15 characters.
up vote 19 down vote
14 A possible solution in such cases is to write your answer in German. mathoverflow.net/questions/62218/… – danseetea Apr 26 '11 at 16:15
Or include dollar sign, backslash, space, backslash, space, ..., dollar sign. Oops, now I've done it. – Allen Knutson Oct 19 '11 at 10:27
add comment
Freudenthal's magic square of Lie algebras
and the corresponding square of projective planes:
ℝP^2 ℂP^2 ℍP^2 ↀP^2
ℂP^2 (ℂ⊗ℂ)P^2 (ℂ⊗ℍ)P^2 (ℂ⊗ↀ)P^2
up vote 19 down vote ℍP^2 (ℍ⊗ℂ)P^2 (ℍ⊗ℍ)P^2$ (ℍ⊗ↀ)P^2
ↀP^2 (ↀ⊗ℂ)P^2 (ↀ⊗ℍ)P^2 (ↀ⊗ↀ)P^2
Also: have a look at John Baez's cheat sheet,
and at the subsections G[2], F[4], E[6], E[7], E[8].
add comment
up vote The Vogel plane.
14 down
3 I'm flattered! Please don't take this picture too seriously. – Bruce Westbury Apr 26 '11 at 19:04
From Bruce's website: "The idea is that this should be thought of as a two-parameter family of Lie algebras which contains every simple Lie algebra. The three vertical lines are the
three families of classical simple Lie algebras. The other two lines are the last two rows of the Freudenthal magic square." Is that really true? What about this paper of Dylan Thurston
math.columbia.edu/~dpt/writing/F4E6.ps that claims to have evidence against that conjecture? – André Henriques Apr 26 '11 at 20:11
So there's two issues here: one is how meaningful the points in the plane are, the other is how meaningful the lines are. Dylan's paper actually gives (a very small amount of) evidence
1 in support of the conjecture that every point in the plane determines at most one Lie algebra object. It also gives (a very small amount of) evidence against the idea that every point
actually gives a Lie algebra object. What it conclusively shows is that there's no line going through F4 or E6 consisting of Lie algebras objects whose representation theory looks like
F4 or E6. – Noah Snyder Apr 26 '11 at 20:32
It looks like some exceptional low-rank isomorphisms aren't identified. Am I missing something? – S. Carnahan♦ Apr 28 '11 at 6:25
Hrm, good question... The description of the Vogel plane that I'm most familiar with seems like it should identify sl(2) and so(3) at the same point. My guess is that what Bruce has
written down there is actually a ramified cover of that P^2 where he's also keeping track of the choice of defining representation. As such so(3) might better be labelled SO(3). But I'm
not totally sure. – Noah Snyder Apr 28 '11 at 16:36
show 1 more comment
Maximal subgroups
up vote 12 down vote
add comment
Finite subgroups
up vote 11 down
Is there something you mean here other than listing them for the first couple small groups? For large rank this is hopeless, right? Or are there specific kinds of finite subgroups
that you had in mind? – Noah Snyder Apr 26 '11 at 16:27
add comment
Handy identities for rings associated to Lie groups and algebraic groups, such as the enveloping algebra, hyperalgebra, or quantum groups of various kinds. For example, I often find
up vote 11 myself needing to look up commutation relations between generators for these rings as in Lusztig's papers.
down vote
add comment
Besides a magnifying glass and a very large empty wall for the poster, I'd like to have all relevant data (dimensions, Dynkin and Bala-Carter labels, ... ) for nilpotent orbits of the five
exceptional Lie types including correct versions of the intricate closure ordering graphs worked out by Spaltenstein and others (reproduced in Carter's 1985 book with apparently some
omitted edges for types $E_7, E_8$). And of course designation of the *special" nilpotent orbits, Richardson orbits, etc.
up vote 11
down vote Actually, a reliable online database for all items mentioned in the answers here would be even better than a poster.
1 Most of what you ask for is here: liegroups.org/tables/unipotentOrbits – Jeffrey Adams May 17 '11 at 1:44
add comment
Types of the fundamental representations (real, complex, or quaternionic)
up vote 10 down vote
I had so much trouble finding this info a couple years ago. But I managed to find the answer for E7 somewhere on John Baez's site. – Noah Snyder Apr 26 '11 at 20:58
I usually just run it through LiE since it is able to calculate the decomposition of the symmetric and exterior squares into irreducibles. – ARupinski Apr 26 '11 at 21:07
add comment
Combinatorial presentations of the cohomology rings of $G/P$'s. (And equivariant cohomology, and $K$-theory, and quantum cohomology ... Maybe this should be a second poster?)
up vote 9 down vote
add comment
Minuscule representations
up vote 7
down vote
Obviously there's going to be dependency between these answers, but what more do you want in this direction once you've given the coefficients of the simple roots in the highest root,
and the dimensions of all fundamental representations? – Allen Knutson Apr 27 '11 at 2:03
(Insofar as the minuscules correspond to the coefficients "1") – Allen Knutson Apr 28 '11 at 1:44
add comment
Affine Dynkin diagrams (with the linear combination of simple roots that has norm 0)
up vote 7 down vote
add comment
Description of the fibres of the Springer resolution. Classification of nilpotent orbits.
up vote 6 down vote
add comment
I am surprised nobody so far has mentioned Hecke algebras. I don't know what exactly to write about them -- defnition? finite, affine and double Hecke algebras?, Kazhdan-Lusztig basis
up vote 6 down (in some form)? geometric realization (say via the Steinberg variety)?
add comment
Probably you want the dual Coxeter number as well.
up vote 5 down vote
add comment
Weyl Character formula and multiplicity formula for the weight spaces in a irreducible representation.
up vote 5 down vote
add comment
Moving quite a bit beyond the level of what is contained in Bourbaki's planches, one could ask for information about infinite-dimensional representations of Lie groups.
up vote 4 down Ideally we would include a complete description of the unitary dual, but since this is still an open problem, how about an explicit as possible description of the tempered dual and
vote the Plancherel formula.
add comment
I know of a couple physicists with charts of 6j's hanging in their office.
up vote 3 down vote
6 Are you asking us to pity you? :) – José Figueroa-O'Farrill Apr 26 '11 at 15:18
7 I don't get it. :( – B. Bischof Apr 26 '11 at 18:18
All I meant was that some people might find it unfortunate to have such acquaintances :) – José Figueroa-O'Farrill Apr 27 '11 at 2:30
2 I don't mind having physicist acquaintances, in fact I like it. However I do mind having 6j acquaintances, those things are jerks... – B. Bischof Apr 27 '11 at 5:13
add comment
Simple relationships between the fundamental representations (in terms of exterior powers of extremal fundamental representations based on a result of Adams)
up vote 3 down vote
add comment
Edit: Per Noah's suggestion I have broken up my list into four separate posts, so this is the original post but with only one of the original list items.
For lack of a better term, "Quotient of Faithfulness" indicating for each fundamental representation what quotient of the simply-connected group the representation is a faithful
representation of; this information is useful for questions of distinguishing between actions such as $Spin(k)$ vs. $SO(k)$ and determining whether a given reducible representation of
up vote 3 $Spin(4n)$ is faithful (since it has no faithful irreducibles).
down vote
Awhile back I was also compiling some basics along the lines you are looking for. I also had a few other little facts that aren't widely known or used but that I was using in my research.
My version was going to be in the form of a short pamphlet, but I think a nicely layed-out poster would definitely be very cool and useful.
1 For big list questions you should only have one suggestion per answer so that voting can sort the best suggestions to the top. In this case I really want to vote up your first
suggestion (which I had a really hard time finding the one time I needed to know), but I'd like it to be separated out. – Noah Snyder Apr 26 '11 at 18:52
@Noah: Sorry about that, I started writing down the ideas and completely forgot about the fact that this was wiki mode. – ARupinski Apr 26 '11 at 20:25
It's not too late. You can edit this one down to one answer, and then post other answers with the other points. – Noah Snyder Apr 26 '11 at 20:33
add comment
For each real form, the poset K_C\G/B, as calculable by the Atlas here.
up vote 3 down vote
add comment
Much of the rudimentary stuff requested above is accessible via an online sql database that I set up. It's at
Don't be scared of the phrase "sql database". It is extremely easy to use for simple queries. For example, if you want to know how the Lie algebra E8 decomposes under its various maximal
compact subalgebras you could get a table for that and more by simply filling out a form that passes the question to the database as an SQL query:
SELECT k , vogan_diagram , satake_diagram , k , krep_p , krep_k , rel_root_sys
FROM realform_data
WHERE g='E8'
With just a smidgen of SQL experience it's equally easy to design correlative queries utilizing data from more than one table.
The UMRK database contains following tables:
up vote 3 g_data : data for simple complex Lie algebras (a la Bourbaki)
down vote
orbit_data : data for nilpotent orbits of simple complex Lie algebras
wrep_data : data for the irreducible representations of Weyl groups of simple complex Lie algebras
ss_data : data for the root subsystems of simple complex Lie algebras
levi_data :data for the Levi subsystems of simple complex Lie algebras
wcc_data : data for the conjugacy classes of Weyl groups of simple complex Lie algebras
realform_data : data for the real forms simple complex Lie algebras
(BTW this thread was brought to my attention by Jeff Adams.)
Birne Binegar
Dear Birne. I regret to announce that your webpage is not very user-friendly. I have just spent 10 minutes trying to enter the example SELECT g , common_name , component_group FROM
orbit_data WHERE g='C7' and component_group != '1'" into the UMRK Simple Query Form and repeatedly got error messages like <<Query failed: ERROR: column "component_group" does not
exist>>. I finally just gave up. – André Henriques May 18 '11 at 15:34
Well, the error message is actually explains the problem - it's that the example at the top of the form is obsolete. If you look below the entry fields there's a listing of what fields
are accessible in each table. There is no "component_group" field in the orbit_data table, so that's why you got the error. What you want, presumably is the field A ( = A(O) = component
group of the stabilizer in G_ad of a representative element of the orbit. Thus, you should try SELECT g , common_name , A FROM orbit_data WHERE g='C7' User-friendly-ness will improve with
more feedback. – Birne Binegar May 20 '11 at 17:06
Ok. Now it works. Sorry for not realizing that there was just a little mistake in the example. – André Henriques May 20 '11 at 18:55
add comment
Adding to the listing of maximal subgroups, the decomposition of fundamental representations on restriction to a maximal subgroup
up vote 2 down vote
add comment
Their Stiefel manifold torsors.
up vote 0
down vote
Can you say more about this? I don't understand what you would actually want to see written on a poster. – Allen Knutson Apr 27 '11 at 12:28
I find it useful to use the isomorphisms between the classical compact groups and their Stiefel manifold torsors in certain computations, for example in the integration with respect to
the Haar measure. Thus my suggestion is to include the isomorphisms given in the table of the linked Wikipedia page. – David Bar Moshe Apr 27 '11 at 13:05
The table in the "Special cases" section? So is this any different from specifying the compact forms of the groups? Sorry for being dense about this. – Allen Knutson Apr 28 '11 at 1:46
You are welcome. Yes, I meant the isomorphisms listed in the table in the "special cases" section. I guess that you can take them as definitions of the compact forms, but they are more
useful than just that. They allow performing group averaging by integration over frames. I hope to clarify myself with this example: Suppose one wants to average a Hamiltonian function
over (the invariant measure of a complete) flag manifold $U(n)/T$. The Hamiltonian can be extended to a function over the group $U(n)$ manifold which is polynomial in the frame variables.
– David Bar Moshe Apr 28 '11 at 8:52
The computation can then performed by iterative integrations over spherical fibers of Stiefel manifolds. Thus, one ends up with a sequence of polynomial integrations over spheres. – David
Bar Moshe Apr 28 '11 at 8:52
add comment
For each Dynkin diagram, the involution taking a fundamental weight to its dual. (The main issue being that it's trivial for $D_{even}$, and nontrivial for $D_{odd}$. Mnemonic: what
up vote 0 down could happen with $D_4$?)
add comment
Not the answer you're looking for? Browse other questions tagged lie-groups lie-algebras big-list or ask your own question. | {"url":"http://mathoverflow.net/questions/63033/what-would-you-want-on-a-lie-theory-cheat-poster","timestamp":"2014-04-17T04:53:31Z","content_type":null,"content_length":"170656","record_id":"<urn:uuid:32085da2-2020-4d16-8bc2-6f4fdd14b498>","cc-path":"CC-MAIN-2014-15/segments/1398223203422.8/warc/CC-MAIN-20140423032003-00057-ip-10-147-4-33.ec2.internal.warc.gz"} |
How Many of Each Ticket Were Sold?
Date: 09/06/2001 at 02:59:52
From: Sarah Kromer
Subject: Turning a word problem into an equation
Five hundred tickets were sold for a play, for $8 at the lower level
and $6 at the upper level, totaling $3600. How many of each were sold?
I am having trouble putting this into an equation.
(8x)+ (6Y) = 500
or (8x) + (6y) = 3600
I think it is the latter but I 'm getting stuck on where to go next.
I don't seem to have such a problem working out equations but when it
comes to some - not all - word problems I experience a block.
Please help. Thank you.
Date: 09/06/2001 at 10:49:45
From: Doctor Peterson
Subject: Re: Turning a word problem into an equation
Hi, Sarah.
An important part of writing an equation for a word problem is to be
aware of exactly what everything means. That starts with writing down
explicitly what your variables mean:
x = number of lower level tickets sold
y = number of upper level tickets sold
Now you can write expressions corresponding to particular phrases used
in the problem, or to concepts implied by the problem; again, be as
clear as possible:
8x = cost in dollars for all lower level tickets sold
6y = cost in dollars for all upper level tickets sold
Now the total cost of ALL the tickets is the sum of these:
8x + 6y = cost in dollars of all tickets sold
Now that we see just what this expression means, we know that it has
to be equal to the total cost, $3600:
8x + 6y = 3600
We have one equation and two unknowns; that's not enough in general to
solve the problem. But there's a piece of information we haven't used
yet. Can you see how
500 = total number of tickets sold
can be related to the variables x and y? That will give you a second
equation; you can solve that for y, and then replace y in the first
equation with the resulting expression in order to solve the problem.
I hope this helps. An orderly approach can help a lot in cutting
through the complexities of the English language, as well as those of
- Doctor Peterson, The Math Forum | {"url":"http://mathforum.org/library/drmath/view/57515.html","timestamp":"2014-04-18T04:12:02Z","content_type":null,"content_length":"7091","record_id":"<urn:uuid:464e056d-e610-4c81-921f-19a6928d3df2>","cc-path":"CC-MAIN-2014-15/segments/1397609532480.36/warc/CC-MAIN-20140416005212-00336-ip-10-147-4-33.ec2.internal.warc.gz"} |
No. of days.
August 28th 2008, 04:11 AM #1
Junior Member
Aug 2008
No. of days.
8 children and 10 men complete a certain piece of work in 9 days.Each child takes twice the time taken by a man to finish the work.In how many days will 12 men finish the same work?
Hello !
Usually, you can find in how many days a child finishes the work, etc...
Here's how I'd do. Since each child takes twice the time than a man, we can consider that 2 children are worth 1 man.
Hence 8 children worth 4 men. And thus 14 men would complete the piece of work in 9 days.
14 men -> 9 days
12 men -> ? days
That's a cross product
August 28th 2008, 04:15 AM #2 | {"url":"http://mathhelpforum.com/algebra/46981-no-days.html","timestamp":"2014-04-20T11:54:22Z","content_type":null,"content_length":"31918","record_id":"<urn:uuid:3b5a6317-f194-4d5f-a17a-01cd77e8dfd6>","cc-path":"CC-MAIN-2014-15/segments/1397609539447.23/warc/CC-MAIN-20140416005219-00207-ip-10-147-4-33.ec2.internal.warc.gz"} |
Isomorphism of Rings
May 12th 2010, 08:07 PM
Isomorphism of Rings
I was just trying out problems from Herstein, and I have a doubt in this. I would really appreciate any help.
Let R be a ring with unit element. Using its elements we define a ring R' by defining a (+) b = a+b+1 and a.b=a+b+ab where a,b are in R and where the addition and multiplication on the right hand
side of these relations are those of R.
a) Prove that R' is a ring under (+) and .
This I was able to do.
b) What acts as the zero-element of R'?
The zero-elt of R' is -1
c) What acts as the unit element of R'?
The unit elt of R' is 0.
d) Prove that R is isomorphic to R'.
This is where I am having problems. I need to define a map f:R' --> R which is 1-1, onto and a homomorphism.
I can also see that f shouls map the zero elts (ie) -1~~> 0 and since f should be onto, the unit element to the unit elt of R (ie) 0~~>1
The only map I can think of is f(a) = a+1
Using that f(a(+)b) = f(a + b + 1) = a + b + 1 +1 = a +1 + b +1 = f(a) + f(b)
f(a.b) = f(a + b + ab) = a+b +ab + 1 = . this should be ab to make it f(a)f(b). Where have I gone wrong, is the map itself wrong??
Thank you :))
May 12th 2010, 08:33 PM
Oh i cant believe i missied something simple,
f(a.b) = f(a +b +ab) = a+ b+ ab +1 = (a +1) +b(a+1) =(a+1)(b+1) = f(a)f(b) | {"url":"http://mathhelpforum.com/advanced-algebra/144452-isomorphism-rings-print.html","timestamp":"2014-04-16T19:26:36Z","content_type":null,"content_length":"4571","record_id":"<urn:uuid:a09cdd50-2518-4ee4-ac84-02dd51b96d31>","cc-path":"CC-MAIN-2014-15/segments/1397609524644.38/warc/CC-MAIN-20140416005204-00499-ip-10-147-4-33.ec2.internal.warc.gz"} |
New Exact Solutions of Ion-Acoustic Wave Equations by (
Journal of Applied Mathematics
Volume 2013 (2013), Article ID 810729, 11 pages
Research Article
New Exact Solutions of Ion-Acoustic Wave Equations by ()-Expansion Method
School of Mathematical Sciences, Universiti Kebangsaan Malaysia UKM, 43600 Bangi, Selangor, Malaysia
Received 24 May 2013; Revised 19 September 2013; Accepted 20 September 2013
Academic Editor: Saeid Abbasbandy
Copyright © 2013 Wafaa M. Taha et al. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any
medium, provided the original work is properly cited.
The ()-expansion method is used to study ion-acoustic waves equations in plasma physic for the first time. Many new exact traveling wave solutions of the Schamel equation, Schamel-KdV (S-KdV), and
the two-dimensional modified KP (Kadomtsev-Petviashvili) equation with square root nonlinearity are constructed. The traveling wave solutions obtained via this method are expressed by hyperbolic
functions, the trigonometric functions, and the rational functions. In addition to solitary waves solutions, a variety of special solutions like kink shaped, antikink shaped, and bell type solitary
solutions are obtained when the choice of parameters is taken at special values. Two- and three-dimensional plots are drawn to illustrate the nature of solutions. Moreover, the solution obtained via
this method is in good agreement with previously obtained solutions of other researchers.
1. Introduction
The ion-acoustic solitary wave is one of the fundamental nonlinear waves phenomena appearing in fluid dynamics [1] and plasma physics [2]. To allowing for the trapping of some of the electrons on
ion-acoustic waves, Schamel proposed a modified equation for ion-acoustic waves [3] given by where is the wave potential and is a constant, this equation describing the ion-acoustic wave, where the
electrons do not behave isothermally during their passage of the wave in a cold-ion plasma. Then, combining the equations of Schamel and the KdV equation, one obtains the so-called one-dimensional
form of the Schamel-KdV (S-KdV) equation equation [4, 5]: where , , and are constants. This equation is established in plasma physics in the study of ion acoustic solitons when electron trapping is
present, and also it governs the electrostatic potential for a certain electron distribution in velocity space. Note that we obtain the KdV equation when and the Schamel equation when for . Due to
the wide range of applications of (2), it is important to find new exact wave solutions of the Schamel-KdV (S-KdV) equation. Another equation arising in the study of ion-acoustic waves is the
so-called modified Kadomtsev-Petviashvili (KP) equation given by [6]
Equation (3) was firstly derived by Chakraborty and Das [7]; the modified KP equation containing a square root nonlinearity is a very attractive model for the study of ion-acoustic waves in a
multispecies plasma made up of non-isothermal electrons in plasma physics.
In the literature, the KP equation is also known as the two-dimensional KdV equation [8].
It has lately become more interesting to obtain exact analytical solutions to nonlinear partial differential equations such as the one arising from the ion-acoustic wave phenomena, by using
appropriate techniques. Several important techniques have been developed such as the tanh-method [9, 10], sine-cosine method [11, 12], tanh-coth method [13], exp-function method [14],
homogeneous-balance method [15, 16], Jacobi-elliptic function method [17, 18], and first-integral method [19, 20] to solve analytically nonlinear equations such as the above ion-acoustic wave
Moreover, in the standard tanh method developed by Malfliet in 1992 [21], the tanh is used as a new variable. Since all derivatives of a tanh are represented by tanh itself, the solution obtained by
this method may be solitons in terms of or may be kinks in terms of tanh. We believe that the -expansion method is more efficient than the tanh method. Moreover, the tanh method may yield more than
one soliton solution, a capability which the tanh method does not have. The sine-cosine method yields a solution in trigonometric form. The Exp-function method leads to both generalized solitary
solution and periodic solutions. The homogeneous-balance method is a generalized tanh function method for many nonlinear PDEs. The first integral method, which is based on the ring theory of
commutative algebra, was first proposed by Feng. There is no general theory telling us how to find its first integrals in a systematic way; so, a key idea of this approach to find the first integral
is to utilize the division theorem. The traveling wave solutions expressed by the -expansion method, which was first proposed by Wang et al. [22], transform the given difficult problem into a set of
simple problems which can be solved easily to get solutions in the forms of hyperbolic, trigonometric, and rational functions. The main merits of the -expansion method over the other methods are as
follows. (i)Higher-order nonlinear equations can be reduced to ODEs of order greater than 3. (ii)There is no need to apply the initial and boundary conditions at the outset. The method yields a
general solution with free parameters which can be identified by the above conditions. (iii)The general solution obtained by the -expansion method is without approximation. (iv)The solution procedure
can be easily implemented in Mathematica or Maple.
In fact, the -expansion method has been successfully applied to obtain exact solution for a variety of NLPDE [23–34].
In this paper, the -expansion method is used to study ion-acoustic waves equations in plasma physic for the first time. We obtain many new exact traveling wave solutions for the Schamel equation,
S-KdV, and the two-dimensional modified KP equation. The traveling wave solutions obtained via this method are expressed by hyperbolic functions, the trigonometric functions, and the rational
functions. In addition to solitary waves solutions, a variety of special solutions like kink shaped, antikink shaped, and bell type solitary solutions are obtained when the choice of parameters is
taken at special values. Two- and three-dimensional plots are drawn to illustrate the nature of solutions. Moreover, the solution obtained via this method is in good agreement with previously
obtained solutions of other researchers.
Our paper is organized as follows: in Section 2, we present the summary of the -expansion method, and Section 3 describes the applications of the -expansion method for Schamel equation, S-KdV
equation, and modified KP equation, and lastly, conclusions are given in Section 4.
2. Summary of the -Expansion Method
In this section, we describe the -expansion method for finding traveling wave solutions of NLPDE. Suppose that a nonlinear partial differential equation in two independent variables, and , is given
by where is an unknown function, is a polynomial in and its various partial derivatives, in which highest order derivatives and nonlinear terms are involved.
The summary of the -expansion method can be presented in the following six steps.
Step 1. To find the traveling wave solutions of (4), we introduce the wave variable: where the constant is generally termed the wave velocity. Substituting (5) into (4), we obtain the following
ordinary differential equations (ODE) in (which illustrates a principal advantage of a traveling wave solution; i.e., a PDE is reduced to an ODE):
Step 2. If necessary, we integrate (6) as many times as possible and set the constants of integration to be zero for simplicity.
Step 3. Suppose that the solution of nonlinear partial differential equation can be expressed by a polynomial in as where satisfies the second-order linear ordinary differential equation where , ,
and , , and are real constants with . Here, the prime denotes the derivative with respect to . Using the general solutions of (8), we have
The above results can be written in simplified forms as
Step 4. The positive integer can be accomplished by considering the homogeneous balance between the highest order derivatives and nonlinear terms appearing in (6) as follows: if we define the degree
of as , then the degree of other expressions is defined by
Therefore, we can get the value of in (7).
Step 5. Substituting (7) into (6), useing general solutions of (8), and collecting all terms with the same order of together, then setting each coefficient of this polynomial to zero yields a set of
algebraic equations for , , , and .
Step 6. Substitute , , , and obtained in Step 5 and the general solutions of (8) into (7). Next, depending on the sign of the discriminant , we can obtain the explicit solutions of (4) immediately.
3. Applications of the -Expansion Method
3.1. Schamel Equation
In order to find the solitary wave solution of (1), we use the transformations Then, (1) becomes Integrating (13) with respect to and setting the integration constant equal to zero, we have According
to the previous steps, using the balancing procedure between with in (14), we get so that . Now, assume that (14) has the following solution: where , , and are unknown constants to be determined
later. Substituting (15) along with (8) into (14) and collecting all terms with the same order of , the left hand side of (14) is converted into a polynomial in . Equating each coefficient of the
resulting polynomials to zero yields a set of algebraic equations for , , , , , , , and as follows: On solving the above set of algebraic equations by Maple, we have Now, (15) becomes Substituting
the general solution of (8) into (18), we obtain the three types of traveling wave solutions depending on the sign of .
If , we have the following general hyperbolic traveling wave solutions of (1):where and are arbitrary constants.
If , we have the following general trigonometric function solutions of (1):
If , we have the following general rational function solutions of (1): where .
Writing and setting and in (19), we reproduce the result of Khater and Hassan [35] (see their Equation (4.7)), where .
Note that Khater and Hassan [35] obtained only hyperbolic solutions, but in this work, we found two additional types of solutions, that is, trigonometric and rational solutions.
3.2. S-KdV Equation
To find the general exact solutions of (2), we first write to transform (2) into Assume the traveling wave solution of (23) in the form Hence, (23) becomes Suppose that the solution of (25) can be
expressed by a polynomial in as and satisfies (8). The homogeneous balance between the highest order derivative and the nonlinear term appearing in (25) yields , and hence, we take the following
formal solution: where the positive integers and are to be determined later. Substituting (27) along with (8) into (25), collecting all the terms with the same power of , and equating each
coefficient to zero yield a set of simultaneous algebraic equations for , , , , , , and as follows: The above system admits the following sets of solutions: Now, substituting (29)-(30) into (27)
gives, respectively, When substituting the general solutions (9) into (32), we obtain the following three types of traveling wave solutions:
Case : (hyperbolic type)
If we set and write , then the above solutions can be written as Note that (35) is exactly the same solution of Khater and Hassan [35] as given in their first equation of (3.9) with . Similarly we
can obtain the second solution of (36) in Hassan [5] if we set in our solution (35). The solution (35) represents kink shaped solitary and antikink shaped solitary solutions (depending upon the
choice of sign) which are shown graphically in Figure 1 for the case :
Case : (trigonometric type)
But if and , then trigonometric type solution becomes
Case : (rational type)
As mentioned before, the -expansion method gives more general types of solutions than that found by Khater and Hassan [35] and Hassan [5].
3.3. The Modified Two-Dimensional KP (Kadomtsev-Petviashvili) Equation
The modified KP equation (3) containing a square root nonlinearity is a very attractive model for the study of ion-acoustic waves in plasma physic [8]. We will obtain more general exact solutions of
the modified KP equation. In order to find the traveling wave solution of (3), we let Now, taking , (3) becomes where , , , , and are constants and the prime denotes differentiation with respect to .
Integrating (41) with respect to and setting the integration constant equal to zero, we obtain Balancing with gives . Therefore, we can write the solution of (42) in the form where , , and are
constants to be determined later. Substituting (43) along with (8) into (42) and collecting all terms with the same order of , the left hand sides of (42) are converted into a polynomial in . Setting
each coefficient of each polynomial to zero, we derive a set of algebraic equations for , , , , , , , , , and as follows:
Solving this system by Maple gives two sets of solutions.
Case 1. We have Substituting the above case and the general solution (8) into (43) and according to (42), we obtain three types of traveling wave solutions of (3) as follows.
If , we have the hyperbolic typeIn particular, if , , , and , then becomes If , we have the trigonometric type So, the traveling wave solutions of (3) in this case are
Case 2. We have If , we have the hyperbolic typeHowever, if , , , and , then becomes If , we have the trigonometric type In the particular case when , , , and , becomes If , we have the rational type
where .
We remark that our results in (47) and (52), when , , , and , match those of Khater et al. [6] (2.19) when . In Figure 2, we plot the bell type solitary for 2D profile and the corresponding 3D plot
of (47) for parameters , and , , , and .
4. Conclusion
The -expansion was applied to solve the model of ion-acoustic waves in plasma physics where these equations each contain a square root nonlinearity. The -expansion has been successfully used to
obtain some exact traveling wave solutions of the Schamel equation, Schamel-KdV (S-KdV) equation, and modified KP (Kadomtsev-Petviashvili) equation. Moreover, the reliability of the method and the
reduction in the size of computational domain give this method a wider applicability. This fact shows that our algorithm is effective and more powerful for NLPDE. In all the general solutions (22), (
35), (47), and (52), we have the additional arbitrary constants , , , and . We note that the special case , , , and reproduced the results of Khater and Hassan [35], Hassan [5] and Khater et al. [6].
Many different new forms of traveling wave solutions such as the kink shaped, antikink shaped, and bell type solitary solutions were obtained. Finally, numerical simulations are given to complete the
Moreover, all the methods have some limitations in their applications. In fact, there is no unified method that can be used to handle all types of nonlinear partial differential equations (NLPDE).
Certainly, each investigator in the field of differential equations has his own experience to choose the method depending on form of the nonlinear differential equation and the pole of its solution.
So, the limitations of the -expansion method used a rise only when the equation has the traveling wave and becomes powerful in finding traveling wave solutions of NLPDE only.
In our future works, we can extend our method by introducing a more generalized ansätz , where , to solve Schamel equation, Schamel-KdV (S-KdV) equation, and modified Kadomtsev-Petviashvili (KP)
This work is financially supported by Universiti Kebangsaan Malaysia Grant: UKM-DIP-2012-31.
1. G. B. Whitham, Linear and Nonlinear Waves, Pure and Applied Mathematics, John Wiley & Sons, New York, NY, USA, 1974. View at MathSciNet
2. R. C. Davidson, Methods in Nonlinear Plasma Theory, Academic Press, New York, NY, USA, 1972.
3. H. Schamel, “A modified Korteweg de Vries equation for ion acoustic waves due to resonant electrons,” Journal of Plasma Physics, vol. 9, pp. 377–387, 1973.
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Graphical Gaussian Models for Genome Data
Notes: Graphical Gaussian Models for Genome Data
(back to index)
For software to efficiently identify GGM networks from data visit the
A simple method for inferring the network of (linear) dependencies among a set of variables is to compute all pairwise correlations and subsequently to draw the corresponding graph (for some
specified threshold). While popular and often used on many types of genomic data (e.g. gene expression, metabolite concentrations etc.) the naive correlation approach does not allow to infer the
dependency network. Instead, graphical Gaussians models (GGMs) should be used. These allow to correctly identify direct influences, have close connections with causal graphical models, are
straightforward to interpret, and yet are essentially as easy to compute as naive correlation models. This page lists pointers to learning GGMs from data, including procedures suitable for "small n,
large p" data sets (category iii).
Graphical Gaussian Models (GGMs), also known as "covariance selection" or " concentration graph" models, have recently become a popular tool to study gene association networks. The key idea behind
GGMs is to use partial correlations as a measure of independence of any two genes. This makes it straightforward to distinguish direct from indirect interactions. Note that partial correlations are
related to the inverse of the correlation matrix. Also note that in GGMs missing edges indicate conditional independence.
A related but completely different concept are the so-called gene relevance networks which are based on the "covariance graph" model. In the latter interactions are defined through standard
correlation coefficients so that missing edges denote marginal independence only.
There is a simple reason why GGMs should be preferred over relevance networks for identification of gene networks: the correlation coefficient is weak criterion for measuring dependence, as
marginally, i.e. directly and indirectly, more or less all genes will be correlated. This implies that zero correlation is in fact a strong indicator for independence, i.e. the case of no edge in a
network - but this is of course not what one usually wants to find out by building a relevance network... On the other hand, partial correlation coefficients do provide a strong measure of dependence
and, correspondingly, offer only a weak criterion of independence (as most partial correlations coefficients usually vanish).
The best starting place to learn about GGMs is the classic paper that introduced this concept in the early 1970s. (A.P. Dempster. 1972. Covariance Selection. Biometrics 28:157-175). Further details
can be found in the GGM books by J. Whittaker (1990) and by D. Edwards (1995).
Application of GGMs to Genomic Data:
Application of GGMs to genomic data is quite challenging, as the number of genes (p) is usually much larger than the number of available samples (n), and classical GGM theory is not valid in a small
sample setting. With this page I'd like to provide a commented list of some recent work dealing with GGM gene expression analysis (there are only very few so far). In my understanding, all of these
paper fit in one of three categories:
i. analysis with classic GGM theory,
ii. using limited order partial correlations, and
iii. application of regularized GGMs.
For small n, large p data it seems that methods from section iii. are most suited (see below for references and software).
I. Classic GGM Analysis:
The following papers simply apply classical GGM theory (i.e. with not further modification) to analyze gene expression data. It turns out that such an analysis is necessarily restricted to very small
numbers of genes or gene clusters as to satisfy n > p.
II. Limited Order Partial Correlations:
One way to circumvent the problem of computing full partial correlation coefficients when the sample size is small compared to the number of genes is to use partial correlation coefficients of
limited order. This results in something inbetween a full GGM model (with correlation conditioned on all p-2 remaining genes) and a relevance network model (with unconditioned correlation). This is
the strategy employed in the following papers:
1. A. de la Fuente, N. Bing, I. Hoeschele, and P. Mendes. 2004. Discovery of meaningful associations in genomic data using partial correlation coefficients. Bioinformatics 20:3565-3574.
2. A. Wille, P. Zimmermann et al. 2004. Sparse graphical Gaussian modeling of the isoprenoid gene network in Arabidopsis thaliana. Genome Biology 5:R92
3. P. M. Magwene and J. Kim. 2004. Estimating genomic coexpression networks using first-order conditional independence. Genome Biology 5:R100
4. A. Wille and P. Bühlmann. 2006. Low-order conditional independence graphs for inferring genetic networks. Statist. Appl. Genet. Mol. Biol. 4: 32.
5. R. Castelo and A. Roverato. A robust procedure for Gaussian graphical model search from microarray data with p larger than n. Preprint.
III. Regularized GGMs:
Another possibility (and in my opinion the statistically most sound way) to marry GGMs with small sample modeling is to introduce regularization and moderation. This essentially boils down to finding
suitable estimates for the covariance matrix and its inverse when n < p. This can either be done in a full Bayesian manner, or in an empirical Bayes way via variance reduction, shrinkage estimates
etc. Once regularized estimates of partial correlation are available then heuristic searches can subsequently to be employed to find an optimal graphical model (or set of models).
Outside a genomic context using regularized GGMs was first proposed by F. Wong, C.K. Carter, and R. Kohn. (2003. Efficient estimation of covariance selection models. Biometrika 90:809-830). For gene
expression data this strategy is pursued in the following papers:
2. In these papers a regularized estimate of the correlation matrix is obtained, either by Stein-type shrinkage (3) or by bootstrap variance reduction (2). This estimate is subsequently employed for
computing partial correlation. Network selection is based on false discovery rate multiple testing. This method is implemented in GeneNet.
5. This approach uses lasso regression to induce sparsity on a node level among the partial correlations.
6. These authors regularize the concentration matrix rather than the covariance matrix.
Please drop me me a line (korbinian.strimmer@lmu.de) for suggestions and comments. | {"url":"http://www.strimmerlab.org/notes/ggm.html","timestamp":"2014-04-21T09:53:13Z","content_type":null,"content_length":"14984","record_id":"<urn:uuid:13d66fbe-80d6-4b46-9bb8-20945463bb29>","cc-path":"CC-MAIN-2014-15/segments/1397609539705.42/warc/CC-MAIN-20140416005219-00195-ip-10-147-4-33.ec2.internal.warc.gz"} |
Algebra Help
Yeah, I am having problems with these 3 problems.
14. 6x^−2 y^−4 z
15. (r ^−5 )(s^4 )
16. (x^−3 y ^4 z^−2 ) ÷ (xy^2 z^−2 )
The teacher is saying, Simplify each expression. Do not use negative exponents. So, what do I do with those negatives?
Also, would it be against the rules to post the read of the questions That I have answered, so that somebody can check my work? | {"url":"http://mathhelpforum.com/algebra/104886-algebra-help.html","timestamp":"2014-04-18T10:09:58Z","content_type":null,"content_length":"37712","record_id":"<urn:uuid:3e8e4679-288f-4c31-9f9a-d0da52e5a2b4>","cc-path":"CC-MAIN-2014-15/segments/1397609536300.49/warc/CC-MAIN-20140416005216-00115-ip-10-147-4-33.ec2.internal.warc.gz"} |
Multiply two Numbers Without Using * Operator
Multiply two Numbers Without Using * Operator
Lets have a fun question. Multiply two numbers without using the * operator.
Here’s the code in C/C++:
while (~scanf("%d%d",&a,&b))
while (a)
if (a&1)
The above code is a implementation of an algorithm better known as the Ethiopian Multiplication or the Russian Peasant Multiplication.
Here’s the algorithm :
1. Let the two numbers to be multiplied be a and b.
2. If a is zero then break and print the result.
3. If a is odd then add b to the result.
4. Half a, Double b. Goto Step 2.
7 Responses
nice logic :).I hope that u will come up with some new posts too.
dear i guess this wont work if a is -ve no. so condition to check that must be applied before carrying out multiplication
here is same code with some modification
using namespace std;
int main()
int a=5,b=7,c=0,ch=5,flag;
{ flag=0;
if(a & 1)
cout<<"\n loop executed";
cout<<"\n *** multi plication of the 2 is : "<>a;
return 0;
2. number multiply by 2 with out using * operator
int num = 5;
num = num<<1;
Op : 10
Exp : num number left shift by the 1 position we can get.
Hi gaurav…
Nice coding :) . I enjoyed working around the codes tried an extended version of the code above.. can you suggest me which of the codes will take less resources…?
int a,b;
int m=0;
while (~scanf("%d%d",&a,&b))
while (a)
if (a&2)
if (a&1) m+=b+b+b;//remain 3
else m+=b+b;//remain 2
else if(a&1)//remain 1
//printf("%d,%d,%d\n",a,b,m); //to see intermediate values
As you can see that this algorithm can be extented to higher form in (multiple of 8,16,32,64..). which of the algorithm will be best suited for normal calculations??
[...] this I like Categories: Code LikeBe the first to like this post. Comments (0) Trackbacks (0) Leave a [...]
int x=10, y=10, z;
int mult = 0;
for ( z = 1 ; z <= x ; z++)
mult = mult + y;
printf("The multiplication of two numbers: %ld\n",mult);
return 0;
thanxx for such a easy way… | {"url":"http://comeoncodeon.wordpress.com/2010/04/08/multiply-two-numbers-without-using-operator/","timestamp":"2014-04-17T21:32:07Z","content_type":null,"content_length":"95555","record_id":"<urn:uuid:d27e143f-5857-468b-a357-3f1faeea0f10>","cc-path":"CC-MAIN-2014-15/segments/1397609532128.44/warc/CC-MAIN-20140416005212-00557-ip-10-147-4-33.ec2.internal.warc.gz"} |
parameterization and path integral
October 8th 2010, 01:26 PM
parameterization and path integral
Hey all, I have a question here and I can't seem to get around solving it.
Show that the circle that is the intersection of the plane x + y + z = 0 and the sphere x^2 + y^2 + z^2 = 1 can be expressed as
x(t) = (cos t - sqrt(3)sin t)/sqrt(6)
y(t) = (cos t + sqrt(3)sin t)/sqrt(6)
z(t) = - 2cos t/sqrt(6)
where 0<t<2π
And then, the question continues to...
Calculate the path integral where f(x,y,z)=xy-z and c is a path on the circle in the first part starting from the point (-1/sqrt(2),1/sqrt(2), 0) and finishing at the point (1/sqrt(2),-1/sqrt(2),
0), in the direction of increasing t.
Some help please?
October 8th 2010, 05:46 PM
Prove It
Well, they intersect where they are equal.
So $x+y+z+1 = x^2 + y^2 + z^2$
$1 = x^2 - x + y^2 - y + z^2 - z$
$1 + \left(-\frac{1}{2}\right)^2 + \left(-\frac{1}{2}\right)^2 + \left(-\frac{1}{2}\right)^2 = x^2 - x + \left(-\frac{1}{2}\right)^2 + y^2 - y + \left(-\frac{1}{2}\right)^2 + z^2 - z + \left(-\
$1 + \frac{1}{4} + \frac{1}{4} + \frac{1}{4} = \left(x - \frac{1}{2}\right)^2 + \left(y - \frac{1}{2}\right)^2 + \left(z - \frac{1}{2}\right)^2$
$\frac{7}{4} = \left(x - \frac{1}{2}\right)^2 + \left(y - \frac{1}{2}\right)^2 + \left(z - \frac{1}{2}\right)^2$.
Now work with the right hand side and substitute the suggested parameterisations for $x, y, z$. If you get the left hand side after some simplification, then this parameterisation is acceptable.
October 8th 2010, 06:04 PM
Another approach would be to use rotation matrices. You could assume that, since you have the intersection of a sphere of radius 1 with a plane going through the origin, that that intersection
must be described by a circle (locus of points equidistant from a point, and in a plane = circle). To get the desired parametrization, you could start with a parametrization of the circle of
radius 1 in the xy plane, and apply to that vector the rotations required to transform the z unit vector into a unit vector in the <1,1,1> direction. I think that would do the trick, more or
less. This approach at least has the virtue of being a direct approach.
October 9th 2010, 02:24 AM
Thank you for your inputs! But there are still some stuff I don't understand...
For the first answer by Prove It, I don't quite get why the (-1/2)'s are added to the equation. Also, I don't see how all that will simplify into 7/4 when t does not have an assigned value...
October 9th 2010, 03:03 AM
He doesn't add "-1/2", he adds $(-1/2)^2$. He is completing the square to get the equation of the intersection in standard form for a circle.
October 9th 2010, 06:05 PM
Okay, I understand the completing the square thing. But I still don't get how it gets me the parameterization required...
October 9th 2010, 06:21 PM
Prove It
You're not asked to derive the parameterisation, you're asked to verify it.
So you can substitute the suggested parameterised $x, y, z$ into the RHS.
If you can simplify and get back the LHS, then you can say the parameterisation is appropriate.
October 9th 2010, 06:58 PM
Okay I plugged the suggested x,y,z values into the RHS and got:
7/4 + 2sqrt(3)sin t/sqrt(6)
... which is the LHS with the extra second term. Does it still count as an appropriate parameterisation?
October 9th 2010, 06:59 PM
Okay I plugged the suggested x,y,z values into the RHS and got:
7/4 + 2sqrt(3)sin t/sqrt(6)
... which is the LHS with the extra second term. Does it still count as an appropriate parameterisation?
October 10th 2010, 04:51 AM
You plugged the suggested x, y, z values into the RHS of what? None of the equations you give has x, y, or z on the right hand side.
x(t) = (cos t - sqrt(3)sin t)/sqrt(6)
y(t) = (cos t + sqrt(3)sin t)/sqrt(6)
z(t) = - 2cos t/sqrt(6)
into the LHS of $x^2 + y^2 + z^2 = 1$ gives $\frac{(cos t- sqrt{3}sin t)^2}{6}+ \frac{(cos t+ \sqrt{3}sin t)^2}{6}+ \frac{4cos^2 t}{6}$$= \frac{cos^2 t- 2\sqrt{3}sin(t)cos(t)+ 3sin^2(t)+ cos^2(t)
+ 2\sqrt{3}sin(t)cos(t)+ 3sin^2(t)}{6}+ \frac{4cos^2 t}{6}$$= \frac{2cos^2 t+ 6 sin^2 t+ 4 cos^2 t}{6}= cos^2 t + sin^2 t= 1$ which is a true statement. The curve described by this
parameterization does lie on that sphere.
Putting it into the LHS of x+ y+ z= 0 we have
$\frac{cos t- \sqrt{3}sin t+ cos t+ \sqrt{3}sin t- 2 cos t}{\sqrt{6}}= \frac{2 cos t- 2 cos t}{\sqrt{6}}= 0$
so the curve also lies on that plane.
Another thing that you should check to sure this is "valid" parameterization is that there is no point where all of the derivatives of x, y, and z, with respect to the parameter, t, are
simultaneously 0. | {"url":"http://mathhelpforum.com/calculus/158849-parameterization-path-integral-print.html","timestamp":"2014-04-21T13:15:57Z","content_type":null,"content_length":"13591","record_id":"<urn:uuid:c46ff137-fc4a-49bb-a21a-bed462ee4c54>","cc-path":"CC-MAIN-2014-15/segments/1397609539776.45/warc/CC-MAIN-20140416005219-00595-ip-10-147-4-33.ec2.internal.warc.gz"} |
Short-time Existence/Uniqueness for Non-linear Schrodinger with Loss of Several Derivatives
up vote 1 down vote favorite
I have the following question about short-time existence and uniqueness results for non-linear schrodinger equations (NLSE) where the non-linearity involves a loss of derivatives (in my case, it is a
non-local non-linearity involving a loss of two derivatives.) It seems that most current techniques allow some small number of derivatives in the non-linearity, except for a series of papers by
Poppenberg, which use Nash-Moser. (Unfortunately he seems to have left math.)
Question: are there any short-time existence and uniqueness results for NLSE with loss of several derivatives for a space of initial conditions smaller than $C^\infty$? I have in mind for example a
space introduced by Floer which is a dense subspace of $C^\infty$ that carries the structure of a Banach space whose norm is a combination of all C^k norms. One could imagine doing something similar
by combining all Sobolev norms. I would be quite happy with a short-time existence and uniqueness result for initial conditions in some dense subspace of C^\infty.
p.s. I am posting in a second a related question about Floer's Banach space with a symplectic geometry tag.
Are you looking for an existence theorem that works with only finitely many derivatives? I believe it is possible to use Nash-Moser to start with $C^{k+N}$ (or a Sobolev analogue) initial data and
obtain a $C^k$ (or Sobolev analogue) solution. This might also give what you need for the Floer topology. You might want to look at Hormander's version of the Nash-Moser implicit function theorem.
– Deane Yang Aug 20 '10 at 16:50
add comment
1 Answer
active oldest votes
Just a few thoughts. The answer to your question depends on a number of key factors. To focus, let us consider a nonlinear term like $F(D_x^k u)$ and let us work in one space dimension.
1) How large is $k$? If $k\le2$ you can linearize the equation and work in Sobolev spaces. Of course you need some structural assumption on the nonlinearity (otherwise you may take e.g. $F
(u_{xx})=\pm 2i u_{xx}+...$, and create all sorts of difficulties). There are some classical works by Kenig, Ponce, Vega on this (see "The Cauchy problem for the quasilinear S.e." around
2002 I think) which more or less give the complete picture from the classical point of view i.e. without trying to push below critical Sobolev etc. So if this is the case what are you
exactly looking for? if you prefer to work in smaller spaces, what you need is a 'regularity' result, i.e., if the data are in some smaller space, this additional regularity propagates and
the solution stays in the same space for some time. There are some results of this type, in classes of analytical or Gevrey functions; but see below.
2) If $k\ge3$, then the same remark as in (1) applies, you need some strong structural assumptions on the nonlinearity. Indeed, now the poor $u_{xx}$ is no longer the leading term and the
character of the (linearized) equation is entirely determined by the Taylor coefficient of $D^ku$ in the expansion of $F$. So then a careful case-by-case discussion is necessary. Unless...
3) unless, and we come maybe closer to your question, you decide to work in MUCH smaller spaces than $C^\infty$. Gevrey classes are roughly speaking classes of smooth functions such that
the derivatives of order $j$ grow at most like $j!^s$. For $s=1$ you get analytic functions. For $ 1 < s < \infty $ you get larger classes $G^s$, with quite nice properties (to mention
just one, you have compactly supported functions in these classes). For $ 0 < s < 1 $ the classes $G^s$ are rather small, strictly contained in the space of analytic functions. The only
up vote 5 reason why $G^s$ for $s<1$ are useful is that you can prove a sort of very general Cauchy-Kowalewski theorem, local existence in Gevrey classes, for any evolution equation $u_t=F(D^k_xu)$,
down vote provided $s<1/k$. No structure is required on $F$, only smoothness. Contrary to the appearance this is a weak result, e.g. you can solve locally both $u_t=\Delta u$ and $u_t=-\Delta u$ in
accepted $G^{1/2}$ (and globally in $G^s$ for $s<1/2$ ), so you are essentially trivializing the equation and forgetting all of its structure. But if this is what you need I can give you pointers.
EDIT (since this does not fit in the comments): I am a bit rusty on these topics, now I start to remember more details.
1) $s<1$. There is a problem with working in $G^s$ for $s<1$, and it is that this space is unstable for product of functions. So also in this case you need very special nonlinearities to
work with. But the linear theory is straightforward and you can find an account maybe here. Also Beals in some old paper developed semigroup theory in Gevrey classes. BTW, if you find a
way to handle products this might be quite interesting.
2) $s>1$. The Mizohata school and other japanese mathematicians worked on NLS in Gevrey classes quite a lot, see e.g. this paper
3) The Nash-Moser approach might be useful provided you can prove that the linearized of your operator is solvable in every Sobolev class, with a fixed loss of derivatives. If you want to
try this route, the best introduction to the theory I know of is in Hamilton's 1992 (?) Bull. AMS paper. It's very long but extremely readable, give it a try.
Thanks Piero and Deane! (1) In my particular equation the potential is obtained from an expression involving the fourth derivatives of the square of the wave-function, by solving a
second-order Poisson equation. So if $\psi$ is in $H^s$, then $V(\psi)$ is in $H^{s-2}$ for s sufficiently large, So in this sense $k=2$. But $V(\psi)$ depends non-locally on $\psi$ in
much the same sense as in the paper by Ginibre-Velo on non-local interaction in non-linear Schrodinger. – Chris Woodward Aug 20 '10 at 21:48
(ctd) So it seems my problem is closest to (3) in Piero's answer, and in fact it seems what I need to do is try to find for which Gevrey class of initial conditions is good. Do you have
a recommended reference involving Gevrey initial conditions with loss of derivatives in non-linear Schrodinger? I would be very happy to show uniqueness for $G^s$ initial condition for
some $s > 1$. – Chris Woodward Aug 20 '10 at 21:48
add comment
Not the answer you're looking for? Browse other questions tagged ap.analysis-of-pdes or ask your own question. | {"url":"http://mathoverflow.net/questions/36190/short-time-existence-uniqueness-for-non-linear-schrodinger-with-loss-of-several/36196","timestamp":"2014-04-19T22:12:58Z","content_type":null,"content_length":"59032","record_id":"<urn:uuid:2cc3778f-0efc-4bb2-b6d1-f4cd9493dd7e>","cc-path":"CC-MAIN-2014-15/segments/1397609537754.12/warc/CC-MAIN-20140416005217-00338-ip-10-147-4-33.ec2.internal.warc.gz"} |
Fitting Experimental Data with
Fitting Experimental Data with Mathematica
by Ian Brooks
Download this example as a Mathematica notebook.
Analysis of biomedical data usually involves fitting the measured data to a model in order to make some predictions about the system under investigation. Choosing the right approach to experimental
curve fitting is important if you are to get good results. Sometimes it may be sufficient to use an interpolation method, but often it will be necessary to use nonlinear fitting and all of its
associated diagnostic statistics. For example, kinetic data may be fit to a Michaelis-Menten model in order to gain insight into the efficiency of an enzyme, or data from a baseline control
experiment may be fit to a quadratic in order to predict the correction necessary for the full experiment. While these two examples may seem very similar, they are in fact very different, and it is
important to understand how the differences affect the analysis.
These differences lead to different procedures for performing the fit and to differences in the degree of care that must be taken to ensure good results. Another difference is that in the case of the
Michaelis-Menten example each of the parameters being determined has a definite biological significance, while there is no biological significance to the parameters in the quadratic case. Clearly it
is critical to check the determined parameters for physical reality, good confidence intervals, and so on if they are expected to have biological significance, but it is unimportant if you just want
an approximate curve. The biggest difference though is that the Michaelis-Menten example is a nonlinear model, while the quadratic example is a mathematically linear model.
Linear or Nonlinear?
If you have to do a full curve fit, the critical question becomes whether a model is linear or nonlinear. Unfortunately, this is not always immediately obvious (although, as a general rule of thumb,
biology is nonlinear). A nonlinear model can be defined as one in which at least one of the partial derivatives of the model with respect to its parameters includes at least one of the parameters.
Use of the symbolic capabilities of Mathematica allows the determination of linearity to be done quickly without having to remember any calculus. (See Figure 1.)
Given some data and a linear model, the fitting algorithm solves certain linear equations to determine the best-fit parameters and their confidence intervals. With a nonlinear model, however, it is
not possible to solve the equations directly and so the fitting algorithm has to perform a search. With any search it is possible to get different results depending on where you start and when you
decide to stop. Nonlinear fitting is a little like being placed in the middle of a mountain range and trying to find the lowest point. The point you think is the lowest is dependent on where you are
when you start and how long you are prepared to look. Unless you search the entire area, no matter where you stop, you cannot be certain that you have found the lowest point in the range or just the
lowest point that you have seen. In fitting terms, you can never be sure that you have found the absolute best fit, just that you have found a good fit; worse, you have to make sure that you even
have a good fit. Although this is true with any software, Mathematica's advanced algorithms and extended precision capability enable it to perform more reliably on curve-fitting benchmark tests than
any other program tested so far (McCullough).
Because of these fundamental differences between linear and nonlinear fitting, starting guesses and convergence checking are important for nonlinear models but are irrelevant to linear models. The
fitting algorithm in Mathematica is powerful enough to solve many problems without needing good starting guesses, but, even so, there are inevitably some problems where they are necessary.
Mathematica's algorithm also eliminates the need to consider effects such as parameter scaling by handling a number of these issues invisibly.
Whether you need to perform a complex nonlinear fit or can make do with a simple interpolation, Mathematica provides the functionality and reliability to tackle any fitting task. This flow chart
shows the four main approaches and the Mathematica functions that are most appropriate.
If you are simply trying to obtain a standard curve for comparison or for baseline subtraction, and you have plenty of data to describe the line or curve, it is simplest to use Mathematica's
Interpolation function. Interpolation constructs a curve that gives a good approximation to the data. As a simple example, consider trying to find an approximation to y = Sqrt[x].
data = {{0.0, 0.0}, {1.0, 1.0}, {4.0, 2.0}, {9.0, 3.0}, {16.0, 4.0}, {25.0, 5.0}, {36.0, 6.0}, {49.0, 7.0}, {64.0, 8.0}, {81.0, 9.0}, {100.0, 10.0}};
sq = Interpolation[data]
InterpolatingFunction[{{0., 100.}}, <>]
The returned InterpolatingFunction can be used to calculate the expected values of any input x, but will be most accurate within the specified range, in this case between 0 and 100.
Compare this result with the expected value of Sqrt[50].
InterpolatingFunction will give the exact value of y at the x values used to construct the approximation.
Linear Fitting
Linear fitting is performed in Mathematica with the Fit and Regress commands. If Interpolation is not appropriate for some reason, but you are not trying to determine biologically significant
parameters, then the Fit command (which simply returns the best-fit function) should be sufficient.
As an example, consider fitting the square root data to a quadratic.
Fit[data, {1, x, x^2}, x]
1.01599 + 0.166593x - 0.000801884x^2
If you need diagnostic information to determine the quality of the fit or confidence intervals of the parameters, then it will be necessary to use the function Regress available in the standard
add-on package Statistics`LinearRegression`.
<< Statistics`LinearRegression
Regress[data, {1, x, x^2}, x]
This output is the default. In short, TStat should be high, PValue should be below 0.05, RSquared should be close to 1, and the lower EstimatedVariance, the better. Many more diagnostics can be
generated, and a more complete example is shown in Figure 2.
A complete list of the diagnostics is as follows:
{AdjustedRSquared, ANOVATable, BestFit, BestFitParameters, BestFitParametersDelta, CatcherMatrix, CoefficientOfVariation, CookD, CorrelationMatrix, CovarianceMatrix, CovarianceMatrixDetRatio,
DurbinWatsonD, EigenstructureTable, EstimatedVariance, FitResiduals, HatDiagonal, JackknifedVariance, MeanPredictionCITable, ParameterCITable, ParameterConfidenceRegion, ParameterTable,
PartialSumOfSquares, PredictedResponse, PredictedResponseDelta, RSquared, SequentialSumOfSquares, SinglePredictionCITable, StandardizedResiduals, StudentizedResiduals, SummaryReport,
Nonlinear Fitting
Good nonlinear curve fitting is an art form. Even with the best algorithms, it is often necessary to provide accurate starting guesses for parameters, and it is always necessary to check the results
thoroughly. Fortunately, it is easy to provide starting guesses in Mathematica, and a very extensive list of fit diagnostics may be generated.
<< Statistics`NonlinearFit`
data = {{1.0, 1.0, 0.126}, {2.0, 1.0, 0.219}, {1.0, 2.0, 0.076}, {2.0, 2.0, 0.126}, {0.1, 0.0, 0.186}};
This data set (Meyer and Roth, 1972) gives five values for x1, x2, and y, describing a reaction involving the catalytic dehydration of n-hexyl alcohol. Here x1 is partial pressure of alcohol, x2 is
olefin, and y is the rate of reaction.
model = (t1 t3 x1)/(1 + t1 x1 + t2 x2);
NonlinearRegress[data, model, {x1, x2}, {t1, t2, t3}]
In short, the lower EstimatedVariance is, the better the fit, and the smaller the range of the confidence intervals (CI), the better. Of particular interest is FitCurvatureTable, which may be used to
gauge the reliability of the linear approximation confidence intervals. If the linear approximations are reasonable, the maximum intrinsic curvature will be small compared to the confidence region.
As in the linear case, the default behavior shows only a few of the many diagnostic statistics that may be generated. A more complete example is shown in Figure 3.
{ANOVATable, AsymptoticCorrelationMatrix, AsymptoticCovarianceMatrix, BestFit, BestFitParameters, EstimatedVariance, FitCurvatureTable, FitResiduals, HatDiagonal, MeanPredictionCITable,
ParameterBias, ParameterConfidenceRegion, ParameterCITable, ParameterTable, PredictedResponse, SinglePredictionCITable, StartingParameters, StandardizedResiduals, SummaryReport}
Getting good results from curve fitting is not as simple as having good data and the right model. It is also important to choose the right algorithm. When performing nonlinear fitting, it is critical
to take notice of all of the information that is provided and check that the fit is really a good one that makes physical sense. Remember that the algorithm being used knows nothing about biology,
and it is up to you to accept or reject its results based on your knowledge.
Further examples of data analysis with Mathematica may be found at http://www.wolfram.com/solutions.
McCullough, B. D. "The Accuracy of Mathematica 4 as a Statistical Package," Journal of Computational Statistics (2000). (accepted)
(Ian Brooks is an applications developer at Wolfram Research.) | {"url":"http://library.wolfram.com/examples/curvefit/","timestamp":"2014-04-20T10:57:57Z","content_type":null,"content_length":"32605","record_id":"<urn:uuid:84c5d2ef-2bf3-48e6-bccb-1ae705760359>","cc-path":"CC-MAIN-2014-15/segments/1397609538423.10/warc/CC-MAIN-20140416005218-00547-ip-10-147-4-33.ec2.internal.warc.gz"} |
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Congruent Supp. and Comp.
September 25th 2007, 06:21 PM #1
Sep 2007
The measure of the supp. of an angle exceeds 3 times the measure of the comp. of the angle by 10. Find he measure of the comp.
I do not understand this equation, please help me to understand this and how you get the answer.
supplementary angles are angles that add up to 180 degrees, complimentary angles are angles that add up to 90 degrees.
so, if the original angle is $x$ then it's supplementary angle is $(180 - x)$ and it's complimentary angle is $(90 - x)$.
we are told that the supp. angle is 10 more degrees than 3 times the comp. angle. thus we have:
$180 - x = 3(90 - x) + 10$
i think you can take it from here
Thank you very much for the help...
Maybe if I get stuck on another geometry problem, you could help me.
Thank you for your help.
September 25th 2007, 06:55 PM #2
September 26th 2007, 03:40 PM #3
Sep 2007 | {"url":"http://mathhelpforum.com/geometry/19514-congruent-supp-comp.html","timestamp":"2014-04-18T01:41:49Z","content_type":null,"content_length":"36053","record_id":"<urn:uuid:03293626-7840-49d4-831a-7cdec517138b>","cc-path":"CC-MAIN-2014-15/segments/1397609532374.24/warc/CC-MAIN-20140416005212-00006-ip-10-147-4-33.ec2.internal.warc.gz"} |
Archives of the Caml mailing list > Message from skaller
Date: -- (:)
From: skaller <skaller@u...>
Subject: Re: [Caml-list] Coinductive semantics
On Sat, 2006-01-21 at 19:36 +0100, Andrej Bauer wrote:
> skaller wrote:
> > How would you decode an Andrej sum without a conditional
> > control transfer?
> The control is a consequence of the fact that natural numbers are an
> inital algebra
Hmm ..
> rec x f 0 = x
> rec x f (n+1) = f (rec x f n)
Yeah. Of course, natural numbers subsume bool: the switch
is right there already. Thanks!
FYI in Felix 0,1,2 .. etc are types given by n = 1 + 1 + .. + 1.
Not quite natural numbers (since + isn't strictly associative).
For any sum t a variant has notation like
case 5 of t
in particular the components of bool = 2 are names
false = case 1 of 2 and true = case 2 of 2. The underlying
representation is a tagged pointer, with the pointer optimised
away if no variant has a (non-unit) argument. Since the
tag is an int, the run time representation of a unit sum
is just an int.
I also have arrays t ^ n = t * t * t * .. * t.
In principle, the j'th component is found by indexing
prj: t ^ n * n -> t
Unlike 'normal' array indexing .. this is entirely safe.
BTW I learned this trick from Pascal, which never needs
array bounds checks -- you have to check conversion
int ==> n of the index instead.
What I don't know how to do is extend the collection of
fixed length array types, kind of like
(lim n-> inf) Sum (t ^ n)
Of course I know how to *implement* this infinite sum.
(Which in practice isn't infinite of course)
The problem is that I have to somehow convert from
a run time dynamic type (the array bound is a type!)
to an integer with a 'super switch'. Rougly the
projection for the above limit is
typematch .. with ..
| 5 => let j = check 5 i in a.[j]
in other words its an infinite switch over all the fixed
length array types. Clearly for this particular case
the implementation is just to do an dynamic array bounds
check the way Ocaml does.
But this is very unsatisfying, and in particular provides
no way to optimise away gratuitous checks at compile time.
In other words, you can't compose these things.
I'm not sure my rant really explains what I'm looking for:
basically I don't know enough theory to turn the purely
algebraic finite array types -- fairly useless in practice --
into variable length arrays required in practice.
[By variable I mean the length is an immutable dynamic type,
not that you can change the length of the array]
It would be really nice -- in BOTH Felix and Ocaml --
to have safe arrays. Unlike lists, functions like
combine and split then make sense.
John Skaller <skaller at users dot sf dot net>
Felix, successor to C++: http://felix.sf.net | {"url":"http://caml.inria.fr/pub/ml-archives/caml-list/2006/01/25f4d3ae25d0e3865543d33c31e5a0ee.en.html","timestamp":"2014-04-18T08:31:37Z","content_type":null,"content_length":"15963","record_id":"<urn:uuid:e4a60063-fc33-40da-8afb-996a3c87ee58>","cc-path":"CC-MAIN-2014-15/segments/1398223205375.6/warc/CC-MAIN-20140423032005-00556-ip-10-147-4-33.ec2.internal.warc.gz"} |
Re: First moment of spectrum (Paul Boersma )
Subject: Re: First moment of spectrum
From: Paul Boersma <paul.boersma(at)HUM.UVA.NL>
Date: Tue, 13 Jun 2000 22:08:24 +0200
> I am amazed that no phonetician has so far bothered to weigh in in this
> debate, and I have not seen any psychoacoustic concerns addressed, either.
>Prey tell, what would the first moment tell about an /i/ or /u/ vowel?
[i] has the highest, [u] the lowest first moment of all vowels.
But a better single-valued approximation of these vowels is
the one-formant perceptual equivalent, which is 4000 Hz for /i/ and 300 Hz for /u/.
We can use the first moment, however, to nicely distinguish the retroflex sibilant,
the palatalized palatoalveolar sibilant, and the alveolar sibilant in Polish
(e.g. 3300 Hz, 5000 Hz, and 7400 Hz, respectively).
By the way, the following is a script to compute the first moment (magnitude-wise)
of a selected Spectrum object in the Praat program:
nbin = Get number of bins
Copy... sum
Copy... weightedSum
select Spectrum sum
# Integrate the rms power:
Formula... sqrt (self[1,col]^2 + self[2,col]^2) + self [col-1]
sum = Get real value in bin... nbin
select Spectrum weightedSum
# Integrate the rms power, weighted by frequency:
Formula... x * (sqrt (self[1,col]^2 + self[2,col]^2)) + self [col - 1]
weightedSum = Get real value in bin... nbin
firstMoment = weightedSum / sum
select Spectrum sum
plus Spectrum weightedSum
echo 'firstMoment' Hz
Of course, the values obtained may depend rather critically on the
chosen Nyquist frequency, i.e. white noise has an infinite first moment.
Paul Boersma
Institute of Phonetic Sciences, University of Amsterdam
Herengracht 338, 1016CG Amsterdam, The Netherlands
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English 10 --- Check Please
Number of results: 226,370
Now, let's check how well you can use the expressions you've learned. I will distribute handouts. You have to solve the problems. Are you finished? Then, I'll check the answers. Look at the answers
on the blackboard. Check your answers. How many students have 10 points? Are ...
Sunday, May 27, 2012 at 9:36pm by rfvv
English check please help me out
3 - 10 are fine. Please repost 1-2 with the correct words and endings typed in. It doesn't make sense to me as it is.
Wednesday, September 14, 2011 at 1:35pm by Writeacher
math log PLEAseee helpp pretty please!!!
solve for first x 10^y=(x-2) x= 10^y +2 solve for the second x y-1=log(x+1) 10^(y-1)=x+1 or x= 10^(y-1) +1 the x=x, y=y 10^(y-1)+1=10^y +2 remember a^(b-c)= a^b/a^c so 10^y/10 + 1=10^y +2 10^y (-9/
10)=1 or -9*10^y=10 log of each side -9 +y=1 y=10 so, now figure x.. x= 10^(y-1...
Thursday, January 20, 2011 at 1:00pm by bobpursley
English check please I've corrected it again
4.enhance 5.discern Please check. If its not correct can you give me anotther hint please.
Wednesday, September 14, 2011 at 1:12pm by ME
English 7 - Please Check My Journal Entry
Please check my journal entry. This is a HUGE part of my grade for english. Is it good, any grammer mistakes. Please read it.
Thursday, February 2, 2012 at 5:38pm by Laruen
English 10 --- Check Please
Okay, i will next time...But i promise i didn't plagiarize....
Tuesday, February 14, 2012 at 1:36pm by Mackenzie
Maths;Check Answers Please :P
1-5 , all good #6. (9 × 10^–3)^2 = (9 × 10^–3)(9 × 10^–3) = 81 x 10^-6 = 8.1 x 10^-5 #8 , ok #9 (4xy^2)^3(xy)^5 = (64x^3 y^6)(x^5 y^5) = 64 x^8 y^11 10. (2t^-3)^3 (.4r)^2 = 8 t^-9 (.16r^2 = 12.8 r^2
/ t^9
Friday, January 17, 2014 at 6:37pm by Reiny
3, 5, 6, and 10 are correct. Please check a dictionary for the meanings of these words and try again.
Thursday, December 6, 2012 at 4:30pm by Ms. Sue
TO Ms. Sue
If you have do english tutoring, when you have a chance can you review my "English 10 HELP" posting please? Writeacher helped me but he said to post the answers I chose so someone could check them
and help me further.
Sunday, June 13, 2010 at 5:41pm by Ariel
SS Please check answers
1, 8, and 10 I believe are wrong. I don't know about 9. Please double check your book.
Monday, April 8, 2013 at 5:37pm by Ms. Sue
English-CHECK PLEASE
Ok, I figured this out (I think) but can you please check? Don't worry. I can carry the tent all by myself because it is so light. I do not even need a lantern it is so light outside. In these
sentences, the word "light" is used as a: A:homophone B:homonyme C:synonym D:simile ...
Friday, April 15, 2011 at 6:03pm by Catherine
Write each number in expanded form using exponents. Pls check 1, 234 (1*10^3)+(3*10^2)+(2×10^1)+4 210, 290, 450 (2*10^8)+(1*10^7)+(2*10^6)+(9*10^5)+(4*10^4)+5 7,123,456,089 (7*10^9)+(1*10^8)+(2*10^7)
Monday, September 16, 2013 at 8:23pm by Gio
Help me please I get confused with the positives and negatives. Please be kind to check my work and let me know where am I going wrong -4x + 10 = 58 -4x + 10 -10 = 58 -10 (is it - or +10?) -4x = 48
or 68 (do i add 58&10 or sub?) -4x/-4 = 48/-4 or 68/-4 (is the 4 -or+) x=12 or ...
Saturday, November 13, 2010 at 10:47am by thawk
PLease Check My Work
3. T^2-25/t^2+t-20= (t-5) (t+5)/ (t-10) (t+2) 4. 2x^2+6x+4/4x^2-12x-16= 2(x+2) (x+1)/4 (x-4) (x+1) 5. 6-y/y^2-2y-24= 1/(y+4) *Please check them.- 3. (t-5)(t+5) / (t+4)(t-5) 4. 2((x-1)(x+2)) / 4((x+4)
(x-1)) 5. (6-y) / (y+6)(y-4) i dont really understand what this is... but lol ...
Thursday, February 22, 2007 at 7:31pm by Margie
procedure: Formative Test Now. let's check how well you can use the expressions you've learned. I will distribute handouts. You should pick up a handout on the table. You have to fill in the blanks
with suitable expressions. Are you finished. Two minutes left. We'll check the ...
Monday, June 9, 2008 at 6:58am by John
English 10
Which of the following is a possible setting for works of American literary realism? CHECK ALL THAT APPLY! *a. American West *b. Post-Civil War South c. New England d. Gold Coast I think there is one
more that needs to be check, but I don't know which one it is. Can you please...
Tuesday, December 10, 2013 at 6:15pm by Cassie
Math HELP
I have my answers at the bottom but please do these questions so I can check my answers. Check my answers as well. Subtract these fractions. !Reduce them! 1. 5/7 - 3/10= 2. 11/18 - 5/12= 3. 1/3 - 1/6
= 4. 7/12 - 1/16= 5. 7/8 - 3/10= 6. 7/8 - 1/12= 7. 7/10 - 2/5= 8. 3/4 - 1/2= ...
Tuesday, November 23, 2010 at 9:16pm by Pratigya
math - scientific notation
please check. thanks. (8.3 x 10^-15) (7.7 x 10^4) = 6.391 x 10^-10 Is that right?
Thursday, August 21, 2008 at 5:03pm by anonymous
You are going to take a quiz. I'll distribute exam papers. You should write down the answers. Fill in the blanks with suitable expressions. Take one and pass the others to the other students. First
write down your number and name on the top of the paper. You have 3 minutes ...
Tuesday, June 10, 2008 at 9:27am by John
English Essay Writing MS Sue please check
that ws great but try not to make mistakes and you will be a level 10 writter
Wednesday, May 14, 2008 at 2:11pm by permicouas
english please check
this are my Answer 1.Elicit 2.Engross 3.obsolete 4.Terminate 5.Tangible 6.Acclaim 7.Exploit 8.Escalate 9.Methodical 10. Adjacent
Monday, August 30, 2010 at 5:34pm by jae
Math 157
Can someone please help me. At a quality control checkpoint on a manafacturing assembly line, 10% of the items failed check A, 12% failed check B, and 3% failed both checks A and B. a. If a product
failed check A, what is the probability that it also failed check B? b. If a ...
Wednesday, December 8, 2010 at 9:51pm by Rena
Can someone please help me. At a quality control checkpoint on a manafacturing assembly line, 10% of the items failed check A, 12% failed check B, and 3% failed both checks A and B. a. If a product
failed check A, what is the probability that it also failed check B? b. If a ...
Wednesday, December 8, 2010 at 8:37pm by Rena
Check this problem for me do I have negative sign right with the answer. -1(-1)+-5-5=-2/-10==-1/5 the one on top and the -1(-1) and on the bottom is -5-5. so check this answer for me. Please, Please
Saturday, March 7, 2009 at 12:08am by Elaine
Check this problem for me do I have negative sign right with the answer. -1(-1)+-5-5=-2/-10==-1/5 the one on top and the -1(-1) and on the bottom is -5-5. so check this answer for me. Please, Please
Saturday, March 7, 2009 at 12:08am by Elaine
English 10 --- Check Please
I would give you the link to prove it to you, but it won't let me. I'm not a thief. I have never plagiarized, and won't ever. The above comment was mine too..
Tuesday, February 14, 2012 at 1:36pm by Mackenzie
please check. thanks. (8.3 x 10^-15) (7.7 x 10^4) = 6.391 x 10^-10 Is that right?
Thursday, August 21, 2008 at 9:48pm by anonymous
English please check my answer
please check to see if I have the right answer thank you The patient had a radiogram performed. What doed the root word indicate I think it's b
Thursday, March 20, 2008 at 5:59pm by Dakotah
please check my answer An electron moves north at a velocity of 9.8 * 10^4 m/s and has a magnetic force of 5.6 * 10^-18 N west exerted on it. If the magnetic field points upward, what is the
magnitude of the magnetic field? F=BqV F= 5.6 * 10^-18N q= 16 * 10^-19C V= 9.8 * 10^4 ...
Thursday, March 19, 2009 at 11:42am by physics
English 8R - Homework Check
it said AT LEAST but the question said to write a paragraph..... oops I forgot one detail!!! I'm going to check it, meanwhile please reply my other post (health and english)
Monday, November 12, 2012 at 5:48pm by Laruen
Pre-Calc-Please check my answer-I think its correc
Could someone please check these-I had tried one last night but it was really wrong so I'm been working-hopefully these are correct. What are the roots of the equation and answers have to be
simplified 2x^2 + 5x-10 =0 2x^2 + 5x-10+10=0+10 2x^2 + 5x = 10 2x^2 + 5x/2 = 10/2 2x^2...
Wednesday, September 7, 2011 at 1:47pm by Sheila
Ms.Sue English check
I'd be glad to check it. But first, please proofread it carefully yourself and run it through a spell check. Then post it. My practice is to stop reading after I find three careless mistakes.
Friday, July 31, 2009 at 5:59pm by Ms. Sue
Algebra 1: 8th Grade
Simplify each expression. Write your answers using positive exponents. Check Answers Please: 1) 2^3*2^4 = 2^7 2)10^6/10^9 = 1/10^3 3)3^5*3^1 = 3^6 4)5^7*5^-4 = 5^3 5)9^3/9^-10 = 9^13 6) 4^11/4^8*4^-2
= 4^5 7) a^2*a^3 = a^5 8)b^-9/b^4 = 1/b^13 9)c^-5*c^-2 = 1/c^7 11)(5d^4)(8d^6...
Tuesday, April 28, 2009 at 12:35am by Joy
PreAP English Grade 10
We'll be glad to help you. Please post your questions and what you think the answers may be. We'll be glad to check and critique your thinking.
Monday, October 8, 2007 at 8:45pm by Ms. Sue
Please check for any errors. If there are some errors please correct them for me. When you were in my office last Tuesday, you stated that you wanted to alter your will to include two additional
heirs. I have now prepared a will in compliance with your instructions. Please ...
Wednesday, March 2, 2011 at 9:34pm by shay
Please check for any errors. If there are some errors please correct them for me. When you were in my office last Tuesday, you stated that you wanted to alter your will to include two additional
heirs. I have now prepared a will in compliance with your instructions. Please ...
Wednesday, March 2, 2011 at 9:37pm by shay
Please check for any errors. If there are some errors please correct them for me. When you were in my office last Tuesday, you stated that you wanted to alter your will to include two additional
heirs. I have now prepared a will in compliance with your instructions. Please ...
Wednesday, March 2, 2011 at 9:37pm by shay
frequency = speed/wavelength = c/L = (3*10^8 m/s) / .1*10^-9 m = (3/.1) * 10^17 s = 3 * 10^18 Hz check k = 2 pi/L k = 2 pi / (.1*10^-9) = 6.28 * 10^10 check
Monday, March 31, 2008 at 6:14pm by Damon
7th grade math check my answers please quick
11. 1.72*10^4 10,000 is 10^4 you only have to move the decimal point 4 places to the left to get a number less than 10. 21. correct
Friday, September 28, 2012 at 6:32pm by Steve
A 10.0-kg cylinder rolls without slipping on a rough surface. At the instant its center of amss has a speed of 10m/s, determine (a) the translational kinetic of its center of mass (b) the rotational
kinetic energy about its center of mass and (c) its total kinetic energy. can ...
Monday, January 29, 2007 at 4:18pm by cyndi
Pb3(PO4)2 <==> 3Pb^+2 + 2PO4^-3 Since [Pb3(PO4)2] = 2.3 x 10^-9 (Pb^+2) = 6.9 x 10^-9 (PO4^-3)= 4.6 x 10^-9 Ksp = (Pb^+2)^3*(PO4^-3)^2 = (6.9 x 10^-9)^3*(4.6 X 10^-9)^2 =(3.2850 X 10^-25)(2.116 X 10^
-17) =6.951 X 10^-42 Please check if my answer is correct. It seems...
Saturday, April 19, 2008 at 11:07am by Anna
Physics please check
Use the density of water as 210*10^-6 so (210*10^-6)(0.5*10^-3)(90-19) = 7.455*10^-6 or 0.000007455
Friday, April 13, 2007 at 10:17pm by Summer
English Check
thanks. Here is the rest of the answer please check: 3.fluctuate 4.turmoil 7.venture
Wednesday, September 14, 2011 at 12:31pm by ME
what is the value of the expression (4.32 x 10^-3) + (6.025 x 10^-3) ? i came up with either 0.010345*10^-3 or 10.345*10^-3 and when i check them they're both wrong.
Thursday, February 24, 2011 at 7:34pm by angiie
English (Check Please)
Someone will be glad to check your answers.
Monday, April 8, 2013 at 12:11pm by Ms. Sue
The classroom has 10 short/low chirs, 10 tall/high chairs, 10 small desks, 10 big desks, 5 windows, and 20 lockers. (Did I use correct adjectives? Would you check them?)
Wednesday, April 3, 2013 at 5:11pm by rfvv
English 10 --- Check Please
No, no its for a class, and i was supposed to correct the errors, and i did. And i corrected it, and posted it on here to see if all the errors where out! I didn't plagiarize, i promise!
Tuesday, February 14, 2012 at 1:36pm by Anonymous
pH = 6 = (H^+) = 1 x 10^-6M and OH^- = 1 x 10^-8 You want pH = 7 = (H^+) = 1 x 10^-7 and OH^- = 1 x 10^-7 In 1 L you have 1 x 10^-8 moles and you want 1 x 10^-7 moles so you must add 1 x 10^-7 - 1 x
10^-8 = 9 x 10^-8 mols and that times 6.022 x 10^23 OH^- ions/mole = number of...
Tuesday, September 21, 2010 at 10:21pm by DrBob222
Pre-Calc-Please check my answer-I think I have it
Could someone please check these-I had tried one last night but it was really wrong so I'm been working-hopefully these are correct. What are the roots of the equation and answers have to be
simplified 2x^2 + 5x-10 =0 2x^2 + 5x-10+10=0+10 2x^2 + 5x = 10 2x^2 + 5x/2 = 10/2 2x^2...
Wednesday, September 7, 2011 at 4:07pm by Sheila
Pre-Calc-Please check my answer-I think I have it
Could someone please check these-I had tried one last night but it was really wrong so I'm been working-hopefully these are correct. What are the roots of the equation and answers have to be
simplified 2x^2 + 5x-10 =0 2x^2 + 5x-10+10=0+10 2x^2 + 5x = 10 2x^2 + 5x/2 = 10/2 2x^2...
Wednesday, September 7, 2011 at 4:07pm by Sheila
Math PLEASE HELP
Please finish these questions so I can check my answers. All my answers are at the bottom. Adding and subtracting mixed numbers. Please finish these questions. The answer must be reduced to lowest
terms! 1. 3 2/5 + 6 1/5= 2. 4 2/5 + 1/10= 3. 2 3/12 + 2 3/6= 4. 5 3/9 + 2 1/3= 5...
Wednesday, December 1, 2010 at 9:06pm by Pratigya
language art CHECK
Will somebody please check this for me and see if i used the words correctly please? Having spent many years as political opponents, the two senators have developed a(n)MUTUAL respect for each other.
Please check if it is correct. thank you
Tuesday, January 17, 2012 at 8:51pm by saranghae12
This is an awful lot to check. Please indicate 10 or less that you're really unsure about and post them. Then someone will check those for you.
Saturday, May 25, 2013 at 7:11pm by Ms. Sue
Plz -- Chem Help
What I got is a bit different... Can you please check my work? [H+] = 10^-3.39 = 4.074*10^-4 = [C6H4NO2] [HC6H4NO2] = 0.012-4.074*10^-4 = 0.0116 M Ka = [4.074*10^-4]^2/0.0116 = 1.43*10^-5 pKa =
-logKa = 4.84
Sunday, March 15, 2009 at 3:35pm by Ed
Algebra-Please check
Write y = 10^(x) as a logarithmic function Choices are log(x)10 = y log(y)x = 10 log(10)y = x log(x)y = 10 I think it is log (10)y = x
Tuesday, November 16, 2010 at 10:48am by Annie
English? please check
What do you want us to check?
Monday, August 30, 2010 at 5:19pm by Ms. Sue
The classroom has 10 short/low chairs, 10 tall/high chairs, 10 small desks, 10 big desks, 5 windows, and 20 lockers. (Did I use correct adjectives? Would you check them?)
Wednesday, April 3, 2013 at 5:11pm by rfvv
MATH PLEASE CHECK IT!!2
sqrt2/sqrt10 here is what I got: sqrt (2/10)= sqrt (1/5) sqrt 0.2 * I don't think that it is right but pleasee check it.Thanks Yes, that is one way you could do it, then simplify further by taking
the square root of 0.2, but you are probably supposed to simplify like this: ...
Tuesday, February 6, 2007 at 9:15am by Margie
math-please check my answer
0.6x + 4 < 1.0x - 1 6x + 40 < 10x - 10 4x > 50 x > 12.5 can someone please check my work and tell me if I am right?
Friday, July 30, 2010 at 10:22pm by Michelle
english Please check answer
10. In a research proposal, your sources should not have the potential to answer research questions. A.) true *B.) false Thank You
Thursday, February 21, 2013 at 2:56pm by Mohammed
somebody please tell the answer of b. I tried with all equation. D=D0e^-Ea/RT Created Eq.1 for 35+273K and Eq.2 for 45+273k i got Ea 1.55*10^4J/mol even after conversion with j/mol to EV not getting
green check 1KJ/mol=1.03*10^-2 if some body got green check please suggest me.
Friday, January 11, 2013 at 10:48pm by martin
Write the answer in simplest form: 1. 1/3-2/9=1/9; 2. 3/5-1/10=1/2; 3.1/2-1/8=3/8; 4. 2/3-1/6=1/2; 5. 7/12-1/3=1/4; 6. 4/5-2/10=3/5; 7. 7/15-2/5=1/15; 8. 3/4-3/16=9/16; 9. 5/8-1/2=1/8; 10. 1/2-1/5=3/
10 please check my answers... Thanks
Wednesday, February 6, 2013 at 8:05pm by Thea
English 10 --- Check Please
If all you had to do was correct it, then we'd never be able to tell unless you post the "before" and the "after" (your correction). If you decide to post BOTH versions, be sure to label them
Tuesday, February 14, 2012 at 1:36pm by Writeacher
please check: 10 to the 1st power = 10*1??? 10 to the 0 power = 10*0??? I'm not sure.
Wednesday, January 9, 2013 at 5:24pm by Celest
Ok. I did that and I found that Q= 1.64 x 10 ^ -18. I then divided that by 1.6 x 10 ^ -19. The answer I got was 10.23. I am being told that is the wrong answer. Could you please double check that for
me? Thank you!!
Wednesday, February 11, 2009 at 7:19pm by Jordan
English please check my answer
Please check my answer thank you Cardiology is the study of the heart. Which of the following terms describes the foundation upon which the medical word is built that usually relates to the body
organ involved ? A. Prefix B. Suffix C. Root D. Combining form My answer is Root (c)
Wednesday, March 19, 2008 at 9:39pm by Dakotah
Calculus (Normal Line, please check my work)
Undefined means very large magnitude, like infinite. 1/1 = 1 1/.1 = 10 1/.01 = 100 . . . 1/10^-6 = 1,000,000 1/10^-10 = 10,000,000,000 etc as the denominator goes to zero, the term goes to infinity
Saturday, December 17, 2011 at 3:51pm by Damon
english/check please
subject = one word Please re-vise some of these.
Monday, April 30, 2012 at 8:06pm by Writeacher
Can you please check these few sentences, please? Thank you. 1) I just wanted to know why two meals are not included in your rate and how much your hotel charges for a dinner. 2) Would you please be
so kind and send me the details of your offer again?
Friday, June 3, 2011 at 1:55pm by Henry1
physics-bobpursley/drwls please check
bobpursley/ drwls please help me with my post on Sunday, April 15 at 10:16pm and 10:19pm. Thanks!
Monday, April 16, 2007 at 7:24am by Mary
English composition
I bought a pedometer on the Auction site on-line, so I can measure my steps. If I walk for 10,000 steps a day, walking with the pedometer on my belt will be helpful for my health. This product was
made in China, and it is 1,500 won. The pedometer is rather inexpensive. Because...
Thursday, January 3, 2008 at 5:44pm by John
English (Check Please)
Please see my answers to your previous post. 15 is also right.
Monday, May 28, 2012 at 6:45pm by Ms. Sue
Chemistry(Please check)
1) The Ka for acetic acid is 1.8e-5. What is the pH of a 3.18M solution of this acid? I did 1.8e-5 = x^2/3.18 x=sqrt 1.8e-5 X 3.18 = 7.56e-3 pH=-log(7.56e-3) = 2.12 The pH is 2.12 2) At 25 degrees
celsius, the pH of a 1.75M NaCN solution (the Ka for HCN is 4.0e-10) is ? 4.0e-...
Monday, April 2, 2012 at 3:47pm by Hannah
Algebra- Check Please
Monday, January 7, 2013 at 11:30pm by lorrie
English 10 Check
Someone please check! * is the answer I chose. For each of the following pairs of topics select the one that is more focused. 1.)Topic. a.)Why beaches are fun *b.)Why I love Miami Beach 2.)Topic. a.)
The wonderful world of vegetables *b.)The health benefits of carrots 3.)Topic...
Sunday, June 13, 2010 at 3:02pm by Ariel
Thank you very much for your last corrections. Can you please check the following sentences I wrote on the Canterbury tales, please?
Saturday, April 10, 2010 at 6:38am by Franco
Writeacher, could you please check the last two paragraphs I posted. I need to check everything for tomorrow. There is still one more paragraph after those and then we will have finished! Thank you
for your invaluable help!
Wednesday, February 8, 2012 at 6:32pm by Henry2
chem/some1 check my work plz.
2NH3(g)+3CuO(s)=N2(g)+3Cu(s)+3H2O(g) if 18.1grams of ammonia reacts with 90.4 grams of copper(II) oxide, which is the limitting reagent?? what is the theorectical yield of nitrogen gasd? if 8.52 g of
N2 are formed, what is the % yield of nitrogn gas?? please check if is ...
Tuesday, November 16, 2010 at 7:23pm by julia
please check
1) exploit 2) obsolete 3)elicited 4) acclaim 5) escalated 6)engrossed 7)adjacent 8)terminate 9)methodical 10)tangible 9) 10)
Monday, August 30, 2010 at 5:19pm by jae
PLEASE HELP- spelling
10 places, 10 foods, 10 animals, 10 things, 10 people, and 10 games with double letters PLEAse!!! I don't have much time
Monday, March 9, 2009 at 7:53pm by ???????????
(ENGLISH) please check over english work.
She is young, I guess she is in her twenties. Instead of writing "I" can I use the word daughter?
Tuesday, October 8, 2013 at 9:22pm by Shannon
8th Grade Algebra: Answer Check
1,2,3,4,5,6,7 are ok. 8. sqrt (3x^2) = x*sqrt(3) Check that. 9,10,11 ok. 12. sqrt(3)*sqrt(27) = sqrt 3*sqrt(3*9) 9*sqrt(9) = 9*3= 27. Check that. check 15. I think you just wrote it wrong. I get 1/
5*sqrt(10). Check that. The others look ok.
Tuesday, January 20, 2009 at 10:42pm by DrBob222
Now.<~~comma, not period let's check how well you can use the expressions you've learned. I will distribute handouts. You should pick up a handout on the table. You have to fill in the blanks with
suitable expressions. Are you finished.<~~question mark needed here Two ...
Monday, June 9, 2008 at 6:58am by Writeacher
Check English Answers
Please check my answers: 1)pathos 2)thesis statements 3)Lincoln promises retribution for war 4)Ethos 5)logos
Monday, April 14, 2014 at 2:26pm by AMK
Please check these 25. If a dress were 20% off of the marked price of $32.00, how much would it cost? $25.60 26. Write 1,000,000,000 in scientific notation 1.0 x 10^9 27. (3.0 x 10^2) x (2.0 x 10^6)
= ? 6.0 x 10^-4
Thursday, June 10, 2010 at 12:39pm by y912f
College English Check
One is wrong. Please check the past tense and past participles of both verbs. Also --it should be TOO soon . . .
Tuesday, June 22, 2010 at 10:11pm by Ms. Sue
No. K1 = 1.82 x 10^-10 K2 = (1/1.15 x 10^-4)^2 Then Keq for the reaction is K1*(1/K2)^2 = 1.82 x 10^-10*(1/1.15 x 10^-4)^2 = 1.82 x 10^-10*(7.56 x 10^7) = 0.01376. Check it for math. Check it for
significant figures.
Thursday, February 5, 2009 at 8:40pm by DrBob222
Physics repost please check
A glass bottle of soda is sealed with a screw cap. The absolute pressure of the carbon dioxide inside the bottle is 1.80 x10^5pa. Assuming that the top and bottom surfaces of the cap each have an
area of .0004M^2, obtain the magnitude of the force that the screw thread exerts ...
Monday, April 9, 2007 at 12:04am by CJ
Physic please check
B = mu H = (4 pi*10^-7)*i N/(2 r) N is the number of turns. r is the coil radius i is the current i = V/R R = N* 2 pi r * (6.5*10^-3 ohm/m) B = [(4 pi*10^-7)*V*N]/[(N*4*pi*r^2*(6.5*10^-3)] = 10^-7 V/
[r^2*(6.5*10^-3 ohm/m)] The 4 pi factors and the number of turns N both cancel...
Monday, November 5, 2007 at 10:09pm by drwls
Can someone check this for me please.. 18 MB of data are to be transmitted over a a megabyte per second connection. How long should the complete transaction take, to the nearest 0.1s. Size of file /
data rate so, 18,874,368 bytes / 4,000,000 = 4.718592 = 4.7 is that correct a ...
Wednesday, January 12, 2011 at 8:10am by Anonymous
physics...please check my answer...
4.A mass of one kg of nickels has a monetary value in the US $ of approximately: a. $1 b. 10 cents c. $10 d. $1000 is it c?
Tuesday, September 11, 2007 at 4:45pm by Mandee
Thanks Anon but 1.25663706*10^-6 is 4*pi*10^-7, please check if you like. It still does not work, any other suggestion, am I missing a bracket somewhere?
Friday, April 26, 2013 at 1:20pm by FLu
Pre-Algebra(please check)
Please check these simplified expressions(my answer is after the ;) 4(2k-6)-8(3-3k); -16k-(-48) 9+(-2k)+(-4)-6k; -8k+5 -x+4x-(-8x)-9+2x+1; 13x+(-8) 5(-3x+6)+8x; -7x+30 8+6x+4-8x; -2x+12 5(-m-3)
+4m-10; -9m+5
Sunday, May 18, 2008 at 8:09pm by Jillian
1. check if 10.25-10 = 10.50625 - 10.25 , if true then AS 2. check if 10.25/10 = 10.50625/10.25 , if true then GS if both are false, then "neither" if one is true, verify with the other given value.
Sunday, October 2, 2011 at 3:03am by Reiny
English 10-Check/Help
Part 1: Poem Structure Check my following answers below, and I need help with theme? Dx In this part, you will analyze the structure of the poem. Please fill out the following. Title of Poem: Annabel
Lee Poet: Edgar Allan Poe Theme: Type of Poem: Narrative poetry Style of ...
Wednesday, May 1, 2013 at 10:21am by Victoria
Chemistry(Please check)
. Ksp for Fe(IO3)3 is 10-14. Two solutions, one being iron(III) nitrate and the other being sodium iodate, were mixed. At the instant of mixing, [Fe3+] = 10-4M and [IO3-] = 10-5M. What happens? My
answer is that a precipitate forms because Qsp > Ksp. Would you agree? ...
Monday, April 9, 2012 at 9:51pm by Hannah
IT computer memory
Can someone check this for me please.. 18 MB of data are to be transmitted over a a megabyte per second connection. How long should the complete transaction take, to the nearest 0.1s. Size of file /
data rate so, 18,874,368 bytes / 4,000,000 = 4.718592 = 4.7 is that correct a ...
Wednesday, January 12, 2011 at 5:40pm by freddo
well check it check #1, yes, length is twice width check #2, yes, the area is 20*10 = 200 however remember the units 20 Yards by 10 yeards
Sunday, May 11, 2008 at 10:27am by Damon
chem can u check answer please
what is the hydrogen ion concentration of an aqueous solution with a pOH of 9.262? I got 5.47X 10 ^-10 is this correct? Thanks Andy
Sunday, May 23, 2010 at 10:36pm by Andy
Science (Please Respond SOON!)
I think I'm using the wrong formulas or SOMETHING... Here's the question: "Two equal charges are separated by 3.7 ´ 10^-10 m. The force between the charges has a magnitude of 2.37 ´ 10^-3 N. What is
the magnitude of q on the charges?" Here's my work: F = kc(q1q2/r^2) 2.37 x 10...
Thursday, May 7, 2009 at 3:54am by Judy
Pages: 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 | Next>> | {"url":"http://www.jiskha.com/search/index.cgi?query=English+10+---+Check+Please","timestamp":"2014-04-20T09:38:00Z","content_type":null,"content_length":"39929","record_id":"<urn:uuid:4a91b6cf-38e9-42fe-b3e8-e69bab868596>","cc-path":"CC-MAIN-2014-15/segments/1397609538110.1/warc/CC-MAIN-20140416005218-00390-ip-10-147-4-33.ec2.internal.warc.gz"} |
real analysis question
May 2nd 2008, 07:56 PM
real analysis question
i need to determine and prove whether or not the annulus in R^2 defined by d((x_1,x_2),(y_1,y_2))=sqrt((x_1-y_1)^2+(x_2-y_2)^2) is path-connected.
i know that it is path-connected (logically) but i got stuck at the part where we need to find a path between two arbitrary points in the annulus. can someone help me? thanks.
May 3rd 2008, 07:45 AM
i need to determine and prove whether or not the annulus in R^2 defined by d((x_1,x_2),(y_1,y_2))=sqrt((x_1-y_1)^2+(x_2-y_2)^2) is path-connected.
i know that it is path-connected (logically) but i got stuck at the part where we need to find a path between two arbitrary points in the annulus. can someone help me? thanks.
Lets assume you mean the region $R=\{(x,y), 1/4\le x^2+y^2 \le 1\}$.
Take any two points, the line segment conecting them lies either entirly within R or it meets the inner boundary at two points. In the latter case connect the two ponits on the inner boundary by
the minor arc.
Thus either two points are connected by the line segment between them if this is within the annulus or by a pair of line segments and an arc otherwise.
Alternativly take the two points as $(r_1, \theta_1)$, $(r_2, \theta_2)$ in polars, then $1/2 \le r_1 \le 1$ and $1/2 \le r_2 \le 1$.
Then the curve:
$<br /> C = \{ (r_1 ,\theta ),\quad \theta \in [\min (\theta _1 ,\theta _2 ),\max (\theta _1 ,\theta _2 )]\} \cup \{ (r_1 + \lambda (r_2 - r_1 ),\theta _2 ),\quad \lambda \in [0,1]\} <br />$
lies entirly within the annulus and conects the two points
Modifying these to apply to any annulus I leave to you.
May 4th 2008, 11:59 AM
hmm can you please explain how that curve works? i'm trying to figure it out and i'm not quite getting the picture. thanks.
May 4th 2008, 12:03 PM
It's an arc of a circle starting from one of the points to the polar angle of the pther point, then proceeds radialy out to the the second point. | {"url":"http://mathhelpforum.com/calculus/36950-real-analysis-question-print.html","timestamp":"2014-04-18T18:39:23Z","content_type":null,"content_length":"7916","record_id":"<urn:uuid:0d162531-c806-4b68-a171-72f098d710ad>","cc-path":"CC-MAIN-2014-15/segments/1397609535095.7/warc/CC-MAIN-20140416005215-00093-ip-10-147-4-33.ec2.internal.warc.gz"} |
Manifold definition
I don't think this is correct, it is coordinate independent.
Nope. Manifolds generally have no coordinate independent vector space structure. Not even locally. The idea lavinia mentioned was to use charts to locally pull the vector space structure on [itex]\
mathbb{R}^n[/itex] back to the manifold, but this structure depends heavily on which coordinates you choose. For example, consider the space [itex]\mathbb{R}[/itex] as a topological manifold. Then
both [itex](\mathbb{R},\mathrm{id})[/itex] and [itex](\mathbb{R},\mathrm{id}^3)[/itex] give us global charts, but they define totally different vector space structures. The same problem occurs in the
smooth case.
Edit: To help clear up confusion, regard the domains of [itex]\mathrm{id}[/itex] and [itex]\mathrm{id}^3[/itex] as purely topological spaces. Do
give them any algebraic structure. View the codomains of these maps as real vector spaces with the natural topology.
Edit II: For further clarification, all this argument shows is that charts cannot be used to give a canonical local vector space structure on a manifold, since each choice of chart will generally
produce a different a different local vector space structure. So it does not matter that [itex]\mathbb{R}[/itex] has a natural vector space structure. | {"url":"http://www.physicsforums.com/showthread.php?t=669909","timestamp":"2014-04-18T10:55:02Z","content_type":null,"content_length":"79548","record_id":"<urn:uuid:a36ec800-0256-45ca-af7c-13ee16ab06f4>","cc-path":"CC-MAIN-2014-15/segments/1397609533308.11/warc/CC-MAIN-20140416005213-00045-ip-10-147-4-33.ec2.internal.warc.gz"} |
My New Twin Prime Numbers
Re: My New Twin Prime Numbers
Ok. If you start calling me a mathematician I'll know our friendship is off.
"The good news about computers is that they do what you tell them to do. The bad news is that they do what you tell them to do." - Ted Nelson
Re: My New Twin Prime Numbers
Mathematics is beautiful. We are lucky to be able to study it from the viewpoint of an amateur.
In mathematics, you don't understand things. You just get used to them.
I have the result, but I do not yet know how to get it.
All physicists, and a good many quite respectable mathematicians are contemptuous about proof.
Re: My New Twin Prime Numbers
Hi bobbym
Amongst of those famous mathematicians there were amateurs like Pierre Fermat (a lawyer), Mersenne (a priest), etc. Here list of them http://en.wikipedia.org/wiki/List_of_am … ematicians
There is a mafia culture in the mathematical fields and those people tends to garbage other people ideas because they are jealous.
Re: My New Twin Prime Numbers
Q:What do men with power want?
A:More power.
It is very easy for men to become power mad. Even easier than becoming greedy. The scientific community in general had to withstand many centuries of persecution. Then they rose to the position of
power where they promptly began to persecute people and differerent ideas. They began to stifle and repress new ideas that might endanger their position of power.
I like Fermat, Ramanujan, Lovelace and Pascal most among that list. And am saddened at the absence of the great Forman S. Acton, chemist. He is widely considered the greatest numerical analyst alive.
In mathematics, you don't understand things. You just get used to them.
I have the result, but I do not yet know how to get it.
All physicists, and a good many quite respectable mathematicians are contemptuous about proof.
Re: My New Twin Prime Numbers
Hi bobbym
Anyway, how long do you think it would take to get 1st 100,000 digits for this prime equation? I hope it won't take too much computing time.
Re: My New Twin Prime Numbers
Please tell me which equation you mean.
In mathematics, you don't understand things. You just get used to them.
I have the result, but I do not yet know how to get it.
All physicists, and a good many quite respectable mathematicians are contemptuous about proof.
Re: My New Twin Prime Numbers
Hi bobbym
Equation in this thread..so far phrontister got 715 digit prime and how about 100,000 digits prime? How long would it take? And would it fry the CPU?
Re: My New Twin Prime Numbers
Nobody even knows where it would be. I would think that it would take a very long time with a desktop,
In mathematics, you don't understand things. You just get used to them.
I have the result, but I do not yet know how to get it.
All physicists, and a good many quite respectable mathematicians are contemptuous about proof.
Re: My New Twin Prime Numbers
I am revising the equation and making it harder to get new resulting prime. I named it Perfect Twin Prime Numbers. The revised equation is given as follows:
Where all Ps are prime numbers.
Re: My New Twin Prime Numbers
Hi Stangerzv,
Yes...that certainly has made it more difficult to find solutions!
"569" should read "659".
I've changed my code for the original problem and have found prime solutions for Pₜ=2 and 3, but not yet for 5.
Pₜ=2: (from 1st 1000 primes) +/- = 7, 139, 199, 463, 877 and 6121
Pₜ=3: (from 1st 1000 primes) +/- = 3
I'm off to bed now, and I've given my computer the task of solving Pₜ=5 while I'm asleep.
Last edited by phrontister (2013-05-31 21:21:43)
"The good news about computers is that they do what you tell them to do. The bad news is that they do what you tell them to do." - Ted Nelson
Re: My New Twin Prime Numbers
Woke up to this:
Pₜ=5: (from 1st 10000 primes) +/- = 17159, 32831, 62011, 70241 and 96053
Since then, this:
Pₜ=7: (from 1st 10000 primes) +/- = 81899 and 104311
Perfect Twin Prime solutions found so far (smallest):
Last edited by phrontister (2013-05-31 21:22:33)
"The good news about computers is that they do what you tell them to do. The bad news is that they do what you tell them to do." - Ted Nelson
Re: My New Twin Prime Numbers
Hi phrontister
Thanks for finding the solutions. I am setting up a website for finding my prime numbers. Maybe in the future I could offer some prize money for the larger prime (>100,000 digits)
Re: My New Twin Prime Numbers
One thing that I notice about these prime pairs is that, all of their digital root for n>1 would be in pair of 2^y (y=1..3) and 1 or 7, example, (2,1), (2,7),(4,1), (4,7),(8,1), (8,7) and special
case (7,7) when n is a square number (i.e n=9). This indicates that the prime distribution is not random but organized. Unless someone could find the counter examples. Perhaps for n=9, adding this
into digital root system won't change anything as 9 has zero value in the decimal system and this is why the prime pairs would have the same digital roots. This applies for all n which has digital
root of 9.
Last edited by Stangerzv (2013-05-26 18:55:09)
Re: My New Twin Prime Numbers
Hi Stangerzv,
I've done some work on what I think you're looking for, and here are pics of my findings.
I hope I've understood you correctly and that this is the info you want, but if not, please give a different example to help me understand and I'll do my best to get the info for you.
The pic with the different Pₜ values (from 2 to 13) in the left-hand column gives the results for the smallest Pₓ of each of those Pₜ values, while the other three (for Pₜ=2, Pₜ=3 and Pₜ=5) give
results from the smallest Pₓ to successively-higher Pₓ values. All pics show digit roots.
I started testing for Pₜ=17, but that may have to be an overnight job while I'm asleep because after about an hour there were no results.
EDIT: The "152" in line 3 of the Sums column should read "157", which gives the digital sum 4.
Last edited by phrontister (2013-05-31 21:23:34)
"The good news about computers is that they do what you tell them to do. The bad news is that they do what you tell them to do." - Ted Nelson
Re: My New Twin Prime Numbers
Hi Stangerzv,
Got an overnight result for Pₜ=17:
Last edited by phrontister (2013-05-31 21:24:14)
"The good news about computers is that they do what you tell them to do. The bad news is that they do what you tell them to do." - Ted Nelson
Re: My New Twin Prime Numbers
Kool, it seems I had overlooked the primes and new pairs of digital roots, (5,7) and (4,8)
Re: My New Twin Prime Numbers
Hi Bobby,
Do you know of a 'proper' function in M to obtain the digital root (sum) of a number?
I couldn't find one anywhere, so I made up this one which seems to work quite well for the number sizes I've worked on:
DRoot[x_] := Total[IntegerDigits[Total[IntegerDigits[Total[IntegerDigits[Total[IntegerDigits[x]]]]]]]]
If there isn't one, could mine be squashed up somehow to eliminate the repeats?
"The good news about computers is that they do what you tell them to do. The bad news is that they do what you tell them to do." - Ted Nelson
Re: My New Twin Prime Numbers
There are simpler alternatives.
How does this do?
NewRoot[n_] := (Mod[n, 9]) /. (0 -> 9)
I do not know how it would perform on decimals or negative integers. I tested it on 10 or 12 million from 0 to 100 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000.
In mathematics, you don't understand things. You just get used to them.
I have the result, but I do not yet know how to get it.
All physicists, and a good many quite respectable mathematicians are contemptuous about proof.
Re: My New Twin Prime Numbers
I tested it on the numbers in the hide box on my post #115.
Could you explain what it does, in particular the "/. (0 -> 9)" part?
Last edited by phrontister (2013-05-30 14:14:41)
"The good news about computers is that they do what you tell them to do. The bad news is that they do what you tell them to do." - Ted Nelson
Re: My New Twin Prime Numbers
That is called a replacement rule. It says turn 0 into 9. Note the syntax because these replacement rules work on anything.
In mathematics, you don't understand things. You just get used to them.
I have the result, but I do not yet know how to get it.
All physicists, and a good many quite respectable mathematicians are contemptuous about proof.
Re: My New Twin Prime Numbers
Ok...thanks for that. Learnt something new.
I've included your function in my code, which is looking much better now.
Here's my new M code:
This code will expand the information given in post #111 to show primes' sums and digital roots:
Last edited by phrontister (2013-05-30 15:42:53)
"The good news about computers is that they do what you tell them to do. The bad news is that they do what you tell them to do." - Ted Nelson
Re: My New Twin Prime Numbers
Hi phrontister
Can you find the digital roots pair (1,5) & (1,7) for the twin primes?
Re: My New Twin Prime Numbers
Hi Stangerzv,
The digital roots pair {1,5} for Pₜ=2 occurs first at:
Then at:
There are also a couple more after that, at Pₓ=70207 and Pₓ=74167. No doubt there will be more.
Still looking for {1,7}, but have to go out now for a while. Maybe there will be an answer waiting for me when I get back.
Last edited by phrontister (2013-05-31 21:25:02)
"The good news about computers is that they do what you tell them to do. The bad news is that they do what you tell them to do." - Ted Nelson
Re: My New Twin Prime Numbers
Thanks for your input.
Re: My New Twin Prime Numbers
Hi phrontister;
Found it under 'Rule' in the M docs.
The replacement rules are a part of the Prolog programming language and make M unique.
In mathematics, you don't understand things. You just get used to them.
I have the result, but I do not yet know how to get it.
All physicists, and a good many quite respectable mathematicians are contemptuous about proof. | {"url":"http://www.mathisfunforum.com/viewtopic.php?id=19254&p=5","timestamp":"2014-04-20T08:36:29Z","content_type":null,"content_length":"52923","record_id":"<urn:uuid:33a382fb-3edd-461d-a88f-1734ec96a82a>","cc-path":"CC-MAIN-2014-15/segments/1397609538110.1/warc/CC-MAIN-20140416005218-00516-ip-10-147-4-33.ec2.internal.warc.gz"} |
Higher homotopy groups of slice disk complement
up vote 9 down vote favorite
Let $K \subseteq{\mathbb{S}}^3=∂\mathbb{D}^4$ be a non-trivial slice knot, i.e. $K$ bounds a slice disk $\Delta$ in $\mathbb{D}^4$. Let $N(\Delta)$ be a regular neighborhood of $\Delta$ in $\mathbb
What is known about $\pi_i(\mathbb{D}^4-N(\Delta))$, for $i\geq2$?
For what it's worth, $\pi_1(\mathbb{D}^4-N(\Delta))$ is known to be normally generated by the meridian of $K$. The knot complement $\mathbb{S}^3 - K$ and $∂(\mathbb{D}^4-N(\Delta))=M_K$ (the
zero-framed surgery on the knot $K$) are both known to be aspherical.
gt.geometric-topology knot-theory
I don't think the non-triviality of $K$ has much impact on your question. For example, connect-sum any slice disc with a $2$-knot such as a Cappell-Shaneson knot. – Ryan Budney Feb 1 '12 at 23:54
Ah, I wanted $K$ to be non-trivial since I believe that $\mathbb{S}^3-K$ is aspherical only for non-trivial knots. – Aru Ray Feb 2 '12 at 2:00
You can also do a boundary connect-sum of two slice discs, that would give the connect sum of the two slice knots on the boundary. – Ryan Budney Feb 2 '12 at 2:26
1 @Arunima Ray: If $K$ is trivial, its complement is an open solid torus, which is aspherical. – Richard Kent Feb 2 '12 at 2:54
oops, my bad, I meant the the zero surgery on the knot $M_K$ is not aspherical for a trivial knot - if I've thought this through correctly, the zero surgery on the trivial knot is $\mathbb{S}^2x\
mathbb{S}^1$ – Aru Ray Feb 2 '12 at 3:17
add comment
2 Answers
active oldest votes
The homotopy groups can be pretty big things. For example, your $D^4 - N(\Delta)$ class of spaces contains the class of all $2$-knot complements -- simply remove a 4-ball neighbourhood
of $S^4$ that intersects the $2$-knot in an unknotted disc.
$2$-knot complements have fairly complicated homotopy groups. For example, Cappell-Shaneson knot complements fiber over $S^1$ with fiber a once-punctured $(S^1)^3$. The universal cover
of this space is $\mathbb R \times (\mathbb R^3 - \mathbb Z^3)$, so by the Hilton-Milnor theorem, rationally the homotopy groups are a free lie algebra with countably-infinite many
up vote 8 down generators (up to the action of $\pi_1$ there's just one generator, though).
vote accepted
Off the top of my head I don't know if slice disc groups are any more general than knot groups, they're probably not very far from each other. I think Kawauchi may not cover this but
the references in his survey book should mention something.
add comment
To add to Ryan's answer,
$2$--knots usually don't have aspherical complements, see Dyer & Vasquez, The sphericity of higher dimensional knots, Canad. J. Math. 25(1973), 1132-1136. This suggests a
up vote 10 down complicated answer for slice disks in general.
On the other hand, if the disk is ribbon, then the complement is aspherical, see Asano, Marumoto, and Yanagawa, T, Ribbon knots and ribbon disks, Osaka J. Math. 18 (1981), 161-174
add comment
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st: re: double if-sentences in Stata
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st: re: double if-sentences in Stata
From Christopher Baum <baum@bc.edu>
To statalist@hsphsun2.harvard.edu
Subject st: re: double if-sentences in Stata
Date Thu, 27 Nov 2008 14:00:32 -0500
Linn said
Here is what I write. ...
I thus try to write a command where the executed command if the result (if shift==-1) is true also includes an if-formulation, and the same with the executed command in the case where the result of
the initial criterion is false. If the shift value is -1, what I do with the variable flow depends on whether the sum (flow[_n-1]+`cap') is larger or smaller than the value given in `add'. Similarly,
if the shift variable isn't -1 but 1, what I do with flow depends on the sum (flow[_n-1]-`cap') and whether this sum is smaller or larger than the value given by `add'.
Others have pointed out that unlike other programming languages, the - if- command is rarely the answer to a problem where you want to test the value of each observation. I believe that this code
does what you descirbed in your original posting.
set obs 10
set seed 20081127
g t = _n
g cap = log(t+1)
g add = _pi + runiform()
tsset t
g byte shift = cond(runiform() > 0.5, 1, -1)
g flow = 5 in 1
replace flow = ///
(shift == -1 ) * cond(L.flow + cap < add, -cap, L.flow - add) ///
+ (shift == 1) * cond(L.flow - cap > add, cap, L.flow + add) in 2/l
In this case, I have avoided any use of -if- (as a command or qualifier!) by using the cond() function and the Boolean test, i.e. (foo == bar) evaluates to 1 if the condition is true and 0 if it is
false. Also note that the construction of the running variable -flow- must avoid the first observation, lest a missing value be propagated.
Kit Baum, Boston College Economics and DIW Berlin
An Introduction to Modern Econometrics Using Stata:
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FOM: Re: Reuben Hersh: Mitteilungen der DMV (fwd)
Ching-Tsun Chou ctchou at mipos2.intel.com
Fri Feb 27 23:06:39 EST 1998
Let me first say that I am neither a mathematician nor a philosopher;
I'm just an engineer. So what I say below is definitely not up to the
professional standards.
It seems to me that it is perhaps not very interesting to ask:
WHAT IS MATHEMATICS, REALLY?
Instead, a better question may be:
Why and how does mathematics work?
By "work" I mean that using the machinery of mathematics, we can reach
conclusions that can be verified empirically. Professor Hersh listed
quite a few "social-cultural-historical" ("social" for short)
constructs. Most of them (especially the religions and the -isms ;-)
do not seem to "work" in this sense. If some guru tells me that if I
intone certain magical formulas in a certain way then I shall get a
raise, I will be very skeptical. But if some physicist tells me that,
according to her calculation, if I squeeze a certain amount of a
certain type of uranium in a certain way then there will be a gigantic
explosion, I will take her much more seriously. Indeed, if I don't
believe her assertion, she can dispel my doubts by demonstration. I
don't think this kind of demonstration is available in most of the
"social" constructs that Professor Hersh mentioned.
It also seems to me that the line that Professor Hersh drew between
things real and things unreal is somewhat arbitrary. He wrote:
Here is an important objection.
Aren't some mathematical concepts grounded in physical
For instance, N, the natural numbers?
I surely have five fingers on my left hand, so "five" has
a physical meaning. On the other hand, N includes some very large
numbers, ((2 to a very high power) raised to a very high power) raised to
a very high power. It is questionable what physical meaning this
big number has. So the natural numbers as describing physical objects
are not the same as the natural numbers in pure mathematics.
The fact that I have five fingers on my left hand is an empirical
observation. "Five" in that usage is an adjective. There is no
conceptual difficulty there, any more than in saying my fingers are long
or short. But five in pure mathematics is less than the big number I just
defined, and is relatively prime to it, and so on. It possesses an
endless list of properties and relationships, not only in N, but also in
R, in C, and beyond. It's part of an abstract theory.
As such, it is not a material object, not a mental object, but a shared
concept, existing in the social consciousness of mathematicians and
I guess Professor Hersh considers the five fingers on his left hand
real, perhaps the number 5 too, but certainly not some very large
numbers. But would he consider the electromagnetic field real? Or
the genes? Or the water molecules? What is the fundamental
difference between these quite abstract objects and the five fingers
on his left hand?
I submit that there is no essential difference between them. They are
all useful (so far at least) in constructing an intelligible account
of "the physical reality" (whatever that term means) THAT WORKS. But
perhaps these questions about what is real and what is not, are not
very interesting after all. Perhaps the really interesting question
is whether Professor Hersh, or anyone else, can construct a coherent
body of mathematics that does not contain those very large numbers,
BUT STILL WORKS. Before anyone can do that, I shall consider (((2
raised to a very high power) raised to a very high power) raised to a
very high power) as real as the five fingers on my left hand.
Locating mathematics in the social-cultural realm
means that it is human. For example, there is no sense to
talking about mathematics existing before the human race existed or after
it has vanished.
Some people find this conclusion shocking. One may point to
some beautiful theorem of Lagrange and say, "Isn't it obvious that this
was always true, before there were any humans?"
You could just as well cry, "Listen to this wonderful symphony of
Beethoven! Isn't it clear that this was always beautiful, even before
Beethoven was born?"
Indeed, I am rather doubtful whether my goldfish finds Beethoven's
symphonies beautiful. However, if some beautiful theorem of Lagrange
has some implications about the motion of my goldfish's body, I have
very little doubt that it would indeed be the case. Similarly, I have
very little doubt that many of the conclusions about the physical
world that we can reach using the machinery of mathematics were true
long before the rise of the humankind, are true now, and will continue
to be true long after its demise. I wonder whether one can say the
same about the beauty of Beethoven's symphonies.
Ching-Tsun Chou
Ching-Tsun Chou E-mail: ctchou at mipos2.intel.com
Intel Corporation, RN6-16 Tel: (408) 765-5468
2200 Mission College Blvd. Fax: (408) 653-7933
Santa Clara, CA 95052, U.S.A. Sec: (408) 653-8849
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Mathematics 120 > Crazy > Flashcards > Math Mid Term | StudyBlue
There is no proof that p is true therefore p is false
P is associated with a positive emotional response therefore P is true
P is true and P is restated in different ways
P is related to Q and I have an argument concerning Q therefore P is true
I have an argument concerning or distorted version of P therefore I hope you are fooled into concluding I have an argument concerning the real version of P
Given two propositions p and q, the statement p and q is called their conjunction. It is true only if p and q are both true.
Given two propositions p and q, the statement p or q is called their disjunction. In logic, we assume that or is inclusive, so the disjunction is true if either or both propositions are true, and
false only if both propositions are false.
A statement of the form if p, then q is called a conditional proposition. Proposition p is called hypothesis and q is called the conclusion. The conditional if p, then q is true in all cases except
when p is true and q is false.
A may be a subset of B or vice versa, meaning that all members of A are also members of B.
A may be disjoint from B meaning that the two sets have no members in common.
A and B may be overlapping sets meaning that the two sets share some of the same members.
All S are P; No S are P; Some S are P; Some S are not P
estimate specifies only a broad range of values, such as "in the millions"
simple way to compare measurements made at different times or in different places
represents an average of prices in a sample of more than 60,000 goods, services, and households
relative change in CPI from one year to the next
interest is paid both on the original principal and all the interest that has been added to the original principal | {"url":"http://www.studyblue.com/notes/note/n/math-mid-term/deck/706174","timestamp":"2014-04-16T13:19:10Z","content_type":null,"content_length":"59195","record_id":"<urn:uuid:3cb504cb-d336-422f-ac1f-5e79d3ce4702>","cc-path":"CC-MAIN-2014-15/segments/1397609523429.20/warc/CC-MAIN-20140416005203-00543-ip-10-147-4-33.ec2.internal.warc.gz"} |
Introduction to stochastic control theory
, 1977
"... The paper derives a general form of the term structure of interest rates. The following assumptions are made: (A.l) The instantaneous (spot) interest rate follows a diffusion process; (A.2) the
price of a discount bond depends only on the spot rate over its term; and (A.3) the market is efficient. U ..."
Cited by 603 (0 self)
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The paper derives a general form of the term structure of interest rates. The following assumptions are made: (A.l) The instantaneous (spot) interest rate follows a diffusion process; (A.2) the price
of a discount bond depends only on the spot rate over its term; and (A.3) the market is efficient. Under these assumptions, it is shown by means of an arbitrage argument that the expected rate of
return on any bond in excess of the spot rate is proportional to its standard deviation. This property is then used to derive a partial differential equation for bond prices. The solution to that
equation is given in the form of a stochastic integral representation. An interpretation of the bond pricing formula is provided. The model is illustrated on a specific case. | {"url":"http://citeseerx.ist.psu.edu/showciting?cid=11497140","timestamp":"2014-04-21T12:21:19Z","content_type":null,"content_length":"11989","record_id":"<urn:uuid:41bd1371-79b5-4cc7-82ea-5be3d5ab83d4>","cc-path":"CC-MAIN-2014-15/segments/1397609539776.45/warc/CC-MAIN-20140416005219-00314-ip-10-147-4-33.ec2.internal.warc.gz"} |
Cell nuclei extraction from breast cancer histopathologyimages using colour, texture, scale and shape information
Standard cancer diagnosis and prognosis procedures such as the Nottingham Grading System for breast cancer incorporate a criterion based on cell morphology known as cytonuclear atypia. Therefore,
algorithms able to precisely extract the cell nuclei are a requirement in computer-aided diagnosis applications.
However, unlike other modalities such as needle aspiration biopsy images, H&E stained surgical breast cancer slides are a particularly challenging image modality due to the heterogeneity of both the
objects and background, low object-background contrast and frequent overlaps as illustrated in Figure 1. As a consequence, existing extraction methods which are largely reliant on color intensities
do not perform well on such images.
Figure 1. Ground truth (Left) magnified 250 × 250 region of a frame and (right) the same region with the nuclei delineated by a pathologist. Nuclei delineated with a thinner outline are hard to
distinguish from the background. Dark and bright areas can indiscriminately occur inside and outside nuclei.
Materials and methods
We propose a method based on the creation of a new image modality consisting in a grayscale map where the value of each pixel indicates its probability of belonging to a cell nuclei. This probability
map is calculated from texture and scale information in addition to simple pixel color intensities. The resulting modality has a strong object-background contrast and evens out the irregularities
within the nuclei or the background. The actual extraction is performed using an AC model with a nuclei shape prior included to deal with overlapping nuclei.
Feature model
First, a color deconvolution [1] is applied in order to separate the immunohistochemical stains from which 3 grayscale images are produced: a haematoxilin image, an eosin image and a third residual
component orthogonal in RGB space. Next, local features based on Laws’ texture measures [2] are computed for each pixel of the 3 obtained images. 5 different 1-dimensional convolution kernels (L5 = (
1, 4, 6, 4, 1), E[5] = (–1, –2, 0, 2, 1), W[5] = (–1, 2, 0, –2, 1), S[5] = (–1, 0, 2, 0, –1) and R[5] = (1, –4, 6, –4, 1)) are used to compute 25 different 5 × 5 kernels by convolving avertical
1-dimensional kernel with a horizontal one. The 5 × 5 kernels are applied at every pixel to extract 25 features which are then combined into 15 rotationally invariant features after normalizing by
the output of the L[5]^T × L[5] kernel and smoothing with a Gaussian kernel of standard deviation σ = 1.5 pixels.
The same process is repeated at 4 different scales after locally re-sampling the image using Lanczos-3 sinc kernels. Re-sampled images are locally computed around each pixel to allow the computation
of the 15 texture features for the same pixel at different scales. Local texture features are computed at 1:1, 1:2, 1:4 and 1:8 scales for every pixel.
Probability map
The resulting 180-dimensional feature vector x is used to compute the probability p[n](x) of each pixel to belong to a cell nuclei. Let μ[n] (resp. μ[b]) be the mean of the feature vectors for the
pixels belonging to the nuclei (resp. to the background). A class dependent LDA is performed in order to find two directions in the feature space, w[n] and w[b], such that the projection of the
classes on these directions has a maximum inter-class scatter over within-class scatter ratio. The estimated class probability associated with the feature vector x is then calculated from the linear
scores l[n] = (x – μ[n]) · w[n] and l[b] = (x – μ[b]) · w[b] using the softmax function:
The resulting probability map exhibits strong contrast between the objects and the background. Moreover, nuclei and background appear more homogeneous than in the original image. A post processing
step is also applied to fill small holes still remaining in nuclei (larger holes are not removed to prevent the unintended deletion of interstices between different nuclei).
AC model including shape prior
The actual extraction of cell nuclei is performed from the probability map with an AC model with shape prior information. The total energy E(γ) associated to a contour γ is a weighted sum of an image
term E[i](γ) and a shape term E[s](γ). The latter is itself the weighted sum of a smoothing term E[sm](γ) and a shape prior term E[sp](γ).
The shape prior term δr(t) of a contour around a circle at different frequencies k of the Fourier components by adjusting the coefficients f[k]. Detailed formulas and explanations for this and the
other energy terms can be found in the work of Kulikova et al. [3]. The shape prior information allows to properly extract overlapping nuclei according to their expected shape without arbitrarily
discarding the overlapping parts.
The detection of nuclei is performed by a marked point process model the details about which the interested reader can find in [4]. An empirical study in [5] shows that this particular combination of
MPP and AC over-performs other state-of-the-art methods for nuclei detection and extraction.
Results and discussion
The training set used for the LDA consists of 6 1024×1024 images where the nuclei have been manually delineated by a pathologist. Object and background parameters used in the AC model are also
calculated from the training set. Weight parameters for the energy terms in the AC model are adjusted with a grid search. Images used for training are distinct from the images used for validation.
Figure 2 shows results obtained with the AC model applied to the probability map side-by-side with results obtained with the same AC model applied to the original image (in fact, the slightly better
performing haematoxilin image from the color deconvolution was used instead of the red channel from the RGB image commonly used in other methods [6]). On the original image, the contours have a
tendency to match irregularities within the cell nuclei rather than their actual boundaries. This problem is largely improved by using the probability map where the nuclei boundaries are much more
salient and other irrelevant features are smoothed out.
Figure 2. Extraction results The top row is obtained using the probability map and the bottom row is obtained using the haematoxilin channel after the image color deconvolution.
1. Ruifrok AC, Johnston DA: Quantification of histochemical staining by color deconvolution.
Analytical and Quantitative Cytology and Histology 2001, 23:291-299. PubMed Abstract
2. Kulikova M, Jermyn I, Descombes X, Zhizhina E, Zerubia J: A marked point process model with strong prior shape information for extraction of multiple, arbitrarily-shaped objects.
Proc. Signal-Image Technology and Internet-Based Systems 2009.
3. Descombes X, Minlos R, Zhizhina E: Object extractionusing a stochastic birth-and-death dynamics in continuum.
J. Math. Imaging Vis 2009, 33(3):347-359. Publisher Full Text
4. Kulikova MS, Veillard A, Roux L, Racoceanu D: Nuclei extraction from histopathological images using a marked point process approach.
Proc. SPIE Medical Imaging, San Diego, California, USA, 2012
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Apply IF Formulas And Conditional Formatting in Excel 2010
Formulas are the heart of Excel. With huge list of built-in formula and functions, Excel stands out from other datasheet handling applications. Formulas are used for calculating/analyzing data based
on values in designated cells. It supports trigonometric, statistical and other functions. You can also create a new rule, or constraint to apply over your datasheet. This post covers writing
formulas and applying conditional formatting on a basic level.
For Instance: We will start off with creating simple worksheet of students, which includes; Name of the student, and obtained Marks in their respective courses.
We need to add two new columns which shows which grade the student has secured and status of the student which shows whether he is pass or fail. So we will add two new columns having names Grade and
Status respectively.
We will write a formula which evaluates the grade student has secured. By considering if secured marks are greater than/equal to 80, then he falls in A grade, greater than or equal to 75 secured B+
grade and so on and student who obtained less than 60 marks secured D grade.
The basic syntax of the formula is this:
=IF(Criteria, Action if Criteria Met, Other Action)
At the first row of Grade column, formula goes like this.
=IF(D2>=80,”A”, IF(D2>=75, “B”, IF(D2>=70, “B+”, IF(D2>=70,”B”, IF(D2>=65,”C+”, IF(D2>=60, “C”,”D”))))))
In the line of code, we are writing formula for D2 position which actually contains 60 marks in Grade column. From the formula (=IF D2 >=80,”A”,…..) inferring that if D2 position contains value which
is greater than or equal to 80. the Grade ‘A’ would be printed in the designated cell and if marks are greater than or equal to 75, ‘B’ would show and so on. When we put in this formula, it will show
the grade student has secured according to the formula constraints. When you will drag down the cross symbol to the end of the column, the formula will be applied to all respective cells.
Now at the first row of the Status column, we will write formula which goes like this:
From the above formula we can infer that if E2 position contains D, the word Fail will be seen in the Statuscolumn and in all other conditions Pass would show. Drag down the cross symbol to the end
of the column to apply the formula over all the respective cells.
As shown in the above screenshot, where the grade is ‘D’ the word ‘Fail’ can be seen in the corresponding cell.
Conditional Formatting allow users to show only specific data that meets a certain criteria or condition. Navigate to Home tab and click Conditional Formatting button, you will see list of different
Select the Marks column, go to conditional formatting and in Highlight Cells Rules, click Less than. Enter the value 60 and select any formatting style. On writing valid value, the selected column
will automatically be updated with selected formatting style. Click OK to apply.
Apply conditional formatting styles over all the data to make portion of datasheet more prominent and clear. | {"url":"http://wizapps.org/excel/apply-if-formulas-and-conditional-formatting-in-excel-2010/","timestamp":"2014-04-17T18:47:25Z","content_type":null,"content_length":"10808","record_id":"<urn:uuid:2817be01-3d99-43ab-9d8f-df6ae1563d8f>","cc-path":"CC-MAIN-2014-15/segments/1398223203422.8/warc/CC-MAIN-20140423032003-00528-ip-10-147-4-33.ec2.internal.warc.gz"} |
How to convert VA to kW
How to convert apparent power in volt-amps (VA) to real power in kilowatts (kW).
Volt-amps to kW calculation formula
The real power P in kilowatts (kW) is equal to the apparent power S in volt-amps (VA), times the power factor PF divided by 1000:
P[(kW)] = S[(VA)] × PF / 1000
So kilowatts are equal to volt-amps times the power factor divided by 1000.
kilowatts = volt-amps × PF / 1000
kW = VA × PF / 1000
What is the real power in kilowatts when the apparent power is 3000 VA and the power factor is 0.8?
P = 3000VA × 0.8 / 1000 = 2.4kW
See also | {"url":"http://www.rapidtables.com/convert/electric/va-to-kw.htm","timestamp":"2014-04-20T13:20:42Z","content_type":null,"content_length":"9611","record_id":"<urn:uuid:527e9900-3b33-4b1a-978d-f7d8572d77a8>","cc-path":"CC-MAIN-2014-15/segments/1397609538787.31/warc/CC-MAIN-20140416005218-00306-ip-10-147-4-33.ec2.internal.warc.gz"} |
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Pi question [Archive] - Straight Dope Message Board
11-23-1999, 09:48 AM
I have two questions.
1. What two numbers do you divide to get pi? I thought it was something like 22/7 but when I did some calculater calculations it did not come out exactly right.
2. If we used another system of counting (not base 10) could we get pi to be a definite number? I think it would work in theory, but practice is something completely different.
Just some things that I think about.
Gasoline: As an accompaniement to cereal it made a refreshing change. Glen Baxter | {"url":"http://boards.straightdope.com/sdmb/archive/index.php/t-8570.html","timestamp":"2014-04-19T22:34:53Z","content_type":null,"content_length":"30792","record_id":"<urn:uuid:ff237f96-0dd0-401f-bf5b-6d90373fef5b>","cc-path":"CC-MAIN-2014-15/segments/1398223205137.4/warc/CC-MAIN-20140423032005-00331-ip-10-147-4-33.ec2.internal.warc.gz"} |
Uniqueness of the Limit.
December 23rd 2005, 10:29 AM #1
Global Moderator
Nov 2005
New York City
Uniqueness of the Limit.
I was thinking perhaps the limit of a function at a constant or at infinity is not unique. Thus by delta-epsilon I mean there exists another number L' such that satisfies the condition of the
limit. Thus, prove that if the limit exists it is unique. I was able to show that if L is one limit and L' is another limit then (L+L')/2 is also a limit. Thus, this shows that if a limit is not
unique there exists infinitely many limits for that function. But this is not true I was trying to proof that because my books on advanced calculus did not consider this concept.
I was thinking perhaps the limit of a function at a constant or at infinity is not unique. Thus by delta-epsilon I mean there exists another number L' such that satisfies the condition of the
limit. Thus, prove that if the limit exists it is unique. I was able to show that if L is one limit and L' is another limit then (L+L')/2 is also a limit. Thus, this shows that if a limit is not
unique there exists infinitely many limits for that function. But this is not true I was trying to proof that because my books on advanced calculus did not consider this concept.
The limit of a real function $f(x)$ as $x \rightarrow a$ if it exists is unique. Suppose otherwise, then there exist $L_1$ and $L_2$ such that given any $\varepsilon >0$ there exist $\delta_1>0$
and $\delta_2>0$ such that when:
$|x-a|< \delta_1\ \Rightarrow \ |f(x)-L_1|<\varepsilon$
$|x-a|< \delta_2\ \Rightarrow \ |f(x)-L_2|<\varepsilon$.
So let $\delta=min(\delta_1,\delta_2)$, then we have:
$|x-a|< \delta\ \Rightarrow \ |f(x)-L_1|<\varepsilon$,
$|x-a|< \delta\ \Rightarrow \ |f(x)-L_2|<\varepsilon$.
But $|L_1-L_2|=|(L_1-f(x))-(L_2-f(x))|$, then by the triangle inequality:
$|L_1-L_2| \leq |L_1-f(x)|+|L_2-f(x)| < 2 \varepsilon$.
Hence $|L_1-L_2|$ is less than any positive real number, so $L_1=L_2$.
Thank you CaptainBlack, is that your own proof because if it is you used the triangle inequality nicely.
I also have a question how can I use Equation Editor like you did.
It's called Latex and you type in these tags to use it on this site: [ math ] code here [ /math ] (but without the spaces). There's a tutorial on how to use it somewhere on this site, I'll do
some searching.
Thank you CaptainBlack, is that your own proof because if it is you used the triangle inequality nicely.
I also have a question how can I use Equation Editor like you did.
Is it my own proof: yes and no, the ideas are recycled from things I have
see probably including proofs of this result.
Equation formatting uses this sites built in LaTeX facilties, see:
Thank you, the tutorial is rather complicated. Time for me to master it. Is Latex also used in other sites?
Funny if mathemations would have used latex code for real math (not interent) how messy it would have been.
Thank you, the tutorial is rather complicated. Time for me to master it. Is Latex also used in other sites?
Funny if mathemations would have used latex code for real math (not interent) how messy it would have been.
The LaTeX used here is a version of what is the de facto standard for writing
mathematical text. Most papers are produced in some version of TeX.
December 23rd 2005, 12:06 PM #2
Grand Panjandrum
Nov 2005
December 24th 2005, 02:37 PM #3
Global Moderator
Nov 2005
New York City
December 24th 2005, 04:27 PM #4
MHF Contributor
Oct 2005
December 24th 2005, 11:41 PM #5
Grand Panjandrum
Nov 2005
December 25th 2005, 12:33 PM #6
Global Moderator
Nov 2005
New York City
December 25th 2005, 01:15 PM #7
Grand Panjandrum
Nov 2005 | {"url":"http://mathhelpforum.com/calculus/1501-uniqueness-limit.html","timestamp":"2014-04-19T14:45:49Z","content_type":null,"content_length":"56590","record_id":"<urn:uuid:5bb5caec-fb63-4f11-8c90-21693b3e7890>","cc-path":"CC-MAIN-2014-15/segments/1397609539447.23/warc/CC-MAIN-20140416005219-00379-ip-10-147-4-33.ec2.internal.warc.gz"} |
Multiple Regression Assumptions and Outliers
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1. SW388R7 Data Analysis & Computers II Slide 1 Multiple Regression ? Assumptions and Outliers Multiple Regression and Assumptions Multiple Regression and Outliers Strategy for Solving Problems
Practice Problems
2. SW388R7 Data Analysis & Computers II Slide 2 Multiple Regression and Assumptions Multiple regression is most effect at identifying relationship between a dependent variable and a combination of
independent variables when its underlying assumptions are satisfied: each of the metric variables are normally distributed, the relationships between metric variables are linear, and the relationship
between metric and dichotomous variables is homoscedastic. Failing to satisfy the assumptions does not mean that our answer is wrong. It means that our solution may under-report the strength of the
3. SW388R7 Data Analysis & Computers II Slide 3 Multiple Regression and Outliers Outliers can distort the regression results. When an outlier is included in the analysis, it pulls the regression line
towards itself. This can result in a solution that is more accurate for the outlier, but less accurate for all of the other cases in the data set. We will check for univariate outliers on the
dependent variable and multivariate outliers on the independent variables.
4. SW388R7 Data Analysis & Computers II Slide 4 Relationship between assumptions and outliers The problems of satisfying assumptions and detecting outliers are intertwined. For example, if a case has
a value on the dependent variable that is an outlier, it will affect the skew, and hence, the normality of the distribution. Removing an outlier may improve the distribution of a variable.
Transforming a variable may reduce the likelihood that the value for a case will be characterized as an outlier.
5. SW388R7 Data Analysis & Computers II Slide 5 Order of analysis is important The order in which we check assumptions and detect outliers will affect our results because we may get a different
subset of cases in the final analysis. In order to maximize the number of cases available to the analysis, we will evaluate assumptions first. We will substitute any transformations of variable that
enable us to satisfy the assumptions. We will use any transformed variables that are required in our analysis to detect outliers.
6. SW388R7 Data Analysis & Computers II Slide 6 Strategy for solving problems Run type of regression specified in problem statement on variables using full data set. Test the dependent variable for
normality. If it does not satisfy the criteria for normality unless transformed, substitute the transformed variable in the remaining tests that call for the use of the dependent variable. Test for
normality, linearity, homoscedasticity using scripts. Decide which transformations should be used. Substitute transformations and run regression entering all independent variables, saving studentized
residuals and Mahalanobis distance scores. Compute probabilities for D?. Remove the outliers (studentized residual greater than 3 or Mahalanobis D? with p <= 0.001), and run regression with the
method and variables specified in the problem. Compare R? for analysis using transformed variables and omitting outliers (step 5) to R? obtained for model using all data and original variables (step
7. SW388R7 Data Analysis & Computers II Slide 7 Transforming dependent variables If dependent variable is not normally distributed: Try log, square root, and inverse transformation. Use first
transformed variable that satisfies normality criteria. If no transformation satisfies normality criteria, use untransformed variable and add caution for violation of assumption. If a transformation
satisfies normality, use the transformed variable in the tests of the independent variables.
8. SW388R7 Data Analysis & Computers II Slide 8 Transforming independent variables - 1 If independent variable is normally distributed and linearly related to dependent variable, use as is. If
independent variable is normally distributed but not linearly related to dependent variable: Try log, square root, square, and inverse transformation. Use first transformed variable that satisfies
linearity criteria and does not violate normality criteria If no transformation satisfies linearity criteria and does not violate normality criteria, use untransformed variable and add caution for
violation of assumption
9. SW388R7 Data Analysis & Computers II Slide 9 Transforming independent variables - 2 If independent variable is linearly related to dependent variable but not normally distributed: Try log, square
root, and inverse transformation. Use first transformed variable that satisfies normality criteria and does not reduce correlation. Try log, square root, and inverse transformation. Use first
transformed variable that satisfies normality criteria and has significant correlation. If no transformation satisfies normality criteria with a significant correlation, use untransformed variable
and add caution for violation of assumption
10. SW388R7 Data Analysis & Computers II Slide 10 Transforming independent variables - 3 If independent variable is not linearly related to dependent variable and not normally distributed: Try log,
square root, square, and inverse transformation. Use first transformed variable that satisfies normality criteria and has significant correlation. If no transformation satisfies normality criteria
with a significant correlation, used untransformed variable and add caution for violation of assumption
11. SW388R7 Data Analysis & Computers II Slide 11 Impact of transformations and omitting outliers We evaluate the regression assumptions and detect outliers with a view toward strengthening the
relationship. This may not happen. The regression may be the same, it may be weaker, and it may be stronger. We cannot be certain of the impact until we run the regression again. In the end, we may
opt not to exclude outliers and not to employ transformations; the analysis informs us of the consequences of doing either.
12. SW388R7 Data Analysis & Computers II Slide 12 Notes Whenever you start a new problem, make sure you have removed variables created for previous analysis and have included all cases back into the
data set. I have added the square transformation to the checkboxes for transformations in the normality script. Since this is an option for linearity, we need to be able to evaluate its impact on
normality. If you change the options for output in pivot tables from labels to names, you will get an error message when you use the linearity script. To solve the problem, change the option for
output in pivot tables back to labels.
13. SW388R7 Data Analysis & Computers II Slide 13 Problem 1 In the dataset GSS2000.sav, is the following statement true, false, or an incorrect application of a statistic? Assume that there is no
problem with missing data. Use a level of significance of 0.01 for the regression analysis. Use a level of significance of 0.01 for evaluating assumptions. The research question requires us to
identify the best subset of predictors of "total family income" [income98] from the list: "sex" [sex], "how many in family earned money" [earnrs], and "income" [rincom98]. After substituting
transformed variables to satisfy regression assumptions and removing outliers, the total proportion of variance explained by the regression analysis increased by 10.8%. 1. True 2. True with caution
3. False 4. Inappropriate application of a statistic
14. SW388R7 Data Analysis & Computers II Slide 14 Dissecting problem 1 - 1 In the dataset GSS2000.sav, is the following statement true, false, or an incorrect application of a statistic? Assume that
there is no problem with missing data. Use a level of significance of 0.01 for the regression analysis. Use a level of significance of 0.01 for evaluating assumptions. The research question requires
us to identify the best subset of predictors of "total family income" [income98] from the list: "sex" [sex], "how many in family earned money" [earnrs], and "income" [rincom98]. After substituting
transformed variables to satisfy regression assumptions and removing outliers, the total proportion of variance explained by the regression analysis increased by 10.8%. 1. True 2. True with caution
3. False 4. Inappropriate application of a statistic
15. SW388R7 Data Analysis & Computers II Slide 15 Dissecting problem 1 - 2 In the dataset GSS2000.sav, is the following statement true, false, or an incorrect application of a statistic? Assume that
there is no problem with missing data. Use a level of significance of 0.01 for the regression analysis. Use a level of significance of 0.01 for evaluating assumptions. The research question requires
us to identify the best subset of predictors of "total family income" [income98] from the list: "sex" [sex], "how many in family earned money" [earnrs], and "income" [rincom98]. After substituting
transformed variables to satisfy regression assumptions and removing outliers, the total proportion of variance explained by the regression analysis increased by 10.8%. 1. True 2. True with caution
3. False 4. Inappropriate application of a statistic
16. SW388R7 Data Analysis & Computers II Slide 16 Dissecting problem 1 - 3 In the dataset GSS2000.sav, is the following statement true, false, or an incorrect application of a statistic? Assume that
there is no problem with missing data. Use a level of significance of 0.01 for the regression analysis. Use a level of significance of 0.01 for evaluating assumptions. The research question requires
us to identify the best subset of predictors of "total family income" [income98] from the list: "sex" [sex], "how many in family earned money" [earnrs], and "income" [rincom98]. After substituting
transformed variables to satisfy regression assumptions and removing outliers, the total proportion of variance explained by the regression analysis increased by 10.8%. 1. True 2. True with caution
3. False 4. Inappropriate application of a statistic
17. SW388R7 Data Analysis & Computers II Slide 17 R? before transformations or removing outliers
18. SW388R7 Data Analysis & Computers II Slide 18 R? before transformations or removing outliers
19. SW388R7 Data Analysis & Computers II Slide 19 R? before transformations or removing outliers
20. SW388R7 Data Analysis & Computers II Slide 20 Normality of the dependent variable: total family income
21. SW388R7 Data Analysis & Computers II Slide 21 Normality of the dependent variable: total family income
22. SW388R7 Data Analysis & Computers II Slide 22 Linearity and independent variable: how many in family earned money
23. SW388R7 Data Analysis & Computers II Slide 23 Linearity and independent variable: how many in family earned money
24. SW388R7 Data Analysis & Computers II Slide 24 Normality of independent variable:how many in family earned money
25. SW388R7 Data Analysis & Computers II Slide 25 Normality of independent variable:how many in family earned money
26. SW388R7 Data Analysis & Computers II Slide 26 Normality of independent variable:how many in family earned money
27. SW388R7 Data Analysis & Computers II Slide 27 Transformation for how many in family earned money The independent variable, how many in family earned money, had a linear relationship to the
dependent variable, total family income. The logarithmic transformation improves the normality of "how many in family earned money" [earnrs] without a reduction in the strength of the relationship to
"total family income" [income98]. We will substitute the logarithmic transformation of how many in family earned money in the regression analysis.
28. SW388R7 Data Analysis & Computers II Slide 28 Normality of independent variable:respondent?s income
29. SW388R7 Data Analysis & Computers II Slide 29 Normality of independent variable: respondent?s income
30. SW388R7 Data Analysis & Computers II Slide 30 Linearity and independent variable: respondent?s income
31. SW388R7 Data Analysis & Computers II Slide 31 Linearity and independent variable: respondent?s income
32. SW388R7 Data Analysis & Computers II Slide 32 Homoscedasticity: sex
33. SW388R7 Data Analysis & Computers II Slide 33 Homoscedasticity: sex
34. SW388R7 Data Analysis & Computers II Slide 34 Adding a transformed variable
35. SW388R7 Data Analysis & Computers II Slide 35 The transformed variable in the data editor
36. SW388R7 Data Analysis & Computers II Slide 36 The regression to identify outliers
37. SW388R7 Data Analysis & Computers II Slide 37 Saving the measures of outliers
38. SW388R7 Data Analysis & Computers II Slide 38 The variables for identifying outliers
39. SW388R7 Data Analysis & Computers II Slide 39 Computing the probability for Mahalanobis D?
40. SW388R7 Data Analysis & Computers II Slide 40 Formula for probability for Mahalanobis D?
41. SW388R7 Data Analysis & Computers II Slide 41 Multivariate outliers
42. SW388R7 Data Analysis & Computers II Slide 42 Univariate outliers
43. SW388R7 Data Analysis & Computers II Slide 43 Omitting the outliers
44. SW388R7 Data Analysis & Computers II Slide 44 Specifying the condition to omit outliers
45. SW388R7 Data Analysis & Computers II Slide 45 The formula for omitting outliers
46. SW388R7 Data Analysis & Computers II Slide 46 Completing the request for the selection
47. SW388R7 Data Analysis & Computers II Slide 47 The omitted multivariate outlier
48. SW388R7 Data Analysis & Computers II Slide 48 Running the regression without outliers
49. SW388R7 Data Analysis & Computers II Slide 49 Opening the save options dialog
50. SW388R7 Data Analysis & Computers II Slide 50 Clearing the request to save outlier data
51. SW388R7 Data Analysis & Computers II Slide 51 Opening the statistics options dialog
52. SW388R7 Data Analysis & Computers II Slide 52 Requesting descriptive statistics
53. SW388R7 Data Analysis & Computers II Slide 53 Requesting the output
54. SW388R7 Data Analysis & Computers II Slide 54 Sample size requirement
55. SW388R7 Data Analysis & Computers II Slide 55 Significance of regression relationship
56. SW388R7 Data Analysis & Computers II Slide 56 Increase in proportion of variance
57. SW388R7 Data Analysis & Computers II Slide 57 Problem 2 In the dataset GSS2000.sav, is the following statement true, false, or an incorrect application of a statistic? Assume that there is no
problem with missing data. Use a level of significance of 0.05 for the regression analysis. Use a level of significance of 0.01 for evaluating assumptions. The research question requires us to
examine the relationship of "age" [age], "highest year of school completed" [educ], and "sex" [sex] to the dependent variable "occupational prestige score" [prestg80]. After substituting transformed
variables to satisfy regression assumptions and removing outliers, the proportion of variance explained by the regression analysis increased by 3.6%. 1. True 2. True with caution 3. False 4.
Inappropriate application of a statistic
58. SW388R7 Data Analysis & Computers II Slide 58 Dissecting problem 2 - 1 In the dataset GSS2000.sav, is the following statement true, false, or an incorrect application of a statistic? Assume that
there is no problem with missing data. Use a level of significance of 0.05 for the regression analysis. Use a level of significance of 0.01 for evaluating assumptions. The research question requires
us to examine the relationship of "age" [age], "highest year of school completed" [educ], and "sex" [sex] to the dependent variable "occupational prestige score" [prestg80]. After substituting
transformed variables to satisfy regression assumptions and removing outliers, the proportion of variance explained by the regression analysis increased by 3.6%. 1. True 2. True with caution 3. False
4. Inappropriate application of a statistic
59. SW388R7 Data Analysis & Computers II Slide 59 Dissecting problem 2 - 2 In the dataset GSS2000.sav, is the following statement true, false, or an incorrect application of a statistic? Assume that
there is no problem with missing data. Use a level of significance of 0.05 for the regression analysis. Use a level of significance of 0.01 for evaluating assumptions. The research question requires
us to examine the relationship of "age" [age], "highest year of school completed" [educ], and "sex" [sex] to the dependent variable "occupational prestige score" [prestg80]. After substituting
transformed variables to satisfy regression assumptions and removing outliers, the proportion of variance explained by the regression analysis increased by 3.6%. 1. True 2. True with caution 3. False
4. Inappropriate application of a statistic
60. SW388R7 Data Analysis & Computers II Slide 60 In the dataset GSS2000.sav, is the following statement true, false, or an incorrect application of a statistic? Assume that there is no problem with
missing data. Use a level of significance of 0.05 for the regression analysis. Use a level of significance of 0.01 for evaluating assumptions. The research question requires us to examine the
relationship of "age" [age], "highest year of school completed" [educ], and "sex" [sex] to the dependent variable "occupational prestige score" [prestg80]. After substituting transformed variables to
satisfy regression assumptions and removing outliers, the proportion of variance explained by the regression analysis increased by 3.6%. 1. True 2. True with caution 3. False 4. Inappropriate
application of a statistic Dissecting problem 2 - 3
61. SW388R7 Data Analysis & Computers II Slide 61 R? before transformations or removing outliers
62. SW388R7 Data Analysis & Computers II Slide 62 R? before transformations or removing outliers
63. SW388R7 Data Analysis & Computers II Slide 63 Normality of the dependent variable
64. SW388R7 Data Analysis & Computers II Slide 64 Normality of the dependent variable
65. SW388R7 Data Analysis & Computers II Slide 65 Normality of independent variable: Age
66. SW388R7 Data Analysis & Computers II Slide 66 Normality of independent variable: Age
67. SW388R7 Data Analysis & Computers II Slide 67 Linearity and independent variable: Age
68. SW388R7 Data Analysis & Computers II Slide 68 Linearity and independent variable: Age
69. SW388R7 Data Analysis & Computers II Slide 69 Transformation for Age The independent variable age satisfied the criteria for normality. The independent variable age did not have a linear
relationship to the dependent variable occupational prestige. However, none of the transformations linearized the relationship. No transformation will be used - it would not help linearity and is not
needed for normality.
70. SW388R7 Data Analysis & Computers II Slide 70 Linearity and independent variable: Highest year of school completed
71. SW388R7 Data Analysis & Computers II Slide 71 Linearity and independent variable: Highest year of school completed
72. SW388R7 Data Analysis & Computers II Slide 72 Normality of independent variable: Highest year of school completed
73. SW388R7 Data Analysis & Computers II Slide 73 Normality of independent variable: Highest year of school completed
74. SW388R7 Data Analysis & Computers II Slide 74 Transformation for highest year of school The independent variable, highest year of school, had a linear relationship to the dependent variable,
occupational prestige. The independent variable, highest year of school, did not satisfy the criteria for normality. None of the transformations for normalizing the distribution of "highest year of
school completed" [educ] were effective. No transformation will be used - it would not help normality and is not needed for linearity. A caution should be added to any findings.
75. SW388R7 Data Analysis & Computers II Slide 75 Homoscedasticity: sex
76. SW388R7 Data Analysis & Computers II Slide 76 Homoscedasticity: sex
77. SW388R7 Data Analysis & Computers II Slide 77 Adding a transformed variable
78. SW388R7 Data Analysis & Computers II Slide 78 The transformed variable in the data editor
79. SW388R7 Data Analysis & Computers II Slide 79 The regression to identify outliers
80. SW388R7 Data Analysis & Computers II Slide 80 Saving the measures of outliers
81. SW388R7 Data Analysis & Computers II Slide 81 The variables for identifying outliers
82. SW388R7 Data Analysis & Computers II Slide 82 Computing the probability for Mahalanobis D?
83. SW388R7 Data Analysis & Computers II Slide 83 Formula for probability for Mahalanobis D?
84. SW388R7 Data Analysis & Computers II Slide 84 The multivariate outlier
85. SW388R7 Data Analysis & Computers II Slide 85 The univariate outlier
86. SW388R7 Data Analysis & Computers II Slide 86 Omitting the outliers
87. SW388R7 Data Analysis & Computers II Slide 87 Specifying the condition to omit outliers
88. SW388R7 Data Analysis & Computers II Slide 88 The formula for omitting outliers
89. SW388R7 Data Analysis & Computers II Slide 89 Completing the request for the selection
90. SW388R7 Data Analysis & Computers II Slide 90 The omitted multivariate outlier
91. SW388R7 Data Analysis & Computers II Slide 91 Running the regression without outliers
92. SW388R7 Data Analysis & Computers II Slide 92 Opening the save options dialog
93. SW388R7 Data Analysis & Computers II Slide 93 Clearing the request to save outlier data
94. SW388R7 Data Analysis & Computers II Slide 94 Opening the statistics options dialog
95. SW388R7 Data Analysis & Computers II Slide 95 Requesting descriptive statistics
96. SW388R7 Data Analysis & Computers II Slide 96 Requesting the output
97. SW388R7 Data Analysis & Computers II Slide 97 Sample size requirement
98. SW388R7 Data Analysis & Computers II Slide 98 Significance of regression relationship
99. SW388R7 Data Analysis & Computers II Slide 99 Increase in proportion of variance
100. SW388R7 Data Analysis & Computers II Slide 100 Impact of assumptions and outliers - 1
101. SW388R7 Data Analysis & Computers II Slide 101 Impact of assumptions and outliers - 2
102. SW388R7 Data Analysis & Computers II Slide 102 Impact of assumptions and outliers - 3
103. SW388R7 Data Analysis & Computers II Slide 103 Impact of assumptions and outliers - 4
104. SW388R7 Data Analysis & Computers II Slide 104 Impact of assumptions and outliers - 5 | {"url":"http://www.slideserve.com/Jims/multiple-regression-assumptions-and-outliers","timestamp":"2014-04-19T02:56:23Z","content_type":null,"content_length":"66691","record_id":"<urn:uuid:5fed2310-bfeb-4077-a941-68173ea7da87>","cc-path":"CC-MAIN-2014-15/segments/1398223207046.13/warc/CC-MAIN-20140423032007-00105-ip-10-147-4-33.ec2.internal.warc.gz"} |
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[Haskell-cafe] Gentle introduction questions / comments
Cristiano Paris frodo at theshire.org
Tue Jan 27 09:10:47 EST 2009
2009/1/27 Matthijs Kooijman <matthijs at stdin.nl>
> Hi all,
> I've been reading the gentle introduction to Haskell a bit more closely today
> and there a few things which I can't quite understand (presumably because they
> are typo's). I've found two issues with the "Using monads" section [1]. Not
> sure if this is the right place to report this, but there's probably somewhere
> here who can fix it :-)
> The first one is in the state monad data type. It reads:
> data SM a = SM (S -> (a,S)) -- The monadic type
> I can't seem to find out what "S" is exactly here. It seems to be a
> universally quantified type variable, but then it should be lowercase.
> Considering that it is referred to later on as "s", I suspect this is a type?
I guess it's wrong.
S should be the type of the state but then it should've been declared
as a type variable. I think a more readable declaration could be:
data SM s a = SM { runState :: s -> (a,s) }
> Further on, a resource monad is described. Here, a lift1 function is given
> to allow one to lift a normal computation into a resource aware computation,
> with the following type signature:
> lift1 :: (a -> b) -> (R a -> R b)
Mmmmhhh... this seems the signature of the liftM function, whose
purpose is to make a function operate on monadic values instead of
pure values. Notice that this is different from the "lift" function
you described above. A computation is a monadic value (i.e. an object
of the type "Monad m => m a") and the "lift" function wraps that value
into a monad transformer, making it a new monadic value (i.e. an
object of the type "MonadTrans t, Monad m => t m a").
> A few lines down, the inc operation is defined as follows:
> inc i = lift1 (i+1)
This is clearly wrong, as the expression "i+1" is not of the function
type, as required by the signature of lift1 (i.e. a -> b).
> To me, this seems wrong. Even without going into the semantics, it seems that
> the (i+1) expression does not satisfy the type of the first argument to lift1.
> I suspect this should have been:
> inc = lift1 (+1)
You're right.
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Topic: Finishing up, explaining FLT Proof conclusion
Replies: 45 Last Post: Aug 1, 2001 2:53 AM
Messages: [ Previous | Next ]
Re: Finishing up, explaining FLT Proof conclusion
Posted: Jul 26, 2001 6:26 PM
James Harris wrote:
> Paul Sperry <plsperry@sc.rr.com> wrote in message news:<plsperry-3C9F3E.00043026072001@news2.southeast.rr.com>...
> > In article <3c65f87.0107251356.16778672@posting.google.com>,
> > jstevh@msn.com (James Harris) wrote:
> > [...]
> > > So, let me re-emphasize my point, by asking you a simple question:
> > >
> > > Can you start with integers and end up in the field of rationals with
> > > fractions...your regular old garden variety fraction like 1/2...using
> > > only addition and multiplication?
> > >
> >
> > Yes, I can. The construction is standard, routine and well known:
> > On the subset S of the set Z x Z of ordered pairs of integers consisting
> > of all pairs (a,b) with b != 0, define (a,b)==(a',b') iff
> > a * b' = a' * b where "*" is integer multiplicaion.
> "standard", "well known" and CIRCULAR.
Well known, yes, but not by you it seems.
> Folks, he's saying that in integers, just say that 2b = 3, claim that
> b "exists" (which he can't prove, so he must base it on some
No, Paul is not bringing b into existence by fiat. He is defining it
from the raw ingredients: integers, ordered pair, and equivalence
class. Specifically, b is the equivalence class containing the pair of
integers 3, 2 (in that order). Let's denote that equivalence class by
[3,2]. There are also equivalence classes [2,1] and [3,1]. We can
define an operation of multiplication (let's call it x) on the
equivalence classes such that [2,1]x[3,2] = [3,1]. Now, note that we
are talking about the things that Paul _constructed_ NOT something that
he magic'd into existence. Note also that with a change of notation:
write [u,v] as u/v and u/v as u if v = 1, we get
2/1 x 3/2 = 3/1 or 2 x 3/2 = 3
which is _still_ a statement about arithmetic with those equivalence
classes. Let's make a further change of notation and write the
operation of multiplication of those equivalence classes (that's x)
using the symbol for the operation of multiplication of integers: *. We
2 * 3/2 = 3.
So, by an abuse of notation, the left-most 2 and the right-most 3 _look_
like integers, but they are not. Is that why you thought that the
definition was circular? Finally, using the embedding a -> [a,1] of the
integers into the rationals, we _can_ take the left-most 2 and the
right-most 3 to be the integers they look like. This isomorphism makes
official, so to speak, the re-writing of [u,1] as u.
You may have this worry: 3/2 corresponds to the ordered pair <3,2> in
the equivalence class [3,2] but what if another representative, say
<6,4> had been chosen? Luckily one can prove that the choice of
representative makes no difference.
At this point it will do no harm if you go back and read Paul's
construction of the rationals.
> principle, which basically just says it exists), and presto
> magico!--you have a fraction.
> Of course, in the real world we use fractions as an intermediate step.
> The silliness of a circular definition like what he just gave is that
> you can use if for just about ANYTHING.
> And then argue that people who question it are crazy.
> Oh yeah, speaking of "crazy" stuff. Since I'm downplaying fractions,
> how do I handle transcendentals? I skipped past that before to say
> they lie at infinity.
> So what does that mean?
> First off, remember how I said that decimals are humanity's way of
> using really big counting numbers. Well, you could say that human
> beings got tired of things like 1414, and realized it was more
> convenient to *write* 1.414, because you see, the front of it doesn't
> move so much.
Your language is so strange sometimes! I picture a car's odometer
reading 1414, as we drive along the least significant 4 quickly changes
to 5. The most significant 1 only changes at one thousandth the speed
of that 4. A child (bright, but lacking the vocabulary) points to the
odometer and announces "The front of it doesn't move so much".
> Well, what pi actually IS, is the count of the minimum number of
> elements necessary to make half the circumference of a circle.
> And you're going, hey, I knew that.
Are they? Then they understand you better than I do. Are you thinking
of small measuring rods (called "elements"?) laid out end-to-end along
the half-circumference and ditto along half the diameter (dunno why
we've got two halves)? Let's say that there are 987 along the
half-circumference and 314 along the radius ignoring in each case
fractions. So we *almost* say that pi is 987/314 but not really because
we know we've only done the job approximately. So we use a smaller
measuring rod to do it more accurately. Do you want to use a measuring
rod that fits along the radius some-power-of-ten times? Lets say 1000
and lets say the half-circumference is 3143. So pi is approximately
3143/1000 but since fractions are fictitious we'll just say 3143.
Didn't Simon Stevin travel this route, didn't the Pythagoreans before
him? Don't reinvent the wheel James, or even the half-wheel. What I
don't understand is "_minimum_ number of elements".
> Well, then then think about this:
> We know that there are other numbers besides 1. Think of pi as the
> count with 1, and yes it's that count because human beings LOVE
I've changed my mind about what you might mean. Are you thinking: let
the measuring rod be flexible and let it be as long as half the diameter
and deem its length to be 1. Now as we lay it along the
half-circumference we find it goes three times and falls short by a
bit. Cut the measuring rod into ten equal pieces and use one of those
pieces to measure that bit. One, and still a bit short. Cut into ten
again and get four. And so on. But "_minimum_ number of elements"?
And in this version what is an element?
> counting numbers. You see, they're really good with them.
> And they get started early with their fingers and toes.
> And their brains are PROGRAMMED to feel good with counting numbers, so
Do you think that what Chomsky claimed of a "deep structure" of language
has a counter-part in mathematics? Are counting numbers and their
"grammar" the deep structure of mathematics as a whole? Is the actual
performance of mathematicians a flawed version of the "surface
> they feel COMFORTABLE with counting numbers :-).
> How about something outside of your regular programming? How many NON
> discrete elements does it take to make half the circumference of a
> circle?
I don't know! What's a non discrete element? Is it an infinitesimal by
any chance?
> James Harris
Date Subject Author
7/24/01 Finishing up, explaining FLT Proof conclusion JAMES HARRIS
7/25/01 Re: Finishing up, explaining FLT Proof conclusion The Scarlet Manuka
7/25/01 Re: Finishing up, explaining FLT Proof conclusion Dave Pritchard
7/25/01 Re: Finishing up, explaining FLT Proof conclusion Yimin Rong
7/25/01 Re: Finishing up, explaining FLT Proof conclusion pritchar@ca.ibm.com
7/25/01 Re: Finishing up, explaining FLT Proof conclusion Elaine Jackson
7/25/01 Re: Finishing up, explaining FLT Proof conclusion The Scarlet Manuka
7/25/01 Re: Finishing up, explaining FLT Proof conclusion Jan Kristian Haugland
7/25/01 Re: Finishing up, explaining FLT Proof conclusion Douglas
7/25/01 Re: Finishing up, explaining FLT Proof conclusion Randy Poe
7/25/01 Re: Finishing up, explaining FLT Proof conclusion David C. Ullrich
7/25/01 Re: Finishing up, explaining FLT Proof conclusion Rick Decker
7/25/01 Re: Finishing up, explaining FLT Proof conclusion JAMES HARRIS
7/25/01 Re: Finishing up, explaining FLT Proof conclusion Peter Percival
7/25/01 Re: Finishing up, explaining FLT Proof conclusion Dik T. Winter
7/26/01 Re: Finishing up, explaining FLT Proof conclusion Paul Sperry
7/26/01 Re: Finishing up, explaining FLT Proof conclusion JAMES HARRIS
7/26/01 Re: Finishing up, explaining FLT Proof conclusion Phil Carmody
7/26/01 Re: Finishing up, explaining FLT Proof conclusion Torkel Franzen
7/26/01 Re: Finishing up, explaining FLT Proof conclusion Phil Carmody
7/26/01 Re: Finishing up, explaining FLT Proof conclusion mareg@mimosa.csv.warwick.ac.uk
7/26/01 Re: Finishing up, explaining FLT Proof conclusion Lee Rudolph
7/26/01 Re: Finishing up, explaining FLT Proof conclusion Phil Carmody
7/26/01 Re: Finishing up, explaining FLT Proof conclusion moesyzlak@hotmail.com
7/26/01 Re: Finishing up, explaining FLT Proof conclusion JAMES HARRIS
7/26/01 Re: Finishing up, explaining FLT Proof conclusion magidin@math.berkeley.edu
7/26/01 Re: Finishing up, explaining FLT Proof conclusion JAMES HARRIS
7/26/01 Re: Finishing up, explaining FLT Proof conclusion magidin@math.berkeley.edu
7/27/01 Re: Finishing up, explaining FLT Proof conclusion Richard Tobin
7/27/01 Re: Finishing up, explaining FLT Proof conclusion pritchar@ca.ibm.com
7/27/01 Re: Finishing up, explaining FLT Proof conclusion David C. Ullrich
7/26/01 Re: Finishing up, explaining FLT Proof conclusion Peter Percival
7/27/01 Re: Finishing up, explaining FLT Proof conclusion Paul Sperry
7/28/01 Re: Finishing up, explaining FLT Proof conclusion Jan Kristian Haugland
7/28/01 Re: Finishing up, explaining FLT Proof conclusion JAMES HARRIS
7/28/01 Re: Finishing up, explaining FLT Proof conclusion magidin@math.berkeley.edu
7/26/01 Re: Finishing up, explaining FLT Proof conclusion The Scarlet Manuka
7/26/01 Re: Finishing up, explaining FLT Proof conclusion magidin@math.berkeley.edu
7/27/01 Re: Finishing up, explaining FLT Proof conclusion The Scarlet Manuka
7/26/01 Re: Finishing up, explaining FLT Proof conclusion Rick Decker
7/26/01 Re: Finishing up, explaining FLT Proof conclusion JAMES HARRIS
7/26/01 Re: Finishing up, explaining FLT Proof conclusion Rick Decker
7/27/01 Re: Finishing up, explaining FLT Proof conclusion Andy Spragg
7/28/01 Re: Finishing up, explaining FLT Proof conclusion David C. Ullrich
8/1/01 Re: Finishing up, explaining FLT Proof conclusion The Scarlet Manuka
7/26/01 Re: Finishing up, explaining FLT Proof conclusion Dave Pritchard | {"url":"http://mathforum.org/kb/thread.jspa?threadID=73092&messageID=325761","timestamp":"2014-04-19T12:25:03Z","content_type":null,"content_length":"77624","record_id":"<urn:uuid:9a686075-8350-43c1-a5ef-5e41dea2fdcc>","cc-path":"CC-MAIN-2014-15/segments/1397609537186.46/warc/CC-MAIN-20140416005217-00288-ip-10-147-4-33.ec2.internal.warc.gz"} |
Markov Random Fields Without Data Term
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Maybe this is a very elementary question. But, I'm trying to learn how to work with Markov Random Fields for computer vision applications. Most, MRF Problems involve a data term and a pairwise
smoothness term. But I came across a paper in which the data term was missing. So, from the point of view of implementation, I assume that their effect is nullified by setting them to the same value.
Now, the inference algorithms that I know of (alpha-expansion and alpha-beta search and other move-making ones) adjust the labels and check for a decrease in the energy of the MRF. Now, assuming that
I'm using a Potts model for the smoothness prior, where 0 energy is assigned to an edge if both labels are the same and positive if different. Wouldn't I end up getting an MRF whose labels are
identical? Or is there something wrong in my data terms being set to the same value assumption.
machine-learning computer-vision
asked Oct 17 '13 at 6:31
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(a) Its Notion
§ 497
Quantum alters and becomes another quantum; the further determination of this alteration, namely, that it goes on to infinity, lies in the circumstance that quantum is established as being immanently
self-contradictory. Quantum becomes an other; but it continues itself into its otherness; the other is thus also a quantum. This, however, is not only the other of a particular quantum, but of
quantum itself, the negative of quantum as limited; hence it is the unlimitedness of quantum, its infinity. Quantum is an ought-to-be; it is by implication determined as being for itself, and this
being-determined-for-itself is rather the being-determined-in-an-other, and, conversely, it is the sublation of being-determined-in-an-other, is an indifferent subsisting for itself.
§ 498
In this way, finitude and infinity each acquire in themselves a dual, and indeed, an opposite meaning. The quantum is finite, in the first place simply as limited, and secondly, as impelled beyond
itself, as being determined in an other. But the infinity of quantum is first, its unlimitedness, and secondly, its returnedness into itself, its indifferent being-for-self. If we now compare these
moments with each other, we find that the determination of the finitude of quantum, the impulse to go beyond itself to an other in which its determination lies, is equally the determination of the
infinite; the negation of the limit is the same impulsion beyond the determinateness, so that in this negation, in the infinite, quantum possesses its final determinateness. The other moment of
infinity is the being-for-self which is indifferent to the limit; but the limiting of quantum itself is such that quantum is explicitly indifferent to its limit, and hence to other quanta and to its
beyond. In quantum, finitude and infinity (the spurious infinity supposedly separate from the finite) each already has within it the moment of the other.
§ 499
The difference between the qualitative and quantitative infinite is that in the former the finite and infinite are qualitatively opposed and the transition of the finite into the infinite, or the
relation of each to the other, lies only in the in-itself, in their Notion. Qualitative determinateness, as an immediacy, is related to otherness essentially as to an alien being; it is not posited
as having its negation, its other within it. Quantity, on the other hand, is, as such, sublated determinateness; it is posited as being unlike and indifferent to itself, consequently as alterable.
Therefore the qualitative finite and infinite stand absolutely, that is abstractly, opposed to each other; their unity is their underlying inner relation; and therefore the finite continues itself
into its other only implicitly, not affirmatively. The quantitative finite, on the other hand, is self-related in its infinite, in which it has its absolute determinateness. This their relation is
displayed in the first place in the quantitative infinite progress.
(b) The Quantitative Infinite Progress
§ 500
The progress to infinity is in general the expression of contradiction, here, of that which is implicit in the quantitative finite, or quantum as such. It is the reciprocal determining of the finite
and infinite which was considered in the sphere of quality, with the difference that, as just remarked, in the sphere of quantity the limit in its own self dispatches and continues itself into its
beyond and hence, conversely, the quantitative infinite too is posited as having quantum within it; for quantum in its self-externality is also its own self, its externality belongs to its
§ 501
Now the infinite progress is only the expression of this contradiction, not its resolution; but because the one determinateness is continued into its other, the progress gives rise to the show of a
solution in a union of both. As at first posed, it is the problem of attaining the infinite, not the actual reaching of it; it is the perpetual generation of the infinite, but it does not get beyond
quantum, nor does the infinite become positively present. It belongs to the Notion of quantum to have a beyond of itself. This beyond is first, the abstract moment of the non-being of quantum: the
vanishing of quantum is its own act; it is thus related to its beyond as to its infinity, in accordance with the qualitative moment of the opposition. Secondly, however, quantum is continuous with
its beyond; quantum consists precisely in being the other of itself, in being external to itself; this externality is, therefore, no more an other than quantum itself; the beyond or the infinite is,
therefore, itself a quantum. In this way, the beyond is recalled from its flight and the infinite is attained. But because the infinite now affirmatively present is again a quantum, what has been
posited is only a fresh limit; this, too, as a quantum, has again fled from itself, is as such beyond itself and has repelled itself into its non-being, into its own beyond, and as it thus repels
itself into the beyond, so equally does the beyond perpetually become a quantum.
§ 502
The continuity of quantum with its other produces the conjunction of both in the expression of an infinitely great or infinitely small. Since both still bear the character of quantum they remain
alterable, and the absolute determinateness which would be a being-for-self is, therefore, not attained. This self-externality of the determination is posited in the dual infinite — which is opposed
to itself as a 'more' and a 'less' — in the infinitely great and infinitely small. In each, the quantum is maintained in perpetual opposition to its beyond. No matter how much the quantum is
increased, it shrinks to insignificance; because quantum is related to the infinite as to its non-being, the opposition is qualitative; the increased quantum has therefore gained nothing from the
infinite, which is now, as before, the non-being of quantum. In other words, the increase of quantum brings it no nearer to the infinite; for the difference between quantum and its infinity is
essentially not a quantitative difference. The expression 'the infinitely great' only throws the contradiction into sharper relief; it is supposed to be great, that is, a quantum, and infinite, that
is, not a quantum. Similarly, the infinitely small is, as small, a quantum, and therefore remains absolutely, that is, qualitatively, too great for the infinite and is opposed to it. In both, there
remains the contradiction of the infinite progress which in them should have reached its goal.
§ 503
This infinity which is perpetually determined as the beyond of the finite is to be described as the spurious quantitative infinite. Like the qualitative spurious infinite, it is the perpetual
movement to and fro from one term of the lasting contradiction to the other, from the limit to its non-being, and from this back again to the limit. It is true that in the quantitative progress the
movement is not simply towards an abstract other in general, but towards an explicitly different quantum; but this remains in the same way opposed to its negation. The progress, too, is therefore not
a real advance but a repetition of one and the same thing, a positing, a sublating, and then again a positing and again a sublating, an impotence of the negative, for what it sublates is continuous
with it, and in the very act of being sublated returns to it. Thus there are two terms, the bond between which is such that they simply flee from each other; and in fleeing from each other they
cannot become separated but are joined together even in their flight from each other.
§ 504
The spurious infinite, especially in the form of the quantitative progress to infinity which continually surmounts the limit it is powerless to remove, and perpetually falls back into it, is commonly
held to be something sublime and a kind of divine worship, while in philosophy it has been regarded as ultimate. This progression has often been the theme of tirades which have been admired as
sublime productions. As a matter of fact, however, this modern sublimity does not magnify the object — rather does this take flight — but only the subject which assimilates such vast quantities. The
hollowness of this exaltation, which in scaling the ladder of the quantitative still remains subjective, finds expression in its own admission of the futility of its efforts to get nearer to the
infinite goal, the attainment of which must, indeed, be achieved by a quite different method.
§ 505
In the following tirades of this kind it is also stated what becomes of such exaltation and how it finishes. Kant, for example, at the close of the Critique of Practical Reason, represents it as
sublime 'when the subject raises himself in thought above the place he occupies in the world of sense, reaching out to infinity, to stars beyond stars, worlds beyond worlds, systems beyond systems,
and then also to the limitless times of their periodic motion, their beginning and duration. Imagination fails before this progress into the infinitely remote, where beyond the most distant world
there is a still more distant one, and the past, however remote, has a still remoter past behind it, the future, however distant, a still more distant future beyond it; thought fails in the face of
this conception of the immeasurable, just as a dream, in which one goes on and on down a corridor which stretches away endlessly out of sight, finishes with falling or fainting.'
§ 506
This exposition, besides giving a concise yet rich description of such quantitative exaltation, deserves praise mainly on account of the truthfulness with which it states how it fares finally with
this exaltation: thought succumbs, the end is falling and faintness. What makes thought succumb, what causes falling and faintness, is nothing else but the wearisome repetition which makes a limit
vanish, reappear, and then vanish again, so that there is a perpetual arising and passing away of the one after the other and of the one in the other, of the beyond in the here and now, and of the
here and now in the beyond, giving only the feeling of the impotence of this infinite or this ought-to-be, which would be master of the finite and cannot.
§ 507
Also Haller's description of eternity, called by Kant terrifying, is usually specially admired, but often just not for that very reason which constitutes its true merit:
'I heap up monstrous numbers,
Pile millions upon millions,
I put aeon upon aeon and world upon world,
And when from that awful height
Reeling, again I seek thee,
All the might of number increased a thousandfold
Is still not a fragment of thee.
I remove them and thou nest wholly before me.'
§ 508
When this heaping and piling up of numbers is regarded as what is valuable in a description of eternity, it is overlooked that the poet himself declares this so-called terrifying journey into the
beyond to be futile and empty, and that he closes by saying that only by giving up this empty, infinite progression can the genuine infinite itself become present to him.
§ 509
There have been astronomers who liked to pride themselves on the sublimity of their science because it had to deal with an innumerable host of stars, with such immeasurable spaces and times in which
distances and periods, already vast in themselves, serve as units which, in whatever multiples taken, are again abbreviated to insignificance. The shallow astonishment to which they surrender
themselves, the absurd hopes of wandering in another life from one star to another and into immeasurable space to acquire fresh facts of the same kind, this they declare to be a cardinal factor in
the excellence of their science — a science which is admirable not on account of such quantitative infinitude but, on the contrary, on account of the relations of measure and the laws which reason
recognises in these objects and which are the infinite of reason in contrast to that other, irrational infinite.
§ 510
To the infinity of outer, sensuous intuition, Kant opposes the other infinite, when 'the individual withdraws into his invisible ego and opposes the absolute freedom of his will as a pure ego to all
the terrors of fate and tyranny, and starting with his immediate surroundings, lets them vanish before him, and even what seems enduring, worlds upon worlds, collapse into ruins, and, alone, knows
himself as equal to himself.'
§ 511
The ego in being thus alone with itself is, it is true, the reached beyond; it has come to itself, is with itself, here and now; the absolute negativity which in the progress beyond the quantum of
sense was only a flight, in pure self-consciousness becomes affirmative and present. But this pure ego, because it has fixed itself in its abstraction and emptiness, has determinate reality, the
fulness of the universe of nature and mind, confronting it as a beyond. We are faced with that same contradiction which lies at the base of the infinite progress, namely a returnedness-into-self
which is at the same time immediately an out-of-selfness, a relation to its other as to its non-being; and this relation remains a longing, because on the one side is the unsubstantial, untenable
void of the ego fixed as such by the ego itself, and on the other, the fulness which though negated remains present, but is fixed by the ego as its beyond.
§ 512
On these two sublimes Kant remarks 'that admiration (for the first, outer) and reverence (for the second, inner) do indeed stimulate inquiry but cannot be a substitute for their defect'. Thus he
declares those exaltations to be unsatisfying for reason, which cannot stop at them and the feelings associated with them, nor can it let the beyond and the void rank as ultimate.
§ 513
But it is specially in its application to morality that the infinite progress has been taken as ultimate. The just quoted antithesis of finite and infinite in the shape of the manifold world and the
ego raised to its freedom, is primarily qualitative. The ego in its self-determining forthwith proceeds to determine nature and to liberate itself therefrom; it thus connects itself through itself
with its other which, as an external reality, is manifold and quantitative. The relation to the quantitative becomes itself quantitative; the negative relation of the ego to it, the power of the ego
over the non-ego, over sense and outer nature, is consequently so conceived that morality can and ought continually to increase, and the power of sense continually to diminish. But the perfect
adequacy of the will to the moral law is placed in the unending progress to infinity, that is, is represented as an absolutely unattainable beyond, and this very unattainableness is supposed to be
the true sheet-anchor and fitting consolation; for morality is supposed to be a struggle, but such it can be oily if the will is inadequate to the moral law which thus becomes a sheer beyond for it.
§ 514
In this opposition, ego and non-ego or the pure will and the moral law, and nature and the sensuousness of the will, are presupposed as completely self-subsistent and mutually indifferent. The pure
will has its own appropriate law which stands in an essential relationship to the sphere of sense; and nature and sense on its side has laws which neither stem from nor are conformable to the will
nor, although distinct from it, have they even in principle an essential connection with it but are determined independently, are finished and complete in themselves. At the same time, however, both
are moments of one and the same simple being, the ego; the will is determined as the negative in relation to nature so that the will only is in so far as there is a sphere distinct from it which it
sublates, but with which it thereby comes into contact and by which it is itself affected. Nature itself and nature as the sensuous sphere of man, as an independent system of laws, is indifferent to
limitation by an other; it preserves itself in this process of limitation, enters into the relation as an independent factor and limits the will of law just as much as this limits it. The two
processes comprise a single act: the self-determining of the will with the sublating of the otherness of nature, and the positing of this otherness as continuing itself as a reality in the process of
being sublated, so that the otherness is not sublated. The contradiction involved in this is not resolved in the infinite progress: on the contrary, it is represented and affirmed as unresolved and
unresolvable; the conflict of morality and sense is represented as the ultimate, absolute relation.
§ 515
This standpoint which is powerless to overcome the qualitative opposition between the finite and infinite and to grasp the idea of the true will which is substantial freedom, has recourse to quantity
in order to use it as a mediator, because it is sublated quality, the difference which has become indifferent. But since both members of the antithesis remain implied as qualitatively distinct, the
fact is rather that each is straightway made indifferent to this alteration because it is as quanta that they are related to each other. Nature is determined by the ego, sense by the will of the
good; the alteration produced in sense by the will is only a quantitative difference, one which leaves sense itself unchanged.
§ 516
In the more abstract exposition of the Kantian philosophy, or at least of its principles, namely in Fichte's Theory of Science, the infinite progress in the same way constitutes the foundation and
the ultimate. In this exposition, the first axiom, ego = ego, is followed by a second, independent of it, the opposition of the non-ego; the relation between the two is also directly assumed as a
quantitative difference, that is, the non-ego is partly determined by the ego, and partly not. In this way, the non-ego is continued into its non-being in such wise that in its non-being it remains
opposed as something not sublated. Consequently, after the contradictions contained in this have been developed in the system, the final result is that relationship which formed the beginning: the
non-ego remains an infinite obstacle, an absolute other; the final relation of the non-ego and the ego to each other is the infinite progress, a longing and aspiration — the same contradiction with
which the system began.
§ 517
Because the quantitative is determinateness posited as sublated it was thought that much, or rather everything, had been gained for the unity of the absolute, for the one substantiality, when
opposition generally had been reduced to a merely quantitative difference. That all opposition is only quantitative was for some time a cardinal thesis of recent philosophy; the opposed
determinations have the same nature, the same content; they are real sides of the opposition in so far as each of them has within it both determinations, both factors of the opposition, only that on
one side one of the factors preponderates, on the other side the other, that is, one of the factors, a material substance or activity, is present in a greater quantity or in an intenser degree in one
side than in the other. But in so far as substances or activities are presupposed, the quantitative difference rather confirms and completes their externality and indifference to each other and to
their unity. The difference of the absolute unity is supposed to be only quantitative; the quantitative, it is true, is immediate, sublated determinateness, but only the imperfect, as yet only first,
negation, not the infinite, not the negation of the negation. When being and thought are represented as quantitative determinations of absolute substance they too, as quanta, become completely
external to each other and unrelated as, in a subordinate sphere, do carbon, nitrogen, etc. It is a third, an external reflection, which abstracts from their difference and recognises their unity,
but a unity which is inner, implicit only, not for itself. This unity is, therefore, in fact conceived only as a first, immediate unity, or only as being, which in its quantitative difference remains
like itself, but does not of itself posit itself as like itself; hence it is not grasped as a negation of the negation, as an infinite unity. Only in the qualitative opposition does the posited
infinitude, being-for-self, emerge and the quantitative determination itself pass over into the qualitative, as we shall presently find.
§ 518
It was remarked above that the Kantian antinomies are expositions of the opposition of finite and infinite in a more concrete shape, applied to more specific substrata of conception. The antinomy
there considered contained the opposition of qualitative finitude and infinitude. In another, the first of the four cosmological antinomics, it is the conflict arising rather from the quantitative
limit which is considered. I shall therefore proceed to examine this antinomy here.
§ 519
It concerns the limitation or non-limitation of the world in time and space. This antithesis could be considered equally well with reference to time and space themselves, for whether time and space
are relations of things themselves or are only forms of intuition, the antinomy based on limitation or non-limitation in them is not affected thereby.
§ 520
The detailed analysis of this antinomy will likewise show that both statements and equally their proofs (which, like those already considered, are conducted apagogically) amount to nothing more than
the two simple opposite assertions: (1) there is a limit, and (2) the limit must be transcended.
The thesis is:
'The world has a beginning in time and is also enclosed within spatial limits.'
That part of the proof which concerns time assumes the opposite:
'The world has no beginning in time; therefore, up to any given point of time, an eternity has elapsed and consequently an infinite series of successive states of things in the world has passed away.
Now the infinity of a series consists precisely in the impossibility of ever completing it by successive synthesis. Therefore an infinite world series which has passed away is impossible and
consequently a beginning of the world is a necessary condition of its existence — which was to be proved.'
§ 521
The other part of the proof which concerns space is based on time. To comprehend a spatially infinite world would require an infinite time and this time must be regarded as having already elapsed in
so far as the world in space is to be regarded not as gradually coming to be but as completely given. But it was shown of time in the first part of the proof that it is impossible to assume an
infinite time as elapsed.
§ 522
But it is at once evident that it was unnecessary to make the proof apagogical, or even to carry out a proof at all, since the basis of the proof itself is the direct assertion of what was to be
proved. Namely, there is assumed some or any given point of time up to which an eternity has elapsed (eternity here has only the trivial meaning of a simply endless time). Now a given point of time
means nothing else than a definite limit in time. In the proof therefore, a limit of time is presupposed as actual; but that is just what was to be proved. For the thesis is, that the world has a
beginning in time.
§ 523
There is only this difference, that the assumed limit of time is a now as end of the time already elapsed, but the limit which is to be proved is a now as beginning of a future. But this difference
is immaterial. The now is taken as the point in which an infinite series of successive states of things in the world is supposed to have passed away, therefore as end, a qualitative limit. If this
now were considered to be merely a quantitative limit which flows on and which not only must be transcended but is only as the transcending of itself, then the infinite time series would not have
passed away in it, but would continue to flow on, and so the argument of the proof would vanish. On the other hand, if the point of time is assumed as a qualitative limit for the past, in which case
it is also a beginning for the future (for each point of time is in itself the connection of the past and the future), then it is also an absolute, that is, abstract beginning for the future — and it
was this that was to be proved. The fact that its future and this its beginning is already preceded by a past does not affect the argument; because this point of time is a qualitative limit — and
that it is to be taken as qualitative is implied in the description of it as completed, elapsed, and therefore as not continuing — therefore in it time is broken off and the past lacks a connection
with this time which could only be called future with reference to that past and, consequently, without such connection is only time as such, which has an absolute beginning. But if — as is, then,
the case — it were related to the past through the now, the given point of time, and were thus determined as a future, then this point of time, too, regarded from the other side, would not be a
limit; the infinite time series would continue itself in what was called future and would not be, as was assumed, completed.
§ 524
In truth, time is pure quantity; the point of time in which it is supposed to be interrupted, which is employed in the proof, is really only the self-sublating being-for-self of the now. All that the
proof does is to represent the absolute limit of time asserted in the thesis as a given point of time, and then straightway to assume it as a completed, that is, abstract point — a popular
determination which sensuous conception readily lets pass as a limit, thus allowing as an assumption in the proof what had been put forward as the thing to be proved.
§ 525
The antithesis runs:
'The world has no beginning and no limits in space but is infinite with reference both to time and space.'
The proof likewise assumes the opposite:
'The world has a beginning. Since the beginning is an existence preceded by a time in which the thing is not, there must have been a preceding time in which the world was not, that is, an empty time.
Now no originating of anything is possible in an empty time; because no part of such a time possesses in itself and in preference to any other, any distinguishing condition of existence or
non-existence. In the world, therefore, many groups of things can indeed begin, but the world itself can have no beginning and with respect to past time is infinite.'
§ 526
This apagogical proof, like the others, contains the direct and unproved assertion of what it was supposed to prove. That is, it first assumes a beyond of the existing world, an empty time; but it
also equally continues the existence of the world beyond itself into this empty time which is thereby sublated, with the result that the existence of the world is continued into infinity. The world
is an existence; the proof presupposes that this existence comes into being and that the coming-to-be has an antecedent condition which is in time. But the antithesis itself consists in the very fact
that there is no unconditioned existence, no absolute limit, but that the existence of the world always requires an antecedent condition. Thus, what was to be proved is found as an assumption in the
proof. Further, the condition is sought in empty time, which means in effect that it is taken as temporal and therefore as an existence and as limited. Altogether, then, the assumption is made that
the world as an existence presupposes another conditioned existence in time, and so on, therefore, to infinity.
§ 527
The proof regarding the infinity of the world in space is the same. Apagogically, the spatial finiteness of the world is assumed; 'this (the world) would therefore exist in an empty unlimited space
and would stand in a relation to it; but such a relation of the world to no object is a nullity'.
Here, too, what was supposed to be proved is directly presupposed in the proof. It is directly assumed that the spatially limited world exists in an empty space and is supposed to stand in a relation
to it, that is, there must be a movement out beyond it — on the one hand into the void, into the beyond and non-being of the world, but on the other hand, in order that it be in relation with its
beyond, that is, continue itself into it, the beyond must be imagined as filled with the existence of the world. The infinity of the world in space which is asserted in the antithesis is nothing else
than, on the one hand, empty space, and on the other the relation of the world to it, that is, the continuity of the world in empty space or the filling of space — which contradiction, namely, space
as simultaneously empty and also filled, is the infinite progress of existence in space. This very contradiction, the relation of the world to empty space, is directly made the basis of the proof.
§ 528
The thesis and antithesis and their proofs therefore represent nothing but the opposite assertions, that a limit is, and that the limit equally is only a sublated one; that the limit has a beyond,
with which however it stands in relation, and beyond which it must pass, but that in doing so there arises another such limit, which is no limit.
§ 529
The solution of these antinomies, as of those previously mentioned, is transcendental, that is, it consists in the assertion of the ideality of space and time as forms of intuition — in the sense
that the world is in its own self not self-contradictory, not self-sublating, but that it is only consciousness in its intuition and in the relation of intuition to understanding and reason that is a
self-contradictory being. It shows an excessive tenderness for the world to remove contradiction from it and then to transfer the contradiction to spirit, to reason, where it is allowed to remain
unresolved. In point of fact it is spirit which is so strong that it can endure contradiction, but it is spirit, too, that knows how to resolve it. But the so-called world (whether it be called an
objective, real world or, according to transcendental idealism, a subjective intuition and a sphere of sense determined by the categories of the understanding) is never and nowhere without
contradiction, but it is unable to endure it and is, therefore, subject to coming-to-be and ceasing-to-be.
(c) The Infinity of Quantum
§ 530
1. The infinite quantum as infinitely great or infinitely small is itself implicitly the infinite progress; as great or small it is a quantum and at the same time it is the non-being of quantum. The
infinitely great and infinitely small are therefore pictorial conceptions which, when looked at more closely, turn out to be nebulous shadowy nullities. But in the infinite progress, this
contradiction is explicitly present and with it that which is the nature of quantum which, as an intensive magnitude, has attained its reality and now in its determinate being is posited as it is in
its Notion. It is this identity which we have now to consider.
§ 531
Quantum as degree is unitary, self-related and determinate within itself. Through this unitary nature, the otherness and determinateness in quantum are sublated, so that the determinatensess is
external to it; it has it determinateness outside it. This its self-externality is in the first place the abstract non-being of quantum generally, the spurious infinity. But, further, this non-being
is also quantitative and this continues itself into its non-being, for it is in its externality that quantum has its determinateness; this its externality is, therefore, itself equally a quantum.,
this non-being of quantum, infinity, is thus limited, that is, this beyond is sublated, is itself determined as quantum which, therefore, in its negation is with itself.
§ 532
But this is what quantum as such is in itself. For it is itself just by being external to itself; externality constitutes that whereby it is quantum and is with itself. In the infinite progress,
therefore, the Notion of quantum is posited.
§ 533
Let us take the progress at first in its abstract determinations as we find them; then in it we have the sublating of quantum, but equally too of its beyond, therefore the negation of quantum as well
as the negation of this negation. Its truth is their unity in which they are, but only as moments. It is the resolution of the contradiction of which it is the expression, and its immediate
significance is, therefore, the restoration of the Notion of quantity, namely, that quantity is an indifferent or external limit. In the infinite progress as such, the only reflection usually made is
that every quantum, however great or small, must be capable of vanishing, of being surpassed; but not that this self-sublating of quantum, the beyond, the spurious infinite itself also vanishes.
§ 534
Even the first sublation, the negation of quality as such whereby quantum is posited, is in principle [an sich] the sublating of the negation — the quantum is sublated qualitative limit, hence
sublated negation — but at the same time it is this only in principle; it is posited as a determinate being, and then its negation is fixed as the infinite, as the beyond of quantum, which remains on
this side as an immediate; thus the infinite is determined only as a first negation and it appears as such in the infinite progress. But we have seen that in this something more is present, the
negation of the negation, or that which the infinite in truth is. We regarded this previously as the restoration of the Notion of quantity; this restoration means in the first place, that its
determinate being has received a more precise determination; we now have quantum determined in conformity with its Notion, which is different from quantum in its immediacy; externality is now the
opposite of itself, posited as a moment of quantity itself — quantum is posited as having its determinateness in another quantum by means of its non-being, of infinity; that is, it is qualitatively
that which it is. However, this comparison of the Notion of quantum with its determinate being belongs more to our reflection, to a relationship which is not yet present here. The immediately
following determination is that the quantum has reverted to quality, is from now on qualitatively determined. For its peculiarity, its quality, is the externality, the indifference of the
determinateness; and quantum is now posited as being in fact itself in its externality, as self-related therein, in simple unity with itself, that is, qualitatively determined. This qualitative
moment is still more closely determined, namely as being-for-itself; for the self-relation to which it has attained has proceeded from mediation, from the negation of the negation. Quantum has
infinity, self-determinedness, no longer outside it but within itself.
§ 535
The infinite, which in the infinite progress has only the empty meaning of a non-being, of an unattained but sought beyond, is in fact nothing else than quality. Quantum as an indifferent limit goes
out beyond itself to infinity; in doing so it seeks nothing else than to be determined for itself, the qualitative moment, which, however, is thus only an ought-to-be. Its indifference to limit, and
hence its lack of an explicit determinateness of its own and its passage away from and beyond itself, is that which makes quantum what it is; this its passage into the beyond is to be negated and
quantum is to find in the infinite its absolute determinateness.
§ 536
Quite generally: quantum is sublated quality; but quantum is infinite, goes beyond itself, is the negation of itself. Thus its passage beyond itself is, therefore, in itself the negation of the
negated quality, the restoration of it; and thus quantum is explicitly determined as possessing as its own moment, the externality which formerly appeared as a beyond.
§ 537
Quantum is thus posited as repelled from itself, with the result that there are two quanta which, however, are sublated, are only as moments of one unity, and this unity is the determinateness of
quantum. Quantum as thus self-related as an indifferent limit in its externality and therefore posited as qualitative, is quantitative ratio. In the ratio, quantum is external to itself, is
distinguished from itself; this its externality is the relation of one quantum to another, each of which has meaning only in this its relation to its other; and this relation constitutes the
determinateness of the quantum, which is as such a unity. It has in this unity not an indifferent, but a qualitative, determination; in this its externality it has returned into itself, and in it
quantum is that which it is.
Remark 1: The Mathematical Infinite - next section | {"url":"http://www.marxists.org/reference/archive/hegel/works/hl/hl225.htm","timestamp":"2014-04-17T13:12:53Z","content_type":null,"content_length":"41503","record_id":"<urn:uuid:1ee2219a-58ed-45ce-8c15-1e77adc0324a>","cc-path":"CC-MAIN-2014-15/segments/1397609530131.27/warc/CC-MAIN-20140416005210-00579-ip-10-147-4-33.ec2.internal.warc.gz"} |
A small C program to print the biggest prime number
Here it is (447 bytes):
int m=1711276033,N=1,t[1<<25]={2},a,*p,i,e=39717691,s,c,U=1;g(d,h){for(i=s;i<1<<
This program computes 2^57885161-1, which is the biggest known prime number (about 17 million digits !). For more information about how it was found and who found it, look at the GIMPS Project .
I compiled it successfully with gcc with Linux. In order to compile it, your C compiler must support the 64 bit long long type. The running time is less than 2 minutes on a 2.33 GHz Core2 CPU.
In order to have a small and yet asymptotically efficient code, I decided to do the computation of 2^N-1 directly in base 10. The power involves repeated squarings and multiplications by two. The
squarings are implemented by doing fast convolutions using a Number Theoretic Transform.
A previous version of this program to compute 2^6972593-1 won the International Obfuscated C Code Contest of Year 2000.
Thanks to Marco Cecchi who suggested some syntactic changes to save a few characters.
This program is Freeware.
Fabrice Bellard - http://bellard.org/
last update: Sep 09, 2013 | {"url":"http://www.bellard.org/mersenne.html","timestamp":"2014-04-20T00:37:31Z","content_type":null,"content_length":"2252","record_id":"<urn:uuid:e6e7e387-f517-45c9-80e0-9c2d141851f4>","cc-path":"CC-MAIN-2014-15/segments/1397609537804.4/warc/CC-MAIN-20140416005217-00015-ip-10-147-4-33.ec2.internal.warc.gz"} |
[Numpy-discussion] convolving (or correlating) with sliding windows
[Numpy-discussion] convolving (or correlating) with sliding windows
josef.pktd@gmai... josef.pktd@gmai...
Tue Feb 15 11:27:51 CST 2011
On Tue, Feb 15, 2011 at 11:42 AM, Davide Cittaro
<davide.cittaro@ifom-ieo-campus.it> wrote:
> Hi all,
> I have to work with huge numpy.array (i.e. up to 250 M long) and I have to perform either np.correlate or np.convolve between those.
> The process can only work on big memory machines but it takes ages. I'm writing to get some hint on how to speed up things (at cost of precision, maybe...), possibly using a moving window... is it correct to perform this:
> - given a window W and a step size S
> - given data1 and data2
> - pad with zeros data1 and data2 by adding W/2 0-arrays
> - get the np.correlation like
> y = np.array([np.correlate(data1[x:x+W], data2[x:x+W], mode='valid') for x in np.arange(0, len(data1) - W, S)]).ravel()
> instead of np.correlate(data1, data2, mode='same')
If data2 is of similar length as data1, then you should use
fftconvolve which is much faster for long arrays than np.correlate or
convolve. I'm not sure about the rest.
> - interpolate the correlation using scipy.interpolate
> ?
> Thanks
> d
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Launchings by David Bressoud
In May this year, the National Center on Education and the Economy (NCEE) released its report, "What Does It Really Mean to Be College and Work Ready?" [1]. The report is in two parts: Mathematics
and English Literacy. It is based on a national study of the proficiencies required and actually used for the most popular associate’s degree programs at two-year colleges. Just a few weeks earlier,
Jordan Weissman published a piece in The Atlantic, "Here’s How Little Math Americans Actually Use at Work" [2]. That was based on a 2010 report written by Michael Handel at Northeastern University, "
What Do People Do at Work? A Profile of U.S. Jobs from the Survey of Workplace Skills, Technology, and Management Practices" [3]. This large-scale survey includes an assessment of what mathematics is
actually used in the workplace.
I hope that it will come as a surprise to no one that not everyone actually uses the contents of Algebra II in their work or that College Algebra taught in two-year colleges is essentially high
school Algebra II. Weissman highlights Handel’s data that less than a quarter of all workers use any mathematics that is more advanced than fractions, ratios, and percentages. He raises the question
whether requiring Algebra II for high school graduation is placing an unnecessary roadblock in the way of too many students. NCEE poses more nuanced questions. Why are so many students being hurried
through the critical early mathematics that they will need to be work and college ready, especially fractions, ratios, and percentages, just so that they can get to Algebra II? Isn’t there a better
way to prepare them for what they will need?
I will begin with the data. Handel divided the workforce into five categories:
• Upper White Collar (management, professional, technical occupations)
• Lower White Collar (clerical, sales)
• Upper Blue Collar (craft and repair workers, construction trades, mechanics)
• Lower Blue Collar (factory workers, truck drivers)
• Service (food service, home health care, child care, janitors)
Generally, the best paying and most desirable jobs are Upper White Collar (UWC) and Upper Blue Collar (UBC). We should be equipping our students so that they can aspire to such jobs. It’s still not
true that everyone needs Algebra II, but 35% of UWC workers reported using basic algebra, geometry, and/or statistics in their work. This level of mathematics is even more important for UBC workers,
with 41% reporting using mathematics at the level of basic algebra, geometry, and/or statistics. This is still not Algebra II (which Handel lists as “complex Algebra” as opposed to “basic Algebra”),
which was reported being used by 14% of UWC workers and 16% of UBC workers. Much less is it Calculus, which was reported by 8% of both UWC and UBC workers. But about 40% of those working in UWC or
UBC jobs need a working knowledge of some high school mathematics, a higher bar than simply having passed the relevant courses. It is interesting to observe that UBC workers are more likely to use
mathematics than UWC workers.
The NCEE report looked at the mathematics required for the nine most popular associate’s degree programs at two-year colleges: Accounting, Automotive Technology, Biotech/Electrical Technology,
Business, Computer Programming, Criminal Justice, Early Childhood Education, Information Technology, and Nursing, as well as the General Track. This ties nicely to the Handel study because the nine
are generally seen as preparation for UWC or UBC careers. NCEE selected seven two-year colleges in seven states and examined the texts, assignments, and exams in the introductory courses for these
disciplines as well as for the mathematics courses required for these fields. There were three notable insights:
First, except for some work on geometric visualization, NCEE found no content in either College Algebra or Statistics, two college-credit bearing courses, that goes beyond the high school curriculum
described in the Common Core State Standards in Mathematics (CCSS-M). They found that College Algebra had a large component of middle school topics, especially CCSS-M for grades 6–8 in Expressions
and Equations, Functions, Number Systems, Geometry, and Ratios and Proportions. Statistics was a mix of CCSS-M middle and high school level statistics, with a significant component of grades 6–8
Ratios and Proportions and Expressions and Equations.
Second, the introductory textbooks in the disciplinary fields used nothing beyond Algebra I. Ratios and proportions are important as well as interpreting quantitative relationship expressed in
tables, graphs, and formulae, but, as the report says,
When mathematics is present in the texts, equations are not solved, quadratics are absent, and functions are present but not named or analyzed, just treated as formulae. […] Students do not have
to perform algebraic manipulations nor construct graphs or tables. […] The area of high school content with the highest representation in the texts, Number Systems, is found in six percent of the
text chapters. [p. 16]
Third, the mathematical knowledge that was tested in these introductory courses in the disciplinary fields was far lower than what was in the textbooks. Not only was there nothing requiring Algebra
II on the exams, the NCEE team could find nothing, or almost nothing, that reflected knowledge of Algebra I. Furthermore, the questions that were asked on examinations were of low difficulty. The
NCEE team used the PISA (Program for International Student Assessment) Item-Difficulty Coding Framework with four levels. Examples of what is expected at each level include
• Level 0: perform simple calculations and make direct inferences;
• Level 1: use simple functional relationships and formal mathematical symbols, interpret models;
• Level 2: use multiple relationships, manipulate mathematical symbols, modify existing models; and
• Level 3: solve multi-step application of formal procedures, evaluate arguments, create models.
The team found that over 60% of the mathematical questions on the examinations given in introductory courses were at Level 0. Few rose to Level 2, much less Level 3. (This was not the case in College
Algebra and Statistics where most of the examination items were at Level 1 or 2 and some attained Level 3. This suggests that even though the material of College Algebra and Statistics does not go
beyond topics covered in CCSS-M, the level of expected proficiency may be higher than what is typically encountered in high school.)
NCEE did find three mathematical topics required for the introductory courses that are not covered in CCSS-M nor in the College Algebra or Statistics classes: complex applications of measurements,
schematic diagrams (2-D schematics of 3-D objects and flow charts), and geometric visualization. They also found a much greater demand for knowledge of statistics, probability, and modeling (“how to
frame a real-world problem in mathematical terms”) than is commonly taught in most mainstream high school mathematics programs today.
What makes the NCEE report even more depressing is that it restricted its attention to college-credit bearing courses. Most of the mathematics taught at two-year colleges is below the level of
College Algebra (see Figure 1). The mathematical requirements for UWC and UBC jobs may not be high, but we do not seem to be doing a very good job of preparing students even for what they will need.
Figure 1. Fall term Mathematics course enrollments (thousands). “Introductory” includes College Algebra, Trigonometry, and Precalculus.
Source: CBMS.
All of this raises serious questions about whether Algebra II should be expected of all graduating high school students. This parallels the situation that has been my primary concern: Should Calculus
be expected of all graduating high school students who are going directly into a four-year undergraduate program, especially those who may need to take Calculus in college? I would far prefer a
student who can operate at PISA Level 3 in Algebra I over a student who cannot handle problems above Level 1 in Algebra II. I would prefer Level 3 in Precalculus over Level 1 in Calculus. When
students are short-changed in their mathematical preparation simply so that Algebra II or Calculus appears on the high school transcript, with little regard to what that actually means, then neither
they nor society as a whole are well served.
It also raises questions about what mathematics should be required for an associate’s degree. College Algebra constitutes a significant hurdle for most two-year college students. Should there be
alternatives? In this case, I believe that most two-year college students would be better served with a program that combines demanding use of the topics of Algebra I with a college-level
introduction to Statistics.
We are not at the point where we can demand Algebra II for high school graduation. To do so would either create unacceptable rates of high school failure or force us to change what we mean by
“understanding Algebra II.” But I worry that if we simply lower our sights and decide that, since few of our students actually will use anything from Algebra II once they have graduated, it should
not be expected for graduation, then that will actually weaken the preparation that occurs in the earlier grades. Elementary and middle school mathematics should be laying the foundation for a
student to succeed in Algebra II. If we want our students to have a strong working knowledge of the high school mathematics that is needed for 40% of the UWC and UBC jobs, then we want them to have
the mathematical preparation that would enable them to succeed in Algebra II.
[1] National Center on Education and the Economy. 2013. What does it really mean to be college and work ready? The mathematics requirements of first year community college students. Washington, DC.
Available at http://www.ncee.org/college-and-work-ready/
[3] Michael Handel. 2010. What do people do at work? A Profile of U.S. jobs from the Survey of Workplace Skills, Technology, and Management Practices. OECD (forthcoming). Available at http://
3 comments:
1. There are a number of practical problems with Professor Bressoud's recommendations in the real world of the high school student. I will focus on just two:
(1) Omit the second year algebra requirement and what do you have left: Nothing is said in his essay about geometry but, assuming (hopefully) that subject remains, you require just two years of
math content. That would be followed for many students by two years with no math at all.
(2) Most of algebra 2 is simply algebra 1 again. That may seem to a college teachers (and other outside curriculum critics) to represent a waste of time, but in the real world of the high school
classroom it upgrades the level of understanding of that content. If you want to raise that PISA level of algebra 1 understanding, teach algebra 2.
2. Richard PainterJune 15, 2013 at 6:47 PM
In my experience (I am a retired high school math teacher), too often the first semester of Algebra II has devolved into covering material that was once covered in the second semester of Algebra
I. I am specifically thinking of quadratic functions, factoring, and solving quadratic equations. Where the goal is to get everyone through Algebra II, the courses are watered down at least going
all the way back to 7th grade math.
3. Keith StroyanJune 19, 2013 at 8:00 AM
I admire Prof. Bressoud's efforts in these matters. For years I have worked to try to help high school teachers develop materials that would show students how advanced high school math (I don't
like to call it "pre-calculus") can solve real problems. Perhaps if training included that, more could be used in the workplace.
My efforts in calculus were similarly directed:
Why do so many students take calculus?, Notices of the AMS, Sept. 2011, pp.1122-1124
Some time ago a group of us from all of Iowa's public Universities tried to develop a large project to work with high schools. Jerry Mathews and Elgin Johnston at ISU collected a number of real
problems from Iowa industry that could be solved with algebra and trig. (Design a conference table, a ground silo for cattle, ...) When we tried for NSF funding we were told our problems weren't
Calculus teachers often think they should teach from Spivak's "Calculus" (that isn't calculus!, but IS clever) - perhaps great for math majors, not so much for the "workplace."
I worry about "downstream" consequences of changes. When you don't teach fractions in late grade school because calculators can add 0.25 + 0.5, students have more trouble with 1/x+1/y in algebra
and in calculus. The course I took in thermodynamics in engineering school (a LONG time ago) had a lot more math than the one our students take at U of I now. (Too bad, some great math!)
We DO want more students to succeed in Algebra 2. A little time spent on real workplace problems could help.
(Thanks David for making me lose sleep another night. Keep it up! It IS important.) | {"url":"http://launchings.blogspot.com/2013/06/who-needs-algebra-ii.html","timestamp":"2014-04-21T05:39:24Z","content_type":null,"content_length":"118010","record_id":"<urn:uuid:2f9a9be6-73aa-4698-8346-277d605ccd19>","cc-path":"CC-MAIN-2014-15/segments/1397609539493.17/warc/CC-MAIN-20140416005219-00332-ip-10-147-4-33.ec2.internal.warc.gz"} |
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\[{x1}, ...., {x6} = 15\]
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\[\left(\begin{matrix}15 + 6 - 1 \\ 15\end{matrix}\right)\] is this the answer?
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what kind operations there, add, subtrac, or multiplication or division ?
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is it like a+b+c+d+e+f=15 ? with a,b,c,d,e, and f are natural numbers
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hello ?
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discrete math, ow sorry its addition
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hehe... i have guessed actually, i will use the formula : |dw:1355741391689:dw|
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sorry if im wrong...
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why did you use (15 - 1)?
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wait... looks i have mistaken
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what did you do wrong then, exactly?
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ok, for the first... i have mistaken after i do some experiments, it shoulde be |dw:1355743741310:dw|
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let's discuss for simple example below : a+b+c=6, for a,b,c are natural numbers. # if a=1, then b+c=5 or c=5-b to get c be integer , so satisfied for b=1,2,3,4 (there are 4 ways) # if a = 2, then
b+c=4, or c=4-b to get c be integer , so satisfied for b=1,2,3, (there are 3 ways) # if a = 3, then b+c=3, or c=2-b to get c be integer , so satisfied for b=1,2, (there are 2 ways) # if a = 4,
then b+c=2, satisfied just for b=1 and c=1 ( 1 way) so, the total ways = 4+3+2+1 = 10 it just similar by C(6-1, 3-1) = C(5,2) = 10
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there are 6 x's, from x1 till x6 and together they add up to 15
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with same idea, if given the problem : a+b+c+d=8 so, the total ways = C(8-1,4-1) = C(7,3) = 35 ways btw, i have done it by manual also like the first ^ (the answer is same)
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yeah, like i said before it will be C(15-1,6-1) = C(14,5) = 14!/(9!5!) = ....
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but, this method will be different, if a,b,c,d,... are the whole numbers :) thanks for ur question, i become more knowing for this problems ......
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thank you very much for the explanation
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you're welcome :)
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is replying to Can someone tell me what button the professor is hitting...
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Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.
This is the testimonial you wrote.
You haven't written a testimonial for Owlfred. | {"url":"http://openstudy.com/updates/50cef676e4b0031882dca985","timestamp":"2014-04-19T02:09:38Z","content_type":null,"content_length":"138181","record_id":"<urn:uuid:80d8e524-341a-4375-b32a-8d545b25ecbb>","cc-path":"CC-MAIN-2014-15/segments/1397609535745.0/warc/CC-MAIN-20140416005215-00436-ip-10-147-4-33.ec2.internal.warc.gz"} |
limit and integration
Hi, I will continue mr fantastic idea... Why you can apply l'Hospital's rule? Let we see what happens to the whole expession when $x\to 0$: $\frac{\int_0^0\,e^{t^2}(e^t-1)^2\, dt}{0(1-cos 0)}=\frac
{0}{0}$ Next, ${x (1-cos(x))}' = {x sin(x)-cos(x)+1}eq 0$ for every $xeq0$ Applying l'Hospital's rule: Now by the Fundamental Theorem of Calculus: ${\int_0^x\,e^{t^2}(e^t-1)^2\ dt}}'=e^{x^2}(e^x-1)^
2$ ${\lim }\limits_{x \to 0} {\frac{(\int_0^x\,e^{t^2}(e^t-1)^2\, dt)'}{(x(1-cos x))'}={\lim }\limits_{x \to 0} \frac{e^{x^2}(e^x-1)^2}{x sin(x)-cos(x)+1}$ The limit is still from the form of $\frac
{0}{0}$, applying l'Hospital's rule again...
Last edited by mr fantastic; October 20th 2010 at 04:04 AM. Reason: I would like to see the OP do some work before the complete solution is posted.
...............in this manner, if L'Hospitals rule is applied, at the $3rd$ oreder differentiation, we get:- $\frac{((8x^3+24x^2+36x+20)e^{2x}+(-16x^3-24x^2-36x-14)e^x+8x^3+12x)e^{x^2}}{3cos x-xsin
x}$ putting x=0, we get:- $6/3 = 2$ am i right?
oh yes... actually my denominator was right, but i differentiated the numerator 4 times instead of 3 times....now i have got it correct... | {"url":"http://mathhelpforum.com/calculus/160337-limit-integration.html","timestamp":"2014-04-16T05:29:56Z","content_type":null,"content_length":"51605","record_id":"<urn:uuid:0675ce83-794f-4b70-9607-7bc9d952bc4b>","cc-path":"CC-MAIN-2014-15/segments/1397609521512.15/warc/CC-MAIN-20140416005201-00344-ip-10-147-4-33.ec2.internal.warc.gz"} |
Easy kinematics problem
December 4th 2012, 01:38 PM
Easy kinematics problem
Hi! I'm dealing with this problem that seems to be very easy, however the teacher says I'm wrong, and I don't know where's my mistake (She didn't tell me). So I hope you can tell me where did I
go wrong(Wink) Here's the problem:
A falling sky diver travels 64m in 2 seconds. How much will he travel in 8 seconds? (Ignore the air resistance)
My solution:
I used the equation y=yo+vot+1/2gt^2
g=9.8 m/s^2
Then I solved it for vo, obtaining vo=22.2 m/s. And finally obtained:
y=(22.2)(8)+1/2g(8)^2=491.2 m
I don't know what did I do wrong, and the teacher didn't even tell me what was the right answer(Wondering). I hope you guys help me figure it out. Thanks!
December 8th 2012, 01:55 AM
Re: Easy kinematics problem
Hi! I'm dealing with this problem that seems to be very easy, however the teacher says I'm wrong, and I don't know where's my mistake (She didn't tell me). So I hope you can tell me where did I
go wrong(Wink) Here's the problem:
A falling sky diver travels 64m in 2 seconds. How much will he travel in 8 seconds? (Ignore the air resistance)
My solution:
I used the equation y=yo+vot+1/2gt^2
g=9.8 m/s^2
Then I solved it for vo, obtaining vo=22.2 m/s. And finally obtained:
y=(22.2)(8)+1/2g(8)^2=491.2 m
I don't know what did I do wrong, and the teacher didn't even tell me what was the right answer(Wondering). I hope you guys help me figure it out. Thanks!
What I understand from the question is that the diver has traveled in 2 seconds 64m then for another 8 seconds how much will he travel..
So we can find his velocity after 2 seconds and then use it to find the distance he traveled in 8s..
What you did is that u found his velocity when he just started travelling the 64 m..
After that you need to find the velocity that he was traveling with after 2 seconds and substitute that in the second equation-which you used to find the distance..
December 11th 2012, 01:34 PM
Re: Easy kinematics problem
What I understand from the question is that the diver has traveled in 2 seconds 64m then for another 8 seconds how much will he travel..
So we can find his velocity after 2 seconds and then use it to find the distance he traveled in 8s..
What you did is that u found his velocity when he just started travelling the 64 m..
After that you need to find the velocity that he was traveling with after 2 seconds and substitute that in the second equation-which you used to find the distance..
That's exactly what I had to do. Thanks! | {"url":"http://mathhelpforum.com/math-topics/209081-easy-kinematics-problem-print.html","timestamp":"2014-04-21T09:47:27Z","content_type":null,"content_length":"7685","record_id":"<urn:uuid:2daa3422-f580-47fe-9398-007d90625749>","cc-path":"CC-MAIN-2014-15/segments/1397609539705.42/warc/CC-MAIN-20140416005219-00562-ip-10-147-4-33.ec2.internal.warc.gz"} |
Please Excuse My Dear Aunt Sally
Posted by Bruce Kahl on July 24, 2000
In Reply to: Aunt Sally ?! posted by kts on July 24, 2000
: Does anyone know what an 'Aunt Sally' list is ?
In a former life back in the late sixties thru the mid eighties I taught mathematics to junior high school children in NYC, Bronx and Brooklyn.
The only Aunt Sally thingy I know has to do with mathematics.It is a mnemonic aid.
Please Excuse My Dear Aunt Sally.
This mnemonic helps us remember which order to complete mathematical operations. The order in which a problem is solved is crucial. Without having a set order, we would arrive at different results.
2 + 3 x 6
Is the answer 30? 2 + 3 = 5 5 x 6 = 30??
This is why we need to learn the mnemonic Please Excuse My Dear Aunt Sally. If not, we would not know the correct answer to the above is 20!
The answer is 20. Our mnemonic tells us which order to solve the problem. We do not just read left to right as if it were an English sentence. Multiplication comes before addition in the mnemonic (My
comes before Aunt), therefore, we must multiply before we add.
If we do that, then the answer is 20! 3 x 6 = 18. 18 + 2 = 20. | {"url":"http://www.phrases.org.uk/bulletin_board/5/messages/845.html","timestamp":"2014-04-21T04:34:06Z","content_type":null,"content_length":"8542","record_id":"<urn:uuid:7c7cff1e-4075-49d9-bec7-c2cb1599e492>","cc-path":"CC-MAIN-2014-15/segments/1397609539493.17/warc/CC-MAIN-20140416005219-00371-ip-10-147-4-33.ec2.internal.warc.gz"} |
sequentially compact
sequentially compact
Every sequentially compact space is countably compact. Conversely, every first countable countably compact space is sequentially compact. The ordinal space $W(2\omega_{1})$ is sequentially compact
but not first countable, since $\omega_{1}$ has not countable local basis.
Next, compactness and sequential compactness are not compatible. In other words, neither one implies the other. Here’s an example of a compact space that is not sequentially compact. Let $X=I^{I}$,
where $I$ is the closed unit interval (with the usual topology), and $X$ is equipped with the product topology. Then $X$ is compact (since $I$ is, together with Tychonoff theorem). However, $X$ is
not sequentially compact. To see this, let $f_{n}:I\to I$ be the function such that for any $r\in I$, $f(r)$ is the $n$-th digit of $r$ in its binary expansion. But the sequence $f_{1},\ldots,f_{n},\
ldots$ has no convergent subsequences: if $f_{{n_{1}}},\ldots,f_{{n_{k}}},\ldots$ is a subsequence, let $r\in I$ such that its binary expansion has its $k$-th digit $0$iff $k$ is odd, and $1$
otherwise. Then $f_{{n_{1}}}(r),\ldots,f_{{n_{k}}}(r),\ldots$ is the sequence $0,1,0,1,\ldots$, and is clearly not convergent. The ordinal space $\Omega_{0}:=W(\omega_{1})$ is an example of a
sequentially compact space that is not compact, since the cover $\{W(\alpha)\mid\alpha\in\Omega_{0}\}$ has no finite subcover.
topology, sequence, convergence
Compact, LimitPointCompact, BolzanoWeierstrassTheorem, Net
Mathematics Subject Classification
no label found
no label found
Are you sure that
"Sequential compactness is equivalent to compactness when $X$ is a metric space."?
I can prove this claim only when $X$ is metric and separable.
Munkres proves this as part of Theorem 28.2 in Topology, 2nd ed. (pp. 179-180).
In particular, Munkres proves that if $X$ is sequentially compact metric, then $X$ is totally bounded, that is, given a positive $\epsilon$, there is a finite covering of $X$ by $\epsilon$-balls.
Hence $X$ is Lindel\"{o}f. Because $X$ is also metric, this implies that $X$ is separable. So the assumption that $X$ is separable is not necessarily to show that sequential compactness implies
The proof that compactness implies sequential compactness does not use separability.
I try to prove that "Sequentially compactness is equivalent to compactness when X is metric space", if you interested in this problem, can you give me some help?
Added: 2002-07-07 - 02:50
Attached Articles | {"url":"http://planetmath.org/SequentiallyCompact","timestamp":"2014-04-18T00:26:07Z","content_type":null,"content_length":"68382","record_id":"<urn:uuid:71adc5c1-04ad-4cb5-832d-584dae2e5103>","cc-path":"CC-MAIN-2014-15/segments/1398223211700.16/warc/CC-MAIN-20140423032011-00399-ip-10-147-4-33.ec2.internal.warc.gz"} |
Re: C code .vs. Assembly code for Microcontrollers/DSPs ?
torbenm@diku.dk (Torben AEgidius Mogensen)
25 Mar 1996 21:49:01 -0500
From comp.compilers
| List of all articles for this month |
From: torbenm@diku.dk (Torben AEgidius Mogensen)
Newsgroups: comp.compilers,comp.dsp
Date: 25 Mar 1996 21:49:01 -0500
Organization: Department of Computer Science, U of Copenhagen
References: 96-03-006 96-03-044 96-03-078 96-03-098 96-03-144
Keywords: architecture
Robert A Duff <bobduff@world.std.com> wrote:
>The size in bits of the result
>should be determined by the size of the operands, and not their "type".
jfc@mit.edu (John Carr) writes:
>Unless you move away from a static typed language, the C model has a
>great advantage. If the result of an operation is defined to have
>infinite precision a long expression gains bits for every operator and
>the code to implement it is big and slow.
>To get the exact result for
> (x + y) * (a + b) / d
>where all variables are 32 bits requires 66 bit arithmetic. Addition
>adds 1 bit and multiplication doubles the number of bits. The quotient
>has the same number of bits as the dividend.
Robert Duff's proposal should be refined: the size in bits of the
result should be determined by the size of the operands and the type
(size) of the variable that gets the result. Hence, multiplying two
n-bit numbers and putting the result in an n-bit variable doesn't need
a 2n-bit intermediate result. Hence, the compiler should move size
information both forwards (upwards in the expression tree) and
backwards (downwards in the expression tree): The forwards information
describes the maximum number of bits required to hold the full
precision, the backwards information the number of bits required by
the destination. The compiler can choose any number of bits that is at
least the _minimum_ of these two numbers. Hence, you don't get the
explosion in the size of intermediate results that John Carr fears.
I much favour the Pascal idea of specifying exact number ranges, e.g.
var x : -13342..275249;
The compiler can choose any word size that contains this range for
implementing x. This will typically be the register size if this is
large enough, as this would give the best speed. For arrays, the
'packed' keyword tells the compiler that it should try to optimize the
array for size rather than speed, but it gives no guarantee that the
absolute minimum space is used.
Pascal does have an 'integer' type that is machine specific. I would
actually prefer omitting this altogether. If you think it is tedious
to write -32768..32767 every time you declare a variable, you are free
to declare
type integer = -32768..32767;
which gives the compiler useful information when moving to a different
platform. Note that nothing prevents the compiler from using 32-bit
words to implement this type. If you want to have a type that is
exactly 16 bits (for bitwise operations etc.), you could add an
'exact' keyword to specify that the machine should use the least
number of bits that can hold the range.
Torben Mogensen (torbenm@diku.dk)
Post a followup to this message
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Dimension of Chow groups
up vote 1 down vote favorite
The integral Chow groups are infinite dimensional. Can we say something about their dimension, for example, how many elements are required to generate them? Or their vector space dimension after
tensoring with $\mathbb{Q}$. What is known in this direction? I recall hearing some statements whose proof uses the theory of Chow varieties.
Also, is there a good reference for Chow varieties. I am looking for something which will assume familiarity with algebraic geometry at the level of Hartshorne and which uses that language. Since I
am reading Vistoli's notes on stacks, reference to some construction using the language of stacks might also be helpful.
EDIT: Removed the remarks on countable generation.
1 A countably generated abelian group is a quotient of a direct sum of countably many abelian groups, and hence is countable. – Emerton May 19 '11 at 12:23
Thanks for pointing that out. It was really stupid of me. – Rex May 19 '11 at 13:15
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1 Answer
active oldest votes
Chow groups are not in general countable. For example, $CH^1(X) = Pic(X)$ is uncountable for a smooth projective curve $X$ of positive genus over any uncountable algebraically
closed field.
up vote 5 down vote The theory of Chow varieties allows one to prove the countability of Chow groups modulo algebraic equivalence (this follows easily from the definitions). A nice reference for Chow
accepted varieties is Chapter I of the book "Rational curves on algebraic varieties" by Kollar.
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Function from Metric Space to R is Continuous
February 28th 2011, 09:28 AM
Function from Metric Space to R is Continuous
Let $(M, d)$ be a metric space. Let $f:M\longrightarrow M$ be a continuous function. Let $g:M\longrightarrow\mathbb{R}$ be defined by $x\longmapsto d(x,f(x))$. Show that $g$ is continuous.
So what we know is that $f$ is continuous, so for any $\eta>0$, there exists $\mu>0$ such that:
$d(x,y)<\mu \Rightarrow d(f(x),f(y))<\eta$
But what we want to do for any $\epsilon>0$ is to produce $\delta>0$ such that:
$d(x,y)<\delta \Rightarrow |g(x)-g(y)|=|d(x,f(x))-d(y,f(y))|<\epsilon$
I have a feeling we'll want to produce $\delta$ in terms of $\epsilon, \eta, \mu$. I was thinking breaking something up using a triangle inequality, but I've tried a few different ways that have
all led to dead ends.
February 28th 2011, 09:56 AM
You can use the inequality $|d(x,z)-d(y,z)|\leq d(x,y)$ and write $d(x,f(x))-d(y,f(y))=d(x,f(x))-d(x,f(y))+d(x,f(y))-d(y,f(y))$.
February 28th 2011, 10:24 AM
I already tried that and came up with a dead end. The result of that yields:
$|g(x)-g(y)|\leq d(f(x),f(y))+d(x,y)$
Which if we let those be less than $\eta, \mu$ respectively from the previous part, we get:
Which I just don't see how you could possibly get $\delta$ out of something of that sort.
February 28th 2011, 10:29 AM
Given $\eta$, we can find $\mu$ such that $d(f(x),f(y))\leq \frac{\eta}2$ if $d(x,y)\leq \mu$. We can choose this $\mu$ smaller than $\frac{\eta}2$. | {"url":"http://mathhelpforum.com/differential-geometry/172937-function-metric-space-r-continuous-print.html","timestamp":"2014-04-20T09:11:54Z","content_type":null,"content_length":"9890","record_id":"<urn:uuid:577c6187-6a9b-4299-bd77-457276e1148f>","cc-path":"CC-MAIN-2014-15/segments/1397609538110.1/warc/CC-MAIN-20140416005218-00516-ip-10-147-4-33.ec2.internal.warc.gz"} |
Modulus of continuity is continuous?
June 7th 2012, 11:53 AM #1
Modulus of continuity is continuous?
I'm currently reading a book on convergence of probability measures, and there is a property that they assert without too many details that I can't manage to work out for myself.
To put you in context, we're in the space $C[0,1]$ of functions $f:[0,1]\to\mathbb{R}$ that are continuous with respect to the standard euclidean metric $d(x,y)=|x-y|$, and we define the metric
on $C[0,1]$ to be $\rho(f,g)=\sup_{t\in[0,1]}|f(t)-g(t)|$.
For every $\delta>0$ and $f\in C[0,1]$, define the modulus of continuity $w(f,\delta)$ as
In the book, they say that for any fixed $\delta>0$, the function $w(\cdot,\delta):C[0,1]\to\mathbb{R}$ is continuous. Their only argument is that
for any $f,g\in C[0,1]$, we have $|w(f,\delta)-w(g,\delta)|\leq 2\rho(f,g)$. It's easy to see how this implies continuity, but I can't manage to show this inequality myself. Any help or hints
would be greatly appreciated.
Re: Modulus of continuity is continuous?
We have $|f(x)-f(y)|\leq 2\rho(f,g)+|g(x)-g(y)|$ then take supremum. Switch the roles of $f$ and $g$ to get what you want.
P.S. Which book are you studying?
Re: Modulus of continuity is continuous?
Thanks for the answer, I'll try this out.
I'm studying Convergence of Probability Measures 2nd edition by Patrick Billingsley, I'm trying to gain a better understanding of Donsker's Theorem, which is basically a version of the Central
Limit Theorem for random variables taking values in C[0,1].
June 7th 2012, 01:32 PM #2
June 7th 2012, 02:18 PM #3 | {"url":"http://mathhelpforum.com/differential-geometry/199767-modulus-continuity-continuous.html","timestamp":"2014-04-19T07:26:24Z","content_type":null,"content_length":"40371","record_id":"<urn:uuid:4ec48619-1edc-4e03-9a4d-4bdf30110d74>","cc-path":"CC-MAIN-2014-15/segments/1397609536300.49/warc/CC-MAIN-20140416005216-00321-ip-10-147-4-33.ec2.internal.warc.gz"} |
Angle Sum of a Polygon
The sum of the interior angles is [tex]180n - 360[/tex]. Let a set Q denote n angles whose sum is 180n - 360: [tex][ {a_{1}, a_{2} , ... , a_{n} ][/tex]. For any of these angles, the external angle
is equal to [tex]180 - a_{k} [/tex]. Since there are n angles, the sum of all external angles is
[tex]S = \sum_{k = 1}^{n} 180 - a_{k} [/tex]
S is obviously equal to [tex] 180n - (180n - 360) = 360 [/tex] | {"url":"http://www.physicsforums.com/showthread.php?t=171474","timestamp":"2014-04-21T02:16:49Z","content_type":null,"content_length":"28765","record_id":"<urn:uuid:302bed2d-0a83-4def-ac50-ec9209787c4f>","cc-path":"CC-MAIN-2014-15/segments/1397609539447.23/warc/CC-MAIN-20140416005219-00074-ip-10-147-4-33.ec2.internal.warc.gz"} |
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Nonlinear Approximation Techniques in General Relativity and Geometric Analysis
Seminar Room 1, Newton Institute
In this lecture, we consider nonlinear approximation techniques for treating singularities in geometric analysis and general relativity. We first review approximation theory for Petrov-Galerkin and
Galerkin techniques for nonlinear variational problems. We then examine the use of a posteriori error estimation for adaptive construction of discrete (finite element, wavelet, spectral) spaces for
deriving nonlinear approximation techniques; these techniques attempt to meet a target approximation quality using discrete spaces of minimal dimension, and are of increasing importance in modeling
and computational science.
We then turn to nonlinear elliptic problems in geometric analysis, and focus on the constraints in Einstein flow. We look briefly at weak solution theory on manifolds with boundary, and various Lp
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How to calculate the Factorial of Numbers > 170
04-25-2011 02:05 PM #2
04-25-2011 01:57 PM #1
Thanked 0 Times in 0 Posts
Re: How to calculate the Factorial of Numbers > 170
How to calculate the Factorial of Numbers > 170
I use factorial() function, but it doesn't work for numbers more than 170. I also use lfactorial() but doesn't work too.
I put simple codes below :
> factorial(200)
[1] Inf
Warning message:
In factorial(200) : value out of range in 'gammafn'
> exp(lfactorial(200))
[1] Inf
> gamma(201)
[1] Inf
Warning message:
value out of range in 'gammafn'
What are you trying to do? Do you need that exact value or are you using it for further calculations? For example: if you're using binomial coefficients these have factorials in the definition
but you can simplify the calculations so you don't need to compute the full factorials (otherwise it would be impossible to compute binomial probabilities for large sample sizes).
Re: How to calculate the Factorial of Numbers > 170
The factorial of x, x! is given by the gamma function +1 ; gamma (x+1).
Therefore use ?gamma in r.
factorial of 100;
>170 is likely too large for the standard gamma function so use the log gamma function (for example the factorial of 200):
lgamma(200+1) is 863.232, so the factorial of 200 is e^863.232. Does this help?
The true ideals of great philosophies always seem to get lost somewhere along the road..
Re: How to calculate the Factorial of Numbers > 170
Otherwise you can use this (from Martin Maechler´s reply);
#1 'mpfr' number of precision 128 bits
# 4.023872600770937735437024339230039857186 e 2564
Last edited by TheEcologist; 04-27-2011 at 09:44 AM. Reason: second evaluation was a little much, removed it
The true ideals of great philosophies always seem to get lost somewhere along the road..
Re: How to calculate the Factorial of Numbers > 170
I didn't know about
I was considering writing a little bit of code to do arbitrary length integer calculations. Then I thought it had probably been done before so I just stopped and since it wasn't pressing to any
of my research I just forgot about it.
But I still say the question remains as to what the numbers are being used for because a lot of times you can reduce the calculations down a lot if lots of things are going to cancel.
Re: How to calculate the Factorial of Numbers > 170
In the case of binomial probabilities much will indeed 'cancel out' and such algorithms already exist :Catherine Loader (2000). Fast and Accurate Computation of Binomial Probabilities. And
therefore have already been implemented in functions as dbinom - I just automatically assumed DataMiner needs this for something that isn't already implemented... but you are correct.. it would
be way better if DataMiner "Describes the goal and not only the step".
The true ideals of great philosophies always seem to get lost somewhere along the road..
Re: How to calculate the Factorial of Numbers > 170
Isn't there a "Stirling's approximation" to calculate large factorials?
Re: How to calculate the Factorial of Numbers > 170
Yes. And it's quite good for large n in a relative error sense. So depending on what you're doing it could be a decent approximation. But even using the approximation... 170! is too large for R
to handle using the base numeric types it has built in so you'd either need an arbitrary length integer or you need to specify what you're doing and figure out a better way to do it than to do
the full computations. So it all really depends on if the OP ever shows up again.
Re: How to calculate the Factorial of Numbers > 170
Stirling<- function (n)((2*pi*n)^.5)*(n/exp(1))^n
I think this is the Stirling approximation.
Re: How to calculate the Factorial of Numbers > 170
Dason, I love mpfr. It has proven very useful in the past ! (when correct rounding becomes important)
For those of you who dont know about it:
and http://www.mpfr.org/
The true ideals of great philosophies always seem to get lost somewhere along the road..
Re: How to calculate the Factorial of Numbers > 170
Oh I didn't realize it was actually a C library. That might come in even more useful than I thought before...
Re: How to calculate the Factorial of Numbers > 170
Exactly. Depending on what you're doing you might be more concerned that the absolute error gets QUITE large.
Re: How to calculate the Factorial of Numbers > 170
While we're on the subject, how many significant figures should be carried with stats work? Is it different for p < .05 than for p < .001?
04-25-2011 02:35 PM #3
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04-27-2011 09:53 AM #6
04-27-2011 07:37 PM #7
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References for Madhava
References for Madhava of Sangamagramma
Version for printing
1. G G Joseph, The crest of the peacock (London, 1991).
2. K V Sarma, A History of the Kerala School of Hindu Astronomy (Hoshiarpur, 1972).
3. A K Bag, Madhava's sine and cosine series, Indian J. History Sci. 11 (1) (1976), 54-57.
4. D Gold and D Pingree, A hitherto unknown Sanskrit work concerning Madhava's derivation of the power series for sine and cosine, Historia Sci. No. 42 (1991), 49-65.
5. R C Gupta, Madhava's and other medieval Indian values of pi, Math. Education 9 (3) (1975), B45-B48.
6. R C Gupta, Madhava's power series computation of the sine, Ganita 27 (1-2) (1976), 19-24.
7. R C Gupta, Madhava's rule for finding angle between the ecliptic and the horizon and Aryabhata's knowledge of it, in History of oriental astronomy, New Delhi, 1985 (Cambridge, 1987), 197-202.
8. R C Gupta, On the remainder term in the Madhava-Leibniz's series, Ganita Bharati 14 (1-4) (1992), 68-71.
9. R C Gupta, The Madhava-Gregory series, Math. Education 7 (1973), B67-B70.
10. T Hayashi, T Kusuba and M Yano, The correction of the Madhava series for the circumference of a circle, Centaurus 33 (2-3) (1990), 149-174.
11. C T Rajagopal and M S Rangachari, On an untapped source of medieval Keralese mathematics, Arch. History Exact Sci. 18 (1978), 89-102.
12. C T Rajagopal and M S Rangachari, On medieval Keralese mathematics, Arch. History Exact Sci. 35 (1986), 91-99.
JOC/EFR November 2000
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9500 us dollars to pounds
You asked:
9500 us dollars to pounds
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If you set it up as a mortgage, where each payment reduces principal until payment no 40 takes it to zero, the yield on payment no 21 (the median payment) is 2.90%.
However, if a would-be buyer of TINTs does the math as it should be done, then the CAGR is merely 1.28%. I could be wrong, of course. But allow me to work through another example, and let’s make the
same, counter-factual assumption that we can trade for free.
E*Trade is offering 20-year Treasury derivatives at 57.155, the prin STRIPS of 08/15/32. Let’s make settlement T-1. Thus, the holding-period would be 19.86 years. ET says the YTM would be 2.879%.
Excel’s YIELD function returns 2.84%. So the numbers are ballpark enough. If TINTs of comparable maturity are priced as ‘PAR minus the STRIP price’, then a TINT due in 08/15/32 would be priced at
42.845. The question now becomes, “What does the buyer get for that $428.45? According to your example, he/she would get 39.72 payments of approximately $14.30, or a total of approximately $567.85.
Do the math. The would-be buyer pays $428.45 to receive $567.85 prorated over a 19.86-year holding-period, or a very underwhelming, nominal CARG of 1.43% and a very serious loss of purchasing-power
once taxes are paid and inflation is subtracted. If an investor wants to screw around with STRIPs, TINTs, and zeros, then at least they should be buying them when they are attractively priced, and
they should be calculating their yields correctly, as this representative sample of my own holdings suggests.
Issue CPN Due Traded Price CAGR
FICO 0.000 08/08/16 05/08/06 53.160 6.36%
Cabco 0.000 10/01/30 07/11/08 23.112 6.81%
TVA 0.000 11/01/25 07/01/09 41.914 5.47%
FNMA 0.000 10/09/19 12/21/09 55.587 6.17%
Merrill 0.000 09/25/18 12/29/09 61.795 5.66%
Israel 0.000 02/15/24 01/06/10 48.976 5.19%
Intl Bk 0.000 05/01/30 03/31/10 33.284 5.63%
Where CAGR = (PAR/price)^ (1/Holding-period) -1, which would then also have to be discounted for taxes and inflation to estimate what the effective-yield would really be (aka, money that is spendable
at the grocery store and not accounting tricks based on a return of principal that should never have been surrendered in the first place).
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Discussion about math, puzzles, games and fun. Useful symbols: ÷ × ½ √ ∞ ≠ ≤ ≥ ≈ ⇒ ± ∈ Δ θ ∴ ∑ ∫ • π ƒ -¹ ² ³ °
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#1 2005-12-15 00:39:15
Since I could not solve the following equation though applying all information I have about trigonometry, I would very appreciate if anybody could help or make some suggestion.
Thanks for your time.
(Answer: -1/2)
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#2 2005-12-15 05:19:28
Power Member
Re: Trigonometry
Those angles be not on the unit circle, matey. Aaargh!
Ahem. I don't think they have coterminal angles on the unit circle, either.
Which means...use a calculator! Right? Don't tell me there's a goofy trig identity that will come to the rescue. I hates trig identities.
El que pega primero pega dos veces.
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#3 2005-12-15 06:02:19
Re: Trigonometry
I didn't know it be talk like a pirate day. Yarr!
That scurvy dog, ryos, be correct. There be no identity that will rescue ye, landlubber.
At least, none that me or my parrot know. Arr!
Why did the vector cross the road?
It wanted to be normal.
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#4 2005-12-15 06:51:34
Full Member
Re: Trigonometry
Aye aye, captain Rottingham! Calculators at deck, rrreform in rradials if ye think it will make ye life easier.
As at night, we realise we will never Know and further thinking is pointless....Yet we start again and again
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#5 2005-12-15 07:29:52
Re: Trigonometry
Me only contribution, mateys, is this ere lubberly sketch-ning
"The physicists defer only to mathematicians, and the mathematicians defer only to God ..." - Leon M. Lederman
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Discussion about math, puzzles, games and fun. Useful symbols: ÷ × ½ √ ∞ ≠ ≤ ≥ ≈ ⇒ ± ∈ Δ θ ∴ ∑ ∫ • π ƒ -¹ ² ³ °
Hello,Since I could not solve the following equation though applying all information I have about trigonometry, I would very appreciate if anybody could help or make some suggestion. Thanks for your
time.cos(2pi/11)+cos(4pi/11)+cos(6pi/11)+cos(8pi/11)+cos(10pi/11)=?(Answer: -1/2)
Those angles be not on the unit circle, matey. Aaargh!Ahem. I don't think they have coterminal angles on the unit circle, either.Which means...use a calculator! Right? Don't tell me there's a goofy
trig identity that will come to the rescue. I hates trig identities.
I didn't know it be talk like a pirate day. Yarr!That scurvy dog, ryos, be correct. There be no identity that will rescue ye, landlubber. At least, none that me or my parrot know. Arr!
Aye aye, captain Rottingham! Calculators at deck, rrreform in rradials if ye think it will make ye life easier. | {"url":"http://www.mathisfunforum.com/viewtopic.php?pid=20627","timestamp":"2014-04-19T19:40:40Z","content_type":null,"content_length":"13035","record_id":"<urn:uuid:671c41a9-9396-4908-9198-6a228c1364d6>","cc-path":"CC-MAIN-2014-15/segments/1398223202774.3/warc/CC-MAIN-20140423032002-00336-ip-10-147-4-33.ec2.internal.warc.gz"} |
Flight through a Wormhole
Wormholes are traversable connections between two universes or between two distant regions of the same universe. This contribution shows a flight through the wormhole that connects Tübingen
University with Boulogne sur Mer in the north of France.
What is a wormhole?
Wormholes are traversable connections between two universes or between two distant regions of the same universe. They are a popular feature of science fiction stories where they allow travelling to
extraterrestrial cultures.
The wormholes described here are exact solutions of Einstein's field equations; they were developed by K. Thorne to give a scientific foundation to the novel Contact by C. Sagan ([1]) (The story of
this development is told in [2]; [3] gives the theoretical background).
Even though these wormholes are in agreement with the theory of relativity, their implementation in practice meets a little problem: In order to curve spacetime in such a way that a wormhole forms,
an exotic type of matter is required with negative energy density. In the case of the wormhole described here, the absolute values of this energy density would have to be extremely high, up to a
billion times the density of a neutron star. Such a type of matter is not known in our universe. Also, to make it possible to fly and to look through the wormhole, this exotic matter must not
interact with normal matter or with electromagnetic radiation.
But these technical problems cannot prevent us from simulating, on the computer, a flight through such a wormhole.
The wormhole shown here connects the place in front of the physics institutes of Tübingen University with the beautiful sand dunes near Boulogne sur Mer in the north of France (image source [4]). For
the theoretical astrophysics personnel it is handy for short trips in their lunch hour.
In order to clearly show the properties of the curved space (that itself is invisible), a cubic lattice of 12 yellow rods has been built around the wormhole. The rods follow geodesics, i.e. lines
that are as straight as possible. The visible distortions, especially of the rods behind the wormhole, come from light deflection in the curved space.
Eight green rods point radially inwards from the corners of the yellow cube. They approach each other and in flat space they would meet in the middle. In the curved space of the wormhole they do not.
At the throat, the interface between the two universes connected by the wormhole, the diagonal rods meet the corners of a red cube the edges of which are also geodesics. The diagonal rods continue
straight-lined into the other universe. There, they move apart and meet the edges of the third, blue cube from the inside. The blue cube has the same size as the yellow one.
The following movies show a flight from Tübingen to Boulogne sur Mer and back.
Movie: Flight through a wormhole (short)
The movie shows a short trip through the wormhole (section of the second movie)
MPEG1 320×240 (2.1 MB), MPEG1 640×480 (20.1 MB)
Movie: Flight through a wormhole (long)
Flight through a wormhole and back with an extra round trip at the throat. (Attention: big!)
MPEG1 320×240 (9.6 MB), MPEG4 640×480 (47.2 MB)
A few scenes of the movie
The wormhole from the outside
The wormhole in front of the physics building of Tübingen University.
Approaching the throat
Approaching the throat we see the landscape on the other side of the wormhole. The green diagonal rods continue straight on.
Almost at the throat
We see that the green diagonal rods meet the corners of the blue cube from the inside.
On the other side
Looking back through the wormhole, we perceive the buildings of Tübingen University on the other side.
The wormhole from the outside
The wormhole is seen to hover above a sand dune near Boulogne sur Mer.
Flying back
Round trip at the throat
Before traversing the throat we turn upwards and move over the red cube that marks the plane of the throat.
Above the plane of the throat
The red cube appears to be an infinite plane. Its edges are flat and the quadratic side faces have corners of 120 degrees. "Cubes" of this kind, contradictory to everyday geometry, are possible only
in a curved space.
Looking down
Through the side faces of the red cube, we can look into the other universe. Here we look down on the floor below the wormhole in Tübingen.
Looking to the side
When looking through the next red square, we see the Tübingen scene from the side.
Looking up
On to the next red square: Through it we are looking up to the sky above the wormhole.
Looking to the other side
The view through the next square shows a number of benches, upside down.
Another look downwards
We have completed the tour around the wormhole and look once again through the first red square, downwards.
Back to Tübingen
We leave the wormhole and come back to Tübingen.
The wormhole from the outside
There, once again we look back towards the wormhole.
Metric and topology
Morris and Thorne ([3]) describe a wormhole as a solution of Einstein's field equation with the following metric:
radius of the throat
Energy momentum tensor
In an orthonormal frame
and the components that follow from these by symmetry. Substituting the metric into the field equations we obtain the energy momentum tensor:
It has the unpleasant property of a negative energy density
Embedding diagram
Due to the spherical symmetry one can use a two-dimensional plane through the origin to describe the main properties of the metric and the photon paths. This plane can be embedded in a
three-dimensional Euclidean space. The spacelike metric in the equatorial plane reads
the embedding surface (top) consists of the points with the cartesian coordinates
[1] C. Sagan, Contact, Simon and Schuster, 1985.
[2] K. Thorne, Black Holes and Time Warps: Einstein's Outrageous Legacy, W. W. Norton & Company, 1995.
[3] M. S. Morris, K. S. Thorne, Wormholes in spacetime and their use for interstellar travel: A tool for teaching general relativity, Am. J. Phys. 56, 395-412, 1988.
Contact: Would you like to be notified of new contributions or to send us a message?
Survey: How do you use this site and how would you like it to develop?
Corvin Zahn
, Date: March 13, 2008
About Us
All contents copyright (C) 2001-2013 Ute Kraus, Corvin Zahn. All rights reserved. For more information see | {"url":"http://www.spacetimetravel.org/wurmlochflug/wurmlochflug.html","timestamp":"2014-04-17T21:55:21Z","content_type":null,"content_length":"20235","record_id":"<urn:uuid:b9a22180-3057-4131-8b28-f87ebc8629a7>","cc-path":"CC-MAIN-2014-15/segments/1397609532128.44/warc/CC-MAIN-20140416005212-00142-ip-10-147-4-33.ec2.internal.warc.gz"} |
rotation matrices
September 25th 2011, 06:24 AM #1
Jul 2011
rotation matrices
Hi, I am trying to find some rotation matrices of $R^3$.
So for angle $\frac{\pi}{2}$ and axis $e_2$ I think the rotation matrix is
$R_{e_2} (\frac{\pi}{2}) = \left(\begin{matrix} cos(\frac{\pi}{2}) &0 &sin(\frac{\pi}{2})\\0 &1 &0\\-sin(\frac{\pi}{2}) &0 &cos(\frac{\pi}{2})\end{matrix}\right) = \left(\begin{matrix} 0 &0 &1\\0
&1 &0\\-1 &0 &0\end{matrix}\right)$
I just used the given rotation matrix in $R^3$ from the text with no real working out needed. Is this correct?
Need help on this one:
For the angle $\frac{\pi}{6}$ and axis containing the vector $(1,1,1)^t$.
Not sure how to use the $(1,1,1)^t$ in this question.
Thanks for any help.
Re: rotation matrices
I think I have misunderstood what the axis e2 is and I think my working out is wrong.
Any help would be nice.
Re: rotation matrices
Anyone have any ideas as to how I should interpret the axis $e_2$ and use $(1,1,1)^t$?
I have thought about this question but still can't get it.
And sorry for making another post, wanted to edit a previous post but I couldn't.
Re: rotation matrices
Hi, I am trying to find some rotation matrices of $R^3$.
So for angle $\frac{\pi}{2}$ and axis $e_2$ I think the rotation matrix is
$R_{e_2} (\frac{\pi}{2}) = \left(\begin{matrix} cos(\frac{\pi}{2}) &0 &sin(\frac{\pi}{2})\\0 &1 &0\\-sin(\frac{\pi}{2}) &0 &cos(\frac{\pi}{2})\end{matrix}\right) = \left(\begin{matrix} 0 &0 &1\\0
&1 &0\\-1 &0 &0\end{matrix}\right)$
I just used the given rotation matrix in $R^3$ from the text with no real working out needed. Is this correct?
Need help on this one:
For the angle $\frac{\pi}{6}$ and axis containing the vector $(1,1,1)^t$.
Not sure how to use the $(1,1,1)^t$ in this question.
Thanks for any help.
This is a much more complicated question, and requires more advanced treatment. The best information I can give you is to go to this website and use that procedure. As you can see, it's a
multi-step procedure.
September 27th 2011, 06:30 PM #2
Jul 2011
September 29th 2011, 03:38 AM #3
Jul 2011
September 29th 2011, 07:33 AM #4 | {"url":"http://mathhelpforum.com/advanced-algebra/188778-rotation-matrices.html","timestamp":"2014-04-23T13:02:17Z","content_type":null,"content_length":"43161","record_id":"<urn:uuid:7ca0e24e-7cd8-4a53-bab7-4ee133877de5>","cc-path":"CC-MAIN-2014-15/segments/1398223202548.14/warc/CC-MAIN-20140423032002-00465-ip-10-147-4-33.ec2.internal.warc.gz"} |
Getting Week Numbers in Access
July 7th, 2009 by datapig Leave a reply »
It’s always a little embarrassing when you roll your own Function to achieve some functionality that already exists in Excel or Access.Â
Recently, I stumbled across a user-defined function designed to find a week number in Access. I asked the owner of this code why he wrote a function to do this when Access has one. He gave me
the ‘thousand-yard stare’, then he went on to explain that he couldn’t find the WEEKNUM function in Access.Â
In Excel, the WEEKNUM function converts a given date into a week number. For example, this would give you the week number in which the current date falls.Â
There is no such function in Access. You won’t find WEEKNUM in the list of available Access Functions. There’s is WEEKDAY and WEEKDAYNAME, but no WEEKNUM.
So he naturally started to write his own function.Â
Well, it turns out you can get to a week number in Access by using the DATEPART function. Like so:
You can also use this in Excel VBA to get around calling Application.WorksheetFunction.WeekNum
Today, my friend still uses his own Week Number user-defined function – mainly out of spite. I suggested that he create a function that returns the average of range of numbers. He told me to
shut up.
Have you ever rolled your own user-defined function only to find out that the functionality already existed?
I’m waiting for someone to point out the difference between US and European definitions for Week Numbers.
Hi Jon
Why not
Ron: Interesting. I did not know there was an international impact using WEEKNUM. Good link.
Ron: Wouldn’t this give you an ISO week number?
DatePart(“ww”, Date, vbMonday, vbFirstFullWeek)
Hi Mike
There are a few bugs when you use this so this is not a option
Submiting this test procedure to prove out the bug:
Sub Test2()
Dim GL_Dt As Date
Dim I As Long
For I = 39814 To 100000 ‘ test for more then 27 years
GL_Dt = I
If Int(([GL_Dt] – DateSerial(Year([GL_Dt] – Weekday([GL_Dt] – 1) + 4), 1, 3) + Weekday(DateSerial(Year([GL_Dt] – Weekday([GL_Dt] – 1) + 4), 1, 3)) + 5) / 7) _
<> DatePart(“ww”, [GL_Dt], vbMonday, vbFirstFourDays) Then MsgBox “Error in date ” & I
Next I
End Sub
Public Function IsoWeekNumber(d1 As Date) As Integer
‘ Attributed to Daniel Maher
Dim d2 As Long
d2 = DateSerial(Year(d1 – Weekday(d1 – 1) + 4), 1, 3)
IsoWeekNumber = Int((d1 – d2 + Weekday(d2) + 5) / 7)
End Function
In Access:
Format ([YourField],”ww”)
This will not return the correct weeknumber for all days. Check out this KB article and you will understand why he told….(DATAPIG SAYS PLAY NICE HENRY).
I can hardly call this a PowerTip. A good function in Access for ISO (european) weeknumbers and the year that goes with it, that would be a PowerTip.
Wow, many rather logical tips! I appreciate you crafting this posting and the remainder of your internet site is outstanding! | {"url":"http://datapigtechnologies.com/blog/index.php/getting-week-number-in-access/","timestamp":"2014-04-19T22:22:53Z","content_type":null,"content_length":"40477","record_id":"<urn:uuid:fec8ff2f-eb70-401b-9fd7-b1d8a63b7d97>","cc-path":"CC-MAIN-2014-15/segments/1398223205137.4/warc/CC-MAIN-20140423032005-00184-ip-10-147-4-33.ec2.internal.warc.gz"} |