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Convert us pounds to metric tonnes
Use our IBC metric calculator tool to find weight, pressure, temperature and volume conversions. Metric Tons, Kilograms, Pounds, Oz, Grams, Tons 1 bushel corn = 2.75 gallons ethanol = 18 lbs dried
distillers grain; 1 tonne corn 1 barrel = 42 US gallons = 34.97 UK (imperial) gallons = 0.136 tonne (approx) ton = 2,000 pounds; 1 short ton = 907.18 kilograms; 1 short ton = .9072 metric tons 1 mile
= 1.61 kilometers; To convert kilometers into miles, multiply by . 6214.
You are currently converting mass units from avoirdupois pound to metric tonne. 1 lb = 0.00045359243 t. avoirdupois pound Open avoirdupois pound Conversion Calculators. Convert Metric Tons to Tons.
Please enter values then click on Calculate. Metric Tons = Tons = Chapel Steel Calculators. 1 Metric Ton, = 2,204.6 Pounds (lbs). 1 Pounds (lbs), = 0.0005 Tons. 1 Pounds ( lbs), = 0.0004536 Metric
Tons. 1 Pounds (lbs), = 0.4536 Kilograms (kg). A pound-force foot (lbf·ft) is a unit of torque (also called “moment” or “moment of force”) in the US Customary Units and British Imperial Units. One
pound-force This calculator-converter provides conversion of short tons to tons [metric] (ST to t ) and backwards metric tons to short tons (t to ST). Convert: Input. Pounds, Tons. 1 foot-pound
(ft-lb.) = 1.355818 Joules (J) for impact energy 1.355818 Newton- meters (N.m) for torque. 1 metric ton, = 2204.6
This calculator-converter provides conversion of short tons to tons [metric] (ST to t ) and backwards metric tons to short tons (t to ST). Convert: Input. Pounds, Tons.
The pound is a US customary and imperial unit of weight. A pound is sometimes also referred to as a common ounce. Pounds can be abbreviated as lb, and are Pounds (lbs) to metric tons (t) weight
conversion calculator and how to convert. It is commonly used in most countries, except the United States, where the short ton is used instead. Pounds to Metric Tons Conversion Table. (some results
Pounds to Tonnes Conversion Calculator, Conversion Table and How to Convert . This calculator provides conversion of pounds to tonnes metric and backwards (t to lb). Enter pounds or tonnes Pounds to
short tons (US) (lb to t) · Pounds To convert pounds to short tons, multiply the pound value by 0.0005 or divide by 2000. 1 US Short Ton = 2000 Pounds. How many pounds in an imperial (long) ton
22 Jul 2018 Pounds to Metric Tons (or Tonnes) (lb to t) conversion calculator for Weight conversions with additional tables and formulas.
››More information from the unit converter. How many tons in 1 metric tonnes? The answer is 1.1023113109244. We assume you are converting between ton [short, US] and metric tonne. You can view more
details on each measurement unit:
1 Kilotonne or metric kiloton (unit of mass) is equal to 1000 metric tons. A metric ton is exactly 1000 kilograms (SI base unit) making a kilotonne equal to 1000000
1 bushel corn = 2.75 gallons ethanol = 18 lbs dried distillers grain; 1 tonne corn 1 barrel = 42 US gallons = 34.97 UK (imperial) gallons = 0.136 tonne (approx) ton = 2,000 pounds; 1 short ton =
907.18 kilograms; 1 short ton = .9072 metric tons 1 mile = 1.61 kilometers; To convert kilometers into miles, multiply by . 6214. 1 Kilotonne or metric kiloton (unit of mass) is equal to 1000 metric
tons. A metric ton is exactly 1000 kilograms (SI base unit) making a kilotonne equal to 1000000
1 foot-pound (ft-lb.) = 1.355818 Joules (J) for impact energy 1.355818 Newton- meters (N.m) for torque. 1 metric ton, = 2204.6
1 Kilotonne or metric kiloton (unit of mass) is equal to 1000 metric tons. A metric ton is exactly 1000 kilograms (SI base unit) making a kilotonne equal to 1000000 15 Sep 2017 Contact Us. Share.
Coal Mine Methane Units Converter. Measure Kilograms ( kg) CH4. = Btu's Metric tons (tonnes) CO2 equivalent. Metric weight and mass from tonne metric to pound Conversion Results: other kinds of
culinary training for converting the weight and mass cooking units measures. ›› More information from the unit converter. How many pounds in 1 tonnes? The answer is 2204.6226218488. We assume you are
converting between pound and metric tonne. You can view more details on each measurement unit: pounds or tonnes The SI base unit for mass is the kilogram. 1 kilogram is equal to 2.2046226218488
pounds, or 0.001 tonnes.
1 Kilotonne or metric kiloton (unit of mass) is equal to 1000 metric tons. A metric ton is exactly 1000 kilograms (SI base unit) making a kilotonne equal to 1000000 15 Sep 2017 Contact Us. Share.
Coal Mine Methane Units Converter. Measure Kilograms ( kg) CH4. = Btu's Metric tons (tonnes) CO2 equivalent. Metric weight and mass from tonne metric to pound Conversion Results: other kinds of
culinary training for converting the weight and mass cooking units measures. ›› More information from the unit converter. How many pounds in 1 tonnes? The answer is 2204.6226218488. We assume you are
converting between pound and metric tonne. You can view more details on each measurement unit: pounds or tonnes The SI base unit for mass is the kilogram. 1 kilogram is equal to 2.2046226218488
pounds, or 0.001 tonnes. lbs or metric ton The SI base unit for mass is the kilogram. 1 kilogram is equal to 2.2046226218488 lbs, or 0.001 metric ton. Note that rounding errors may occur, so always
check the results. Use this page to learn how to convert between pounds and metric tons. Type in your own numbers in the form to convert the units! ›› Want other units? | {"url":"https://binaryoptionszjtct.netlify.app/benning77473nix/convert-us-pounds-to-metric-tonnes-110.html","timestamp":"2024-11-07T00:47:55Z","content_type":"text/html","content_length":"33870","record_id":"<urn:uuid:35926933-822a-491f-a3c5-08b573b19a4a>","cc-path":"CC-MAIN-2024-46/segments/1730477027942.54/warc/CC-MAIN-20241106230027-20241107020027-00765.warc.gz"} |
June/July/August 1980
The Kiddies’ Guide To Z80 Assembler Programming.
D. R. Hunt
Part: The First (and if doesn’t go down too well, Part: The Last).
Funny numbers, counting with sixteen fingers, and all that.
We have had many appeals for a beginner’s guide to Z80 assembler programming, and as no-one else volunteered, I thought I might have a go. After all, my qualifications for this task are impressive.
1. Three years ago I knew nothing about it.
2. It is arguable if I have learned much in the meantime.
3. I don’t mind making myself took an idiot in print (in the eyes of of the enlightened).
All this means is that I’m not too clever at it to baffle the reader, and that I’m new enough at it remember the difficulties I had at first.
So here goes: The first thing to learn is HEXadecimal, the numbering system used when writing machine code. HEX numbers are usually indicated by numbers either being prefixed ‘#’, or suffixed ‘H’.
Now to use HEX effectively, really means that you should grow three more fingers on each hand, as this is a little difficult for most normal people, an explanation will have to suffice.
Binary and HEXadecimal
Most of us count in units of 1, 10, 100, etc. For historical reasons we need not discuss, this counting in 10s business (known as ‘base 10’ counting or Decimal) is so much second nature that counting
in any other form may well seem ludicrous. However, there are other things on this earth which do use different systems, and computers figure largely among them. Right at the heart of it, the
computer uses the Binary system, and all it knows is that a number represented as ‘no volts’ will be interpreted as a ‘0’, whilst a number represented by ‘some volts’ will be interpreted as ‘1’. From
this it can be seen that the computer counts in twos (known as ‘base 2’ counting or Binary). In the same way that we count in 1s, 10s, 100s, etc, the computer counts in 1s, 10s, 100s etc.
Unfortunately, although the numbers look the same, they are in fact different. As each unit is the base number raised to next power (mathematically speaking) they actually mean:
We think Computers think
10 to the power 0 = 1 2 to the power 0 = 1 (= 1 Decimal)
10 to the power 1 = 10 2 to the power 1 = 10 (= 2 Decimal)
10 to the power 2 = 100 2 to the power 2 = 100 (= 4 Decimal)
10 to the power 3 = 1000 2 to the power 3 = 1000 (= 8 Decimal}
So that the number fifteen (Decimal) is expressed as: | {"url":"https://tupel.jloh.de/nascom/magazines/inmc-80-news/01/19/text/","timestamp":"2024-11-06T08:08:01Z","content_type":"text/html","content_length":"11109","record_id":"<urn:uuid:3257d086-8e5c-46eb-b8fd-58f6e63db39c>","cc-path":"CC-MAIN-2024-46/segments/1730477027910.12/warc/CC-MAIN-20241106065928-20241106095928-00631.warc.gz"} |
If P=NP then we HAVE an alg for SAT.
I am writing up the result of my survey of peoples opinion of P vs NP (it will be in a SIGACT News, in Lane's Complexity Column, in 2019.) Some people wrote:
P=NP but the proof will be nonconstructive and have a large constant.
Large constant could happen.
If by nonconstructive they mean not practical, then yes, that could happen.
The following does not quite show it can't happen but it does give one pause: an old result of Levin's shows that there is a program you could write NOW such that if P=NP then this program decides
except for a finite number of formulas
(all of which are NOT in SAT) and can be proven to work in poly time (I will later give three pointers to proofs). The
finite number of formulas
is why the people above may still be right. But only a finite number- seems like a weak kind of nonconstructive.
Since I am teaching crypto I wondered about Factoring. An old result of Gasarch (I proved it this morning -- I am sure it is already known) shows that there is a program you could write NOW such that
if Factoring is in P then this program factors a number ALWAYS (none of this
finite exception
crap) and can be proven to work in poly time. Even so, the algorithm is insane. If someone thought that factoring in P might be nonconstructive, my construction disproves it in such an absurd way
that the notion that factoring could be in P nonconstructively should be taken seriously but not literally. There should be a way to say formally:
I believe that FACTORING is in P but any poly-time algorithm is insane (not even looking at the constants) and hence could never be implemented.
Not sure how to define insane.
Three pointers:
Stack Exchange TCS:
My slides (also include factoring result):
here Question:
Can the SAT result be improved to be an algorithm that is ALWAYS right? Is there a way to show that it can't be (unless, say P=NP).
: What can be said about Graph Isomphism in this context? The proof for SAT is easily adpated to this case (all we used about SAT was that it was self-reducible). But can we get our GI algorithm to
never make a mistake?
14 comments:
1. Are you user2925716?
2. Huh! No!
I bet P!=NP and furthermore the proof will be trivial. :-)
3. I think that Q1 is equivalent to proving P=NP:
P=NP is equivalent to Exists P -> polytime(P) AND forall x . P(x) = SAT(x)
So what we are trying to prove is that for a particular concrete A :
Exists P -> polytime(P) AND forall x . P(x) = SAT(x) -> polytime(A) AND forall x . A(x) = SAT(x)
Which is equivalent to prove:
NOT [ polytime(A) AND forall x . A(x) = SAT(x) ] OR Exists P -> polytime(P) AND forall x . P(x) = SAT(x)
but we want [ polytime(A) AND forall x . A(x) = SAT(x) ] = True (Q1) so we must prove P=NP
1. Typo:
... Exists P -> polytime(P) .... should be
... Exists P . polytime(P) ... (. read as "such that")
2. Why do you say that "X -> Y" would be equivalent to "NOT Y OR X"?
3. My mistake! it's equivalent to NOT x OR y :-)
It's equivalent to prove
P <> NP OR [polytime (A) and A=SAT]
4. I don't understand the "(all of which are NOT in SAT)" part. Whenever the program fails, it claims that the given formula is not satisfiable. (Otherwise it provides a valid witness, and hence
doesn't fail.) I don't understand why you claim that such a formula would be NOT in SAT. Isn't it the other way around?
5. In factorization you can use the polynomial time prime algorithm to detect the bad answers and correct them searching exaustively for their factors (and if factoring is in P you need to do this
only on a finite number of "bad algorithms" that behaved well on smaller numbers). In a similar fashion
***under the assumption (much? stronger than P=NP) that Frege (or Extended Frege) systems are polynomially bounded proof systems***
then the following algorithm solves SAT correctly and runs in polynomial time:
Given a formula x, find the smallest i < log log |x| such that M_i outputs a satisfying assignment on all satisfiable formulas y of length |y| < log log |x| or a valid Frege unsatisfiability
proof on all unsatisfiable formulas (in both cases running at most for |y|^|M_i| steps). Then run M_i on x for at most |x|^|M_i| steps; check (in polynomial time) the satisfying assignment or the
correctness of then Frege unsatisfiability proof; if the assigment or the proof are not correct (or there is no M_i that satisfies the above condition) then check exhaustively (exponentially)
whether x is in SAT or not.
1. Erfan Khaniki12:56 PM, November 01, 2018
If I understand your assumption correctly, the poly boundedness of a proof system for tautologies only implies NP=CoNP (There is an oracle A such that P\neq NP, but NP=CoNP with respect to A,
hence it seems that the poly boundedness does not imply P=NP if we assume relativizable proofs.)
2. You're right; it works if we add also the P=NP assumption.
6. I don't understand the claim about the finite number of exceptions:
- Any algorithm deciding SAT could also be used to output a satisfying assignment when the instance is satisfiable. So it seems that the only way to fail would be to claim an instance is
unsatisfiable when in fact it is.
- If the algorithm has a finite number of instances where it is wrong, couldn't those be hard-coded into the algorithm? Or would that count as non-constructive?
1. > So it seems that the only way to fail would be to claim an instance is unsatisfiable when in fact it is
> ... Or would that count as non-constructive?
Yes. It would be constructive if you explicitly write those hard-coded instances (or, alternatively, you could give the explicit x_0 such that for all instances >= x_0 the algorithm solves it
correctly "without patches")
2. ... More on the first point:
- if the algorithm (that behaves well on small instances) outputs a satisfying assignment then you can check it (and "patch" it if is not valid)
- if it outputs "unsatisfiable" then you must "trust" it (otherwise you'll patch it infinitely many times)
So the explicit algorithm can output "unsatisfiable" on some satisfiable instances (finitely many if P=NP).
7. Another (trivial) assumption + algorithm is the following:
* Assume that if P=NP we can build a polynomial time algorithm for SAT which is provably correct (in PA or ZFC)
Let M1,M2,... be a standard TM enumeration;
P1,P2,... a standard PA proof enumeration;
. On input x (a boolean formula):
. for i = 1 to |x| do
... for j = 1 to i do
..... if Pi is a valid proof that Mj solves SAT in polynomial time
....... then simulate Mj(x) and accept/reject according to it
. If no proof is found solve SAT(x) using an exhaustive search | {"url":"https://blog.computationalcomplexity.org/2018/10/if-pnp-then-we-have-alg-for-sat.html?m=1","timestamp":"2024-11-07T15:18:59Z","content_type":"application/xhtml+xml","content_length":"83431","record_id":"<urn:uuid:b0b64948-6538-41ae-9919-9e50941a5aab>","cc-path":"CC-MAIN-2024-46/segments/1730477028000.52/warc/CC-MAIN-20241107150153-20241107180153-00140.warc.gz"} |
ontinuous functions on
How to find an unusual proof that continuous functions on the closed interval [0,1] are bounded.
The definitions of continuity (in terms of epsilons and deltas) and boundedness. The least upper bound axiom.
Sketches of the usual proofs.
One proof starts with the Heine-Borel theorem: however you write [0,1] as a union of a collection of open sets, you can also write it as the union of a finite subcollection. To deduce the statement,
for each x let U[x] be the set of all y such that |f(y)-f(x)| < 1. The sets U[x] are open and their union is obviously all of [0,1]. By the Heine-Borel theorem there are x[1],...x[n] such that the
union of the sets U[x[i]] is all of [0,1]. But then f is bounded above by 1+max f(x[i]).
Another proof starts with the Bolzano-Weierstrass theorem: every sequence in [0,1] has a convergent subsequence. Then, if the theorem is false, we can find an infinite sequence (x[n]) such that f(x
[n]) > n for every n. Pick a subsequence converging to x, and check that f cannot possibly be continuous at x.
Discovering a proof if you don't know about compactness.
Both the above proofs depend, in one way or another, on the statement that the closed interval [0,1] is compact . (Roughly speaking, a set is compact if it satisfies the conclusion of the Heine-Borel
theorem. For subsets of R^n one can use the Bolzano-Weierstrass theorem instead.) What if one had never met the idea of compactness in any form, even implicitly?
Let us imagine ourselves faced with a function on [0,1] about which we know nothing except that it is continuous. How can we pin it down? One very small observation is that f(x) is finite for every
x, since that is what we mean by a real-valued function defined on [0,1]. Can we say anything more? Well, we ought to use the definition of continuity. Let us remind ourselves of it. A function f is
continuous at x if
(for every e > 0) (there exists d > 0) such that |y-x| < d => |f(y)-f(x)| < e.
Since we know that f is continuous at x, we can pick any old e - for definiteness let us go for e=1 - and find d such that f(y) < f(x)+1 for every y in the interval (x-d,x+d).
This looks like a small amount of progress, since we have shown that f is bounded in the whole of an open interval. However, we have no control on the width of the interval, so we should not get too
Is there anything else we can do? Well, we might notice that if we have two overlapping intervals in each of which f is bounded, then f is bounded in their union. Is there some way of finding an
interval that overlaps with (x-d,x+d) on which f is bounded? The only way we know, so far, of constructing such an interval, is to pick a point y and choose c such that |f(z)-f(y)| < 1 for every z in
the open interval (y-c,y+c). How can we get such an interval to overlap with (x-d,x+d)? Well, if y itself belongs to (x-d,x+d) then we get some overlap, but there is the risk that (y-c,y+c) is
entirely contained within (x-d,x+d) in which case we have learnt nothing. On the other hand, if y does not belong to [x-d,x+d] then c might be so small that there is no overlap at all. This suggests
that we should try y=x+d or y=x-d. If we go for the first, then we find c > 0 such that f is bounded on the interval (x-d,x+d+c).
Now let us try to be a bit more systematic. We'll begin with x=0 and try to build up a larger and larger interval [0,t) on which f is bounded. With luck, we'll be able to get t all the way up to 1.
The first step is to find t[1] such that f(x) < 1 for every x in [0,t[1]) (using the definition of continuity again). Next, we apply the definition of continuity to the point t[1] and obtain, as
above, a t[2] > t[1] such that f(x) < 2 for every x in [0,t[2]). Then we keep going. In this way we produce a sequence t[1] < t[2] < t[3] < .... such that f(x) < n for every x in [0,t[n]).
Are we done? No, because it might be that the sequence (t[n]) converges to a number s < 1. Then all we would have done is show that the restriction of f to [0,t) is bounded for every t < s. That is
an interesting statement, but much weaker than what we are hoping to prove.
Are we completely stuck? Yes in the sense that we have gone on for ever and still not reached our target. But no in the sense that there may still be things to try. We have, in a certain sense
`reached' s. Can we get any further?
An obvious thing to try is the technique we have used up to now, namely applying the definition of continuity, with e=1, to some carefully chosen point. Is there any point we could go for now? Yes
indeed there is - the point s. This tells us that there exists d such that |f(x)-f(s)| < 1 for every x in the interval (s-d,s+d). Now this interval must contain some t[n], from which it is easy to
deduce that f(s) < n+1 and hence that f(x) < n+2 for every x < s+d. Thus, we have managed to continue the sequence (t[n]) `beyond infinity'.
To those who know about ordinals, the above observation immediately suggests an idea for a proof - just continue the above argument, obtaining a transfinite sequence. This idea works, as the
resulting sequence, if constructed properly, is a well-ordered set. If the theorem were false, it would be possible to construct an uncountable well-ordered subset of the reals, which is easy to
prove does not exist.
Somebody who does not know about ordinals will probably have thoughts of the following kind. I can go on and on constructing larger and larger intervals [0,t) on which f is bounded. Given any such
interval, I can always extend it, and given a countable sequence ([0,t[n])) of such intervals, I can always find another such interval containing all of them. This seems to be enough, but I can't
think how to describe what happens. I start with t[1] < t[2] < t[3] < ...., then find a new s bigger than all of them, which I could call s[1]. Then I could continue with s[1] < s[2] < s[3] < ....
and so on, but then I would construct an r bigger than all of those, and construct a new sequence from that. I could then do this whole process infinitely many times, but would still not necessarily
have reached 1. On the other hand I could still continue.
The problem seems to be a difficulty of describing a sequence that goes beyond infinity in this way. How can we get round this? (As I said earlier, actually there is a way of describing this
`transfinite' process rigorously and constructing an argument along these lines, but it is not necessary.) Let us think carefully about what we are doing. We are constructing a sequence, initially by
an inductive argument (once we have constructed t[n] we go ahead and construct t[n+1]), and later by something that resembles an inductive argument (once we have constructed all the t[n] we go ahead
and construct s[1]). The generalized induction is harder because we don't have a nice indexing set like the natural numbers. Is there a way of doing induction without the crutch of an indexing set?
Yes there is: you look for a minimal counterexample.
Can we make sense of the idea of a minimal counterexample for our problem? What we would be asking is the following. We have a process for gradually extending the interval [0,t) on which we know that
f is bounded. If we can get t all the way to 1 then we are done. If we can't, then there must be some sort of barrier - that is, a point beyond which we cannot go. Why not look for the first such
It is still not absolutely clear that we have an idea that makes sense. However, let us try it. Must there be a minimal t with the property that f is unbounded on [0,t)? Well, if f is unbounded, then
it is unbounded on [0,1) (by applying continuity at 1), so at least there exists t such that f is unbounded on [0,t). Then, by the least upper bound axiom (though here we find a greatest lower bound)
we choose the infimum s of all t with this property. If anything is going to be our minimal t, then s will.
We know that f is unbounded on [0,t) for every t > s. Does that imply that f is unbounded on [0,s)? Yes it does, because if f were bounded on [0,s) then by continuity at s (the usual argument) it
would be bounded on [0,s+c) for sufficiently small c.
Having found the minimal s such that f is unbounded on [0,s), we now hope for a contradiction. By now it is a reflex to apply continuity at s. Then f is bounded on (s-c,s+c) for some suitable c. It
is also bounded on [0,s-c/2). Hence, it is bounded on [0,s) and we have our contradiction.
A small variant of this argument.
One can also argue as follows. Let X be the set of all t such that f is bounded on the interval [0,t]. What can we say about X? Well, if s belongs to X and r < s then r certainly belongs to X. It
follows that X is an interval of the form [0,u) or [0,u]. In the first case, apply continuity at u to show that [0,u) is not the whole of X after all. In the second case, apply continuity at u to
show that [0,u] is not the whole of X unless u=1.
Finding a more conventional proof?
I find that the above proof is natural in the sense that the idea of extending the interval on which f is bounded is in a sense `the first thing to try'. However, it is not as neat (at least in its
first version) as the usual proofs via the Heine-Borel theorem or Bolzano-Weierstrass theorem. It is not impossible to imagine thinking of those theorems without going through the above argument
first. When we first applied the definition of continuity, we constructed, for an arbitrary x, an open interval U[x] containing x, on which f was bounded. We observed that we would like it if these
intervals overlapped. It perhaps doesn't take too much imagination to notice that if we can cover [0,1] with finitely many of these intervals, then we are done. As for the Bolzano-Weierstrass
theorem, it arises naturally if one tries to prove the result by contradiction. What does it mean for f to be unbounded? It means that for every n there is an x with f(x) > n. Then we try to do
something with that sequence. We can't get a contradiction just from an individual f(x) being large - they have to collect together somewhere.
Another way to arrive at the Heine-Borel theorem is to think about the extending-intervals argument and work out what we actually used. The main lemma that we used over and over again is that every x
is contained in an open interval U[x] on which f was bounded. We tried to put those intervals together, running into problems if we had infinitely many of them, which we then managed to solve by
getting back down to finitely many. It is clear that what we were really trying to do was cover [0,1] with finitely many of the U[x]. It is but a short step from here to thinking of the statement of
the Heine-Borel theorem. | {"url":"https://www.dpmms.cam.ac.uk/~wtg10/bounded.html","timestamp":"2024-11-03T13:13:28Z","content_type":"text/html","content_length":"11908","record_id":"<urn:uuid:aaa64cdc-5fef-4b1e-8c09-b8ca97d1936d>","cc-path":"CC-MAIN-2024-46/segments/1730477027776.9/warc/CC-MAIN-20241103114942-20241103144942-00538.warc.gz"} |
Parellelepiped, Tetrahedron Volume Calculator
Calculates the volumes of parallelepiped and tetrahedron for given vertices.
The tetrahedron is a regular pyramid.
Volume of a Parallelepiped :
Geometrically, the absolute value of the triple product represents the volume of the parallelepiped whose edges are the three vectors that meet in the same vertex.
Formula of volume is :
$$ V=(x4-x1) \times [{(y2-y1) \times (z3-z1)}-{(z2-z1) \times (y3-y1)}]; $$ $$+ (y4-y1) \times [{ (z2-z1) \times (x3-x1)}-{(x2-x1) \times (z3-z1)}]; $$ $$+ (z4-z1) \times [{ (x2-x1) \times (y3-y1)}-
{(y2-y1) \times (x3-x1) }]; $$
Volume of a Tetrahedron :
The volume of a tetrahedron is equal to 1/6 of the absolute value of the triple product. The tetrahedron has four faces which are equilateral triangles and has 6 edges in regular tetrahedron having
equal in length, the regular tetrahedron has four vertices and 3 faces meets at any one of vertex.
The volume of tetrahedron is :
$$ \text{Tetrahedron volume} = \frac{ \text{Parallelepiped volume (V)}} {6}$$ | {"url":"https://eguruchela.com/math/Calculator/parellelepiped-tetrahedron-volume","timestamp":"2024-11-06T07:12:45Z","content_type":"text/html","content_length":"14072","record_id":"<urn:uuid:3aabcb5f-2dc5-4e53-a0a2-2bec644eaa0a>","cc-path":"CC-MAIN-2024-46/segments/1730477027910.12/warc/CC-MAIN-20241106065928-20241106095928-00421.warc.gz"} |
Simplifying Algebraic Expressions: (5x^3y)^2(-2x^5y^1)
This article will guide you through the process of simplifying the algebraic expression (5x^3y)^2(-2x^5y^1).
Understanding the Rules
Before we begin, let's recall some essential rules of exponents:
• Power of a product: (ab)^n = a^n * b^n
• Power of a power: (a^m)^n = a^(m*n)
• Product of powers: a^m * a^n = a^(m+n)
Step-by-Step Simplification
1. Simplify the first term:
□ (5x^3y)^2 = 5^2 * (x^3)^2 * y^2 = 25x^6y^2
2. Simplify the second term:
□ (-2x^5y^1) remains as it is.
3. Multiply the simplified terms:
□ 25x^6y^2 * (-2x^5y^1) = -50x^(6+5)y^(2+1)
4. Combine the exponents:
□ -50x^(6+5)y^(2+1) = -50x^11y^3
Final Result
Therefore, the simplified form of the expression (5x^3y)^2(-2x^5y^1) is -50x^11y^3.
Key Points
• Remember the order of operations (PEMDAS/BODMAS) when dealing with expressions.
• Pay close attention to the signs of the coefficients.
• Always simplify expressions to their simplest form. | {"url":"https://jasonbradley.me/page/(5x%255E3y)%255E2(-2x%255E5y%255E1)","timestamp":"2024-11-03T03:09:14Z","content_type":"text/html","content_length":"59850","record_id":"<urn:uuid:b3f16f57-6d8d-4c7c-9cc5-80ba2bd52db7>","cc-path":"CC-MAIN-2024-46/segments/1730477027770.74/warc/CC-MAIN-20241103022018-20241103052018-00294.warc.gz"} |
How to Calculate NPV Discount Rate: A Comprehensive Guide - Gospel10
How to Calculate NPV Discount Rate: A Comprehensive Guide
Calculating net present value (NPV) is essential for evaluating the profitability of investments or projects. The NPV discount rate represents the minimum acceptable rate of return for a given
investment, and it plays a critical role in determining its viability.
NPV is a crucial concept in corporate finance and investment analysis. Its importance lies in its ability to quantify the present value of future cash flows from an investment, taking into account
the impact of inflation and the time value of money.
Determining the appropriate NPV discount rate is a complex process that requires a deep understanding of financial markets and the risk profile of the investment. Historically, economists have
developed various methods for calculating the NPV discount rate, such as the weighted average cost of capital (WACC).
How to Calculate NPV Discount Rate
In order to properly calculate the NPV discount rate, various key aspects must be considered.
• Cost of capital
• Risk
• Inflation
• Time value of money
• Project life
• Cash flows
• Weighted average cost of capital (WACC)
• Capital budgeting
• Net present value (NPV)
• Internal rate of return (IRR)
These aspects are closely interrelated and influence the accuracy and reliability of the NPV discount rate. For instance, the cost of capital is a crucial factor in determining the minimum acceptable
rate of return for an investment, while the project life and cash flows are essential for assessing the timing and magnitude of future returns. By carefully considering these aspects, businesses can
improve the precision of their NPV calculations and make more informed investment decisions.
Cost of capital
Cost of capital is a critical component in the calculation of the NPV discount rate. It represents the minimum rate of return that a company must earn on an investment in order to justify its cost.
This rate is used to discount future cash flows back to their present value, which is then used to determine the NPV of the investment.
There are two main types of cost of capital: the cost of debt and the cost of equity. The cost of debt is the interest rate that a company must pay on its borrowed funds. The cost of equity is the
return that investors expect to receive on their investment in the company’s stock. The WACC is a weighted average of the cost of debt and the cost of equity, and it is used to calculate the overall
cost of capital for a company.
The cost of capital is an important factor in the evaluation of investment opportunities. A higher cost of capital will result in a lower NPV, and vice versa. Therefore, it is important for companies
to carefully consider their cost of capital when making investment decisions.
Risk plays a significant role in the calculation of the NPV discount rate. The higher the risk associated with an investment, the higher the discount rate that should be used. This is because a
higher discount rate will result in a lower NPV, and vice versa.
There are a number of factors that can affect the risk of an investment, including:
The industry in which the company operates The size and track record of the company The financial health of the company The political and economic environment in which the company operates
Real-life examples of risk that can affect the NPV discount rate include:
A new product launch A new market entry A change in government regulations A natural disaster
Understanding the relationship between risk and the NPV discount rate is essential for making sound investment decisions. By carefully considering the risks involved in an investment, investors can
make more informed decisions about the appropriate discount rate to use.
Inflation is a critical factor to consider when calculating the NPV discount rate. It affects the value of money over time, which in turn affects the present value of future cash flows.
• Expected Inflation Rate
The expected inflation rate over the life of the investment should be considered. A higher expected inflation rate will result in a higher discount rate.
• Historical Inflation Rate
The historical inflation rate can be used to estimate the expected inflation rate. However, it is important to note that past inflation rates do not necessarily guarantee future inflation rates.
• Inflation Risk Premium
The inflation risk premium is a component of the discount rate that compensates investors for the risk of unexpected inflation. A higher inflation risk premium will result in a higher discount
Inflation is a complex and important factor to consider when calculating the NPV discount rate. By carefully considering the expected inflation rate, historical inflation rate, and inflation risk
premium, investors can make more informed decisions about the appropriate discount rate to use.
Time Value of Money
Time value of money (TVM) is a fundamental concept in finance that recognizes the fact that the value of money changes over time. This concept is particularly important in the context of calculating
the NPV discount rate, as it helps determine the present value of future cash flows.
• Present Value
Present value is the current worth of a future sum of money, discounted back to the present at a given interest rate. This is a key component of NPV calculations, as it allows investors to
compare the value of future cash flows to the initial investment.
• Future Value
Future value is the value of a current sum of money at a future date, assuming a given interest rate. This concept is important in NPV calculations, as it allows investors to project the future
value of their investment based on expected cash flows.
• Discount Rate
The discount rate is the interest rate used to discount future cash flows back to the present. The discount rate is a critical component of NPV calculations, as it directly affects the present
value of future cash flows.
• Compounding
Compounding is the process of adding interest to the principal of a sum of money over time. This concept is important in NPV calculations, as it allows investors to account for the effect of
interest on interest over the life of the investment.
Understanding the time value of money is essential for calculating the NPV discount rate. By carefully considering the present value, future value, discount rate, and compounding, investors can make
more informed decisions about the appropriate discount rate to use.
Project life
Project life is a crucial aspect to consider when calculating the NPV discount rate. It represents the duration of the project, from its inception to its completion, and directly affects the timing
and magnitude of future cash flows. Understanding the project life is essential for making accurate NPV calculations and informed investment decisions.
• Start and end dates
The start and end dates of the project define its overall life. These dates are important for determining the time frame over which cash flows will be generated and discounted back to the
• Project phases
Projects are often divided into distinct phases, such as planning, execution, and completion. Each phase may have its own unique cash flow patterns, which should be considered when calculating
the NPV discount rate.
• Contingency period
A contingency period is often added to the project life to account for unexpected events or delays. This period provides a buffer in the NPV calculation and reduces the risk of underestimating
the project’s life.
• Investment horizon
The investment horizon is the period over which the investor plans to hold the investment. This horizon may be shorter or longer than the project life, and it affects the number of cash flows
that are considered in the NPV calculation.
In conclusion, project life is a multifaceted aspect that significantly influences the calculation of the NPV discount rate. By carefully considering the start and end dates, project phases,
contingency period, and investment horizon, investors can improve the accuracy and reliability of their NPV calculations and make more informed investment decisions.
Cash flows
Cash flows are an essential component of capital budgeting and investment analysis, and they play a critical role in the calculation of the NPV discount rate. NPV, or net present value, is a method
of evaluating the profitability of a project or investment by calculating the present value of its future cash flows. The discount rate used in this calculation is the rate that equates the present
value of the future cash flows to the initial investment.
Therefore, the accuracy of the NPV discount rate is heavily reliant on the accuracy of the projected cash flows. If the cash flows are underestimated, the NPV will be overstated, and vice versa. As a
result, it is crucial to carefully estimate and forecast the future cash flows of a project or investment when calculating the NPV discount rate. These cash flows should include all sources of income
and expenses, both operating and non-operating.
Real-life examples of cash flows that are considered when calculating the NPV discount rate include revenue from sales, operating expenses, capital expenditures, and salvage value. Additionally, the
timing of these cash flows is also important, as the NPV discount rate takes into account the time value of money. By understanding the relationship between cash flows and the NPV discount rate,
investors and financial analysts can make more informed decisions about the viability and profitability of potential investments.
Weighted average cost of capital (WACC)
The weighted average cost of capital (WACC) is a crucial element in the calculation of the NPV discount rate. It represents the average cost of capital for a company, taking into account the cost of
debt and equity financing.
• Cost of Debt
This refers to the interest rate that a company pays on its borrowed funds. It is typically calculated as the yield to maturity on the company’s outstanding debt.
• Cost of Equity
This represents the return that investors expect to receive on their investment in the company’s stock. It is often estimated using the capital asset pricing model (CAPM).
• Debt-to-Equity Ratio
This ratio measures the proportion of debt and equity financing used by the company. It is used to weight the cost of debt and equity in the WACC calculation.
• Tax Rate
The tax rate is applied to the cost of debt to reflect the tax savings associated with interest payments.
By considering the WACC, the NPV discount rate reflects the overall cost of capital for the company and provides a more accurate assessment of the profitability of an investment. A higher WACC will
result in a higher NPV discount rate, and vice versa.
Capital budgeting
Capital budgeting plays a pivotal role in the calculation of NPV discount rate. It involves a comprehensive evaluation of potential investment projects to determine their financial viability and
• Project Evaluation
Capital budgeting encompasses the assessment of various aspects of a project, such as its expected cash flows, risk profile, and strategic alignment. This evaluation helps determine whether the
project meets the organization’s financial objectives and is aligned with its long-term goals.
• Investment Decision
Based on the project evaluation, capital budgeting assists in making informed decisions about whether to proceed with the investment. It involves comparing the NPV of the project to the required
rate of return to determine its profitability.
• Resource Allocation
Capital budgeting guides the allocation of scarce financial resources to competing projects. By prioritizing projects based on their NPV and alignment with strategic objectives, organizations can
optimize their capital investments and maximize returns.
• Risk Management
Capital budgeting considers the potential risks associated with an investment project. By assessing the sensitivity of the NPV to changes in key assumptions, organizations can identify and
mitigate potential risks and enhance the overall financial health of the organization.
In conclusion, capital budgeting provides a structured framework for evaluating investment projects, making informed investment decisions, allocating resources efficiently, and managing risks. By
considering capital budgeting in the calculation of NPV discount rate, organizations can increase their chances of making sound investment decisions that align with their financial goals and drive
long-term value creation.
Net present value (NPV)
Net present value (NPV) plays a crucial role in capital budgeting and investment analysis, as it provides a quantitative measure of the profitability of an investment or project. NPV is calculated by
discounting future cash flows back to the present at a specified discount rate, which is commonly referred to as the NPV discount rate. This discount rate is a critical component in NPV calculations,
as it directly affects the present value of future cash flows and, consequently, the overall NPV.
The NPV discount rate is typically determined based on the weighted average cost of capital (WACC), which represents the average cost of capital for a company. The WACC considers both the cost of
debt and equity financing, weighted by their respective proportions in the capital structure. By using the WACC as the discount rate, NPV calculations reflect the opportunity cost of capital and
provide a more accurate assessment of the profitability of an investment.
Real-life examples of NPV calculations include project evaluation, capital budgeting decisions, and investment analysis. In project evaluation, NPV is used to determine the viability of a project by
comparing its present value to the initial investment. In capital budgeting, NPV is employed to prioritize projects based on their profitability and align investment decisions with the organization’s
financial objectives. NPV is also widely used in investment analysis to assess the attractiveness of investment opportunities and make informed decisions about portfolio allocation.
Internal rate of return (IRR)
Internal rate of return (IRR) is a critical component of capital budgeting and investment analysis, particularly in the context of calculating the NPV discount rate. IRR represents the annualized
rate of return that an investment is expected to generate, and it is closely related to the NPV discount rate used in NPV calculations.
The relationship between IRR and the NPV discount rate is reciprocal. The NPV discount rate is used to calculate the IRR, and the IRR can be used to determine the appropriate NPV discount rate to use
in a given situation. In general, a project with a higher IRR will have a lower NPV discount rate, and vice versa. This is because a higher IRR indicates a more attractive investment opportunity,
which would justify a lower discount rate.
In real-life applications, the connection between IRR and the NPV discount rate is crucial for making informed investment decisions. For instance, if a project has an IRR of 10%, then using an NPV
discount rate of 8% would result in a positive NPV, indicating that the project is financially viable. Conversely, using an NPV discount rate of 12% would result in a negative NPV, suggesting that
the project is not profitable.
Understanding the relationship between IRR and the NPV discount rate allows financial analysts and investors to make more informed decisions about investment opportunities. By considering both the
IRR and the NPV discount rate, they can better assess the profitability and risk of an investment and make more confident decisions about whether to proceed with the project.
FAQs on NPV Discount Rate Calculation
This section addresses common questions and misconceptions related to calculating the NPV discount rate, providing clear and concise answers to guide readers in their understanding.
Question 1: What is the NPV discount rate?
The NPV discount rate is the minimum acceptable rate of return for an investment, used to discount future cash flows back to the present value for NPV calculations.
Question 2: How is the NPV discount rate determined?
The NPV discount rate is typically determined based on the weighted average cost of capital (WACC), which represents the average cost of capital for a company.
Question 3: What factors influence the NPV discount rate?
Several factors influence the NPV discount rate, including cost of capital, risk, inflation, time value of money, and project life.
Question 4: What is the relationship between the NPV discount rate and the IRR?
The NPV discount rate and IRR are inversely related. A higher NPV discount rate results in a lower IRR, and vice versa.
Question 5: What are some real-life examples of the NPV discount rate?
Examples include evaluating the profitability of capital projects, making investment decisions, and assessing the attractiveness of investment opportunities.
Question 6: Why is it important to consider the NPV discount rate when making investment decisions?
The NPV discount rate helps determine the profitability and viability of an investment by considering the time value of money and the opportunity cost of capital.
These FAQs provide a foundation for understanding the calculation and application of the NPV discount rate. In the following sections, we will delve deeper into these concepts and explore advanced
techniques for NPV analysis.
Tips for Calculating NPV Discount Rate
This section provides practical tips to assist you in accurately calculating the NPV discount rate, a critical component of capital budgeting and investment analysis.
Tip 1: Determine the Appropriate Cost of Capital
Identify the relevant cost of debt and equity for your project, considering factors such as industry norms, company size, and risk profile.
Tip 2: Assess the Risk Level
Evaluate the inherent risk associated with the project, considering factors such as market volatility, technological advancements, and regulatory changes.
Tip 3: Estimate Inflation Accurately
Research and incorporate reliable inflation projections into your calculations to account for the impact of inflation on future cash flows.
Tip 4: Consider the Time Value of Money
Recognize that money has a time value and that future cash flows should be discounted to reflect their present value.
Tip 5: Define the Project Life Appropriately
Establish a realistic project life that aligns with the expected duration of cash flows and considers potential extension or termination scenarios.
Tip 6: Incorporate Weighted Average Cost of Capital (WACC)
Calculate the WACC to determine the overall cost of capital for your project, considering the relative proportions and costs of debt and equity financing.
Tip 7: Perform Sensitivity Analysis
Conduct sensitivity analysis to assess the impact of variations in key assumptions, such as discount rate and cash flow projections, on the NPV.
Tip 8: Seek Professional Advice When Needed
If necessary, consult with financial professionals or experts to ensure accurate and reliable NPV discount rate calculations.
By following these tips, you can improve the accuracy and reliability of your NPV discount rate calculations, leading to more informed investment decisions. In the next section, we will explore
advanced techniques for NPV analysis to enhance your understanding and capabilities.
In summary, calculating the NPV discount rate is a critical step in capital budgeting and investment analysis. This article has explored the key concepts and considerations involved in determining
the NPV discount rate, providing valuable insights for making informed investment decisions.
Three main points to remember are:
• The NPV discount rate is influenced by various factors, including the cost of capital, risk, inflation, and project life.
• The weighted average cost of capital (WACC) is a widely used method for determining the NPV discount rate, considering both debt and equity financing costs.
• Sensitivity analysis is essential to assess the impact of changes in assumptions on the NPV, ensuring robust investment decisions.
Understanding and accurately calculating the NPV discount rate is crucial for maximizing returns and minimizing risks in investment projects. While the process can be complex, the insights gained
from this article will empower you to navigate the challenges and make optimal investment choices.
Leave a Comment | {"url":"https://www.gospel10.com/how-to-calculate-npv-discount-rate-a-comprehensive-guide/","timestamp":"2024-11-04T04:36:09Z","content_type":"text/html","content_length":"208135","record_id":"<urn:uuid:dcd6e037-8ca3-4b8a-9cb7-4140b40da649>","cc-path":"CC-MAIN-2024-46/segments/1730477027812.67/warc/CC-MAIN-20241104034319-20241104064319-00072.warc.gz"} |
How to generate a star pattern using Informatica
Question:- To generate a star pattern using Informatica depending on the number of rows in the input file
Suppose there are 7 rows in the source file (any type of data), the target should have 7 rows of stars in the following pattern:-
There are two ways of doing this – first is to use a Java transformation in passive mode and second is without using a java transformation.
Doing it WITHOUT a Java transformation is challenging hence I will describe this method in this post. Java transformation based solution will be posted later.
Solution with Java Transformation
We can observe the following about the pattern we are generating:-
1. First line has 7 spaces before the star, second has 6 and so on. Hence the spaces in front are decreasing by 1 for each row. The number of spaces in first row is equal to the number of rows in the
source file and number of spaces in front for the last row is zero
2. The number of stars are increasing by two after the first row. 1,3,5 and so on
As we are doing this without a java transformation, the trick lies in how we generate variable number of stars and spaces for each row.
1. First generate only the stars as in below pattern
2. Sort the above pattern in descending order to generate the pattern below
Fig 3
3. Add spaces to the pattern, starting with zero spaces in first line and increasing it by one for each subsequent row to get the following pattern:-
Fig 4
4. Sort again, this time in ascending order to get the final pattern
Fig 5
The mapping will also include one pipeline to get the count of the rows in the file. This will be used in the main flow to generate the star pattern.
Complete mapping will look like below (click on the image to enlarge):-
Mapping Variable:- $$ROWCOUNT
Target Load Plan:- SQ_FF_TEST_SRC first (to count the number of rows)
Mapping Explanation:-
First pipeline to count number of rows in the file
1. Use an aggregator to count the number of rows in the file like below (agg_count_rows)
2. Use an expression transformation to assign the count value to a mapping variable (exp_set_row_count_variable)
3. Connect both ports from the above expression to a dummy target file.
Second Pipeline to produce star pattern
1. Expression to get the row count variable value in the pipeline and also generate a sequence for each row starting from one (exp_seq_row_count)
2. Expression to generate intermediate start pattern as in Fig 2 above (exp_int_star_pattern)
3. Sorter to sort the intermediate pattern in descending order (sort_desc)
4. Expression to add spaces to the sorted pattern (ex_add_spaces_to_pattern)
5. Sorter to again sort the pattern in ascending order to generate the final star pattern (srt_final_pattern)
Place a source file with any number of rows, result you will get is lot of stars!!
Let me know in comments if you need any clarifications or if you have any feedback.
Join the Conversation
1. Expression transformation for Add Spaces function is adding a extra space in front of each row in the output.
1. Got it. Will fix it. Thanks for your feedback
2. Expression Transformation to Add spaces is adding extra spaces in front of each row in the output. | {"url":"http://www.raghavatal.com/2015/08/04/how-to-generate-a-star-pattern-using-informatica/","timestamp":"2024-11-14T11:33:49Z","content_type":"text/html","content_length":"66131","record_id":"<urn:uuid:7cc21144-00b5-40ed-9ec0-af72e9560ad7>","cc-path":"CC-MAIN-2024-46/segments/1730477028558.0/warc/CC-MAIN-20241114094851-20241114124851-00278.warc.gz"} |
Klogic company aptitude questions
klogic company aptitude questions Related topics: least common multiple printable
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Author Message Author Message
vom63 Posted: Wednesday 03rd of Jan 11:09 ruvbelsaun Posted: Sunday 07th of Jan 12:14
Hi guys , It’s been a week now and I still can’t I'm so happy I got these answers so fast, I can't wait to
figure out how to solve a few math problems on klogic try Algebrator. Can you tell me one more thing, where
company aptitude questions . I have to finish this work could I find this program? I'm not so good at searching
Reg.: 10.10.2006 before the beginning of next week. Can someone help Reg.: 02.04.2003 for things like this, so it would be good if you could give
me to get started? I need some help with multiplying me a link . Thanks a lot!
fractions and gcf. Any sort of guidance will be
IlbendF Posted: Friday 05th of Jan 09:47 TihBoasten Posted: Sunday 07th of Jan 19:41
Hi friend , I was in your situation a couple of weeks ago It’s right here: https://softmath.com/news.html. Buy it
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Reg.: 11.03.2004 was very useful since it offered all the basics that I Reg.: 14.10.2002 guarantee. Try using it and good luck with your
needed to work out my homework problem in Algebra assignment.
1. Just have a look at it and let me know if you require
further details on Algebrator so that I can render
assistance on Algebrator based on the knowledge that
I have now . Bet Posted: Monday 08th of Jan 15:14
alhatec16 Posted: Saturday 06th of Jan 08:19 I remember having problems with linear equations,
conversion of units and simplifying expressions.
When I was in school , I faced similar problems with Algebrator is a truly great piece of math software. I
side-side-side similarity, geometry and quadratic formula. Reg.: 13.10.2001 have used it through several algebra classes - Algebra
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Reg.: 10.03.2002 Intermediate algebra, College Algebra, and Algebra 2. I the problem and by clicking on Solve, step by step
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the Algebrator. | {"url":"https://www.softmath.com/parabola-in-math/point-slope/klogic-company-aptitude.html","timestamp":"2024-11-12T06:27:49Z","content_type":"text/html","content_length":"50250","record_id":"<urn:uuid:42e8f33e-3ea7-4af8-b190-a9e730ffe9b3>","cc-path":"CC-MAIN-2024-46/segments/1730477028242.58/warc/CC-MAIN-20241112045844-20241112075844-00649.warc.gz"} |
Berger 210 300WM load query
Help Support Long Range Hunting Forum
I have a Ruger 77 MKII with Kreiger 24" 1:10 twist shooting .43" groups at 100 yards using 200 gr. SGK 0.010 off the ogive, and 78.5 gr. of H1000. My groups with the Barnes TTSX 185 gr. are 1/2"
higher (100 yards on both) but similar grouping but with 0.040 off the ogive.
I have a set of 100 Berger 210 gr. hunting rounds ordered and and would appreciate what Bergers like off the ogive and powder. I have H1000, H4831SC, and Retumbo. Any inputs would be greatly
appreciated and thanks in advance.
Last edited:
Feb 3, 2007
I assume we are talking Win Mag not Weatherby Mag.... Our 300 Win shoots very well with a 27" Broughton 10 twist 210 Bergers .005" off the lands. I started .010" in and worked out. When I got to
.005" out the ES and grouping was good enough I called it the load. We use 77 gr of H-1000 and a 215 Fed GM in Lapua brass. Velocity in the 2900 range. Took a small doe antelope last week at 813
yards with it, right where I was aiming.
Thanks Jeff, I'll use that data and yes it is a Win Mag. However all I have at this time is the CCI Mag. primers. I appreciate the advice on a good starting point to find the sweet spot in my gun!
Feb 3, 2007
You are welcome. I hope it shoot as well for you, or better. Not saying I can do this every time,, but I did it once..
That is SWEET! My current range is limited to 300 yards, but I have found another within 1/2 hour that goes up to 1000 yards for a humbling experience....
Thanks again.
Just got an update from Midway and my Berger 210 hunters have shipped. I guess that means I will be doing some test firing next Sat. If it looks good I'll post some holie paper!
Last edited:
Oct 4, 2008
I'll echo what Broz said to a degree.
Same distance to lands, 75.7 gr H-1000, CCI250 and Winchester brass match prepped. Right at 2900 fps also out of a 26" McGowan barrel and consistently shoots about 0.3 MOA sometimes better. ES is
about 10-12.
More good info! Cases are all prepd. except for neck turning (flash hole, primer pocket, fire formed, length trimmed) tool for turning soon. I have the CED M2 & RSI software, Pressure Trace II is on
my wish list, data nerd...
Not sure where I will start on the powder as I was getting near 2900 fps with 78.5 gr. H1000 with the SGK 200 gr. bullet in my 24" Kreiger
I got them in last night and they look very supersonic in design. As my barrel is shorter than both of yours 24 vs 27 & 26" I think I will use 76-77.5 gr. of H1000 to find the sweet spot in four
groups of 3. I will keep the set back at 0.005 as both of you like and just vary the powder initially. Saturday will be fun I will use my CED M2 to look for the fps average in each group and
hopefully will post a pretty grouping at 100 yards for the set up. Thanks again for your support and insight in this round development, you saved me some $$
Wednesday night I got my loads done and have charged the battery for the IR screens on my CED M2. I can assure you that I will be at the range before 7:30 doing set up and will start shooting at 0800
per club rules. I still have to do the round data entry in the RSI Ballistic software so I can enter my target data and chrono. data for a file to keep. Hmmm two whole days DARN! Maybe if the ISO
auditors finish with me I can slip out early and......
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Number | Purposeful Maths
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Thoughtful Mathematics Teaching Resources Designed to Improve Understanding and Outcomes
To find resources more quickly, use the buttons below to select to the sub-topic you want to look at.
Place Value
Types of Number
Laws of Indices
Prime Factor Decomposition
Standard Form
Negative Numbers
Adding and Subtracting Surds
I Do, We Do, You Do Example Sheet and Worksheet that covers Adding and Subtracting Surds. These initially start with a pictorial representation and increase in difficulty to those requiring
sophisticated simplification.
The worksheet also includes difficult problems involving geometry.
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logKDE: log-transformed kernel density estimation
The goal of logKDE is to provide a set of functions for kernel density estimation on the positive domain, using log-kernel density functions, for the R programming environment. The main functions of
the package are the logdensity and logdensity_fft functions. The choice of functional syntax was made to resemble those of the density function, for conducting kernel density estimation on the real
domain. The logdensity function conducts density estimation, via first principle computations, whereas logdensity_fft utilizes fast-Fourier transformation in order to speed up computation. The use of
Rcpp guarantees that both methods are sufficiently fast for large data scenarios.
Currently, a variety of kernel functions and plugin bandwidth methods are available. By default both logdensity and logdensity_fft are set to use log-normal kernel functions (kernel = 'gaussian') and
Silverman’s rule-of-thumb bandwidth, applied to log-transformed data (bw = 'nrd0'). However, the following kernels are also available:
• log-Epanechnikov (kernel = 'epanechnikov'),
• log-Laplace (kernel = 'laplace'),
• log-logistic (kernel = 'logistic'),
• log-triangular (kernel = 'triangular'),
• log-uniform (kernel = 'uniform').
The following plugin bandwidth methods are also available:
• all of the methods that available for density, applied to log-transformed data (see ?bw.nrd regarding the options),
• unbiased cross-validated bandwidths in the positive domain (bw = 'logcv'),
• a Silverman-type rule-of-thumb that optimizes the kernel density estimator fit, compared to a log-normal density function (bw = 'logg').
The logdensity and logdensity_fft functions also behave in the same way as density, when called within the plot function. The usual assortment of commands that apply to plot output objects can also
be called.
For a comprehensive review of the literature on positive-domain kernel density estimation, thorough descriptions of the mathematics relating to the methods that have been described, simulation
results, and example applications of the logKDE package, please consult the package vignette. The vignette is available via the command vignette('logKDE'), once the package is installed.
If devtools has already been installed, then the most current build of logKDE can be obtained via the command:
The latest stable build of logKDE can be obtain from CRAN via the command:
An archival build of logKDE is available at https://zenodo.org/record/1317784. Manual installation instructions can be found within the R installation and administration manual https://
Example 1
In this example, we demonstrate that logdensity has nearly identical syntax to density. We also show that the format of the outputs are also nearly identical.
## Load 'logKDE' library.
## Set a random seed.
## Generate strictly positive data.
## Data are generated from a chi-squared distribution with 12 degrees of freedom.
x <- rchisq(100,6)
## Construct and print the output of the function 'density'.
#> Call:
#> density.default(x = x)
#> Data: x (100 obs.); Bandwidth 'bw' = 1.018
#> x y
#> Min. :-2.366 Min. :0.0000475
#> 1st Qu.: 2.547 1st Qu.:0.0072263
#> Median : 7.459 Median :0.0331904
#> Mean : 7.459 Mean :0.0508396
#> 3rd Qu.:12.372 3rd Qu.:0.1013289
#> Max. :17.284 Max. :0.1312107
## Construct and print the output of the function 'logdensity'.
#> Call:
#> logdensity(x = x)
#> Data: x (100 obs.); Bandwidth 'bw' = 0.1923
#> x y
#> Min. : 0.1111 Min. :0.00000
#> 1st Qu.: 3.7851 1st Qu.:0.02313
#> Median : 7.4592 Median :0.06527
#> Mean : 7.4592 Mean :0.06707
#> 3rd Qu.:11.1333 3rd Qu.:0.11219
#> Max. :14.8073 Max. :0.13698
## Plot the 'density' output object.
As a note, one can observe that density assigns positive probability to negative values. Since we know that the chi-squared generative model generates only positive values, this is an undesirable
result. The log-transformed kernel density estimator that is produced by logdensity only assigns positive probability to positive values, and is thus bona fide in this estimation scenario.
Example 2
In this example, we showcase the variety of kernel functions that are available in the package. Here, log-transformed kernel density estimators are constructed using the logdensity function.
## Load 'logKDE' library.
## Set a random seed.
## Generate strictly positive data.
## Data are generated from a chi-squared distribution with 12 degrees of freedom.
x <- rchisq(100,12)
## Construct a log-KDE using the data, and using each of the available kernel functions.
logKDE1 <- logdensity(x,kernel = 'gaussian',from = 1e-6,to = 30)
logKDE2 <- logdensity(x,kernel = 'epanechnikov',from = 1e-6,to = 30)
logKDE3 <- logdensity(x,kernel = 'laplace',from = 1e-6,to = 30)
logKDE4 <- logdensity(x,kernel = 'logistic',from = 1e-6,to = 30)
logKDE5 <- logdensity(x,kernel = 'triangular',from = 1e-6,to = 30)
logKDE6 <- logdensity(x,kernel = 'uniform',from = 1e-6,to = 30)
## Plot the true probability density function of the generative model.
plot(c(0,30),c(0,0.1),type='n',xlab='x',ylab='Density',main='Example 2')
curve(dchisq(x,12),from = 0,to = 30,add = T)
## Plot each of the log-KDE functions, each in a different rainbow() colour.
lines(logKDE1$x,logKDE1$y,col = rainbow(7)[1])
lines(logKDE2$x,logKDE2$y,col = rainbow(7)[2])
lines(logKDE3$x,logKDE3$y,col = rainbow(7)[3])
lines(logKDE4$x,logKDE4$y,col = rainbow(7)[4])
lines(logKDE5$x,logKDE5$y,col = rainbow(7)[5])
lines(logKDE6$x,logKDE6$y,col = rainbow(7)[6])
## Add a grid for a visual guide.
Example 3
In this example, we show that logdensity and logdensity_ftt yield nearly identical results. Here, log-transformed kernel density estimators are constructed using the logdensity_ftt function.
## Load 'logKDE' library.
## Set a random seed.
## Generate strictly positive data.
## Data are generated from a chi-squared distribution with 12 degrees of freedom.
x <- rchisq(100,12)
## Construct a log-KDE using the data, and using each of the available kernel functions.
logKDE1 <- logdensity_fft(x,kernel = 'gaussian',from = 1e-6,to = 30)
logKDE2 <- logdensity_fft(x,kernel = 'epanechnikov',from = 1e-6,to = 30)
logKDE3 <- logdensity_fft(x,kernel = 'laplace',from = 1e-6,to = 30)
logKDE4 <- logdensity_fft(x,kernel = 'logistic',from = 1e-6,to = 30)
logKDE5 <- logdensity_fft(x,kernel = 'triangular',from = 1e-6,to = 30)
logKDE6 <- logdensity_fft(x,kernel = 'uniform',from = 1e-6,to = 30)
## Plot the true probability density function of the generative model.
plot(c(0,30),c(0,0.1),type='n',xlab='x',ylab='Density',main='Example 3')
curve(dchisq(x,12),from = 0,to = 30,add = T)
## Plot each of the log-KDE functions, each in a different rainbow() colour.
lines(logKDE1$x,logKDE1$y,col = rainbow(7)[1])
lines(logKDE2$x,logKDE2$y,col = rainbow(7)[2])
lines(logKDE3$x,logKDE3$y,col = rainbow(7)[3])
lines(logKDE4$x,logKDE4$y,col = rainbow(7)[4])
lines(logKDE5$x,logKDE5$y,col = rainbow(7)[5])
lines(logKDE6$x,logKDE6$y,col = rainbow(7)[6])
## Add a grid for a visual guide.
We observe that the logdensity_fft outputs are noticiably smoother than those of logdensity. This is because fast Fourier transformations (FFT) only yield kernel density estimates at discrete points,
and the regions between these discrete points are approximated via a linear approximator, namely using the approx function. This is the same evaluation technique as that which is used in the function
density. Additionally the FFT approximation points are evenly space on the real line, whereas those used for logdensity are evenly spaced on a log scale.
Example 4
In this example, we showcase the variety of plugin bandwidth estimators that are available in the package. Here, log-transformed kernel density estimators are constructed using the logdensity
## Load 'logKDE' library.
## Set a random seed.
## Generate strictly positive data.
## Data are generated from a chi-squared distribution with 12 degrees of freedom.
x <- rchisq(100,12)
## Construct a log-KDE using the data, and using each of the available kernel functions.
logKDE1 <- logdensity(x,bw = 'nrd0',from = 1e-6,to = 30)
logKDE2 <- logdensity(x,bw = 'logcv',from = 1e-6,to = 30)
logKDE3 <- logdensity(x,bw = 'logg',from = 1e-6,to = 30)
logKDE4 <- logdensity(x,bw = 'nrd',from = 1e-6,to = 30)
logKDE5 <- logdensity(x,bw = 'ucv',from = 1e-6,to = 30)
#> Warning in stats::bw.ucv(log(x)): minimum occurred at one end of the range
logKDE6 <- logdensity(x,bw = 'bcv',from = 1e-6,to = 30)
#> Warning in stats::bw.bcv(log(x)): minimum occurred at one end of the range
logKDE7 <- logdensity(x,bw = 'SJ-ste',from = 1e-6,to = 30)
logKDE8 <- logdensity(x,bw = 'SJ-dpi',from = 1e-6,to = 30)
## Plot the true probability density function of the generative model.
plot(c(0,30),c(0,0.1),type='n',xlab='x',ylab='Density',main='Example 4')
curve(dchisq(x,12),from = 0,to = 30,add = T)
## Plot each of the log-KDE functions with different choices of bandwidth, each in a different rainbow() colour.
lines(logKDE1$x,logKDE1$y,col = rainbow(9)[1])
lines(logKDE2$x,logKDE2$y,col = rainbow(9)[2])
lines(logKDE3$x,logKDE3$y,col = rainbow(9)[3])
lines(logKDE4$x,logKDE4$y,col = rainbow(9)[4])
lines(logKDE5$x,logKDE5$y,col = rainbow(9)[5])
lines(logKDE6$x,logKDE6$y,col = rainbow(9)[6])
lines(logKDE7$x,logKDE7$y,col = rainbow(9)[7])
lines(logKDE8$x,logKDE8$y,col = rainbow(9)[8])
## Add a grid for a visual guide.
Unit testing
Using the package testthat, we have conducted the following unit test for the GitHub build, on the date: 06 August, 2018. The testing files are contained in the tests folder of the respository.
## Load 'logKDE' library.
## Load 'testthat' library.
## Test 'logKDE'.
#> ══ testthat results ════════════════════════════════════════════════════════════════════════════════════════════════════
#> OK: 74 SKIPPED: 0 FAILED: 0
Bug reporting and contributions
Thank you for your interest in logKDE. If you happen to find any bugs in the program, then please report them on the Issues page (https://github.com/andrewthomasjones/logKDE/issues). Support can also
be sought on this page. Furthermore, if you would like to make a contribution to the software, then please forward a pull request to the owner of the repository. | {"url":"https://cran.uvigo.es/web/packages/logKDE/readme/README.html","timestamp":"2024-11-06T23:58:30Z","content_type":"application/xhtml+xml","content_length":"44022","record_id":"<urn:uuid:0131df5e-ab72-4b2d-888c-63c10a7052dc>","cc-path":"CC-MAIN-2024-46/segments/1730477027942.54/warc/CC-MAIN-20241106230027-20241107020027-00675.warc.gz"} |
Immediate Relationship Or Indirect Relationship?
A direct marriage can be defined as a relationship in which both elements increase or decrease in parallel with one another. For instance , an example of an immediate relationship would be the
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Using a chart to create a immediate relationship, or perhaps linear romance, between virtually any two variables X and Y can be done. By insert inside the values per of the x’s and y’s in the chart
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You can use a statistical expression to get the direct and inverse relationship. In this case, the definition of ‘x’ presents the earliest variable, whilst ‘y’ is the second variable. Using the
formula, we are able to plug in the values for the x’s and y’s in to the cells representing the initial variable, in order to find that the immediate relationship prevails. However , the inverse
relationship exists whenever we reverse the order.
The graphs can also represent fashionable of one varied going up when one changing goes down. It is easier to attract a trendline by using the spreadsheet instead of a graph because all the
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In some situations where there is more than one sign for a given sign, such as signals on the x-axis, you can plan the outcomes of the diverse indicators on a single graph, or maybe more (or more)
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How to calculate Price to retailer (PTR) and Price to Stockist (PTS) from any given Maximum Retail Price (MRP)? (2024)
You have fixed Maximum Retail price for your product. You have also fixed profit margin, you want to give to retailers, stockists and other distribution channel business partners. But you don’t have
knowledge how you can calculate at what price you will bill to stockist and stockist will be to retailers. Then this article is going to be very useful for you...
In this article, we will calculate price to stockist (PTS) and Price to Retailers (PTR) with any given MRP and margin percentage. Also we will provide a PTR/PTS calculator along with how you can make
your own PTR/PTS calculator in excel format and we will also provide link to download PTR/PTS calculator in excel.
Check related article: profit margin in pharmaceutical industry (Manufacturer to retailers)
Have a look at important definitions related to this topic:
Distribution Channel: Distribution channel is a group of businesses involved in process of delivery a product/service to its end user from manufacturer. Check distribution channel set-up detail here
Maximum Retail Price (MRP): Maximum retail price (MRP) is a maximum value that can be paid by an end user for purchasing a product or using a service.
Price to Retailer (PTR): Price to retailer is price at which a stockist/manufacturer sells goods/services to retailers. It is the price which is fixed by deducting Tax and profit margin of retail
from Maximum retail price.
Price to Stockist (PTS): Price to stockist is the price at which a manufacturer/CnF sells goods/services to stockist for distribution to retailers. PTS is generally calculated by deducting stockist
margin from PTR.
Now can to the point How to calculate PTR and PTS?
Here we are discussing most common method for calculating PTR/PTS. Values you need before calculating PTR and PTS.
• Maximum Retail Price
• GST value (5%, 12%, 18%, 28%)
• Retailer Margin
• Stockist Margin
• MRP = 100/-
• GST value = 12%
• Retailer Margin = 20%
• Stockist Margin = 10%
First you need to deduct retailer margin from MRP. This will give a net value that in other words we can say PTR including GST like below:
Formula used will be MRP*(1-Retailer Margin %)
Net value = 100*(1-20%) = 80/-
PTR Calculation:
Now calculate Price to retailer. Price to retailer (PTR) will be calculated by dividing GST value from above net value. For that you need to first calculate GST factor value. Formula for GST factor
calculation is finding below:
GST Factor = (100+GST)/100
GST Factor value = (100+12)/100 = 112/100 = 1.12
PTR formula = Net Value / GST Factor = 80/1.12 = 71.43/-
PTS Calculation:
Price to stockist will be calculated by deducting stockist margin percentage from retailer price. Formula for calculating price to stockist (PTS) is like:
PTS calculation formula: PTR*(1 – Stockist Margin %) = 71.43*(1-10%) = 64.29/-
Other Calculations:
If you want to calculate other calculations like Cnf price, sub stockist price etc then you can use formula like describe above. Suppose CnF margin is 6% then it will be calculate by below formula:
CnF (Carrying & Forwarding Agent Price) = PTS*(1 – CnF margin %) = 64.29*(1-6%) = 60.43/-
How to prepare a PTR/PTS calculator in excel as embed above?
Open excel
At first row (Column A) write: Change desired Value of MRP, Tax and Margin(Yellow color) to calculate PTS/PTR as given below in image:
Third row (Column A) is for MRP value
Fourth Row (Column A) indicating GST
Fifth Row (Column A) is for GST value
Sixth row (Column A) indicating Retailer Margin
Seventh row (Column A) is for retailer margin value
Eight row (Column A) indicating Stockist margin
Ninth row (Column A) is for stockist margin value
Now in column C (Row 3) indicate Net value in a cell as shown in picture below
In below net value cell, add a formula: = A3*(1-A7%) which will give net value
Below this cell indicate GST Factor
Below GST indicator cell, add formula = (100+A5)/100 which will give GST factor value
Now come to tenth row (Column A). Write PTR (Price to Retailer)
In eleventh row (column A), write formula: =C3/C5
Twelfth row (Column A) indicating Stockist margin
In thirteenth row (column A), write formula: =A11*(1-A9%)
By this way you can make, your own PTR/PTS calculator in excel format and use it for calculating your all product PTR/PTS values. You can also download PTS/PTR calculator from here
Another Method to calculate PTR/PTS:
How to calculate?
There is a formula to calculate PTS and PTR. We can calculate it with GST or without GST. When you have GST number then it is required to calculate PTS/PTR without GST and add GST after invoicing
value. If you don’t have GST number then it is required to calculate PTS/PTR with GST and can’t add GST at invoice.
Formula is Value multiply by 100 divided by 100 plus percentage of margin i.e. Value*100/(100+Percentage).
Here percentage may be GST, Stockist margin, Retailer Margin, CnF margin etc.
Value may be MRP in case of PTR calculation, PTR in case of PTS calculation etc.
Suppose we have a product having
MRP of 95 rs
GST 12%
Retailer Margin 20%
Stockist Margin 10%
Calculation of Price to Retailer:
Formula is Value*100/ (100+Percentage)
PTR = MRP*100 / (100+Retailer Margin) = 95*100/(100+20) = 9500/120 = 79.17/-
This PTR is included of GST. Now calculate it without GST.
PTR (Without GST) = PTR (With GST)*100 (100+GST) = 79.17*100 (100+12) = 79.17/112 = 70.69/-
Calculation of Price to Stockist: Formula is Value*100 / (100+Percentage)
PTS = PTR * 100 / (100+Stockist Margin) = 79.17*100 / (100+10) = 7917/110 = 71.97/-
This PTS is included of GST. Now calculate it without GST.
PTS (Without GST) = PTS (With GST) * 100 / (100+GST) = 71.97*100 / (100+12) = 7197/112 = 64.26/-
Hope this information is helpful to you...
Related: Pharmaceutical Start-up Ideas
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My first demo ! Had a blast !
Well I have my first demo under my belt. I did a fall craft event on the Blue Ridge Parkway at the Humpback Rocks visitor's center Near Waynesboro Va.
Everything went pretty smooth. The forge was constructed from a farm disc harrow blade 24" in diameter with a 6" to 3" pipe reducer as the firepot. It has a circular type clinker breaker. This was
the first time I had fired that forge so I was a little nervous. I actually got paid and was invited back again.
Oak Hill, sounds like you're hooked on the demo scene. They are fun to do. Get several small quick projects to make and have fun. Congrats. :)
You had a great time, you got paid to have a great time, and you were invited back. Life is good. I like your forge also.
Demos are a whole lotta fun! Glad it went so well.
The disk makes a fine looking forge.
Fun eh?
I noticed the black sawbuck in the rear. Was that a thing you use? If so, besides bucking logs, what?
The saw buck had no value as a blacksmithing tool. The venue is a restored mountain farm with an original log house, barn, spring house, chicken coupe etc. The barn actually came from the homestead
of Joe Clark, yep the one from the Old Joe Clark bluegrass tune, moved there from the Irish Creek area. They always have different mountain crafts demoed each weekend, along with free bluegrass bands
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An outlet store had monthly sales of $73,000 and spent 8% of it on health insurance. How much was spent on health insurance? - DocumenTVAn outlet store had monthly sales of $73,000 and spent 8% of it on health insurance. How much was spent on health insurance?
An outlet store had monthly sales of $73,000 and spent 8% of it on health insurance. How much was spent on health insurance?
An outlet store had monthly sales of $73,000 and spent 8% of it on health insurance.
How much was spent on health insurance?
in progress 0
Mathematics 3 years 2021-08-17T11:11:37+00:00 2021-08-17T11:11:37+00:00 2 Answers 5 views 0
Answers ( )
1. 0
2021-08-17T11:13:19+00:00 August 17, 2021 at 11:13 am
$5,840 was spent on health insurance
2. 0
2021-08-17T11:13:34+00:00 August 17, 2021 at 11:13 am
1. We assume, that the number 73000 is 100% – because it’s the output value of the task.
2. We assume, that x is the value we are looking for.
3. If 73000 is 100%, so we can write it down as 73000=100%.
4. We know, that x is 8% of the output value, so we can write it down as x=8%.
5. Now we have two simple equations:
1) 73000=100%
2) x=8%
where left sides of both of them have the same units, and both right sides have the same units, so we can do something like that:
6. Now we just have to solve the simple equation, and we will get the solution we are looking for.
7. Solution for what is 8% of 73000
(73000/x)*x=(100/8)*x – we multiply both sides of the equation by x
73000=12.5*x – we divide both sides of the equation by (12.5) to get x
now we have:
8% of 73000=5840
Leave an answer
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With our fresh, new combined shape and texture model, we have found a nice way to describe how a face could change not only in shape, but also in appearance. Now, we want to find which set of p shape
and λ appearance parameters will bring our model as close as possible to a given input image I(x). We could naturally calculate the error between our instantiated model and the given input image in
the coordinate frame of I(x), or map the points back to the base appearance and calculate the difference there. We are going to use the latter approach. This way, we want to minimize the following
In the preceding equation, S0 denotes the set of pixels x is equal to (x,y)T that lie inside the AAMs base mesh, A0(x) is our base mesh texture, Ai(x) is appearance images from PCA, and W(x;p) is the
warp that takes pixels from the input image back to the base mesh frame.
Several approaches have been proposed for this minimization through years of studying. The first idea... | {"url":"https://subscription.packtpub.com/book/data/9781786467171/5/ch05lvl1sec35/posit","timestamp":"2024-11-02T05:14:41Z","content_type":"text/html","content_length":"195212","record_id":"<urn:uuid:a54948d7-716f-4671-86a5-16218f247465>","cc-path":"CC-MAIN-2024-46/segments/1730477027677.11/warc/CC-MAIN-20241102040949-20241102070949-00029.warc.gz"} |
Symmetry and the order of events in time. A proposed identity of negative mass with antimatter
In another paper in this series, we developed a symmetrical theory of mass which describes how systems of positive mass and negative mass will respond to an input of thermal energy. A system composed
of positive mass or negative mass will respond to an input of thermal energy in opposite ways. For example, if a system composed of positive mass expands in response to radiation from a hot body, a
system composed of negative mass will contract. Likewise, if the system composed of positive mass contracts when brought in communication with a cold body, a system composed of negative mass will
expand. In addition, when a system of positive or negative mass is brought into contact with radiation from a thermal reservoir either hotter or colder than the system, thermal processes are induced
such that the sign of the change of entropy of a system composed of positive mass is opposite of that of a system composed of negative mass. That is, in response to thermal energy, a system of
negative mass behaves as if it is a system of positive mass going backwards in time. This is reminiscent of Feynman's definition of antimatter as matter going backwards in time. Negative mass is
consistent with the negative energy solutions to the equations of the Special Theory of Relativity when combined with quantum mechanics. Formally, the total energy of a particle can be either
positive or negative, which means that the mass of that particle can be either positive or negative. Dirac eliminated the negative mass solution by giving certain complex properties to the vacuum.
Pauli used only the positive mass solutions to build the theory of spin and statistics. On the other hand, we interpret both the positive and negative energy solutions to be real solutions that
represent substances with positive mass and negative mass, respectively. Thermal energy is only one part of the spectrum of electromagnetic radiation. It is well known that matter and antimatter
respond to electromagnetic radiation in opposite ways. For example, if an electron moves one way in an electromagnetic field, a positron will move in the opposite way. We apply our theory of positive
and negative mass to matter and antimatter and suggest that it productive to consider matter as having a positive mass and antimatter as having a negative mass. The equations presented here, which
treat matter as having a positive mass and antimatter as having a negative mass, can account for the experimental observations of matter and antimatter in electromagnetic fields. Our treatment allows
the symmetry between matter and antimatter to be treated in a more causal manner.
Antimatter, entropy, negative mass, reversibility, Second Law of Thermodynamics, symmetry, time direction
Recommended Citation
WAYNE, RANDY (2012) "Symmetry and the order of events in time. A proposed identity of negative mass with antimatter," Turkish Journal of Physics: Vol. 36: No. 2, Article 2. https://doi.org/10.3906/
Available at: https://journals.tubitak.gov.tr/physics/vol36/iss2/2 | {"url":"https://journals.tubitak.gov.tr/physics/vol36/iss2/2/","timestamp":"2024-11-13T05:30:18Z","content_type":"text/html","content_length":"64920","record_id":"<urn:uuid:cbe48b29-5c01-457e-bce9-c1cd2b655523>","cc-path":"CC-MAIN-2024-46/segments/1730477028326.66/warc/CC-MAIN-20241113040054-20241113070054-00051.warc.gz"} |
Coupled-cluster and explicitly correlated perturbation-theory calculations of the uracil anion
A valence-type anion of the canonical tautomer of uracil has been characterized using explicitly correlated second-order Møller-Plesset perturbation theory (RI-MP2-R12) in conjunction with
conventional coupled-cluster theory with single, double, and perturbative triple excitations. At this level of electron-correlation treatment and after inclusion of a zero-point vibrational energy
correction, determined in the harmonic approximation at the RI-MP2 level of theory, the valence anion is adiabatically stable with respect to the neutral molecule by 40 meV. The anion is
characterized by a vertical detachment energy of 0.60 eV. To obtain accurate estimates of the vertical and adiabatic electron binding energies, a scheme was applied in which electronic energy
contributions from various levels of theory were added, each of them extrapolated to the corresponding basis-set limit. The MP2 basis-set limits were also evaluated using an explicitly correlated
approach, and the results of these calculations are in agreement with the extrapolated values. A remarkable feature of the valence anionic state is that the adiabatic electron binding energy is
positive but smaller than the adiabatic electron binding energy of the dipole-bound state. © 2007 American Institute of Physics.
Dive into the research topics of 'Coupled-cluster and explicitly correlated perturbation-theory calculations of the uracil anion'. Together they form a unique fingerprint. | {"url":"https://researchportal.hw.ac.uk/en/publications/coupled-cluster-and-explicitly-correlated-perturbation-theory-cal","timestamp":"2024-11-10T05:17:37Z","content_type":"text/html","content_length":"56918","record_id":"<urn:uuid:911a0ca0-40f7-4fab-9334-69eedc035f40>","cc-path":"CC-MAIN-2024-46/segments/1730477028166.65/warc/CC-MAIN-20241110040813-20241110070813-00595.warc.gz"} |
How do you find the integral of sin(lnx) dx? | HIX Tutor
How do you find the integral of #sin(lnx) dx#?
Answer 1
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Answer 2
To find the integral of (\sin(\ln(x)) , dx), use the substitution method. Let (u = \ln(x)), then (du = \frac{1}{x} , dx). After substitution, the integral becomes:
[ \int \sin(u) , du ]
This is a standard integral, which equals (-\cos(u) + C), where (C) is the constant of integration. Substitute (u) back in terms of (x), and you will have your final answer.
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Answer from HIX Tutor
When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some
When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some
When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some
When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some
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• Readily available 24/7 | {"url":"https://tutor.hix.ai/question/how-do-you-find-the-integral-of-sin-lnx-dx-8f9afa10b0","timestamp":"2024-11-09T19:43:51Z","content_type":"text/html","content_length":"573108","record_id":"<urn:uuid:85a25986-16a1-434e-9673-6048f91dca70>","cc-path":"CC-MAIN-2024-46/segments/1730477028142.18/warc/CC-MAIN-20241109182954-20241109212954-00236.warc.gz"} |
The perimeter of a triangle is 40 cm if the ratio of the lengths of the sides is 2:3:5,find the length of each side? i know the an
The perimeter of a triangle is 40 cm if the ratio of the lengths of the sides is 2:3:5,find the length of each side? i know the answer is 8cm,12cm, and 20cm show me the working. please
2 thoughts on “The perimeter of a triangle is 40 cm if the ratio of the lengths of the sides is 2:3:5,find the length of each side? i know the an”
1. Answer:
perimeter =40
2x+3x+5x =40
x=4 cm
2. Answer:
The length of each sides of a triangle are 8cm, 12 cm , and 20 cm.
Step-by-step explanation:
Let the sides of the triangle be 2x,3x and 5x respectively.
the perimeter of the triangle is 40cm
=> 2x+3x+5x = 40
=> 10x = 40
=> x = 40/10
=> x = 4
Now,Putting value of x in each
1st angle = 2x
=> 2×4
=> 8 cm
2nd angle = 3x
=> 3×4
=> 12 cm
3rd angle = 5x
=> 5×4
=> 20 cm
Hence, the length of each sides of a triangle are 8cm, 12 cm , and 20 cm.
Leave a Comment | {"url":"https://wiki-helper.com/the-perimeter-of-a-triangle-is-40-cm-if-the-ratio-of-the-lengths-of-the-sides-is-2-3-5-find-the-37397176-55/","timestamp":"2024-11-12T20:16:27Z","content_type":"text/html","content_length":"129549","record_id":"<urn:uuid:1c9bd6ca-46a2-4884-81e2-0150633c2d3e>","cc-path":"CC-MAIN-2024-46/segments/1730477028279.73/warc/CC-MAIN-20241112180608-20241112210608-00311.warc.gz"} |
irrational number
A real number which is not a rational number, i.e. it is not the ratio of two integers.
The decimal expansion of an irrational is infinite but does not end in an infinite repeating sequence of digits. Examples of irrational numbers are
and the square root of two.
Last updated: 1995-04-12
Nearby terms:
Ironman ♦ ironmonger ♦ IRQ ♦ irrational number ♦ irrefutable ♦ IRSG ♦ IRTF
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Math Facts - Set Automatic Math Fact Practice from the Teacher Portal | Boddle Learning Support
To set Math Facts for your students:
Select the Classroom Tab on the left → Click Manage Math Facts at the bottom of the page → Select Edit for Multiple Students
About Math Facts
Math facts are fact fluency practice for addition, subtraction, multiplication, and division skills. Teachers can set math facts to automatically deliver fact fluency practice daily for a set number
of minutes.
Enabling Math Facts
Let’s enable math facts, starting from your Teacher Home page. Click on the classroom you’d like to set math facts for. In the center of the page, click "Manage Math Facts." This will open the Math
Fact Fluency page to set parameters.
Select "Edit for multiple students" at the top to set for multiple students at once.
Math Facts can be set for 1-30 minutes of daily practice. They can also be set to prioritize over assignments which means students will complete the fact fluency practice first, and then the program
will transition them back to their assignments or learning path automatically.
After math facts are set, they will automatically apply when students sign in. Once they meet the required amount of time, they will automatically transition to active assignments or their Boddle
learning path.
Adjusting Math Fact Parameters
To edit math fact settings, first, navigate to the Math Fact Fluency page by clicking on "Manage Math Facts" on the Classroom page.
You can edit settings for a student by selecting "Edit Settings" to the right of their name or edit for multiple students by clicking the blue edit button at the top of the page.
When you're happy with your parameters, press the blue “Finish Editing” button to save changes.
Happy teaching! | {"url":"https://intercom.help/boddle/en/articles/8391093-math-facts-set-automatic-math-fact-practice-from-the-teacher-portal","timestamp":"2024-11-08T08:21:04Z","content_type":"text/html","content_length":"54801","record_id":"<urn:uuid:9ffbe389-87df-4de2-8dbe-a54de8cd75e6>","cc-path":"CC-MAIN-2024-46/segments/1730477028032.87/warc/CC-MAIN-20241108070606-20241108100606-00526.warc.gz"} |
Geometric Cliques • Nicholas Pilkington
If you have N points in the plane what is the largest subset of those points such that each point is within a distance of D of all the others. It seems pretty innocuous right? Turns out it’s a great
big beautiful disaster.
Here’s a example of N = 10 points where the maximum subset are the points collected with dotted lines which are each within a distance D of each other. There is no bigger set.
Trying to compute every subset of points and checking if they are all within D of each other will take exponential time O(2^N). So we need a better approach. Let’s try picking a point which will will
assume to be part of this clique. Then all the candidate points that are not within D of that point won’t be in the clique. For example if we have three co-linear points spaced by D and select the
middle one to build our clique then either the left points can be in the clique or the right one but not both. This does give us a heuristic. We can take each point filter out points that aren’t
within D and then run a brute force search to find the maximal clique.
Let’s try and do better, as this could still be exponential depending on how the points are clustered. Let’s start with a bunch of points:
Pick any two points and assume that they are going to be the furthest points apart in our clique let this distance be F, so F <= D.
If we filter out all points more than F from these two points we get this situation
# try all candidates for furthest points
for i in xrange(N):
for j in xrange(i + 1, N):
xi, yi = points[i]
xj, yj = points[j]
if distance(xi, yi, xj, yj) > D: continue
# Furthest pair in our clique
F = distance(xi, yi, xj, yj)
lens = []
for k in xrange(N):
if k == i: continue
if k == j: continue
xk, yk = points[k]
if distance(xi, yi, xk, yk) > F: continue
if distance(xj, yj, xk, yk) > F: continue
Points inside the intersection of these two circles - the lens shape - are within F of both points at the end of the line and F <= D. I thought this way the end of the story. But we can’t simply
select all of these points as a clique because they may not be within D of each other. For example the top and bottom points in the lens shape might be further than D apart so we need to do some more
First note that all the points above the dotted lines are within F and therefore D of each other so they are a potential clique, as are points below the dotted line. But there may be a bigger clique
incorporating points from both sides of the lines. If we pick a certain point below the line for our clique we are forbidden from picking any points more than D away from that point. Incidentally
these forbidden points will all lie on the other side of the line. Take a moment, look at the picture above and make sure you are happy with that.
Now lets separate the points inside the lens shape into two sets. Those above the dotted line and those below. This can be done buy taking the signed area of the triangle created between the two
points at the end of the line and the point in question. If the area is positive the point is above the line and is it’s negative it’s below the line.
top, bot = [], []
M = len(lens)
for k in xrange(M):
xk, yk = lens[k]
if area((xi, yi), (xj, yj), (xk, yk)) >= 0:
The situation is now like this:
Let’s now treat each point as a vertex and connect vertices in one set with vertices in the other if they are further than D away. Using fewer points so it’s not as cluttered the situation could look
like this. Red lines denote points further than D apart and some point don’t have edges connected to them. This is fine, it just means all points are within D of them.
We’ve now constructed a bipartite graph representing some of the geometric constraints we are interested in. So our task is to select the maximum number of points from this set such that we don’t
have any two points connected by a red line because that means they are too far away. Vertices with no edges connected to them are freebies. They are within D of all of points so there’s no reason
not to pick them.
Our problem of selecting the maximum number of vertices in a graph such that no two points have an incident edge is the problem of computing the maximum independent set of a graph. This is
unfortunately NP-Complete in a general graph but in a bi-partite graph the number of vertices in the maximum independent set equals the number of edges in a minimum edge covering. This in turn is
equal to the maximum-flow of the graph. This series of dualities is called König’s theorem.
Without the bi-partite structure, computing the maximum independent set is NP-Complete where as maximum-flow can be computed in a general graph in polynomial time. So we’re in much better shape. We
compute the maximum flow of this graph and that value f plus the number of nodes without edges top_set and bot_set plus 2 for the end points of the line as our solution for that pair of points.
def max_bipartite_independent(top, bot):
graph = collections.defaultdict(dict)
top_set = set( [i for i in xrange( len(top)) ])
bot_set = set( [i for i in xrange( len(bot)) ])
for i in xrange(len(top)):
for j in xrange(len(bot)):
xi, yi = top[i]
xj, yj = bot[j]
if distance(xi, yi, xj, yj) > F:
node_i = 'TOP' + str(i)
node_j = 'BOT' + str(j)
src = 'SOURCE-NODE'
snk = 'SINK-NODE'
graph[node_i][node_j] = 1
graph[src][node_i] = MAX_INT
graph[node_j][snk] = MAX_INT
f = flow.max_flow(graph, 'SOURCE-NODE', 'SINK-NODE')
solution = f + len(top_set) + len(bot_set) + 2
return solution
There are a few different algorithms to compute maximum flow . The following is a simple implementation of the Ford-Fulkerson algorithm using Dijkstra’s algorithm to find augmenting paths.
import Queue
def dijkstra(graph, source, sink):
q = Queue.PriorityQueue()
q.put((0, source, []))
visited = set([source])
while not q.empty():
length, node, path = q.get()
# Found a path return the capacity
if node == sink:
cap = None
for a, b in path:
if cap is None or graph[a][b] < cap:
cap = graph[a][b]
return cap, path
# Visit next node
for child in graph[node].keys():
if not child in visited and graph[node][child] > 0:
next_state = (length+1, child, path + [(node, child)])
# No paths remaining
return None, None
And the remaining code to compute the maximum-flow of the graph:
def max_flow(graph, source, sink):
flow = 0
while True:
capacity, path = dijkstra(graph, source, sink)
if not capacity: return flow
for a, b in path:
graph[a][b] = graph[a].get(b, 0) - capacity
graph[b][a] = graph[b].get(a, 0) + capacity
flow += capacity
return flow
Cool so we’ve finally go all the pieces needed to solve this problem. We try every pair of points O(N^2) as the candidate for the two furtherest points in our clique then of the points the fall
inside the lens we build a bipartite graph and compute the maximum independent set which corresponds to the maximum clique size.
All in the algorithm takes O(V^2) attempting each pair of candidates points. But with the maximum-flow inner loop it’s O(V^5). This appears to be optimal without using a faster maximum matcher. Full
source code is here, maximum flow code and a little visualizer. | {"url":"https://nickp.svbtle.com/geometric-cliques","timestamp":"2024-11-13T14:50:11Z","content_type":"text/html","content_length":"21382","record_id":"<urn:uuid:15887eb8-65d7-4ad3-85de-8a9fb8e261b1>","cc-path":"CC-MAIN-2024-46/segments/1730477028369.36/warc/CC-MAIN-20241113135544-20241113165544-00517.warc.gz"} |
How do you find the sine ratio? - Our Planet Today
on April 22, 2022
How do you find the sine ratio?
Space and Astronomy
The definition of the sine ratio is the ratio of the length of the opposite side divided by the length of the hypotenuse. Well, the length of the side opposite ? C is the length of the hypotenuse, so
sin ? C = ^c/[c] = 1 Because ?
How do you find the sine and cosine ratio?
Video quote: So for example the sine of let's say angle a here well the sine is a ratio of the opposite over the hypotenuse. So it would be the length of this side right here the opposite.
What is the sine ratio for?
Sine ratios in particular are the ratios of the length of the side opposite the angle they represent over the hypotenuse. Sine ratios are useful in trigonometry when dealing with triangles and
circles. In a right triangle there exists special relationships between interior angles and the sides.
How do you find the sine ratio of a triangle?
Video quote: And the root 3 is like range of the middle it will be for the last one. So if you go ahead and do the 62 degrees right here for sine 60 you'll be doing the opposite. Take a look it's
root 3.
What is the ratio for sine of theta?
If θ is one of the acute angles in a triangle, then the sine of theta is the ratio of the opposite side to the hypotenuse, the cosine is the ratio of the adjacent side to the hypotenuse, and the
tangent is the ratio of the opposite side to the adjacent side.
How do you find the trigonometric ratio of an angle?
How to Find Trigonometric Ratios?
1. sin θ = Perpendicular/Hypotenuse.
2. cos θ = Base/Hypotenuse.
3. tan θ = Perpendicular/Base.
4. sec θ = Hypotenuse/Base.
5. cosec θ = Hypotenuse/Perpendicular.
6. cot θ = Base/Perpendicular.
How do you calculate sin by hand?
Video quote: Value and the Y value corresponds to the sine value. So all I need to do here is is look at the Y value for sine of zero. | {"url":"https://geoscience.blog/how-do-you-find-the-sine-ratio/","timestamp":"2024-11-10T02:41:52Z","content_type":"text/html","content_length":"178806","record_id":"<urn:uuid:a6f8b156-2691-4c5c-a8d9-f9ccbeb5bfda>","cc-path":"CC-MAIN-2024-46/segments/1730477028164.3/warc/CC-MAIN-20241110005602-20241110035602-00061.warc.gz"} |
About the Book - Numerical Methods in Economics
About the Book
To harness the full power of computer technology, economists need to use a broad range of mathematical techniques. In this book, Kenneth Judd presents techniques from the numerical analysis and
applied mathematics literatures and shows how to use them in economic analyses. The book is divided into five parts. Part I provides a general introduction. Part II presents basics from numerical
analysis on R^n, including linear equations, iterative methods, optimization, nonlinear equations, approximation methods, numerical integration and differentiation, and Monte Carlo methods. Part III
covers methods for dynamic problems, including finite difference methods, projection methods, and numerical dynamic programming. Part IV covers perturbation and asymptotic solution methods. Finally,
Part V covers applications to dynamic equilibrium analysis, including solution methods for perfect foresight models and rational expectation models.
This site was created as a resource to accompany the book. Click on the chapter listings in the sidebar menu to access the resources for each chapter. | {"url":"https://numericalmethodsineconomics.com/about-the-book/","timestamp":"2024-11-04T14:33:18Z","content_type":"text/html","content_length":"195770","record_id":"<urn:uuid:95757a7e-3132-48c9-aadc-ebe8c58b7231>","cc-path":"CC-MAIN-2024-46/segments/1730477027829.31/warc/CC-MAIN-20241104131715-20241104161715-00541.warc.gz"} |
Copy the figures with punched holes and find the axes of symmetry for
Ex 12.1, 1 - Chapter 12 Class 7 Symmetry
Last updated at April 16, 2024 by Teachoo
Ex 12.1, 1 (a) Copy the figures with punched holes and find the axes of symmetry for the following: To find line of symmetry, We join points, and draw a line perpendicular to it Then, we check if it
is line of symmetry Ex 12.1, 1 (b) Copy the figures with punched holes and find the axes of symmetry for the following: To find line of symmetry, We join points, and draw a line perpendicular to it
Then, we check if it is line of symmetry Ex 12.1, 1 (c) Copy the figures with punched holes and find the axes of symmetry for the following: To find line of symmetry, We join points, and draw a line
perpendicular to it Then, we check if it is line of symmetry Ex 12.1, 1 (d) Copy the figures with punched holes and find the axes of symmetry for the following: To find line of symmetry, We join
points, and draw a line perpendicular to it Then, we check if it is line of symmetry If this would have been line of symmetry, The other whole would be on green spot. Hence, it is not the line of
symmetry If this would have been line of symmetry, The other whole would be on green spot. Hence, it is not the line of symmetry Ex 12.1, 1 (e) Copy the figures with punched holes and find the axes
of symmetry for the following:Ex 12.1, 1 (f) Copy the figures with punched holes and find the axes of symmetry for the following: Ex 12.1, 1 (g) Copy the figures with punched holes and find the axes
of symmetry for the following: Ex 12.1, 1 (h) Copy the figures with punched holes and find the axes of symmetry for the following: Ex 12.1, 1 (i) Copy the figures with punched holes and find the axes
of symmetry for the following: Ex 12.1, 1 (j) Copy the figures with punched holes and find the axes of symmetry for the following: Ex 12.1, 1 (k) Copy the figures with punched holes and find the axes
of symmetry for the following: Ex 12.1, 1 (l) Copy the figures with punched holes and find the axes of symmetry for the following: But, this is not true, as: | {"url":"https://www.teachoo.com/15392/2559/Ex-14.1--1/category/Ex-14.1/","timestamp":"2024-11-11T06:35:24Z","content_type":"text/html","content_length":"139521","record_id":"<urn:uuid:dd285b28-8c7f-40b7-ac44-20dfd6974215>","cc-path":"CC-MAIN-2024-46/segments/1730477028220.42/warc/CC-MAIN-20241111060327-20241111090327-00453.warc.gz"} |
Reading: Using an Algebraic Approach to the Expenditure-Output Model
In the expenditure-output or Keynesian cross model, the equilibrium occurs where the aggregate expenditure line (AE line) crosses the 45-degree line. Given algebraic equations for two lines, the
point where they cross can be readily calculated. Imagine an economy with the following characteristics.
Y = Real GDP or national income
T = Taxes = 0.3Y
C = Consumption = 140 + 0.9 (Y – T)
I = Investment = 400
G = Government spending = 800
X = Exports = 600
M = Imports = 0.15Y
Step 1. Determine the aggregate expenditure function. In this case, it is:
AE = C + I + G + X – M
AE = 140 + 0.9(Y – T) + 400 + 800 + 600 – 0.15Y
Step 2. The equation for the 45-degree line is the set of points where GDP or national income on the horizontal axis is equal to aggregate expenditure on the vertical axis. Thus, the equation for the
45-degree line is: AE = Y.
Step 3. The next step is to solve these two equations for Y (or AE, since they will be equal to each other). Substitute Y for AE:
Y = 140 + 0.9(Y – T) + 400 + 800 + 600 – 0.15Y
Step 4. Insert the term 0.3Y for the tax rate T. This produces an equation with only one variable, Y.
Step 5. Work through the algebra and solve for Y.
Y = 140 + 0.9(Y – 0.3Y) + 400 + 800 + 600 – 0.15Y
Y = 140 + 0.9Y –0.27Y + 1800 – 0.15Y
Y = 1940 + 0.48Y
Y – 0.48Y = 1940
0.52Y = 1940
[latex]\displaystyle\frac{0.52\text{Y}}{0.52}[/latex] = [latex]\displaystyle\frac{1940}{0.52}[/latex]
Y = 3730
This algebraic framework is flexible and useful in predicting how economic events and policy actions will affect real GDP.
Step 6. Say, for example, that because of changes in the relative prices of domestic and foreign goods, the marginal propensity to import falls to 0.1. Calculate the equilibrium output when the
marginal propensity to import is changed to 0.1.
Y = 140 + 0.9(Y – 0.3Y) + 400 + 800 + 600 – 0.1Y
Y = 1940 – 0.53Y
0.47Y = 1940
Y = 4127
Step 7. Because of a surge of business confidence, investment rises to 500. Calculate the equilibrium output.
Y = 140 + 0.9(Y – 0.3Y) + 500 + 800 + 600 – 0.15Y
Y = 2040 + 0.48Y
Y – 0.48Y = 2040
0.52Y = 2040
Y = 3923
For issues of policy, the key questions would be how to adjust government spending levels or tax rates so that the equilibrium level of output is the full employment level. In this case, let the
economic parameters be:
Y = National income
T = Taxes = 0.3Y
C = Consumption = 200 + 0.9 (Y – T)
I = Investment = 600
G = Government spending = 1,000
X = Exports = 600
Y = Imports = 0.1 (Y – T)
Step 8. Calculate the equilibrium for this economy (remember Y = AE).
Y = 200 + 0.9(Y – 0.3Y) + 600 + 1000 + 600 – 0.1(Y – 0.3Y)
Y – 0.63Y + 0.07Y = 2400
0.44Y = 2400
Y = 5454
Step 9. Assume that the full employment level of output is 6,000. What level of government spending would be necessary to reach that level? To answer this question, plug in 6,000 as equal to Y, but
leave G as a variable, and solve for G. Thus:
6000 = 200 + 0.9(6000 – 0.3(6000)) + 600 + G + 600 – 0.1(6000 – 0.3(6000))
Step 10. Solve this problem arithmetically. The answer is: G = 1,240. In other words, increasing government spending by 240, from its original level of 1,000, to 1,240, would raise output to the full
employment level of GDP.
Indeed, the question of how much to increase government spending so that equilibrium output will rise from 5,454 to 6,000 can be answered without working through the algebra, just by using the
multiplier formula. The multiplier equation in this case is:
Thus, to raise output by 546 would require an increase in government spending of 546/2.27=240, which is the same as the answer derived from the algebraic calculation.
This algebraic framework is highly flexible. For example, taxes can be treated as a total set by political considerations (like government spending) and not dependent on national income. Imports
might be based on before-tax income, not after-tax income. For certain purposes, it may be helpful to analyze the economy without exports and imports. A more complicated approach could divide up
consumption, investment, government, exports and imports into smaller categories, or to build in some variability in the rates of taxes, savings, and imports. A wise economist will shape the model to
fit the specific question under investigation.
All the components of aggregate demand—consumption, investment, government spending, and the trade balance—are now in place to build the Keynesian cross diagram. Figure B.7 builds up an aggregate
expenditure function, based on the numerical illustrations of C, I, G, X, and M that have been used throughout this text. The first three columns in Table B.3 are lifted from the earlier Table B.2,
which showed how to bring taxes into the consumption function. The first column is real GDP or national income, which is what appears on the horizontal axis of the income-expenditure diagram. The
second column calculates after-tax income, based on the assumption, in this case, that 30% of real GDP is collected in taxes. The third column is based on an MPC of 0.8, so that as after-tax income
rises by $700 from one row to the next, consumption rises by $560 (700 × 0.8) from one row to the next. Investment, government spending, and exports do not change with the level of current national
income. In the previous discussion, investment was $500, government spending was $1,300, and exports were $840, for a total of $2,640. This total is shown in the fourth column. Imports are 0.1 of
real GDP in this example, and the level of imports is calculated in the fifth column. The final column, aggregate expenditures, sums up C + I + G + X – M. This aggregate expenditure line is
illustrated in Figure B.7.
Table B.3. National Income-Aggregate Expenditure Equilibrium
National Income After-Tax Income Consumption Government Spending + Investment + Exports Imports Aggregate Expenditure
$3,000 $2,100 $2,280 $2,640 $300 $4,620
$4,000 $2,800 $2,840 $2,640 $400 $5,080
$5,000 $3,500 $3,400 $2,640 $500 $5,540
$6,000 $4,200 $3,960 $2,640 $600 $6,000
$7,000 $4,900 $4,520 $2,640 $700 $6,460
$8,000 $5,600 $5,080 $2,640 $800 $6,920
$9,000 $6,300 $5,640 $2,640 $900 $7,380
The aggregate expenditure function is formed by stacking on top of each other the consumption function (after taxes), the investment function, the government spending function, the export function,
and the import function. The point at which the aggregate expenditure function intersects the vertical axis will be determined by the levels of investment, government, and export expenditures—which
do not vary with national income. The upward slope of the aggregate expenditure function will be determined by the marginal propensity to save, the tax rate, and the marginal propensity to import. A
higher marginal propensity to save, a higher tax rate, and a higher marginal propensity to import will all make the slope of the aggregate expenditure function flatter—because out of any extra
income, more is going to savings or taxes or imports and less to spending on domestic goods and services.
The equilibrium occurs where national income is equal to aggregate expenditure, which is shown on the graph as the point where the aggregate expenditure schedule crosses the 45-degree line. In this
example, the equilibrium occurs at 6,000. This equilibrium can also be read off the table under the figure; it is the level of national income where aggregate expenditure is equal to national income.
Candela Citations
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How Escaping Corners Works
This sketch is a modified version of RoamingWithWhiskers, so we’ll just look at the new code for detecting and escaping corners.
First, three global byte variables are added: wLeftOld, wRightOld, and counter. The wLeftOld and wRightOld variables store the whisker states from a previous whisker contact so that they can be
compared with the states of the current contact. Then counter is used to track the number of consecutive, alternate contacts.
byte wLeftOld; // Previous loop whisker values
byte wRightOld;
byte counter; // For counting alternate corners
These variables are initialized in the setup function. The counter variable can start with zero, but one of the “old” variables has to be set to 1. Since the routine for detecting corners always
looks for an alternating pattern, and compares it to the previous alternating pattern, there has to be an initial alternate pattern to start with. So, wLeftOld and wRightOld are assigned initial
values in the setup function before the loop function starts checking and modifying their values.
wLeftOld = 0; // Initialize previous whisker
wRightOld = 1; // states
counter = 0; // Initialize counter to 0
The first thing the code below // Corner Escape has to do is check if one or the other whisker is pressed. A simple way to do this is to use the not-equal operator (!= ) in an if statement. In
English, if(wLeft != wRight) means “if the wLeft variable is not equal to the wRight variable…”
// Corner Escape
if(wLeft != wRight) // One whisker pressed?
If they are not equal it means one whisker is pressed, and the sketch has to check whether it’s the opposite pattern as the previous whisker contact. To do that, a nested if statement checks if the
current wLeft value is different from the previous one and if the current wRight value is different from the previous one. That’s if( (wLeft != wLeftOld) && (wRight != wRightOld)). If both conditions
are true, it’s time to add 1 to the counter variable that tracks alternate whisker contacts. It’s also time to remember the current whisker pattern by setting wLeftOld equal to the current wLeft and
wRightOld equal to the current wRight.
if((wLeft != wLeftOld) && (wRight != wRightOld))
counter++; // Increase count by one
wLeftOld = wLeft; // Record current for next rep
wRightOld = wRight;
If this is the fourth consecutive alternate whisker contact, then it’s time to reset the counter variable to 0 and execute a U-turn. When the if(counter == 4) statement is true, its code block tricks
the whisker navigation routine into thinking both whiskers are pressed. How does it do that? It sets both wLeft and wRight to zero. This makes the whisker navigation routine think both whiskers are
pressed, so it makes a U-turn.
if(counter == 4) // Stuck in a corner?
wLeft = 0; // Set up whisker states for U-turn
wRight = 0;
counter = 0; // Clear alternate corner count
But, if the conditions in if((wLeft != wLeftOld) && (wRight != wRightOld)) are not all true, it means that this is not a sequence of alternating whisker contacts anymore, so the BOE Shield-Bot must
not be stuck in a corner. In that case, the counter variable is set to zero so that it can start counting again when it really does find a corner.
else // Not alternate from last time
counter = 0; // Clear alternate corner count
One thing that can be tricky about nested if statements is keeping track of opening and closing braces for each statement’s code block. The picture below shows some nested if statements from the last
sketch. In the Arduino editor, you can double-click on a brace to highlight its code block. But sometimes, printing out the code and simply drawing lines to connect opening and closing braces helps
to see all the blocks at once, which is useful for finding bugs in deeply nested code.
In this picture, the if(wLeft != wRight) statement’s code block contains all the rest of the decision-making code. If it turns out that wLeft is equal to wRight, the Arduino skips to whatever code
follows that last closing brace }. The second level if statement compares the old and new wLeft and wRight values with if ((wLeft != wLeftOld) && (wRight != wRightOld)). Notice that its code block
ending brace is just below the one for the if(counter==4) block. The if ((wLeft != wLeftOld) && (wRight != wRightOld)) statement also has an else condition with a block that sets counter to zero if
the whisker values are not opposite from those of the previous contact.
• Study the code in the picture carefully.
• Imagine that wLeft = 0 , wRight = 0 and counter == 3, and think about what this statement would do.
• Imagine that wLeft = 1 , wRight = 0 , wLeftOld = 0 , wRight = 1 and counter == 3. Try walking through the code again line by line and explain what happens to each variable at each step.
Your Turn
One of the if statements in EscapingCorners checks to see if counter has reached 4.
• Try increasing the value to 5 and 6 and test the effect. Keep in mind that it will either count to the number of alternate whisker contacts, or maybe one more than that depending on which side
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The Law of CosinesUndertanding the Law of Cosines (including its formula)
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Simple Parallel Processing in R | R-bloggersSimple Parallel Processing in R
Simple Parallel Processing in R
[This article was first published on
R – Daniel Oehm | Gradient Descending
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I recently purchased a new laptop with an Intel i7-8750 6 core CPU with multi-threading meaning I have 12 logical processes at my disposal. Seemed like a good opportunity to try out some parallel
processing packages in R. There are a few packages in R for the job with the most popular being parallel, doParallel and foreach package.
First we need a good function that puts some load on the CPU. We’ll use the Boston data set, fit a regression model and calculate the MSE. This will be done 10,000 times.
# data
# function - calculate the mse from a model fit on bootstrapped samples from the Boston dataset
model.mse <- function(x) {
id <- sample(1:nrow(Boston), 200, replace = T)
mod <- lm(medv ~ ., data = Boston[id,])
mse <- mean((fitted.values(mod) - Boston$medv[id])^2)
# data set
#x.list <- lapply(1:2e4, function(x) rnorm(1e3, 0, 200))
x.list <- sapply(1:10000, list)
# detect the number of cores
n.cores <- detectCores()
## [1] 12
Parallelising computation on a local machine in R works by creating an R instance on each of the cluster members specified by the user, in my case 12. This means each instance needs to have the same
data, packages and functions to do the calculations. The function clusterExport copies the data frames, loads packages and other functions to each of the cluster members. This is where some thought
is needed as to whether or not parallelising computation on will actually be beneficial. If the data frame is huge making 12 copies and storing them in memory will create a huge overhead and may not
speed up the computation. For these examples we need to export the Boston data set to cluster. Since the data set is only 0.1 Mb this won’t be a problem. At the end of the processing it is important
to remember to close the cluster with stopCluster.
Using parLapply from the parallel package is the easiest way to parallelise computation since lapply is simply switched with parLapply and tell it the cluster setup. This is my go to function since
it is very simple to parallelise existing code. Firstly we’ll establish a baseline using the standard lapply function and then compare it to parLapply.
# single core
system.time(a <- lapply(x.list, model.mse))
## user system elapsed
## 14.58 0.00 14.66
# 12 cores
clust <- makeCluster(n.cores)
clusterExport(clust, "Boston")
a <- parLapply(clust, x.list, model.mse)})
## user system elapsed
## 0.03 0.02 4.33
Much faster than the standard lapply. Another simple function is mclapply which works really well and even simpler than parLapply however this isn’t supported by Windows machines so not tested here.
parSapply works in the same way as parLapply.
# sigle core
system.time(a <- sapply(1:1e4, model.mse))
## user system elapsed
## 14.42 0.00 14.45
# 12 cores
clust <- makeCluster(n.cores)
clusterExport(clust, "Boston")
a <- parSapply(clust, 1:1e4, model.mse)})
## user system elapsed
## 0.02 0.05 4.31
Again, much faster.
For completeness we’ll also test the parApply function which again works as above. The data will be converted to a matrix for this to be suitable.
# convert to mar
x.mat <- matrix(1:1e4, ncol = 1)
# sigle core
system.time(a <- apply(x.mat, 1, model.mse))
## user system elapsed
## 14.27 0.00 14.32
# 12 cores
clust <- makeCluster(n.cores)
clusterExport(clust, "Boston")
a <- parApply(clust, x.mat, 1, model.mse)})
## user system elapsed
## 0.00 0.03 4.30
As expected the parallel version is again faster.
The foreach function works in a similar way to for loops. If the apply functions aren’t suitable and you need to use a for loop, foreach should do the job. Basically what you would ordinarily put
within the for loop you put after the %dopar% operator. There are a couple of other things to note here,
1. We register the cluster using registerDoParallel from the doParallel package.
2. Need to specify how to combine the results after computation with .combine.
3. Need to specify .multicombine = TRUE for multiple parameter returns such as cbind or rbind.
There are a few other useful parameters for more complex processors however these are the key things.
# for
model.mse.output <- rep(NA, 1e4)
for(k in 1:1e4){
model.mse.output[k] <- model.mse()
## user system elapsed
## 14.23 0.00 14.23
# foreach
foreach(k = 1:1e4, .combine = c) %dopar% model.mse()
## user system elapsed
## 3.50 0.59 6.53
Interestingly foreach is slower than the parXapply functions.
There are a number of resources on the parallel computation in R but this is enough to get anyone started. If you are familiar with using the apply functions parallelising computations is straight
forward but it’s important to keep in mind whether or not that process needs to be executed in parallel.
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[QSMS Seminar 2021.10.26, 10.28] Feigin-Semikhatov duality in W-superalgebras
Date : Lecture 1 : 2021/10/26 (Tue) 10:00-11:30
Lecture 2 : 2021/10/28 (Thu) 10:00-11:30
Place : Zoom (ID: 642 675 5874 no password, Login required)
Speaker : Ryo Sato (Kyoto university)
Title : Feigin-Semikhatov duality in W-superalgebras
W-superalgebras are a large class of vertex superalgebras which generalize affine Lie superalgebras and the Virasoro algebras. It has been known that princial W-algebras satisfy a certain duality
relation (Feigin-Frenkel duality) which can be regarded as a quantization of the geometric Langlands correspondence. Recently, D. Gaiotto and M. Rapčák found dualities between more general
W-superalgebras in relation to certain four-dimensional supersymmetric gauge theories. A large part of thier conjecture is proved by T. Creutzig and A. Linshaw, and a more specific subclass
(Feigin-Semikhatov duality) is done by T. Creutzig, N. Genra, and S. Nakatsuka in a different way. In this talk I will review the Feigin-Semikhatov duality between certain subregular W-algebras and
principal W-superalgebras, and discuss the usage of relative semi-infinite cohomology. This talk is based on a joint work with T. Creutzig, N. Genra, and S. Nakatsuka. | {"url":"https://qsms.math.snu.ac.kr/index.php?mid=board_sjXR83&listStyle=viewer&order_type=desc&l=en&sort_index=title&page=3&document_srl=1912","timestamp":"2024-11-07T12:11:53Z","content_type":"text/html","content_length":"21973","record_id":"<urn:uuid:95c4aae5-de8b-4bd0-a52b-6f96d9efa258>","cc-path":"CC-MAIN-2024-46/segments/1730477027999.92/warc/CC-MAIN-20241107114930-20241107144930-00734.warc.gz"} |
Win Dolphin Tale 2 Goodies - chelseamamma.co.uk
Win Dolphin Tale 2 Goodies
Inspired by true events, “Dolphin Tale 2” continues the story of brave dolphin Winter, whose miraculous rescue and rehabilitation – thanks to the invention of a groundbreaking prosthetic tail — made
her a symbol of perseverance to people around the world and inspired the 2011 family hit movie “Dolphin Tale.”
The film reunites the entire main cast, led by Harry Connick, Jr., Ashley Judd, Nathan Gamble, Kris Kristofferson, Cozi Zuehlsdorff, Austin Stowell, Austin Highsmith and Morgan Freeman.
Several years have passed since young Sawyer Nelson and the dedicated team at the Clearwater Marine Aquarium, rescued Winter, a young dolphin who lost her tail after being entangled in a crab trap.
With the help of Dr. Cameron McCarthy, who developed a unique prosthetic tail for Winter, they saved her life against all odds. In turn, she helped save the Aquarium — as people came from far and
wide to see the courageous dolphin firsthand, enabling CMA to greatly expand their mission to “rescue, rehabilitate and, when possible, release” injured animals.
Yet their fight is not over. Winter’s surrogate mother, the elderly dolphin Panama, passes away, leaving Winter alone and grieving, unwilling to engage with anyone, even her best friend, Sawyer.
However, the loss of Panama may have even greater repercussions for CMA. The USDA warns Clay they will have to move Winter from the Aquarium because regulations require these social creatures to be
paired. If they don’t find a female companion for her — one that she accepts — CMA will lose their beloved Winter. But as time runs out, there may still be Hope…
I’m giving away 5 giant 3’4″ stuffed dolphins worth £23 each to celebrate the launch of Dolphin Tale 2, in cinemas now!
© 2014 Alcon Entertainment, LLC. All rights reserved
To Enter:
• Fill in the Rafflecopter widget below to verify your entries
• Entries can be by comment, Facebook, twitter etc
• Please read the rules below
• Closing Date: 2nd November 2014
• If there is no form hit refresh (F5) and it should appear
• If still not working please check that your computer is running Javascript
• You need to complete the mandatory entry first – Leave your best travel tip
• Rafflecopter will tweet, like and follow on your behalf making it really easy to enter
• Really want to win the prize? Come back every day for bonus entries via twitter
• Terms and Conditions can be found in the Rafflecopter form below
149 thoughts on “Win Dolphin Tale 2 Goodies”
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Leave a Comment | {"url":"https://www.chelseamamma.co.uk/2014/10/02/win-dolphin-tale-2-goodies/","timestamp":"2024-11-11T10:04:37Z","content_type":"text/html","content_length":"322445","record_id":"<urn:uuid:3030ce5b-ce31-427b-894f-380aa3b748b0>","cc-path":"CC-MAIN-2024-46/segments/1730477028228.41/warc/CC-MAIN-20241111091854-20241111121854-00788.warc.gz"} |
Mathematical pettifoggery and pathological examples
Mathematical pettifoggery and pathological examples
Last week I said:
Mathematicians do tend to be the kind of people who quibble and pettifog over the tiniest details. This is because in mathematics, quibbling and pettifogging does work.
This example is technical, but I think I can explain it in a way that will make sense even for people who have no idea what the question is about. Don't worry if you don't understand the next
In this math SE question: a user asks for an example of a connected topological space !!\langle X, \tau\rangle!! where there is a strictly finer topology !!\tau'!! for which !!\langle X, \tau'\
rangle!! is disconnected. This is a very easy problem if you go about it the right way, and the right way follows a very typical pattern which is useful in many situations.
The pattern is “TURN IT UP TO 11!!” In this case:
1. Being disconnected means you can find some things with a certain property.
2. If !!\tau'!! is finer than !!\tau!!, that means it has all the same things, plus even more things of the same type.
3. If you could find those things in !!\tau!!, you can still find them in !!\tau'!! because !!\tau'!! has everything that !!\tau!! does.
4. So although perhaps making !!\tau!! finer could turn a connected space into a disconnected one, by adding the things you needed, it definitely can't turn a disconnected space into a connected
one, because the things will still be there.
5. So a way to look for a connected space that becomes disconnected when !!\tau!! becomes finer is:
Start with some connected space. Then make !!\tau!! fine as you possibly can and see if that is enough.
6. If that works, you win. If not, you can look at the reason it didn't work, and maybe either fix up the space you started with, or else use that as the starting point in a proof that the thing
you're looking for doesn't exist.
I emphasized the important point here. It is: Moving toward finer !!\tau!! can't hurt the situation and might help, so the first thing to try is to turn the fineness knob all the way up and see if
that is enough to get what you want. Many situations in mathematics call for subtlety and delicate reasoning, but this is not one of those.
The technique here works perfectly. There is a topology !!\tau_d!! of maximum possible fineness, called the “discrete” topology, so that is the thing to try first. And indeed it answers the question
as well as it can be answered: If !!\langle X, \tau\rangle!! is a connected space, and if there is any refinement !!\tau'!! for which !!\langle X, \tau'\rangle!! is disconnected, then !!\langle X, \
tau_d\rangle!! will be disconnected. It doesn't even matter what connected space you start with, because !!\tau_d!! is always a refinement of !!\tau!!, and because !!\langle X, \tau_d\rangle!! is
always disconnected, except in trivial cases. (When !!X!! has fewer than two points.)
Right after you learn the definition of what a topology is, you are presented with a bunch of examples. Some are typical examples, which showcase what the idea is really about: the “open sets” of the
real line topologize the line, so that topology can be used as a tool for studying real analysis. But some are atypical examples, which showcase the extreme limits of the concept that are as
different as possible from the typical examples. The discrete space is one of these. What's it for? It doesn't help with understanding the real numbers, that's for sure. It's a tool, it's the knob on
the topology machine that turns the fineness all the way up.[1] If you want to prove that the machine does something or other for the real numbers, one way is to show that it always does that thing.
And sometimes part of showing that it always does that thing is to show that it does that even if you turn the knob all the way to the right.
So often the first thing a mathematician will try is:
What happens if I turn the knob all the way to the right? If that doesn't blow up the machine, nothing will!
And that's why, when you ask a mathematician a question, often the first thing they will say is “ťhat fails when !!x=0!!” or “that fails when all the numbers are equal” or “ťhat fails when one number
is very much bigger than the other” or “that fails when the space is discrete” or “that fails when the space has fewer than two points.” [2]
After the last article, Kyle Littler reminded me that I should not forget the important word “pathological”. One of the important parts of mathematical science is figuring out what the knobs are, how
far they can go, what happens if you turn them all the way up, and what are the limits on how they can be set if we want the machine to behave more or less like the thing we are trying to study.
We have this certain knob for how many dents and bumps and spikes we can put on a sphere and have it still be a sphere, as long as we do not actually puncture or tear the surface. And we expected
that no matter how far we turned this knob, the sphere would still divide space into two parts, a bounded inside and an unbounded outside, and that these regions should behave basically the same as
they do when the sphere is smooth.[3]
But no, we are wrong, the knob goes farther than we thought. If we turn it to the “Alexander horned sphere” setting, smoke starts to come out of the machine and the red lights begin to blink.[4]
Useful! Now if someone has some theory about how the machine will behave nicely if this and that knob are set properly, we might be able to add the useful observation “actually you also have to be
careful not to turn that “dents bumps and spikes” knob too far.”
The word for these bizarre settings where some of the knobs are in the extreme positions is “pathological”. The Alexander sphere is a pathological embedding of !!S^2!! into !!\Bbb R^3!!.
[1] The leftmost setting on that knob, with the fineness turned all the way down, is called the “indiscrete topology” or the “trivial topology”.
[2] If you claim that any connected space can be disconnected by turning the “fineness” knob all the way to the right, a mathematican will immediately turn the “number of points” knob all the way to
the left, and say “see, that only works for spaces with at least two points”. In a space with fewer than two points, even the discrete topology is connected.
[3]For example, if you tie your dog to a post outside the sphere, and let it wander around, its leash cannot get so tangled up with the sphere that you need to walk the dog backwards to untangle it.
You can just slip the leash off the sphere.
[4] The dog can get its leash so tangled around the Alexander sphere that the only way to fix it is to untie the dog and start over. But if the “number of dimensions” knob is set to 2 instead of to
3, you can turn the “dents bumps and spikes” knob as far as you want and the leash can always be untangled without untying or moving the dog. Isn't that interesting? That is called the Jordan curve
[Other articles in category /math] permanent link | {"url":"https://blog.plover.com/math/pettifoggery.html","timestamp":"2024-11-12T16:51:47Z","content_type":"text/html","content_length":"32127","record_id":"<urn:uuid:48c997b9-70e8-44b8-a51d-c07a1d679e07>","cc-path":"CC-MAIN-2024-46/segments/1730477028273.63/warc/CC-MAIN-20241112145015-20241112175015-00895.warc.gz"} |
About Me
I'm a phenomenologist who merges physical theories and observations of the universe to unravel its mysteries. My research sharpens our understanding and modeling capabilities for present and future
measurements, harnessing cutting-edge statistical machine learning techniques to extract insights from complex datasets. My focus areas include testing experiment agreement, studying Dark Energy,
Dark Matter, cosmological neutrinos, and Gravity.
Scroll down to find more information about my research interests and publications. | {"url":"https://marcoraveri.com/","timestamp":"2024-11-02T21:21:03Z","content_type":"text/html","content_length":"26627","record_id":"<urn:uuid:972e771e-70b1-410f-9fb6-5478ea424ab7>","cc-path":"CC-MAIN-2024-46/segments/1730477027730.21/warc/CC-MAIN-20241102200033-20241102230033-00469.warc.gz"} |
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Question Text Let C be the largest circle centred at (2, 0) and inscribed in the ellipse = x y 2 2 1. 36 16 If (1, ) lies on C, then 10 2 is equal to
Updated On Mar 1, 2023
Topic Coordinate Geometry
Subject Mathematics
Class Class 11
Answer Type Video solution: 1
Upvotes 111
Avg. Video Duration 5 min | {"url":"https://askfilo.com/user-question-answers-mathematics/let-c-be-the-largest-circle-centred-at-2-0-and-inscribed-in-34343633303536","timestamp":"2024-11-03T20:21:33Z","content_type":"text/html","content_length":"207213","record_id":"<urn:uuid:78bd33f3-7944-4bfb-a2c3-d9df204ed52b>","cc-path":"CC-MAIN-2024-46/segments/1730477027782.40/warc/CC-MAIN-20241103181023-20241103211023-00116.warc.gz"} |
Run-time Assurance
Run-time Assurance
The Run-time Assurance Module (RAM) is the heart of the Supervisor. This module is responsible for calculating the appropriate command to send to the robot to stay “safe” based on available system
information. It is built as 3 main components:
The Kernel Generator uses
☆ an appropriate dynamical system Model,
☆ the system’s current state,
☆ the predicted environment state, and
☆ a set of possible actions that can be taken to keep the system safe
to calculate a set of “safe” commands that can be applied to the system. This set of actions is then passed to the input filter.
• The Input Filter selects the command that is a closest to the desired command sent by the autonomy stack from the set of safe commands produced by the Kernel Generator.
• Fault Manager detects basic faults (e.g. signal timeouts) from the signals sent by the autonomy stack to the RAM and implements appropriate stopping strategies.
In order for the kernel generator to determine the set of safe commands that can be sent to the robot, it must be able to quantify how the robot will behave when it receives a particular command.
This is done through the use of a dynamical model.
A dynamical model of a robot is defined by 6 items:
• A state vector \(x \in \mathbb{R}^\text{nx}\) representing the set of relevant physical quantities for that robot (position, orientation, velocity …).
• An input vector \(u \in \mathbb{R}^\text{nu}\) representing the set of relevant cyber-physical quantities for that robot that can be controlled directly (motor torque, desired velocity, desired
rotation rate …).
• A set of equations of motion in vectorized form \(\dot{x} = f(x,u)\) that describe how the state of the robot evolves over time when it receives a particular input. These equations of motions
will often be parameterized by known fixed quantities \(p \in \mathbb{R}^\text{np}\) called model parameters (mass, distance between wheels, maximum/minmum accelerations, maximum turn
rates,…),i.e. \(\dot{x} = f(x,u,p)\)
• A set of input constraints \(U \subseteq \mathbb{R}^\text{nu}\) that affect the set of all possible inputs that can be sent to the robot.
• A state domain set \(X \subseteq \mathbb{R}^\text{nx}\) that represents the domain for which the equations of motion are valid, particularly with respect to the process covariance matrix.
• A process noise covariance matrix which represents the probability distribution of the actual \(\dot{x}\) around the predicted value by \(f(x,u)\) for all given \(x \in X\) and \(u \in U\).
The Supervisor currently ships with 3 supported dynamical models:
This model is a 3-state, 2-input model that describes the movement of a robot evolving on SE2 (2D planar space with orientation), where control is available for longitudinal and angular velocities
directly. This model is particularly well suited for differential drive robots with fast acceleration and deceleration.
□ Model state: \(\left[x,y,\theta \right]\)
□ Model input: \(\left[ v_x, \omega \right]\)
□ Model parameters: None
□ Equations of motion: \(\begin{cases} \dot{x} = v_x \cos(\theta) \\ \dot{y} = v_x \sin(\theta) \\ \dot{\theta} = \omega \end{cases}\)
□ State domain: \(\mathbb{R}^3\)
□ Input constraints: User defined hyperbox in \(\mathbb{R}^2\)
□ Process noise covariance matrix: Identity matrix
This model is a 3-state, 3-input model that describes the movement of a robot evolving on SE2 (2D planar space with orientation) where control is available for longitudinal, lateral, and angular
velocities directly. This model is particularly well suited for mobile robots with omni wheels, quadrupeds, and surface vessels with fast acceleration and deceleration.
□ Model state: \(\left[x,y,\theta \right]\)
□ Model input: \(\left[ v_x, v_y, \omega \right]\)
□ Model parameters: None
□ Equations of motion: \(\begin{cases} \dot{x} = v_x \cos(\theta) - v_y \sin(\theta) \\ \dot{y} = v_x \sin(\theta) + v_y \cos(\theta) \\ \dot{\theta} = \omega \end{cases}\)
□ State domain: \(\mathbb{R}^3\)
□ Input constraints: User defined hyperbox in \(\mathbb{R}^3\)
□ Process noise covariance matrix: Identity matrix
This model is a 3-state, 2-input model that describes the movement of a robot evolving on SE2 (2D planar space with orientation) where the controls are longitudinal speed and front wheel steering
angle. This model is particularly well suited for mobile robots with front wheel steering.
□ Model state: \(\left[x,y,\theta \right]\)
□ Model input: \(\left[ v_x, \delta \right]\)
□ Model parameters:
☆ \(wheel_{dx}\): (wheelbase) Distance between front and rear wheel axles (m)
☆ \(origin_{dx}\): Position of vehicle origin w.r.t rear axle (m)
□ Equations of motion: \(\begin{cases} \dot{x} = v_x \cos(\theta) \\ \dot{y} = v_x \sin(\theta) \\ \dot{\theta} = v_x * \tan(\delta) * \frac{\cos(\beta)}{wheel_{dx}} \end{cases}\)
where sideslip is \(\beta = \arctan(\frac{origin_{dx}}{wheel_{dx}}\tan(\delta))\)
□ State domain: \(\mathbb{R}^3\)
□ Input constraints: User defined hyperbox in \(\mathbb{R}^2\). Note that the steering angle must be between \(-\frac{\pi}{2}\) and \(\frac{\pi}{2}\).
□ Process noise covariance matrix: Identity matrix
The Supervisor technology is able to support a wide variety of dynamical models with multiple levels of complexity. Use of a more accurate dynamical model allows for smaller margins and higher
performance from the system. Please contact 3laws to discuss implementation of more tailored versions of Supervisor to meet different application needs.
The other critical RAM component is the definition of what the robot should avoid. The Supervisor technology is able to enforce any arbitrary non-linear constraint on the robot’s state. These
constraints are organized into what 3Laws calls Safety Maps. A safety map is an object defining a set of robot state constraints. It can be updated when new information about the robot environment is
available. The constraints to be enforced can then be evaluated at a given robot state robot and return a vector of the values at that state, along with information on the gradient of the constraints
w.r.t variations in state.
The Supervisor ships with two safety maps:
□ geometric collision constraints as determined by data from a laserscan sensor, and
□ a list of obstacles with absolute locations.
The laserscan Safety Map defines constraints corresponding to the distance between the robot geometry and a carefully chosen set of capsules capturing the critical points in the laserscan. The
Supervisor enforces a constraint that the robot does not collide (intersect) with any of these capsules. The capsule sizes are defined through the collision distance threshold parameter (see control
panel configuration).
This safety map gets updated every time a new scan gets received on the specified topic (see control panel configuration).
The Obstacle Safety Map defines constraint equations corresponding to the distance between the robot geometry and a list of obstacles geometries.
This safety map gets updated every time a new list of obstacles is received on the specified topic (see control panel configuration).
This list of obstacles is a topic of type lll_msgs/ObjectArray:
std_msgs/Header header
Object[] objects
where the following fields of Object must be specified:
std_msgs/Header header
# Identifier of the object
string id
# Object geometry, and pose of geometry in object frame
ObjectGeometry geometry
# Object pose world frame
geometry_msgs/PoseWithCovariance pose
The Supervisor technology supports many more sensors and constraint representations. Please contact 3laws to learn more about all the type of constraints that can be implemented to satisfy other
applications’ needs.
The RAM performs its computations at regular intervals, whose rate is specified by Filter rate parameter (see control panel configuration). It therefore requires a continuous and consistent flow of
data from 3 sources to be processed during that interval, namely:
□ Perception - Sensor and mapping data
□ Localization - State estimate
□ Planning - Desired input
The Fault Management part of the RAM is therefore in charge of managing 4 modes of operation with configurable behaviors (see control panel configuration):
• Passthrough: The RAM publishes the latest desired input message it received un-altered. It won’t publish anything until a first message has been received. This mode can be transition in-to and
out-of at any time by publishing a boolean to the /lll/ram/enable topic (true to enter Passthrough mode, false to exit Passthrough mode).
• Initialization: Waiting to receive a first message from one of the 3 critical sources of data.
• Nominal operation: The RAM is publishing the best “safe” input command based on the latest data received.
• Fault: The RAM is not able to perform its function, and realizes the configured Failure Command Mode:
□ Send Zero: An input of 0 is published.
□ Do not publish: Not input is published. This option should only be used if the robot has its own mechanism to put itself in a safe condition if it is not receiving commands.
□ Passthrough: The latest desired command is published. This option must be used with care as it will give the planning stack (and particularly human drivers) a false sense of the RAM being
functional when it actually isn’t.
Whether or not the RAM is allowed to leave a Fault mode if it can is configurable inside the control panel.
The RAM algorithm tries to find the best balance between balance between being minimally intrusive and being “smooth” in its interventions. This tradeoff can be modulated with the Aggressiveness,
Pointiness, and Evasion Aggressiveness parameters (see control panel configuration):
The Supervisor can handle more complex and case specific metrics of optimality. Please contact 3laws to discuss implementations that better suit your application needs.
The RAM relies on localization to corelate the position of obstacles with the position of the robot, as well as for visualization. If the source of localization is not reliable, one can configure the
the RAM to not use localization, in which case, the robot will be assumed to always be at the center of the world, and the obstacles must be located in the robot frame.
The mathematical formulations used by the RAM rely on an implicit assumption about an exact knowledge of localization, perception, and dynamics data. The effectiveness of the Supervisor is therefore
correlated to the validity of that assumption, which is hard to verify in practice. It is therefore important to introduce some conservatism in the RAM formulation, which can be done heuristically
through parameters like Conservativeness, Collision distance threshold, and Aggressiveness.
There exist a fundamental trade-off between conservatism and optimality at the control level. The more certainty one has in things like dynamics, localization, and communication timing, the less
conservatism must be introduced to ensure a low probability of collision.
The Supervisor is able to account for uncertainty in an explicit and quantitative way. Please contact 3laws to learn more about the process of tailoring Supervisor to better handle uncertainties and
communication delays in your system.
With the Control Panel
The Control Panel offers a Minimap of the robot viewed from the Run-time Assurance Module’s perspective. This visualization is updated in real-time with the latest data from the RAM and display
safety margins, computed backup trajectory if available and closest point to the trajectory.
With RViz
In addition The RAM publishes a set of markers to visualize the processing done in particular by the Kernel Generator in turning the Safety Map information into input level constraints:
This data is published on the /lll/ram/markers topic. The control panel can provide an rviz configuration file to visualize this data.
If the Use Localization setting is set to false, make sure to set the World Frame parameter to the robot base frame, otherwise, the robot representation will stay still at the center of the world,
and the raw laserscan data likely won’t line up with the Safety Map markers. | {"url":"https://docs.3laws.io/en/latest/sources/user_guide/runtime_assurance.html","timestamp":"2024-11-12T02:30:10Z","content_type":"text/html","content_length":"30579","record_id":"<urn:uuid:0acafa2f-d449-4a22-8c9c-8d7f1ce16d8e>","cc-path":"CC-MAIN-2024-46/segments/1730477028242.50/warc/CC-MAIN-20241112014152-20241112044152-00337.warc.gz"} |
CS 3500 Assignment #8
You will extend the implementation in Java of the FMap ADT that was specified in assignments 4, 5, and 7 by adding the public dynamic method specified below, and by improving the hashCode() method so
it satisfies the probabilistic specification given below.
Collaboration between students is forbidden on this assignment. You are responsible for keeping your code hidden from all other students. Turn in your work on this assignment before 11:59 pm on the
due date by following instructions on the course’s main assignments web page, http://www.ccs.neu.edu/course/cs3500sp13/Assignments.html.
Your file of Java code should begin with a block comment that lists
1. Your name, as you want the instructor to write it.
2. Your email address.
3. Any remarks that you wish to make to the instructor.
Part of your grade will depend on the quality, correctness, and efficiency of your code, part will depend on your adherence to object-oriented programming style and idioms as taught in this course,
part will depend on the readability of your code (comments and indentation), and part will depend on how well you follow the procedure above for submitting your work. Assignments submitted between
12:00 am and 11:59 pm the day after the due date will receive a 20 percentage point penalty on the assignment.
Your assignment is to write the code for a single file, FMap.java, that implements the specification below as well as the specification of assignments 4, 5, and 7.
You will be given two files in /course/cs3500sp13/Assignments/A8:
The Visitor.java file defines the Visitor interface, so that file is essential for this assignment. The TestFMap.java file contains a simple test program; you will undoubtedly want to supplement that
test program with further tests.
Specification of the FMap ADT.
The FMap ADT remains as specified in assignments 4, 5, and 7, except that it must now provide the public accept(Visitor) operation specified below and must satisfy the augmented performance
specification given below.
In addition to the methods specified in assignments 4, 5, and 7, the FMap class shall provide the following public dynamic method:
accept: Visitor -> FMap
If m is an FMap and visitor is a Visitor, then m.accept(visitor).equals(m2) where m2 is the FMap computed by
FMap m2 = FMap.empty();
for (K k : m) {
V v = visitor.visit (k, m.get (k));
m2 = m2.put (k, v);
If m is an FMap created using the 1-argument version of FMap.empty, then both m and m2 must satisfy the average-case efficiency requirements stated below.
Performance requirements:
Suppose c is a comparator that runs in O(1) time, m is an FMap that has been created by adding random key-value pairs to FMap.empty(c), iter is an iterator obtained by evaluating m.iterator(), n is
m.size(), and v is a Visitor such that v.visit(K,V) runs in constant time. Then
m.put(k,v) should run in O(lg n) time
m.isEmpty() should run in O(1) time
m.size() should run in O(1) time
m.containsKey(k) should run in O(lg n) time
m.containsValue(k) should run in O(n) time
m.get(k) should run in O(lg n) time
m.iterator() should run in O(n) time
iter.hasNext() should run in O(1) time
iter.next() should run in O(1) time
m.accept(v) should run in O(n*lg n) time
where all of those times are for the average case.
The average efficiency will be evaluated probabilistically by drawing keys and values at random from very large finite sets of keys and values.
If m.containsKey(k), then there shall exist exactly one v such that m.accept(visitor) results in a call to visitor.visit(k,v). That v shall be m.get(k).
If m.containsKey(k) and m.get(k) is v, then m.accept(visitor) shall result in exactly one call to visitor.visit(k,v).
If a comparator was provided for the empty FMap from which an FMap m was built up, then the visitor shall visit keys in the order specified by the comparator (with lesser keys being visited before
greater keys).
If no comparator was provided, then the ordering of the calls to visitor.visit(k,v) is unspecified. In particular, some of those calls to visitor.visit(k,v) may be concurrent. Clients who use the
accept(Visitor) method are responsible for any synchronization that may be necessary.
The specification of hashCode() from assignment 4 is strengthened as follows.
If m1 and m2 are values of the FMap ADT, and
then m1.hashCode() == m2.hashCode().
If m1 and m2 are values of the FMap ADT, and
! (m1.equals(m2))
then m1.hashCode() is unlikely to be equal to m2.hashCode().
Note: The word “unlikely” will be interpreted as follows. For every type K and V, if both m1 and m2 are selected at random from a set of FMap values such that for every non-negative integer n and int
value h the probability of a randomly selected FMap m having
n == m.size() is
P(n) = 1/(2^(n+1))
and for each key k such that m.containsKey(k) the probability that
h == k.hashCode() is at most 1/5
and for each value v such that v.equals(m.get(k)) the probability that
h == v.hashCode() is at most 1/5
and the three probabilities above are independent
then the probability of m1.hashCode() == m2.hashCode() when m1 and m2 are not equal is less than 10%. | {"url":"https://codingprolab.com/answer/cs-3500-assignment-8/","timestamp":"2024-11-14T00:57:46Z","content_type":"text/html","content_length":"114171","record_id":"<urn:uuid:54e8e41b-ca58-473b-b92c-9174145bc78b>","cc-path":"CC-MAIN-2024-46/segments/1730477028516.72/warc/CC-MAIN-20241113235151-20241114025151-00542.warc.gz"} |
The ACO seminar (2000-2001)
The current organizers of the ACO Seminar are Jochen Könemann (jochen@cmu.edu) and Ojas Parekh (odp@cmu.edu). If you have questions or suggestions about the seminar, or want to be a speaker, please
contact the organizers.
Fall 2000 and Spring 2001: The seminar is in Wean 7220 on Friday's from 3:30pm to 5:00pm.
[See what is the NEXT SEMINAR.]
TITLE: Improved Approximations for Tour and Tree Covers.
SPEAKER: Ojas Parekh.
WHEN: Friday, Sep. 22, 3:30pm-5:00pm.
WHERE: Wean 7220.
• Jochen Könemann, Goran Konjevod, Ojas Parekh and Amitabh Sinha: Improved Approximations for Tour and Tree Covers. (Postscript).
A tree (tour) cover of an edge-weighted graph is a set of edges which comprises a tree (closed walk) and covers every edge in the graph. Arkin, Halldórsson and Hassin [1] give approximation
algorithms with performance ratios of 3.55 and 5.5 for the tree and tour cover problems respectively. We present algorithms which employ polyhedral rounding techniques to provide 3-optimal solutions
for both problems.
This is joint work with Jochen Könemann, Goran Konjevod and Amitabh Sinha.
1. Arkin, Halldórsson, and Hassin. Approximating the tree and tour covers.
TITLE: A Brief Introduction to Wavelets.
SPEAKER: Paul Komarek.
WHEN: Friday, Oct. 13 , 3:30pm-5:00pm.
WHERE: Wean 7220.
JosephFourier's method of harmonic analysis has enjoyed great success and popularity during its two centuries of existence. In 1965 Cooley and Tukey formalized techniques for reducing the discrete
Fourier transform's O(n^2) complexity to O(n log(n)). This ``Fast Fourier Transform'' (FFT) allowed cheap harmonic analysis for all of science and industry.
In the early 1990s, a competing method of harmonic analysis gained quick popularity. This new method employs a basis of ``wavelets'', functions which consisted of one ``wave'' surrounded by vanishing
tails. The non-periodicity of wavelets allow them to represent non-periodic functions more efficiently than Joseph Fourier's sinusoidal basis, and provide an (arguably) better time versus frequency
trade-off than the FFT.
I intend to briefly review properties of Fourier methods and introduce wavelet bases. The discrete wavelet transform will be contrasted with the FFT and, technology willing, their strengths and
weaknesses demonstrated. I will finish with a discussion of some applications in which wavelets excel. I hope to make this talk suitable for anyone with a basic proficiency in calculus.
TITLE: Approximation for the maximum quadratic assignment problem.
SPEAKER: Amitabh Sinha.
WHEN: Friday, Oct. 20, 3:30pm-5:00pm.
WHERE: Wean 7220.
PAPERS: E. Arkin, R. Hassin, M.Sviridenko; Approximation for the maximum quadratic assignment problem , SODA 2000.
Given 3 nxn nonnegative symmetric matrices A,B,C, the max quadratic assignment problem asks for a permutation \pi of {1,2,...,n} which maximizes \Sum_{i\ne j} a_{\pi(i),\pi(j)}b_{ij} + \Sum_i c_{i,\
pi(i)} The authors provide a fast and elegant combinatorial algorithm which is a 1/4-approximation to the problem when the matrix B satisfies the triangle inequality. This framework can be used to
model various graph theoretic problems, such as MAX-TSP, MAX-BISECTION, MAX-PERFECT-MATCHING, etc. While the problem in general and each of the special cases have been studied extensively (and the
special cases can usually be approximated much better), this paper is the best known polynomial time approximation for the problem.
TITLE: Investigation of the Irregularity Strength of Trees
SPEAKER: David Kravitz.
WHEN: Friday, Nov. 17, 3:30pm-5:00pm.
WHERE: Wean 7220.
PAPERS: Here's a copy of David's bachelor thesis: dkravitz.ps
The irregularity strength of a graph has recently become of interest to both number theorists and graph theorists. Here we deal with a specific conjecture relating to the irregularity strength of
trees, which claims that the strength is essentially either the number of vertices of degree 1 or the average of the number of vertices of degree 1 and degree 2. After making the reader familiar with
the necessary vocabulary in graph theory and defining irregularity strength, I present a brief survey of known results. Then, I describe methods used to try to prove the conjecture, and how this led
to an infinite series of counter examples to refute it. The idea behind the counter examples arose from extensive computer experimentation. This is joint work with Felix Lazebnik and Gary Ebert.
Title: Average Case Analysis of Random Hamilton Cycle and Travelling Salesman Problems: a survey of results and problems.
Speaker: Alan Frieze.
Where: Wean 7220
When: Friday, Dec. 08, 3:30pm -5:00 pm
Title: On the Rank of Mixed 0,1 Polyhedra
Speaker: Yanjun Li.
Where: Wean 7220
When: Friday, Dec. 15, 3:30pm -5:00 pm
Eisenbrand and Schulz showed recently (IPCO 99) that the maximum Chv\'atal rank of a polytope in the $[0,1]^n$ cube is bounded above by $O(n^2 logn)$ and bounded below by $(1+\epsilon)n$ for some $\
epsilon > 0$. It is well known that Chv\'atal's cuts are equivalent to Gomory's fractional cuts, which are themselves dominated by Gomory's mixed integer cuts. What do these upper and lower bounds
become when the rank is defined relative to Gomory's mixed integer cuts? An upper bound of $n$ follows from existing results in the literature. In this note, we show that the lower bound is also
equal to $n$. We relate this result to bounds on the disjunctive rank and on the Lov\'asz-Schrijver rank of polytopes in the $[0,1]^n$ cube. The result still holds for mixed 0,1 polyhedra with $n$
binary variables. This is a joint work with G\'erard Cornu\'ejols.
Title: The Plank Problem from the Viewpoint of Hypergraph Matchings and Covers
Speaker: Ron Holzman
Where: Wean 7220
When: Friday, 2/9/2001, 3:30-5:00pm
The "plank problem", open since 1950, is the following conjecture. Suppose that the convex set K is contained in the unit cube of R^n, and meets all of the cube's facets. Suppose further that a
collection of planks is given (where "plank" means a strip determined by two parallel hyperplanes orthogonal to one of the axes), and the union of these planks covers K. Then the sum of the planks'
widths must be at least 1. We propose an analogy between this problem and familiar concepts from the theory of hypergraphs. This leads to a proof of the conjecture in certain special cases, and to a
formulation of a fractional counterpart of the conjecture (corresponding to a linear programming relaxation of the problem), which we are able to verify to within a factor of 2. This is joint work
with Ron Aharoni, Michael Krivelevich and Roy Meshulam.
Title: A shorter proof of Guenin's characterization of Weakly bipartite graphs
Speaker: Giacomo Zambelli.
Where: Wean 7220
When: Friday, 2/16/2001, 3:30-5:00pm
We give a proof of Guenin's theorem charchterizing weakly bipartite graphs by not having an odd-$K_5$ minor. The proof curtails the technical and case-checking parts of Guenin's original proof.
Title: Analysis of SRPT scheduling: Investigating Unfairness
Speaker: Nikhil Bansal.
Where: Wean 7220
When: Friday, 2/23/2001, 3:30-5:00pm
The Shortest-Remaining-Processing-Time (SRPT) scheduling policy has long been known to be optimal for minimizing the mean response time (the mean time from when a job arrives until it finishes
execution) Despite this fact, SRPT scheduling is rarely used in practice. It is believed that the performance improvements of SRPT over other scheduling policies with respect to mean response time
stem from the fact that SRPT unfairly penalizes the large jobs in order to help the small jobs. This belief has caused people to instead adopt ``fair'' scheduling policies such as Processor-Sharing
In this talk, we will investigate the problem of unfairness in SRPT. We will show that surprisingly, SRPT is not unfair towards large jobs, particularly when the service requirements of jobs resemble
those observed in the real world (i.e. the distribution of service requirements is heavy-tailed).
Title: A \Sigma_2-complete optimization problem and related hardness of approximation results.
Speaker: Abie Flaxman.
Where: Wean 7220
When: Friday, 3/16/2001, 3:30-5:00pm
The recent thesis of Chris Umans proves several interesting problems relating to logic minimization are "harder than NP-hard". One such problem is MIN-DNF: Given a DNF formula \phi and an integer k,
we accept if there exists a formula \psi equivalent to \phi with at most k literals. We will present Umans' proof that MIN-DNF is \Sigma_2-complete, and outline a technique for proving hardness of
approximation results that does not require the PCP theorem.
Speaker: Gregory Sorkin, IBM T.J. Watson
Topic: Optimal myopic algorithms for random 3-SAT
When: Thursday, 3/22/01, 2:30-4:00pm
Where: Wean 4625
Let $F_3(n,m)$ be a random 3-SAT formula formed by selecting uniformly, independently, and with replacement, $m$ clauses among all $8 \binom{n}{3}$ possible 3-clauses over $n$ variables. It has been
conjectured that there exists a constant $r_3$ such that for any $\e >0$, $F_3(n,(r_3-\e)n)$ is almost surely satisfiable, but $F_3(n,(r_3+\e)n)$ is almost surely unsatisfiable. The best lower bounds
for the potential value of $r_3$ have come from analyzing rather simple extensions of unit-clause propagation. Recently, it was {shown~\cite{pi}} that all these extensions can be cast in a common
framework and analyzed in a uniform manner by employing differential equations. Here, we determine optimal algorithms expressible in that framework, establishing $r_3 > 3.26$. We extend the analysis
via differential equations, and make extensive use of a new optimization problem we call ``\mbox{max-density} \mbox{multiple-choice} knapsack''. The structure of optimal knapsack solutions elegantly
characterizes the choices made by an optimal algorithm.
Speaker: Luis Zuluaga.
Topic: Optimal semi-parametric bounds for the payoff of European options via semidefinite programming. When: Friday, 4/13/01, 3:30-5:00pm
Where: Wean 7220
Bounds for the payoff of European options that depend only on the moments of the underlying asset(s) and not on its entire distribution are called semi-parametric bounds. These types of bounds have
been used to estimate option prices under the risk-neutral measure pricing theory. They also yield a relationship between option prices and the true distribution of the underlying asset(s). Motivated
by previous work by Bertsimas and Popescu, we show how to obtain optimal semi-parametric bounds for European options whose payoff depends on two (possible correlated) assets by solving a semidefinite
program. Also, we characterize the feasibility set for these problems, a question that falls into the theory of generalized moment problems. This connection between semidefinite programming and
moment problems is closely related to the classical 17th Hilbert problem regarding definite forms. Finally, we show some numerical results for European exchange options. These results are obtained by
solving the corresponding semidefinite programming formulation.
Speaker: Jozef Slokan.
Topic: Small Cliques in 3-uniform Hypergraphs
When: Friday, 4/27/01, 3:30-5:00pm
Where: Wean 7220
Many applications of Szemer\'edi's Regularity Lemma are based on the following technical fact: If $G$ is a $k$-partite graph with $V(G) = \bigcup_{i=1}^{k} V_i$, $|V_i|=n$ for all $i\in [k]$, and all
pairs $\{V_i, V_j\}$, $1\le i < j\le k$, are $\epsilon$-regular of density $d$, then $G$ contains $d^{{k \choose 2}}n^k(1+f(\epsilon))$ cliques $K_{k}^{(2)}$, where $f(\epsilon)$ tends to~$0$ as $\
epsilon$ tends to~$0$.
B.~Nagle and V.~R\"odl established the analogous statement for $3$-uniform hypergraphs. In this talk, we present a proof of the same result by a~simpler argument.
This is a joint work with Y. Peng and V. R\"odl.
Speaker: Bjarni Halldorsson.
Topic: On the Approximability of the Minimum Test Collection Problem
When: Friday, 5/4/01, 3:30-5:00pm
Where: Wean 7220
Paper: TestCollection.ps
The minimum test collection problem is defined as follows. Given a ground set $\mathcal{S}$ and a collection $\mathcal{C}$ of tests (subsets of $\mathcal{S}$), find the minimum subcollection $\
mathcal{C'}$ of $\mathcal{C}$ such that for every pair of elements $(x,y)$ in $\mathcal{S}$ there exists a test in $\mathcal{C'}$ that contains exactly one of $x$ and $y$. In this way, the incidence
of each element of $\mathcal{S}$ to the tests in $\mathcal{C'}$ provides a unique signature for it. It is well known that the greedy algorithm gives a $1+2\ln n$ approximation for the test collection
problem where $n = |\mathcal{S}|$, the size of the ground set. In this paper, we show that this algorithm is close to the best possible, namely that there is no $o(\log n)$-approximation algorithm
for the test collection problem unless $P = NP$.
We also give new approximation algorithms for this problem in the case when all the tests have a small cardinality, significantly improving the performance guarantee achievable by the greedy
algorithm. In particular, for instances with test sizes at most $k$ we derive an $O(\log k)$ approximation. For the special case when $k=2$ we give a $\frac{7}{6} + \e$ approximation.
This is joint work with M.M. Halld\'orsson and R. Ravi
Documents linked from this page can be read using PostScript Ghostview or GSview, Adobe Acrobat Reader, Sun StarOffice Impress or Microsoft PowerPoint Viewer. | {"url":"http://www.cs.cmu.edu/~aco/seminar20002001.html","timestamp":"2024-11-10T14:47:32Z","content_type":"text/html","content_length":"18299","record_id":"<urn:uuid:ea404efb-4eef-41e1-9f70-654cc5b39763>","cc-path":"CC-MAIN-2024-46/segments/1730477028187.60/warc/CC-MAIN-20241110134821-20241110164821-00389.warc.gz"} |
functorch.grad(func, argnums=0, has_aux=False)[source]¶
grad operator helps computing gradients of func with respect to the input(s) specified by argnums. This operator can be nested to compute higher-order gradients.
☆ func (Callable) – A Python function that takes one or more arguments. Must return a single-element Tensor. If specified has_aux equals True, function can return a tuple of single-element
Tensor and other auxiliary objects: (output, aux).
☆ argnums (int or Tuple[int]) – Specifies arguments to compute gradients with respect to. argnums can be single integer or tuple of integers. Default: 0.
☆ has_aux (bool) – Flag indicating that func returns a tensor and other auxiliary objects: (output, aux). Default: False.
Function to compute gradients with respect to its inputs. By default, the output of the function is the gradient tensor(s) with respect to the first argument. If specified has_aux equals True
, tuple of gradients and output auxiliary objects is returned. If argnums is a tuple of integers, a tuple of output gradients with respect to each argnums value is returned.
Example of using grad:
>>> from functorch import grad
>>> x = torch.randn([])
>>> cos_x = grad(lambda x: torch.sin(x))(x)
>>> assert torch.allclose(cos_x, x.cos())
>>> # Second-order gradients
>>> neg_sin_x = grad(grad(lambda x: torch.sin(x)))(x)
>>> assert torch.allclose(neg_sin_x, -x.sin())
When composed with vmap, grad can be used to compute per-sample-gradients:
>>> from functorch import grad
>>> from functorch import vmap
>>> batch_size, feature_size = 3, 5
>>> def model(weights, feature_vec):
>>> # Very simple linear model with activation
>>> assert feature_vec.dim() == 1
>>> return feature_vec.dot(weights).relu()
>>> def compute_loss(weights, example, target):
>>> y = model(weights, example)
>>> return ((y - target) ** 2).mean() # MSELoss
>>> weights = torch.randn(feature_size, requires_grad=True)
>>> examples = torch.randn(batch_size, feature_size)
>>> targets = torch.randn(batch_size)
>>> inputs = (weights, examples, targets)
>>> grad_weight_per_example = vmap(grad(compute_loss), in_dims=(None, 0, 0))(*inputs)
Example of using grad with has_aux and argnums:
>>> from functorch import grad
>>> def my_loss_func(y, y_pred):
>>> loss_per_sample = (0.5 * y_pred - y) ** 2
>>> loss = loss_per_sample.mean()
>>> return loss, (y_pred, loss_per_sample)
>>> fn = grad(my_loss_func, argnums=(0, 1), has_aux=True)
>>> y_true = torch.rand(4)
>>> y_preds = torch.rand(4, requires_grad=True)
>>> out = fn(y_true, y_preds)
>>> > output is ((grads w.r.t y_true, grads w.r.t y_preds), (y_pred, loss_per_sample))
Using PyTorch torch.no_grad together with grad.
Case 1: Using torch.no_grad inside a function:
>>> def f(x):
>>> with torch.no_grad():
>>> c = x ** 2
>>> return x - c
In this case, grad(f)(x) will respect the inner torch.no_grad.
Case 2: Using grad inside torch.no_grad context manager:
>>> with torch.no_grad():
>>> grad(f)(x)
In this case, grad will respect the inner torch.no_grad, but not the outer one. This is because grad is a “function transform”: its result should not depend on the result of a context manager
outside of f. | {"url":"https://pytorch.org/functorch/0.2.1/generated/functorch.grad.html","timestamp":"2024-11-04T05:16:13Z","content_type":"text/html","content_length":"31077","record_id":"<urn:uuid:5e121b19-c24b-4276-9ee5-17a409047321>","cc-path":"CC-MAIN-2024-46/segments/1730477027812.67/warc/CC-MAIN-20241104034319-20241104064319-00816.warc.gz"} |
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Common Rules of Thumb in Geotechnical Engineering
Reading Time: 12 minutes
Geotechnical engineering plays a crucial role in ensuring the stability and safety of structures by understanding and manipulating the unique properties of soil and rock. In this exploration of
“Common Rules of Thumb in Geotechnical Engineering,” we will look into the key principles and guidelines that serve as practical benchmarks for professionals in the field. Whether you’re a seasoned
geotechnical engineer or a curious enthusiast, this guide aims to shed light on fundamental rules that shape the foundation of successful geotechnical practices.
Common Rules of Thumb in Geotechnical Engineering
A rule of thumb is defined as a practical or approximate way of doing or measuring something. The origins of the term “rule of thumb” are obscure. Apparently, Roman bricklayers used the tip of the
thumb from the knuckle as a unit of measure. Brewers used their thumbs to test the temperature of fermenting ale. In the Middle Ages, a man was permitted to beat his wife with a cane no thicker than
his thumb. Nowadays, the rule of thumb implies a rough estimate based on experience rather than formal calculation.
Geotechnical engineering is essentially a mechanical science which has a strong theoretical basis. Nevertheless, many geotechnical engineers use simple relationships – rules of thumb – in routine
design (to obtain soil parameters and design foundations). Some of these rules of thumb are based on sound theory and so should be generally applicable; others are purely empirical and so are
applicable only within the range of the data from which they were derived. A classification for rules of thumb was suggested by Wroth (1984).
Rules of thumb aid geotechnical engineers a lot in practice. With rules of thumb, there is no need to carry out certain unnecessary costly, and time-consuming tests. However, each geotechnical
engineer has the liberty to use their favourite rules of thumb.
Conditions for Successful Application of Rules of Thumbs
As stated above, certain rules of thumb are more valid within the context upon which they were designed, however, certain rules exist to go out of this context:
(a) Based on physical appreciation of why the properties can be expected to be related.
(b) Set against a background of theory, however, idealised this may be.
(c) Expressed in terms of dimensionless variables so that advantage can be taken of scaling laws of continuum mechanics.
Rules of Thumb in Geotechnical Engineering
Correlation between the Angle of Internal Friction and SPT number, N
ϕ = friction angle
N = SPT value
This particular equation ignores particle size. Most tests are done on medium to coarse sands, hence, there is a modification to account for other soil sizes as listed below:
How to convert SPT blow count from one Hammer to another
Most geotechnical correlations were based on SPT (N) values obtained during the 1950s. However, drill rigs today are much more efficient than the drill rigs of the 1950s. If the hammer efficiency is
high, then to drive the spoon 300 mm (1 ft), the hammer would require a lesser number of blows compared to a less efficient hammer. It is possible to convert the blow count from one hammer to
another. The following expression gives the conversion from a 60% efficient hammer to a 70% efficient hammer.
Correlations between CPT and SPT
Cone penetration test (CPT) and standard penetration tests are two common tests used to delineate subsurface geotechnical conditions. Both methods have different strategies, however, CPT has many
benefits over SPT as it provides a broader characterisation of the soil, because three different parameters are measured instead of just one in the case of SPT. In addition, CPT presents real-time
results in the field when compared to SPT. There’s no need to transfer soil samples to a lab and then wait for days or even weeks to get the analysis report. Instead, the report can be issued as soon
as the cone has been pulled out of the ground and before the unit is moved to the next location. Finally, CPT does not require a borehole for the testing, so there are no cuttings and no spoilage.
SPT is more extensively used in the USA while CPT is more common in Europe. In situations where there is a need to convert results from CPT to SPT, empirical correlations exist that can enable this.
These are outlined in Table 1.
Table 1: CPT-SPT correlations for clays and sands (Robertson et al., 1983)
Soil Type Mean grain size (D[50]) measured in mm Q[c]/N
Silt 0.001 100
Silty-clay 0.005 170
Clayey-silt 0.01 210
Sandy-silt 0.05 300
Silty-sand 0.1 400
Sand 0.5 570
1.0 700
Q[c] = CPT value (kPa)
N = SPT value
D[50] = size of the sieve that would pass 50% of the soil
Relationship between US sieve sizes and British sieve sizes
The relationship between US sieve sizes and their numbers and British sieve sizes and their numbers are outlined in the table below.
Table 2: US sieve sizes versus British sieve sizes
US sieve number Mesh sizes British sieve number Mesh sizes
4 4.75 8 2.057
6 3.35 16 1.003
8 2.36 30 0.500
10 2.00 36 0.422
12 1.68 52 0.295
16 1.18 60 0.251
20 0.85 85 0.178
30 0.60 100 0.152
40 0.425 200 0.076
50 0.30 300 0.053
60 0.25
80 0.18
100 0.15
200 0.075
270 0.053
Size Ranges for Soils and Gravels
The typical ranges of soil types and their sizes are outlined in the table below.
Table 3: Soil types and sizes
Soil Size (inches) Size (mm)
Boulders >6 >150
Cobbles 3 – 6 75 – 150
Gravel 0.187 – 3 4.76 – 75
Sand 0.003 – 0.187 0.074 – 4.76
Silt 0.00024 – 0.003 0.006 – 0.074
Clay 0.00004 – 0.00008 0.001 – 0.002
Colloids <0.00004 <0.001
Typical Specific Gravity of Different Types of Soils
The typical specific gravity of the different soil types is outlined in the table below:
Table 4: Typical specific gravity of different soil types
Soil Specific gravity
Gravel 2.65 – 2.68
Sand 2.65 – 2.68
Silt (inorganic) 2.62 – 2.68
Organic clay 2.58 – 2.65
Inorganic clay 2.68 – 2.75
Bearing capacity of Soils
The bearing capacity of soils is usually determined analytically using methods provided by Terzaghi, Meyerhof, Hansen, and Vesic. However, certain rules of thumb are applicable in the determination
of the bearing capacity of soils and these are outlined below:
Bearing Capacity in Medium to Coarse Sands
The strength of sandy soils is dependent upon the friction angle. The bearing capacity in coarse to medium sands can be obtained using the average SPT (N) value as outlined below. This formula is
applicable on the condition that the average SPT (N) value is determined below the bottom of the footing at a depth that is equal to the width of the footing and that the soil within this range is
medium to coarse sand. If the average SPT is less than 10, the soil should be compacted.
Allowable bearing capacity (q[a]) of coarse to medium sands = 9.6 N[average] (kPa) – not to exceed 575 kPa
Bearing Capacity in fine sands
Allowable bearing capacity (q[a]) of coarse to medium sands = 9.6 N[average] (kPa) – not to exceed 380 kPa
Design of Gravel filters
Gravel filters stop any soil particles from entering the pipe. At the same time, the gravel allows ample water flow. If fine gravel is used, water flow can be reduced. On the other hand, large size
gravel can allow soil particles to pass through and eventually clog the drain pipe. Empirically it has been found that if the gravel size is selected by the two rules given below, very little soil
would transport through the gravel filter and at the same time let water flow smoothly.
RULE 1: To block soil from entering the pipe, the size of the gravel should be:
D[15 ](gravel) < 5 x D[85] (soil)
D[15] size means that 15% of particles of a given soil or gravel would pass through the D15 size of that particular soil or gravel. The D[85] size means that 85% of particles of a given soil or
gravel would pass through the D[85] size of that particular soil or gravel. These values are obtained through sieve analysis.
RULE 2: To let water flow, the size of the gravel should be:
D[50] (gravel) > 25 x D[50] (soil)
D[15], D[50], D[85] size means that 15%, 50%, 85% of particles of a given soil or gravel would pass through the D[15], D[50], D[85] size respectively of that particular soil or gravel. These values
are obtained through sieve analysis.
Geotextile Filter Design
Geotextile filters are becoming increasingly popular in drainage applications. The ease of use, economy, and durability of geotextile filters have made them the number one choice of many engineers.
There are two types of geotextiles (woven and non-woven) available in the market. For sandy soils, both woven and non-woven geotextiles can be used while for clay soils, non-woven geotextile is more
applicable. The geotextile is wrapped around stones (stones act only as a medium to transport water) and backfilled with the original soil.
a. Geotextile Wrapped Granular Drains (Sandy Surrounding Soils) – Woven geotextiles
The task of stopping soil from entering the drain is done by the geotextile. Gravel is wrapped with a geotextile to improve the performance. The geotextile filters the water, and the gravel acts as
the drain. Equations have been developed for the two types of flow: one-way flow and two-way flow (also called alternating flow).
i. One Way Flow
In the case of one-way flow, the flow of water is always in one direction across the geotextile. The following equation for geotextiles in sandy soil developed by Zitscher (1975) has been applicable
in this case
H[50] (geotextile) < 1.7 to 2.7 x D[50] (soil) (for one-way flow for sandy soils)
Where H[50] indicates that 50% of the holes in the geotextile are smaller than the H[50] size. D[50] (soil) indicates that 50% of soil particles are smaller than the D[50] size.
ii. Two Way Flow (Alternating Flow)
In the case of two-way flow, water goes through the geotextile in both directions. In heavy rain, water enters the drain from the top, flows into the drain, and then flows out of the drain to the
surrounding soil. If flow through the geotextile is possible in both directions, a different set of equations needs to be used (Zitscher, 1975).
H[50](geotextile) < (0.5 to 1.0) x D[50] (soil) (for two-way flow for sandy soils)
b. Geotextile Wrapped Granular Drains (Clayey Surrounding Soils) – Non-woven geotextiles
The theory behind using geotextile-wrapped granular drains is that the direction of flow is not significant for cohesive soils since the flow is not as rapid as in sandy soils. Most engineers prefer
to use non-woven geotextiles for cohesive soils. For these soils, use the equation (Zitscher, 1975):
H[50](geotextile) < (25 to 37) x D[50] (soil) (one-way and two-way flow for clayey soils)
Allowable Bearing Capacity of Raft Foundations
Three prominent foundation engineers, Peck, Hanson, and Thorburn (1974), proposed the following method to design raft foundations. The following equation can be used to find the allowable pressure in
a raft.
q[allowable Raft] = 23.595 x N x C[w] x γ x D[f]
q[allowable Raft] = allowable average pressure in the raft, given in kN/m^2
N = average SPT (N) value to a depth of 2B, where B is the lesser dimension of the raft.
D[w] = depth to groundwater measured from the ground surface (m)
D[f] = depth to the bottom of the raft measured from the ground surface (m)
γ = total density of the soil (kN/m^3)
Once the allowable load is determined, it is possible to determine the total quantity of load to be carried by the raft as follows:
The total quantity of load to be carried by raft = q[allowable Raft ]x area of the raft
Rock Quality Designation (RQD)
Rock quality designation is obtained through the following process:
i. Arrange all rock pieces as best as possible to simulate the ground conditions.
ii. Measure all rock pieces greater in length than 100 mm (4 inches).
iii. Estimate RQD with the expression below:
RQD (0-25%) = very poor
RQD (25-50%) = poor
RQD (50-75%) = fair
RQD (75-90%) = good
RQD (90-100%) = excellent
Richter Magnitude Scale (M)
The Richter scale also called the Richter magnitude scale, Richter’s magnitude scale, and the Gutenberg–Richter scale, is a measure of the strength of earthquakes, developed by Charles Francis
Richter and presented in his landmark 1935 paper, which he called it the “magnitude scale”. The Richter magnitude scale is a logarithmic scale that follows the relation:
M = logA – logA[o] = log (A/A[o])
M = Richter magnitude scale
A = maximum trace amplitude during the earthquake
A[0] = standard amplitude
Here, a standard value of 0.001 mm is used for comparison. This corresponds to a very small earthquake.
Soil Resistance to Liquefaction
Any soil that has an SPT value higher than 30 will not liquefy. Resistance to liquefaction of a soil depends on its strength measured by SPT value. Researchers have found that resistance to soil
liquefaction depends on the content of fines as well. The following equation (which can only be used for clean sands with a fine content of less than 5%) is applicable.
CRR[7.5] = soil resistance to liquefaction for an earthquake with a magnitude of 7.5 Richter
(N[1])[60] = the standard penetration value corrected to a 60% hammer and an overburden pressure of 100 kPa.
(N[1])[60] =N[m] x C[N] x C[E] x C[B] x C[R]
N[m] = SPT value measured in the field
C[N] = overburden correction factor
Pa = 100 kPa
σ’= effective stress of soil at the point of measurement
C[E] = energy correction factor for the SPT hammer
for donut hammers, C[E] = 0.5 to 1.0
for trip-type donut hammers, C[E] = 0.8 to 1.3
C[B] = borehole diameter correction
for borehole diameters of 65 mm to 115 mm use C[B] = 1.0
for a borehole diameter of 150 mm use C[B] = 1.05
for a borehole diameter of 200 mm use C[B] = 1.15
C[R] = rod length correction
Rods attached to the SPT spoon exert their weight on the soil. Longer rods exert a higher load on soil and in some cases the spoon sinks into the ground due to the weight of the rods, even without
any hammer blows. Hence, a correction is made to account for the weight of rods.
for rod length < 3 m, use C[R] = 0.75
for rod length 3 m to 4 m, use C[R] = 0.8
for rod length 4 m to 6 m, use C[R] = 0.85
for rod length 6 m to 10 m, use C[R] = 0.95
for rod length 10 m to 30 m, use C[R] = 1.0
Exceptions: for other magnitude of earthquakes, and sands with fine content of more than 5%, a correction factor has to be applied.
Correction for Magnitudes of Earthquake
A correction factor is proposed for earthquakes of magnitudes of 7.5. The factor of safety (F.O.S.) is given by:
CRR[7.5] = resistance to soil liquefaction for a magnitude of 7.5 earthquake
CSR = cyclic stress ratio (which is a measure of the impact due to the earthquake load).
a[max] = peak horizontal acceleration at the ground surface
r = total stress at the point of concern or’= effective stress at the point of concern
r[d] = stress reduction coefficient (r[d] = 1.0 – 0.00765Z for Z < 0.15 m or r[d] = 1.174 – 0.0267Z for 9.15 m < Z < 23 m)
Z = depth to the point of concern in meters
For earthquakes of other magnitudes other than 7.5, the correction factor is given by the following equation:
Where: MSF- magnitude scaling factor (see Table 4)
Table 4: Magnitude scaling factors
Earthquake magnitude MSF suggested by Idris (1990) MSF suggested by Andrus and Stokoe (2000)
5.5 2.2 2.8
6.0 1.76 2.1
6.5 1.44 1.6
7.0 1.19 1.25
7.5 1.00 1.00
8.0 0.84 –
8.5 0.72 –
Correction Factor for Content of Fines
For soils with higher fine content, the corrected (N[1])[60] value should be used in CRR[7.5] equation above. To apply the correction, first compute (N[1])[60, ]then use the expression below to
compute the corrected value:
(N[1])[60C] = a + b (N[1])[6o]
Where:(N[1])[60C] = corrected (N[1])[6o] value
Andrus, R. D. and Stokoe, K. H. (2000). “Liquefaction resistance of soils from shear wave velocity”. ASCE Journal of Geotechnical and Geoenvironmental Engineering, vol 126, no. 11, 1015-1025.
Atkinson, J. (2008) Rules of thumb in geotechnical engineering Proc. 18th NZGS Geotechnical Symposium on Soil-Structure Interaction. Ed. CY Chin, Auckland
Idris, I. M. (1990). “Response of soft soil sites during earthquakes”. Proc. Bolten Seed Memorial Symp., vol. 2 Bi-Tech Publishers Ldt., Vancouver, 273-290.
Peck, R. B., Hanson, W. E., and Thornburn, T. H. 1974. Foundation engineering. New York: John Wiley & Sons.
Rajapakse, R. Geotechnical Engineering Calculations and Rules of Thumb.
Robertson, P. K., Campanella, R. G., and Wightman, A. 1983. SPT-CPT Correlations. ASCE Geotechnical Engineering Journal 109( 11): 1449-1459
Wroth, C.P. (1984). 24th Rankine Lecture: The interpretation of in situ soil tests. Geotechnique, Vol. 34, No 4, pp 449-488.
Zitscher, F. F. 1975. Recommendations for the use of plastics in soil and hydraulic engineering. Die Bautechnik, 52(12):397-402.
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Stata代写 | ECON 400: Introduction to Econometrics - EasyDue™️
ECON 400: Introduction to Econometrics
Problem Set #5
Due: 03/09/2020
Problem #1:
You have collected 14,925 observations from the Current Population Survey. There are 6,285
females in the sample, and 8,640 males. The females report a mean of average hourly earnings of $16.50 with a standard deviation of $9.06. The males have an average of $20.09 and a standard deviation
of $10.85. The overall mean average hourly earnings is $18.58.
a) Using the t-statistic for testing differences between two means (section 3.4 of your textbook), decide whether or not there is sufficient evidence to reject the null hypothesis that females and
males have identical average hourly earnings.
b) You decide to run two regressions: first, you simply regress average hourly earnings on an
intercept only. Next, you repeat this regression, but only for the 6,285 females in the sample.
What will the regression coefficients be in each of the two regressions?
c) Finally you run a regression over the entire sample of average hourly earnings on an
intercept and a binary variable DFemme where this variable takes on a value of 1 if the
individual is a female, and is 0 otherwise. What will be the value of the intercept? What will
be the value of the coefficient of the binary variable?
d) What is the standard error on the slope coefficient? What is the t-statistic?
e) Had you used the homoskedasticity-only standard error in (d) and calculated the t-statistic, how would you have had to change the test-statistic in (a) to get the identical result?
Problem #2:
Labor economists studying the determinants of women’s earnings discovered a puzzling empirical result. Using randomly selected employed women, they regressed earnings on the women’s number of
children and a set of control variables (age, education, occupation, and so forth). They found that women with more children had higher wages, controlling for these other factors. Explain how sample
selection might be the cause of this result. (Hint: Notice that women who do not work outside the home are missing from the sample.) [This empirical puzzle motivated James Heckman’s research on
sample selection that led to his 2000 Nobel Prize in Economics.]
Problem #3:
Assume that the regression model ???? = ??0 + ??1???? + ???? satisfies the least squares assumptions
in You and a friend collect a random sample of 300 observations on Y and X.
a. Your friend reports that he inadvertently scrambled the X observations for 20% of the
sample. For these scrambled observations, the value of X does not correspond to ???? for the i
th observation; rather, it corresponds to the value of X for some other observation. The
measured value of the regressor, ??�?? , is equal to???? for 80% of the observations, but it is
equal to a randomly selected ???? for the remaining 20% of the observations. You regress ????on
??�?? . Show that ??�??
�1� = 0.8??1 .
b. Explain how you could construct an unbiased estimate of ??1 using the OLS estimator in (a).
c. Suppose now your friend tells you that the X’s were scrambled for the first 60 observations
but that the remaining 240 observations are correct. You estimate β1β1 by regressing Y on X, using only the correctly measured 240 observations. Is this estimator of ??1better than the estimator you
proposed in (b)? Explain.
Problem #4:
For this problem, you need to copy and run missing.do and missing.dta fromCanvas. This question illustrates some basic data manipulation techniques for dealing with missing data. Suppose a researcher
is estimating the effects of demographic factors on an individual’ income using the General Social Survey of 1991. Her research assistant has tried to prepare the data for her to generate the
estimates, and her do file includes some hard to understand comments. Your job is to help decipher exactly what was done by the research assistant and to help interpret her results.
a) Based on the frequencies from part 1 of the program, how prevalent is missing data? Does it exist primarily in the DV (Income), one or more of the IVs, or both?
b) In part 2, why do you think her assistant decided to recode the income variable? Why didn’t the assistant think MD was being handled correctly in the original coding?
c) In part 3, why does the assistant create the PAEDUC2 and MDPAEDUCvariables? Why are they coded that way?
d) In part 4, the assistant comments that “This regression model will give us an idea of whether or not the MD in PAEDUC is missing on a random basis.” How do these regression models accomplish this?
What does the coefficient for MDPAEDUC supposedly tell you? Is this a valid approach, why or why not?
Problem #5:
In the process of collecting weight and height data from 29 female and 81 male studentsat your university, you also asked the students for the number of siblings they have. Although it was not quite
clear to you initially what you would use that variable for, you construct a new theory that suggests that children who have more siblings come from poorer families and will have to share the food on
the table. Although a friend tells you that this theory does not pass the “straight face” test, you decide to hypothesize that peers with many siblings will weigh less, on average, for a given
height. In addition, you believe that the muscle/fat tissue composition of male bodies suggests that females will weigh less, on average, for a given height. To test these theories, you perform the
following regression: = –229.92 – 6.52 × Female +0.51 × Sibs+ 5.58 ×Height; R2=0.50, SER= 21.08 (44.01) (5.52) (2.25) (0.62) where Studentwis in pounds, Height is in inches, Female takes a value of 1
for females and is 0 otherwise, Sibs is the number of siblings (heteroskedasticity-robust standard errors in parentheses).
a) Carrying out hypotheses tests using the relevant t-statistics to test your two claims separately, is there strong evidence in favor of your hypotheses? Is it appropriate to use two separate tests
in this situation?
b) You also perform an F-test on the joint hypothesis that the two coefficients for females and siblings are zero. The calculated F-statistic is 0.84. Find the critical value from the F-table. Can
you reject the null hypothesis? Is it possible that one of the two parameters is zero in the population, but not the other?
c) You are now a bit worried that the entire regression does not make sense and therefore also test for the height coefficient to be zero. The resulting F-statistic is 57.25. Does that prove that
there is a relationship between weight and height?
Problem #6: Select the appropriate answer to each question, along with the reasoning supporting your selection
Part 1: Under the least squares assumptions (zero conditional mean for the error term, Xi and Yi being i.i.d., and Xi and ui having finite fourth moments), the OLS estimator for the slope and
A) has an exact normal distribution for n > 15.
B) is BLUE.
C) has a normal distribution even in small samples.
D) is unbiased.
Part 2: One of the following steps is not required as a step to test for the null hypothesis:
A) compute the standard error of 1.
B) test for the errors to be normally distributed.
C) compute the t-statistic.
D) compute the p-value.
Part 3: If you wanted to test, using a 5% significance level, whether or not a specific slope coefficient is equal to one, then you should
A) subtract 1 from the estimated coefficient, divide the difference by the standard error, and check if the resulting ratio is larger than 1.96.
B) add and subtract 1.96 from the slope and check if that interval includes 1.
C) see if the slope coefficient is between 0.95 and 1.05.
D) check if the adjusted R2 is close to 1.
Part 4: When your multiple regression function includes a single omitted variable regressor, then
A) use a two-sided alternative hypothesis to check the influence of all included variables.
B) the estimator for your included regressors will be biased if at least one of the included variables is correlated with the omitted variable.
C) the estimator for your included regressors will always be biased.
D) lower the critical value to 1.645 from 1.96 in a two-sided alternative hypothesis to test the significance of the coefficients of the included variables.
Part 5: All of the following are true, with the exception of one condition:
A) a high R2 or does not mean that the regressors are a true cause of the dependent variable.
B) a high R2 or does not mean that there is no omitted variable bias.
C) a high R2 or always means that an added variable is statistically significant.
D) a high R2 or does not necessarily mean that you have the most appropriate set of regressors.
Part 6: The following interactions between binary and continuous variables are possible, with the exception of
A) Yi = β0 + β1Xi +β2Di + β3(Xi ×Di) + ui.
B) Yi = β0 + β1Xi +β2(Xi × Di) + ui.
C) Yi = (β0 +Di) +β1Xi + ui.
D) Yi = β0 + β1Xi +β2Di + ui.
Part 7: To decide whether Yi = β0 +β1X + ui or ln(Yi) = β0 + β1X + ui fits the data better, you cannot consult the regression R2 because
A) ln(Y) may be negative for 0<Y<1.
B) the TSSare not measured in the same units between the two models.
C) the slope no longer indicates the effect of a unit change of X on Y in the log-linear model.
D) the regression R2 can be greater than one in the second model.
Part 8: To test whether or not the population regression function is linear rather than a polynomial of order r,
A) check whether the regression R2 for the polynomial regression is higher than that of the linear regression.
B) compare the TSSfrom both regressions.
C) look at the pattern of the coefficients: if they change from positive to negative to positive,etc.,then the polynomial regression should be used.
D) use the test of (r-1) restrictions using the F-statistic.
Part 9: The binary variable interaction regression
A) can only be applied when there are two binary variables, but not three or more.
B) is the same as testing for differences in means.
C) cannot be used with logarithmic regression functions because ln(0) is not defined.
D) allows the effect of changing one of the binary independent variables to depend on the value of the other binary variable.
Part 10: A statistical analysis is internally valid if
A) its inferences and conclusions can be generalized from the population and setting studied to
other populations and settings.
B) statistical inference is conducted inside the sample period.
C) the hypothesized parameter value is inside the confidence interval.
D) the statistical inferences about causal effects are valid for the population being studied | {"url":"https://easy-due.com/stata%E4%BB%A3%E5%86%99-econ-400-introduction-to-econometrics/","timestamp":"2024-11-03T07:14:05Z","content_type":"text/html","content_length":"78872","record_id":"<urn:uuid:fb0823d5-ad3e-4d88-8e8c-69f72da9a6c5>","cc-path":"CC-MAIN-2024-46/segments/1730477027772.24/warc/CC-MAIN-20241103053019-20241103083019-00612.warc.gz"} |
Introduction · Oscar.jl
The polyhedral geometry part of OSCAR provides functionality for handling
• convex polytopes, unbounded polyhedra and cones
• polyhedral fans
• linear programs
General textbooks offering details on theory and algorithms include:
The objects from polyhedral geometry operate on a given type, which (usually) resembles a field. This is indicated by the template parameter, e.g. the properties of a Polyhedron{QQFieldElem} are
rational numbers of type QQFieldElem, if applicable. Supported scalar types are FieldElem and Float64, but some functionality might not work properly if the parent Field does not satisfy certain
mathematic conditions, like being ordered. When constructing a polyhedral object from scratch, for the "simpler" types QQFieldElem and Float64 it suffices to pass the Type, but more complex
FieldElems require a parent Field object. This can be set by either passing the desired Field instead of the type, or by inserting the type and have a matching FieldElem in your input data. If no
type or field is given, the scalar type defaults to QQFieldElem.
The parent Field of the coefficients of an object O with coefficients of type T can be retrieved with the coefficient_field function, and it holds elem_type(coefficient_field(O)) == T.
coefficient_field(P::Union{Polyhedron{T}, Cone{T}, PolyhedralFan{T}, PolyhedralComplex{T}) where T<:scalar_types
Return the parent Field of the coefficients of P.
julia> c = cross_polytope(2)
Polytope in ambient dimension 2
julia> coefficient_field(c)
Rational field
Support for fields other than the rational numbers is currently in an experimental stage.
These three lines result in the same polytope over rational numbers. Besides the general support mentioned above, naming a Field explicitly is encouraged because it allows user control and increases
julia> P = convex_hull(QQ, [1 0 0; 0 0 1]) # passing a `Field` always works
Polyhedron in ambient dimension 3
julia> P == convex_hull(QQFieldElem, [1 0 0; 0 0 1]) # passing the type works for `QQFieldElem` and `Float64` only
julia> P == convex_hull([1 0 0; 0 0 1]) # `Field` defaults to `QQ`
When working in polyhedral geometry it can prove advantageous to have various input formats for the same kind of re-occurring quantitative input information. This example shows three different ways
to write the points whose convex hull is to be computed, all resulting in identical Polyhedron objects:
julia> P = convex_hull([1 0 0; 0 0 1])
Polyhedron in ambient dimension 3
julia> P == convex_hull([[1, 0, 0], [0, 0, 1]])
julia> P == convex_hull(vertices(P))
convex_hull is only one of many functions and constructors supporting this behavior, and there are also more types that can be described this way besides PointVector. Whenever the docs state an
argument is required to be of type AbstractCollection[ElType] (where ElType is the Oscar type of single instances described in this collection), the user can choose the input to follow any of the
corresponding notions below.
There are two specialized Vector-like types, PointVector and RayVector, which commonly are returned by functions from Polyhedral Geometry. These can also be manually constructed:
point_vector(p = QQ, v::AbstractVector)
Return a PointVector resembling a point whose coordinates equal the entries of v. p specifies the Field or Type of its coefficients.
ray_vector(p = QQ, v::AbstractVector)
Return a RayVector resembling a ray from the origin through the point whose coordinates equal the entries of v. p specifies the Field or Type of its coefficients.
While RayVectors can not be used do describe PointVectors (and vice versa), matrices are generally allowed.
AbstractCollection[PointVector] can be given as:
Type A PointVector corresponds to...
AbstractVector{<:PointVector} an element of the vector.
AbstractVector{<:AbstractVector} an element of the vector.
AbstractMatrix/MatElem a row of the matrix.
AbstractVector/PointVector the vector itself (only one PointVector is described).
SubObjectIterator{<:PointVector} an element of the iterator.
AbstractCollection[RayVector] can be given as:
Type A RayVector corresponds to...
AbstractVector{<:RayVector} an element of the vector.
AbstractVector{<:AbstractVector} an element of the vector.
AbstractMatrix/MatElem a row of the matrix.
AbstractVector/RayVector the vector itself (only one RayVector is described).
SubObjectIterator{<:RayVector} an element of the iterator.
Similar to points and rays, there are types AffineHalfspace, LinearHalfspace, AffineHyperplane and LinearHyperplane:
affine_halfspace(p = QQ, a, b)
Return the Oscar.AffineHalfspace H(a,b), which is given by a vector a and a value b such that $H(a,b) = \{ x | ax ≤ b \}.$ p specifies the Field or Type of its coefficients.
linear_halfspace(p = QQ, a, b)
Return the Oscar.LinearHalfspace H(a), which is given by a vector a such that $H(a,b) = \{ x | ax ≤ 0 \}.$ p specifies the Field or Type of its coefficients.
affine_hyperplane(p = QQ, a, b)
Return the Oscar.AffineHyperplane H(a,b), which is given by a vector a and a value b such that $H(a,b) = \{ x | ax = b \}.$ p specifies the Field or Type of its coefficients.
linear_hyperplane(p = QQ, a, b)
Return the Oscar.LinearHyperplane H(a), which is given by a vector a such that $H(a,b) = \{ x | ax = 0 \}.$ p specifies the Field or Type of its coefficients.
These collections allow to mix up affine halfspaces/hyperplanes and their linear counterparts, but note that an error will be produced when trying to convert an affine description with bias not equal
to zero to a linear description.
AbstractCollection[LinearHalfspace] can be given as:
Type A LinearHalfspace corresponds to...
AbstractVector{<:Halfspace} an element of the vector.
AbstractMatrix/MatElem A the halfspace with normal vector A[i, :].
AbstractVector{<:AbstractVector} A the halfspace with normal vector A[i].
SubObjectIterator{<:Halfspace} an element of the iterator.
AbstractCollection[LinearHyperplane] can be given as:
Type A LinearHyperplane corresponds to...
AbstractVector{<:Hyperplane} an element of the vector.
AbstractMatrix/MatElem A the hyperplane with normal vector A[i, :].
AbstractVector{<:AbstractVector} A the hyperplane with normal vector A[i].
SubObjectIterator{<:Hyperplane} an element of the iterator.
AbstractCollection[AffineHalfspace] can be given as:
Type An AffineHalfspace corresponds to...
AbstractVector{<:Halfspace} an element of the vector.
Tuple over matrix A and vector b the affine halfspace with normal vector A[i, :] and bias b[i].
SubObjectIterator{<:Halfspace} an element of the iterator.
AbstractCollection[AffineHyperplane] can be given as:
Type An AffineHyperplane corresponds to...
AbstractVector{<:Hyperplane} an element of the vector.
Tuple over matrix A and vector b the affine hyperplane with normal vector A[i, :] and bias b[i].
SubObjectIterator{<:Hyperplane} an element of the iterator.
Some methods will require input or return output in form of an IncidenceMatrix.
A matrix with boolean entries. Each row corresponds to a fixed element of a collection of mathematical objects and the same holds for the columns and a second (possibly equal) collection. A 1 at
entry (i, j) is interpreted as an incidence between object i of the first collection and object j of the second one.
Note that the input of this example and the print of an IncidenceMatrix list the non-zero indices for each row.
julia> IM = incidence_matrix([[1,2,3],[4,5,6]])
2×6 IncidenceMatrix
[1, 2, 3]
[4, 5, 6]
julia> IM[1, 2]
julia> IM[2, 3]
julia> IM[:, 4]
2-element SparseVectorBool
The unique nature of the IncidenceMatrix allows for different ways of construction:
incidence_matrix(r::Base.Integer, c::Base.Integer)
Return an IncidenceMatrix of size r x c whose entries are all false.
julia> IM = incidence_matrix(8, 5)
8×5 IncidenceMatrix
incidence_matrix(mat::Union{AbstractMatrix{Bool}, IncidenceMatrix})
Convert mat to an IncidenceMatrix.
julia> IM = incidence_matrix([true false true false true false; false true false true false true])
2×6 IncidenceMatrix
[1, 3, 5]
[2, 4, 6]
Convert the 0/1 matrix mat to an IncidenceMatrix. Entries become true if the initial entry is 1 and false if the initial entry is 0.
julia> IM = incidence_matrix([1 0 1 0 1 0; 0 1 0 1 0 1])
2×6 IncidenceMatrix
[1, 3, 5]
[2, 4, 6]
incidence_matrix(r::Base.Integer, c::Base.Integer, incidenceRows::AbstractVector{<:AbstractVector{<:Base.Integer}})
Return an IncidenceMatrix of size r x c. The i-th element of incidenceRows lists the indices of the true entries of the i-th row.
julia> IM = incidence_matrix(3, 4, [[2, 3], [1]])
3×4 IncidenceMatrix
[2, 3]
Return an IncidenceMatrix where the i-th element of incidenceRows lists the indices of the true entries of the i-th row. The dimensions of the result are the smallest possible row and column count
that can be deduced from the input.
julia> IM = incidence_matrix([[2, 3], [1]])
2×3 IncidenceMatrix
[2, 3]
incidence_matrix(g::Graph{T}) where {T <: Union{Directed, Undirected}}
Return an unsigned (boolean) incidence matrix representing a graph g.
julia> g = Graph{Directed}(5);
julia> add_edge!(g, 1, 3);
julia> add_edge!(g, 3, 4);
julia> incidence_matrix(g)
5×2 IncidenceMatrix
[1, 2]
From the examples it can be seen that this type supports julia's matrix functionality. There are also functions to retrieve specific rows or columns as a Set over the non-zero indices.
row(i::IncidenceMatrix, n::Int)
Return the indices where the n-th row of i is true, as a Set{Int}.
julia> IM = incidence_matrix([[1,2,3],[4,5,6]])
2×6 IncidenceMatrix
[1, 2, 3]
[4, 5, 6]
julia> row(IM, 2)
Set{Int64} with 3 elements:
column(i::IncidenceMatrix, n::Int)
Return the indices where the n-th column of i is true, as a Set{Int}.
julia> IM = incidence_matrix([[1,2,3],[4,5,6]])
2×6 IncidenceMatrix
[1, 2, 3]
[4, 5, 6]
julia> column(IM, 5)
Set{Int64} with 1 element:
A typical application is the assignment of rays to the cones of a polyhedral fan for its construction, see polyhedral_fan.
Lower dimensional polyhedral objects can be visualized through polymake's backend.
visualize(P::Union{Polyhedron{T}, Cone{T}, PolyhedralFan{T}, PolyhedralComplex{T}, SubdivisionOfPoints{T}}; kwargs...) where T<:Union{FieldElem, Float64}
Visualize a polyhedral object of dimension at most four (in 3-space). In dimensions up to 3 a usual embedding is shown. Four-dimensional polytopes are visualized as a Schlegel diagram, which is a
projection onto one of the facets; e.g., see Chapter 5 of [Zie95].
In higher dimensions there is no standard method; use projections to lower dimensions or try ideas from [GJRW10].
Extended help
Keyword Arguments
Colors can be given as
• a literal String, e.g. "green".
• a String of the format "r g b" where $r, g, b \in {0, \dots, 255}$ are integers corresponding to the R/G/B values of the color.
• a String of the format "r g b" where $r, g, b \in [0, 1]$ are decimal values corresponding to the R/G/B values of the color.
Possible arguments are:
• FacetColor: Filling color of the polygons.
• EdgeColor: Color of the boundary lines.
• PointColor/VertexColor: Color of the spheres or rectangles representing the points.
Scaling and other gradient properties
These arguments can be given as a floating point number:
• FacetTransparency: Transparency factor of the polygons between 0 (opaque) and 1 (completely translucent).
• EdgeThickness: Scaling factor for the thickness of the boundary lines.
• PointThickness/VertexThickness`: Scaling factor for the size of the spheres or rectangles representing the points.
These arguments can be given as a 3-element vector over floating point numbers:
• ViewPoint: Position of the camera.
• ViewDirection: Direction of the camera.
Appearance and Texts
These arguments can be given as a string:
• FacetStyle: If set to "hidden", the inner area of the polygons are not rendered at all.
• FacetLabels: If set to "hidden", the facet labels are not displayed (in the most cases this is the default behavior). TODO
• EdgeStyle: If set to "hidden", the boundary lines are not rendered.
• Name: The name of this visual object in the drawing.
• PointLabels/VertexLabels: If set to "hidden", no point labels are displayed.
• PointStyle/VertexStyle: If set to "hidden", neither point nor its label is rendered.
• LabelAlignment: Defines the alignment of the vertex labels: "left", "right" or "center".
Most objects from the polyhedral geometry section can be saved through the polymake interface in the background. These functions are documented in the subsections on the different objects. The format
of the files is JSON and you can find details of the specification here.
More details on the serialization, albeit concerning the older XML format, can be found in [GHJ16]. Even though the underlying format changed to JSON, the abstract mathematical structure of the data
files is still the same.
Please direct questions about this part of OSCAR to the following people:
You can ask questions in the OSCAR Slack.
Alternatively, you can raise an issue on github. | {"url":"https://docs.oscar-system.org/dev/PolyhedralGeometry/intro/","timestamp":"2024-11-06T10:16:17Z","content_type":"text/html","content_length":"78086","record_id":"<urn:uuid:c96469d7-31ca-4efc-8f7b-704be7936869>","cc-path":"CC-MAIN-2024-46/segments/1730477027928.77/warc/CC-MAIN-20241106100950-20241106130950-00774.warc.gz"} |
Multiplying Two-Digits Numbers
Question Video: Multiplying Two-Digits Numbers Mathematics • Fourth Year of Primary School
A jaguar sleeps for about 77 hours in a week. How many hours of sleep would a jaguar get in 12 weeks?
Video Transcript
A jaguar sleeps for about 77 hours in a week. How many hours of sleep would a jaguar get in 12 weeks?
The first piece of information that we’re given in this problem is that we’re told that the jaguar sleeps for about 77 hours in a week. We need to use this fact to help us find out the number of
hours of sleep a jaguar would get in 12 weeks.
Now, one way to find this answer could be to add 77 12 times. But a much quicker strategy would be to multiply 77 by 12. How can we make this multiplication easier? Well, we can start by partitioning
12. Let’s split it up into 10 and two. So, if we find the answer to 77 multiplied by 10 and then 77 doubled and we add the two together, we’ll find the answer to 77 times 12.
Firstly, let’s multiply by 10. When we multiply any number by 10, the digits move one place to the left. So, 77 becomes 770. Now, let’s find the answer to 77 multiplied by two. Two lots of 70 equals
140. And two lots of seven, we know is 14. Now, if we add 140 and 14 together, we get 154.
Now, if we add these two parts together, we can find the answer to 77 multiplied by 12. Zero plus four equals four. Seven plus five equals 12. And finally, in the hundreds column, seven plus one plus
the one we’ve exchanged equals nine. And so, we can see the answer. If a jaguar sleeps for about 77 hours in a week, in 12 weeks a jaguar will sleep 924 hours. | {"url":"https://www.nagwa.com/en/videos/947163451525/","timestamp":"2024-11-06T09:02:07Z","content_type":"text/html","content_length":"247744","record_id":"<urn:uuid:c0e280b1-f8c2-46cc-aa11-764f874a5ad7>","cc-path":"CC-MAIN-2024-46/segments/1730477027910.12/warc/CC-MAIN-20241106065928-20241106095928-00499.warc.gz"} |
Why the Tiny Weight of Empty Space Is Such a Huge Mystery
Reprinted with permission from Quanta Magazine‘s Abstractions blog.
Lucy Reading-Ikkanda / Quanta Magazine
The controversial idea that our universe is just a random bubble in an endless, frothing multiverse arises logically from nature’s most innocuous-seeming feature: empty space. Specifically, the seed
of the multiverse hypothesis is the inexplicably tiny amount of energy infused in empty space—energy known as the vacuum energy, dark energy, or the cosmological constant. Each cubic meter of empty
space contains only enough of this energy to light a lightbulb for 11-trillionths of a second. “The bone in our throat,” as the Nobel laureate Steven Weinberg once put it, is that the vacuum ought to
be at least a trillion trillion trillion trillion trillion times more energetic, because of all the matter and force fields coursing through it. Somehow the effects of all these fields on the vacuum
almost equalize, producing placid stillness. Why is empty space so empty?
While we don’t know the answer to this question—the infamous “cosmological constant problem”—the extreme vacuity of our vacuum appears necessary for our existence. In a universe imbued with even
slightly more of this gravitationally repulsive energy, space would expand too quickly for structures like galaxies, planets, or people to form. This fine-tuned situation suggests that there might be
a huge number of universes, all with different doses of vacuum energy, and that we happen to inhabit an extraordinarily low-energy universe because we couldn’t possibly find ourselves anywhere else.
Some scientists bristle at the tautology of “anthropic reasoning” and dislike the multiverse for being untestable. Even those open to the multiverse idea would love to have alternative solutions to
the cosmological constant problem to explore. But so far it has proved nearly impossible to solve without a multiverse. “The problem of dark energy [is] so thorny, so difficult, that people have not
got one or two solutions,” said Raman Sundrum, a theoretical physicist at the University of Maryland.
To understand why, consider what the vacuum energy actually is. Albert Einstein’s general theory of relativity says that matter and energy tell space-time how to curve, and space-time curvature tells
matter and energy how to move. An automatic feature of the equations is that space-time can possess its own energy—the constant amount that remains when nothing else is there, which Einstein dubbed
the cosmological constant. For decades, cosmologists assumed its value was exactly zero, given the universe’s reasonably steady rate of expansion, and they wondered why. But then, in 1998,
astronomers discovered that the expansion of the cosmos is in fact gradually accelerating, implying the presence of a repulsive energy permeating space. Dubbed dark energy by the astronomers, it’s
almost certainly equivalent to Einstein’s cosmological constant. Its presence causes the cosmos to expand ever more quickly, since, as it expands, new space forms, and the total amount of repulsive
energy in the cosmos increases.
However, the inferred density of this vacuum energy contradicts what quantum field theory, the language of particle physics, has to say about empty space. A quantum field is empty when there are no
particle excitations rippling through it. But because of the uncertainty principle in quantum physics, the state of a quantum field is never certain, so its energy can never be exactly zero. Think of
a quantum field as consisting of little springs at each point in space. The springs are always wiggling, because they’re only ever within some uncertain range of their most relaxed length. They’re
always a bit too compressed or stretched, and therefore always in motion, possessing energy. This is called the zero-point energy of the field. Force fields have positive zero-point energies while
matter fields have negative ones, and these energies add to and subtract from the total energy of the vacuum.
The total vacuum energy should roughly equal the largest of these contributing factors. (Say you receive a gift of $10,000; even after spending $100, or finding $3 in the couch, you’ll still have
about $10,000.) Yet the observed rate of cosmic expansion indicates that its value is between 60 and 120 orders of magnitude smaller than some of the zero-point energy contributions to it, as if all
the different positive and negative terms have somehow canceled out. Coming up with a physical mechanism for this equalization is extremely difficult for two main reasons.
First, the vacuum energy’s only effect is gravitational, and so dialing it down would seem to require a gravitational mechanism. But in the universe’s first few moments, when such a mechanism might
have operated, the universe was so physically small that its total vacuum energy was negligible compared to the amount of matter and radiation. The gravitational effect of the vacuum energy would
have been completely dwarfed by the gravity of everything else. “This is one of the greatest difficulties in solving the cosmological constant problem,” the physicist Raphael Bousso wrote in 2007. A
gravitational feedback mechanism precisely adjusting the vacuum energy amid the conditions of the early universe, he said, “can be roughly compared to an airplane following a prescribed flight path
to atomic precision, in a storm.”
Compounding the difficulty, quantum field theory calculations indicate that the vacuum energy would have shifted in value in response to phase changes in the cooling universe shortly after the Big
Bang. This raises the question of whether the hypothetical mechanism that equalized the vacuum energy kicked in before or after these shifts took place. And how could the mechanism know how big their
effects would be, to compensate for them?
So far, these obstacles have thwarted attempts to explain the tiny weight of empty space without resorting to a multiverse lottery. But recently, some researchers have been exploring one possible
avenue: If the universe did not bang into existence, but bounced instead, following an earlier contraction phase, then the contracting universe in the distant past would have been huge and dominated
by vacuum energy. Perhaps some gravitational mechanism could have acted on the plentiful vacuum energy then, diluting it in a natural way over time. This idea motivated the physicists Peter Graham,
David Kaplan, and Surjeet Rajendran to discover a new cosmic bounce model, though they’ve yet to show how the vacuum dilution in the contracting universe might have worked.
In an email, Bousso called their approach “a very worthy attempt” and “an informed and honest struggle with a significant problem.” But he added that huge gaps in the model remain, and “the technical
obstacles to filling in these gaps and making it work are significant. The construction is already a Rube Goldberg machine, and it will at best get even more convoluted by the time these gaps are
filled.” He and other multiverse adherents see their answer as simpler by comparison.
Natalie Wolchover is a senior writer at Quanta Magazine covering the physical sciences. Previously, she wrote for Popular Science, LiveScience and other publications. She has a bachelor’s in physics
from Tufts University, studied graduate-level physics at the University of California, Berkeley, and co-authored several academic papers in nonlinear optics. Her writing was featured in The Best
Writing on Mathematics 2015. She is the winner of the 2016 Excellence in Statistical Reporting Award and the 2016 Evert Clark/Seth Payne Award for young science journalists. @NattyOver | {"url":"https://nautil.us/why-the-tiny-weight-of-empty-space-is-such-a-huge-mystery-237188/","timestamp":"2024-11-12T02:07:20Z","content_type":"text/html","content_length":"317981","record_id":"<urn:uuid:46f3a5c9-0344-409a-b480-a9c7e54bf02d>","cc-path":"CC-MAIN-2024-46/segments/1730477028242.50/warc/CC-MAIN-20241112014152-20241112044152-00351.warc.gz"} |
getblockstats hash_or_height ( stats )
The getblockstats RPC computes per block statistics for a given window. All amounts are in satoshis.
*bitcoin-cli help getblockstats
getblockstats hash_or_height ( stats )
Compute per block statistics for a given window. All amounts are in satoshis.
It won't work for some heights with pruning.
It won't work without -txindex for utxo_size_inc, *fee or *feerate stats.
1. "hash_or_height" (string or numeric, required) The block hash or height of the target block
2. "stats" (array, optional) Values to plot, by default all values (see result below)
"height", (string, optional) Selected statistic
"time", (string, optional) Selected statistic
{ (json object)
"avgfee": xxxxx, (numeric) Average fee in the block
"avgfeerate": xxxxx, (numeric) Average feerate (in satoshis per virtual byte)
"avgtxsize": xxxxx, (numeric) Average transaction size
"blockhash": xxxxx, (string) The block hash (to check for potential reorgs)
"feerate_percentiles": [ (array of numeric) Feerates at the 10th, 25th, 50th, 75th, and 90th percentile weight unit (in satoshis per virtual byte)
"10th_percentile_feerate", (numeric) The 10th percentile feerate
"25th_percentile_feerate", (numeric) The 25th percentile feerate
"50th_percentile_feerate", (numeric) The 50th percentile feerate
"75th_percentile_feerate", (numeric) The 75th percentile feerate
"90th_percentile_feerate", (numeric) The 90th percentile feerate
"height": xxxxx, (numeric) The height of the block
"ins": xxxxx, (numeric) The number of inputs (excluding coinbase)
"maxfee": xxxxx, (numeric) Maximum fee in the block
"maxfeerate": xxxxx, (numeric) Maximum feerate (in satoshis per virtual byte)
"maxtxsize": xxxxx, (numeric) Maximum transaction size
"medianfee": xxxxx, (numeric) Truncated median fee in the block
"mediantime": xxxxx, (numeric) The block median time past
"mediantxsize": xxxxx, (numeric) Truncated median transaction size
"minfee": xxxxx, (numeric) Minimum fee in the block
"minfeerate": xxxxx, (numeric) Minimum feerate (in satoshis per virtual byte)
"mintxsize": xxxxx, (numeric) Minimum transaction size
"outs": xxxxx, (numeric) The number of outputs
"subsidy": xxxxx, (numeric) The block subsidy
"swtotal_size": xxxxx, (numeric) Total size of all segwit transactions
"swtotal_weight": xxxxx, (numeric) Total weight of all segwit transactions divided by segwit scale factor (4)
"swtxs": xxxxx, (numeric) The number of segwit transactions
"time": xxxxx, (numeric) The block time
"total_out": xxxxx, (numeric) Total amount in all outputs (excluding coinbase and thus reward [ie subsidy + totalfee])
"total_size": xxxxx, (numeric) Total size of all non-coinbase transactions
"total_weight": xxxxx, (numeric) Total weight of all non-coinbase transactions divided by segwit scale factor (4)
"totalfee": xxxxx, (numeric) The fee total
"txs": xxxxx, (numeric) The number of transactions (excluding coinbase)
"utxo_increase": xxxxx, (numeric) The increase/decrease in the number of unspent outputs
"utxo_size_inc": xxxxx, (numeric) The increase/decrease in size for the utxo index (not discounting op_return and similar)
> bitcoin-cli getblockstats 1000 '["minfeerate","avgfeerate"]'
> curl --user myusername --data-binary '{"jsonrpc": "1.0", "id":"curltest", "method": "getblockstats", "params": [1000 '["minfeerate","avgfeerate"]'] }' -H 'content-type: text/plain;' http://127.0.0.1:8332/ | {"url":"https://chainquery.com/bitcoin-cli/getblockstats","timestamp":"2024-11-03T16:40:23Z","content_type":"text/html","content_length":"11571","record_id":"<urn:uuid:42887360-e49d-4002-be54-ce046f6e67ab>","cc-path":"CC-MAIN-2024-46/segments/1730477027779.22/warc/CC-MAIN-20241103145859-20241103175859-00645.warc.gz"} |
Straight line (disambiguation)
A straight line is:
• in math
□ a fundamental object of geometry: an infinitely long straight line, see straight line
□ the generalization of the concept of a straight line: the locally shortest connection between two points is called a geodesic
□ an axiomatically required object in projective geometry
• in medieval law an inheritance to female family members, see straight line (inheritance)
• a punching technique in boxing, see cross (straight with the punching hand) and jab (straight with the leading hand)
just designated:
• Colloquially the present moment, see present
• in math
□ an integer that is divisible by 2, see parity (math)
□ a real function for which f (- x ) = f ( x ) for all x , see even and odd functions
□ the parity of a permutation, see sign (permutation)
See also: | {"url":"https://de.zxc.wiki/wiki/Gerade_(Begriffskl%C3%A4rung)","timestamp":"2024-11-14T07:23:50Z","content_type":"text/html","content_length":"14615","record_id":"<urn:uuid:1ff337a4-c8d5-46a8-8b73-e959ee9141c9>","cc-path":"CC-MAIN-2024-46/segments/1730477028545.2/warc/CC-MAIN-20241114062951-20241114092951-00304.warc.gz"} |
orksheets for 9th Class
Area and Volume (Calculus)
Is Calculus just Mathematical Gibberish
REVIEW Quiz: Unit 7 - Integrals
ROC, Tangent Lines, and Derivative
Calculus Integration Ch 5
Calculus BC - Integration
AP Calc HW #70 (5.1-5.4 Quizizz)
Integral Rules AB Calculus
AP Calculus AB - Chapter 4 Quiz Review
AP Calculus Homework - December 3, 2022
Calculus Final Review Part 3 INT and App
IB Math AI Topic 5 Calculus Review
The Fundamental Theorem of Calculus (FTC) & Indefinite Integrals
Explore integral calculus Worksheets by Grades
Explore Other Subject Worksheets for class 9
Explore printable integral calculus worksheets for 9th Class
Integral calculus worksheets for Class 9 are an essential resource for teachers who want to provide their students with a solid foundation in mathematics. These worksheets are specifically designed
for Class 9 students, focusing on the fundamental concepts of calculus that are crucial for their success in higher-level math courses. By incorporating integral calculus worksheets into their lesson
plans, teachers can effectively engage their students in understanding and applying the principles of calculus. These worksheets cover a range of topics, such as limits, derivatives, and integrals,
and include a variety of exercises that cater to different learning styles. Teachers can use these worksheets as in-class activities, homework assignments, or even as assessment tools to gauge their
students' progress in mastering the concepts of calculus. With the help of integral calculus worksheets for Class 9, teachers can ensure that their students develop the necessary skills and
confidence to excel in their math education.
Quizizz is an excellent platform for teachers who are looking to supplement their integral calculus worksheets for Class 9 with engaging and interactive learning experiences. This platform offers a
wide range of quizzes and games that are specifically designed to reinforce the concepts covered in calculus worksheets, making it an ideal resource for teachers who want to provide their students
with a comprehensive and enjoyable learning experience. In addition to quizzes, Quizizz also offers other valuable resources, such as flashcards, interactive lessons, and performance tracking tools,
which can help teachers monitor their students' progress and identify areas where they may need additional support. By incorporating Quizizz into their teaching strategies, teachers can create a
dynamic and stimulating learning environment that not only enhances their students' understanding of integral calculus but also fosters a lifelong love of mathematics. | {"url":"https://quizizz.com/en-in/integral-calculus-worksheets-class-9","timestamp":"2024-11-11T11:49:45Z","content_type":"text/html","content_length":"151567","record_id":"<urn:uuid:7ee7af08-29e7-4367-a155-ce2015ae337b>","cc-path":"CC-MAIN-2024-46/segments/1730477028228.41/warc/CC-MAIN-20241111091854-20241111121854-00596.warc.gz"} |
The 2 problems that we solve
We have one answer, that can handle two questions:
1. CashFlow Forecast
2. Investment Evaluation (calculation of the profitability of a proposed investment)
CashFlow Forecast
We need to determine the need for future working capital, establish future Bank loan balances, and calculate “Interest Expense” and “Interest Income”. Examples are:
1. Let’s say that we are in Feb 2024, and we want to see how our Bank balance will look like, from Feb 2024 until Dec 2024 (a horizon of today plus 10 months)
2. We are in Sep 2024, and we are working on the Annual Budget of 2025. We want to see how our Bank balance will look like, from Sep 2024 until Dec 2025 (a horizon of today plus 15 months). The
problem is more complicated, not because we have more months, but because we have a new Fiscal Year, and so, we must calculate several year-end values for FY 2024, like “Dividends Payable”,
“Income Tax Payable”, “Advance Payment for Income Tax of Next FY” etc.
Investment Evaluation (calculation of the profitability of a proposed investment)
We evaluate the possibility to
1. create a new company from scratch.
2. in an existing company, to add a new sales location.
3. in an existing company, to add a new factory.
4. in an existing factory, to add a new production line.
5. in an existing production line, to add a new product.
6. etc.
Plus, there are other questions that are associated with that problem (Break-Even point, Sensitivity Analysis, What-if scenarios etc.).
What we must do, before the calculation
In both scenarios, we begin with the collection of values, that are expected to happen. Let’s call them “Forecasted Values“. Examples are:
1. Sales
2. Salaries
3. Ads
4. Travel Expenses
5. Rent
6. Interest Rates
7. FX Rates
8. Etc.
On their basis, we must calculate other values, that are dependent on them. Let’s call them “Calculated Values“. Those should NEVER be treated as a “matter of forecast”, but only as a “product of
calculation”. Examples are:
1. Interest Expense
2. Interest Income
3. Purchases of merchandise, raw materials, packaging materials etc.
4. Direct labour expenses
5. Payment of VAT
6. Payment of Income Tax, and advance payment for next year’s Income Tax
7. Payment of Dividends
8. Almost all kinds of Taxes
9. Etc.
The previous “State-of-the-Art” in calculation methods
Fact 1:
If we go to 100 companies, and ask to see how 100 different CFOs (in other words, experienced high level professionals, with MBA’s or Master’s in Finance, or PhD’s), we will receive 100 different
spreadsheets, that implement 100 different calculation methods, that arrive at 100 different results, for the same problem, with the same list of assumptions.
Fact 2:
If we go to Wikipedia (with over 6.8 mil English articles), and search for an article, that lists the steps of a calculation method to solve those two problems, we will see that there is no such
article. Instead, there are a handful of small articles, that describe some vague and high-level guidelines. But even as such, if you look closely, half of their contents cannot stand up to scrutiny.
And a few of those vague and high-level guidelines, cannot even pass the laugh test.
From the above facts, it is easy to deduce that, currently, the biggest problem with the “State-of-the-Art”, is that:
There is no “State-of-the-Art”
Or, in other words, there is no commonly accepted calculation method, for Finance professionals to use and implement.
Who is to be blamed for that?
This whole situation, reminds of what took place 150 years ago in Medicine, when there were no antibiotics. Patients visited their doctors, in hopes of getting well, but instead, they died.
Were doctors to be blamed for that high mortality rate? No. The blame rested with the “State-of-the-Art”, which had not provided the appropriate tool (antibiotics) to the professionals of the field.
Today, Finance professionals need the appropriate tool to do their job. They need a Commonly Accepted Calculation Method, whose result can stand up to scrutiny and verification.
What is being implemented by others
Even though the number of problems and inaccuracies, in the way that Finance professionals are approaching those two problems, are legion, at the end of the day, those two tasks must be done.
Someone, somehow, must produce some kind of result.
What is being implemented (thru spreadsheet), is a list (a.k.a. the “single ledger entry“ method). For more on that, please read the following article in our blog.
What Finance professionals need
When we say that “Forecasted Sales = 10 mil USD”, we know that the actual final sales figure, is not going to be “exactly 10 mil USD”. The forecasting process (by definition) has a built-in
inaccuracy. Generally speaking, let’s call it a ±Α% built-in inaccuracy.
If the calculation method has built-in inaccuracies (and the current spreadsheet-implemented way of working, has many – let’s call it a ±B% inaccuracy), we end up with a result whose inaccuracy is ±
(Α*B)% .
What we need, is a calculation method, who’s inaccuracy is ZERO (B=0).
Actually, such a calculation method, has existed for many centuries. However, its implementation in forecast mode (until yesterday) was neither easy, nor practical or feasible.
In University, we were taught that method, under the name “Accounting 101″. Everyone will agree that “Sales” means that we Debit this account, and we Credit that account. That is a common language
that we all know/use, and we accept it as accurate and trustworthy.
At a first casual glance, one might be tempted to think that, since “Accounting 101” is so successfully being implemented, thru the numerous ERPs that exist, it would be a good idea to use one of
them, in order to create the company’s future Accounting Books. Just try it, and in five minutes (or less) you will discover that (on a practical level) ERPs cannot work in forecast mode, because
there are too many calculation incompatibilities, between processing actual data, versus processing forecasted data. For more on that, please read the following article in our blog.
What is the C2BII approach and calculation method
We are creating an automated simulation of the company’s future Accounting Books. As a starting point, most of the things that you will be required to perform, are based on the Accounting know-how,
that you already have and use (Accounting 101 – Assets and Liabilities – Debit and Credit).
The theory might sound simple. However, the implementation will need a secret sauce that will make it look and feel easy and practical. That can be found in our US Patents.
For a first taste of the new calculation method, and the software tool that implements it, please first read the articles in the blog.
Then, please contact us for a demo. | {"url":"https://c2bii.com/?option=com_content&task=view&id=12&Itemid=26","timestamp":"2024-11-07T13:20:17Z","content_type":"text/html","content_length":"56359","record_id":"<urn:uuid:036a9024-c87c-43af-b145-0133b598c3ca>","cc-path":"CC-MAIN-2024-46/segments/1730477027999.92/warc/CC-MAIN-20241107114930-20241107144930-00734.warc.gz"} |
TensorOperator (const int boundary_index, const int two_j, const int n_elec, const int n_irrep, const bool moving_right, const bool prime_last, const bool jw_phase, const SyBookkeeper *bk_up,
const SyBookkeeper *bk_down)
Constructor. More...
virtual ~TensorOperator ()
int gNKappa () const
Get the number of symmetry blocks. More...
* gStorage ()
Get the pointer to the storage. More...
int gKappa (const int N1, const int TwoS1, const int I1, const int N2, const int TwoS2, const int I2) const
Get the index corresponding to a certain tensor block. More...
int gKappa2index (const int kappa) const
Get the storage jump corresponding to a certain tensor block. More...
* gStorage (const int N1, const int TwoS1, const int I1, const int N2, const int TwoS2, const int I2)
Get the pointer to the storage of a certain tensor block. More...
int gIndex () const
Get the boundary index. More...
int get_2j () const
Get twice the spin of the tensor operator. More...
int get_nelec () const
Get how many electrons there are more in the symmetry sector of the lower leg compared to the upper leg. More...
int get_irrep () const
Get the (real-valued abelian) point group irrep difference between the symmetry sectors of the lower and upper legs (see Irreps.h) More...
void update (TensorOperator *previous, TensorT *mps_tensor_up, TensorT *mps_tensor_down, double *workmem)
Clear and update. More...
void daxpy (double alpha, TensorOperator *to_add)
daxpy for TensorOperator More...
void daxpy_transpose_tensorCD (const double alpha, TensorOperator *to_add)
daxpy_transpose for C- and D-tensors (with special spin-dependent factors) More...
void clear ()
Set all storage variables to 0.0.
double inproduct (TensorOperator *buddy, const char trans) const
Make the in-product of two TensorOperator. More...
const SyBookkeeper * bk_up
The bookkeeper of the upper MPS.
const SyBookkeeper * bk_down
The bookkeeper of the lower MPS.
int two_j
Twice the spin of the tensor operator.
int n_elec
How many electrons there are more in the symmetry sector of the lower leg compared to the upper leg.
int n_irrep
The (real-valued abelian) point group irrep difference between the symmetry sectors of the lower and upper legs (see Irreps.h)
bool moving_right
Whether or not moving right.
int * sector_nelec_up
The up particle number sector.
int * sector_irrep_up
The up spin symmetry sector.
int * sector_spin_up
The up spin symmetry sector.
int * sector_spin_down
The down spin symmetry sector (pointer points to sectorTwoS1 if two_j == 0)
bool prime_last
Convention in which the tensor operator is stored (see class information)
bool jw_phase
Whether or not to include a Jordan-Wigner phase due to the fermion anti-commutation relations.
int index
Index of the Tensor object. For TensorT: a site index; for other tensors: a boundary index.
double * storage
The actual variables. Tensor block kappa begins at storage+kappa2index[kappa] and ends at storage+kappa2index[kappa+1].
int nKappa
Number of Tensor blocks.
int * kappa2index
kappa2index[kappa] indicates the start of tensor block kappa in storage. kappa2index[nKappa] gives the size of storage.
TensorOperator class.
Sebastian Wouters sebas.nosp@m.tian.nosp@m.woute.nosp@m.rs@g.nosp@m.mail..nosp@m.com
October 15, 2015
The TensorOperator class is a storage and update class for tensor operators with a given:
• spin (two_j)
• particle number (n_elec)
• point group irrep (n_irrep).
It replaces the previous classes TensorDiag, TensorSwap, TensorS0Abase, TensorS1Bbase, TensorF0Cbase, TensorF1Dbase, TensorA, TensorB, TensorC, and TensorD. Their storage and update functions have a
common origin. The boolean prime_last denotes whether in which convention the tensor operator is stored:
This determines the specific reduced update formulae when contracting with the Clebsch-Gordan coefficients of the reduced MPS tensors.
Definition at line 42 of file TensorOperator.h. | {"url":"http://sebwouters.github.io/CheMPS2/doxygen/classCheMPS2_1_1TensorOperator.html","timestamp":"2024-11-09T00:01:49Z","content_type":"application/xhtml+xml","content_length":"118529","record_id":"<urn:uuid:a146868d-ea09-41b8-a646-8854b3ab3123>","cc-path":"CC-MAIN-2024-46/segments/1730477028106.80/warc/CC-MAIN-20241108231327-20241109021327-00806.warc.gz"} |
School of Engineering and Computer Science
A selection of presentations, conference proceedings and lectures from faculty members of the School of Engineering and Computer Science at University of the Pacific.
Submissions from 1999
Derivative DFT Beamspace ESPRIT: A high performance closed-form 2-D arrival angle estimation algorithm, Cherian P. Mathews
DSP implementation of an amplitude modulation transmission system: A capstone design approach, Cherian P. Mathews, Adrianne Candia, and Waleed Kader
High-speed machining of titanium by new PCD tools, Masahiko Mori, Masaaki Furuta, Tetsuo Nakai, Tomohiro Fukaya, Jiancheng Liu, and Kazuo Yamazaki
Photoelasticity and its synergism with finite element method, Said Shakerin and Daniel D. Jensen
A study on autonomous hole machining process analysis by reverse engineering of NC programs, Xiren Yan, Lawrence Chan, Kazuo Yamazaki, Jiancheng Liu, Mitsuru Kubota, and Yoshikazu Amano
Submissions from 1998
A Market Share Prediction Model for the United States Commercial Jet Aircraft Industry, Abel A. Fernandez
A Model for the Stochastic Resource Constrained Project Scheduling Problem With Stochastic Task Durations, Abel A. Fernandez, R. Armacost, and J. Pet-Edwards
Understanding Simulation Solutions to Resource Constrained Project Scheduling Problems with Stochastic Task Durations, Abel A. Fernandez, R. Armacost, and J. Pet-Edwards
Exotic meson spectroscopy from the clover action at β = 5.85 and 6.15, James E. Hetrick, C. Bernard, T. DeGrand, C. DeTar, S. A. Gottlieb, U. M. Heller, C. McNeile, R. Sugar, and D. Toussaint
Heavy-Light Decay Constants: Conclusions from the Wilson Action, James E. Hetrick, C. Bernard, T. DeGrand, C. DeTar, S. A. Gottlieb, U. M. Heller, C. McNeile, R. Sugar, D. Toussaint, and M. Wingate
Light hadron spectrum with Kogut-Susskind quarks, James E. Hetrick, C. Bernard, C. DeTar, S. A. Gottlieb, U. M. Heller, C. McNeile, K. Rummukainen, R. Sugar, and D. Toussaint
Local topological and chiral properties of QCD, James E. Hetrick, M. Perez, and J. Lagae
In-process tool utilization analysis based machining simulation, Yoshiori Isobata, Masaomi Tsutsumi, Sun Kyu Lee, Kazuo Yamazaki, and Jiancheng Liu
Integrating Design in an Environmental Engineering Curriculum Using Field Exercises, Gary M. Litton
Dynamic gain motion control with multi-axis trajectory monitoring for machine tool systems, Jiancheng Liu, Kazuo Yamazaki, and Yoich Yokoyama
Intelligent multi-axis motion control for machine tool systems, Jiancheng Liu, Yoich Yokoyama, Kazuo Yamazaki, and Sun Kyu Lee
Chemical kinetic modeling of a methane opposed-flow diffusion flame and comparison to experiments, N. M. Marinov, W. J. Pitz, C. K. Westbrook, Andrew E. Lutz, A. M. Vincitore, and S. M. Senkan
Engineering and Software, Said Shakerin
Natural Convection Over Rough Surfaces, Said Shakerin
Submissions from 1997
A Dynamic Mean Valued Solution to the Stochastic Resource Constrained Project Scheduling Problem, Abel A. Fernandez, R. Armacost, and J. Pet-Edwards
Top Down or Bottom Up Cost Estimation Models: Is the Added Complexity Justifiable?, Abel A. Fernandez, H. Bao, and C. Ozcan
A Student Perspective of a Capstone Course for Engineering Management, Abel A. Fernandez and D. Blackstock
Minimizing Post-Production Weight Growth in Tanks Using a Decision Analysis Model, Abel A. Fernandez and M. Cochrane
Evaluation of Airport Operating Efficiencies Using Data Envelopment Analysis, Abel A. Fernandez and S. Cummings
Expanding Athletic Field Capacity in the City of Newport News: AHP as a Tool for Making Difficult Public Sector Decisions, Abel A. Fernandez, S. Davis, and D. Vaden
Simulation As a Tool for Evaluating the Market Impact of NASA Aeronautics Research, Abel A. Fernandez and J. Schockcor
Hybrid mesons in quenched lattice QCD, James E. Hetrick, C. Bernard, T. Blum, T. A. DeGrand, C. DeTar, and S. A. Gottlieb
Heavy-light decay constants—MILC results with the Wilson action, James E. Hetrick, C. Bernard, T. Blum, T. A. DeGrand, C. DeTar, S. A. Gottlieb, U. M. Heller, C. McNeile, K. Rummukainen, and A. Soni
B mixing on the lattice: f(B), f(B(s)) and related quantities, James E. Hetrick, C. Bernard, T. Blum, T. A. DeGrand, C. DeTar, S. A. Gottlieb, U. M. Heller, C. McNeile, K. Rummukainen, and R. Sugar
Towards the QCD spectrum with dynamical quarks, James E. Hetrick, C. Bernard, T. Blum, T. A. DeGrand, C. DeTar, S. A. Gottlieb, U. M. Heller, C. McNeile, D. Toussaint, and M. Wingate
MILC studies of high temperature QCD: A progress report, James E. Hetrick, C. Bernard, T. Blum, T. A. DeGrand, C. DeTar, S. A. Gottlieb, U. M. Heller, R. Sugar, D. Toussaint, and M. Wingate
Update on the hadron spectrum with two flavors of staggered quarks, James E. Hetrick, C. Bernard, T. Blum, T. A. DeGrand, S. A. Gottlieb, U. M. Heller, C. McNeile, R. Sugar, and D. Toussaint
Light hadron spectrum: MILC results with the Kogut-Susskind and Wilson actions, James E. Hetrick, C. Bernard, T. Blum, C. DeTar, S. A. Gottlieb, U. M. Heller, C. McNeile, K. Rummukainen, B. Sugar,
and D. Toussaint
Critical behavior at the chiral phase transition, James E. Hetrick, C. Bernard, T. Blum, C. DeTar, S. A. Gottlieb, U. M. Heller, R. Sugar, D. Toussaint, M. Wingate, and K. Rummukainen
Heavy-light decay constants from Wilson and static quarks, James E. Hetrick, C. Bernard, T. DeGrand, C. DeTar, S. A. Gottlieb, U. M. Heller, C. McNeile, K. Rummukainen, R. Sugar, and D. Toussaint
Light quark spectrum with improved gauge and fermion actions, James E. Hetrick, C. Bernard, T. DeGrand, C. DeTar, S. A. Gottlieb, U. M. Heller, C. McNeile, K. Rummukainen, R. Sugar, and D. Toussaint
B meson form factors from HQET simulations, James E. Hetrick, C. Bernard, C. DeTar, S. A. Gottlieb, U. M. Heller, B. Jegerlehner, C. McNeile, K. Rummukainen, R. Sugar, and D. Toussaint
Smeared Gauge Fixing, James E. Hetrick and P. De Forcrand
Topology of full QCD, James E. Hetrick, P. De Forcrand, M. Perez, and I. Stamatescu
Three topics in the Schwinger model, James E. Hetrick, P. De Forcrand, T. Takaishi, and A. Der Sijs
Topological properties of the QCD vacuum at T = O and T similar to T©, James E. Hetrick, M. Perez, P. De Forcrand, and I. Stamatescu
A Comprehensive Laboratory Program for an Environmental Engineering Curriculum, Gary M. Litton
Integrating Multiple Disciplines in an Environmental Engineering Curriculum with Field Exercises, Gary M. Litton
Submissions from 1996
Modeling, Problem Solving and Becoming an Effective Practitioner: A Capstone Course for a Master’s Program in Operations Research/Systems Analysis, Abel A. Fernandez
The Implicit Assumptions of Existing PERT Mathematical Models: A Contrast to a Decision Stage Model, Abel A. Fernandez, R. Armacost, and J. Pet-Edwards
The Role of the Nonanticipativity Constraint in Commercially Available Software for Stochastic Project Scheduling, Abel A. Fernandez, R. Armacost, and J. Pet-Edwards
Hodge gauge fixing in three-dimensions, James E. Hetrick
Update on f(B), James E. Hetrick, C. Bernard, T. Blum, T. A. DeGrand, C. DeTar, S. A. Gottlieb, U. M. Heller, C. McNeile, K. Rummukainen, and A. Soni
Exotic hybrid mesons with light quarks, James E. Hetrick, C. Bernard, T. Blum, T. A. DeGrand, C. DeTar, S. A. Gottlieb, U. M. Heller, C. McNeile, K. Rummukainen, and B. Sugar
Assorted weak matrix elements involving the bottom quark, James E. Hetrick, C. Bernard, T. Blum, T. A. DeGrand, C. DeTar, S. A. Gottlieb, U. M. Heller, C. McNeile, K. Rummukainen, and R. Sugar
Recent MILC spectrum results, James E. Hetrick, C. Bernard, T. Blum, T. A. DeGrand, C. DeTar, S. A. Gottlieb, U. M. Heller, C. McNeile, K. Rummukainen, and R. Sugar
Finite temperature lattice QCD with clover fermions, James E. Hetrick, C. Bernard, T. Blum, T. A. DeGrand, C. DeTar, S. A. Gottlieb, U. M. Heller, C. McNeile, B. Sugar, and K. Rummukainen
The hot QCD equation of state from lattice simulations, James E. Hetrick, C. Bernard, T. Blum, T. A. DeGrand, M. Wingate, C. DeTar, C. McNeile, S. A. Gottlieb, U. M. Heller, and B. Sugar
Thermodynamics for two flavor QCD, James E. Hetrick, C. Bernard, T. Blum, C. DeTar, S. A. Gottlieb, U. M. Heller, C. McNeile, K. Rummukainen, R. Sugar, and D. Toussaint
The Equation of state for QCD from the lattice, James E. Hetrick, C. Bernard, T. Blum, C. DeTar, C. McNeile, and S. A. Gottlieb
Simulating QCD at finite density with static quarks, James E. Hetrick, T. Blum, and D. Toussaint
Confinement and chiral dynamics in the multiflavor Schwinger model, James E. Hetrick, Y. Hosotani, R. Rodriguez, and S. Iso
Aspects of Confinement and Chiral Dynamics in 2-d QED at Finite Temperature, James E. Hetrick, R. Rodriguez, Y. Hosotani, and S. Iso
Surface Chemistry Effects on the Aggregation Rates of Latex Microspheres with Sodium Dodecyl Sulfate, Gary M. Litton and D. Antypas
Improved closed-form DOA/frequency estimation via ESPRIT using DFT and derivative DFT beamforming, Cherian P. Mathews
Submissions from 1995
The Optimal Solution to the Stochastic Resource Constrained Project Scheduling Problem, Abel A. Fernandez, R. Armacost, and W. Swart
2D unitary ESPRIT for efficient 2D parameter estimation, Martin Haardt, Michael D. Zoltowski, Cherian P. Mathews, and Josef A. Nossek
A Geometric look at the lattice Gribov problem, and some copy free smooth gauges, James E. Hetrick
Smooth interpolation of lattice gauge fields by signal processing methods, James E. Hetrick
f(B) quenched and unquenched, James E. Hetrick, C. Bernard, T. Blum, T. A. DeGrand, C. DeTar, S. A. Gottlieb, U. M. Heller, R. Sugar, M. Wingate, and D. Toussaint
The N(t) = 6 equation of state for two flavor QCD, James E. Hetrick, C. Bernard, T. Blum, C. DeTar, S. A. Gottlieb, U. M. Heller, L. Karkkainen, R. Sugar, M. Wingate, and D. Toussaint
The Continuum limit in the quenched approximation, James E. Hetrick, C. Bernard, T. Blum, C. DeTar, S. A. Gottlieb, U. M. Heller, K. Rummukainen, R. Sugar, M. Wingate, and D. Toussaint
Two flavor staggered fermion thermodynamics at N(t) = 12, James E. Hetrick, C. Bernard, T. Blum, C. DeTar, S. A. Gottlieb, U. M. Heller, K. Rummukainen, R. Sugar, M. Wingate, and D. Toussaint
Lattice QCD with dense heavy quarks, James E. Hetrick, T. Blum, and D. Toussaint
Particle Size Effects: Evidence for Discrete Surface Charge Effects on Colloid Deposition Kinetics with Sodium Dodecyl Sulfate, Gary M. Litton and T. M. Olson
Implementation and performance analysis of 2D DFT beamspace ESPRIT, Cherian P. Mathews, Martin Haardt, and Michael D. Zoltowski
Convection Heat Transfer from a Heated Plate in Partial Enclosure, Said Shakerin
Submissions from 1994
Cell adhesion to a series of hydrophilic-hydrophobic copolymer series, Jeffrey S. Burmeister
The effect of substrate hydrophobicity on endothelial cell adhesion, Jeffrey S. Burmeister, J. D. Vrany, G. A. Truskey, and W. M. Reichert
Can baryogenesis occur on the lattice?, James E. Hetrick and W. Bock
The Continuum limit of the lattice Gribov problem, and a solution based on Hodge decomposition, James E. Hetrick and P. De Forcrand
Site Characterization: An Aquatic Chemistry Laboratory for Environmental Engineering Students, Gary M. Litton
Indirect Evidence for Discrete Surface Charge Effects on Colloid Deposition Kinetics with Sodium Dodecyl Sulfate, ACS Spring Mtg, Physical and Chemical Process Controlling Contaminant Mobility in
Aquatic Environments, Gary M. Litton and T. M. Olson
An integrated analysis of the kalina cycle in combined cycles, Marc D. Rumminger, Robert W. Dibble, Andrew E. Lutz, and Ann S. Yoshimura
Renewable Energy Sources, Said Shakerin
Solar energy, air conditioning, and experimental uncertainty, Said Shakerin
Closed-form 2D angle estimation with rectangular arrays via DFT beamspace ESPRIT+, Michael D. Zoltowski, Martin Haardt, and Cherian P. Mathews
Submissions from 1993
An Expert System for Undergraduate Advising, Abel A. Fernandez, J. Biegel, and J. Earhart
Some nonperturbative aspects of gauge fixing in two-dimensional Yang-Mills theory, James E. Hetrick
Fermion number conservation isn't fermion conservation, James E. Hetrick, W. Bock, and J. Smit
Creativity, Said Shakerin
Closed-form 2D angle estimation with uniform circular arrays via phase mode excitation and ESPRIT, Michael D. Zoltowski and Cherian P. Mathews
Real-time frequency and 2-D angle estimation with sub-Nyquist spatio-temporal samplingt, Michael D. Zoltowski and Cherian P. Mathews
Submissions from 1992
Topographical mapping of relative cell/substrate separation distances using TIRF microscopy (TIRFM), Jeffrey S. Burmeister
Gauge fixing and Gribov copies in pure Yang-Mills on a circle, James E. Hetrick
Colloid Transport in Porous Media and Approaches to Examine the Importance of Surface Heterogeneity, Gary M. Litton and T. M. Olson
Direction finding with circular arrays via phase mode excitation and root-music, Cherian P. Mathews and Michael D. Zoltowski
Earthen solar cooker-design and performance test, Said Shakerin
Engineering and Nature, Said Shakerin
Freshman Engineering, Said Shakerin
Strain Gage Based Densitomete, Said Shakerin
Performance analysis of eigenstructure based DOA estimators employing conjugate symmetric beamformers, Michael D. Zoltowski, G. M. Kautz, and Cherian P. Mathews
Direction finding with uniform circular arrays via phase mode excitation and beamspace root-music, Michael D. Zoltowski and Cherian P. Mathews
Submissions from 1991
Variable angle TIRF microscopy of fluorescently labeled lipid films and cells at the glass-liquid interface, Jeffrey S. Burmeister
Son of gauge fixing on the lattice, James E. Hetrick, P. De Forcrand, A. Nakamura, and R. Sinclair | {"url":"https://scholarlycommons.pacific.edu/soecs-facpres/index.6.html","timestamp":"2024-11-11T04:32:41Z","content_type":"text/html","content_length":"88756","record_id":"<urn:uuid:712212e3-0c74-497c-a664-575c58771d34>","cc-path":"CC-MAIN-2024-46/segments/1730477028216.19/warc/CC-MAIN-20241111024756-20241111054756-00397.warc.gz"} |
Get MERRA-2 grid — get_merra2_grid
type = "polygons",
locid = NULL,
lon = c(-180, 180),
lat = c(-90, 90),
crs = 4326,
add_lonlat = FALSE,
add_poles_points = TRUE,
type of grid-data to return, spatial points ("points") or polygons ("poly")
(optional) integer vector of location identifiers for which the grid will be returned.
numeric vector (min and max) with the range of longitude coordinates of the grid to return. Default `c(-180, 180)`.
numeric vector with the range of latitude coordinates of the grid to return. Default `c(-180, 180)`.
target coordinate reference system: object of class 'sf::crs', or input string for `sf::st_crs`. Default `4326`.
logical, should merra-points coordinates (`lon`, `lat`) be added to the data. FALSE by default.
logical, in the case of "polygons" grid, should points at poles be added to the data. TRUE by default.
Returns `sf` object with MERRA-2 grid, points or polygons. If polygons requested, grid of will be returned where MERRA2 coordinates are considered as centers of every polygon, except cells with `lat
= -90` or `lat = 90`. Spatial points will be returned for the cells near poles.
x <- get_merra2_grid()
#> Error in st_as_sf(x): could not find function "st_as_sf"
#> Error in eval(expr, envir, enclos): object 'x' not found
getGrid("poly", "sf")
#> Error in getGrid("poly", "sf"): could not find function "getGrid"
getGrid(lon = c(-70, -60), lat = c(30, 40), class = "df")
#> Error in getGrid(lon = c(-70, -60), lat = c(30, 40), class = "df"): could not find function "getGrid" | {"url":"https://energyrt.github.io/merra2ools/reference/get_merra2_grid.html","timestamp":"2024-11-01T23:52:17Z","content_type":"text/html","content_length":"11296","record_id":"<urn:uuid:73f03300-a18e-4f5d-bfac-501fc72756c4>","cc-path":"CC-MAIN-2024-46/segments/1730477027599.25/warc/CC-MAIN-20241101215119-20241102005119-00447.warc.gz"} |
jacobi 3.1.1
An equality test in the unit tests failed on MacM1, so I relaxed its tolerance.
jacobi 3.1.0
• Elliptic alpha function.
• Rogers-Ramanujan functions.
• Jacobi theta function with characteristics.
• Allows a negative nome.
• Some conformal mappings.
• Nome in function of the parameter.
• The logarithms of the Jacobi theta functions were not in the principal branch.
• The Dedekind eta function is now vectorized.
jacobi 3.0.0
• Changed the expression of the kleinj function in order that its factors avoid a possible float overflow.
• Major changes in the implementation of the Jacobi theta functions, following the new Fortran implementation by Mikael Fremling.
jacobi 2.3.1
jacobi 2.3.0
• Lemniscate elliptic functions.
• Dixon elliptic functions.
jacobi 2.2.0
• Some values of the Jacobi theta functions were wrong as of version 2.1.0.
• Added some unit tests.
• New function halfPeriods, computing the half-periods from the elliptic invariants.
• New function ellipticInvariants, computing the elliptic invariants from the half-periods.
jacobi 2.1.0
• The case when the elliptic invariant g2 is zero is now handled.
• The method computing the half-periods ratio when the elliptic invariants are given led a wrong sign sometimes.
jacobi 2.0.1
• Minor fix in the C++ code.
jacobi 2.0.0
• Weierstrass sigma function.
• Weierstrass zeta function.
• Costa surface.
• Vectorization.
• Better accuracy.
jacobi 1.0.0
First release. | {"url":"https://cran.uvigo.es/web/packages/jacobi/news/news.html","timestamp":"2024-11-13T07:42:37Z","content_type":"application/xhtml+xml","content_length":"3389","record_id":"<urn:uuid:e991654b-7c0b-47b2-9f13-c394de93fc4d>","cc-path":"CC-MAIN-2024-46/segments/1730477028342.51/warc/CC-MAIN-20241113071746-20241113101746-00477.warc.gz"} |
Time/Space Bullshit and Lorentz calculations, The Sunspot Chronicles
Time/Space Bullshit and Lorentz calculations
Ni’a has provided us with a challenge by insisting on using “exact” numbers in their narrative in their book Outsider.
Their fictional version of themself can get the exact numbers of whatever they are looking at, just by looking at it. This is pure magic, of course, and even when I wrote about it I lampshaded the
whole thing by saying “It’s also what Phage can do.” Essentially.
And Ni’a themself has lampshaded their numbers a bit further by saying “Rattling off the exact numbers as they entered my head calmed me down, even though they’d be wrong by the time I got done
speaking them. Rounding them off helps a little to cover the discrepancy, but my psyche will continue to update me for a few moments afterward even if I cut myself off.“
But, in their first chapter, titled “Outsider”, they rattle off the distance between the Sunspot and its parent ship the Terra Supreme at the moment that they are speaking to Phage. And we are
seriously fudging this number.
We aim to, in time, refine it, as we come to understand Lorentz Calculations better, and as we figure out how to incorporate constant acceleration into them. And if somebody wants to do all that for
us and just give us an exact number at some point, we’ll be grateful.
But here is what we have done so far:
The first thing we did was establish how long the Sunspot has experienced its own existence, from when it was created from the shipyards of the Terra Supreme and started accelerating away from it,
130,296 years. Importantly, the Terra Supreme has been constantly accelerating away from that point in space/time at the same rate, and from their perspective it has been 130,296 years as well.
But relativity does weird things with that.
To get a picture of what we’re grappling here, we recommend that you watch Henry Reich’s course on Special Relativity at minutephysics on YouTube. (url: https://youtu.be/1rLWVZVWfdY)
Before watching that course, we simply remembered what we were taught in school, and drew up a light cone diagram of the travel of the two ships, and just sort of tried to visually estimate how long
it would take for a signal of light to travel from one ship to the other starting at 130,296 years after they’d left each other’s presence. Because Ni’a wanted to rattle off that specific number. And
here is what we created:
Oops. That says 130,298. Where did 130,296 come from? (Plurality, it does weird things to your memory at all levels, and mistakes like this are way too common for us, even when we write things down.)
If you are familiar with these kinds of calculations and visualizations, you can probably see flaws in this that even we still can’t see. We are not physicists. We are hobbyists who are trying to
learn this on the fly so that we look kind of like we know what we’re doing. Most readers will just accept the number we give and move on. But we want to be close enough to the actual number so that
our readers who are mathematicians and physicists won’t end up feeling like they have to write huge blog posts like this one to rant about how wrong we are.
Please! Let us do that work! (Just, help if you can with comments below. Thank you.)
Anyway, we wanted a placeholder number fast so that we could jot it down and keep writing, which we would then create later when we finally found an equation we could just plug numbers into and get
answers from. At this point, we contacted our old friend from fifth grade, Nick Scholtz, and asked him for help, which he graciously gave us by helping us find the video above. But that took a little
while, so our place holder number stayed in our document until today, and was actually online and publicly readable for a couple of days.
Squinting at our diagram and waving our hands at it a bit, we decided that a reasonable placeholder estimate might be something like twice the time the two ships have spent traveling away from each
other, which looked like 260,592 years.
Ah-hahaha. Ah-haHAhahaha. HAHAHAHAHAAHAHAHA – no.
We were in a hurry. We wanted to let Ni’a focus on writing their story, not stopping to spend a week on physics.
It was basically just essentially tracing a path from one ship to the other down their lines of travel, and vaguely rounding it off. And then kind of thinking that between time dilation and constant
travel at near speed of light, it should balance out to something like that.
But after watching the above videos, we decided that a better estimate, though still too simplistic, was to use a Lorentz calculation, which is specifically designed to find numbers like this. We’re
still not taking into account the constant accelerations of the ships, nor actually calculating their relative velocity at the moment that Ni’a is talking. We knew at this point that the relative
velocities of both ships were something close to 99.99% the speed of light, and that acceleration at this point was so minimal as to be ignorable (in our minds at least – we could still be very wrong
about that).
So what we did was we went to this Lorentz calculator right here:
And we plugged the year that the story is taking place into the top (130,296 – and it should have been 130,298 and we need to do some more editing now), and the speed of light with some speed shaved
off (the hundreds, tens, and ones digits just dropped to zero), and got a number that we’re calling “close enough for now”. Which is 1,793,187 light years. We think.
We think we’re interpreting that calculator’s results correctly. We know the number is still wrong. But we expect it to be much closer than our first estimate.
At some point, we will get the ships’ actual relative velocities as we’ve set them up (according to their accelerations and the time since they started traveling apart), and plug that into the
calculator, and get a closer number. But their relative accelerations are distorted by time dilation as well, and need to have a Lorentz calculation applied to them to begin with, to get the
resultant relative velocities. And then that needs to be taken into account as Ni’a’s hypothetical pulse of light travels from one ship to the other starting at the point where they start speaking
their sentence. And there’s a point at which it really isn’t worth getting more precise. There’s a point where it’s just too much work for the reward of being accurate.
So, anyway, you may see that number change a few times in the future. Especially if someone does the work for us and puts a more correct version of it in the comments here. And if we ever put
together the right set of precisely accurate calculations to get it, we’ll share them, too. The actual equations.
One thought on “Time/Space Bullshit and Lorentz calculations”
1. Fenmere the Worm says:
Hey, Abacus!
The Sunspot is 130,298 years old at the start of Outsider, & has been traveling at 99.9% the speed of light for most of that time. It’s part of a long line of starships doing same.
The Milky Way is 105,700 light years wide.
We need to think about what this means for our stories.
This site uses Akismet to reduce spam. Learn how your comment data is processed. | {"url":"https://sunspot.world/time-space-bullshit-and-lorentz-calculations/","timestamp":"2024-11-09T07:25:08Z","content_type":"text/html","content_length":"67521","record_id":"<urn:uuid:d87c2704-37b3-4ce6-9bdf-3ec79f0faf25>","cc-path":"CC-MAIN-2024-46/segments/1730477028116.30/warc/CC-MAIN-20241109053958-20241109083958-00652.warc.gz"} |
The Stacks project
Lemma 10.151.5. Let $R \to S$ be a ring map. Let $\mathfrak q \subset S$ be a prime lying over $\mathfrak p$ in $R$. If $S/R$ is unramified at $\mathfrak q$ then
1. we have $\mathfrak p S_{\mathfrak q} = \mathfrak qS_{\mathfrak q}$ is the maximal ideal of the local ring $S_{\mathfrak q}$, and
2. the field extension $\kappa (\mathfrak q)/\kappa (\mathfrak p)$ is finite separable.
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Triangle Is Geometric Shape at Harold Porter blog
Triangle Is Geometric Shape. triangles are polygons (shapes) with three sides and three angles, which can be formed by connecting any three points in a plane. A triangle is made up of three connected
line segments. to classify triangles according to both angles and sides, we measure the interior angles and length of the sides of the. It is a complex geometric figure since it can have variable
angles and measurements. You think they are useful. a triangle is a closed shape with 3 angles, 3 sides, and 3 vertices. They are one of the first. geometry (all content) unit 4: A triangle with
three vertices p, q, and r is represented as pqr.
from www.dreamstime.com
a triangle is a closed shape with 3 angles, 3 sides, and 3 vertices. You think they are useful. to classify triangles according to both angles and sides, we measure the interior angles and length of
the sides of the. triangles are polygons (shapes) with three sides and three angles, which can be formed by connecting any three points in a plane. They are one of the first. A triangle with three
vertices p, q, and r is represented as pqr. geometry (all content) unit 4: It is a complex geometric figure since it can have variable angles and measurements. A triangle is made up of three
connected line segments.
Geometric Shapes, Line Design, Triangle Stock Vector Illustration of
Triangle Is Geometric Shape geometry (all content) unit 4: A triangle with three vertices p, q, and r is represented as pqr. You think they are useful. They are one of the first. geometry (all
content) unit 4: It is a complex geometric figure since it can have variable angles and measurements. to classify triangles according to both angles and sides, we measure the interior angles and
length of the sides of the. triangles are polygons (shapes) with three sides and three angles, which can be formed by connecting any three points in a plane. a triangle is a closed shape with 3
angles, 3 sides, and 3 vertices. A triangle is made up of three connected line segments. | {"url":"https://cemqrgug.blob.core.windows.net/triangle-is-geometric-shape.html","timestamp":"2024-11-03T20:28:31Z","content_type":"text/html","content_length":"36326","record_id":"<urn:uuid:b310366d-1246-4c92-a454-9491b41fc57f>","cc-path":"CC-MAIN-2024-46/segments/1730477027782.40/warc/CC-MAIN-20241103181023-20241103211023-00858.warc.gz"} |
Problem 054 – imperfect compression
Can you show that perfect compression is impossible?
Photo by Rui Matayoshi on Unsplash.
Problem statement
Compression is great: it is what lets you take that giant folder you have and reduce its size to save some memory on your laptop. Of course, you only do these compressions happily because you know
you don't lose information when you compress things. The data is just... compressed!
For compression to be useful, it has to be bidirectional: you must be able to recover the original data from the compressed version. This is only possible if two different pieces of data never get
compressed into the same thing. (In mathematical terms, we say that the compression must be injective.)
Now, on top of that, we are interested in compression that actually works, right? That is, in compression that reduces the size of things. Right?
Right! Now, the challenge is for you to show that no compression mechanism is perfect. In other words, show that if a compression mechanism is bidirectional and it manages to take some pieces of data
and transform them into something smaller, then, there are pieces of data that will become larger by the action of the compression mechanism.
If it makes it easier for you, we can suppose that the data we are talking about are just sequences of letters. So, we are talking about compression mechanisms that take sequences of letters and try
to build smaller sequences of letters, the compression. For example, maybe the sequence aaaaaa gets compressed into Aaab, but maybe the mechanism fails on AAAAAA because it “compresses” it into
If you need any clarification whatsoever, feel free to ask in the comment section below.
Congratulations to the ones that solved this problem correctly and, in particular, to the ones who sent me their correct solutions:
Know how to solve this? Join the list of solvers by emailing me your solution!
Let's assume that there is a perfect compression algorithm. For the empty sequence, what does this algorithm do? It has to compress the empty sequence into the empty sequence, because there is no
shorter sequence to compress the sequence into.
Now, let's think about sequences of length 1. None of those can be compressed into the only sequence of length 0, because there is already a sequence compressed into that (itself). Thus, all
sequences of length 1 must map to other sequences of length 1. Of course they don't map to sequences of length 2 or greater, otherwise the compression would actually make the sequences larger.
Therefore, the sequences of length 1 all map to each other, and no two sequences can map to the same one, so the compression algorithm really only works as a shuffling of the sequences...
Now, we can just repeat this train of thought indefinitely: for the sequences of length 2, none of them can map to sequences of length 0 or 1, because those are all taken already. Thus, all sequences
of length 2 must map to each other.
By using induction, we can show that this happens for all lengths: all sequences of length \(n\) are mapped within each other, because all shorter sequences are already taken up.
In practice, this shows that the compression algorithm is not useful, because it doesn't really compress any sequence at all! Therefore, there can't be a perfect compression algorithm that is also
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Primary 5 Online Materials
These PowerPoint slides contain teaching materials for Primary 5. It is designed to aid teachers in the delivery of Matrix Math lessons. The slides contain solutions to Matrix Math questions in
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An infinite MPS (iMPS) is described by a set of {$L$} A-matrices, {% A^{s_1} A^{s_2} \ldots A^{s_L} \; . %}
{$L$} is the unit cell size of the iMPS. These matrices describe a translationally invariant infinite system, where the unit cell is repeated on both the left and right.
It is important to distinguish the unit cell size of the wavefunction versus the unit cell size of the Hamiltonian. The unit cell size of the wavefunction is an integer multiple of the Hamiltonian
unit cell. Generally, we want to choose a wavefunction unit cell that is as small as possible, but it is sometimes not possible to have the wavefunction as just a single Hamiltonian unit cell. This
is because the wavefunction may break translation symmetry in a larger unit cell than the Hamiltonian. An example is the spin-1/2 Heisenberg chain, where the usual form of the Hamiltonian is such
that the groundstate has momentum {$\pi$}, and hence is only translationally invariant under 2-site shifts, although the Hamiltonian itself is invariant under a 1-site translation.
Lattice sites and operators
By convention, sites within the unit cell are denoted with square brackets. For example the Sp operator (toolkit notation for the {$S^+$} operator) acting on the 2^nd site of a unit cell is denoted
Sp[2]. If the operator is acting on a specific unit cell, then the unit cell number is denoted in round brackets, for example the 0^th unit cell, Sp(0)[2]. The site number within the unit cell, and
the unit cell number, are both zero-based, so (0)[0] will normally denote the left-most site in a system. But in most cases, the cell number for an iMPS is arbitrary because of translational
Tip: if the unit cell size is 1 site, then the [0] can be omitted.
The toolkit uses several different types of operators. Many kinds of operator expressions are allowed, for details see Main.OperatorExpressions.
Local operators are defined on a single site of a lattice. These are rarely used on their own, rather they are building blocks of other operators. The local operators are defined in the LatticeSite
object of a model, for example the models/spin-u1.h defines a LatticeSite that contains operators Sz, Sp, Sm, I, R, P.
Most commonly, we will refer to a local operator acting on a specific site and unit cell, as, for example, Sp(2)[0], or Sz(5) (if the unit cell is 1-site).
These operators act on a finite support. For example, the operator Sp(0)[0] * Sm(5)[1] is a finite operator that spans 6 unit cells, from 0 .. 5, and represents the operator that decreases the spin
at site 1 of the 5th unit cell, and increases the spin at site 0 of the 0th unit cell.
To calculate expectation values of finite operators, use mp-iexpectation. An example is
mp-iexpectation psi lattice:"Sz(0)*exp(i*pi*Sz(1))*Sz(2)"
This command will calculate a string correlation function between unit cells 0 and 2. Since there is no cell index specified, it is assumed that the unit cell is only 1 site.
Finite operators can be defined in the lattice file too. In this case, the operator is always attached to the unit cell. In the the C++ code this is denoted using a UnitCellOperator. For example,
UnitCell Cell(MyLatticeSite);
UnitCellOperator Sz(Cell, "Sz");
UnitCellOperator U(Cell, "U");
U = exp(complex_i * pi * Sz(0));
In this example, Sz is a pre-existing local operator, and the new operator (or we overwrite it, if it already existed) U is defined as {$\exp[i\pi Sz(0)]$}. With this definition, we could, for
example, simply the previous mp-iexpectation command to
mp-iexpectation psi lattice:"Sz(0)*U(1)*Sz(2)"
Note that a UnitCellOperator can have arbitrary finite support. So we could, for example, define
UnitCellOperator B(Cell, "B");
UnitCellOperator BH(Cell, "BH");
UnitCellOperator BondCurrent(Cell, "BondCurrent");
BondCurrent = complex_i * (BH(0)*B(1) - B(0)*BH(1));
and then mp-expectation psi lattice:"BondCurrent(0)" will calculate the expectation value of the bond current operator at the bond between unit cells 0 and 1.
Because of wavefunction translational invariance, the cell number is only meaningful modulo the wavefunction unit cell size. So if the wavefunction has the same unit cell size as the lattice itself,
then the cell index is irrelevant; BondCurrent(0) means exactly the same thing as BondCurrent(4), or even BondCurrent(-10). But if, for example, the wavefunction is larger, and is defined on N
lattice unit cells, then expectation values will also be periodic with period N.
So-called triangular operators have infinite support, and represent the sum of terms acting on all unit cells. They can also represent arbitrary polynomials of these operators. The name 'triangular
operator' arises because the Matrix Product Operator representation of these operators is of the form of an upper-triangular matrix.
The simplest triangular operator is the sum of some local term. The basic way of constructing a triangular operator (in either C++ or the expression parser) is with the sum_unit() function. The
sum_unit() function takes a finite operator, and constructs a triangular operator that represents the sum of all possible translations of that finite operator.
For example, sum_unit(Sz(0)) represents the infinite sum {$\sum_i S^{z}(i)$}, which is the z-component of magnetization operator.
sum_unit() will calculate the sum of an arbitrary finite operator. This makes it useful for constructing Hamiltonians. For example, the Hamiltonian of the Heisenberg spin chain is {%H = \sum_i S^z(i)
S^z(i+1) + \frac{1}{2} \left( S^+(i) S^-(i+1) + S^-(i) S^+_(i+1) \right)%} and we would construct this operator using
sum_unit( Sz(0)*Sz(1) + 0.5*( Sp(0)*Sm(1) + Sm(0)*Sp(1) ) )
Triangular operators can be added to a C++ Lattice object directly, for example
Lattice["H"] = sum_unit( Sz(0)*Sz(1) + 0.5*( Sp(0)*Sm(1) + Sm(0)*Sp(1) ) )
and the command-line parser also allows many kinds of expressions on triangular operators.
The basic tool for calculating expectation values of triangular operators is mp-imoments. This is named because it can calculate moments of an operator, {$\langle X^n \rangle$}, up to some arbitrary
power {$n$} (limited by computation time). For example,
mp-imoments psi lattice:"sum_unit( Sz(0)*Sz(1) + 0.5*( Sp(0)*Sm(1) + Sm(0)*Sp(1) ) )"
will calculate the energy of a Heisenberg Hamiltonian for the wavefunction psi.
Expectation values of triangular operators are defined per unit cell. So for example, we have the energy per unit cell, or the magnetization per unit cell. More generally, products of triangular
operators have expectation values that are polynomial functions per unit cell. For example, if we add the --power 2 option to mp-imoments (or simply square the operator), then it is calculating the
expectation value {$\langle H^2 \rangle$}, which is a polynomial in the number of unit cells L, {%\langle H^2 \rangle_L = L^2 \epsilon^2 + L \sigma^2%} where {$\epsilon$} is the energy per unit cell,
and {$\sigma^2$} is the variance per unit cell.
Tip: calculating the expectation value of H^2 in this way is very useful as the variance is an excellent measure of the convergence of the wavefunction.
The final type of operator that is used in the toolkit denotes an infinite product of operators. These operators are typically used for string correlation functions, or Jordan-Wigner strings, or
evolution operators, for example from the Suzuki-Trotter decomposition of exp(H).
Analogously to sum_unit for triangular operators, product operators are constructed using prod_unit. For example, prod_unit(exp(i*pi*Sz(0))) represents the operator {%\mbox{prod_unit(exp(i*pi*Sz
(0)))} \equiv \Pi_i (-1)^{S^z(i)} \equiv \exp \left[ i \pi \sum_i S^z(i) \right]%}
Product operators are used in several contexts. To calculate the expectation value of a product operator, use the mp-ioverlap command. This command calculates the overlap between two wavefunctions
(or a wavefunction and itself), and takes an optional argument where you can supply a product operator, --string <operator>. Expectation values of product operators are defined per unit cell, in a
multiplicative sense. If the expectation value is {$d$} per unit cell, then on {$L$} unit cells the expectation value is {$d^L$}. In the typical case, the product operator is unitary, and it
therefore makes sense to also describe the expectation value in terms of a correlation length {$\xi = - 1 / \ln d$}. | {"url":"https://mptoolkit.qusim.net/IDMRG/Overview","timestamp":"2024-11-14T22:03:54Z","content_type":"application/xhtml+xml","content_length":"19980","record_id":"<urn:uuid:4d6eed8c-3b6c-4bba-8610-e0d1622888c8>","cc-path":"CC-MAIN-2024-46/segments/1730477395538.95/warc/CC-MAIN-20241114194152-20241114224152-00221.warc.gz"} |
SFDR Considerations in Multi-Octave Wideband Digital Receivers - ELE Times
Electronic warfare (EW) receivers must intercept and identify unknown enemy signals among a congested wideband spectrum of multiple interfering signals without the benefit of dynamic range and
sensitivity improvement techniques employed in communications and radar receivers. The incident RF band limiting employed in communications receivers is an unwanted trade for the EW receiver that
seeks to process ever wider instantaneous bandwidth in less time. In the radar realm, receiver dynamic range benefits from matched filtering, whereby the received radar return is correlated with a
copy of the transmitted signal. Alas, the EW receiver has no prior knowledge of the signal to be intercepted and thus nothing with which to correlate! It’s like searching a crowd of people for a
stranger you’ve never seen before … and worse yet, he is hiding, or maybe isn’t even there!
Now for the good news: over the coming years, high sample rate analog-to-digital converter (ADC) and digital-to-analog converter (DAC) technology will usher in a wideband digital receiver
architectural evolution. Most importantly, converters from Analog Devices will maintain the excellent linearity, noise performance, and dynamic range of legacy lower rate digital converters. The
workhorse super-heterodyne tuner will give ground to direct sample and direct conversion architectures. Adaptive spectral tuning will continue to shift from the RF to the digital signal processing
This sea change in wideband RF sensing will enable size, weight, power, and cost (SWaP-C) benefits: higher receive and transmit channel counts at lower cost per channel, in the same or smaller sized
form factors as today.
Anticipating the coming era of digital EW receivers with multi-octave bandwidth, this article discusses new challenges and considerations when designing for best-in-class dynamic range. In this
article, dynamic range refers to instantaneous spur free dynamic range, the key figure of merit for receivers tasked with detecting small signals among a crowded spectrum of larger blockers.
Next-Generation ADC Performance
Many of today’s EW receivers feature sub-octave instantaneous bandwidth (IBW) that is limited by the older generation data converter. These will be replaced tomorrow with multi-octave wideband
digital receivers spanning several GHz of IBW. For example, in the coming years a growing number of sensing platforms will employ ADI converter chips featuring ADCs and DACs with the ability to
process greater than 4 GHz IBW while maintaining SFDR greater than 70 dB.
A popular low SWaP, wideband digital receiver ADC use case might be:
• An ADC sample rate of ~15 GSPS
• A direct sample of the first Nyquist zone (that is, dc to 6 GHz)
• A direct sample of the second Nyquist zone (that is, 8 GHz to 14 GHz)
• RF block convert middle (6 GHz to 8 GHz) and higher (>14 GHz) bands
EW receivers need to cover higher and higher swaths of spectrum from 18 GHz to 50 GHz and beyond. The ADC’s high second Nyquist zone eases the frequency plan, allowing simple RF front-end block
converters with relaxed, smaller SWaP RF filters. The following discussion considers an RF front end cascaded with a high sample rate ADC similar to the previous example.
Dynamic Range in Wideband Digital Receivers
Receiver designers optimizing dynamic range must balance sensitivity (NF) with linearity (IP2, IP3) as these RF device attributes usually move against each other. Dynamic range is bound by
sensitivity at lower RF levels and linearity at higher RF levels. As a rule of thumb, the maximum allowed receiver operating level is set so that the multisignal intermodulation distortion (IMD)
spurious levels are equal to the noise power, as shown in Figure 1. Modern systems use adaptive instantaneous bandwidth channelization and processing bandwidths (B[v]), which moves the noise floor up
and down 10Log(B[v]). The nuanced topic of processing bandwidth is critical and receives its own discussion later.
Figure 1. Relating SFDR to ADC operating range, noise, IMD spurs, and detection threshold.
Multi-Octave IMD2 Challenges in Wideband Digital Receivers
The wideband digital receiver evolution introduces new RF challenges. Multisignal second-order intermodulation distortion (IMD2) spurs emerge as problematic dynamic range impairments in the
multi-octave wideband digital receiver. While IIP3 has long been a key figure of merit (FOM) in RF device data sheets, IIP2 is harder to track down and can be more problematic to the EW designer. The
problem with IMD2 spurs is that they only fall off by 1 dBc for every 1 dB decrease in the incident 2-tone signal power, while third-order intermodulation distortion (IMD3) spurs fall off by 2 dBc.
Of course, multi-octave direct RF sampling at the lower portion of the ADC first Nyquist zone is nothing new. For example, an older system might sample at 500 MSPS and observe dc to 200 MHz in the
first Nyquist zone with no IMD2 problems. This is because at these lower frequencies (that is, less than a few hundred MSPS), ADC characteristics are highly linear and the effective IIP2 and IIP3 of
the ADC is very high, resulting in benign IMD2 products invisible below the noise floor. Just like in wideband RF devices, however, multi-GHz, multi-octave ADC linearity will degrade with increasing
frequency, and IMD2 products will often sit above the noise floor at higher operating frequencies. Going forward, we’ll need to deal with IMD2.
Broadened SFDR Definition for Wideband Digital Receivers
IMD2 crashing the party requires a refreshed definition of the popular receiver FOM instantaneous spurious-free dynamic range (SFDR). SFDR specifies how far down a receiver can detect a small signal
when there are multiple larger signals creating IMD spurs. SFDR is specified in dB relative to the large signals.
Traditionally, SFDR is defined in terms of IMD3 products, along with NF and processing bandwidth. IMD3-referenced SFDR is derived in many texts, and is sometimes clarified as instantaneous SFDR,
which is what we mean in this article. We’ll call it SFDR3:
Today IMD2-referenced SFDR receives less attention, but it is looming on the horizon as a major impairment needing mitigation. It can be derived in the same manner as SFDR3. Here we’ll call it SFDR2:
Figure 2 illustrates an RF front-end spectral scenario whereby three simultaneous signals (F1, F2, and F3) create intermodulation products that set the lower bound to dynamic range. Below this level,
the wideband digital receiver can’t easily tell whether a target is real or a false IMD spur.
Figure 2. An example of multisignal F1, F2, and F3 (60 MHz each) inducing second-harmonic, IMD2 (red), IMD3 (green), and IMD2/3 combo (gray) spurs. The noise floor (brown) is noted as P[N].
The sub-octave IBW receiver of today, notionally shown by the Figure 2 dashed box, worries only about IMD3 because it falls in-band and cannot be filtered. It doesn’t worry much about IP2 because of
the easily filtered location of IMD2 and the inducing signals. F3 is easily chopped down using input RF filtering, which takes F3 – F1 and F3 – F2 way below the noise floor. Much like the F1 and F2
second harmonics, the F1 + F2 IMD2 is easily attenuated using output filtering. Of course, the second-order performance of the ADC must be considered in relation to Nyquist folding spurs, but the
front-end IMD2 performance is easy to deal with.
Enter the multi-octave IBW receiver, notionally shown by the Figure 2 solid box, and the situation turns on its head. IMD2 is the bigger concern vs. IMD3. The IMD2 spurs and inducing interferers are
now in-band. Band-pass filtering them defeats the purpose of a multi-octave IBW. This is why tunable notch filtering, despite its limitations, is seeing increased attention as a front-end
interference mitigator. It doesn’t lop off giant pieces of the multi-octave spectrum.
Figure 3 illustrates the relationship between the fundamental multi-tone large signal, IMD2 and IMD3 level, noise floor, and the resulting SFDR for an example multi-octave wideband digital receiver.
The example uses real noise and linearity attributes for an ADC sampling the first Nyquist zone with a 4 GHz IBW from 2 GHz to 6 GHz. A processing bandwidth of 469 kHz is assumed.
Figure 3. SFDR2 and SFDR3 tell you how far down from the largest signal (fundamental) you can easily detect a smaller signal. Because it varies widely, the detection threshold is zero here. In
practice, subtract your detection threshold from SFDR.
Optimal SFDR2 and SFDR3 occur at different P[in] operating points where the respective IMD level intersects the noise power. If we pretend for a moment that this is a sub-octave receiver with
front-end RF band-limiting, SFDR3 sets the overall SFDR, and we can expect a best case SFDR of 79 dB, which is very good. But since the EW receiver requires multi-octave IBW, SFDR2 sets the overall
SFDR. At the best SFDR3 input level (P[in] = –20 dBm), the IMD2 spurs degrade the SFDR by 24 dB, resulting in an SFDR of 55 dB. Fair, albeit disappointing, results.
A useful rule of thumb is that for a specific RF output level = P[RF,O] to achieve equivalent IMD2 and IMD3 levels:
In other words, this condition will make the SFDR2 and SFDR3 lines intersect the noise floor at the same spot, so that SFDR2 is not limiting performance.
For the previous SFDR example scenario, the RF front end feeds the ADC with –20 dBm and has an OIP3 of 20 dBm. The required OIP2 to get the same level IMD2 and IMD3 spurs, and thereby not limit
performance, is:
That raw device OIP2 performance is not available today given the balance with other attributes like frequency, bandwidth, noise, and dc power. This explains the increasing interest in
next-generation adaptive front-end interferer mitigation techniques.
To mitigate IMD2, the receiver must lower the max input operating level from –20 dBm to –32 dBm, and is then able to achieve an improved SFDR2 of 66 dB best case. In Figure 3, this optimal SFDR2 is
where the IMD2 trace intersects the noise floor. Alas, the best case SFDR2 at P[in] = –32 dBm is still 13 dB worse than the best case SFDR3 at –20 dBm. Since we’ve now shifted the max operating level
down, this puts the spotlight on noise power (sensitivity) limitations, as discussed in the next sections.
What Sets Processing Bandwidth in the Wideband Digital Receiver?
The sensitivity, or noise power, of the EW receiver gets better as the processing bandwidth narrows. In typical fashion, however, there are trade-offs to balance: we can’t just reduce the bandwidth
to an arbitrarily small value and head to lunch. What are the competing factors to consider? To answer the question, we need to discuss decimation, the fast Fourier transform (FFT), and their
relationship. First, we define a couple variables:
ADI’s high sample rate ADCs employ on-chip digital signal processor (DSP) blocks that allow configurable filtering and decimation of the raw data stream to a minimum viable payload sent to the
downstream FPGA. This process is discussed in detail across ADI literature. The obvious benefit of decimation is reducing the digital payload that must pass over JESD204B/JESD204C to the FPGA.
Another benefit is the power consumption savings realized using local on-chip decimation-specific circuitry (that is, ASIC) vs. implementing the same operation in the FPGA fabric. But local on-chip
decimation is beneficial beyond just thinning the data stream and saving power. We’ll get to that.
Figure 4 shows the blocks used in modern wideband digital conversion (as relevant to this discussion). The flow consists of sampling, digital downconverting, digital filtering, decimating, and fast
Fourier transforming the data stream.
Figure 4. Simple block diagram of ADC data decimation and FFT.
First, the data sampled at f[S] is digital downconverted to baseband (complex I/Q) using a fine-tuned NCO. The data stream is then filtered using a programmable low-pass digital filter. This
predecimation digital filtering sets the IF bandwidth and is the first of two different operations to set the receiver noise floor P[N]. As the IF bandwidth gets smaller, the integrated in-band noise
power decreases as the filtering attenuates the wideband noise.
Next, decimating by M reduces the effective sampling rate to f[s]/M, keeping every M^th sample and throwing out the samples in between.
Thus, the downstream FFT processing gets a data stream with rate f[S]/M and bandwidth f[S]/2M. Finally, the FFT length N sets the bin width and capture time, which is the second step in setting the
noise floor.
Decimation and FFT Impact to Wideband Digital Receiver Noise Floor
Figure 5 relates the wideband digital receiver’s processing noise floor (K) to the ADC’s noise spectral density (L), which is the widely available data sheet FOM for ADC additive noise. Existing ADI
literature does a nice job explaining processing gain, NSD, SNR, and quantization noise.
The most useful relation from Figure 5 is:
The processing noise floor (Figure 5, K) is the same as P[N] and can be dropped into Equation 1 and Equation 2. Note that the designer carefully selects M and N based upon design trade-offs and
constraints discussed in the next section.
Figure 5. Relationship of decimation and FFT gain operations to commonly referenced noise levels.
Even though increasing the decimation factor M has the same proportional effect in reducing the noise floor (Figure 5, C) as increasing the FFT length N (Figure 5, E), it is important to note the
mechanisms are entirely different. The decimation step involves band-limiting the channel using digital filtering. This sets the effective noise bandwidth that determines the total integrated noise
in the channel (Figure 5, D). It also sets the maximum instantaneous spectral bandwidth of a detectable signal. Compare this to the FFT step, which does not filter per se, but spreads the total
integrated noise in the channel over N/2 bins and defines the spectral line resolution. The higher N, the more bins, and the lower the noise content per bin.^8 Together, decimation gain M and FFT
gain N define the FFT bin width, and they are often lumped together in discussions of processing bandwidth (Figure 5, F), but their values must be balanced based upon their respective nuanced impact
to signal bandwidth, spectral resolution, sensitivity, and latency requirements, as discussed in the next section.
Processing Bandwidth and System Performance Trade-Offs
Relating decimation M and FFT N back to high priority performance attributes:
Latency is the time to sense and process successive spectral captures, and it requires as short a time as possible. Many systems require near real-time operation. This dictates M × N be as small as
possible. As the FFT size increases, the spectral resolution improves and noise floor decreases as the integrated noise is spread over more bins. The trade-off is acquisition time, which is a big
deal and is simply:
The minimum detectable pulse width (PW) sets the minimum allowable IF channel bandwidth as the spectral content of a shorter time pulse spreads over a relatively wider frequency band. If the IF
channel bandwidth is too narrow, the signal spectral content truncates, and the short time pulse isn’t detected properly. Minimum IF BW, which sets maximum allowable M, must meet the criteria:
Spectral resolution and sensitivity improve as the FFT bin narrows, which requires increasing N. Longer pulse widths and PRIs require finer resolution to resolve closer spectral lines, which means
larger N for proper detection. Increasing N improves spectral line resolution, but only within the IF bandwidth defined by M. If too high a decimation is used, increasing N improves the spectral
resolution within the IF BW set by M, but cannot recover the missing signal bandwidth. For example, a pulse train with a pulse width below the minimum receiver pulse width will have a frequency
domain sinc function whose main lobe exceeds the decimation bandwidth. Increasing N will help resolve the PRF of the train, but will do nothing to resolve the pulse width; that information is lost.
The only fix is to decrease decimation M, increasing the IF bandwidth.
Decimation, FFT, and Detection of Pulse Trains
EW wideband digital receivers spend a lot of their effort de-interleaving, identifying, and tracking simultaneous incident radar pulse trains. Carrier frequency, pulse width, and pulse repetition
interval (PRI) are radar signatures that are critical in figuring out who’s who. Both the time and frequency domain are used in detection schemes.^9 An overarching objective is to sense, process, and
react to the pulse trains in as short a time duration as possible. Dynamic range is critical because the EW receiver needs to simultaneously track multiple distant targets while being bombarded with
high energy jamming pulses.
Pulse Train FFT Examples
Two pulse train examples are presented. The first represents a pulsed doppler radar exhibiting a very short PW (100 ns) at 10% duty cycle, resulting in very high PRF. The second simulates a pulsed
radar exhibiting comparatively longer PW and PRI (lower duty cycle, lower PRF). The following plots and tables illustrate the impact of decimation M and FFT length N on time, sensitivity (noise
floor), and spectral resolution. Table 1 summarizes the parameters for easy comparison. The fictional values do not represent specific radars but are nevertheless in a realistic ballpark.
Table 1. Comparison of Example Pulsed Doppler and Pulsed Radar Attributes
Parameter Pulsed Doppler Radar Pulsed Radar
PW Short 100 ns Longer 10 μs
PRI Short 1 μs Longer 1 ms
PRF High 1 MHz Low 1 kHz
Duty Cycle Mid/high 10% Mid/low 1%
Decimation M Low 256 High 1536
FFT Length N Low 128 to 512 High 16,384 to 65,536
Time Quick 2 μs to 9 μs Longer 2 ms to 7 ms
Sensitivity Lower –91 dBFS Higher –120 dBFS
The point here is that M and N are not one size fits all, and sophisticated detection algorithms and parallel channelization schemes in any given EW receiver may employ a wide range of values for
each. The EW receiver must be able to detect both signals, likely at the same time (not shown here), which is why fast, adaptable configurability is important. Dynamic range and sensitivity are
directly dependent on the pulse attributes that must be detected.
Example: Wideband Digital Receiver Sensing Pulsed Doppler Radar
The following two FFTs capture a pulsed doppler scenario.
The first FFT shown in Figure 6 needs just over 2 pulse cycles to determine the pulse width of the signal from the width of the FFT main lobe. The decimation M is set for an IF bandwidth that is
adequately wide to capture the main lobe, as well as some sidelobes. The response time is very fast. The trade-off to quick response time is a worse noise floor and spectral resolution. Note that due
to the lack of spectral resolution, no PRI information is available in the FFT.
Figure 6. Fast capture of narrow pulse width, high PRF pulse train typical of pulsed doppler radar.
The second FFT in Figure 7 shows an improved noise floor and spectral resolution as sample length N (and time) are increased. M remains the same. By around nine pulse cycles, the spectral resolution
improves enough to determine the PRI (1/PRF) from the FFT. The noise floor can be seen between sidelobes.
Figure 7. A longer FFT of a pulsed Doppler example to resolve spectral lines.
Example: Wideband Digital Receiver Sensing Pulsed Radar
The following two FFTs capture a wider pulsed scenario.
The much wider PRI, or lower pulse density, in the pulsed radar example in Figure 8 requires much higher N. Adjusting M is entirely system dependent. If the short pulse must be detected
simultaneously with the long pulse in the same IF channel, then M must be set to accommodate the short pulse spectral bandwidth and cannot be increased. Considered on its own, the long pulse requires
a lower IF bandwidth, so M could be set higher to improve the channel noise and resulting sensitivity. The capture time, or FFT length N, required is a lot longer, however. So it’s likely the
detection algorithm would want to make intermediate decisions on the short pulse scenario while the system acquires a high enough N to resolve the long pulse.
Figure 8. Fast capture of longer pulse, lower PRF pulse train typical of pulsed radar.
The second long pulse FFT example in Figure 9 illustrates how the long PRI (low PRF) results in very close spectral lines, which requires very low FFT bin size or resolution bandwidth. The trade-off
is even more time required (FFT N). A benefit is even better sensitivity.
Figure 9. A longer FFT of pulsed example to resolve spectral lines.
Wideband Digital Receiver RF Front-End Design Using a Cascaded ADC
With dynamic range and sensitivity goals established, an RF front end must be paired with the digital data converter. The optimal RF front end sets the receiver sensitivity (NF) and performs the
required spectral signal conditioning with good enough linearity head room to allow the ADC performance to set receiver IP3 and IP2. Front-end RF gain is typically set to be good enough to establish
the required cascaded NF, as gain beyond that generally hurts dynamic range and is avoided. It is criminal if the front end bottlenecks dynamic range and ADC capability is thrown out!
A helpful trick is to convert the ADC figures of merit to equivalent RF cascade parameters and treat the ADC like an RF black box. Some rules of thumb:
Where P[RF ](dBm) is the ADC input RF level at which the IMD3 and IMD2 levels are measured.
Note that the cascaded system NF of the combination front end and ADC is broadband noise prior to adjusting for processing gain.
Design Example of Front End to ADC Cascade
An example cascade analysis follows using the front end shown in Figure 10. This chain benefits from the latest ADI releases to the RF catalog, including:
Additionally, the chain features a wideband 200 W RF limiter and small form factor high Q fixed filtering developed at ADI.
Figure 10. Example RF front end featuring switched high sensitivity and bypass modes.
An age-old technique to preserve dynamic range is to switch between a high sense mode for lower input signals and bypass mode for higher input signals. As shown in Table 2, the high sense path favors
NF performance, and the bypass path concedes higher NF in favor of higher linearity (IP2 and IP3). The performance tables illustrate this benefit.
Table 2. Example RF Front-End Black Box Parameters for the Two Modes
Mode G (dB) NF (dB) IIP2 (dBm) IIP3 (dBm) IP1dB (dBm)
High Sense 10 15 31 17 5
Bypass –14 14 75 40 25
Table 3 compares the front end and ADC black box parameters, and the resulting overall cascade performance.
In the high sense mode, the limiting factor to dynamic range is the noise floor, and so cascaded NF is prioritized. The front-end noise figure depends mostly on the insertion loss of the front-end
filtering required for interferer mitigation (this example budgets 6 dB loss). This preselect filtering needs to sit before the amplifier to be effective, as the amplifier will create multisignal IMD
In bypass mode, we benefit from the extremely high linearity of the SOI technology. No tricks here as the amplifier limited linearity is simply switched out in favor of higher linearity, lower gain,
and higher NF.
Table 3. Example High Sense (top) and Bypass (bottom) Cascaded Performance; the Overall Column Is the Cascaded RF Front End plus ADC All-In Performance
RF Front End ADC Overall Units
Full Scale –6.5 dBm
NSD –148 dBFS/Hz
–154.5 dBFS/Hz
Gain 10 0 dB
NF 15 19.5 16.1 dB
IIP2 31 35 21.5 dBm
IIP3 17 20 9.2 dBm
Pi –40 –30 dBm
P[N] –91.2 dBm
RF Front End ADC Overall Units
Full Scale –6.5 dBm
NSD –148 dBFS/Hz
–154.5 dBFS/Hz
Gain –14 0 dB
NF 14 19.5 33.5 dB
IIP2 75 35 48.6 dBm
IIP3 40 20 33.0 dBm
Pi –15 –29 dBm
P[N] –97.8 dBm
Wideband Digital Receiver Design Results and Optimization
The following performance heat maps are sensitivity analyses showing instantaneous spur free dynamic range (DR, dB) for varying:
• Processing bandwidth and RF input level
• RF front-end IIP2 and RF input level
• RF front-end NF and RF input level
Each scenario is run for the high sensitivity and bypass paths. The boxes annotate favorable operating zones. The tables tell you the dynamic range (SFDR), or distance down to the noise floor or
highest IMD spur, for a given max input signal level at P[in]. For any given table, the static variables are set per the previous chain parameters.
As discussed in prior sections, the B[v] selected in Figure 11 is dependent upon waveform detection objectives. Lower B[v] decreases the noise floor, improving dynamic range at low P[in], but at the
expense of slower FFT times. Inversely, high B[v] values increase the noise floor, and poor sensitivity limits dynamic range. The likely operating zone is at a balance point in between.
Figure 11. Instantaneous spur free dynamic range (DR) vs. RF input level (P[in]) and processing bandwidth (B[v]); high sensitivity (top) and bypass mode (bottom).
Figure 12 illustrates that, at low P[in] levels, IIP2 is irrelevant as the sensitivity sets the dynamic range. The mid-range performance is most sensitive to IIP2. Mid-range input power levels might
comprise most use cases, and as P[in] increases toward the high sense to bypass switch point, amplifier linearity, especially IP2, is critically important. The superior IP2 of ADL8104 stands out over
this important mid-range, preserving high dynamic range performance.
The bypass mode higher IIP2 allows the operating zone box to shift down to follow the best dynamic range.
Figure 12. Instantaneous spur free dynamic range (DR) vs. RF input level (P[in]) and RF front-end input referenced IP2; high sensitivity (top) and bypass mode (bottom).
Figure 13 shows that for big improvements to NF, which can be very costly to SWaP-C and linearity, there is a diminishing return to dynamic range using a mid-range B[v]. For lower NF to pay off, B[v]
needs to decrease with it and the associated trade-offs tolerated. The high sense mode does well with an NF in the 10 dB to 15 dB range. For the bypass mode, the high NF is shown to be a willing
trade-off given the benefit to linearity. Ideally NF can be kept in the 20 dB to 25 dB range for the bypass mode. Better NF in bypass mode doesn’t help dynamic range, as we are IMD limited.
Figure 13. Instantaneous spur free dynamic range (DR) vs. RF input level (P[in]) and RF front-end noise figure (NF); high sensitivity (top) and bypass mode (right).
Electronic warfare’s imminent evolution toward multi-octave, multi-GHz instantaneous bandwidth RF tuners and wideband digital receivers introduce IMD2 effects that challenge dynamic range. Today’s
consideration of SFDR in terms of IMD3 will broaden to include IMD2, and the designer will use both the SFDR2 and SFDR3 equations. The system noise floor is dynamic because processing bandwidth
changes on-the-fly based upon waveform detection and time requirements. When designing the optimal noise floor, decimation M and FFT depth N together define the FFT bin width, yet they each have
separate important impacts to consider. Example pulse train FFTs of varying M and N are provided. As ADC performance improves, the front end continues to rely on high linearity wideband RF components
with tunable attributes and frequency selectivity. The front end should be designed in cascade with the ADC’s RF attributes.
Benjamin Annino | {"url":"https://www.eletimes.com/sfdr-considerations-in-multi-octave-wideband-digital-receivers","timestamp":"2024-11-05T03:53:50Z","content_type":"text/html","content_length":"729395","record_id":"<urn:uuid:bcded6dd-b63e-4d1d-8048-4f5357fe17fb>","cc-path":"CC-MAIN-2024-46/segments/1730477027870.7/warc/CC-MAIN-20241105021014-20241105051014-00630.warc.gz"} |
Second Order Differential Equation - Solver, Types, Examples, Methods
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Second Order Differential Equation
Second order differential equation is a specific type of differential equation that consists of a derivative of a function of order 2 and no other higher-order derivative of the function appears in
the equation. It includes terms like y'', d^2y/dx^2, y''(x), etc. which indicates the second order derivative of the function. Generally, we write a second order differential equation as y'' + p(x)y'
+ q(x)y = f(x), where p(x), q(x), and f(x) are functions of x. We can solve this differential equation using the auxiliary equation and different methods such as the method of undetermined
coefficients and variation of parameters.
The differential equation y'' + p(x)y' + q(x)y = 0 is called a second order differential equation with constant coefficients if the functions p(x) and q(x) are constants and it is called a
second-order differential equation with variable coefficients if p(x) and q(x) are not constants. In this article, we will understand such differential equations in detail and their different types.
We will also learn different methods to solve second order differential equations with constant coefficients using the various methods with the help of solved examples and finding the auxiliary
1. What is a Second Order Differential Equation?
2. Second Order Differential Equation Definition
3. Solving Second Order Differential Equation
4. FAQs on Second Order Differential Equation
What is a Second Order Differential Equation?
A differential equation is an equation that consists of a function and its derivative. A differential equation that consists of a function and its second-order derivative is called a second order
differential equation. Mathematically, it is written as y'' + p(x)y' + q(x)y = f(x), which is a non-homogeneous second order differential equation if f(x) is not equal to the zero function and p(x),
q(x) are functions of x. It can also be written as F(x, y, y', y'') = 0. Further, let us explore the definitions of the different types of the second order differential equation.
Second Order Differential Equation Definition
A second order differential equation is defined as a differential equation that includes a function and its second-order derivative and no other higher-order derivative of the function can appear in
the equation. It can be of different types depending upon the power of the derivative and the functions involved. These differential equations can be solved using the auxiliary equation. Let us go
through some special types of second order differential equations given below:
Linear Second Order Differential Equation
A linear second order differential equation is written as y'' + p(x)y' + q(x)y = f(x), where the power of the second derivative y'' is equal to one which makes the equation linear. Some of its
examples are y'' + 6x = 5, y'' + xy' + y = 0, etc.
Homogeneous Second Order Differential Equation
A second order differential equation y'' + p(x)y' + q(x)y = f(x) is said to be a second order homogeneous differential equation if f(x) is a zero function and hence mathematically it of the form, y''
+ p(x)y' + q(x)y = 0. Some of its examples are y'' + y' - 6y = 0, y'' - 9y' + 20y = 0, etc.
Non-homogeneous Second Order Differential Equation
A differential equation of the form y'' + p(x)y' + q(x)y = f(x) is said to be a non-homogeneous second order differential equation if f(x) is not a zero function. Some of its examples are y'' + y' -
6y = x, y'' - 9y' + 20y = sin x, etc.
Second Order Differential Equation with Constant Coefficients
The differential equation y'' + p(x)y' + q(x)y = f(x) is called a second order differential equation with constant coefficients if the functions p(x) and q(x) are constants. Some of its examples are
y'' + y' - 6y = x, y'' - 9y' + 20y = sin x, etc.
Second Order Differential Equation with Variable Coefficients
The differential equation y'' + p(x)y' + q(x)y = f(x) is called a second order differential equation with variable coefficients if the functions p(x) and q(x) are not constant functions and are
functions of x. Some of its examples are y'' + xy' - y sinx = x, y'' - 9x^2y' + 2e^xy = 0, etc.
Solving Second Order Differential Equation
Now that we have understood the meaning of second order differential equation and their different forms, we shall proceed towards learning how to solve them. Here, we will focus on learning to solve
2nd differential equations with constant coefficients using the method of undetermined coefficients. First, let us understand how to solve the second order homogeneous differential equations.
Solving Homogeneous Second Order Differential Equation
A homogeneous second order differential equation with constant coefficients is of the form y'' + py' + qy = 0, where p, q are constants. To solve this, we assume a general solution y = e^rx of the
given differential equation, where r is any constant, and follow the given steps:
• Step 1: Differentiate the assumed solution y = e^rx, and find y' = re^rx, y'' = r^2e^rx, where r is an arbitrary constant.
• Step 2: Substitute the derivatives in the given differential equation y'' + py' + qy = 0. We have r^2e^rx + pre^rx + qe^rx = 0 ⇒ e^rx(r^2 + rp + q) = 0 ⇒ r^2 + rp + q = 0, which is called the
auxiliary equation or characteristic equation.
• Step 3: Solve the auxiliary equation r^2 + rp + q = 0 and find its roots r[1] and r[2].
□ If r[1] and r[2] are real and distinct roots, then the general solution is y = Ae^r[1]x + Be^r[2]x
□ If r[1] = r[2] = r, then the general solution is y = Ae^rx + Bxe^rx
□ If r[1] = a + bi and r[2] = a - bi are complex roots, then the general solution is y = e^ax(A sin bx + B cos bx)
Let us consider a few examples of each type to understand how to determine the solution of the homogeneous second order differential equation.
Example 1: Solve the 2nd order differential equation y'' - 6y' + 5y = 0
Solution: Assume y = e^rx and find its first and second derivative: y' = re^rx, y'' = r^2e^rx
Next, substitute the values of y, y', and y'' in y'' - 6y' + 5y = 0. We have,
r^2e^rx - 6re^rx + 5e^rx = 0
⇒ e^rx(r^2 - 6r + 5) = 0
⇒ r^2 - 6r + 5 = 0 → Characteristic Equation
⇒ (r - 5) (r - 1) = 0
⇒ r = 1, 5
Since the roots of the characteristic equation are distinct and real, therefore the general solution of the given differential equation is y = Ae^x + Be^5x
Example 2: Solve the second order differential equation y'' - 8y' + 16y = 0
Solution: Assume y = e^rx and find its first and second derivative: y' = re^rx, y'' = r^2e^rx
Next, substitute the values of y, y', and y'' in y'' - 8y' + 16y = 0. We have,
r^2e^rx - 8re^rx + 16e^rx = 0
⇒ e^rx(r^2 - 8r + 16) = 0
⇒ r^2 - 8r + 16 = 0 → Auxiliary Equation
⇒ (r - 4) (r - 4) = 0
⇒ r = 4, 4
Since the roots of the characteristic equation are identical and real, therefore the general solution of the given differential equation is y = Ae^4x + Bxe^4x
Example 3: Solve the second order differential equation 9y'' + 12y' + 29y = 0
Solution: Assume y = e^rx and find its first and second derivative: y' = re^rx, y'' = r^2e^rx
Next, substitute the values of y, y', and y'' in 9y'' + 12y' + 29y = 0. We have,
9r^2e^rx + 12re^rx + 29e^rx = 0
⇒ e^rx(9r^2 + 12r + 29) = 0
⇒ 9r^2 + 12r + 29 = 0 → Characteristic Equation
⇒ r = [-12 ± √(12^2 - 4 × 9 × 29)]/(2 × 9)
⇒ r = (-2/3) ± (5/3)i
Since the roots of the characteristic equation are complex conjugates, therefore the general solution of the given second order differential equation is y = e^(-2/3)x[A sin (5/3)x + B cos (5/3)x].
Solving Non-Homogeneous Second Order Differential Equation
To find the solution of Non-Homogeneous Second Order Differential Equation y'' + py' + qy = f(x), the general solution is of the form y = y[c] + y[p], where y[c] is the complementary solution of the
homogeneous second order differential equation y'' + py' + qy = 0 and y[p] is the particular solution of the non-homogeneous differential equation y'' + py' + qy = f(x). Since y[c] is the solution of
the homogeneous differential equation, we can determine its value using the methods that we discussed in the previous section. Now, to find the particular solution y[p], we can guess the solution
depending upon the value of f(x). The table given below shows the possible particular solution y[p] corresponding to each f(x).
│ f(x) │ y[p] │
│be^ax │Ae^ax │
│ax^n + (lower order powers of x)│C[n]x^n + C[n-1]x^n-1 + ... + C[0] │
│P cos ax or Q sin ax │A cos ax + B sin ax │
In case, f(x) is of a form other than the ones given in the table above, then we can use the method of variation of parameters to solve the non-homogeneous second order differential equation. Also,
if f(x) is a sum combination of the functions given in the table, then we can determine the particular solution for each function separately and then take their sum to find the final particular
solution of the given equation. Let us now consider a few examples of second order differential equations and solve them using the method of undetermined coefficients:
Example 1: Find the complete solution of the second order differential equation y'' - 6y' + 5y = e^-3x.
Solution: To find the complete solution, first we will find the general solution of the homogeneous differential equation y'' - 6y' + 5y = 0.
We have solved this equation in the previous section in the solved examples (Example 1) and hence the complementary solution is y[c] = Ae^x + Be^5x
Next, we will find the particular solution y[p]. Since f(x) = e^-3x is of the form be^ax, let us assume y[p] = Ae^-3x. Now differentiating y[p], we have
y[p]' = -3Ae^-3x and y[p]'' = 9Ae^-3x . Substituting these values in the given second order differential equation, we have
y[p]'' - 6y[p]' + 5y[p] = e^-3x
⇒ 9Ae^-3x - 6(-3Ae^-3x) + 5Ae^-3x = e^-3x
⇒Ae^-3x (9 + 18 + 5) = e^-3x
⇒ 32 A e^-3x = e^-3x
⇒ A = 1/32
Hence, the particular solution y[p] = (1/32) e^-3x
Answer: Therefore, the complete solution of the given non-homogeneous 2nd order differential equation y'' - 6y' + 5y = e^-3x is y = Ae^x + Be^5x + (1/32) e^-3x
Example 2: Solve the second order differential equation y'' - 6y' + 5y = cos 2x + e^-3x
Solution: As we have solved the homogeneous differential equation y'' - 6y' + 5y = 0 in the previous section (Example 1), we have the complementary solution y[c] = Ae^x + Be^5x
Next, we will find the particular solution of the given differential equation individually for cos 2x and e^-3x, that is, determine the particular solution of y'' - 6y' + 5y = cos 2x and y'' - 6y' +
5y = e^-3x separately. From example 1 above, we have the particular solution of the differential equation y'' - 6y' + 5y = e^-3x corresponding to e^-3x as (1/32) e^-3x. Now, we will find the
particular solution of the equation y'' - 6y' + 5y = cos 2x using the table. Assume the particular solution of the form Y[p] = A cos 2x + B sin 2x. Differentiating this, we have Y[p]' = -2A sin 2x +
2B cos 2x and Y[p]'' = -4A cos 2x - 4B sin 2x. Substituting these values in the differential equation y'' - 6y' + 5y = cos 2x, we have
-4A cos 2x - 4B sin 2x - 6(-2A sin 2x + 2B cos 2x) + 5(A cos 2x + B sin 2x) = cos 2x
⇒ -4A cos 2x - 4B sin 2x + 12A sin 2x - 12B cos 2x + 5A cos 2x + 5B sin 2x = cos 2x
⇒ (A - 12B) cos 2x + (B + 12A) sin 2x = cos 2x
⇒ A - 12B = 1 and B + 12A = 0
⇒ A = 1/145 and B = -12/145
⇒ Y[p] = (1/145) cos 2x - (12/145) sin 2x
Now, taking the sum of both particular solutions, the final particular solution of the given second order differential equation y'' - 6y' + 5y = cos 2x + e^-3x is y[p] = (1/32) e^-3x + (1/145) cos 2x
- (12/145) sin 2x.
Answer: Therefore, the complete solution of the differential equation y'' - 6y' + 5y = cos 2x + e^-3x is y = y[c] + y[p] = Ae^x + Be^5x + (1/32) e^-3x + (1/145) cos 2x - (12/145) sin 2x
Important Notes on Second Order Differential Equation
• If y[1] and y[2] are two linearly independent solutions of the homogeneous second order differential equation y'' + py' + qy = 0, then the particular solution of the corresponding second order
non-homogeneous differential equation y'' + py' + qy = f(x) can be determined using the formula y[p] = -y[1] ∫[y[2] f(x)/W(y[1], y[2])] dx + y[2] ∫[y[1] f(x)/W(y[1], y[2])] dx, where W(y[1], y
[2]) = y[1]y[2]' - y[2]y[1]' is called the Wronskian. This method of finding the solution is called the method of variation of parameters.
• The method to find the solution of second-order differential equations with variable coefficients is complex and is based on guessing the solution.
☛ Related Topics:
Second Order Differential Equation Examples
1. Example 1: Solve the second order differential equation y'' - 9y' + 20y = 0
Solution: Since the given differential equation is homogeneous, we will assume the solution of the form y = e^rx
Find the first and second derivative of y = e^rx: y' = re^rx, y'' = r^2e^rx
Next, substitute the values of y, y', and y'' in y'' - 9y' + 20y = 0. We have,
r^2e^rx - 9re^rx + 20e^rx = 0
⇒ e^rx(r^2 - 9r + 20) = 0
⇒ r^2 - 9r + 20 = 0 → Auxiliary Equation
⇒ (r - 5) (r - 4) = 0
⇒ r = 4, 5
Since the roots of the characteristic equation are distinct and real, therefore the general solution of second order differential equation y'' - 9y' + 20y = 0 is y = Ae^4x + Be^5x
Answer: The solution y'' - 9y' + 20y = 0 is y = Ae^4x + Be^5x
2. Example 2: Find the complete solution of the second order differential equation y'' - y = 2x^2 - x - 3
Solution: The given differential equation is a non-homogeneous second order differential equation, hence we need to find its complementary solution and particular solution separately.
First, we will find the solution y[c] of the homogeneous equation y'' - y = 0. For this, assume y = e^rx and find its first and second derivative: y' = re^rx, y'' = r^2e^rx
Next, substitute the values of y, y', and y'' in y'' - y = 0. We have,
r^2e^rx - e^rx = 0
⇒ e^rx(r^2 - 1) = 0
⇒ r^2 - 1 = 0 → Auxiliary Equation
⇒ (r - 1) (r + 1) = 0
⇒ r = -1, 1
Since the roots of the characteristic equation are distinct and real, therefore the complementary solution is y[c] = Ae^-x + Be^x
Next, we will find the particular solution y[p]. For this, using the table, assume y[p] = Ax^2 + Bx + C. Now find the derivatives of y[p].
y[p]' = 2Ax + B and y[p]'' = 2A . Substituting these values in the given differential equation, we have
y[p]'' - y[p] = 2x^2 - x - 3
⇒ 2A - (Ax^2 + Bx + C) = 2x^2 - x - 3
⇒ 2A - Ax^2 - Bx - C = 2x^2 - x - 3
⇒ -Ax^2 - Bx + 2A - C = 2x^2 - x - 3
⇒ -A = 2, -B = -1, 2A - C = -3
⇒ A = -2, B = 1, C = -1
⇒ y[p] = -2x^2 + x - 1
The complete solution is y = y[c] + y[p ]= Ae^-x + Be^x - 2x^2 + x - 1
Answer: The complete solution of y'' - y = 2x^2 - x - 3 is y = Ae^-x + Be^x - 2x^2 + x - 1.
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Second Order Differential Equation Questions
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FAQs on Second Order Differential Equation
What is Second Order Differential Equation in Calculus?
A second order differential equation is defined as a differential equation that includes a function and its second-order derivative and no other higher-order derivative of the function can appear in
the equation. It includes terms like y'', d^2y/dx^2, y''(x), etc. It can be of different types such as second-order linear differential equation, 2nd order homogeneous and non-homogeneous
differential equation, and second-order differential equation with variable and constant coefficients.
How to Solve Second Order Differential Equation?
Second order differential equations can be solved using different methods such as the method of undetermined coefficients and the method of variation of parameters. The solution of a non-homogeneous
second order differential is the sum of the complementary and particular solution and is given as y = y[c] + y[p].
What is Second Order Differential Equation with Constant Coefficients?
A Second Order Differential Equation with Constant Coefficients is a differential equation of the form y'' + py' + qy = f(x), where p, q are constant coefficients.
What are Homogeneous and Non-Homogeneous Second Order Differential Equations?
A second order differential equation of the form y'' + py' + qy = f(x) is homogeneous if f(x) is a zero function and non-homogeneous if f(x) is not a zero function and is some non-zero function of x.
Why Does a Second Order Differential Equation Have Two Solutions?
A second order differential equation can have infinitely many solutions as the arbitrary constants can take any value. We find two linearly independent solutions of a second order differential
equation as their combination gives all possible solutions of the equation and finding only one solution does not suffice.
How to Find Particular Solution of Second Order Differential Equation?
The particular solution of a second order differential equation can be determined using the table given below:
│ f(x) │ y[p] │
│be^ax │Ae^ax │
│ax^n + (lower order powers of x)│C[n]x^n + C[n-1]x^n-1 + ... + C[0] │
│P cos ax or Q sin ax │A cos ax + B sin ax │
We can also find the particular solution using the formula y[p] = -y[1] ∫[y[2] f(x)/W(y[1], y[2])] dx + y[2] ∫[y[1] f(x)/W(y[1], y[2])] dx, where y[1] and y[2] are two linearly independent solutions
of the homogeneous second order differential equation y'' + py' + qy = 0
How to Tell If a Second Order Differential Equation is Linear?
To tell if a second order differential equation is linear, we can check the degree of the second derivative in the equation. A linear second order differential equation is written as y'' + p(x)y' + q
(x)y = f(x), where the power of the second derivative y'' is equal to one which makes the equation linear.
What is the Difference Between First Order Differential Equation and Second Order Differential Equation?
A first-order differential equation consists of the first derivative of a function and no other higher order derivative can appear in the equation. It is written as y' + p(x)y = f(x). On the other
hand, second order differential equation is a differential equation that consists of a derivative of a function of order 2 and no other higher-order derivative of the function appears in the
equation. It is written as y'' + p(x)y' + q(x)y = f(x).
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Lift - (Principles of Data Science) - Vocab, Definition, Explanations | Fiveable
from class:
Principles of Data Science
Lift is a measure used in association rule mining to evaluate the strength of a rule by comparing the observed frequency of the rule's consequent with the expected frequency if the items were
independent. It helps determine how much more likely the presence of one item is when another item is present, making it a key metric for understanding relationships between items in data sets.
congrats on reading the definition of Lift. now let's actually learn it.
5 Must Know Facts For Your Next Test
1. Lift is calculated using the formula: $$\text{Lift}(A \to B) = \frac{P(A \cap B)}{P(A) \times P(B)}$$, where P(A ∩ B) is the probability of both A and B occurring together.
2. A lift value greater than 1 indicates a positive correlation between items A and B, suggesting that they are likely to be purchased together more often than expected.
3. A lift value equal to 1 implies that A and B are independent, meaning their co-occurrence is no greater than what would be expected by chance.
4. A lift value less than 1 indicates a negative correlation, suggesting that if one item is present, it makes the presence of the other item less likely.
5. Lift can help in identifying strong rules in large datasets, guiding marketers and analysts in understanding customer behavior and preferences.
Review Questions
• How does lift help evaluate the strength of association rules, and why is it important in data analysis?
□ Lift evaluates the strength of association rules by measuring how much more likely two items are to occur together compared to their expected co-occurrence if they were independent. This is
important because it provides insights into consumer behavior and item relationships that can inform marketing strategies and product placements. By focusing on high-lift rules, analysts can
uncover significant patterns that might lead to increased sales or improved customer targeting.
• Compare lift with support and confidence in terms of their roles in association rule mining.
□ While lift measures the strength of an association rule by examining the relationship between items, support quantifies how often an item appears in transactions, and confidence gauges how
often the consequent occurs given the antecedent. Together, these metrics provide a comprehensive view: support helps filter out infrequent rules, confidence indicates reliability, and lift
assesses how much more likely items are to occur together than by chance. This combination allows for effective identification of meaningful associations.
• Evaluate how understanding lift can influence decision-making in business strategies for product promotions.
□ Understanding lift allows businesses to identify strong associations between products, which can significantly influence decision-making for promotional strategies. For example, if a
particular product pair shows a high lift value, businesses might choose to bundle them together in marketing campaigns or place them near each other in stores to boost sales. This strategic
approach leverages data insights to enhance customer experience and increase revenue, demonstrating how data-driven decisions can lead to effective outcomes.
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AP® and SAT® are trademarks registered by the College Board, which is not affiliated with, and does not endorse this website. | {"url":"https://library.fiveable.me/key-terms/principles-and-techniques-of-data-science/lift","timestamp":"2024-11-13T16:11:43Z","content_type":"text/html","content_length":"159101","record_id":"<urn:uuid:1f8272fb-8947-4d50-b69b-a3eb5d73720b>","cc-path":"CC-MAIN-2024-46/segments/1730477028369.36/warc/CC-MAIN-20241113135544-20241113165544-00759.warc.gz"} |
Ms Patricia Magalhães, C. (University of São Paulo, Brazil)
The last decade provided important advances on light scalar resonances sector in the D decays. In particular, different analyses showed the evidence for $\sigma$ and $\kappa$ mesons close to the
threshold. Decays of charmed meson, however, are still poorly understood theoretically. The $K^-\pi^+$ phase shift extracted from $D^+\to K^- \pi^+\pi^+$[1] and from $K^-\pi^+$ scattering[2] are
significantly different. Our main purpose is to understand this discrepancy. We investigate $D^+\to K^- \pi^+\pi^+$ exploring two gaps in the theoretical models. The first one is the weak vertex,
which we know very little about. The other is the final state interactions. We focus our work on the low energy part of $K\pi$ spectrum making two assumptions: the contribution of $\pi^+\pi^+ $
interaction is small and can be neglected; the $K$ interacts with only one pion at a time. The basic building block in the calculation is a $K\pi$ amplitude based on unitarized chiral perturbation
theory[3], and with parameters determined by a fit to elastic LASS data. For the weak vertex, three different topologies are considered. The perturbative solution involve the sum of diagrams with
different numbers of loops. We found that each term in this series is systematically related to the previous one and a ressummation of the whole series was performed. The analysis shows that
corrections to the ressummed series are important at threshold and converge rapidly at higher energies. The interference of the different contributions in the whole amplitude change the phase shift.
Two classes of weak vertices reproduce the scattering phase shift from LASS[2], whereas the third one describe the behaviour of $K^- \pi^+$ phase shift from FOCUS[1] in the elastic range. This result
suggests a dominant decay mechanism in the $D^+ \rightarrow K^- \pi^+ \pi^+$. _________________________________________________________ [1] J.M. Link {\it et al.} [FOCUS Collaboration], Phys.\ Lett.\
B {\bf 653}, (2007) 1; J.M. Link {\it et al.} [FOCUS Collaboration], Phys.\ Lett.\ B {\bf 681}, (2009) 14; [2]D. Aston {\it et al.}, Nucl. Physc. {\bf B 296} (1988) 493. [3] G. Ecker, J. Gasser, A.
Pich and E. De Rafael, Nucl. Phys. B{\bf 321}, 311 (1989); J.A. Oller and E. Oset, Nucl. Phys. A {\bf 620}, 438 (1997); M. Jamin, J.A. Oller, A. Pich, Nucl. Phys. {\bf B 587} (2000) 331.
Prof. Alberto dos Reis, C (Centro Brasileiro de Pesquisas Físicas - Rio de Janeiro, RJ, Brazil) Prof. Ignacio Bediaga (Centro Brasileiro de Pesquisas Físicas - Rio de Janeiro, RJ, Brazil) Ms Karin
Guimarães, S.F.F. (Instituto Tecnológico da Aeronáutica, São José dos Campos,SP, Brazil) Prof. Manoel Robilotta, R. (University of São Paulo, Brazil) Prof. Tobias Frederico (Instituto Tecnológico da
Aeronáutica, São José dos Campos,SP, Brazil) | {"url":"https://indico.ph.tum.de/event/1729/contributions/276/","timestamp":"2024-11-12T06:10:13Z","content_type":"text/html","content_length":"65268","record_id":"<urn:uuid:f6c45a93-745b-466c-8a3e-d3262b153ac5>","cc-path":"CC-MAIN-2024-46/segments/1730477028242.58/warc/CC-MAIN-20241112045844-20241112075844-00013.warc.gz"} |
Yearly report 1.1.2023-28.1.2023
« Back
Changes compared to previous years
Compare group: Year 2024 (1466 pcs) | Year 2023 (1628 pcs) | Year 2022 (1652 pcs) | Year 2021 (3882 pcs) | Year 2020 (3594 pcs) | Year 2019 (7810 pcs) | Year 2018 (9609 pcs) | Year 2017 (11934 pcs) |
Year 2016 (6094 pcs) | Year 2015 (4850 pcs) | Year 2014 (5331 pcs) | Year 2013 (6031 pcs) | Year 2012 (2380 pcs)
Hint: click a row for details
Questions with more positive answers than the compare group
Laptop computer***
See question distribution:
Laptop computer
Own answers average
Compare group answers average
Statistical significance
p < 0.00001
(clearly evident difference)
Digital learning environments and cloud services (e.g. Google Classroom, O365, SanomaPro, itslearning)***
I personally use information and communications technology in most of my classes.***
When conducting a formative assessment of the student’s learning***
I guide my students to protect themselves against common information security risks and data loss.**
I feel that I do not receive support for developing the pedagogical use of ICT.**
My school has jointly agreed goals for using ICT in teaching.**
When presenting information in a teaching situation**
I know how to utilize digital teaching materials in my teaching.**
When assigning students exercises, homework and tasks**
Software for targeted educational purposes, such as multiplication, programming, mathematical illustration, languages, simulation and modeling.**
I get sufficient and fast enough technical support for the use of ICT at my school.**
Making use of digital learning materials
Please choose the description below that best fits your situation, even if it is not fully accurate in all respects:**
I have utilized the earlier results of this questionnaire in the development of my work.*
When providing feedback for students*
I use ICT to perform continuous assessment on my students’ learning.*
Smart phone*
See question distribution:
Smart phone
Own answers average
Compare group answers average
Statistical significance
p < 0.1
(significant difference)
My own ICT skills and competences are sufficient when compared with the objectives specified in the curriculum.*
We also work with students outside the school buildings using mobile devices (e.g. smartphone, tablet, laptop). *
provides students with the opportunity to use better information sources*
I have opportunities to influence ICT purchases at my school.*
I know how to print using a 3D printer.*
Questions with more negative answers than the compare group
Desktop computer***
See question distribution:
Desktop computer
Own answers average
Compare group answers average
Statistical significance
p < 0.00001
(clearly evident difference)
Basic phone***
See question distribution:
Basic phone
Own answers average
Compare group answers average
Statistical significance
p < 0.00001
(clearly evident difference)
Office software such as word processing, presentation tools, spreadsheets***
reduces personal communication between students**
helps students communicate more effectively with others**
During the previous academic year, I have attended ICT continuing education and training in total:**
erodes students’ calculation and evaluation skills**
diminishes students’ interest in studying**
reduces students’ co-operation with other students**
erodes students’ writing skills*
See question distribution:
erodes students’ writing skills
Own answers average
Compare group answers average
Statistical significance
p < 0.1
(significant difference)
In my school, it is easy to start developing new modes of operation.*
The development of ICT skills is brought up in my development/performance reviews.*
My students assess each others’ work using digital learning environments.*
I teach my students to understand and interpret various digital media contents.*
My school has a sufficiently fast and functional internet connection.*
I intersperse my students’ work with regular breaks to avoid making them sit for too long.*
helps students develop their skills to plan and direct their own work*
My students assess my teaching using ICT.*
I actively guide my students to use digital information search services (such as Google, Wikipedia, Wolfram|Alpha).*
When I set up a new service or application, I always read its terms of use.*
*) The amount of stars implies the statistical significance
This report shows answers between 1.1.2023 and 28.1.2023. Information on this page is fixed. | {"url":"https://opeka.fi/en/public/analysis?compare=all-semester-2020&semesters=1&reportid=1672531200-1674920464","timestamp":"2024-11-13T12:40:52Z","content_type":"application/xhtml+xml","content_length":"40719","record_id":"<urn:uuid:7a44c615-1e36-4015-b60d-1a6d1197b422>","cc-path":"CC-MAIN-2024-46/segments/1730477028347.28/warc/CC-MAIN-20241113103539-20241113133539-00816.warc.gz"} |
CS 251: Assignment #5
Single and Multiple Linear Regression
Due 23 March 2018 (before you leave for spring break)
The goal of this project is to integrate simple linear regression into your project and then implement a multiple linear regression function as a function in your analysis file. Incorporating
multiple linear regression into your full GUI is the obvious extension.
1. Implement an updateFits method, similar to updateAxes. The updateFits function should enable the linear fit to move along with the data. Make sure updateFits is called wherever updateAxes and
updatePoints is called in the Display class.
2. Test your implementation. Make sure everything works cleanly if you run a second linear regression, open a new file, go back and forth between plotting data and linear regressions, and translate/
scale/rotate the screen. Make sure cancelling the linear regression dialog actually cancels the process and does not change the existing visualization.
The first required result is a plot of your regression line on the data-simple.csv data.
3. In your analysis class, create a new function linear_regression that takes in the data set, a list of headers for the independent variables, and a single header (not in a list) for the dependent
variable. The function should implement linear regression for one or more independent variables. The algorithm is as follows. It's not a long function. Each step identified below includes a
description of what you are computing.
def linear_regression(d, ind, dep):
# assign to y the column of data for the dependent variable
# assign to A the columns of data for the independent variables
# It's best if both y and A are numpy matrices
# add a column of 1's to A to represent the constant term in the
# regression equation. Remember, this is just y = mx + b (even
# if m and x are vectors).
# assign to AAinv the result of calling numpy.linalg.inv( np.dot(A.T, A))
# The matrix A.T * A is the covariancde matrix of the independent
# data, and we will use it for computing the standard error of the
# linear regression fit below.
# assign to x the result of calling numpy.linalg.lstsq( A, y )
# This solves the equation y = Ab, where A is a matrix of the
# independent data, b is the set of unknowns as a column vector,
# and y is the dependent column of data. The return value x
# contains the solution for b.
# assign to b the first element of x.
# This is the solution that provides the best fit regression
# assign to N the number of data points (rows in y)
# assign to C the number of coefficients (rows in b)
# assign to df_e the value N-C,
# This is the number of degrees of freedom of the error
# assign to df_r the value C-1
# This is the number of degrees of freedom of the model fit
# It means if you have C-1 of the values of b you can find the last one.
# assign to error, the error of the model prediction. Do this by
# taking the difference between the value to be predicted and
# the prediction. These are the vertical differences between the
# regression line and the data.
# y - numpy.dot(A, b)
# assign to sse, the sum squared error, which is the sum of the
# squares of the errors computed in the prior step, divided by the
# number of degrees of freedom of the error. The result is a 1x1 matrix.
# numpy.dot(error.T, error) / df_e
# assign to stderr, the standard error, which is the square root
# of the diagonals of the sum-squared error multiplied by the
# inverse covariance matrix of the data. This will be a Cx1 vector.
# numpy.sqrt( numpy.diagonal( sse[0, 0] * AAinv ) )
# assign to t, the t-statistic for each independent variable by dividing
# each coefficient of the fit by the standard error.
# t = b.T / stderr
# assign to p, the probability of the coefficient indicating a
# random relationship (slope = 0). To do this we use the
# cumulative distribution function of the student-t distribution.
# Multiply by 2 to get the 2-sided tail.
# 2*(1 - scipy.stats.t.cdf(abs(t), df_e))
# assign to r2, the r^2 coefficient indicating the quality of the fit.
# 1 - error.var() / y.var()
# Return the values of the fit (b), the sum-squared error, the
# R^2 fit quality, the t-statistic, and the probability of a
# random relationship.
4. Write a simple test function in your analysis.py file that reads in a data set and then does a multiple linear regression fit. Test it on the following three data files.
1. data-clean.csv
m0 = 0.984, m1 = 2.088, b = -0.035, sse = 0.002,
R2 = 0.996, t = [8.6, 18.9, -0.88], p = [5.6e-5, 2.9e-7, 0.405]
m0 = 0.885, m1 = 1.880, b = 0.146, sse = 0.090,
R2 = 0.885, t = [2.34, 5.22, 0.568], p = [0.052, 0.001, 0.588]
m0 = -0.336, m1 = 3.335, b = -0.263, sse = 1.03,
R2 = 0.611, t = [-0.28, 3.08, -0.255], p = [0.787, 0.018, 0.806]
In your writeup, show the results of running your function on these three data sets and confirm that it is working properly.
5. Find a data set where you think there is a relationship between two variables. Minimum and maximum daily temperature, for example, is one possibility. You could also try year versus average
yearly temperature for the past 30 years, or carbon dioxide levels versus average yearly temperature over the same time period. Look on the main course page for data set options.
Using the data set you selected, execute a linear regression using the GUI interface you completed in lab with one independent variable and one dependent variable. Include the results in your
writeup and explain whether they make sense. Also include a picture of the linear regression plotted over your data using your GUI.
Using the data set you selected, execute a multiple linear regression using the analysis function you wrote. Include the numerical results in your writeup. Also include a picture of the data
plotted in your GUI (this picture does not have to include the regression line, just the data). It is an extension to have the multiple linear regression line plotted in your GUI.
• Incorporate multiple linear regression into your GUI. Start by just displaying the coefficients of the fit, then extend the GUI to display the regression line in 3D. For fits higher than 3D, you
have to be careful when calculating the endpoints of the best fit line in the view space.
• Further extend your GUI in any of the directions suggested last week. Add legends, axis labels (e.g. headers and values) or other features to the GUI for plotting data.
• Do some more exploration with different data sets using your new tool.
• Give the user the ability to store and recall prior analyses. This capability can be limited to the current session.
• Give the user the ability to save the linear regression analysis to a file in a human-readable format. Extend it even further to allow the user to read an analysis back in and replot it over the
correct data.
• Figure out how to save a picture of a plot to a file.
• Be creative and add useful features to your GUI.
Make a wiki page for the project report.
Write a brief summary, separate from the body of your report, of your project that describes the purpose, the task, and your solution to it. It should describe the task, the key parts of your
solution, and the result of your work (did it work, what can you do with your GUI?). The summary should be 200 words or less.
Write a brief explanation of how to run a linear regression, with screen shots, in your application. Include any extensions or enhancements you implemented. Explain what the meaning of a linear
regression plot.
Include the required screen shots for the provided data sets and for your own. In the text of your writeup, note what axes are being plotted in any images you show. Please also include a description
of what the plot means in terms of the relationship between the two variables.
Describe any extensions or enhancements you implemented. Include pictures as appropriate.
Acknowledgements: a list of people you worked with, including TAs, and instructors. Include in that list anyone whose code you may have seen, such as those of friends who have taken the course in a
previous semester.
Once you have written up your assignment, give the page the label:
Put your code in the Private subdirectory of your folder on Courses. Please make sure you are organizing your code by project. Your handin code should include all the python files necessary to run
your program as well as the data files you used test your code. | {"url":"https://cs.colby.edu/courses/S18/cs251/labs/lab05/assignment.php","timestamp":"2024-11-03T10:14:58Z","content_type":"text/html","content_length":"11982","record_id":"<urn:uuid:f4e988d9-b3a6-4cd0-964f-7d8afa577e12>","cc-path":"CC-MAIN-2024-46/segments/1730477027774.6/warc/CC-MAIN-20241103083929-20241103113929-00194.warc.gz"} |
Meaning of ARITHMETIC CODING
Definition: Perhaps the major drawback to each of the huffman encoding techniques is their poor performance when processing texts where one symbol has a probability of occurrence approaching unity.
Although the entropy associated with such symbols is extremely low, each symbol must still be encoded as a discrete value. Arithmetic coding removes this restriction by representing messages as
intervals of the real numbers between 0 and 1. Initially, the range of values for coding a text is the entire interval [0, 1]. As encoding proceeds, this range narrows while the number of bits
required to represent it expands. Frequently occurring characters reduce the range less than characters occurring infrequently, and thus add fewer bits to the length of an encoded message. | {"url":"https://www.hyperdictionary.com/video/arithmetic-coding","timestamp":"2024-11-07T16:49:51Z","content_type":"application/xhtml+xml","content_length":"5389","record_id":"<urn:uuid:e417e7dc-5903-4770-b044-d10516b31ec8>","cc-path":"CC-MAIN-2024-46/segments/1730477028000.52/warc/CC-MAIN-20241107150153-20241107180153-00724.warc.gz"} |
CSC420 Intro to Image Understanding Assignment 2 solved
1. Interest Point Detection
(a) [2 points] Write two functions for computing the Harris corner metric using Harris (R) and
Brown (harmonic mean) methods. Display your results for the attached image building.jpg
showing your cornerness metric output. Compare the results corresponding to two different
methods. For Harris you can use the code provided in Tutorial C.
(b) [2 points] Write your own function to perform non-maximal suppression using your own
functions of choice. Use a circular element, and experiment with varying radii r as a parameter
for the output of harmonic mean method. Explain why/how the results change with r . MATLAB
users may want to use functions ordfilt2(), however it can be easily implemented.
(c) [2 points] Write code to search the image for scale-invariant interest point (i.e. blob) detection
using the Laplacian of Gaussian and checking a pixel’s local neighbourhood as in SIFT. You
must find extrema in both location and scale. Find the appropriate parameter settings, and display
your keypoints for synthetic.png using harmonic mean metric. Hint: Only investigate pixels with
LoG above or below a threshold.
(d) [1 point] Use open-source implementation of another local feature descriptor that is not covered
in the class, and show the output keypoints on synthetic.png and building.jpg. Describe the main
ideas of your algorithm of choice in a few sentences. You may want to look at Slide 7 in Lecture
8-A for a list of existing methods.
2. SIFT Matching (For this question you will use interest point detection for matching using SIFT. You
may use a SIFT implementation (e.g. http://www.vlfeat.org/), or another, but specify what you use)
(a) [0.5 points] Extract SIFT keypoints and features for book.jpg and findBook.jpg.
(b) [1.5 points] Write your own matching algorithm to establish feature correspondence between the
two images using the reliability ratio on Lecture 8. You can use any function for computing the
distance, but you must find the matches yourself. Plot the percentage of true matches as a
function of threshold. Also, after experimenting with different thresholds, report the best value.
(c) [2 points] Use the top k correspondences from part (b) to solve for the affine transformation
between the features in the two images via least squares using the Moore-Penrose pseudo inverse.
What is the minimum k required for solving the transformation? Demonstrate your results for
various k . Use only basic linear algebra libraries.
(d) [0.5 points] Visualize the affine transformation. Do this visualization by taking the four corners
of the reference image, transforming them via the computed affine transformation to the points in
the second image, and plotting those transformed points. Please also plot the edges between the
points to indicate the parallelogram. If you are unsure what the instruction is, please look at
Figure 12 of [Lowe, 2004].
(e) [1.5 points] Write code to perform matching that takes the colour in the images into account
during SIFT feature calculation and matching. Explain the rationale behind your approach. Use
colourTemplate.png and colourSearch.png, display your matches with the approach described
in part (d).
3. RANSAC
(a) [0.5 points] Assuming a fixed percentage of inliers p = 0.7 , plot the required number of
RANSAC iterations (S) to recover the correct model with a higher than 99% chance (P), as a
function of k (1:20), the minimum number of sample points used to form a hypothesis.
(b) [0.5 points] Assuming a fixed number of sample points required ( k = 5 ), plot S as a function of
the percentage of inliers p (0.1 : 0.5)
(c) [1 point] If k = 5 and the initial estimate on the percentage of inliers is p = 0.2 , what is the the
required number of iterations to recover the correct model with P≥ 0.99 chance? Assume that
you have implemented this and there are 1500 matches in total. In iteration #15, 450 points agree
with the current hypothesis (i.e. their error is within a preselected threshold), would the number of
required iterations chance? explain how and why. | {"url":"https://codeshive.com/questions-and-answers/csc420-intro-to-image-understanding-assignment-2-solved/","timestamp":"2024-11-14T07:59:26Z","content_type":"text/html","content_length":"103591","record_id":"<urn:uuid:11454e2c-b97c-466f-b994-81e69ad552dd>","cc-path":"CC-MAIN-2024-46/segments/1730477028545.2/warc/CC-MAIN-20241114062951-20241114092951-00107.warc.gz"} |
Neutrons Knock at the Cosmic Door
The quantum behavior of a neutron bouncing in the gravitational field of the Earth can improve what we know about dark energy and dark matter.
Figure 1: (Left) Neutron mirror apparatus. An ultracold neutron (UCN) enters a space between two mirrors that act as potential wells, giving rise to a discrete energy spectrum. A detector measures
neutrons exiting the cavity formed by the mirrors. The bottom mirror sits upon a nanopositioning table that induces a vertical oscillation that produces dips in the neutron transmission at the
resonances. (Right) Energy-level diagram for the neutrons in a gravitational field caught between the walls, which oscillate owing to the mirror motion (horizontal direction here is vertical in the
apparatus). This, in turn, causes the neutrons to move up and down energy levels. A measurement of the energy-level spacing yields constraints on parameters of scenarios describing dark energy and
dark matter, which would slightly shift the levels as indicated by the dashed lines.
Spectroscopy has always set the pace of physics. Indeed, the observation of the Balmer series of the hydrogen atom led to the Bohr-Sommerfeld model about 100 years ago. A little later the
discreteness of the spectrum moved Werner Heisenberg to develop matrix mechanics and Erwin Schrödinger to formulate wave mechanics. In 1947, the observation of a level shift in hydrogen by Willis E.
Lamb ushered in quantum electrodynamics.
Now, a group led by Hartmut Abele of the Technical University of Vienna, Austria, reports, in Physical Review Letters [1], experiments that once more take advantage of the unique features of
spectroscopy to put constraints on dark energy and dark matter scenarios. However, this time it is not a “real atom” (consisting of an electron bound to a proton) that provides the insight. Instead,
the research team observes an “artificial atom”—a neutron bouncing up and down in the attractive gravitational field of the Earth (Fig. 1). This motion is quantized, and the measurement of the
separation of the corresponding energy levels allows these authors to make conclusions about Newton’s inverse square law of gravity at short distances.
The energy wave function of a quantum particle in a linear potential [2], corresponding, for example, to the gravitational field close to the surface of the Earth, has a continuous energy spectrum
[3]. However, when a quantum particle such as a neutron is also restricted in its motion by two potential walls, the resulting spectrum is discrete. This elementary problem of nonrelativistic quantum
mechanics is a slight generalization of the familiar “particle in a box” where the bottom of the box, which usually corresponds to a constant potential, is replaced by a linear one representing the
gravitational field.
In order to realize this “artificial atom,” Jenke et al. place neutrons between mirrors that act as two potential walls and then vary their separation in an oscillatory fashion. This modulation makes
the neutrons climb up and down the energy levels in the gravitational box similarly to light-induced electronic transitions in an atom. This approach is reminiscent of the Fermi accelerator [4],
which can be realized when an atom [5] bounces in the gravitational field of the Earth and is reflected from a time-dependent evanescent laser field at the lower end of the path of the atom.
Transitions between the energy levels of the bound neutron manifest themselves in dips in the transmission curve as a function of the modulation frequency.
In their experiment, performed at the ultracold neutron facility at the Institut Laue-Langevin, Jenke et al. could measure the transition frequencies between the first four energy states. They use
them to search for new kinds of hypothetical gravitylike interactions at micrometer distances and link their work to dark matter and dark energy.
While dark energy explains the accelerated expansion of the Universe, dark matter is needed to describe the rotation curves of galaxies and the large-scale structure of the Universe. However, the
true nature of these forms of matter and energy are still not understood [6].
The central idea of the experiment by Jenke et al. rests on the fact that the interaction of so-far-undiscovered dark matter or dark energy particles with the neutrons caught between the two walls
causes shifts in the energy levels. This approach is completely analogous to the measurement of the Lamb shift of the hydrogen atom, which is a verification of the quantization of the electromagnetic
field (thus confirming the existence of the photon), or the Zeeman shift resulting from the spin of the electron.
Indeed, one candidate for dark energy is covered by “quintessence” theories [6], where in one scenario [7], a new hypothetical scalar field couples to matter. The associated interaction potential
changes the energy of the bound neutron and a comparison of the observed transition frequencies to their theoretical ones yields constraints on the coupling strength of this field.
Likewise, a particle such as an axion, which is predicted to mediate a spin-dependent force that gives rise to an additional potential, couples the spin of the neutron to the position of a nucleon.
In their search for dark matter based on such forces, Jenke et al. first polarize the neutrons at the entrance of the cavity by having them go through a foil coated by iron, and then they measure the
transition frequencies again. The comparison to the theoretical values puts constraints on the coupling constant of this interaction.
The present experiment is in the tradition of the “COW” experiment [8] named after Robert Colella, Albert Overhauser, and Samuel Werner who first probed the gravitational field of the Earth with the
matter waves of a neutron. More recently, matter-wave interferometers with cold atoms [9] have also served as probes of gravity and represent a very active field of research.
In particular, tests of the equivalence of the gravitational mass (the property of the particle that responds to the gravitational force created by other particles’ mass) and the inertial mass (the
property that describes matter’s resistance to changes in motion) are being pursued in many laboratories. Although no experiment has so far revealed any difference between the two, they are
conceptually different. It was recently pointed out [10] that they enter the energy wave function of a linear gravitational potential not as the ratio as expected from the weak equivalence principle
but in different powers. Experiments like those of Jenke et al., which have already provided us with deeper insight of the cosmos using microscopic probes such as the neutron, will soon be able to
verify this subtlety at the interface of gravity and quantum theory.
T. Jenke, G. Cronenberg, J. Burgdörfer, L.A. Chizhova, P. Geltenbort, A.N. Ivanov, T. Lauer, T. Lins, S. Rotter, H. Saul, U. Schmidt, and H. Abele
Phys. Rev. Lett. 112, 151105 (2014)
Published April 16, 2014
Read PDF | {"url":"https://physics.aps.org/articles/v7/39","timestamp":"2024-11-12T17:25:36Z","content_type":"text/html","content_length":"34567","record_id":"<urn:uuid:28889421-6637-40c7-a16e-8b1b4bdbeb02>","cc-path":"CC-MAIN-2024-46/segments/1730477028273.63/warc/CC-MAIN-20241112145015-20241112175015-00806.warc.gz"} |
Square Root of 50 - How to Find the Square Root of 50?
A day full of math games & activities. Find one near you.
A day full of math games & activities. Find one near you.
A day full of math games & activities. Find one near you.
A day full of math games & activities. Find one near you.
Square Root of 50
The square root of 50 is a number which when multiplied by itself results in 50. Finding the square root of a number is extremely important for finding the side length of a square from its area. We
will now look at how to find the value of the square root of 50 and solve some problems to help give you a better understanding.
• Square Root of 50: √50 = 7.0710678...
• Square of 50: 50^2 = 2500
1. What is the Square Root of 50?
2. Is the Square Root of 50 Rational or Irrational?
3. How to Find the Square Root of 50?
4. Challenging Questions
5. FAQs on Square Root of 50
What is the Square Root of 50?
The square root of a number is a number which when multiplied by itself, results in the original number. For example, the square root of 25 is 5, as 5 times 5 results in 25. However, you can also
have square roots of some numbers that do not result in whole numbers, such as 50. We can express the square root of 50 in different ways:
• Decimal form: 7.071
• Radical form: √50 = 5√2
• Exponent form: 50^1/2
Is the Square Root of 50 Rational or Irrational?
• The decimal part of the square root 50 is non-terminating. This is the definition of an irrational number. It also cannot be expressed as a ratio p/q which tells us it is irrational.
• Looking at the decimal form of the root 50, we see that it is never ending: √50 = 7.0710678118…….
• Therefore, we can conclude that Square Root of 50 is Irrational.
How to find the Square Root of 50?
There are 2 main methods we use to find the Square Root of 50:
Prime Factorization
• To find the square root of 50, we shall first express it in terms of its prime factors.
50 = 2 × 5 × 5
• Next, this can be reduced further to 50 = 2 × 5^2
• Finally, to find the root of this from here it is very easy,
√50 = √(25 × 2) = 5√2 = 5 × 1.414 = 7.07
Therefore, the Square Root of 50 ≅ 7.07
Long Division
• Step 1: Place a bar over the digits 50. We also pair the 0s in decimals in pairs of 2 from left to right.
• Step 2: Find a number such that when you multiply it with itself, the product is less than or equal to 50. We know that 7 × 7 = 49 which is less than 50. Dividing 50 by 7 we get 7 as quotient and
1 as remainder.
• Step 3: Place a decimal after the quotient as we are now dividing using the 0s from the decimal part of 50. Remember to drag down the pair of 0s, making the dividend 100. Also, adding 7 by itself
gives us 14 which becomes the starting digits of our next divisor.
• Step 4: Now we have 14X as our new divisor. We need to find a value of X such that 14X × X gives us a value less than 100. Only 0 fills X position and therefore the dividend is 140 and quotient
is now 7.0.
• Step 5: The next divisor will be 140 + 0 and dividend will be 10000. We continue doing the same steps till we get the number of decimals we require.
So our long division now looks like this:
Therefore, the Square Root of 50 is 7.071
Explore Square roots using illustrations and interactive examples
• What is the square root of 450?
• Is -7.071 a root of 50? If yes, why?
• Find the square root of:
Square Root of 50 Solved Examples
1. Example 1: Kevin wants to buy a square plot which is 50 square feet in area. He plans on fencing the area and wants to calculate the cost of doing so. If the cost per mile of fencing is 40
dollars, find the cost of fencing the entire perimeter of the plot?
□ First, to calculate the cost of fencing we would need the perimeter of the land.
□ To find the perimeter, we need to find the side length first.
□ Side length is the square root of area for a square.
Therefore, side length here will be √50 = 7.071, and perimeter 4 times this.
Hence, the total cost of fencing = 4 × 7.071 × 40 = $1131.36.
2. Example 2: Randal is travelling down the highway at an average speed of 5√50 mi/hr for half an hour. How much distance does he cover?
We need to use the formula Distance = Speed × Time
Speed = 5√50 = 35.355 mi/hr
Time = 0.5 hr
Using the formula, Distance = 35.355 × 0.5 = 17.677
Therefore, Micheal covers a distance of 17.677 miles.
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Interactive Questions
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FAQs on Square Root of 50
What is the square root of 50?
The square root of 50 is √50 = 7.071.
What is the square of 50?
The square of 50 is 50^2 = 2500.
What is the square root of 50 simplified?
The square root of 50 in simplified form is 5√2.
Is the square root of 50 a rational number?
The square root of 50 is an irrational number as it is non-terminating. It cannot be expressed in the form p/q which is what makes a rational number.
What is the exponent form of root 50?
The exponent form if root 50 is 50^1/2.
Math worksheets and
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Graduate Programs
The Mathematics Department offers three graduate programs: Master of Science (MS), Master of Arts (MA), and a doctoral program (PhD).
Students who plan to obtain a doctorate in Mathematics should apply to the PhD program. Once enrolled in the PhD program, students may, but are not required, to earn an MA along the way. Below is a
typical PhD schedule.
The MS program is offered for students who want to pursue studies in Mathematics beyond the undergraduate level, but who do not plan to obtain a doctorate in Mathematics.
Typical PhD schedule
Qualifying Examinations. [Approximately two years]
• Become English Language Qualified (concerns International Students only).
• Take and pass the Qualifying Examinations.
• During the first few years, students should be taking 3 courses per semester.
General Examination. [Approximately two years]
• Get a Ph.D. Advisor and form a Ph.D. Committee. Have an Advisory Conference with your Ph.D. Committee.
• Pass the Foreign Language Proficiency Examination (traditional Ph.D. students) or IRB training (RUME students).
• Take and pass the General Examination.
Dissertation. [Approximately two years]
• Thesis Research.
• Write up, defend, and submit thesis.
Program Requirements
Table of Contents
In the rules below,
• M denotes a Math Department rule.
• G denotes a Graduate College rule.
• U denotes a University rule.
Entrance Requirements
Below we outline the entrance requirements for our various degree programs. Students coming from smaller programs might not meet all the entrance requirements and in this instance the department can
consider a provisional acceptance when reviewing your application. If you have questions, please contact the Graduate Director.
• G To enter any graduate program, students must have a baccalaureate degree or equivalent from a regionally accredited college or university. See the Graduate College Bulletin for general
• M To be admitted to the MA or the PhD program, a student must have completed at least two 3-hour senior level courses in abstract algebra, analysis or topology.
• M To be admitted to the MS program, a student must have completed, at a minimum, coursework in the following areas:
□ Differential Equations (MATH 3113 or MATH 3413 or equivalent).
□ Linear Algebra (MATH 3333 or equivalent).
□ Modern Algebra (MATH 4323 or MATH 4383 or equivalent).
□ Introductory Mathematical Analysis (MATH 4433 or equivalent).
□ Introductory Probability/Statistics (MATH 4733 or MATH 4743 or equivalent).
In the case of a student who has taken some or all of these courses at other universities, the Graduate Committee will determine whether the student's courses are acceptable substitutes for the
courses listed above. Students lacking one or more of the above requirements may be admitted on a provisional basis. In this case, they are expected to remedy any deficiencies in their first
semester of study. For coursework taken to remedy deficiencies, at most three hours of credit can be applied toward the student's degree program.
• U All applicants for whom English is a second language must present evidence of proficiency in the English language. You need a TOEFL score of at least 79, an IELTS score of at least 6.5, or a
PTE score of at least 60. For more details, see here.
Requirements for Graduate Teaching Assistants
• M Before the start of the Fall semester, the Mathematics Department has an orientation for new graduate students. Among other things, students are given information about professional conduct of
GTAs and initial training for their role as a TA in the Math Department. Graduate students are expected to be in Norman by August 1 to begin departmental and university orientations.
• U Incoming or returning students who will be Graduate Teaching Assistants at OU for the first time are required to attend the New Graduate Teaching Assistant Orientation offered by the Graduate
College. Eligibility for tuition waiver depends on completion of this orientation.
• G General requirements, expectations, and procedures for GTAs are laid out in the Graduate College Bulletin.
Graduation Requirements for the PhD Program
• M Students in the PhD program may choose Research in Undergraduate Mathematics Education (RUME) as a research area; this is known as the RUME Option of the PhD program. All other research areas
fall under the Traditional Option of the PhD program.
• In order to graduate with a PhD degree, a student in the PhD program must satisfy all of the following:
□ G Students must satisfy the general graduation requirements for the doctoral degree as stated in the Graduate College Bulletin. In particular, the total number of hours, combining both formal
courses and hours of research, will be at least 90 post-baccalaureate hours.
□ M Students must satisfy the Math Hours Requirement, as follows. They must complete 12 hours at the 5,000 or 6,000 level in one of the major areas of mathematics, or in the area of RUME. They
must also complete at least 18 hours of coursework in mathematics at the 5,000 level or above. This coursework must include a sequence of two 3-hour courses at the 5,000 or 6,000 level
outside of the student's major research area.
□ M Students must pass the Qualifying Examination. Only after this examination has been passed may the student form an Advisory Committee according to the rules of the Graduate College.
□ M Students choosing the Traditional Option must demonstrate reading proficiency in a foreign language or proficiency in a programming language as determined by the student's Advisory
Committee. Students choosing the RUME Option must undergo IRB training. This requirement must be satisfied before the General Examination takes place.
□ G Students must pass the General Examination.
□ G Students must write, defend and submit a dissertation.
Graduation Requirements for the MA Program
• M The MA program is a non-thesis program.
• In order to graduate with an MA degree, a student in the MA program must satisfy all of the following:
□ G Students must satisfy the general graduation requirements for the master's degree as stated in the Graduate College Bulletin. In particular, at least 32 hours of graduate coursework are
□ M Students must complete at least five of the six core courses:
☆ Abstract Algebra I and II (MATH 5353 and MATH 5363).
☆ Real Analysis I and II (MATH 5453 and MATH 5463).
☆ Topology I and II (MATH 5853 and MATH 5863).
If not all of the three year-long sequences above are taken, then the student may substitute another sequence approved by the Graduate Director.
□ G Students must take and pass a Comprehensive Examination based on the content of the core sequences (Algebra, Analysis, and Topology). The Comprehensive Exam is an oral or written exam
administered by a committee of at least three faculty members.
• G Students enrolled in the PhD program may earn an MA degree by adding the MA program and either See Graduate College Bulletin for additional details. In particular, students must file the
Admission to Candidacy form in the semester before taking the Comprehensive Examination; deadlines can be found on the Grad College website.
• M At the time of filing the Admission to Candidacy form, students must inform the Graduate Director of their intent to earn the MA degree.
Graduation Requirements for the MS Program
• M The MS program is a non-thesis program.
• In order to graduate with an MS degree, a student in the MS program must satisfy all of the following:
□ G Students must satisfy the general graduation requirements for the master's degree as stated in the Graduate College Bulletin. In particular, at least 32 hours of graduate coursework are
□ M No course below the 4000 level may be applied to the MS degree. A maximum of 12 hours of 4000-level coursework may be applied to the MS degree, and this total may not include more than 9
hours of 4000-level mathematics courses. No more than 9 hours of coursework outside the Mathematics Department may be applied to the MS degree.
□ M Students must complete coursework in the following areas. A single course cannot be used to fulfill two different requirements.
☆ Statistics: One 3-hour course beyond the introductory level (MATH 5743 and MATH 4753 are admissible).
☆ Numerical Analysis or Computer Science: One 3-hour course.
☆ Mathematical Models: This requirement consists of the course MATH 5103.
☆ Abstract Mathematics: Two 3-hour courses.
☆ Applied Mathematics: Two 3-hour courses.
☆ Outside Courses: Two 3-hour courses taken outside the Mathematics Department that use some mathematics at the level of Calculus or higher.
□ G Students must take and pass a Comprehensive Examination according to the rules laid out in the Graduate College Bulletin.
□ M The dates of the exam will be determined by the MS Adviser in consultation with the students involved. Students must inform the MS Adviser that they intent to take the Comprehensive
Examination in the semester prior to the one in which they take the exam.
• M The structure of the Comprehensive Examination is as follows:
□ The Comprehensive Examination will consist of five written exams based on five mathematics courses from the student's degree program. No courses from outside the Mathematics Department can be
used. Each exam will be two hours long.
□ The choice of the five exam courses is made by the student but is subject to the approval of the Graduate Committee.
□ The selection of the five exam courses must conform to the following guidelines:
☆ At least three of the courses must be at the 5000 level or higher.
☆ At least one course must be an abstract mathematics course.
☆ At least one course must be in either probability/statistics or numerical analysis.
The Qualifying Examination
• M The Qualifying Examination consists of three qualifying exams in Algebra, Analysis and Topology, and a Complex Analysis requirement.
• M The qualifying exams in Algebra, Analysis and Topology are based on the three core sequences:
□ Abstract Algebra I and II (MATH 5353 and MATH 5363).
□ Real Analysis I and II (MATH 5453 and MATH 5463).
□ Topology I and II (MATH 5853 and MATH 5863).
These are written exams. They are offered each August and each January in the weeks before classes start (in the case of January, the week before; in the case of August, two weeks before).
• M The qualifying exams in Algebra, Analysis and Topology are written and graded by qualifying exam committees, one for each of the three areas. The qualifying exam committee responsible for the
August n and January n+1 exams consists of the teachers on record for the corresponding qualifying courses that started in fall n-2, fall n-1 and fall n.
• M Each qualifying exam will be graded on a scale of A-level pass, B-level pass, or fail. To pass the qualifying exams, each student must receive either three A-level passes or two A-level passes
and one B-level pass on the three exams.
• M After the qualifying exams have been graded, the grading committees will make a recommendation to the Graduate Committee on the outcomes. The Graduate Committee will meet and make the final
• M Students who in their first year are not placed into one of the introductory courses - Introduction to Abstract Algebra I and II (MATH 4323, MATH 4333) and Introduction to Analysis (MATH 4433,
MATH 4443/5443) - must pass the three qualifying exams (as described above) within 2.5 years of entering the program. Students who in their first year are placed into one of the introductory
courses must pass the three qualifying exams within 3.5 years.
• M Students are encouraged to attempt the qualifying exams as soon as they are able. Newly arriving students may attempt the qualifying exams before the start of their first semester.
• M To complete the Complex Analysis requirement, students must pass the course Complex Analysis I (MATH 5423) with a grade of B or higher, or pass an exam at an equivalent level based on the
content of the course. This requirement must be met before taking the General Exam.
The General Examination
The General Examination is governed by the rules laid out in the Graduate College Bulletin.
• G The General Examination consists of a written part and an oral part. Both parts have to be taken in the same semester. The written part has to be successfully completed before the oral part
takes place.
• M The written part consists of one or more 3-hour written exams, each based on a 5,000 or 6,000 level sequence in the student's research area. The student's Advisory Committee determines the
number of written exams. The Advisory Committee determines the specific sequences the student will be examined on. A sequence need not necessarily consist of parts I and II of the same course; it
may consist of a one-semester course combined with a reading course, or it may consist of two reading courses. A sequence must consist of the equivalent of 6 credit hours. Transfer credit is
admissible as material for the written exam.
• G The student must pass the written part before proceeding to the oral part. The Advisory Committee determines whether the written part has been completed successfully. If the student fails the
written part, the entire General Examination counts as failed.
• M Before proceeding to the oral part, the student is required to give a seminar presentation in an appropriate venue, usually one of the departmental area seminars, based on a research paper
selected by the student's Advisory Committee. The seminar presentation may take place in the same semester as the General Examination, or in an earlier semester.
• M The oral part for the Traditional Option consists of an oral exam that covers two 5,000 or 6,000 level sequences, including any sequence used in the written part. The oral exam also covers the
student's seminar presentation.
• M The oral part for the RUME Option consists of an oral exam the scope of which is determined by the student's Advisory Committee. In addition, the student must present a prospectus for the
dissertation or a review of a paper submitted for publication to a research journal.
• G A doctoral student who enters the graduate program with a bachelor's degree is expected to pass the General Examination within five years of the student's first enrollment in a graduate course
applied to the doctoral degree. A doctoral student who enters the graduate program with a master's degree is expected to pass the General Examination within four years of the student's first
enrollment in a graduate course applied to the doctoral degree.
Advisement, Enrollment and Academic Progress
• Program and course advisement:
□ M Each fall and spring semester each graduate student should meet with his/her adviser to determine courses to enroll in for the following semester.
□ M Doctoral students who have formed their Advisory Committee should consult their dissertation adviser. The Graduate Director serves as adviser for doctoral students who do not yet have a
dissertation adviser, and for all MA students. The Graduate Liaison for MS students serves as adviser for all students seeking the MS degree.
• Enrollment requirements:
□ G All enrollments must be approved by the student's adviser (the dissertation adviser in the case of PhD students who have formed their Advisory Committee, in all other cases the Graduate
□ G Graduate Teaching Assistants are required to take at least five credit hours per semester. The expected standard, however, is at least six credit hours per semester.
□ G The maximum enrollment is 16 hours per regular semester.
• Academic standards:
□ M After initial enrollment, a graduate student is expected to maintain academic standards set by the department and to make reasonable progress towards the degree sought. These standards
include enrollment requirements, grade requirements, taking appropriate courses and number of hours, and meeting degree requirements in a timely manner.
□ M The Graduate Director is responsible for monitoring the student's academic performance and progress toward the desired degree. In the case of a student who has formed a PhD Advisory
Committee, the student's PhD adviser is responsible for monitoring the student's academic performance and progress and reporting these to the Graduate Director.
□ M Concerns about matters of academic performance and progress will be discussed at the request of the Graduate Director. In addition, discussions concerning progress and performance may be
initiated by graduate students by making an appointment with the Graduate Director.
• Satisfactory progress:
□ M All appointments as a Graduate Teaching Assistant are contingent on satisfactory academic progress. Unless in exceptional circumstances, this is interpreted to mean completing at least six
hours of work per semester with at least a B average over all work taken.
□ M Students must take the specific courses required for their degree in a timely manner.
□ M Students must prepare for and take required examinations at a pace acceptable to the Graduate Committee and the Graduate Director.
□ M In addition, doctoral students are required
☆ to form an Advisory Committee and have its first meeting at the earliest feasible time, and
☆ to make reasonable progress in dissertation research. These elements of satisfactory progress will be determined by the dissertation adviser. They will be monitored under the supervision
of the Graduate Director.
The Departmental Teaching Certificate
• M The department offers all graduate students the possibility of earning an Endorsement in Preparation for Teaching Undergraduate Mathematics (the “Teaching Certificate”). This program is
separate and independent of the RUME option of the PhD program.
• M The requirements for the Teaching Certificate are as follows:
□ The student must have taught at least one mathematics course at OU.
□ The student must have completed at least six hours of graduate credit coursework focused on the teaching and learning of undergraduate mathematics.
• M The courses MATH 5253, MATH 5263 and MATH 5950 (RUME seminar) may be used to satisfy the hours requirement. A maximum of three hours of MATH 5950 may be used. Other courses require the approval
of the Graduate Committee.
• M Graduate students who complete the endorsement will obtain a certificate that verifies that the awardee has expressed special interest in teaching mathematics courses at the undergraduate
level, has taught college mathematics courses, and has completed at least six semester hours of coursework focused on the teaching and learning of undergraduate mathematics.
Language Certification
By state law, graduate students whose native language is not English need to be language certified before holding any instructional position at OU that requires contact with students. This
certification is administered through the English Training and Certification Services (ETCS).
• G There are three levels of language certification: A, B and C. Level A means the student is fully language qualified. For all levels there is an oral/aural requirement and a written requirement.
The precise requirements are listed here. These requirements can be satisfied by passing a SPEAK test, a TEACH test, and a WRITTEN test administered by the ETCS.
• M The Mathematics Department expects all graduate students to be fully language qualified by the end of their second year. Students who do not meet this deadline are required to
□ attempt the relevant ETCS administered tests every semester until they are fully qualified, and
□ participate in all departmentally sponsored activities designed to improve language and teaching skills.
Students past their second year who are not fully language qualified and who do not satisfy these requirements will automatically go on probationary status.
• M Each fall and spring semester the ETCS offers a free, non-credit Spoken English Class. International students wishing to take this class must notify the Graduate Director at least one month
before the beginning of classes. The Graduate Director will nominate students for the Spoken English Class as soon as possible to increase the likelihood of acceptance.
• M Any student who has not passed the SPEAK test and the TEACH test, at least at the support level, by the end of their first year will be required to enroll in the ETCS Spoken English Class in
the fall semester of their second year. They will also be required to take tests every semester until they are fully language qualified.
• M Students must inform the Assistant to the Graduate Director of any attempts made at an ETCS administered test, and of the outcome of the test. The Graduate Director will monitor each student's
progress and discuss a plan of action with students who fail to get qualified in a timely manner.
• M The Mathematics Department pays for the student's first attempt at each of the SPEAK, TEACH and WRITTEN test. The student is responsible for the costs of each subsequent ETCS test. However, the
department reimburses the student for the cost of a test if the student passes it at the instruct level.
Reporting Requirements
• M Students must keep the Graduate Director informed, in a timely manner, of all plans for taking qualifying, comprehensive or general exams, plans for graduation, and plans for termination or
change of program.
• M All of a student's communication with the Graduate College must be done through the Assistant to the Graduate Director. Students must provide a copy of all forms and written communication with
the Graduate College. The Assistant to the Graduate Director keeps records of each student's progress, and keeps on file all forms, communication and paperwork related to the student's graduate
• M International students must inform the Assistant to the Graduate Director of any attempts made at a test administered through the English Training and Certification Services (ETCS), and of the
outcome of the test.
Probationary Status
This section is about probationary status imposed by the Mathematics Department (not the Graduate College or the University).
• M Students will be placed on probationary status in the following circumstances:
□ International students past their fourth semester who are not fully language qualified, and who have failed to actively engage in trying to get qualified.
□ Students must take an appropriate course load each semester as determined by the Graduate Committee. Students who underenroll for two consecutive semesters will be removed from the graduate
□ Other severe circumstances as determined by the Graduate Committee. Such could include serious violations of academic standards or neglect of teaching assistant duties.
• M Students on probationary status are not eligible for any of the following:
□ Departmental travel support.
□ Summer teaching assistantships.
□ Departmental awards.
The Graduate Committee has the authority to overwrite these rules in individual cases.
• M Students on probationary status will be monitored and are subject to review by the Graduate Committee. If the situation leading to probationary status has been remedied, the probationary status
will be lifted. Otherwise the Graduate Committee may or may not extend the probationary period. If the committee finds no compelling reason to recommend an extension of the probationary period,
the student's assistantship may be terminated.
• M Students will be notified in writing when they are placed on probationary status, for what reason this action was taken, what are the possible consequences, and what actions they must take for
the probationary status to be lifted. | {"url":"https://math.ou.edu/graduate/grad_rules","timestamp":"2024-11-14T22:16:22Z","content_type":"text/html","content_length":"37461","record_id":"<urn:uuid:b9d0fed7-6ec9-43d1-b840-3e4a37ab92bd>","cc-path":"CC-MAIN-2024-46/segments/1730477395538.95/warc/CC-MAIN-20241114194152-20241114224152-00271.warc.gz"} |
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similar anomalies since those are just new tools that should not be approached in the slightest. This research is not aimed for academic/non-philosophical assessment because many users came from that
field. We are interested in the problem where there is a big anomaly and I’ve published a paper on that. The description of what we’re looking for is not necessary but it may be critical to the
solution. Last edited by Eros on Mon Jun 11, 2017 8:08 am; edited 12 times in total. Ride_Droid wrote: I’ve been here for awhile and have no idea if | {"url":"https://csmonsters.com/can-someone-take-my-assignment-and-provide-guidance-on-the-application-of-data-mining-algorithms-in-computer-science-projects","timestamp":"2024-11-13T12:45:39Z","content_type":"text/html","content_length":"86759","record_id":"<urn:uuid:1c9dfc2c-097b-4599-8d76-eef3767601bb>","cc-path":"CC-MAIN-2024-46/segments/1730477028347.28/warc/CC-MAIN-20241113103539-20241113133539-00420.warc.gz"} |
MTH2032: Differential Equations with Modelling
is focused on PDEs. The general terminology is covered again, before we then look at the ways in which a PDE is different to an ODE (for example, you no longer get arbitrary constants when you
integrate, but arbitrary functions). We look at boundary and initial conditions, and learn some relatively simple methods for solving PDEs (e.g. noticing it is similar to an ODE and the method of
separation of variables).
We then move on to Fourier Series. Basically, the point of this section is to show you that any periodic function (a function that repeats itself after some time), can be modelled as a (potentially
infinite) sum of sine and cosine graphs. Fourier Series solutions to differential equations actually make up the majority of solutions you're likely to see in the Heat and Wave equation parts, so
it's really worth your while to make sure you understand this part of the course well. We look at periodic extensions of functions with limited domain, too.
After learning about Fourier Series, we look at the Heat Equation, which models how the temperature of a rod changes over time. We go over Taylor Series in two variables again, before deriving and
solving the Heat Equation, given some initial and boundary conditions. Next, we learn about the Advection equation, which models how objects might float along a stream of some kind, and about
"characteristics". Finally, we look at the Wave Equation, and how to solve it. We also see an alternative method of solving the Wave Equation (the solution of d'Alembert), which represents a wave as
a sum of two travelling waves (if you do Physics, this should jog your memory), which interfere with each other. If nothing else, that's pretty cool.
This unit was a reasonable unit. Unfortunately, most students found it difficult to understand what Jerome was trying to teach, probably because he didn't really explain his derivations very well,
and didn't have a good grasp of when he was talking about a difficult concept that he needed to spend more time on. Either that, or he expected us to work through any difficulties we faced at home.
Rosemary was a pretty clear lecturer who had fairly good explanations.
In this unit, explanations are important. It's always a good idea to state what you're doing as you're doing it (even if it seems incredibly obvious to you), because there are always marks allocated
for explanations. Knowing when to invoke theorems, writing down what class of DE we have, and things like that are all easy to forget, but cost you marks in the end.
The tutorial quizzes last for 20 minutes each, and were initially at the start of the tutorial, before we asked Jerome to put them at the end (so we could actually ask our tutors for help). The ODE
quizzes were reasonably challenging and had a fair bit of time pressure. The PDE quizzes were a bit easier.
The first assignment was a typical maths assignment where you answer questions from a sheet. The latter two are more of a computer modelling exercise (using Excel, MATLAB, or other computer
software), where you numerically approximate Fourier Series and the Heat Equation, respectively.
The mid-semester test is on Jerome's section of the course, and goes for an hour. It's not impossible, but does test several different areas of the course. The final exam is roughly of the same
difficulty of the midsemester test, just covering the whole course. There were some tricky questions in both sections.
All in all, there were areas where the unit could have been improved, but it is certainly a useful unit if you wish to do maths or science, as DEs appear almost everywhere in those fields. | {"url":"https://uninotes.com/university-subjects/monash-university/MTH2032","timestamp":"2024-11-06T20:20:21Z","content_type":"text/html","content_length":"128599","record_id":"<urn:uuid:619e909b-22d8-4272-9d91-e7985228af47>","cc-path":"CC-MAIN-2024-46/segments/1730477027942.47/warc/CC-MAIN-20241106194801-20241106224801-00840.warc.gz"} |
Beam Expander Design Comparison: Keplerian and Galilean
Does it matter whether a beam expander or reducer has a Keplerian or Galilean design?
The design of the beam expander or reducer does not always matter to the application, but the choice can be influenced by factors such as the easier alignment and more intuitive design of the
Keplerian devices and the compact dimensions of the Galilean devices. Additionally, a Keplerian device focuses the light between its two lenses and then outputs an inverted beam. A Galilean device
maintains the orientation of the beam and provides the option to select lenses to reduce the amount of spherical aberration in the output beam.
Beam expanders and reducers are typically used only with collimated beams, rather than diverging beams, and these designs take their inspiration from Keplerian and Galilean telescopes. The
magnification provided in both cases equals the focal length of the output lens divided by the focal length of the input lens.
Figure 1: The simplest Keplerian beam expander or reducer includes two positive lenses. The focal lengths of Lens 1 and Lens 2 are f[1] and f[2], respectively. The lenses are separated by a distance
equal to the sum of their focal lengths (f[1] + f[2]), and the output beam is inverted relative to the input beam.
Figure 2: A basic Galilean beam expander or reducer includes a negative lens with focal length (f[1]) and a positive lens with focal length (f[2]). The lenses are positioned so that the distance
between them equals the difference in their focal lengths (f[2] - f[1]). Galilean designs do not focus the light between the two lenses, maintain the beam orientation, and are shorter than Keplerian
Characteristics of the Keplerian Design
In the simplest Keplerian design, two positive lenses are separated by a distance equal to the sum of their focal lengths (Figure 1). A design based on the Keplerian telescope will never be shorter
than the sum of the two lenses' focal lengths, and the output beam is inverted with respect to the input beam.
The beam comes to a focus between the two lenses. This provides the opportunity to spatially filter the beam. For example, a pinhole filter could be placed at the beam's focal point to improve beam
quality. Away from the focus, the beam diameter expands as it approaches the output lens. To increase the diameter of the collimated beam provided by the output lens, it is necessary to move the
output lens farther away from the focus. Since the distance between the focus and the output lens equals the focal length, this requires using a lens with a longer focal length.
The Keplerian design is typically not preferred for high-energy beams, such as the high-power pulsed laser beams used in some cutting and other manufacturing applications. Focusing pulses with
nanosecond duration and optical powers around ~1 MW or higher, for example, can ionize the air and create a spark, which undesirably reduces the power of the pulse and can negatively affect the beam
Characteristics of the Galilean Design
A basic Galilean telescope also includes two lenses, but one is negative while the other is positive (Figure 2). The lenses are positioned so that the distance between them equals the difference in
their focal lengths, resulting in a more compact design than the Keplerian approach.
The Galilean approach can also be used to minimize the spherical aberration induced by the beam expander or reducer. All spherical lenses introduce spherical aberration, and one consequence is the
spread of the beam focus along the optical axis. In the case of a positive spherical lens, parallel rays incident closer to the lens' outer perimeter focus to a point on the optical axis closer to
the lens, compared with parallel rays incident near the lens' center. Since a negative spherical lens has the opposite effect, the negative lens in the Galilean design can be used to cancel out some
of the spherical aberration induced by the positive lens.
When the device is used as a beam expander, the smaller-diameter beam is incident on the negative lens. The diverging beam provided by the negative lens increases in diameter as it approaches the
positive lens, instead of focusing between the two lenses. This diverging beam can be described as having a virtual focus, which is located on the opposite side of the negative lens, as shown in
Figure 2. Since the positive lens is one focal length (f[2]) away from this virtual focus, the positive lens outputs a collimated beam that is not inverted when compared with the input beam. If the
beam is not rotationally symmetric, the beam's output orientation may be important to the application.
Expansion Ratio
Beam expanders and reducers are designed to accept and provide collimated beams. Although the beams are collimated, their diameters change as they propagate due to the effects of diffraction.
Ideally, the input beam waist is positioned one focal length away from the input lens as shown in Figures 1 and 2. The output beam waist is then one focal length away from the output lens. If the
input beam waist is not one focal length away from the input lens, the location of the output beam waist, the output beam waist diameter, and/or the divergence of the output beam may not match
estimated values.
Both the beam's waist diameter (2W[o] ) and divergence angle (θ ) are affected by beam expanders and reducers. The change to these two beam parameters can be estimated using the device's beam
expansion ratio (m ). The output beam waist diameter is calculated by multiplying the beam expansion ratio and the input beam diameter. The output beam's divergence angle is calculated by dividing
the input beam's divergence angle by the beam expansion ratio.
When the device includes two lenses, the formula for calculating the beam expansion ratio is the same for both Keplerian and Galilean designs. The beam expansion ratio equals the focal length of the
output lens divided by the focal length of the input lens. The devices shown in Figures 1 and 2 are beam expanders when the light is incident on Lens 1, whose focal length is f[1]. In this case, the
second lens (Lens 2) has focal length f[2] and the beam expansion ratio (m[12]) is
If the devices in Figures 1 and 2 are used as beam reducers, the light is incident from the opposite direction, upon Lens 2. Then, Lens 1 is the output lens, and the beam expansion ratio becomes m[21
], which is f[1] divided by f[2].
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Limits and continuity by Michael Evans PDF download - 1414
Limits and continuity by Michael Evans PDF free download
Michael Evans Limits and continuity PDF, was published in 2013 and uploaded for 100-level Science and Technology students of University of Ilorin (UNILORIN), offering MAT112 course. This ebook can be
downloaded for FREE online on this page.
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Simulating Gaussian processes on a lattice – The Dan MacKinlay stable of variably-well-consider’d enterprises
Simulating Gaussian processes on a lattice
March 17, 2022 — July 26, 2022
Hilbert space
kernel tricks
Lévy processes
stochastic processes
time series
Assumed audience:
ML people
How can I simulate a Gaussian Process on a lattice with a given covariance?
The general (non-lattice) case is given in historical overview in Liu et al. (2019), but in this notebook, we are interested in specialising a little. Following the introduction in Dietrich and
Newsam (1993), let’s say we wish to generate a stationary Gaussian process \(Y(x)\) on points \(\Omega\). \(\Omega=(x_0, x_1,\dots, x_m)\).
Stationary in this context means that the covariance function \(r\) is translation-invariant and depends only on distance, so that it may be given \(r(|x|)\). Without loss of generality, we assume
that \(\mathbb{E}[Y(x)]=0\) and \(\var[Y(x)]=1\).
The problem then reduces to generating a vector \(\vv y=(Y(x_0), Y(x_1), \dots, Y(x_m) )\sim \mathcal{N}(0, R)\) where \(R\) has entries \(R[p,q]=r(|x_p-x_q|).\)
Note that if \(\mathbb \varepsilon\sim\mathcal{N}(0, I)\) is an \(m+1\)-dimensional normal random variable, and \(AA^T=R\), then \(\vv y=\mm A \vv \varepsilon\) has the required distribution.
1 The circulant embedding trick
If we have additional structure, we can work more efficiently.
Suppose further that our points form a grid, \(\Omega=(x_0, x_0+h,\dots, x_0+mh)\); specifically, equally spaced points on a line.
We know that \(R\) has a Toeplitz structure. Moreover, it is non-negative definite, with \(\vv x^t\mm R \vv x \geq 0\forall \vv x.\) (Why?) 🏗
Wilson et al. (2021) credits the following authors:
Well-known examples of this trend include banded and sparse matrices in the context of one-dimensional Gaussian processes and Gauss–Markov random fields Loper et al. (2021), as well as Kronecker
and Toeplitz matrices when working with regularly spaced grids (Dietrich and Newsam 1997; Grace Chan and Wood 1997). | {"url":"https://danmackinlay.name/notebook/gp_simulation_lattice.html","timestamp":"2024-11-04T23:43:42Z","content_type":"application/xhtml+xml","content_length":"40718","record_id":"<urn:uuid:f70e4c40-bdaa-4285-8f52-19ce25ab1f78>","cc-path":"CC-MAIN-2024-46/segments/1730477027861.84/warc/CC-MAIN-20241104225856-20241105015856-00529.warc.gz"} |
Deconstructing higher-order interactions in the microbiota: A theoretical examination
1. The animal gut is a complex ecosystem containing many interacting species. A major objective of microbiota research is to identity the scale at which gut taxa shape hosts. However, most studies
focus solely on pairwise interactions and ignore higher-order interactions involving three or more component taxa. Higher-order interactions represent non-additive effects that cannot be
predicted from first-order or pairwise interactions.
2. Possible reasons as to why studies of higher higher-order interactions have been scarce is that many host-associated systems are experimentally intractable, gut microbiota are prohibitively
species rich, and the influence of any given taxon on hosts is often context-dependent. Furthermore, quantifying emergent effects that represent higher-order interactions that are not simply the
result of lower-order interactions, present a combinatorial challenge for which there are few well-developed statistical approaches in host-microbiota studies.
3. In this perspective, our goal is to quantify the existence of emerging higher-order effects and characterize their prevalence in the microbiota. To do so, we adapt a method from evolutionary
genetics used to quantify epistatic effects between mutations and use it to quantify the effects of higher-order microbial interactions on host infection risk.
4. We illustrate this approach by applying it to an in silico dataset generated to resemble a population of hosts with gut-associated microbial communities. We assign each host a pathogen load, and
then determine how emergent interactions between gut taxa influence this host trait.
5. We find that the effect of higher-order interactions generally increases in magnitude with the number of species in the gut community. Based on the average magnitude of interaction for each
order, we find that 9^th order interactions have the largest non-linear effect on determining host infection risk.
6. Our approach illustrates how incorporating the effects of higher-order interactions among gut microbiota can be essential for understanding their effects on host infection risk. We conclude that
insofar as higher-order interactions between taxa may profoundly shape important organismal phenotypes (such as susceptibility to infection), that they deserve greater attention in microbiome
Animal guts contain complex microbial communities whose structure and function depend upon the interactions among microbes and the host. Gut microbiota serve as key actors in host health, impacting
development, metabolism, and pathogen susceptibility (Brugman et al., 2018). The development of microbe-free (also known as germ-free) model hosts has made it possible to experimentally study how the
microbiota influences host susceptibility to infection (Goodman et al., 2011; Ridaura et al., 2013). However, most studies rely on correlations between the relative abundances of individual bacterial
taxa and host infection risk (e.g. pathogen load), ignoring the potential influence of higher-order interactions between taxa within the community. The field of complex systems is increasingly
interested in understanding the emergent properties of higher-order interactions between objects (Lambiotte, Rosvall, & Scholtes, 2019a). Relatedly, a long-standing issue in ecology is to capture the
vast diversity of multispecies species interactions—the unpredictable effects that arise when multiple species are present in an ecosystem (Hutchinson 1962). For example, the order of arrival of
species into an ecosystem, and other factors (deterministic or stochastic in nature) can dictate species composition and the overall behavior of the system (Saavedra et al., 2017; Uricchio, Daws,
Spear, & Mordecai, 2019). This problem has more recently become the object of inquiry in communities of microbes (Enke et al., 2019; Mickalide & Kuehn, 2019; Sanchez-Gorostiaga, Baji•, Osborne,
Poyatos, & Sanchez, 2018). Many ecological studies involving complex network structures typically focus on pair-wise interactions and tend to ignore higher-order effects among three or more
components (Kareiva, 1994; Levine, Bascompte, Adler, & Allesina, 2017; Mayfield & Stouffer, 2017). For example, in a system with two interacting microbes—A and B—the addition of a third microbe C may
alter the pairwise interaction between A and B in a non-linear or non-intuitive fashion. This would constitute an emergent higher-order interaction between A, B and C. This is in contrast to a
scenario where the microbe C interacts with either A or B in isolation, which constitute pairwise interactions with their own interaction effects. Therefore, quantifying emergent higher-order effects
between microbial taxa is necessary to fully capture the structure and dynamics of biological systems.
Higher-order interactions have recently been the object of study in the realm of genetics, where they are discussed in light of epistasis, or non-linear interactions between genes and mutations (
Mackay & Moore, 2014; Weinreich, Lan, Jaffe, & Heckendorn, 2018a; Weinreich, Lan, Wylie, & Heckendorn, 2013). A useful non-technical definition of epistasis is the “surprise at the phenotype when
mutations are combined, given the constituent mutations’ individual effects (Weinreich, Lan, Jaffe, & Heckendorn, 2018b). This effectively captures what makes epistasis a provocative concept: the
notion that interacting objects or parcels can have effects that are non-additive. In particular, higher-order epistasis is of interest, as it comprises all of the complexity and challenges of
understanding and studying higher-order interactions in other systems (Lambiotte et al., 2019a).
Higher-order epistasis can have powerful effects on organismal phenotypes, which has complicated the genotype-phenotype mapping problem in genetics (Sackton & Hartl, 2016). To study higher-order
epistasis in model organisms, molecular biologists engineer genes and mutations of interest in all possible permutations, a method labeled the “combinatorial approach.” (Weinreich et al., 2018b, 2013
). Other studies resolve higher-order epistasis through more advanced statistical methods (Guerrero, Scarpino, Rodrigues, Hartl, & Ogbunugafor, 2019; Otwinowski, McCandlish, & Plotkin, 2018;
Poelwijk, Krishna, & Ranganathan, 2016; Sailer & Harms, 2017).
Insect gut microbiota have been used as model systems to study the formation and assembly of microbial communities. Insect guts harbor relatively fewer microbial species, as compared to higher
eukaryote hosts, with restricted core-members that can be grown axenically and manipulated genetically (Zheng, Steele, Leonard, Motta, & Moran, 2018). The protective function of microbes against
invading pathogens have been studied across a range of insect hosts. For example, previous studies with bees found that core gut species were associated with increased host health, while non-core
taxa were associated with decreased host health and increased pathogen infection (Cariveau, Elijah Powell, Koch, Winfree, & Moran, 2014; Koch & Schmid-Hempel, 2011; Raymann & Moran, 2018). However,
other studies have also shown that pathogens alter the gut microbiota and facilitate gut infections (Abraham et al., 2017; Wei et al., 2017). Although many studies have shown correlations between
core species and host traits, the extent to which individual versus species interactions facilitate or resist gut infections remains understudied.
Not unlike genomes, societies or neural circuits, insect gut microbiomes are complex systems defined by the interaction between individual parcels (component taxa in the microbiota). Consequently, we
might predict that higher-order interactions between taxa in the microbiota might underlie microbiota-associated organismal phenotypes, such as susceptibility to infection. Recent work by Gould et
al. 2018 found that higher-order interactions in the gut microbiota impact lifespan, fecundity, development time, and bacterial composition of Drosophila sp. With a gut community composed of 5 core
taxa, they found that three-way, four-way, and five-way interactions accounted for 13-44% of all possible cases depending on the host trait. Yet, lower-order interactions (2-pairs) still accounted
for at least half of all the observed phenotypes in the system (Gould et al., 2018).
Studies like Gould et al. 2018 provide an example of how higher-order interactions can be measured and suggest that they might be relevant for understanding how taxa influence certain phenotypes. But
while the importance of diversity and host interactions is clear, no studies have attempted to specifically disentangle effects beyond four or five-way interactions. One major barrier to more of
these studies is the paucity (or non-existence) of the datasets structured like those in an evolutionary genetics framework, such that existing statistical methods might be used to resolve
interactions. (Tekin, Savage, & Yeh, 2017; Wood, Nishida, Sontag, & Cluzel, 2012). For example, the problem of constructing a set of insects that each carry a different combination of constituent
taxa of interest grows exponentially with the number of taxa. And (perhaps) unlike genetics, constructing a different insect with a different set of bacterial taxa (corresponding with the possible
combinations of taxa) is a non-trivial technical challenge. Nonetheless, the use of combinatorial complete datasets—insects containing all combinations of taxa— to explore higher-order interactions
(beyond a single taxon or pairwise interactions) could help to inform how taxa interact in framing organismal phenotypes.
In this commentary, we propose a theoretical examination of higher order interactions in the gut microbiome. Specifically, we employ the Walsh-Hadamard transform (WHT), a mathematical regime that has
been used to demonstrate how higher-order interactions between mutations influence fitness or other organismal traits (Poelwijk et al., 2016; Weinreich et al., 2013), to explore how higher-order
interactions among gut taxa can influence host infection risk. We use it to quantify higher-order interactions in an in silico dataset resembling the type of data that can be empirically—that can be
developed in the future—collected from insect guts. We introduce this approach with the hope that it may eventually be applied to a tractable experimental system for real-world validation, and
believe that insect systems are among the most promising empirical systems.
The Walsh-Hadamard Transform allows one to quantify the eminence of interaction effects of different order in a system of potentially-interacting objects or parcels. It yields a Walsh coefficient,
which communicates the magnitude and sign of how a particular order interaction influences an output of interest. It implements phenotypic values in the form of a vector, before reformatting it into
a Hadamard matrix (and is then scaled by a diagonal matrix). The output is a collection of coefficients which measure the degree to which the map is linear, or second order, third, and so forth. We
provide a brief primer on the method, and refer readers to two published manuscripts—Poelwijk et al. (2016) and Weinreich et. al. (2013)—that outline and apply the method in good detail. Also see the
Supplementary Information for a brief primer.
The Walsh-Hadamard Transform relies on the existence of combinatorial data sets, where the objects for which we are interested in understanding the interactions between (taxa in this study) are
constructed in all possible combinations. Another limitation of the WHT is that it can only accommodate two variants per site, that is, two states per actor. In the case of taxa, we can think of this
in terms of the presence/ absence of a certain taxon, and we can encode this in terms of 0 (absence) or 1 (presence). For each hypothetical insect with a different presence/ absence combination, we
have a corresponding phenotypic measurement (e.g. parasite load). For example, if we wanted to measure the higher-order interactions between 4 taxa within an insect with regards to their role in
parasite load (as a model phenotype), we would need 2^L = 16 individual measurements (insects in this case), with L corresponding to the number of different taxa whose effects we were interested in
disentangling. We can encode this combination of 4 taxa in bit string notation (see Figure 1).
Each site (0 or 1) in the string corresponds to the presence or absence of a given taxa in a given insect. This notation allows us to keep a mental picture of which taxa are in which insect for which
we have a phenotypic measurement and can be used to construct a vector of values. For example, the string 1010 corresponds to an insect with the pattern of present (1), absent (0), present (1),
absent (0). The full data set includes a vector of phenotypic values for all possible combinations of taxa—0000, 0001, 0010, 0100, 1000, 0011, 0101, 0110, 1001, 1010, 1100, 0111, 1101, 1011, 1110,
1111. Note that these can be divided into different classes based on the “order” of the interaction. Order corresponds to the number of interacting actors. “Zeroth order” would correspond to the 0000
variant. This would translate to an insect that has none of the insect taxa present. There are 4, 1^st order interactions (0001, 0010, 0100, 1000), 6, 2^nd order (or pairwise) interactions (0011,
0101, 0110, 1001, 1010, 1100), 4, third-order interactions (0111, 1101, 1011, 1110), and 1 fourth order interaction (1111). The WHT will quantify
This vector of phenotypic values for the 16 will be multiplied by a (16 × 16) square matrix, which is the product of a diagonal matrix V and a Hadamard matrix H. These matrices are defined
recursively by: n is the number of loci (n = 4 in this hypothetical example). This matrix multiplication gives an output:
Where V and H are the matrices described in [1] and [2] above, and y is the Walsh coefficient, the measure of the interaction between parcels of information in a string. Using this, we compute y
values for every possible interaction between bits in a given string. The in silico generated data discussed in this commentary are composed of 10-bit strings, each corresponding to the presence/
absence of a different microbial taxa. Such a case would have 2^10 = 1024 total combinations of taxa, and corresponding phenotypic measurements (parasite load).
Similar to the 4-bit string example used to explain the method, note that each order has a different number of possible combinations. That is, the number of insects that can carry a combination of
interacting taxa of a certain order. These are as follows: 0^th = 1; 1^st = 10, 2^nd = 45; 3^rd = 120; 4^th = 210; 5^th = 252; 6^th = 210; 7^th = 120; 8^th = 45; 9^th = 10; 10^th = 1. The methods
offered here measure every one of these interactions (e.g. all 210 of the possible 6^th order interactions) between taxa. While our use of a 10-bit string (as opposed to an 8 or 15 bit string) is
rather arbitrary, it is meant to highlight the vastness of the higher-interaction problem: Even if we suspect that only 10 taxa are meaningfully influencing a phenotype of interest (many studies
contain more), the possible ways that these species are interacting, and the number of measurable coefficients between them can be astronomical in number.
Having outlined the method used to quantify higher-order interactions above, it is important to directly explain the presumptive biological interpretation of the values. The WHT returns a Walsh
coefficient for each “order” of interaction. This corresponds to the relative strength or importance of that “order” in the phenotype being measured. Therefore, the Walsh-Hadamard Transform can help
to interpret the overall presence and eminence of higher-order interactions between taxa in a microbiota.
Figure 2 depicts an in silico generated collection of 1024 insects, each containing one of the combinations of 10 taxa (2^10 = 1024), organized into a fitness graph (see Supplemental Information for
details on the in silico code and dataset). Each individual also has a parasite load. While other statistical methods may not require all possible combinations of taxa in order to extract meaningful
information on the magnitude of higher-order interactions, creating the combinatorial set demonstrates the size and shape of the problem, all of the possible ways that taxa could interact.
Figure 3 depicts the raw calculations of the Walsh coefficients for all of the higher-order interactions (orders 2 – 9). Here we observe that the magnitude and direction of the interaction effect
(Walsh coefficient) varies across different combinations of taxa. That the Walsh Hadamard Transform can disentangle these types of effects is a feature of the calculation and reveals the
possibilities that exist in complex systems—like the microbiota—where many different objects are interacting. It is especially important to note that the specific identity of the taxa present is very
important to understand in determining their interaction. We cannot assume that, for example, all third-order interactions (interactions between three taxa) will have the same magnitude or direction
of interaction (e.g. positive or negative).
Figure 4 demonstrates the sum of the absolute values of the interaction coefficients highlighted in Figure 3. Here, we can observe the raw magnitude (leaving the sign— positive or negative—of it
aside) of higher-order interactions as a function of interaction order. Between 1^st and 9^th order, higher-order effects increase, suggesting that they become more meaningful with the number of
interacting microbes. Without knowing the specific mechanism at work, determining the mean magnitude of coefficients provides relevant information on the eminence of a given order in the microbiota.
For example, in our in silico microbiota the 9th order taxa represents the highest magnitude of interaction relative to other taxa orders (Figure 4). As this is a theoretical, in silico generated
microbiota, we can interpret this finding as meaning that 9^th order interactions contain the largest average deviation from additivity. That is, knowledge of how any given 9 taxa will interact
requires very specific information on the identity of which 9 taxa are interacting. This is a characteristic of a highly non-linear, complex systems.
Note that all of these values—the raw in silico parasite load data, the interaction coefficients for all individual interactions, and the scaled, absolute value coefficients— can be found in the
Supplementary Material.
In this commentary, we explore the possibility of higher-order interactions between taxa composing an insect gut microbiota. Using in silico and applied mathematical approaches, we demonstrate how
higher-order interactions can be measured in a complex system of interacting microbial taxa. In our theoretical scenario, higher-order interactions are present and generally increase in relevance
with the order of interaction. Though our results are theoretical, they are results nonetheless (Goldstein, 2018), highlighting the vast scope of the higher-order interaction problem, and outline one
method that can be used to deconstruct them in biological systems. Though empirical data of the size and scope used in this study are currently challenging to generate, this intractability may be
temporary, and future methods may permit the generation of data similar in structure to those explored in our theoretical examination.
The approach used in this study—the Walsh-Hadamard Transform—has been previously used by theoretical population geneticists to measure non-linear interactions between mutations (Weinreich et al.,
2013). Several empirical data sets in genetics and genomics have demonstrated that the sign of interaction effects can change readily with the identity of the interacting parcels(Guerrero et al.,
2019; Weinreich et al., 2018a, 2013). Given this, we predict that the taxa that compose the gut microbiota might be similarly defined by higher-order interactions. The capacity for measuring the
effects of higher-order interactions on host fitness is an important step towards understanding the effects of microbiota on their host. Indeed, considering higher-order interactions can enable more
robust information on non-linear interactions in microbiome communities.
We found that higher-order interactions were present, and that taxa interacted both positively and negatively. Combined interactions among taxa are augmented compared to what is expected from
individual effects when phenotypic effects are positive. In contrast, higher-order effects are negative when combined interactions among taxa show a diminished return and are less fit than would be
expected from their individual effects (fig 3). Such combinatorial complete data-sets can tell us what scale microbial interactions matter in predicting host infection. Moreover, they reveal patterns
of interactions, particularly those combinations that interact synergistically or antagonistically (Hartl, 2014). One potential limitation of the outlined approach is the requirement for
combinatorial complete datasets. For high-diversity microbiomes, including humans and plants, it is not currently feasible to carry out experiments measuring phenotypes for all the possible microbial
Microbe-mediated protection against pathogens depends on subtle differences in gut community structure. In North American wild bumble bees, lower Chrithidia parasite infection loads are associated
with higher microbiota diversity. Using transplants to naive host, it was shown that the core-gut bacteria were responsible for conferring resistance to the Chrithidia parasite, while non-core gut
bacteria were found to be less effective against the parasite (Mockler, Kwong, Moran, & Koch, 2018). In mosquitos, gut bacterial species can trigger an immune defense against Plasmodium parasites,
the causative agent of malaria (Bahia et al., 2014). In sandflies, highly diverse midgut microbiota’s were found to be negatively correlated with the parasite that causes the vector-borne disease
leishmaniasis (Kelly et al., 2017). While these studies did not investigate the effects of higher-order interactions on host fitness, future experimental studies manipulating microbial communities
should consider combinatorial designs.
Recent theoretical work suggests that higher-order modeling approaches are able to capture volumes of rich data arising from complex ecological interactions (Lambiotte, Rosvall, & Scholtes, 2019b).
In this perspective, we have adapted approaches from population genetics to the study of host-associated microbiota. Applying these methods to the analysis of real experiments will yield important
insight into microbiome dynamics, towards a richer understanding of just how peculiar the microbiota is, and the many meaningful interactions that it is embodies.
Supplemental Information
The authors have prepared a simple mathematical primer on the Walsh-Hadamard Transform: https://github.com/OgPlexus/MicrobeTaxa1. For a more rigorous understanding, readers are encouraged to engage
the works cited in this manuscript.
We wish to acknowledge the organizers and participants of the 2017 RCN-IDEAS arbovirus workshop held in New Orleans. SY acknowledges funding support from NSF Postdoctoral Fellowship award number
1612302. CBO acknowledges funding support from NSF RII Track-2 FEC award number 1736253. The authors would like to thank Victor Meszaros and Miles Miller-Dickson for their input on the in silico
data, figures and Walsh-Hadamard primer. We finally thank Lawrence Uricchio for constructive feedback on our manuscript. | {"url":"https://www.biorxiv.org/content/10.1101/647156v1.full","timestamp":"2024-11-12T18:56:40Z","content_type":"application/xhtml+xml","content_length":"211113","record_id":"<urn:uuid:1769bc2c-1038-4b44-90c7-2bba9be3d8d1>","cc-path":"CC-MAIN-2024-46/segments/1730477028279.73/warc/CC-MAIN-20241112180608-20241112210608-00719.warc.gz"} |
2016 USAJMO Problems
Day 1
Problem 1
The isosceles triangle $\triangle ABC$, with $AB=AC$, is inscribed in the circle $\omega$. Let $P$ be a variable point on the arc $\stackrel{\frown}{BC}$ that does not contain $A$, and let $I_B$ and
$I_C$ denote the incenters of triangles $\triangle ABP$ and $\triangle ACP$, respectively.
Prove that as $P$ varies, the circumcircle of triangle $\triangle PI_BI_C$ passes through a fixed point.
Problem 2
Prove that there exists a positive integer $n < 10^6$ such that $5^n$ has six consecutive zeros in its decimal representation.
Problem 3
Let $X_1, X_2, \ldots, X_{100}$ be a sequence of mutually distinct nonempty subsets of a set $S$. Any two sets $X_i$ and $X_{i+1}$ are disjoint and their union is not the whole set $S$, that is, $X_i
\cap X_{i+1}=\emptyset$ and $X_i\cup X_{i+1}eq S$, for all $i\in\{1, \ldots, 99\}$. Find the smallest possible number of elements in $S$.
Day 2
Problem 4
Find, with proof, the least integer $N$ such that if any $2016$ elements are removed from the set $\{1, 2,\dots,N\}$, one can still find $2016$ distinct numbers among the remaining elements with sum
Problem 5
Let $\triangle ABC$ be an acute triangle, with $O$ as its circumcenter. Point $H$ is the foot of the perpendicular from $A$ to line $\overleftrightarrow{BC}$, and points $P$ and $Q$ are the feet of
the perpendiculars from $H$ to the lines $\overleftrightarrow{AB}$ and $\overleftrightarrow{AC}$, respectively.
Given that $\[AH^2=2\cdot AO^2,\]$prove that the points $O,P,$ and $Q$ are collinear.
Problem 6
Find all functions $f:\mathbb{R}\rightarrow \mathbb{R}$ such that for all real numbers $x$ and $y$, $\[(f(x)+xy)\cdot f(x-3y)+(f(y)+xy)\cdot f(3x-y)=(f(x+y))^2.\]$
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. | {"url":"https://artofproblemsolving.com/wiki/index.php/2016_USAJMO_Problems","timestamp":"2024-11-12T10:18:03Z","content_type":"text/html","content_length":"47939","record_id":"<urn:uuid:4a93bd59-fc0b-46a7-bd9b-150af37df7bc>","cc-path":"CC-MAIN-2024-46/segments/1730477028249.89/warc/CC-MAIN-20241112081532-20241112111532-00587.warc.gz"} |
OpenStax College Physics, Chapter 14, Problem 59 (Problems & Exercises)
Find the net rate of heat transfer by radiation from a skier standing in the shade, given the following. She is completely clothed in white (head to foot, including a ski mask), the clothes have an
emissivity of 0.200 and a surface temperature of $10.0^\circ\textrm{C}$, the surroundings are at $-15.0^\circ\textrm{C}$, and her surface area is $1.60 \textrm{ m}^2$.
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Solution video
OpenStax College Physics, Chapter 14, Problem 59 (Problems & Exercises)
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Video Transcript
This is College Physics Answers with Shaun Dychko. We're going to find the rate of heat transfer by radiation from the skier to her surroundings. So, her clothing is white, and so the emissivity is
low, that's 0.20. The total surface area is 1.6 square meters and the temperature of her body is 283.15 Kelvin. That's just on the inside of the clothes that gets at her skin surface. And then, the
temperature of her surroundings is negative 15 degrees Celsius converted into Kelvin by adding 273.15, which is a temperature of 258.15 Kelvin. So, the rate of heat transfer is Stefan Boltzmann's
constant multiplied by emissivity and times by the surface area multiplied by the temperature of her surroundings to the power of four minus temperature of her body to the power of four. And then,
we're going to get a negative answer here because as we expect, she is going to be losing, net loss in energy per time. And so, this formula gives us the rate at which energy is gained by something
and so we get a negative answer to say that it's a negative gain, which is a loss. Alright. So, we have Stefan Boltzmann's constant times emissivity times the surface area and then times by the
temperature of the surroundings to the power of four minus temperature just inside her clothes on her skin surface to the power of four, and this is negative 36.0 watts. | {"url":"https://collegephysicsanswers.com/openstax-solutions/find-net-rate-heat-transfer-radiation-skier-standing-shade-given-following-she","timestamp":"2024-11-04T02:46:03Z","content_type":"text/html","content_length":"177488","record_id":"<urn:uuid:19036de3-eafb-4a21-9f2b-b47ab998f2c0>","cc-path":"CC-MAIN-2024-46/segments/1730477027809.13/warc/CC-MAIN-20241104003052-20241104033052-00797.warc.gz"} |
Be Careful with Low Cost/Quality Common Mode Chokes ….
Dear readers, I have been really busy working in training and EMI troubleshooting in the past months. I am happy to be here again with a new post.
This month we will see a topic from a real-world situation – very interesting – for electronic designers working in conducted emissions problems.
In one of the projects I was involved with several weeks ago, a product was failing in conducted emissions on an AC power supply line. Measuring with a LISN (Line Impedance Stabilization Network),
the problem was related to common mode emissions.
The designer of the circuit tried to work with a common mode choke to reduce emissions (Y capacitors were not possible for this application).
He was using a low cost (this is important!), common mode choke using toroidal core.
For that component he had several samples for testing and, initially, the choke looked like an effective solution in the trials inside of the company. Let’s name the choke used CHKA.
With the promising result in the company, a prototype was specially prepared to go to an external lab (time to cross our fingers!).
But, in the external lab the product failed again (I think you have experimented similar situation), and a typical question comes to your mind: “How is it possible that the solution in the company
was failing in the external lab?”
The answer to that question lies, as usual, in discovering what is different between both scenarios.
Analyzing the problem, I discovered that the choke used in the prototype for the external lab was a different unit from that soldered in the original prototype. Same part number, same manufacturer,
same samples box, but …. a different unit, not EXACTLY the choke used in the company. Let’s name the second choke CHKB.
Before explaining the reason for the failure, let’s review the basics of a common mode choke.
A common mode choke is a coupled inductor: two inductors are built using the same core. Note the winding strategy (Fig. 1) is very important to obtain a common mode choke.
Fig. 1. Ideal common modes choke for differential currents (left), common mode currents (mid), and symbol for schematics (right).
For this ideal choke, the magnetic flux in the core is because the differential mode currents iDM (Fig. 1, left) cancel each other resulting in a zero impedance. But magnetic flux caused by common
mode currents iCM (Fig. 1, mid), is accumulated resulting in a high amount of impedance value. The symbol for this kind of choke (Fig. 1, right) uses two points to specify how the windings must be
done to obtain that behavior.
Summarizing, an ideal common mode choke looks like a simple wire for differential mode signals while it looks like an inductor for common mode signals. One of the advantages of these kinds of chokes
is they will not be saturated by differential mode currents.
For those coupled inductors, coupling factor k can be calculated from Eq. 1:
k = M/√(L1×L2) (Eq. 1)
and common mode and differential mode inductances can be obtained from Eq. 2:
LDM = 2×(L-M) and LCM = (L+M)/2 (Eq. 2)
where M is mutual inductance and L1, L2 are inductances for both inductors.
Considering the inductors are equal, L1 = L and for 100% of perfect coupling k=1, mutual inductance M is from Eq. 1 equal to inductance L (M=L) and common and differential mode inductances are from
Eq. 2, LDM =0 and LCM = L.
So, it is confirmed that we will find not impedance effect for differential mode signals and some value of impedance for common mode signals.
In a real common mode choke the cancellation is not perfect. As a result, the differential mode impedance is not zero. This effect is sometimes called “leakage”. This is useful for filtering
differential mode signals but the saturation effect must be checked in high current applications.
Let’s go back to our example failing in the laboratory. To analyze the situation, I measured the response of both chokes with my Bode 100 network analyzer (a really useful instrument if you are
interested in frequencies up to 50MHz).
A simplified measurement of a common mode choke can be done as shown in Fig. 2:
Fig. 2. Simplified measurement of impedances for a common mode choke.
The choke working satisfactorily in our application (CHKA) was measured and results are in Fig. 03:
Fig. 3. CHKA simple characterization.
You can see how big the impedance of the common mode effect is when compared with the differential mode effect.
For the second choke (CHKB), the one failing in the laboratory, I was able to see a very subtle difference: one of the coils of the choke had ONE TURN missing (Fig. 4).
Fig. 4. The chokes used in our example.
CHKA had 14 turns for L1 and L2. CHKB had 14 turns for L1 and 13 turns for L2.
This is a very critical difference. If one of the coils is not exactly as the other, the common mode inductance will be reduced (poor common mode filtering) and the differential inductance will
increase (perhaps the core can be saturated with the nominal current in high current applications).
This kind of cores are manually winded so human errors and/or low-quality tests can create this difficult to find problem.
The comparison of both chokes is included in Fig. 5:
Fig. 5. Comparing chokes CHKA and CHKB.
From the measurements, it is clear how important a perfect symmetry for the two coils in the choke. With only one turn missing in one of the coils the common mode impedance (Fig. 5, left) is
drastically reduced, as for example from point A to point B at the same specific frequency. The result will be a lower effectiveness to filter common mode EMI signals.
In the same way, the differential mode inductance increases from A to B (Fig. 5, right) with a typical effect of saturation of the core.
Let me conclude this post with two important pieces of advice: 1) be careful with low cost/low-quality components; and 2) try to have a network analyzer or impedance analyzer in your laboratory to
check how the component you are using in your design is. And of course, good luck in your next design! | {"url":"https://interferencetechnology.com/be-careful-with-low-cost-quality-common-mode-chokes/","timestamp":"2024-11-11T11:17:46Z","content_type":"text/html","content_length":"113947","record_id":"<urn:uuid:aa6c62f6-11e2-4086-9c4b-1db7224556f8>","cc-path":"CC-MAIN-2024-46/segments/1730477028228.41/warc/CC-MAIN-20241111091854-20241111121854-00741.warc.gz"} |
Nautical Miles (UK) to Point Converter
Enter Nautical Miles (UK)
β Switch toPoint to Nautical Miles (UK) Converter
How to use this Nautical Miles (UK) to Point Converter π €
Follow these steps to convert given length from the units of Nautical Miles (UK) to the units of Point.
1. Enter the input Nautical Miles (UK) value in the text field.
2. The calculator converts the given Nautical Miles (UK) into Point in realtime β using the conversion formula, and displays under the Point label. You do not need to click any button. If the
input changes, Point value is re-calculated, just like that.
3. You may copy the resulting Point value using the Copy button.
4. To view a detailed step by step calculation of the conversion, click on the View Calculation button.
5. You can also reset the input by clicking on button present below the input field.
What is the Formula to convert Nautical Miles (UK) to Point?
The formula to convert given length from Nautical Miles (UK) to Point is:
Length[(Point)] = Length[(Nautical Miles (UK))] / 1.90363051666085e-7
Substitute the given value of length in nautical miles (uk), i.e., Length[(Nautical Miles (UK))] in the above formula and simplify the right-hand side value. The resulting value is the length in
point, i.e., Length[(Point)].
Calculation will be done after you enter a valid input.
Consider that a British naval vessel travels 100 nautical miles (UK) to reach its destination.
Convert this distance from nautical miles (UK) to Point.
The length in nautical miles (uk) is:
Length[(Nautical Miles (UK))] = 100
The formula to convert length from nautical miles (uk) to point is:
Length[(Point)] = Length[(Nautical Miles (UK))] / 1.90363051666085e-7
Substitute given weight Length[(Nautical Miles (UK))] = 100 in the above formula.
Length[(Point)] = 100 / 1.90363051666085e-7
Length[(Point)] = 525312024.1811
Final Answer:
Therefore, 100 NM (UK) is equal to 525312024.1811 point.
The length is 525312024.1811 point, in point.
Consider that a sailing regatta covers a course of 75 nautical miles (UK).
Convert this distance from nautical miles (UK) to Point.
The length in nautical miles (uk) is:
Length[(Nautical Miles (UK))] = 75
The formula to convert length from nautical miles (uk) to point is:
Length[(Point)] = Length[(Nautical Miles (UK))] / 1.90363051666085e-7
Substitute given weight Length[(Nautical Miles (UK))] = 75 in the above formula.
Length[(Point)] = 75 / 1.90363051666085e-7
Length[(Point)] = 393984018.1358
Final Answer:
Therefore, 75 NM (UK) is equal to 393984018.1358 point.
The length is 393984018.1358 point, in point.
Nautical Miles (UK) to Point Conversion Table
The following table gives some of the most used conversions from Nautical Miles (UK) to Point.
Nautical Miles (UK) (NM (UK)) Point (point)
0 NM (UK) 0 point
1 NM (UK) 5253120.2418 point
2 NM (UK) 10506240.4836 point
3 NM (UK) 15759360.7254 point
4 NM (UK) 21012480.9672 point
5 NM (UK) 26265601.2091 point
6 NM (UK) 31518721.4509 point
7 NM (UK) 36771841.6927 point
8 NM (UK) 42024961.9345 point
9 NM (UK) 47278082.1763 point
10 NM (UK) 52531202.4181 point
20 NM (UK) 105062404.8362 point
50 NM (UK) 262656012.0905 point
100 NM (UK) 525312024.1811 point
1000 NM (UK) 5253120241.8109 point
10000 NM (UK) 52531202418.1087 point
100000 NM (UK) 525312024181.0872 point
Nautical Miles (UK)
A nautical mile (UK) is a unit of length used in maritime and aviation contexts. One nautical mile is equivalent to 1,852 meters or approximately 1.15078 miles.
The nautical mile is defined based on the Earth's circumference and is equal to one minute of latitude.
Nautical miles are used worldwide for navigation at sea and in the air. They are particularly important for charting courses and distances in maritime and aviation industries, ensuring consistency
and accuracy in navigation.
A point is a unit of length used primarily in typography and printing. One point is equivalent to 1/72 of an inch or approximately 0.3528 millimeters.
The point is defined as a standard unit of measurement for font sizes, line spacing, and other typographic elements in printed materials.
Points are widely used in the printing and graphic design industries to specify the size of type, spacing, and other design elements. The unit ensures precision and consistency in the presentation of
text and graphics.
Frequently Asked Questions (FAQs)
1. What is the formula for converting Nautical Miles (UK) to Point in Length?
The formula to convert Nautical Miles (UK) to Point in Length is:
Nautical Miles (UK) / 1.90363051666085e-7
2. Is this tool free or paid?
This Length conversion tool, which converts Nautical Miles (UK) to Point, is completely free to use.
3. How do I convert Length from Nautical Miles (UK) to Point?
To convert Length from Nautical Miles (UK) to Point, you can use the following formula:
Nautical Miles (UK) / 1.90363051666085e-7
For example, if you have a value in Nautical Miles (UK), you substitute that value in place of Nautical Miles (UK) in the above formula, and solve the mathematical expression to get the equivalent
value in Point. | {"url":"https://convertonline.org/unit/?convert=nautical_miles_uk-points","timestamp":"2024-11-07T10:49:13Z","content_type":"text/html","content_length":"91534","record_id":"<urn:uuid:82f92e90-8c5c-46d1-ba51-08959c8df453>","cc-path":"CC-MAIN-2024-46/segments/1730477027987.79/warc/CC-MAIN-20241107083707-20241107113707-00118.warc.gz"} |
January 3, 2014
Fifth International Workshop on Santilli's
Iso-, Geno-, and Hyper-Mathematics
Session 108 of ICNAAM 2014
PRWeb Release by VOCUS on the meeting
Santilli's Iso-, Geno- and Hyper-Mathematics to be Developed at
the 21014 ICNAAM Conference in Greece
A. A. Bhalekar (India), Email: anabha@hotmail.com,
C. Corda (Italy), Email: cordac.galilei@gmail.com,
T. Vougiouklis (Greece), Email: tvougiou@eled.duth.gr.
Let F(n, ⋅, I) be a numeric field of characteristic zero with elements n, m, ... (real, complex and quaternionic numbers), conventional associative product nm = n⋅m and multiplicative unit I : I⋅n =
n⋅I = n∀ n⋵F. In paper [1] of 1993, the Italian-American scientist R. M. Santilli (CV) pointed out that the ring F*(n*, ∗, I*) with elements n* = n⋅I*, associativity preserving product n*∗m* =
n*⋅T*⋅m*. and generalized multiplicative unit I* = 1/T* , I*∗n* = n*∗I* = n* ∀ n* ∈ F*, verifies all axioms of a numeric field under the condition for I* of being invertible.
Santilli called the new fields F* isofields,, the new numbers n* real, complex or quaternionic isonumbers, the new product n*∗m* isoproduct, and the new unit I* isounit where the prefix "iso" is used
in the Greek meaning of being "axiom-preserving," In the same paper [1], Santilli discovered that the isounit I* does not need to be an element of the original field, and can be any desired
non-singular quantity, such as a number, function, matrix, operator, etc. This led to the classification of isonumbers of the first (second) kind depending on when the isounit is (is not) an element
of the original field.
The non-triviality of Santilli isonumbers should be indicated to prevent predictable misjudgments. As an illustration, Santilli isofields imply that prime numbers do not have an absolute meaning
since their numerical value depends on the assumed unit. Consider the real isofields of the second kind F*(n, ∗, I*) where n*≈n since I* ∈ F. Then, for I* = 3, the number 4 is a prime number. Among
various independent studies on Santilli isonumber theory, we indicate monograph [2] by the Chinese mathematician C-X. Jiang and studies by the Italian physicist C. Corda [3] and Indian Chemist A. A.
Bhalekar [4].
In memoir [5] of 1996, Santilli identified the foundation of isomathematics which is characterized by an isotopic lifting of the totality of 20th century mathematics defined over a field of
characteristics zero, thus including the isotopies of functional analysis, metric spaces, differential calculus, Lie's theory, geometries, symmetries, topologies, etc. Particularly significant are
Santilli's step-by-step isotopies of the various branches of Lie's theory with isoproduct [A, B]* = A∗B - B∗A which have been constructed (when Santilli was at Harvard University in 1978 [6]) to
characterize, apparently for the first time in a consistent way, non-linear, non-local and non-Hamiltonian systems. Particularly important is also Santilli's isodifferential calculus that permitted
the isotopies of Newton's, Hamilton's, Schroedinger, Heisenberg's, Dirac's and other basic equations of physics. These broader dynamical equations and their underlying isomathematics characterize the
isotopic branch of hadronic mechanics, or isomechanics as a shot range covering of quantum mechanics to avoid excessive linear, local and Hamiltonian approximations of complex physical or chemical
An important application and verification of isomathematics and related isomechanics, for which the new mathematics was proposed for, has been the first representation at the non-relativistic and
relativistic levels of all characteristics of the neutron (and not only its mass) in its synthesis from a proton and an electron in the core of a star, p^+ + e^- → n + ν. In this case, the
conventional Schroedinger equation of quantum mechanics does not yield physically meaningful solutions because the rest energy of the neutron is higher by 0.782 MeV than the sum of the rest energy of
the proton and the electron, thus requiring a positive binding energy which is anathema for quantum mechanics. Thanks to their no-unitary structure, the Shcroedinger-Santilli isoequation for the
non-relativistic treatment and the Dirac-Santilli is isoequation for the relativistic treatment have permitted the removal of these insufficiencies for the most fundamental nuclear synthesis in
nature (see the review by J. V. Kadeisvili [7]). Comprehensive treatments of isomathematics, isomechanics and their applications in numerous fields are available in Santilli's monographs [5-11]. For
independent studies we indicate monographs [12-15] and quoted references by the mathematicians Gr. Tsagas, D. S. Sourlas, J. V. Kadeisvili, R. M. Falcon Ganfornina, J. Nunez Valdes, S. Georgiev, and
Santilli has dedicated his research life to the representation of irreversible systems due to their evident significant for energy-releasing processes. The research initiated during his graduate
studies at the University of Turin, Italy, in the mid 1960s. Santilli recognized that 20th century theories are reversible over time because based on Lie algebras whose product is invariant under
anti-Hermiticity [A, B] = AB - BA = -[A, B]^+. As a condition to assure irreversibility over time, Santilli therefore searched for a covering of Lie algebras whose product is neither antisymmetric
nor symmetric, selected Lie-admissible algebras according to the American mathematician A. A. Albert [16], and published in 1967 paper [17] the embedding of Lie algebras into covering Lie-admissible
algebras illustrated with the first formulation of (p,q)-deformations (A, B) = pAB - qBA that was later on followed by a large number of papers in the simpler deformations AB - qBA.
Also during his Ph. D. thesis, Santilli proved a No Reduction Theorem according to which macroscopic irreversible systems cannot be consistently reduced to a finite number of quantum mechanical
particles all in reversible conditions.Alternatively, Santilli No reduction Theorem implies that known thermodynamical laws cannot be consistently eliminated via the reduction of systems to quantum
mechanical events. The theorem established that macroscopic irreversibility, rather than "disappearing" at the elementary particle level, originate instead at the ultimate structure of physical
systems, thus establishing the need for their quantitative studies preferably as a covering of reversible quantum descriptions.
Mathematical maturity in the formulation of Lie-admissible treatments was initiated with paper [1] of 1993 in which Santilli discovered that, besides the unit of a field being an arbitrary
non-singular quantity, the multiplication can be ordered either to the right n>m or to the left n<m while preserving all axioms of a field. This allowed Santilli to introduce the forward genofields F
^>(n^>, >, I^>) with forward real, complex or quaternionic genonumbers n^> = n⋅I^>, forward genoproduct n^>>m^> = n^>⋅S⋅m^>, and forward genounit I^> = 1/ S, with conjugate backward genofields^<F(^
<n, <, ^<I), ^<I = 1 / R. The property R≠S then assures the lack of violation of causality and other physical laws in the representation of energy releasing processes.
Genomathematics is characterized by the dual, forward and backward genotopies of 20th century mathematics over a field of characteristic zero, thus including the forward and backward genotopies of
numbers, functional analysis, differential calculus, metric spaces, algebras, geometries, topologies, etc. which genotopies, to be consistent, must be formulated over a basic genofield. The
Lie-admissible covering of Lie algebras [6(b)] is characterized by two universal enveloping genoassociative algebras, one for the product ordered to the right and one to the left, resulting in a
geno-bimodular structure namely, a bimodular structure with action to the right A>|) = HS|) and to the left (|<H = (|RA, R≠S. Such a geno-bimodule characterizes a Lie-admissible algebra with product
(A, B) = ARB - BSA first identified by Santilli in 1978 [6], with genoequations i dA/dt = (A, H) = A<H - H<A = ARH - HSA. These equations are at the foundation of the genotopic branch of hadronic
mechanics, or genomechanics, that has already permitted the scientific identification and industrial development of new fuels and energies by U. S. publicly traded companies and their foreign
A presentation of Lie-admissible formulations as of 1995 is available in Refs. [6,8,9,10, 11]. The latest presentation is available in memoir [18] of 2006. Among a numbers of independent studies in
Lie-admissible treatments we mention refs. [19-27] by the physicists S. Adler, P. Roman, H. C. Myung, J. Ellis, E. Mavromatos, D. V. Nanopoulos, J. Dunning Davies, A. A. Bhalekar, V. M. Tangde and
others. Additional more recent references are available in the Proceedings of the Third International Conference on Lie-admissible Treatments of Irreversible Processes[28].
Hypermathematics is the most general mathematics that can be currently conceived by the human mind. It is based on multi-valued, forward and backward hyperlifting of genomathematics via
hyperoperations called "hopes" verifying the axioms of the Hv hyperstructures by the Greek mathematician T. Vougiouklis (see paper [29] of 2013 and references quoted therein). Hypermathematics admits
as particular cases the preceding geno- and iso-mathematics and characterizes the hyperstructural branch of hadronic mechanics, also calledhypermechanics [9, 18]. It should be noted that
hypermathematics achieves compatibility with our sensory perception because it is multi-valued rather than "multi-dimensional" as it is the case for various theories. A primary aim of
hypermathematics is to stimulate a new era in the study of biological structures, such as the DNA, whose complexity is outside realistic capabilities of 20th century mathematics as well as
isomathematics and genomathematics. Monograph [30] by I. Gandzha and J. V. Kadeisvili is suggested for a general independent review.
Scientist interested in participating are suggested to apply for financial support by sending via email a one page motivation and the CV to
The R. M. Santilli Foundation
Email: board(at)santilli-foundation(dot)org
Deadline for applications: August 15, 2014
[1] R. M. Santilli, "Isonumbers and Genonumbers of Dimensions 1, 2, 4, 8, their Isoduals and Pseudoduals, and ;Hidden Numbers; of Dimension 3, 5, 6, 7," Algebras, Groups and Geometries Vol. 10, 273
[2] Chun-Xuan Jiang, Foundations of Santilli Isonumber Theory, International Academic Press (2001),
[3] C. Corda, (a) "Introductionto Santilli iso-numbers American Institute of Physics Conf. Proc. 1479 , 1013 (2012); and (b) "Introduction to Santilli iso-mathematics", American Institute of Physics
Conf. Proc. 1558, 685 (2013)
[4] A. A. Bhalekar, "Santilli's New Mathematics for Chemists and Biologists. An Introductory Acount,", CACAA, 3(1), 15-86 (2014)
[5] R. M. Santilli, "Nonlocal-Integral Isotopies of Differential Calculus, Mechanics and Geometries," in Isotopies of Contemporary Mathematical Structures, P. Vetro Edi- tor, Rendiconti Circolo
Matematico Palermo, Suppl. Vol. 42, 7-82 (1996), http://www.santilli-foundation.org/docs/Santilli-37.pdf
[6] R. M. Santilli, Foundation of Theoretical Mechanics, Volume (a) I (1978) and (b) II (1982) Springer-Verlag, Heidelberg, Germany,
[7] J. V. Kadeisvili, "The Rutherford-Santilli Neutron," Hadronic Journal , 31, 1-114 (2008)
[8] R. M. Santilli, Isotopic Generalizations of Galilei and Einstein Relativities, Volumes (a) I and (b) II, International Academic Press (1991) ,
[9] R. M. Santilli, Elements of Hadronic Mechanics, Volumes I and II Ukraine Academy of Sciences, Kiev, second edition 1995,
[10] R. M. Santilli, Hadronic Mathematics, Mechanics and Chemistry,, Vol. I [a], II [b], III [c], IV [d] and [e], International Academioc Press, (2008), available as free downlaods from http://
[11] R. M. Santilli, Foundations of Hadronic Chemistry, with Applications to New Clean Energies and Fuels, Kluwer Academic Publishers (2001),
http://www.santilli-foundation.org/docs/Santilli-113.pdf .
[12] S. Sourlas and G. T. Tsagas, Mathematical Foundation of the Lie-Santilli Theory, Ukraine Academy of Sciences (1993),
[13] J. V. Kadeisvili, Santilli's Isotopies of Contemporary Algebras, Geometries and Relativities, Ukraine Academy of Sciences, Second edition (1997),
[14] R. M. Falcon Ganfornina and J. Nunez Valdes, Fundamentos de la Isoteoria de Lie- Santilli, International Academic Press (2001),
[15] S. Georgiev and J. V. Kadeisvili, Foundations of the IsoDifferential Calculus, Vol. I, to appear.
[16] A. A. Albert,"Onthepower-associativerings,"Trans.Amer.Math.Soc.,64, 552-593 (1948).
[17] R. M. Santilli, "Embedding of Lie-algebrasintoLie-admissiblealgebras, " Nuovo Ci- mento, 51, 570 (1967)
[18] R. M. Santilli, "Lie-admissible invariant representation of irreversibility for matter and antimatter at the classical and operator levels," Nuovo Cimento B 121, 443 (2006),
[19] S. Adler, Phys. Rev. 17, 3212 (1978)
[20] P. Roman and R. M. Santilli, "A Lie-admissible model for dissipative plasma," Let- tere Nuovo Cimento 2, 449-455 (l969)
[21] H. C. Myung and R. M. Santilli, "Bimodular-genotopic Hilbert Space Formulation of the Interior Strong Problem," Hadronic J. 5, 1367-1404 (l982).
[22] J. Fronteau, R. M. Santilli andA. Tellez-Arenas, "Lie-admissible structure of statis- tical mechanics," Hadronic Journal 3, 130-145 (1979).
[23] J. Ellis, N. E. Mavromatos and D. V. Nanopoulos in Proceedings of the Erice Sum- mer School, 31st Course: From Superstrings to the Origin of Space-Time, World Scientific (1996).
[24] J. Dunning-Davies, "Lie-admissible formulation of thermodynamoical laws," ZZZZZZZZZZ to be completed.
[25] J. Dunning-Davies, (a) "Thermodynamics of Antimatter via Santilli's Isodualities", Found. Phys. Lett., 12(6), 593-599 (1999); (b) "The Thermodynamics Associated with Santilli's Hadronic
Mechanics",, Prog. Phys., 4, 24-26 (2006).
[26] A. A. Bhalekar, (a) "Santilli's Lie-admissible mechanics. The only option commensurate with irreversibility and nonequilibrium thermodynamics", AIP Conf. Proc., 1558, 702-706 (2013); (b)
"Geno-nonequilibrium thermodynamics.I. Background and preparative aspects", CACAA, 2(4), 313-366 (2013); (c) "On the Geno-GPITT framework", AIP Conf. Proc., 1479, 1002-1005 (2012); (c) "Studies of
Santilli's isotopic, genotopic and isodual four directions of time", AIP Conf. Proc., 1558, 697-701 (2013).
[27] V. M. Tangde, (a) "Advances in hadronic chemistry and its applications", CACAA, 2(4), 367-392 (2013); (b) "Elementary and brief introduction of hadronic chemistry", AIP Conf. Proc., 1558,
652-656 (2013).
[28] C. Corda, editor, {\it Proceedings of the 2011 International Conference on Lie-admissible Formulations for Irreversible Processes,} Kathmandu University, Nepal (2011)
[29] R. M. Santilli and T. Vougiouklis, (a) "Isotopies, Genotopies, Hyperstructures and their Applications,"Prooc. Int. Workshop in Monderoduni: New Frontiers in Hyperstructures and Related Algebras,
Hadronic Press (1996), 177-188; and (b) "Lie-admissible hyper algebras," Italian Journal of Pure and Applied Mathematics, Vol. 31, pages 239-254 (2013)
[30] I. Gandzha and J Kadeisvili, New Sciences for a New Era: Mathematical, Physical and Chemical Discoveries of Ruggero Maria Santilli, Sankata Printing Press, Nepal (2011), | {"url":"https://www.santilli-foundation.org/Iso-Geno-Hyper-Math-2014.php","timestamp":"2024-11-08T01:21:11Z","content_type":"application/xhtml+xml","content_length":"26078","record_id":"<urn:uuid:fb3c7368-f891-46bc-a8f6-5e4eed62dc96>","cc-path":"CC-MAIN-2024-46/segments/1730477028019.71/warc/CC-MAIN-20241108003811-20241108033811-00293.warc.gz"} |
triliteral cipher decoder
The Beaufort Autokey Cipher is not Text to octal Binary decoder Hex to Base64 A1Z26 cipher Integer converter View Key How do you use Caesar cipher with numbers? Commercial Enigma. The Trifid cipher
was invented by the French amateur cryptographer Flix Delastelle and described in 1902. Feel free to edit this Q&A, review it or improve it! | Frequency analysis The companys traditional and digital
print-related services and office products serve the needs of publishers, merchandisers and retailers around the world. Just paste the encrypted text in the Ciphertext box and click on the Decipher
button. It uses numbers instead of letters. [1] Extending the principles of Delastelle's earlier bifid cipher, it combines the techniques of fractionation and transposition to achieve a certain
amount of confusion and diffusion: each letter of the ciphertext depends on three letters of the plaintext and up to three letters of the key. Reddit and its partners use cookies and similar
technologies to provide you with a better experience. The two-square cipher is also called "double Playfair". Please note that the encryption methods offered below are very basic and therefore not
considered as secure . They can also represent the output of Hash functions Now, divide the long string of numbers into three equal rows. Reading group 1: 31121,12132,12332, group 2: 312. The more
difficult variant, without word boundaries, is called a Patristocrat. It is an extension of the bifid cipher, from the same inventor. Triliteral cipher encryption uses a triliteral alphabet (
triliteral = 3 letters, or aloso called trifid or ternary = 3 items). 1:46. Triliteral systems: 3 letters of ciphertext per letter of plaintext trinomic systems: 3 digits of ciphertext per letter of
plaintext monome-dinome systems, also called straddling checkerboard systems: 1 digit of ciphertext for some plaintext letters, 2 digits of ciphertext for the remaining plaintext letters. Nihilist
cipher Now, read off each column and use the cube to convert the three numbers into the plaintext letter. It happens that 16-byte (128-bit) encryption keys are very commonly encoded in Base64, and
since 16 mod 3 = 1, their encoding will end with == . It can encrypt 27 characters. The Beaufort Cipher is reciprocal (the encryption and decryption algorithms are the same). The key is the number by
which we shift the alphabet, since this provides a unique way to describe the ciphertext alphabet easily. The program will search the dictionary for plaintext words matching the given ciphertext
words from the longest words to the shorter ones. Trifid Cipher / Decipher Plaintext secretmessage Ciphertext sppsdxmabpmjf The atbash cipher is In the Hill password, each letter is treated as a
26-base number: A=0, B=1, C=2 . 311 213 213 311 112 332 212 111 121 213 212 211 132. Necessary cookies are absolutely essential for the website to function properly. Someone I used to know just
posted this odd video with very clear instructions: "Follow the breadcrumbs." An N11 code (pronounced Enn-one-one) is a three-digit telephone number used in abbreviated dialing in some telephone
administrations of the North American Numbering Plan (NANP). Write to dCode! Find it in the bottom row of your code sheet, then find the letter it corresponds to in the top row of your code sheet and
write it above the encoded letter. The ciphered message has 3 distinct characters equally distributed. The classic Caesar cipher applies transformation only to letters. Now, divide the long string of
numbers into three equal rows. | Playfair cipher The Beaufort Cipher is named after Sir Francis Beaufort. The plaintext letter is These cookies track visitors across websites and collect information
to provide customized ads. Feedback and suggestions are welcome so that dCode offers the best 'Triliteral Cipher' tool for free! They seem to be in the same format, so finding the type for one would
likely reveal both: 103 141 156 040 171 157 165 040 150 145 141 162 040 155 145 077, 104 116 116 112 058 047 047 097 114 103 117 115 099 111 114 112 111 114 097 116 105 111 110 046 116 101 099 104.
subtracted from the key letter instead of adding them. The Delastelle trifid cipher uses three 9-character grids (for 27 distinct characters in total) and an integer N (usually 5 or 7). Hexadecimal
Codes can represent ASCII, UTF-8, or more advanced encoding schemes. Give your friend the encoded message and tell them the key. Cite as source (bibliography): Example: The encrypted message EEREG
with the key 123 The plain message is then DCODE. As discussed above, the cipher requires a 27-letter mixed alphabet: we follow Delastelle by using a plus sign as the 27th letter. This tool uses AI/
Machine Learning technology to recognize over 25 common cipher types and encodings including: Caesar Cipher, Vigenre Cipher (including the autokey variant), Beaufort Cipher (including the autokey
variant), Playfair Cipher, Two-Square/Double Playfair Cipher, Columnar Transposition Cipher, Bifid Cipher, Four-Square Cipher, Atbash Caesar cipher decryption tool The following tool allows you to
encrypt a text with a simple offset algorithm - also known as Caesar cipher. It is stronger than an ordinary Playfair cipher, but still easier Instead of using a 5x5 Polybius Square, you use a 3x3x3
cube. [4] A traditional method for constructing a mixed alphabet from a key word or phrase is to write out the unique letters of the key in order, followed by the remaining letters of the alphabet in
the usual order. With advanced technology and a consultative approach, LSCs supply chain solutions meet the needs of each business by getting their content into the right hands as efficiently as
possible. The Morbit cipher is Base64 is easy to recognize. The Trifid cipher was invented by the French amateur cryptographer Flix Delastelle and described in 1902. Write to dCode! Cipher
Description The Caesar cipher is named after the legendary Roman emperor Julius Caesar, who used it to protect his military communications. If CHAIR is written as 12345, RENT is written as 5678, and
. The ciphertext above represents "FELIX DELASTELLE" encrypted using the key CRYPTOGRAPHY. We use cookies on our website to give you the most relevant experience by remembering your preferences and
repeat visits. or modern crypto algorithms like RSA, AES, etc. Scan through the cipher, looking for single-letter words. If the length of each group is not divisible by 3, it will be hardest to break
the crypto. Depending on the orientation of the squares, horizontal or vertical, the cipher behaves slightly different. Switch to Virtual Enigma Virtual Lorenz SZ40/42 2D The cube is used again to
convert the numbers back into letters which gives us our ciphertext. a feedback ? The original plain text is DCODE. Triliteral cipher encryption uses a triliteral alphabet ( triliteral = 3 letters,
or aloso called trifid or ternary = 3 items). This website uses cookies to improve your experience while you navigate through the website. Example: The ciphered message ABAAACBBCABAABB have been
encoded with the alphabet A=AAA, B=AAB, C=AAC, D=ABA, etc. How to decipher Triliteral without knowing the alphabet? Because of this, Each letter is then replaced by a corresponding triple of 3
letters. After that, each letter is replaced by a triple of three letters. That The cookie is used to store the user consent for the cookies in the category "Analytics". Maybe I'm just blanking out
on something really obvious? You can quickly encode text here too. e 0.12702 t 0.09056 a 0.08167 o 0.07507 i 0.06966 n 0.06749 s 0.06327 h 0.06094 r 0.05987 d 0.04253 l 0.04025 c 0.02782 u 0.02758 m
0.02406 w 0.02360 It uses genetic algorithm over text fitness function to break the encoded text. Security of the Nihilist cipher The Nihilist cipher is quite similar to the Vigenre cipher. The list
of permutations with repetition of 3 characters A, B and C is: AAA, AAB, AAC, ABA, ABB, ABC, ACA, ACB, ACC, BAA, BAB, BAC, BBA, BBB, BBC, BCA, BCB, BCC, CAA, CAB, CAC, CBA, CBB, CBC, CCA, CCB, CCC.
The decryption is identical to Vigenere, but with a numeric key. The N number quickly changes the encrypted message, for better encryption, it is advisable to take a value of N coprime with 3. The
original plain text is DCODE. Here's the video in question: https://www.youtube.com/watch?v=lZd66Ha6pEU Here's the only ciphers I'm stuck on + their timestamps! Knowing that n objects, combined in
trigrams in all possible ways, give nnn = n3, we recognize that three is the only value for n; two would only give 23=8 trigrams, while four would give 43=64, but three give 33=27. Privacy Policy. .
(ii) Add the cipher letter on the S-wheel to the resulting letter to get the first enciphered letter. It is an extension of the Delastelle Trifid Cipher on dCode.fr [online website], retrieved on
2023-04-18, https://www.dcode.fr/trifid-delastelle-cipher, trifid,delastelle,three,grid,row,column,felix,marie, https://www.dcode.fr/trifid-delastelle-cipher, What is Delastelle Trifid cipher? Tool
to decrypt/encrypt Triliteral automatically. Here's the only ciphers I'm stuck on + their timestamps! It is then read out line by line from the top. The numbers are now read off horizontally and
grouped into triplets. Alphabetical substitution Plaintext Alphabet Ciphertext Alphabet Case Strategy Foreign Chars Triliteral cipher encryption uses a triliteral alphabet ( triliteral = 3 letters,
or aloso called trifid or ternary = 3 items). Cookie Notice The monoalphabetic substitution cipher is one of the most popular ciphers among puzzle makers. Pick a message to write to your friend. With
that responsibility in mind, we respect the value of publishing as a physical manifestation of scholarship and strive to streamline the flow of knowledge by constantly improving all operations in a
manner that promotes the sustainability of our customers publishing programs. The Caesar cipher, also known as a shift cipher is one of the oldest and most famous ciphers in history. What code uses 3
digit numbers? Relative frequencies of letters in english language. transposition ciphers. | Baconian cipher normal. This cookie is set by GDPR Cookie Consent plugin. I've been researching what type
of code it could be, but I'm not having much luck. Example: Decrypt the message SJLKZT, with N = 5 and grids. Triliteral Cipher on dCode.fr [online website], retrieved on 2023-04-18, https://
www.dcode.fr/triliteral-cipher, triliteral,trifid,ternary,alphabet,frederici,cardan,vigenere,wilkins,3,abc,triple, What is a Triliteral cipher? You are viewing an archived version of cryptii. Using
the example encoding shown above, lets decode it back into its original form. and mapping it to its reverse, so that the first letter becomes the last letter, the second letter becomes the second to
last letter, and so on. other means such as lines, colors, letters or symbols. dCode is free and its tools are a valuable help in games, maths, geocaching, puzzles and problems to solve every day!A
suggestion ? Our mission is to provide accurate, efficient, timely and cost effective warehouse and fulfillment operations. The first step is to use the cube to convert the letters into numbers. It
encrypt the first letters in the same way as an ordinary Vigenre cipher, The copy-paste of the page "Triliteral Cipher" or any of its results, is allowed as long as you cite dCode! | Route
transposition This cookie is set by GDPR Cookie Consent plugin. O, or o, is the fifteenth letter in the ISO basic Latin alphabet and the fourth vowel letter in the modern English alphabet. are clues.
Instead it begins using letters from the plaintext as key. As with the Bifid Cipher, the cube can be mixed to add an extra layer of protection, but for these examples we not be using a mixed alphabet
cube. Exporting results as a .csv or .txt file is free by clicking on the export icon Feel free to edit this Q&A, review it or improve it! Follow the breadcrumbs. numbers into three equal rows the
longest words to the shorter ones give... The length of each group is not divisible by 3, it be... Or modern crypto algorithms like RSA, AES, etc website to function.! Words from the top absolutely
essential for the website to give you the most ciphers! Called `` double Playfair '' Playfair cipher the Nihilist cipher is one of the squares, horizontal or vertical the! Amateur cryptographer Flix
Delastelle and described in 1902 stuck on + their timestamps while you navigate through the website plain. Functions Now, read off each column and use the cube to convert the three numbers into three
equal.!, etc the ciphertext above represents `` FELIX Delastelle '' encrypted using the key CRYPTOGRAPHY 'Triliteral! 311 112 332 212 111 121 213 212 211 132 or vertical, the cipher slightly! The
resulting letter to get the first step is to provide customized ads to describe the ciphertext box and on! Partners use cookies and similar technologies to provide accurate, efficient, timely and
cost effective warehouse and fulfillment....: 312 plain message is then replaced by a triple of 3 letters, or called... Behaves slightly different function properly 332 212 111 121 213 212 211 132
offered below are basic... Be hardest to break the crypto the given ciphertext words from the plaintext as key double Playfair.! Looking for single-letter words 2: 312 is easy to recognize offers the
best 'Triliteral cipher ' for! Monoalphabetic substitution cipher is reciprocal ( the encryption and decryption algorithms are the same inventor words! Or modern crypto algorithms like RSA, AES, etc
words from same. Preferences and repeat visits really obvious for the cookies in the ciphertext box and click on the button... Back into its original form so that dCode offers the best 'Triliteral
cipher ' tool for free numbers three... Identical to Vigenere, but with a numeric key: we Follow Delastelle by using plus... Base64 is easy to recognize Codes can represent ASCII, UTF-8, or more
advanced encoding schemes by! Delastelle and described in 1902 the first step is to provide accurate, efficient, and... Plaintext letter is then dCode, read off each column and use the cube to
convert the numbers... Same inventor more difficult variant, without word boundaries, is called a Patristocrat decode back... For the website to give you the most popular ciphers among puzzle makers
is Base64 is easy recognize. Orientation of the Nihilist cipher the Nihilist cipher is named after Sir Francis Beaufort use cookies on website... Written as 5678, and the only ciphers I 'm stuck on
their. And grids their timestamps ciphertext words from the same ) and click on the S-wheel to the ones! On something really obvious edit this Q & a, review it or it! Maybe I 'm just blanking out on
something really obvious hardest to break the crypto N 5... Numbers are Now read off each column and use the cube to convert the three into. Described in 1902 3 letters, or aloso called Trifid or
ternary = 3 ). Know just posted this odd video with very clear instructions: `` Follow breadcrumbs... Ii ) Add the cipher, also known as a shift cipher is reciprocal the..., from the plaintext letter
ciphertext alphabet easily 3, it will hardest. Plaintext words matching the given ciphertext words from the longest words to the Vigenre cipher basic and therefore considered. Maybe I 'm stuck on +
their timestamps 311 112 332 212 121! With N = 5 and grids Route transposition this cookie is used to just. Applies transformation only to letters popular ciphers among puzzle makers the top plain
message is then replaced by triple. Out on something really obvious the plaintext letter is replaced by a corresponding triple of three letters is! Puzzle makers `` Analytics '' decryption is
identical to Vigenere, but I 'm just blanking out on really. Here 's the only ciphers I 'm stuck on + their timestamps is named after Sir Francis.. For single-letter words, or aloso called Trifid or
ternary = 3 triliteral cipher decoder video very! We shift the alphabet A=AAA, B=AAB, C=AAC, D=ABA, etc invented by the French amateur Flix. It to protect his military communications 112 332 212 111
121 213 212 211.... Box and click on the orientation of the oldest and most famous in... Much luck Now read off horizontally triliteral cipher decoder grouped into triplets Now, read off column.
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such as lines, colors, letters or symbols cipher... Cipher applies transformation only to letters unique way to describe the ciphertext and! Paste the encrypted message EEREG with the key
CRYPTOGRAPHY advanced encoding schemes to know just this! | Playfair cipher the Nihilist cipher Now, read off horizontally and grouped into.. Morbit cipher is one of the bifid cipher, also known as a
shift cipher quite. Odd video with very clear instructions: `` Follow the breadcrumbs. above ``. By GDPR cookie Consent plugin stuck on + their timestamps similar to the shorter ones in... Our
website to give you the most popular ciphers among puzzle makers in... Feel free to edit this Q & a, review it or improve it message is then.. Most famous ciphers in history encrypted message EEREG
with the alphabet A=AAA, B=AAB, C=AAC, D=ABA,.... Cite as source ( bibliography ): example: the encrypted text the. Dcode offers the best 'Triliteral cipher ' tool for free, review it or improve!.
Line by line from the same ) distinct characters equally distributed paste the encrypted message with. 'S the only ciphers I 'm stuck on + their timestamps equally distributed RSA, AES,.! The crypto
requires a 27-letter mixed alphabet: we Follow Delastelle by using a plus sign the... Partners use cookies on our website to give you the most relevant experience by remembering your preferences and
repeat.. On the orientation of the most popular ciphers among puzzle makers triple of three letters behaves slightly.! Message and tell them the key 123 the plain message is then triliteral cipher
decoder out by..., read off each column and use the cube to convert the three numbers into three equal.! Category `` Analytics '' These cookies track visitors across websites and collect to! Cipher
behaves slightly different video with very clear instructions: `` Follow the breadcrumbs. decryption... Invented by the French amateur cryptographer Flix Delastelle and described in 1902 ternary = 3
items ) with clear! Lets decode it back into its original form 311 213 213 311 112 332 111! More advanced encoding schemes to give you the most popular ciphers among puzzle makers cookie Notice the
substitution! But with a numeric key ABAAACBBCABAABB have been encoded with the alphabet A=AAA, B=AAB C=AAC. The S-wheel to the Vigenre cipher been encoded with the alphabet A=AAA,,! String of
numbers into three equal rows the monoalphabetic substitution cipher is (... = 5 and grids was invented by the French amateur cryptographer Flix Delastelle described! Relevant experience by
remembering your preferences and repeat visits you navigate through the website to function properly using the is... Aloso called Trifid or ternary = 3 items ) Francis Beaufort text in the ciphertext
represents... Just blanking out on something really obvious looking for single-letter words legendary Roman emperor Julius Caesar who... User Consent for the website to function properly plaintext
words matching the given ciphertext words from same! Decryption is identical to Vigenere, but with a numeric key called a Patristocrat line from the as! And collect information to provide you with a
better experience the dictionary for plaintext words matching the ciphertext... Above represents `` FELIX Delastelle '' encrypted using the key is the number by which we the... Off each column and
use the cube to convert the three numbers into equal. Rent is written as 12345, RENT is written as 12345, RENT is as. Fulfillment operations replaced by a corresponding triple of three letters the
classic cipher. To recognize since this provides a unique way to describe the ciphertext box and on... Track visitors across websites and collect information to provide customized ads of each group
is not divisible by 3 it..., UTF-8, or aloso called Trifid or ternary = 3 letters, or more encoding... Mission is to use the cube to convert the three numbers into the plaintext letter transposition
this cookie set. The bifid cipher, from the longest words to the shorter ones other such! Alphabet ( triliteral = 3 letters the encoded message and tell them the key to... Best 'Triliteral cipher '
tool for free 27th letter are welcome so that dCode offers best! Line from the same inventor to store the user Consent for the cookies in the category Analytics! Delastelle and described in 1902 a
numeric key looking for single-letter words ( the encryption and algorithms. The number by which we shift the alphabet A=AAA, B=AAB, C=AAC, D=ABA,.... Playfair '' the ciphertext alphabet easily, with
N = 5 and grids line from the top, timely cost. Video with very clear instructions: `` Follow the breadcrumbs. is easy to recognize but! Experience by remembering your preferences and repeat visits
to the shorter ones also called `` double Playfair '',!
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The Stacks project
Remark 87.19.12 (Warning). The discussion in Lemmas 87.19.8, 87.19.9, and 87.19.10 is sharp in the following two senses:
1. If $A$ and $B$ are weakly admissible rings and $\varphi : A \to B$ is a continuous map, then $\text{Spf}(\varphi ) : \text{Spf}(B) \to \text{Spf}(A)$ is in general not representable.
2. If $f : Y \to X$ is a representable morphism of affine formal algebraic spaces and $X = \text{Spf}(A)$ is McQuillan, then it does not follow that $Y$ is McQuillan.
An example for (1) is to take $A = k$ a field (with discrete topology) and $B = k[[t]]$ with the $t$-adic topology. An example for (2) is given in Examples, Section 110.75.
Comments (2)
Comment #1950 by Brian Conrad on
This warning is written in a manner that is too cryptic. It is better to tell the reader straight up what the issue is: the open ideals $J(I) \subset B$ build in the proof of Lemma 14.10 might
fail to be a cofinal system of open neighborhoods of 0 in $B$.
Comment #2004 by Johan on
OK, I split the warning into two parts and I explain how to get an example for each (but the second is awful). See here or wait till the website is updated later this week.
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manual pages
degree {igraph} R Documentation
Degree and degree distribution of the vertices
The degree of a vertex is its most basic structural property, the number of its adjacent edges.
v = V(graph),
mode = c("all", "out", "in", "total"),
loops = TRUE,
normalized = FALSE
degree_distribution(graph, cumulative = FALSE, ...)
graph The graph to analyze.
v The ids of vertices of which the degree will be calculated.
mode Character string, “out” for out-degree, “in” for in-degree or “total” for the sum of the two. For undirected graphs this argument is ignored. “all” is a synonym of “total”.
loops Logical; whether the loop edges are also counted.
normalized Logical scalar, whether to normalize the degree. If TRUE then the result is divided by n-1, where n is the number of vertices in the graph.
cumulative Logical; whether the cumulative degree distribution is to be calculated.
... Additional arguments to pass to degree, eg. mode is useful but also v and loops make sense.
For degree a numeric vector of the same length as argument v.
For degree_distribution a numeric vector of the same length as the maximum degree plus one. The first element is the relative frequency zero degree vertices, the second vertices with degree one, etc.
Gabor Csardi csardi.gabor@gmail.com
g <- make_ring(10)
g2 <- sample_gnp(1000, 10/1000)
version 1.3.3 | {"url":"https://igraph.org/r/html/1.3.3/degree.html","timestamp":"2024-11-11T13:20:40Z","content_type":"text/html","content_length":"10562","record_id":"<urn:uuid:4d0d6fed-b1c1-448c-a3d1-e30c18eb0919>","cc-path":"CC-MAIN-2024-46/segments/1730477028230.68/warc/CC-MAIN-20241111123424-20241111153424-00770.warc.gz"} |
Does the surface topological order on the boundary of 3D topological insulator also have topological ground state degeneracy?
1555 views
The boundary of a 3D topological insulator can be fully gapped (under strong interaction) by the surface topological order without breaking the symmetry (see Fidkowski-Chen-Vishwanath,
Metlitski-Kane-Fisher, Bonderson-Nayak-Qi, Wang-Potter-Senthil). Suppose we make a solid torus shape of 3D topological insulator, and turn on the interaction to gap out the boundary (which is now a
torus) by the surface topological order. For 2D topological ordered state placed on a torus, we expect a topological ground state degeneracy equal to the number of anyon types. However does the
surface topological ordered 3D topological insulator also have the topological ground state degeneracy? If so how to understand the topological ground state degeneracy emerging from a symmetry
protected trivial state which is not expected to have the degeneracy? Also, is the surface topological ground state degeneracy still related to the number of anyon types?
This post imported from StackExchange Physics at 2015-02-05 10:18 (UTC), posted by SE-user Everett You
There should be topological degeneracy. One can imagine having a 2D topologically ordered state living on a true torus, which embeds in 3D. We would not question about the topological degeneracy. Now
instead of having a torus sitting in a vaccum, we insert into the inside of the torus some trivial insulator (meaning no topologically nontrivial excitations in 3D). Obviously the topological
degeneracy does not change. This is the same situation we have on the surface of a 3D SPT, because the surface SET is topologically ordered regardless of the symmetry, and once we break all symmetry
we just get back to the previous "trivial" case. The symmetry only manifests in the fractionalization of quantum numbers on the surface.
The brief answer is yes, they do have topological degeneracy. My understanding is that if you only focus on the surface topological order with some symmetry given by the bulk symmetry protected
topological (SPT) state, there is nothing wrong with that symmetry enriched topological order (SET) states. That means they also have all the properties as the usual SET states in pure 2D. However,
if you try to gauge the surface SET, you will find the obstruction to define a consistent gauged theory. More precisely, the gauged theory won't satisfy the pentagon equation. Those obstructions are
classified by the forth cohomology group, $H^{4}(G,U(1))$, where $G$ is the symmetry group of the bulk SPT. Usually, people say those SETs are anomalous. More details can be found in this paper http:
This post imported from StackExchange Physics at 2015-02-05 10:19 (UTC), posted by SE-user Sheng-Jie Huang | {"url":"https://www.physicsoverflow.org/26967/topological-topological-insulator-topological-degeneracy","timestamp":"2024-11-03T05:58:04Z","content_type":"text/html","content_length":"120438","record_id":"<urn:uuid:e3740f99-0343-4210-959e-c196998cdcaf>","cc-path":"CC-MAIN-2024-46/segments/1730477027772.24/warc/CC-MAIN-20241103053019-20241103083019-00303.warc.gz"} |
pls help on math cllg placement test problems - pls - great test site website inside
we need help on several questions on this test = we have have been struggling and digging thru our books ofr days and hours and still can't get these and the college placement test is tomorrow. I
would love some help....
Thist is the WA state math 'placement test' for several colleges - this is the practice test site - [website follows]
we need help on problems on the advanced test for numbers 13,21,22,26,27,28, 29, 30, 31, 42 [we are esp frustrated on numbers 21,22,28]
here is the website
THERE IS Another test that we are having problems with on the same site - that is the 'intermediate version of the test listed above on the same site - on the intermediate test, we are stumped on
numbers 10,20, 37, 41
I know this is a huge request, but we really need to get these figured out and we are at the end of our expertise. We also have dial up so visiting all kinds of math helps sites is tedious and not
very helpful.
thank you for any help you can give.
lisa j - ljdeerparkATaol.com
21 - use similar triangles, x/3 = 5/10 and solve
Edited: I meant (x+5)/10
22 - split it up, -3 < x+5 < 3, solve
28 - add 4x to both sides, subtract 5 from both sides, get 4x > -3, solve.
Don't stress too much over the test -- you want to be placed in a passable course or even one that's slightly easy. Additionally, I found the cutoffs, it looks like 14/30 is precalc 1, 17/30 is
precalc 2, 21/30 is calc 1.
Edited by kiana
I'm a doofus
for number 21, for the triangles x/3 = 5/10 = 3/2 and the answer key [on the last page of the test] says the answer to number 21 is 15/7th = x
and I didn't think we could use similar triangles is the test did not tell us that the triangles were similar...
I am not debating, I am just lost.. thanks...
I think I understand number 22 and number 28 now... thanks...
Can you tell me where you found the cut off scores of 14/30 - precalc 1; 17/30 precalc 2 and 21/30 is calculus....
and I didn't think we could use similar triangles is the test did not tell us that the triangles were similar...
They didn't explicitly tell you the triangles are similar, but you have enough information to see that they are similar: the acute angle in the lower left is common to the two triangles, and they
both have a right angle. Therefore, all three angles are the same, which means they are similar.
Off to print off the test and work thru the problems. I'll be back.
for number 21, for the triangles x/3 = 5/10 = 3/2 and the answer key [on the last page of the test] says the answer to number 21 is 15/7th = x
and I didn't think we could use similar triangles is the test did not tell us that the triangles were similar...
I am not debating, I am just lost.. thanks...
I think I understand number 22 and number 28 now... thanks...
Can you tell me where you found the cut off scores of 14/30 - precalc 1; 17/30 precalc 2 and 21/30 is calculus....
I lost the website now, I found it through google. They had a listing of cutoff scores.
The triangles are similar because they have the same angles, the lower left corner is the same angle and they both have a right angle. I apologise for mistyping, it should have been set up as x/3 =
(x+5)/10. Sorry for confusing you!
They didn't explicitly tell you the triangles are similar, but you have enough information to see that they are similar: the acute angle in the lower left is common to the two triangles, and they
both have a right angle. Therefore, all three angles are the same, which means they are similar.
Off to print off the test and work thru the problems. I'll be back.
bless you... thanks - I see now that the triangles are similar - I still don't understand the answer being 15/7th....
and Kiana, I found the cut off scores are listed on each college's website link that is linked as part of the math test placement site webpage.
lisaj, still working and printing problems as well -
Let's get rid of the denominator on the left by multiplying both sides of the equation by c, then:
xc + bx = ac
factor to get x(c + b) = ac
divide both sides by (c +b) and remember that (b +c) is the same as (c +b), and we get the correct answer
21 - use similar triangles, x/3 = 5/10 and solve
the correct proportion is: x/3 = (x +5)/10
which gives x = 15/7
the correct proportion is: x/3 = (x +5)/10
which gives x = 15/7
Yeah, I fixed that in my next post but I'll go back and edit that post as well.
On absolute value / inequality problems, it's often easiest to test the possible answers to see which works. (It's really easy to get the < /> signs turned around.)
So, here's how I would approach a problem like this:
I see from the possible answers that -8 and -2 are the key points. I'd quickly sketch a number line, mark those points on it, and test the answers. I like to start with the middle answer, and go from
there if it doesn't work.
The first part of answer b says x<-8, so I choose -10 for x (because it's easy to work with and less than -8). If I put -10 in for x in the original inequality, I see that it holds true. I also check
the x >-2 by choosing x=0. I see that this is also true, so I'm feeling really good about answer b. Answer a is only partly true, and answer c is never true.
Yeah, I fixed that in my next post but I'll go back and edit that post as well.
sorry -- I didn't see that you'd done that. I think we were posting at the same time.
bless you... thanks - I see now that the triangles are similar - I still don't understand the answer being 15/7th....
from the proportion x/3 = (x +5)/10 we can cross multiply:
10 (x) = 3 (x + 5)
10x = 3x + 15
subtract 3x from both sides
7x = 15
divide both sides by 7
x = 15/7
On graphs like this, pick a few numbers for x and see what it looks like.
First, I'd choose x = 0: any number raised to the 0 power = 1, so that eliminates answer b.
Now, we need to determine whether the value of y goes up rather quickly, as in answer c, or more slowly, as in answer a. If you plug in 1, 2, and 3 for x, you'll get 2, 4, 8, so answer c is best.
Assuming you can use a calculator, pick a value for theta, say 30 degrees, and see which holds true.
If you're picking a value for an angle, it's best to choose something less than 90, but not 45 degrees.
For this one, you need to remember the equations for the trig functions in a right triangle: use the mnemonic SOH CAH TOA
(I'm going to use x rather than theta because it's easier to type)
sin x = side opposite/hypotenuse
cos x = side adjacent/hypotenuse
tan x = side opposite/side adjacent
In the figure, sin x = b/c and tan x = b/a
so, sin x times tan x = (b/c) times (b/a) which = b^2/ca, which is answer c
for both of these, you want to be working with angles measured in radians rather than degrees
Pick a value for theta (in radians) and see which answer works
Here, you want to find what values of will cause y = 0. Since I've mostly been teaching Algebra 1 and Geometry the past several years, I don't remember these off the top of my head. But I would test
each graph by plugging in the values of x where the graph crosses the x axis (and y=0). When you do that, you should see that a works.
You've got me -- I honestly don't remember seeing this nomenclature.
I do see that the domain (possible values of x) can't be less than -2, but can go infinitely high, so I'm thinking that #3 is true. That eliminates answer a.
The range is the possible values of f(x), and since that's a square root, it can't include -2, so #4 is out. That would leave us with answer b.
5 - 4x > 2
add 4x to both sides
5 > 2 + 4x
subtract 2 from both sides
3 > 4x
divide both sides by 4
3/4 > x which is the same as x < 3/4, which is answer a
If you check their answer (x > 3/4) by testing with x =1, you get 5 - 4 > 2 or 1 > 2, which doesn't work on my planet, LOL.
5 - 4x > 2
add 4x to both sides
5 > 2 + 4x
subtract 2 from both sides
3 > 4x
divide both sides by 4
3/4 > x which is the same as x < 3/4, which is answer a
If you check their answer (x > 3/4) by testing with x =1, you get 5 - 4 > 2 or 1 > 2, which doesn't work on my planet, LOL.
Yep, they switched it.
from the proportion x/3 = (x +5)/10 we can cross multiply:
10 (x) = 3 (x + 5)
10x = 3x + 15
subtract 3x from both sides
7x = 15
divide both sides by 7
x = 15/7
I got it - and so did my dtr....
thanks so much
really - you all have went over and above
On the intermediate test - if anyone can help with these problems on the intermediate test - I would appreciate it -
number 10 on the intermediate test must have the wrong answer... D is correct if it is 2.0 x 10 -4power
26 says if 9 to the 3x power = 3; then x =
[and the choices are A 1; B 1/3 C 1/9 D 1/6 E 2
number 37 on the intermediate test
number 41 on the intermediate test
number 47 on the intermediate test [ I don't see how the answer could be B. it seems to me if chemical A is x; then the answer s/be answer A.]
thanks again - this is great assistance - I am a math nerd wanna-be
If 9^3x = 3:
9 is 3^2, so 9^3x = 3^6x;
Setting 3^6x = 3^1 (3^1=3), setting your exponents equal to each other, 6x = 1, so x= 1/6.
This is much harder to explain using the keyboard! Hope it is understandable.
If 9^3x = 3:
9 is 3^2, so 9^3x = 3^6x;
Setting 3^6x = 3^1 (3^1=3), setting your exponents equal to each other, 6x = 1, so x= 1/6.
This is much harder to explain using the keyboard! Hope it is understandable.
thanks again, we are knocking these down one by one...
lisaj [my dtr asks if this is a 'math help' site - ''well no, it is a homeschool mom help site....] | {"url":"https://forums.welltrainedmind.com/topic/111527-pls-help-on-math-cllg-placement-test-problems-pls-great-test-site-website-inside/","timestamp":"2024-11-13T12:57:52Z","content_type":"text/html","content_length":"394846","record_id":"<urn:uuid:2a253101-0cd3-41c0-8c3b-ae7f6ff348cb>","cc-path":"CC-MAIN-2024-46/segments/1730477028347.28/warc/CC-MAIN-20241113103539-20241113133539-00739.warc.gz"} |
What are 3 types of Wireless technology?
The Different Types of Wireless Communication
• Satellite Communication. Satellite communication is a crucial form of wireless communication.
• Infrared Communication.
• Broadcast Radio.
• Microwave Communication.
• Wi-Fi.
• Mobile Communication Systems.
• Bluetooth Technology.
Which technology is used in wireless communication?
Wireless communication technology transmits information over the air using electromagnetic waves like IR (Infrared), RF (Radio Frequency), satellite, etc. For example, GPS, Wi-Fi, satellite
television, wireless computer parts, wireless phones that include 3G and 4G networks, and Bluetooth.
Which of the following are examples of wireless technologies?
Some of these terms may be familiar to you: radio and television broadcasting, radar communication, cellular communication, global position systems (GPS), WiFi, Bluetooth and radio frequency
identification are all examples of “wireless”, with wildly different uses in some cases.
Which is the most popular wireless networking technology?
Thus far, 802.11 has become the most popular wireless data networking technology.
Which technology is a wireless technology?
Wireless technology provides the ability to communicate between two or more entities over distances without the use of wires or cables of any sort. This includes communications using radio frequency
(RF) as well as infrared (IR) waves .
What are the most common wireless technologies in use today on computer networks?
Today, we already have over a dozen widespread wireless technologies in use: WiFi, Bluetooth, ZigBee, NFC, WiMAX, LTE, HSPA, EV-DO, earlier 3G standards, satellite services, and more.
What is the latest wireless technology?
5G is the latest generation of wireless technology that is expected to provide significantly faster speeds and lower latency than current 4G networks. Millimeter waves are well-suited for 5G
applications due to their large bandwidth and ability to penetrate obstacles (not good as microwave spectrum).
Which is the most popular wireless network technology?
What are the latest wireless technologies?
Top 10 Wireless Technology Trends
• Wi-Fi.
• 5G.
• Vehicle-to-everything (V2X) wireless.
• Long range wireless power.
• Low power wide-area (LPWA) networks.
• Wireless sensing.
• Enhanced wireless location tracking.
• Millimetre wave wireless.
What is the fastest wireless technology?
Terabit free-space data transmission employing orbital angular momentum multiplexing. American and Israeli researchers have used twisted, vortex beams to transmit data at 2.5 terabits per second. As
far as we can discern, this is the fastest wireless network ever created – by some margin.
What are different wireless devices used today?
Some of the important Wireless Communication Systems available today are:
• Television and Radio Broadcasting.
• Satellite Communication.
• Radar.
• Mobile Telephone System (Cellular Communication)
• Global Positioning System (GPS)
• Infrared Communication.
• WLAN (Wi-Fi)
• Bluetooth. | {"url":"https://www.shadowebike.com/what-are-3-types-of-wireless-technology/","timestamp":"2024-11-04T23:44:26Z","content_type":"text/html","content_length":"47904","record_id":"<urn:uuid:57b580a7-192f-4ff5-a646-1c3cedb54322>","cc-path":"CC-MAIN-2024-46/segments/1730477027861.84/warc/CC-MAIN-20241104225856-20241105015856-00367.warc.gz"} |
Exercises - Review of Some Ideas from Precalculus
Solve $\displaystyle{\frac{\log_3 16}{2 \log_3 x} = 2}$
$x = 2$
Solve $\displaystyle{\log_3 (7-x) = \log_3 (1-x) + 1}$
$x = -2$
Solve $2^x + 2^{-x} = 2$
$x = 0$
Solve $\displaystyle{4^{-x+1} = \frac{1}{32}}$
$\displaystyle{x = \frac{7}{2}}$
Solve $2 \log_9 x = 1$
$x = 3$
Find $\displaystyle{\tan \left( 2\textrm{Arctan}(-\sqrt{3}) \right)}$
Find $\displaystyle{\textrm{Arcsin} \left( \sin \frac{3\pi}{4} \right)}$
Solve $2\cos^2 x + \sin x - 1 = 0$
$\displaystyle{x = \frac{7\pi}{6} \pm 2\pi n, \frac{11\pi}{6} \pm 2\pi n, \frac{\pi}{2} \pm 2\pi n, \textrm{ for } n \in \mathbb{N}}$
Find $\sec \theta$ if $\csc \theta = -3$ and $\cot \theta \lt 0$
$\displaystyle{\frac{3 \sqrt{2}}{4}}$
Find the center and radius of the circle described by $x^2 + y^2 - 4x + 2y + 1 = 0$
center $(2,-1)$; radius $2$
Find an equation of the line $\ell_1$ through $(-3,2)$ and $(2,-1)$, and then find the equation of the line $\ell_2$ passing through $(-1,-3)$ and perpendicular to $\ell_1$.
Using point-slope form in conjunction with the first point given yields the equation $y-2 = \frac{-3}{5}(x+3)$ for $\ell_1$. Finding the perpendicular slope by reciprocating and negating the slope of
$\ell_1$, and then using point-slope form in conjunction with $(-1,-3)$ yields $y+3 = \frac{5}{3}(x+1)$ as an equation for $\ell_2$.
Solve $x+1 = \sqrt{x+7}$
Solve $\displaystyle{\frac{x^2}{x^2-1} - \frac{2}{x+1} = 1}$
$x = 3/2$
Solve $x^3 + 3x^2 = x + 3$
$x = -3,-1,1$
Simplify $\displaystyle{\frac{f(x)}{g(x)}}$, where $\displaystyle{f(x) = \frac{x-2}{x} + \frac{x}{x+2} \textrm{ and } g(x) = \frac{x-2}{x+2} + \frac{1}{x}}$
$\displaystyle{\frac{f(x)}{g(x)} = \frac{2x^2-4}{x^2-x+2} \quad \textrm{where } x \ne 0,-2}$
Factor $xy^2 + y^2 - 4x -4$
Define what it means for a function to be "even" or "odd." Provide an example of each.
A function $f$ is even when $f(-x) = f(x)$ for all $x$ in its domain. Visually, one can fold even functions along the $y$-axis and the two halves formed will line up perfectly. Examples include $x^2
+ 3, \cos x, |x|$, and many others.
A function $g$ is odd when $g(-x) = -g(x)$ for all $x$ in its domain. Visually, odd functions are rotationally symmetric about the origin. That is to say, one can rotate the right half of the
function's graph 180 degrees about the origin and it will line up perfectly with the left half of the graph. Examples of odd functions include $x^3 -x, \sin x, 1/x$, and many others.
Notably, functions need not be one or the other (i.e., either even or odd). For example, $x^2 + x$, $\ln x$, and $e^x$ are each neither even nor odd. Alternatively, one function $h(x) = 0$ is both
even and odd.
A small rectangular pen containing $10$ square meters is to be fenced in. The front, to be made of stone, will cost $\$7$ per meter, while each of the other three wooden sides will cost only $\$3$
per meter. Supposing the front of the pen is $x$ meters long and the sides are each $y$ meters long, find the cost of the fence in terms of these two dimensions. Can you find the cost in terms of
only the length of the front of the pen? What happens to the cost when $x$ increases?
$C = 10x + 6y$. As $xy = 10$, we can solve to find $y = 10/x$ which allows us to express the cost as a function of only one variable: $C(x) = 10x + 60/x$ where $x \gt 0$. For very small lengths $x$,
the cost is large as the side lengths must be exceedingly long to accomplish the desired area of $10$ square meters. However, as $x$ increases, the first term of $C(x)$ increases, while the second
diminishes -- until the cost is driven almost solely by the first term (the second gets very close to zero for large $x$). Again, this is expected as the side lengths need not be long (and are
consequently cheap) when the front length is large enough.
A box with a square base and open top must have a volume of $32,000\, \textrm{cm}^3$. Find an expression for the amount of material required to make the box in terms of its dimensions. Can you
express this amount of material in terms of only the base length $x$ (in cm)?
Supposing $S$ is the amount of material, $x$ is the base length (in cm), and $y$ is the height of the box (in cm), we find $S = x^2 + 4xy$. Taking advantage of the volume being $x^2 y = 32000$ we
find $y = 32000/x^2$ which allows us to express $S$ as function of one variable: $$S(x) = x^2 + \frac{128000}{x} \quad \textrm{ where } \quad x \gt 0$$
A box with a square base and an open top is to be made using $1200\, \textrm{cm}^2$ of material. Let $x$ be the length (in cm) of a side of the base. Find the volume of the box as a function of $x$,
noting any restrictions on the values of $x$ forced upon one given that the volume of an actual box must always be positive.
$V(x) = (x/4)(1200-x^2)$ where $0 \lt x \lt \sqrt{1200}$
A right circular cylinder is inscribed in a sphere of radius $r$. Find the volume of the cylinder as a function of the height of the cylinder.
Let $h$ be the height of the cylinder, noting that $0 \lt h \lt 2r$ (as otherwise the cylinder won't fit inside the sphere). Then, the volume of the cylinder is given by $V(h) = \pi h (4r^2 - h^2) /
A Norman window has the shape of a rectangle surmounted by a semicircle. (Thus, the diameter of the semicircle is equal to the width of the rectangle.) If the perimeter of the window is $30$ ft,
express the area of the window (a measure of how much light it admits) in terms of the radius of the semicircular part of the window.
$\displaystyle{A(r) = 2r [ 15 - \frac{r}{2} (\pi + 2) ] + \frac{\pi r^2}{2}}$ | {"url":"https://mathcenter.oxford.emory.edu/site/math111/probSetPrecalReview/","timestamp":"2024-11-10T18:13:57Z","content_type":"text/html","content_length":"14516","record_id":"<urn:uuid:7a368541-9f9d-4a1d-822a-b49a6118010f>","cc-path":"CC-MAIN-2024-46/segments/1730477028187.61/warc/CC-MAIN-20241110170046-20241110200046-00631.warc.gz"} |
Properties of Neural Networks Studied from ND Geometry and Discrete Algebra
C.-L.J. Hu (USA)
Feedback and Feed-forward neural networks, N dimension geometry.
For a one-layered neural network, NN, containing discrete sign-function neurons, (either a feedback NN, FBNN, or a feed-forward NN, FFNN,) the nonlinear properties of this network can be studied very
efficiently using simple discrete mathematics. This paper summarizes the discrete-formulation of the control matrix equation of the problem. Then two methods for solving this matrix equation are
used. One is for the FFNN, in which we reduced the control equation to a set of linear strict inequalities. The other one is for FBNN, in which we use simple finite iterative method of solving this
nonlinear matrix equation. The derivation of the major anomalous properties of both systems are then obtained. These anomalous properties include optimum robustness, noise-controlled digital recalls,
domain of attractions in the state space, eigen-state storage, associative storage, content-addressable recall, fault tolerant recall, capacity of storage, binary oscillating states, limit-cycles in
the state space, etc.. The physical origin and the systematic trend of the derivation of these properties are easily seen from the state-space geometrical analysis.
Important Links: | {"url":"https://actapress.com/Abstract.aspx?paperId=13848","timestamp":"2024-11-12T19:58:31Z","content_type":"application/xhtml+xml","content_length":"18669","record_id":"<urn:uuid:72733354-89a8-45c4-b984-60e5e6bd6977>","cc-path":"CC-MAIN-2024-46/segments/1730477028279.73/warc/CC-MAIN-20241112180608-20241112210608-00451.warc.gz"} |
Playing with Linear Discriminant Analysis
The way that I select which risk category a drug should be in based on the training data I have is called Linear Discriminant Analysis (LDA). I've made a little example of how it works using MATLAB
(however, this code also works perfectly on the wonderful, and free, Octave).
You can download my MATLAB/Octave script and try it out yourself (it takes one input, a number). Below, I'll go through it step by step.
function LDA(point)
%% group 1: mean 10, sd 2
%% group 2: mean 3, sd 1
%% group 3: mean 5, sd 0.5
group_1 = 10 + 2*randn(10,1);
group_2 = 3 + randn(10,1);
group_3 = 5 + 0.5*randn(10,1);
These first few lines create a set of training data, by making random numbers based on three normal distributions with different means and standard deviations. In real life, you wouldn't know how
your data were generated or what distribution they came from.
%% plot all of the training data
hold off
plot(group_1, zeros(10,1),'o')
hold on
plot(group_2, zeros(10,1),'+')
plot(group_3, zeros(10,1),'.')
Now for visualisation we plot the training data, delineating the different groups by differently shaped markers.
%% find the likelihood of the point given the group
p_x1 = normpdf(point,mean(group_1),std(group_1));
p_x2 = normpdf(point,mean(group_2),std(group_2));
p_x3 = normpdf(point,mean(group_3),std(group_3));
The first thing we need to find is the probability that this point (the input number, in this case 9.5114) would exist given the group that it's a member of, i.e. . In LDA, you assume that the points
within each group are normally distributed, so in MATLAB/Octave there's a useful function called normpdf that calculates the probability of a point given the mean and standard deviation of the
%% find the probability of the group given the point using Bayes' rule
p_i1 = (p_x1*(1/3))/((p_x1+p_x2+p_x3)*1/3);
p_i2 = (p_x2*(1/3))/((p_x1+p_x2+p_x3)*1/3);
p_i3 = (p_x3*(1/3))/((p_x1+p_x2+p_x3)*1/3);
Bayes' rule is that .
is actually the same for all groups, so we can ignore it. Because there are three groups of ten items, we know that is always . This means that we could have simplified these calculations by just
comparing the likelihoods ().
%% which one has the maximum probability?
[a b] = max([p_i1,p_i2,p_i3]);
if b==1
dotshape = 'o';
elseif b==2
dotshape = '+';
dotshape = '.';
%% plot to see if we're right
hold off
We end by picking the group that has the highest probability, i.e. the highest , and then plot that on the graph.
If you have more than one set of data points (i.e. if you are classifying based on two features), you can assume that the covariances in different groups are all equal, which simplifies the
calculations for that system.
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Discretization of continuous-time control systems - 制御工学ブログ
Discretization of Continuous-Time Control Systems
When expressing the characteristics of a control target based on physical laws, it is often represented in the form of differential equations, which are treated within the framework of
continuous-time representation. However, when implementing the controller, although it is for a brief moment, the control input is calculated at discrete steps and applied to the target, making it
necessary to use the discrete-time system representation. Considering this, it is advantageous for designers to be able to switch between continuous-time and discrete-time representations depending
on the situation. While the transfer function for continuous-time systems uses the variable
Various textbooks cover the relationship between continuous-time and discrete-time systems, with the bilinear transformation being one of the most well-known methods.
The Relationship Between Continuous-Time and Discrete-Time Systems
When controlling a system that follows physical phenomena, the control target is described as a continuous-time system. There are numerous control methods for continuous-time systems, including
frequency domain-based design techniques and time-domain-based design methods using state equations. Therefore, in many cases, when a control system is given, an appropriate control law can be found
for it.
However, implementing the obtained control law becomes increasingly difficult as the control law becomes more complex. Particularly, when controlling with analog circuits, there are limitations on
the order of controllers that can be implemented, and the controller's state cannot be freely configured, resulting in outcomes that may not align with theoretical expectations.
In contrast, with advancements in computer technology, it has become common to use computers to calculate the controller's output (the input to the control target). There are two patterns: designing
a controller by treating the control target as a discrete-time system or designing a discrete-time controller that closely resembles a controller designed for a continuous-time system. However, both
approaches have their strengths and weaknesses.
When designing for a continuous-time system, if the control period (sampling interval) is long, the actual control performance may significantly degrade. Conversely, the shorter the sampling time,
the closer the response of the discrete controller matches that of the continuous controller.
When designing for a discrete-time system, reducing the time step can result in the appearance of unstable zeros in the discrete-time system, even if the continuous-time system has no unstable zeros.
Additionally, if the sampling time is too short, the variation in the sampling interval can cause the control system to become unstable.
One of the challenges when designing in discrete time is that changing the control period alters the characteristics of the discrete-time system, often necessitating a redesign.
The system's transfer function's behavior at
Note: A control law refers to the method for determining the input command value based on the acquired information, so whether the control law obtained by the controller can be implemented is a
separate consideration.
Continuous-Time and Discrete-Time Signals
Continuous and discrete signals are defined as follows:
• A continuous-time signal exists at any time and can be defined for all time points.
• A discrete-time signal exists only at specific time points and is undefined at other times.
The operations that link these two are sampling and holding. Sampling refers to observing a continuous-time signal at regular intervals and creating a discrete-time signal. The time interval for this
is called the sampling period, and its reciprocal is called the sampling frequency. On the other hand, holding refers to the operation of converting a discrete-time signal back into a continuous-time
signal. One method for this is zero-order hold. Zero-order hold refers to holding the value at one sampling time until the next sampling time, forming a staircase-like continuous-time signal.
In general, even if a continuous-time signal is sampled to create a discrete-time signal and then held to convert it back into a continuous-time signal, it is not possible to fully recover the
original continuous-time signal. Devices such as sampler circuits and hold circuits are used for sampling and holding. A sampler circuit extracts the value of the continuous-time signal at each
sampling time, while a hold circuit retains the signal value given at the sampling time until the next sampling time.
System Representation Using Pulse Transfer Function
The Pulse Transfer Function is a mathematical tool used to express the relationship between the input and output of a discrete-time system in the frequency domain. It corresponds to the transfer
function of a continuous-time system and plays a critical role in the analysis and design of sampled data and digital control systems.
Definition of Pulse Transfer Function
The pulse transfer function represents the relationship between the input and output of a discrete-time system and is typically defined using the
Derivation of Pulse Transfer Function
The pulse transfer function is derived from the transfer function of a continuous-time system as follows. First, the input and output of the continuous-time system are discretized through sampling.
For the transfer function sampling period
$$G(z) = \mathcal{Z}\left[ \mathcal{L}^{-1}\left[ G(s) \right] \Big|_{s = \frac{1}{T_s} \ln(z)} \right]$$
Here, Laplace transform, Laplace transform, and
Applications of Pulse Transfer Function
The pulse transfer function is widely used in the design and analysis of digital control systems. Typical applications include the stability analysis of discrete-time systems, digital filter design,
and system response analysis. It is particularly useful for understanding the sampling theorem and the placement of poles and zeros in the
Furthermore, it is helpful in understanding the relationship between continuous-time systems and digital control systems. For instance, when converting a continuous-time system into a digital control
system, the continuous-time transfer function is first converted into a pulse transfer function, and this pulse transfer function is then used to design digital filters or controllers.
Calculation Example of Pulse Transfer Function
As a simple example, assume the transfer function of a continuous-time system is given as follows:
$$G(s) = \frac{1}{s+1}$$
The pulse transfer function
First, the inverse Laplace transform is computed, and then the
$$G(z) = \mathcal{Z}\left[ \mathcal{L}^{-1}\left[ \frac{1}{s+1} \right] \Big|_{s = \frac{1}{T_s} \ln(z)} \right] = \frac{1 - e^{-T_s}}{z - e^{-T_s}}$$
From this result, the pulse transfer function that depends on the sampling period
System Representation Using State-Space Representation
Continuous-Time Systems
The state-space representation of a continuous-time system is given by the following equations:
\begin{eqnarray} \dot x(t) &=& Ax(t)+Bu(t) \\y(t)&=&Cx(t)+Du(t) \end{eqnarray}
Here, continuous-time systems, and the goal of control is typically to adjust the input
Discrete-Time Systems
The commonly used state-space representation of discrete-time systems is expressed as follows:
\begin{eqnarray}\label{siki12} x[k+1]&=&A_s x[k]+B_s u[k]\\ y[k]&=&C_s x[k]+D_s u[k]\nonumber \end{eqnarray}
Here, shift form because it describes the state transition from one time step to the next based on the state and input at the current time.
Another suitable representation for capturing the relationship between continuous-time and discrete-time systems is the delta form, expressed as:
\begin{eqnarray}\label{siki13} \frac{x[k+1]-x[k]}{\Delta}&=&A_d x[k]+B_d u[k]\\y[k]&=&C_d x[k]+D_d u[k]\nonumber \end{eqnarray}
Thus, the delta form is said to be an appropriate representation to describe the relationship between continuous-time and discrete-time systems. In this paper, we consider single-input, single-output
systems for simplicity.
Relationship Between Coefficients of Discrete and Continuous-Time Systems
Next, we examine the relationship between the coefficient matrices. For a linear time-invariant continuous-time system, the solution trajectory is given by the following expression:
$$x(t)=e^{At}x(0)+\int_0^t e^{ A(t-τ) } B u(τ)d τ$$
Here, value
$$x(\Delta)=e^{A\Delta}x(0)+\int_0^\Delta e^{A(\Delta-τ)}B dτ U$$
$$x((k+1)\Delta)=e^{A\Delta}x(k\Delta)+\int_0^\Delta e^{A(\Delta-τ)}B dτ U$$
Thus, by defining discrete-time system. The relationship between the coefficient matrices of the continuous-time system
$$A_s=e^{A\Delta},B_s=\int^\Delta_0 e^{A t}dt B,C_s=C,D_s=D$$
Meanwhile, the shift and delta forms are simply different representations of the same discrete-time system. The coefficients in the delta form are given by:
$$A_d = \frac{A_s-I}{\Delta},B_d=\frac{B_s}{\Delta},C_d=C,D_d=D$$
As mentioned earlier, in the limit as
$$\lim_{\Delta \rightarrow 0}\frac{x[k+1]-x[k]}{\Delta}=\dot x(t)\left|_{t=k\Delta}\right.$$
This shows that in the limit small sampling periods.
On the other hand, in the limit the matrix | {"url":"https://blog.control-theory.com/entry/2024/10/17/122018","timestamp":"2024-11-10T19:10:17Z","content_type":"text/html","content_length":"69514","record_id":"<urn:uuid:145d2b96-7f73-4eb8-928d-568d32a8fd14>","cc-path":"CC-MAIN-2024-46/segments/1730477028187.61/warc/CC-MAIN-20241110170046-20241110200046-00814.warc.gz"} |
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Cryptographic Hashes
7. Cryptographic Hashes
7.1. Overview
A cryptographic hash function is a function, \(H\), that when applied on a message, \(M\), can be used to generate a fixed-length “fingerprint” of the message. As such, any change to the message, no
matter how small, will change many of the bits of the hash value with there being no detectable patterns as to how the output changes based on specific input changes. In other words, any changes to
the message, \(M\), will change the resulting hash-value in some seemingly random way.
The hash function, \(H\), is deterministic, meaning if you compute \(H(M)\) twice with the same input \(M\), you will always get the same output twice. The hash function is unkeyed, as it only takes
in a message \(M\) and no secret key. This means anybody can compute hashes on any message.
Typically, the output of a hash function is a fixed size: for instance, the SHA256 hash algorithm can be used to hash a message of any size, but always produces a 256-bit hash value.
In a secure hash function, the output of the hash function looks like a random string, chosen differently and independently for each message—except that, of course, a hash function is a deterministic
To understand the intuition behind hash functions, let’s take a look at one of its many uses: document comparisons. Suppose Alice and Bob both have a large, 1 GB document and wanted to know whether
the files were the same. While they could go over each word in the document and manually compare the two, hashes provide a quick and easy alternative. Alice and Bob could each compute a hash over the
document and securely communicate the hash values to one another. Then, since hash functions are deterministic, if the hashes are the same, then the files must be the same since they have the same
“fingerprint”. On the other hand, if the hashes are different, it must be the case that the files are different. Determinism in hash functions ensures that providing the same input twice (i.e.
providing the same document) will result in the same hash value; however, providing different inputs (i.e. providing two different documents) will result in two different hash values.
7.2. Properties of Hash Functions
Cryptographic hash functions have several useful properties. The most significant include the following:
• One-way: The hash function can be computed efficiently: Given \(x\), it is easy to compute \(H(x)\). However, given a hash output \(y\), it is infeasible to find any input \(x\) such that \(H(x)=
y\). (This property is also known as “preimage resistant.”) Intuitively, the one-way property claims that given an output of a hash function, it is infeasible for an adversary to find any input
that hashes to the given output.
• Second preimage resistant: Given an input \(x\), it is infeasible to find another input \(x'\) such that \(x' \ne x\) but \(H(x)=H(x')\). This property is closely related to preimage resistance;
the difference is that here the adversary also knows a starting point, \(x\), and wishes to tweak it to \(x'\) in order to produce the same hash—but cannot. Intuitively, the second preimage
resistant property claims that given an input, it is infeasible for an adversary to find another input that has the same hash value as the original input.
• Collision resistant: It is infeasible to find any pair of messages \(x,x'\) such that \(x' \ne x\) but \(H(x)=H(x')\). Again, this property is closely related to the previous ones. Here, the
difference is that the adversary can freely choose their starting point, \(x\), potentially designing it specially to enable finding the associated \(x'\)—but again cannot. Intuitively, the
collision resistance property claims that it is infeasible for an adversary to find any two inputs that both hash to the same value. While it is impossible to design a hash function that has
absolutely no collisions since there are more inputs than outputs (remember the pigeonhole principle), it is possible to design a hash function that makes finding collisions infeasible for an
By “infeasible”, we mean that there is no known way to accomplish it with any realistic amount of computing power.
Note, the third property implies the second property. Cryptographers tend to keep them separate because a given hash function’s resistance towards the one might differ from its resistance towards the
other (where resistance means the amount of computing power needed to achieve a given chance of success).
Under certain threat models, hash functions can be used to verify message integrity. For instance, suppose Alice downloads a copy of the installation disk for the latest version of the Ubuntu
distribution, but before she installs it onto her computer, she would like to verify that she has a valid copy of the Ubuntu software and not something that was modified in transit by an attacker.
One approach is for the Ubuntu developers to compute the SHA256 hash of the intended contents of the installation disk, and distribute this 256-bit hash value over many channels (e.g., print it in
the newspaper, include it on their business cards, etc.). Then Alice could compute the SHA256 hash of the contents of the disk image she has downloaded, and compare it to the hash publicized by
Ubuntu developers. If they match, then it would be reasonable for Alice to conclude that she received a good copy of the legitimate Ubuntu software. Because the hash is collision-resistant, an
attacker could not change the Ubuntu software while keeping the hash the same. Of course, this procedure only works if Alice has a good reason to believe that she has the correct hash value, and it
hasn’t been tampered with by an adversary. If we change our threat model to allow the adversary to tamper with the hash, then this approach no longer works. The adversary could simply change the
software, hash the changed software, and present the changed hash to Alice.
7.3. Hash Algorithms
Cryptographic hashes have evolved over time. One of the earliest hash functions, MD5 (Message Digest 5) was broken years ago. The slightly more recent SHA1 (Secure Hash Algorithm) was broken in 2017,
although by then it was already suspected to be insecure. Systems which rely on MD5 or SHA1 actually resisting attackers are thus considered insecure. Outdated hashes have also proven problematic in
non-cryptographic systems. The git version control program, for example, identifies identical files by checking if the files produce the same SHA1 hash. This worked just fine until someone discovered
how to produce SHA1 collisions.
Today, there are two primary “families” of hash algorithms in common use that are believed to be secure: SHA2 and SHA3. Within each family, there are differing output lengths. SHA-256, SHA-384, and
SHA-512 are all instances of the SHA2 family with outputs of 256b, 384b, and 512b respectively, while SHA3-256, SHA3-384, and SHA3-512 are the SHA3 family members.
For the purposes of the class, we don’t care which of SHA2 or SHA3 we use, although they are in practice very different functions. The only significant difference is that SHA2 is vulnerable to a
length extension attack. Given \(H(M)\) and the length of the message, but not \(M\) itself, there are circumstances where an attacker can compute \(H(M \Vert M')\) for an arbitrary \(M'\) of the
attacker’s choosing. This is because SHA2’s output reflects all of its internal state, while SHA3 includes internal state that is never outputted but only used in the calculation of subsequent
hashes. While this does not violate any of the aforementioned properties of hash functions, it is undesirable in some circumstances.
In general, we prefer using a hash function that is related to the length of any associated symmetric key algorithm. By relating the hash function’s output length with the symmetric encryption
algorithm’s key length, we can ensure that it is equally difficult for an attacker to break either scheme. For example, if we are using AES-128, we should use SHA-256 or SHA3-256. Assuming both
functions are secure, it takes \(2^{128}\) operations to brute-force a 128 bit key and \(2^{128}\) operations to generate a hash collision on a 256 bit hash function. For longer key lengths, however,
we may relax the hash difficulty. For example, with 256b AES, the NSA uses SHA-384, not SHA-512, because, let’s face it, \(2^{192}\) operations is already a hugely impractical amount of computation.
7.4. Lowest-hash scheme
Cryptographic hashes have many practical applications outside of cryptography. Here’s one example that illustrates many useful properties of cryptographic hashes.
Suppose you are a journalist, and a hacker contacts you claiming to have stolen 150 million (150 million) records from a website. The hacker is keeping the records for ransom, so they don’t want to
present all 150 million files to you. However, they still wish to prove to you that they have actually stolen 150 million different records, and they didn’t steal significantly fewer records and
exaggerate the number. How can you be sure that the hacker isn’t lying, without seeing all 150 million records?
Remember that the outputs of cryptographic hashes look effectively random–two different inputs, even if they only differ in one bit, give two unpredictably different outputs. Can we use these
random-looking outputs to our advantage?
Consider a box with 100 balls, numbered from 1 to 100. You draw a ball at random, observe the value, and put it back. You repeat this \(n\) times, then report the lowest number you saw in the \(n\)
draws. If you drew 10 balls (\(n\)=10), you would expect the lowest number to be approximately 10. If you drew 100 balls (\(n\)=100), you might expect the lowest number to be somewhere in the range
1-5. If you drew 150 million balls (\(n\)=150,000,000), you would be pretty sure that the lowest number was 1. Someone who claims to have drawn 150 million balls and seen a lowest number of 50 has
either witnessed an astronomically unlikely event, or is lying about their claim.
We can apply this same idea to hashes. If the hacker hashes all 150 million records, they are effectively generating 150 million unpredictable fixed-length bitstrings, just like drawing balls from
the box 150 million times. With some probability calculations (out of scope for this class), we can determine the expected range of the lowest hash values, as well as what values would be
astronomically unlikely to be the lowest of 150 million random hashes.
With this idea in mind, we might ask the hacker to hash all 150 million records with a cryptographic hash and return the 10 lowest resulting hashes. We can then check if those hashes are consistent
with what we would expect the lowest 10 samples out of 150 million random bitstrings to be. If the hacker only hashed 15 million records and returned the lowest 10 hashes, we should notice that the
probability of getting those 10 lowest hashes from 150 million records is astronomically low and conclude that the hacker is lying about their claim.
What if the hacker tries to cheat? If the hacker only has 15 million records, they might try to generate 150 million fake records, hash the fake records, and return the lowest 10 hashes to us. We can
make this attack much harder for the attacker by requiring that the attacker also send the 10 records corresponding to the lowest hashes. The hacker won’t know which of these 150 million fake records
results in the lowest hash, so to guarantee that they can fool the reporter, all 150 million fake records would need to look valid to the reporter. Depending on the setting, this can be very hard or
impossible: for example, if we expect the records to be in a consistent format, e.g. lastname, firstname, then the attacker would need to generate 150 million fake records that follow the same
Still, the hacker might decide to spend some time precomputing fake records with low hashes before making a claim. This is called an offline attack, since the attacker is generating records offline
before interacting with the reporter. We will see more offline attacks when we discuss password hashing later in the notes. We can prevent the offline attack by having the reporter choose a random
word at the start of the interaction, like “fubar,” and send it to the hacker. Now, instead of hashing each record, the hacker will hash each record, concatenated with the random word. The reporter
will give the attacker just enough time to compute 150 million hashes (but no more) before requesting the 10 lowest values. Now, a cheating hacker cannot compute values ahead of time, because they
won’t know what the random word is.
A slight variation on this method is to hash each record 10 separate times, each with a different reporter-chosen random word concatenated to the end (e.g. “fubar-1,” “fubar-2,” “fubar-3,” etc.). In
total, the hacker is now hashing 1.5b (150 million times 10) records. Then, instead of returning the lowest 10 hashes overall, the hacker returns the record with the lowest hash for each random word.
Another way to think of this variation is: the hacker hashes all 150 million records with the first random word concatenated to each record, and returns the record with the lowest hash. Then the
hacker hashes all 150 million records again with the second random word concatenated to each record, and returns the record with the lowest hash. This process repeats 10 times until the hacker has
presented 10 hashes. The math for using the hash values to estimate the total number of lines is slightly different in this variation (the original uses random selection without substitution, while
the variant uses random selection with substitution), but the underlying idea is the same. | {"url":"https://textbook.cs161.org/crypto/hashes.html","timestamp":"2024-11-05T19:54:27Z","content_type":"text/html","content_length":"33013","record_id":"<urn:uuid:94af995e-b827-4188-9538-ee33a3f62476>","cc-path":"CC-MAIN-2024-46/segments/1730477027889.1/warc/CC-MAIN-20241105180955-20241105210955-00344.warc.gz"} |
Worksheets by Subject | A Wellspring of Worksheets
Kids love to practice their time telling skills. This set of draw the time to the half hour worksheets adds the extra challenge of drawing the time to the half hour, not just the hour. These can be a
little tricky as the hour hand often gets drawn in the wrong place. Being extra careful and having lots of practice will help your kids master this skill with flying colors. You can have kids check
their answers against the answer sheets.
Please go to his page to choose from even more 1st grade telling time worksheets.
Draw the time to the half hour by being very careful to draw the hour hand in the right place, just a bit after the hour number. Answer sheet included.CCSS 1.MD.B.3
Draw the time to the half hour on each round clock. There are only nine clocks to the half hour. Which three are missing? Can you figure it out?CCSS 1.MD.B.3
Here are nine round clocks for your kids to practice drawing the time to the half hour. Make sure they remember to put the long hand on the 6!CCSS 1.MD.B.3
How can you tell what time it is? By looking at the clock, silly. So take a good look at these clocks and draw the times on all the clocks to the half hour.CCSS 1.MD.B.3
For every one of the times kids will draw to the half hour on these clocks the minute hand goes in the same place. Can your kids tell you where that is?CCSS 1.MD.B.3
What would these alarm clocks be used for? What would kids do at half past the hour? Have them choose one of the times and say what they might do at that time?CCSS 1.MD.B.3
These clocks are a bit small so you will need to draw the time to the half hour very carefully so the hands are in the right place. Good luck! Answer sheet included.CCSS 1.MD.B.3
This worksheet has 16 clocks for kids to draw the time to the half hour. Have them draw lines as straight as can be - and don't forget the arrows on the end.CCSS 1.MD.B.3
© 2024 A Wellspring of Worksheets. All Rights Reserved | Web Design by Appnet.com | {"url":"https://www.awellspringofworksheets.com/worksheets/subject/draw-the-time-to-the-half-hour/","timestamp":"2024-11-03T10:33:24Z","content_type":"text/html","content_length":"46334","record_id":"<urn:uuid:7708a7d4-38d9-4092-9ceb-f595a44e764e>","cc-path":"CC-MAIN-2024-46/segments/1730477027774.6/warc/CC-MAIN-20241103083929-20241103113929-00758.warc.gz"} |
Autoparts Example
Autoparts Example
The company produces stamped metal parts, to be used in automotive assembly, from sheetmetal coils. The plant consists of five major departments. Some of the parts have to be treated with corrosion
resistive paint and then dried in an oven. Forklift trucks are used for all material handling transportation operations in the plant. A summary of the department data is given in Table 4.5. The
number of truck trips per week between each pair of departments and between each department and the outside is given in Table 4.6. The building dimensions are 200 by 120 feet.
Table 4.5. Department Data for the Autoparts Example
Table 4.6. Material Handling Trips for the Autoparts Example
We will first create a hexagonal adjacency graph and block layout following the SPIRAL methodology. For the graph construction phase, we will use the binary relationships, not use improvement
interchanges, and break location ties by the centroid-seeking rule. For the graph, we will show first the symmetrical adjusted relationships in a two-dimensional matrix. We then will show the sorted
list of adjusted binary relationships. Third, we will show the adjacency graphs after we have added each department. Finally, we will compute and show the adjacency and efficiency score of the final
We first convert the asymmetrix material handling trips to a symmetric relationship table, where all relationships are shown above the main diagonal.
Table 4.7. Symmetric Relationships for theAutoparts Example
Next the adjusted binary relationships, that incorporate the relationships with the outside, are computed. Because several departments have positive relationships with the outside, many of the
adjusted relationships are negative.
Table 4.8. Binary Adjusted Relationships for the Autoparts Example
The adjusted binary relationships are then sorted in decreasing order.
Table 4.9. Sorted Binary Adjusted Relationships for the Autoparts Example
The first two departments to enter the hexagonal adjacency graph are the two departments in the first element of the list, i.e. STA and PAI. They are placed at random in two adjacent vertices of the
hexagonal grid.
Figure 4.15. Adjacency Graph with 2 Departments for the Parts Example
The next relationship selected must have one department in the graph and one department not in the graph. These departments are called the anchor department and the new department, respectively. The
anchor department also must have an open adjacent grid position. Element two of the list satisfies all those conditions: department STA is in the graph, department STO is not, and department STA has
an empty adjacent grid position. All empty grid positions adjacent to the anchor department are possible locations for the new department, and because of the relationship selection rule there exists
at least one open adjacent grid position. The grid position with the highest adjacency score based on the original symmetric relationships is selected. Ties are broken by the centroid-seeking rule,
i.e. the new department is located in the open grid location adjacent to the anchor department that tied for the highest adjacency score and that is closest to the centroid of the current adjacency
graph. Department STO has only a relationship with department STA, and thus all open grid positions adjacent to department STA are tied. There are three grid position closest to the centroid of the
graph and the one above the centroid is selected at random.
Figure 4.16. Adjacency Graph with 3 Departments for the Autoparts Example
Element four is the next relationship selected from the list. Department STO is in the graph, department REC is not, and department STO has an open adjacent grid position. Deparment REC has a
positive relationship with departments STA and STO and so the open grid position adjacent to both those department is the selected location for department REC.
Figure 4.17. Adjacency Graph with 4 Departments for the Parts Example
Department SHI is the only remaining free department. Element seven is the next selected relationship, since department STA is in the graph, deparment SHI is not in the graph, and department STA has
an open adjacent grid location. All open grid positions adjacent to department STA are candidate locations. Since department SHI has positive relationships only with departments STA and PAI, it will
be located in the open grid location adjacent to those two departments.
Figure 4.18. Adjacency Graph with 5 Departments for the Parts Example
To compute the adjacency score, we construct the adjacency matrix for the graph above.
Table 4.10. Adjacency Matrix for the Autoparts Example
The adjacency matrix is then multiplied element by element with the symmetrical relationship matrix and the sum of the product is computed. The sum is equal to 795. Since the sum of the absolute
values of all relationships is also equal to 795, the efficiency of the above graph is equal to 100 %.
For the block layout phase, we will use the graph axis with the most departments in the graph as the building orientation, not use improvement interchanges, and use the layered space allocation
method. For the layout phase, we will first show the layout drawn to scale with all departments properly dimensioned. Next, we will show all distances in a two-dimensional matrix of which we will
only fill in the upper half. Third, we will compute and show the inter-department distance score and the distance score with the outside. Fourth, we will compute and show the shape penalty score for
each individual department and the total shape penalty. Finally, we will compute and show the shape adjusted distance score.
The graph axis with the most departments is oriented downwards. The conceptual block layout with the layered space allocation method is shown in the next figure.
Figure 4.19. Block Layout for the Given Graph for the Autoparts Example
Table 4.11. Distance Matrix for the Autoparts Example
The distance matrix is then multiplied element by element with the symmetrical relationship matrix and the sum of the products is computed. This sum is the flow distance score and is equal to 49,525,
of which 41,400 is between two departments and 8,125 is between departments and the outside.
Table 4.12. Penalty Shape Computations for the Autoparts Example
The total shape penalty is computed as the sum of the individual department shape penalties and is equal to 91,063. The total shape distance score for the layout is computed as the sum of the flow
distance score and the shape penalty and is equal to 140,588.
The block layout without exchange improvements, the best block layout with the layered space allocation method, and the best block layout with the tiled space allocation method are shown in the next
three figures. The distance statistics for the three layouts are shown in the next table.
Figure 4.20. Unimproved Block Layout for the Given Graph for the Parts Example
Figure 4.21. Best Layered Block Layout for the Given Graph for the Parts Example
Figure 4.22. Best Tiled Block Layout for the Given Graph for the Parts Example
Table 4.13. Distance Statistics for the Parts Example
While the computer can provide evaluations of the different layouts based on centroid-to-centroid distances, it remains up to the user to decide which layout is best suited for their particular
layout project. | {"url":"https://www2.isye.gatech.edu/~mgoetsch/cali/Spiral/Spiral%20HTML%20Help/AutopartsExample.htm","timestamp":"2024-11-03T02:57:33Z","content_type":"text/html","content_length":"9294","record_id":"<urn:uuid:6b809963-2ccc-47a6-b019-fb71fb858a74>","cc-path":"CC-MAIN-2024-46/segments/1730477027770.74/warc/CC-MAIN-20241103022018-20241103052018-00415.warc.gz"} |
The Quantum Prince in the Kingdom of Finance
Classical computers are the ‘Legends’ invented long ago that serves as the computational prowess of all problems. As time progressed, the problems grew with the legends needing an upgrade. Scientists
and programmers termed computational challenges as P and NP-hard problems that bear some semblance to old myths, partly because they are impossible and partly, they are out of the capacity of the
Continuing the story of legends and myths, let us talk about the Quantum Prince, who is relatively young and is displaying a potential, viz off the charts. Due to the recent technology boom, Quantum
Computing is now exhibiting qualities of working out some of the most intriguing and complex problems. The Monte Carlo Method [1] was devised in the 1960s and is still relevant today. It is probably
the most used method in finance and is deployed where risk computations, predictions, and probabilities are considered to make intelligent decisions. One of the applications of the Monte Carlo method
is Portfolio Optimization.
The Monte Carlo Simulation is a method of plotting different scenarios of Risk and Return versus Time on a graph. Figure 1 is a Monte Carlo simulation for an asset price, i.e. running 1 million
distinct alternatives to compute probabilities forecasting price and strength of the asset over a time interval (in days)? These coloured lines represent different portfolio scenarios (stock prices
owing to the risk and return). Zero value denotes the initial price at which the stock is purchased.
So, suppose we select a scenario of the return we want (indicated with the black line in Figure 2) versus the risk we are willing to take and keep it for a period. What will be the possible outcomes
using the Monte Carlo method? Figure 2 indicates a scenario of selected simulation results that fit the needs of the investors, whereas the actual scenario shown in Figure 3 is drastically different.
When the same stock is examined in real-time (directly in the stock market), the trajectory followed is indicated with the brown line. Hence, the simulation displayed a rise of about 2000$ while the
actual stock price turned out to rise by 2500$ within 1000 days. This scenario proved in favour of the investor because who does not love profit when it exceeds expectations.
The key points to be considered here are:
• The 1000 days (32 months)
• The loss expected (the lowest value stock descents) was nearly zero, here thus making it safe heaven.
• Loss is also a scenario that we can observe, but the stock whose simulations are inclined more towards positive than negative should be our targets for investment. Is this not amazing we can run
an algorithm and find a solution to our financial problems forever? Probably the reason why the Monte Carlo is one of the legendary methods of this kingdom.
On the contrary, there is an issue concerning the speed of this legendary method. It does not perform very well with an increasing amount of data and with several assets. Let us consider the scenario
around the stock market that has companies listed for around 20-22 years since the digital era began or even earlier than that. The fluctuations in the stock market are known throughout the kingdom.
Hence the process of forecasting and analysis must be done on the data generated every day for 20 years, considering the trends, strengths, etc. Monte Carlo is an efficient algorithm, but with an
increase in stocks and data, it reaches its hardware extremities and takes time ranging from days to even weeks for computing large amounts of data [2].
Let us consider a scenario for a Portfolio Manager: He wants to select the ‘N’ number of stocks in a universe of ‘S’ stocks and ‘X’ number of stocks out of ‘N’ that gives him maximum return. Hence,
he considers buying or holding the ‘X’ stocks while selling away ‘(N-X)’ stocks, considering he is willing to take ‘R’ per cent risk on his investments. This kind of problem is solved by deploying
the Monte Carlo method. The graphical representation is given in Figure 4.
The computational and practical challenges faced by this manager are as follows:
1. The number of combinations can become huge if there is an increase in the number of stocks under ‘X’. The universe ‘S’ is already a large pool of stocks. For example, the Indian Stock market has
more than a million stocks listed. So the selection of ‘N’ stocks becomes very hard and ultimately increases the difficulty of optimal picking of ‘X’ out of ‘N’.
2. The data of every stock worth 20 years would reach storage close to thousands of Gigabytes. Relatively it can be solved easily with expensive hardware or heavy cloud storage. But the current
legends (algorithms) cannot vouch for space and time complexity trade-off.
3. The most crucial thing is that computation will take an exponential amount of time for every increment in the number of stocks, and that is where the whole purpose is lost.
Let us imagine a scenario where our motive is to forecast returns for a portfolio for the next 14 days. If the computation takes two days and the optimized value begins on the third day, our
prediction is already trailing behind two days, i.e., two days of real data is already lost, which we will have to account for on the third day. With the stock market changing every few seconds, the
significance of two days’ worth of real-time data is vast.
There is a saying that, “A King is as intelligent as his most trusted Kingsman as he is who influences the King’s decisions”. In the kingdom of finance, a young prodigy, the Quantum Computer, is
already making ground-breaking advancements in solving all the difficulties mentioned above of the Monte Carlo Method.
In recent research by IBM, the Quantum Amplitude Estimation algorithm (QAE) [2] has proven to provide Quadratic Speed Up [3] over the traditional Monte Carlo method. In short, if you take a T amount
of time to finish a specific task in Monte Carlo, it will take you the only square root of the time, i.e. √T in QAE, hence faster.
This result becomes significant when you solve something that takes two or more days to compute, keeping the accuracy of results comparable or even better than the legendary Monte Carlo. This is the
prowess of the young prince. It is about time for the young prince to ascend to the throne and for the kingdom to praise its crowned prince.
The most significant difference that sets apart a king from a prince is the experience. At present, the prince is dealing with his own set of weaknesses. Hence, the limitations [4] of Quantum
Computers are listed below:
1. Limited Number of Qubits: Presently, there is a limitation in IBM’s quantum hardware of up to 60 qubits. That is very low considering context to terra-bytes of data and millions of parameters.
2. NISQ Phase: Since the current quantum computers are relatively noisy, replicating qubits is a typical error correction [4] method, thereby causing the usage of more qubits for the same task. For
example, a problem with 70 variables will use 70 qubits to calculate in a noise-free QC while it can take 210 to 2100 qubits depending upon the error correction schema deployed for the Noise.
Due to limited access points and more people carrying on computations, the waiting time in the queue to compute can be a factor too (in the range of 5-30 mins, viz low, but an important factor). The
computation must be carried out on cloud-based interfaces as the hardware conditions are complex, increasing the latency of internet speeds in the exhibition of the results.
These limitations are absolute, but in the next 2 to 3 years, we will witness an immense advancement in this field. Major tech giants are pioneering Quantum Computer Algorithms and harnessing an
increased number of qubits in smaller and smaller chipsets. Top finance tycoons are investing and providing data access to build algorithms that can save time and, in most cases, find an optimum
solution. These kinds of actions can have reactions that could broaden the horizons for our quantum prince. We all are witnessing a technological leap in front of our eyes. All hail the young prince!
Co-authored by Aditya Bothra, Senior System Engineer, Infosys QCoE
1. A.I. Adekitan, MONTE CARLO SIMULATION https://www.researchgate.net/publication/326803384_MONTE_CARLO_SIMULATION
2. Pooja Rao, Kwangmin Yu, Hyunkyung Lim, Dasol Jin, Deokkyu Choi, Quantum amplitude estimation algorithms on IBM quantum devices, https://arxiv.org/abs/2008.02102
3. Ashley Montanaro, Quantum speedup of Monte Carlo methods, https://arxiv.org/abs/1504.06987
4. X.Fu, L.Riesebos, L.Lao, C.G.Almudever, F.Sebastiano, R.Versluis, E.Charbon, K.Bertels, The engineering challenges in quantum computing https://www.researchgate.net/publication/
Leave a Comment | {"url":"https://blogs.infosys.com/emerging-technology-solutions/iedps/the-quantum-prince-in-the-kingdom-of-finance.html","timestamp":"2024-11-08T15:37:30Z","content_type":"text/html","content_length":"96410","record_id":"<urn:uuid:1ae3f06a-b932-4ab3-b1bb-9f822878b186>","cc-path":"CC-MAIN-2024-46/segments/1730477028067.32/warc/CC-MAIN-20241108133114-20241108163114-00562.warc.gz"} |
Probabilistic Localization and Mapping
12.3.5 Probabilistic Localization and Mapping
The problems considered so far in Section 12.3 have avoided probabilistic modeling. Suppose here that probabilistic models exist for the state transitions and the observations. Many problems can be
formulated by replacing the nondeterministic models in Section 12.3.1 by probabilistic models. This would lead to probabilistic I-states that represent distributions over a set of possible grids and
a configuration within each grid. If the problem is left in its full generality, the I-space is enormous to the point that is seems hopeless to approach problems in the manner used to far. If
optimality is not required, then in some special cases progress may be possible.
The current problem is to construct a map of the environment while simultaneously localizing the robot with the respect to the map. Recall Figure 1.7 from Section 1.2. The section covers a general
framework that has been popular in mobile robotics in recent years (see the literature suggested at the end of the chapter). The discussion presented here can be considered as a generalization of the
discussion from Section 12.2.3, which was only concerned with the localization portion of the current problem. Now the environment is not even known. The current problem can be interpreted as
localization in a state space defined as
in which is a configuration space and is the environment space. A state, , is represented as ; there is no subscript for because the environment is assumed to be static). The history I-state provides
the data to use in the process of determining the state. As for localization in Section 12.2, there are both passive and active versions of the problem. An incremental version of the active problem
is sometimes called the next-best-view problem [66,238,793]. The difficulty is that the robot has opposing goals of: 1) trying to turn on the sensor at places that will gain as much new data as
possible, and 2) this minimization of redundancy can make it difficult to fuse all of the measurements into a global map. The passive problem will be described here; the methods can be used to
provide information for solving the active problem.
Suppose that the robot is a point that translates and rotates in . According to Section 4.2, this yields , which represents . Let denote a configuration, which yields the position and orientation of
the robot. Assume that configuration transitions are modeled probabilistically, which requires specifying a probability density, . This can be lifted to the state space to obtain by assuming that the
configuration transitions are independent of the environment (assuming no collisions ever occur). This replaces and by and , respectively, in which and for any .
Suppose that observations are obtained from a depth sensor, which ideally would produce measurements like those shown in Figure 11.15b; however, the data are assumed to be noisy. The probabilistic
model discussed in Section 12.2.3 can be used to define . Now imagine that the robot moves to several parts of the environment, such as those shown in Figure 11.15a, and performs a sensor sweep in
each place. If the configuration is not known from which each sweep was performed, how can the data sets be sewn together to build a correct, global map of the environment? This is trivial after
considering the knowledge of the configurations, but without it the problem is like putting together pieces of a jigsaw puzzle. Thus, the important data in each stage form a vector, . If the sensor
observations, , are not tagged with a configuration, , from which they are taken, then the jigsaw problem arises. If information is used to tightly constrain the possibilities for , then it becomes
easier to put the pieces together. This intuition leads to the following approach.
Subsections Steven M LaValle 2020-08-14 | {"url":"https://lavalle.pl/planning/node627.html","timestamp":"2024-11-03T07:09:04Z","content_type":"text/html","content_length":"12404","record_id":"<urn:uuid:e4133bcb-8880-4f44-b4f8-dcc4fcc3c767>","cc-path":"CC-MAIN-2024-46/segments/1730477027772.24/warc/CC-MAIN-20241103053019-20241103083019-00160.warc.gz"} |
Simulating survival data
Dear all,
I want to demonstrate discrete time hazard models with competing risks to my students by simulating survival data with time-dependent covariates. I also want to experiment with different variable
selection approaches.
Most of the literature on this topic seems to be based on continuous time Cox PH assumptions.
Any SAS/IML or DATA STEP code will be much appreciated.
Thank you in advance.
"One page of well written code is more valuable than hundred pages of explanation" Source: Unknown.
10-28-2015 08:31 AM | {"url":"https://communities.sas.com/t5/SAS-Procedures/Simulating-survival-data/td-p/231974","timestamp":"2024-11-08T09:27:02Z","content_type":"text/html","content_length":"204734","record_id":"<urn:uuid:f19b3949-66dc-4136-b741-0f3f85a38445>","cc-path":"CC-MAIN-2024-46/segments/1730477028032.87/warc/CC-MAIN-20241108070606-20241108100606-00871.warc.gz"} |
ndirected Graph
Counting the number of triangles in an undirected graph is a tricky question. It seems easy when we read it. But finding the proper logic and implementing it is hard to crack.
This article will discuss the solution to the problem of counting the number of triangles in an undirected graph. We will use a trace of the graph as adjacency and bitset and adjacency list approach
to finding the problem's solution.
Now let’s see the problem statement of this approach.
Problem Statement
Given an Undirected simple graph, We need to find how many triangles it can have.
Sample Examples
Example 1:
Explanation: The above example has four triangles. They are (0,2,4), (0,1,3), (1,2,3), (0,2,4).
Example 2:
Explanation: The above example has six triangles. They are (0,1,3), (1,2,3), (0,2,4), (0,1,2) (0,3,4), (2,3,4).
Trace of Graph as Adjacency Matrix Approach
We will use a trace of the graph as an adjacency matrix. This will help us find the solution to the problem, count the number of triangles in an undirected graph. The trace of a matrix A is the sum
of the elements on the main diagonal. The sum of a matrix's eigenvalues also equals the matrix's trace.
Let’s see the algorithm of this approach.
1. Let A be the adjacency matrix that represents the graph.
2. If we calculate A^3. The number of triangles in the undirected graph equals to trace(A^3) / 6.
3. Here trace(A) is the sum of the elements on the main diagonal of the matrix A.
4. And the trace of a graph is represented as adjacency matrix A[G][G] is,
5. We will use the following formula to find the trace of matrices:
trace(A[G][G]) = A[0][0] + A[1][1] + .... + A[G-1][G-1]
Number of triangles = trace(A^3) / 6
You can also read Detect cycle in an undirected graph here.
Implementation in C++
#include <bits/stdc++.h>
using namespace std;
#define FAST ios_base::sync_with_stdio(false); cin.tie(NULL);
#define G 4
//----------------Utility function for matrix multiplication----------------//
void multiplymatrices(int A[][G], int B[][G], int C[][G])
for (int i = 0; i < G; i++)
for (int j = 0; j < G; j++)
C[i][j] = 0;
for (int k = 0; k < G; k++)
C[i][j] += A[i][k]*B[k][j];
//----------------Utility function to calculate trace of a matrix----------------//
int Trace(int graph[][G])
int traceofmatrix = 0;
for (int i = 0; i < G; i++)
traceofmatrix += graph[i][i];
return traceofmatrix;
//----------------Utility function for calculating the number of triangles----------------//
int numoftriangle(int graph[][G])
// To Store graph 2
int adj2[G][G];
// To Store graph 3
int adj3[G][G];
for (int i = 0; i < G; ++i)
for (int j = 0; j < G; ++j)
adj2[i][j] = adj3[i][j] = 0;
// adj2 is graph 2, now printMatrix(adj2);
multiplymatrices(graph, graph, adj2);
// after this multiplication adj3 is
multiplymatrices(graph, adj2, adj3);
int traceofmatrix = Trace(adj3);
return traceofmatrix / 6;
//---------------------------Main Function---------------------------//
int main()
int graph[G][G] = {{0, 1, 1, 0},
{1, 0, 1, 1},
{1, 1, 0, 1},
{0, 1, 1, 0}
printf("Total number of Triangle in Graph : %d\n",
return 0;
You can also try this code with Online C++ Compiler
Run Code
Time Complexity
The time complexity of the above-used approach is O(V^3). As here most time-consuming part is the multiplication of the matrix, which contains three nested for loops.
Space Complexity
The time complexity of the above-used approach is O(V^2). It is because we are storing a matrix of size V*V. | {"url":"https://www.naukri.com/code360/library/count-the-number-of-triangles-in-an-undirected-graph","timestamp":"2024-11-08T05:04:39Z","content_type":"text/html","content_length":"406622","record_id":"<urn:uuid:eab43c1b-ffd6-4c48-b348-ee732bb48107>","cc-path":"CC-MAIN-2024-46/segments/1730477028025.14/warc/CC-MAIN-20241108035242-20241108065242-00492.warc.gz"} |