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https://www.shaalaa.com/question-bank-solutions/x2-x-3-the-concept-of-derivative-algebra-derivative-functions_56350
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# X2 + X + 3 - Mathematics
$\frac{1}{\sqrt{3 - x}}$
#### Solution
$\frac{d}{dx}\left( f(x) \right) = \lim_{h \to 0} \frac{f\left( x + h \right) - f\left( x \right)}{h}$
$= \lim_{h \to 0} \frac{\frac{1}{\sqrt{3 - x - h}} - \frac{1}{\sqrt{3 - x}}}{h}$
$= \lim_{h \to 0} \frac{\left( \sqrt{3 - x} - \sqrt{3 - x - h} \right)}{h\sqrt{3 - x}\sqrt{3 - x - h}}$
$= \lim_{h \to 0} \frac{\left( \sqrt{3 - x} - \sqrt{3 - x - h} \right)}{h\sqrt{3 - x}\sqrt{3 - x - h}} \times \frac{\left( \sqrt{3 - x} + \sqrt{3 - x - h} \right)}{\left( \sqrt{3 - x} + \sqrt{3 - x - h} \right)}$
$= \lim_{h \to 0} \frac{\left( 3 - x - 3 + x + h \right)}{h\sqrt{3 - x}\sqrt{3 - x - h}\left( \sqrt{3 - x} + \sqrt{3 - x - h} \right)}$
$= \lim_{h \to 0} \frac{h}{h\sqrt{3 - x}\sqrt{3 - x - h}\left( \sqrt{3 - x} + \sqrt{3 - x - h} \right)}$
$= \lim_{h \to 0} \frac{1}{\sqrt{3 - x}\sqrt{3 - x - h}\left( \sqrt{3 - x} + \sqrt{3 - x - h} \right)}$
$= \frac{1}{\sqrt{3 - x}\sqrt{3 - x - 0}\left( \sqrt{3 - x} + \sqrt{3 - x - 0} \right)}$
$= \frac{1}{\left( 3 - x \right) \left( 2\sqrt{3 - x} \right)}$
$= \frac{1}{2 \left( 3 - x \right)^\frac{3}{2}}$
Concept: The Concept of Derivative - Algebra of Derivative of Functions
Is there an error in this question or solution?
#### APPEARS IN
RD Sharma Class 11 Mathematics Textbook
Chapter 30 Derivatives
Exercise 30.2 | Q 1.09 | Page 25
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2021-07-31 03:18:49
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https://www.cut-the-knot.org/arithmetic/algebra/ConstraintInequalityInFourVariables4.shtml
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# An Inequality with Constraint in Four Variables IV
### Solution 1
$(x+y)^3=x^3+y^3+3xy(x+y).\,$ With $x=a+b\,$ and $y=c+d\,$ we have
\displaystyle \begin{align} &\left(\sum_{cycl}a\right)^3=(a+b)^3+(c+d)^3+3(a+b)(c+d)\sum_{cycl}a&\Leftrightarrow\\ &27=\sum_{cycl}a^3+3ab(a+b)+3cd(c+d)+9(a+b)(c+d)&\Leftrightarrow\\ &27=\sum_{cycl}a^3+3ab(a+b)+3cd(c+d)+9(ac+ad+bc+bd)&\Leftrightarrow\\ &27+3\sum_{cycl}abc=\sum_{cycl}a^3+3ab\sum_{cycl}a+3cd\sum_{cycl}a+9(ac+ad+bc+bd)&\Leftrightarrow\\ &27+3\sum_{cycl}abc=\sum_{cycl}a^3+9\sum_{all}ab. \end{align}
Suffice it to show that $\displaystyle 9\sum_{all}ab\ge 54\sqrt{abcd}.\,$ But this is true by the AM-GM inequality.
### Solution 2
Denote $\displaystyle s_3=\sum_{cycl}abc\,$ and $\displaystyle S_3=\sum_{cycl}a^3\,$ and homogenize using the constraint:
$\displaystyle \left(\sum_{cycl}a\right)^3+3s_3\ge S_3+18\left(\sum_{cycl}a\right)\sqrt{abcd}.$
Now, recollect one of Newton's identities:
$\displaystyle S_3=\left(\sum_{cycl}a\right)\left(\sum_{cycl}a^2\right)-\left(\sum_{cycl}a\right)\left(\sum_{all}ab\right)+3s_3.$
This leadds to an equivalent inequality:
$\displaystyle \small{\left(\sum_{cycl}a\right)^3+\left(\sum_{cycl}a\right)\left(\sum_{all}ab\right)-\left(\sum_{cycl}a\right)\left(\sum_{cycl}a^2\right)\ge 18\left(\sum_{cycl}a\right)\sqrt{abcd}}.$
This is equivalent to
$\displaystyle \left(\sum_{cycl}a\right)\left[3\left(\sum_{all}ab\right)\right]\ge 18\left(\sum_{cycl}a\right)\sqrt{abcd}$
and, finally, to
$\displaystyle \sum_{all}ab\ge 6\sqrt{abcd}$
which is true by the AM-GM inequality.
Note that equality is reached for $\displaystyle a=b=c=d=\frac{3}{4}\,$ or $(3,0,0,0)\,$ and permutations.
### Acknowledgment
This problem from the Romanian Mathematical Magazine has been kindly posted by Dan Sitaru at the CutTheKnotMath facebook page. Solution 1 is by Kevin Soto Palacios; Solution 2 is by Leo Giugiuc.
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2020-10-31 18:04:05
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http://jtmoulia.pocketknife.io/blog/2015/03/11/elixir-and-distel/
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# jtmoulia
### Elixir and Distel
#### Distel: Emacs talking Erlixir
alchemist.el is a Swiss army knife for working with Elixir from the comfort of Emacs. It can run mix tasks, look up Elixir source documentation, autocomplete, ... etc. Alchemist is awesome.
But, here’s the rub: often alchemist has to boot an Erlang VM to evaluate a single expression. Running tests boots a VM, looking up a function’s source doc boots a VM, each prefix autocompletion boots a VM. This is slow, and if you want the VM to load the code you’re actively developing you better not have any compile-time errors.
Plus, with features like hot code reloading, supervision trees, and remote shells, Erlang loves to be manipulated live. Stuart Sierra, a Clojurian, said:
One of the great pleasures of working with a dynamic language is being able to build a system while simultaneously interacting with it.
Enter distel. Distel implements the Erlang distribution protocol in ELisp. More on the nitty gritty of that later, but long story short: distel allows Emacs to pass as an Erlang node; a gnu in Erlang’s clothing (?!). Instead of booting a new Erlang VM to evaluate an expression, the functions can be called on a running node via RPC. Start your application on the remote node with an autoreloader and you have a long-lived development session.
The rest of this post is a brief overview of how to use distel from an Elixir perspective.
##### Connecting Distel
I’ll leave the installation up to you, clever reader.
For programming with distel, the Emacs info pages are surprisingly mediocre. Instead, see the included Gorrie 02.
First, you need a node running distributed Erlang; booting an Erlang VM running an Elixir console will do:
iex --sname emacs [-S mix]
The first time you run any distel command you’ll be prompted for the name of the Erlang node to connect to. If you used the short name emacs as above, the node name is just emacs
##### RPC against an Erlang node
Distel loads an RPC library, rex, into the remote node. On the Emacs side, distel provides erl-send-rpc as a helper to call into rex. The function signature, (erl-send-rpc NODE MODULE FUNCTION ARGS), is reminiscient to Erlang RPC and apply calls and works similarly.
After handling the call, the target node sends a message back to Emacs wrapped in an rex tuple, which can be caught with erl-receive. Look at the function docs for erl-receive -- between pattern matching and variable binding it does quite a bit.
Putting it all together, here’s a function that will call Enum.count/1 with the arg LIST on the remote node:
(defun elixir-enum-count (node list)
"Use NODE to call Enum.count(LIST)'"
(erl-spawn
(erl-send-rpc node 'Elixir.Enum 'count (list expr))
((['rex results] (message "results: %S" results))))))
Notice that the module, Enum is referenced using its absolute name: Elixir.Enum. You must use the absolute name when referencing an Elixir module.
Another catch: the result of erl-receive isn’t returned to the calling context; you must use the continuation passing-style to handle results. i.e. callbacks, ahoy! Here’s the previous example rewritten with the result being passed to the callback function.
(defun proximel-enum-count (node expr &optional callback)
"Use NODE to generate a list of completions for EXPR. Optionally
call CALLBACK with the completions."
(erl-spawn
(erl-send-rpc node 'Elixir.Proximel 'expand (list expr))
((['rex results]
(progn (message "results: %S" results)
(if callback (funcall callback results))))))))
All iterations of proximel-enum-count accept the Erlang NODE as the first argument. The current value of NODE can be fetched with the function erl-target-node. This makes it easy to define interactive functions that accept a node:
(defun elixir-enum-count (node list)
"Use NODE to call Enum.count(LIST)'"
(interactive (list (erl-target-node)
(erl-spawn
(erl-send-rpc node 'Elixir.Enum 'count (list expr))
((['rex results] (message "results: %S" results))))))
##### Demos: Proximel
I put together a basic repo, proximel, which demos loading beam files compiled from Elixir source into a remote node, and basic Elixir autocompletion.
In a future post I’ll dissect Elixir autocompletion over distel with company-mode.
Written by
Tags:
Published:
Modified:
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2022-12-09 00:23:05
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https://ajitjadhav.wordpress.com/2021/03/10/still-if-ish/
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# Still if-ish…
1. Progress has slowed down:
Yep. … Rather, progress has been coming in the sputters.
I had never anticipated that translating my FDM code (for my new approach to QM) into a coherent set of theoretical statements is going to be so demanding or the progress so uneven. But that’s what has actually occurred.
To be able to focus better on the task at hand, I took this blog and my Twitter account off the ‘net from 26th February through 09th March.[* See the footnote below]
Yes, going off the ‘net did help.
Still, gone is that more of less smooth (or “linear”) flow of progress which I experienced in, say, mid-December 2020 through mid-January 2021 times or so, especially in January. Indeed, looking back at the past couple of weeks or so, I can say that a new pattern seems to have emerged. This pattern goes like this:
• On day 1, I get some good idea about how to capture / encapsulate / present something, or put it in a precise mathematical form. So, I get excited. (I even feel like coming back on the ‘net and saying something.)
• But right on day 2, I begin realizing that it doesn’t capture the truth in sufficient generality, i.e., that the insight is only partial. Or, may be, the idea even has loopholes in it, which come to the light only when I do a quick and dirty simulation about it.
• By the time it’s day 2-end, day 3 or at most day 4, I have become discouraged, and even begin thinking of postponing everything to a June-July 2021-based schedule.
• However, soon enough, I get some idea, hurriedly write it down…
• …But only for the whole cycle to repeat once again!
This kind of a cycle has repeated some 3–4 times within the past 15–20 days alone.
“Tiring” isn’t the right word. “Fatigue” is.
But there is no way out. I don’t have any one to even discuss anything (though I am ready, as always, from my side.)
And, it still isn’t mid-March yet. So, I keep going back to the “drawing board.” Somehow.
[* Footnote: Curiously though, both WordPress and RevolverMaps have reported hits to this blog right in this period—even when it was not available for public viewing! … What’s going on?]
2. Current status:
In a way, persistence does seem to have yielded something on the positive side, though it has not been good enough (and, any progress that did come, has been coming haltingly).
In particular, with persistence, I kept on finding certain loop-holes in my thinking (though not in the special cases which I have implemented in code). These are not major conceptual errors. But errors, they still are. Some of these can be traced back to the June-July times last year. Funny enough, as I flip through my thoughts (and at times through my journal pages), some bits of some ideas regarding how I could possibly get out of these loop-holes, seem to have occurred, in some seed form (or half-baked form), right back to those times. …
Anyway, the current status is that I think that I am nearing completing a correct description, for the new approach, for the linear momentum operator.
This is the most important operator, because in QM, you use this operator, together with the position operators, in order to derive the operators for so many other dynamical quantities, e.g. the total energy, the angular momentum, etc. (See Shankar’s treatment, which was reproduced in the postulates document here [^].)
The biggest source of trouble for the linear momentum operator has been in establishing a mathematically precise pathway (and not just a conceptual one) between my approach and the mainstream QM. What I mean to say is this:
I could have simply postulated an equation (which I used in my code), and presented it as simply coming out of the blue, and be done with it. It would work; many people in QM have followed precisely this path. But I didn’t want to do that.
I also wanted to see if I can make the connections between my new approach and the MSQM as easy to grasp as possible (i.e., for an expert of MSQM). Making people understand wasn’t the only motive, however. I also wanted to anticipate as many objections as I could—apart from spotting errors, that is. Another thing: Given my convictions, I also have to make sure that whatever I propose, there has to be a consistent ontological “picture” which goes with it. I don’t theorize with ontology as an after-thought.
But troubles kept coming up right in the first consideration—in clearly spelling out the precise differences of the basic ideas between my approach and the MSQM.
And yes, MSQM does have a way of suddenly throwing up issues that are quite tricky to handle.
Just for this topic of linear momentum, check out, for instance, this thread at the Physics StackExchange [^] (especially, Dr. Luboš Motl’s answer), and this thread [^] (especially, Dr. Arnold Neumaier’s answer). The more advanced parts of both these threads are, frankly, beyond my capacity. Currently, I only aim for that level of rigour which is at, say, exactly and precisely the first three sentences from Motl’s answer!…
…We the engineers can happily ignore any unpleasant effects that might occur at the singular and boundary points. We simply try and see if we can get away ejecting such isolated domain points from any theoretical consideration! If something workable can still be obtained even after removing such points out of consideration, we go for it. So, that’s the first thing we check. Usually, it turns out we can isolate them out, and so we proceed to do precisely that! And that is precisely the level at which I am operating…
Even then, issues are tricky. And, at least IMO, a good part of the blame must lie with the confusions wrought by the Instrumentalist’s dogma.
… What the hell, if $\Psi(x,t)$ isn’t an observable itself, then why does it find a place in their theory (even if only indirectly, as in Heisenberg’s formulation)? … Why can’t I just talk of a property that exists at each infinitesimal CV (control volume) $\text{d}x$? why must I instead take something of interest, then throw in the middle an operator (say a suitable Dirac’s delta), and then bury it all behind an integral sign? why can’t those guys (I mean the mathematical terms) break the cage of the integral sign, and come out in the open, just to feel some neat fresh air?
… Little wonder these MSQM folks live with an in-principle oscillatory universe. It’s a weird universe they have.
In their universe, Schrodinger’s cat is initially in a superposition of being alive and dead. But that’s not actually the most surprising part. Schrodinger’s cat then momentarily (or for a long but finite time) becomes full dead; but then, immediately, it “returns” from that state (of being actually dead) to once again be in a superposition of dead + alive; it spends some time in that superposition; it then momentarily (or for a long but finite time) becomes fully alive too; but only to return back into that surreal superposition…
And it is this whole cycle which goes on repeating ad infinitum.
… No one tells you. But that’s precisely what the framework of MS QM actually predicts.
MSQM doesn’t predict that once a cat does somehow become dead, it remains dead forever. And that’s because, in the MSQM, the only available mathematical machinery (which has any explanation for the quantum phenomena), in principle, predicts only infinite cycles of superposition–life–superposition–death–superposition–….
The postulates of the MS QM necessarily lead to a forever oscillatory universe! Little wonder they can’t solve the measurement problem!
One consequence of such a state of the MS QM theory is that thinking through any aspect becomes that much harder. It isn’t impossible. But hard, yes, it certainly is, where hard means: “tricky”.
Anyway, since the day before yesterday, it has begun looking like this topic (of linear momentum operator), and to the depth I mentioned above, might get over in a few days’ time. At least, that day 1–day 2–etc. pattern seems to have broken—at least for now!
If things go right at least this time round, then I might be able to finish the linear momentum operator by, say, 15th of March. Or 18th. Or 20th.
Addendum especially for Indians reading this post: No, the oscillatory universe of the MSQM people is not your usual birth-life-death-rebirth cycle as mentioned in the ancient Indian literature. The MSQM kind of “oscillations” aren’t about reincarnations of the same soul but in different bodies. In MSQM, the cat “return”s from being dead with exactly the same physical body. So, it’s not a soul temporarily acquiring one body for a brief while, and then discarding it upon its degeneration, only to get another body eventually (due to “karma” or whatever).
So, the main point is: In MSQM, Schrodinger’s cat not just manages to keep the same body, the physical laws mandate that it be exactly the same body (the same material) too! … And, the MS QM doesn’t talk of a soul anyway; it concerns itself purely with the physical aspects—which is a good thing if you ask me. (Just check the postulates document, and pick up a text book to see their typical implications.)
3. Other major tasks to be done (after the linear momentum operator):
• Write down a brief but sufficiently accurate description of the measurement process following my new approach. This is the easiest task among all the remaining ones, because much of such a description can only be qualitative.
• Translate my ideas for the orbital angular momentum into precise mathematical terms—something to be done, but here I guess that with almost all possible troubles having already shown up right in the linear momentum stage, the angular momentum should proceed relatively smoothly (though it too is going take quite some time).
• Study and take notes on the QM spin.
• Think through and integrate my new approach to it.
• Write down as much using quantitative terms as possible.
At this stage, I don’t know how long it’s going to take. However, for now, I’ve decided on the following plan for now…
4. Plan for now:
If there remain some issues with the linear momentum operator (actually, in respect of its multi-faceted usages in the MSQM, and in explaining these from the PoV of my approach including ontology), and if these still remain not satisfactorily resolved even by 15th or 18th of March (roughly, one week from now), then I will take a temporary (but long) break from QM, and instead turn my attention to Data Science.
However, if my description for $\hat{p}()$ (i.e. the linear momentum operator) does go through smoothly during the next week, then I will immediately proceed with the remaining QM-related tasks too (i.e., only those which are listed above).
5. Bottom-line:
Expect a blog post in a week’s time or so, concerning an update with respect to the linear momentum operator and all. (I will try to keep this blog open for the upcoming week, but I guess my Twitter account is best kept closed for now—I just don’t have the time to keep posting updates there.)
In the meanwhile, take care and bye for now.
A song I like:
(Marathi) ती येते आणिक जाते (“tee yete aaNik jaate…”)
Lyrics: Aaratee Prabhu
Music: Pt. Hridaynath Mangeshkar
Singer: Mahendra Kapoor
[ Mahendra Kapoor has sung this song very well (even if he wasn’t a native Marathi speaker). Hridaynath Mangeshkar’s music, as usual, pays real good attention to words, even as also managing to impart an ingenious melodic quality to the tune—something that’s very rare for pop music in any language.
But still, frankly, this song is almost as nothing if you don’t get the lyrics of it.
And, to get the lyrics here, it’s not enough to know Marathi (the language) alone. You also have to “get” what precisely the poet must have meant when he used some word; for instance, the word “ती” (“she”). [Hint: Well, the hint has already been given. …Notice, I said “what”, and not “who”, in the preceding sentence!]
But yes, once you begin to get the subtle shades of the poetry here, then you can also begin to appreciate Hridaynath’s composition even better—you begin to see the more subtle musical phrases, the twists and turns and twirls in the tune which you had missed earlier. So, there’s a kind of a virtuous feedback circle going on here, between poetry and music… And yes, you also appreciate Mahendra Kapoor’s singing better as you go through the circle.
This song originally appeared as a part of a compilation of Aaratee Prabhu’s poems. If I mistake not (speaking purely from memory, and from a distance of several decades), the book in question was जोगवा (“jogawaa”). I had bought a copy of it during my UG days at COEP, out of my pocket-money.
We in fact had used another poem from this book as a part of our dramatics for the Firodiya Karandak. It was included on my insistence; I was a co-author of the script. As to the competition, we did win the first prize, but not so much because of the script. We won mainly because our singing and music team had such a fantastic, outstanding, class to them. Several of them later on went on to make full-time career in music…. The main judge was the late music composer Anand Modak, who later on went to win National awards too, but back then, he was at a fledgling stage of his career. But yes, talking of the script itself, in the informal chat after the prize announcement ceremony, he did mention, unprompted and on his own, that our script was good too! (Yaaaay!!) …Back then, there was no separate prize for the best script, but if there were to be one, then we would’ve probably won it. During that informal chat, the judges hadn’t bothered to even passingly mention any script by any other team!)
…Coming back to the book of poetry (Aaratee Prabhu’s), I think I still have my copy lying somewhere deep in one of the boxes, though by now, due to too many moves and all (I had also taken it to USA the first time I went there), its cover already had got dislodged from the book itself. Then, a couple of weeks ago, I saw only the title page peeping out of some bunch of unrelated and loose papers, and so, looks like, the book by now has reached a more advanced stage of disrepair! … Doesn’t matter; no one else is going to read it anyway!
A good quality audio is here [^].
]
History:
2021.03.10 20:57 IST: Originally published.
2021.03.10 22.45 IST: Added links to the Physics StackExchange threads and the subsequent comments up to the mention of the measurement problem. Other minor editing. Done with this post now!
2021.03.12 18.43 IST: Some further additions, especially in section 2, including the Addendum written for Indian readers. Also, some further additions in the songs section. Some more editing. Now, am really done with this post!
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2021-05-16 05:56:43
|
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https://www.piping-designer.com/index.php/properties/fluid-mechanics/2464-gas-pressure-loss-through-piping
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# Gas Pressure Loss through Piping
Written by Jerry Ratzlaff on . Posted in Fluid Dynamics
## Gas Pressure Loss through Piping formula
$$\large{ p_l = \frac { \mu \; l \; {v_g}^2 \; \rho } { 2 \; d } }$$
Symbol English Metric
$$\large{ p_l }$$ = gas pressure loss $$\large{\frac{lbf}{in^2}}$$ $$\large{Pa}$$
$$\large{ \rho }$$ (Greek symbol rho) = density of gas $$\large{\frac{lbm}{ft^3}}$$ $$\large{\frac{kg}{m^3}}$$
$$\large{ v_g }$$ = velocity of gas $$\large{\frac{ft}{sec}}$$ $$\large{\frac{m}{s}}$$
$$\large{ \mu }$$ (Greek symbol mu) = friction coefficient $$\large{dimensionless}$$
$$\large{ d }$$ = inside diameter of pipe $$\large{in}$$ $$\large{mm}$$
$$\large{ l }$$ = pipe length $$\large{ft}$$ $$\large{m}$$
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2023-02-05 18:30:11
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https://forums.introversion.co.uk/viewtopic.php?p=237162
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## Let's go.........RANDOM!
The place to hang out and talk about totally anything general.
xander
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Let $(X,d)$ and $(\tilde X, \tilde d)$ be metric spaces. We say that a function $f\colon X\to\tilde X$ is \emph{continuous at $a\in X$} if for every $\varepsilon > 0$, there exists a $\delta > 0$ such that if $x\in X$ and $d(x,a) < \delta$, then $\tilde d(f(x),f(a) < \varepsilon$. If $A\subset X$, then we say that $f$ is \emph{continuous on $A$} if $f$ is continuous at $a$ for all $a\in A$. Prove or disprove: this definition of continuity is equivalent to the topological definition, where the topologies on $X$ and $\tilde X$ are assumed to be those induced by the metric.
xander
Captain Jean-Luc Picard
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Prove or disprove: this definition of continuity is equivalent to the topological definition, where the topologies on $X$ and $\tilde X$ are assumed to be those induced by the metric.
You know what to do.
Captain Jean-Luc Picard
bert_the_turtle
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You forgot to tell us what the topological definition of a continuous function is, and what the topology induced by a metric is.
And you made me hate math with the pure legwork proof you made me go through in my head. Congratulations, that is quite an accomplishment! I'll need to read some Lie Algebra stuff now.
Xocrates
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Today in class we proved Morgan's laws... by using Morgan's laws. I did a double take at that.
Granted, the point was to practice a method that relied on Morgan's laws, so what we were proving was irrelevant. Still weird though.
NeatNit
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Xocrates wrote:Today in class we proved Morgan's laws... by using Morgan's laws. I did a double take at that.
Granted, the point was to practice a method that relied on Morgan's laws, so what we were proving was irrelevant. Still weird though.
What? To prove De Morgan's Laws you just need a quick n' dirty truth table. How could you POSSIBLY mess that up?!
Xocrates
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NeatNit
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I have read said print and deliberately chosen to ignore it.
Mas Tnega
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I had to check to see if there were some other laws to be sure I was at the appropriate level of confusion. I was at the appropriate level of confusion.
How do you even do that without just saying "If A, Q. A. Q. Q.E.D."?
NeatNit
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I am guessing they were using Karnaugh maps or something like that. No wait, that wouldn't... Um.
Xocrates
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It was actually an exercise in propositional logic. The Resolution method, specifically.
And like I said, the method (or more precisely, the conversion of the equation into canonical normal form) assumes the laws are true. So the goal of the exercise wasn't to prove the laws to be true, but to learn to apply the method.
But since you're curious:
Hipothesis:
~( A || B) -> ~A && ~B
proof by contradiction, so we'll deny it:
~(~( A || B) -> ~A && ~B)
Convert into CNF (skipping several steps here - which includes applying de morgan's laws)
~A && ~B && (A || B)
or, alternative representation:
{{~A},{~B},{A,B}} -- this essentially means that in order for the hipothesis to be true, all internal groups must be true
Proof:
1. {~A} Premise
2. {~B} Premise
3. {A,B} Premise
4. {B} Resolution, (1,3) -- this essentially means that for ~A to be true, the remaining symbols in {A,B} - in this case B - must be true
5. {} Resolution, (2,4) -- contradiction, since it means both ~B and B must be true.
So the hipothesis must be true.
NeatNit
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Just thought I'd throw this up here, there is a scary accurate comparison in that half-sentence mention: http://garry.tv/2013/03/12/office-progress/
Edit: Uh, does my avatar work? I think I may have messed something up, it doesn't appear on my screen
xander
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bert_the_turtle wrote:You forgot to tell us what the topological definition of a continuous function is, and what the topology induced by a metric is.
And you made me hate math with the pure legwork proof you made me go through in my head. Congratulations, that is quite an accomplishment! I'll need to read some Lie Algebra stuff now.
It was assumed that you knew what those were. :P And yes, it is a pure legwork proof, and a clever undergrad should be able to figure it out. It was a nice break from covering r-balls with rho-balls.
xander
paktsardines
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Location: Australia
It was a nice break from covering r-balls with rho-balls.
I think they can treat that these days.
MeatNit
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We apparently have a new Cyan.
Any ideas?
Feud
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Reproductive rights portion in constitutional law. People often worry about conservative judges being appointed that will over turn abortion. They really should worry more about the fact that the cases that made abortion a federally protected right are poorly written legal messes. If I turned in a paper written like that I'd probably fail the assignment. The spend half the opinion saying what they can't and won't do, and the other half ignoring what they just said and doing exactly that. Woof.
### Who is online
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2020-02-17 14:10:00
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https://probabilitytopics.wordpress.com/
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# Collapsing states into an absorbing state
This post discusses the technique of collapsing a set of states into an absorbing state to order to make certain Markov chain problems easier. We motivate the technique with the following examples. The method discussed here is to raise the new transition probability (the one with the collapsed absorbing state) to an appropriate $n$th power (Chapman-Kolmogorov equations are discussed here). Other applications of Markov chains are discussed here. Markov chains are applied to the occupancy problem in this previous post.
Example 1
On any given day, the weather condition in a city is either sunny, cloudy (but not rainy) or rainy. The weather condition on a given day is identical to that of the previous day with probability 0.5. The weather condition on a given day is equally likely to be one of the other two weather conditions if it is different from the previous day. Suppose that it is sunny today in this city. Find the probability that none of the next five days will be rainy.
Example 2
A fair coin is flipped repeatedly until a run of 4 consecutive heads appear. We want to know how many flips are required to achieve this goal. Specifically we have the following questions.
• Find the probability that it takes at most 9 flips to get a run of 4 consecutive heads.
• Find the probability that it takes at exactly 9 flips to get a run of 4 consecutive heads.
Example 3
A maze is divided into 9 areas as shown below.
Area 1 = initial position of the mouse
A ouse is placed in area 1 initially. Suppose that areas 3, 7 and 9 contain food and that the other areas in the maze do not contain food. Further suppose that the mouse moves through the areas in the maze at random. That is, whenever the mouse is in an area that has $w$ exits, the next move is to one of these $w$ areas with probability $1/w$.
• Find the probability that after the 5 moves, the mouse is one area from a food source.
• Find the probability that after the 7 moves, the mouse has not found food and is in area 6 or area 8 after the 7th move.
• Find the probability that it takes at most 8 moves for the mouse to find food.
• Find the probability that it takes exactly 8 moves for the mouse to find food.
Working Example 1
We work Example 1 to illustrate the method. The following is the transition probability matrix for Example 1 where state 0 = sunny weather, state 1 = cloudy weather and state 2 = rainy weather.
$\mathbf{P} = \bordermatrix{ & 0 & 1 & 2 \cr 0 & 0.50 & 0.25 & 0.25 \cr 1 & 0.25 & 0.50 & 0.25 \cr 2 & 0.25 & 0.25 & 0.50 \cr } \qquad$
Since it is currently a sunny day, $X_0=0$. The probability of not having a rainy day in the next 5 days is the probability $P[X_5 \le 1, X_4 \le 1, X_3 \le 1, X_2 \le 1, X_1 \le 1 \lvert X_0=0]$. This is the probability that $X_j=0$ or $X_j=1$ for the next 5 periods given the current state is 0. Because the range of movement is restricted, using the Chapman-Kolmogorov equations, i.e. finding the 5th power of the matrix $\mathbf{P}$, does not work. For example, though $P_{01}^5$ is the probability that the chain will end up in state 1 at time 5 given that it is in state 0 at time 0, the probability would include the possibilities of a rainy day (state 2) in between time 0 and time 5. The technique is to make the state 2 (the state that is not to be visited in the next 5 days) an absorbing state. Doing so produces a new transition probability matrix.
$\mathbf{Q} = \bordermatrix{ & 0 & 1 & 2 \cr 0 & 0.50 & 0.25 & 0.25 \cr 1 & 0.25 & 0.50 & 0.25 \cr 2 & 0 & 0 & 1 \cr } \qquad$
The matrix $\mathbf{Q}$ is obtained by changing state 2 in the matrix $\mathbf{P}$ an absorbing state (i.e. the entry in the row for state 2 and in the column for state 2 has probability 1). It is a valid transition probability matrix since the sum of each row is 1. So it describes a Markov chain. How does the new Markov chain relate to the original Markov chain? The following is the 5th power of $\mathbf{Q}$.
$\displaystyle \mathbf{Q}^5 = \bordermatrix{ & 0 & 1 & 2 \cr 0 & \frac{122}{1024} &\frac{121}{1024} & \frac{781}{1024} \cr \text{ } & \text{ } & \text{ } & \text{ } \cr 1 & \frac{121}{1024} & \frac{122}{1024} & \frac{781}{1024} \cr \text{ } & \text{ } & \text{ } & \text{ } \cr 2 & 0 & 0 & 1 \cr } \qquad = \bordermatrix{ & 0 & 1 & 2 \cr 0 & 0.11914 & 0.11816 & 0.76270 \cr \text{ } & \text{ } & \text{ } & \text{ } \cr 1 & 0.11816 & 0.11914 & 0.76270 \cr \text{ } & \text{ } & \text{ } & \text{ } \cr 2 & 0 & 0 & 1 \cr } \qquad$
The transition probability $Q_{02}^5=0.7627$ is the probability of transitioning in the new process from state 0 at time 0 to state 2 at time 5 (the 5th day). Starting at state 0, there is a 76.27% chance that the new process will reach state 2 at time 5. This means that the original process will have reached state 2 at time 5 or prior to time 5. The complement of this probability is $Q_{00}^5+Q_{01}^5=0.2373$. So there is a 23.73% chance the original process will never reach state 2 at any time in the first 5 periods. If the weather is indeed governed the chain described in $\mathbf{P}$, it is likely (76.27% chance) that there will be a rainy day at some point within the next 5 periods (given that today is sunny). In fact, even if today is cloudy, there is also a 76.27% chance that there will be a rainy day within the next 5 days.
Discussion
Before working Example 2, let’s discuss the idea of collapsing states into an absorbing state. Consider a Markov chain with transition probability matrix $\mathbf{P}$. Let $\mathcal{S}$ be the state space of this chain. Let $\mathcal{M}$ be a set of states. Suppose that the states in $\mathcal{M}$ are either desirable to enter in or states that are negative situations to be avoided. In case of $\mathcal{M}$ being desirable or attractive, we wish to find the probability of entering into a state in $\mathcal{M}$ after a number of transitions. In the case of $\mathcal{M}$ being negative situations, we wish to find the probability of not in entering into any state in $\mathcal{M}$ after a number of transitions.
The key to finding these probabilities is to collapse the states in $\mathcal{M}$ into a state called $M$ that is an integer that is different from all states $i$ not in $\mathcal{M}$. For convenience, we let $M$ be the smallest integer that is greater than all states not in $\mathcal{M}$. For example, if $\mathcal{S}=\left\{0,1,2,3,4,5 \right\}$, and $\mathcal{M}=\left\{4,5 \right\}$, then $M=4$.
Now define a new Markov chain by letting all states not in $\mathcal{M}$ remain the same and by collapsing all states in $\mathcal{M}$ into the new state $M$. The new state $M$ is made an absorbing state. The state space of the new Markov chain is the original states not in $\mathcal{M}$ along with the new state $M$. Let $\mathbf{Q}$ be the transition probability matrix of the new Markov chain. The transition probabilities in $\mathbf{Q}$ are defined as follows:
$\displaystyle Q_{ij} = \left\{ \begin{array}{ll} \displaystyle P_{ij} &\ \ \ \ \ \ i \notin \mathcal{M}, j \notin \mathcal{M} \\ \text{ } & \text{ } \\ \displaystyle \sum \limits_{k \in \mathcal{M}} P_{ik} &\ \ \ \ \ \ i \notin \mathcal{M}, j=M \\ \text{ } & \text{ } \\ \displaystyle 1 &\ \ \ \ \ \ i=M, j=M \\ \end{array} \right.$
The transition probability matrix $\mathbf{Q}$ is derived from and is closely related to the matrix $\mathbf{P}$. For states $i,j$ not in $\mathcal{M}$, the transition probabilities are the same in both matrices (the first condition). The second condition says that $Q_{iM}$ is the sum of all $P_{ik}$ where $k \notin \mathcal{M}$. Thus $Q_{iM}$ is the probability of the original process entering one of the states in $\mathcal{M}$ when starting from a state $i \notin \mathcal{M}$. Since $M$ is an absorbing state, $Q_{MM}=1$.
To calculate the two probabilities concerning the process for $\mathbf{P}$ indicated earlier, compute powers of the matrix $\mathbf{Q}$. In the new Markov process for the matrix $\mathbf{Q}$, there are two kinds of states, the states $i \notin \mathcal{M}$ and the collapsed state $M$. Whenever the process is in state $i \ne M$, the process is not yet absorbed (into the state $M$). When the process starts at a state not in $M$ (i.e. not yet absorbed), the process will eventually be absorbed (after some number of transitions).
Consider $Q_{iM}^n$ where $i \ne M$. This is the probability that the process starts in state $i$ (not yet absorbed) and is absorbed at time $n$. This is the connection between the two processes: being absorbed into state $M$ in the new process at time $n$ means that the original process enters a state in $\mathcal{M}$ in some period prior to time $n$ or at time $n$. We have the following summary.
Consider a Markov process $\left\{X_n:n \ge 0 \right\}$ with the transition probability matrix $\mathbf{P}$. Collapse $\mathcal{M}$, a set of states, into an absorbing state called $M$ in the manner described above. Let $\mathbf{Q}$ be the resulting transition probability matrix of the new Markov chain. For $i \ne M$, $Q_{iM}^n$ is the probability of absorption given that the process starts out not being absorbed, i.e. the probability that the original process enters a state in $\mathcal{M}$ at time $n$ or prior to time $n$ given that initially the process is in state $i$. Equivalently $Q_{iM}^n=P[X_k \in \mathcal{M} \text{ for some } k=1,2,\cdots,n \lvert X_0=i]$.
The above observation gives a way to interpret the complement $1-Q_{iM}^n$. It would be the probability of the original process never entering any state in $\mathcal{M}$ during the $n$ time periods (given that the process starts at $i \notin \mathcal{M}$). This is the answer to Example 1. We have the following summary.
The complement of $Q_{iM}^n$ is $1-Q_{iM}^n$. It is the probability that the original process never enters any of the states in $\mathcal{M}$ during the $n$ time periods (given that the process starts at $i \notin \mathcal{M}$). Equivalently, $1-Q_{iM}^n$ can be stated as $1-Q_{iM}^n=P[X_k \notin \mathcal{M} \text{ for all } k=1,2,\cdots,n \lvert X_0=i]$. Note that the probability $1-Q_{iM}^n$ is also the sum $\sum_{j \notin \mathcal{M}} Q_{ij}^n$. Specifically, for $j \notin \mathcal{M}$, $Q_{ij}^n$ can be stated as $Q_{ij}^n=P[X_n=j, X_k \notin \mathcal{M} \text{ for all } k=1,2,\cdots,n-1 \lvert X_0=i]$.
The probability $Q_{iM}^n$ is the probability of absorption on or prior to time $n$. Another probability of interest is the probability of absorption exactly at time $n$.
As described above, let $\mathbf{Q}$ be the resulting transition probability matrix of the new Markov chain after collapsing some states into an absorbing state. For $i \ne M$, $Q_{iM}^n-Q_{iM}^{n-1}$ is the probability of absorption at time $n$ given that the process starts out not being absorbed, i.e. the probability that the original process enters a state in $\mathcal{M}$ at time $n$ given that initially the process is in state $i$. Equivalently $Q_{iM}^n-Q_{iM}^{n-1}=P[X_n \in \mathcal{M} \text{ and } X_k \notin \mathcal{M} \text{ for all } k=1,2,\cdots,n-1 \lvert X_0=i]$.
In a nutshell, the idea of collapsing states into an absorbing state is to inform on two probabilities concerning the original process. To illustrate this process, consider the following example.
Example 0
Consider a Markov chain with the following transition probability matrix.
$\mathbf{P} = \bordermatrix{ & 0 & 1 & 2 & 3 & 4 & 5 \cr 0 & 0.6 & 0.08 & 0.08 & 0.08 & 0.08 & 0.08 \cr 1 & 0.5 & 0.10 & 0.10 & 0.10 & 0.10 & 0.10 \cr 2 & 0.4 & 0.12 & 0.12 & 0.12 & 0.12 & 0.12 \cr 3 & 0.3 & 0.14 & 0.14 & 0.14 & 0.14 & 0.14 \cr 4 & 0.2 & 0.16 & 0.16 & 0.16 & 0.16 & 0.16 \cr 5 & 0.1 & 0.18 & 0.18 & 0.18 & 0.18 & 0.18 \cr } \qquad$
Now collapse state 4 and state 5 into a new state called 4. The following is the resulting matrix $\mathbf{Q}$.
$\mathbf{Q} = \bordermatrix{ & 0 & 1 & 2 & 3 & 4 \cr 0 & 0.6 & 0.08 & 0.08 & 0.08 & 0.16 \cr 1 & 0.5 & 0.10 & 0.10 & 0.10 & 0.20 \cr 2 & 0.4 & 0.12 & 0.12 & 0.12 & 0.24 \cr 3 & 0.3 & 0.14 & 0.14 & 0.14 & 0.28 \cr 4 & 0 & 0 & 0 & 0 & 1 \cr } \qquad$
The following is the 6th power of $\mathbf{Q}$.
$\mathbf{Q}^6 = \bordermatrix{ & 0 & 1 & 2 & 3 & 4 \cr 0 & 0.195466 & 0.034574 & 0.034574 & 0.034574 & 0.700812 \cr 1 & 0.184167 & 0.032578 & 0.032578 & 0.032578 & 0.718099 \cr 2 & 0.172869 & 0.030582 & 0.030582 & 0.030582 & 0.735386 \cr 3 & 0.161570 & 0.028586 & 0.028586 & 0.028586 & 0.752673 \cr 4 & 0 & 0 & 0 & 0 & 1 \cr } \qquad$
As a result, $Q_{04}^6=0.700812$ is the probability that the new process will be in state 4 at time 4 (given the process starts at state 0). This means that it is the probability that the original process will reach state 4 or state 5 by time 6 (at or before time 6) given the original process starts in state 0. Note that regardless of the initial state, the probability of being absorbed into state 4 or state 5 by time 6 is roughly the same (70% to 75%).
On the other hand, $Q_{04}^5=0.631665$. Thus $Q_{04}^6-Q_{04}^5=0.700812-0.631665=0.069147$. Thus there is a 6.9% chance that absorption will take place exactly at time 6.
More Examples
We now work Example 2 and Example 3.
Example 2 (Statement of the example given above)
Define a Markov chain with states 0, 1, 2, 3 and 4 with $i \le 3$ meaning that there is currently a run of $i$ consecutive heads (the last $i$ flips are heads) and with state 4 meaning that a run of 4 consecutive heads has already occurred (the last 4 or more flips are heads). Thus state 4 is an absorbing state. The following is the transition probability matrix.
$\mathbf{P} = \bordermatrix{ & 0 & 1 & 2 & 3 & 4 \cr 0 & 0.5 & 0.5 & 0 & 0 & 0 \cr 1 & 0.5 & 0 & 0.5 & 0 & 0 \cr 2 & 0.5 & 0 & 0 & 0.5 & 0 \cr 3 & 0.5 & 0 & 0 & 0 & 0.5 \cr 4 & 0 & 0 & 0 & 0 & 1 \cr } \qquad$
Here’s how $\mathbf{P}$ is derived. For example, if there is currently a run of 2 heads (the third row), the next state is either 0 (the next flip is a tail) or 3 (the next flip is a head) with equal probability. To find the two probabilities in question, raise $\mathbf{P}$ to the 8th and 9th powers.
$\mathbf{P}^9 = \bordermatrix{ & 0 & 1 & 2 & 3 & 4 \cr 0 & 0.406250 & 0.210938 & 0.109375 & 0.056641 & 0.216797 \cr 1 & 0.376953 & 0.195313 & 0.101563 & 0.052734 & 0.273438 \cr 2 & 0.320313 & 0.166016 & 0.085938 & 0.044922 & 0.382813 \cr 3 & 0.210938 & 0.109375 & 0.056641 & 0.029297 & 0.593750 \cr 4 & 0 & 0 & 0 & 0 & 1 \cr } \qquad$
$\mathbf{P}^8 = \bordermatrix{ & 0 & 1 & 2 & 3 & 4 \cr 0 & 0.421875 & 0.218750 & 0.113281 & 0.058594 & 0.187500 \cr 1 & 0.390625 & 0.203125 & 0.105469 & 0.054688 & 0.246094 \cr 2 & 0.332031 & 0.171875 & 0.089844 & 0.046875 & 0.359375 \cr 3 & 0.218750 & 0.113281 & 0.058594 & 0.031250 & 0.578125 \cr 4 & 0 & 0 & 0 & 0 & 1 \cr } \qquad$
The starting state is 0 since the coin flips are counted from the very first one. With $Q_{04}^9=0.216797$, we know that the probability of taking 9 flips or less to get 4 consecutive heads is 0.216797.
With $Q_{04}^9-Q_{04}^8=0.216797-0.187500=0.029297$, there is a roughly 3% chance that the run of 4 consecutive heads is achieved at the 9th flip.
Example 3 (Statement of the example given above)
Let $X_n$ be the area the mouse is in after making $n$ moves. Then $\left\{X_n: n=0,1,2,\cdots \right\}$ is a Markov chain. The following is the transition probability matrix. To help see the matrix, the diagram of the maze is repeated.
Area 1 = initial position of the mouse
$\mathbf{P} = \bordermatrix{ & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 \cr 1 & 0 & 1/2 & 0 & 0 & 0 & 1/2 & 0 & 0 & 0 \cr 2 & 1/3 & 0 & 1/3 & 0 & 1/3 & 0 & 0 & 0 & 0 \cr 3 & 0 & 1/2 & 0 & 1/2 & 0 & 0 & 0 & 0 & 0 \cr 4 & 0 & 0 & 1/3 & 0 & 1/3 & 0 & 0 & 0 & 1/3 \cr 5 & 0 & 1/4 & 0 & 1/4 & 0 & 1/4 & 0 & 1/4 & 0 \cr 6 & 1/3 & 0 & 0 & 0 & 1/3 & 0 & 1/3 & 0 & 0 \cr 7 & 0 & 0 & 0 & 0 & 0 & 1/2 & 0 & 1/2 & 0 \cr 8 & 0 & 0 & 0 & 0 & 1/3 & 0 & 1/3 & 0 & 1/3 \cr 9 & 0 & 0 & 0 & 1/2 & 0 & 0 & 0 & 1/2 & 0 \cr } \qquad$
States 3, 7 and 9 are desirable states since these are the areas with food. Let $\mathcal{M}=\left\{3,7,9 \right\}$. We then collapse $\mathcal{M}$ into a new state called 9. The state 9 in the new process is an absorbing state, which means that the original process has reached a state with food (states 3, 7 and 9). The following is the transition probability matrix for the new (collapsed) process.
$\mathbf{Q} = \bordermatrix{ & 1 & 2 & 4 & 5 & 6 & 8 & 9 \cr 1 & 0 & 1/2 & 0 & 0 & 1/2 & 0 & 0 \cr 2 & 1/3 & 0 & 0 & 1/3 & 0 & 0 & 1/3 \cr 4 & 0 & 0 & 0 & 1/3 & 0 & 0 & 2/3 \cr 5 & 0 & 1/4 & 1/4 & 0 & 1/4 & 1/4 & 0 \cr 6 & 1/3 & 0 & 0 & 1/3 & 0 & 0 & 1/3 \cr 8 & 0 & 0 & 0 & 1/3 & 0 & 0 & 2/3 \cr 9 & 0 & 0 & 0 & 0 & 0 & 0 & 1 \cr } \qquad$
The state 9 in the matrix $\mathbf{Q}$ refers to the situation in the original process reaching states 3, 7 or 9. Then all the questions in the example can be answered by raising $\mathbf{Q}$ to an appropriate power. In particular, raising $\mathbf{Q}$ to the 5th, 7th and 8th power.
$\mathbf{Q}^5 = \bordermatrix{ & 1 & 2 & 4 & 5 & 6 & 8 & 9 \cr 1 & 0 & 5/36 & 1/18 & 0 & 5/36 & 1/18 & 11/18 \cr 2 & 5/54 & 0 & 0 & 7/54 & 0 & 0 & 7/9 \cr 4 & 1/27 & 0 & 0 & 1/18 & 0 & 0 & 49/54 \cr 5 & 0 & 7/72 & 1/24 & 0 & 7/72 & 1/24 & 13/18 \cr 6 & 5/54 & 0 & 0 & 7/54 & 0 & 0 & 7/9 \cr 8 & 1/27 & 0 & 0 & 1/18 & 0 & 0 & 49/54 \cr 9 & 0 & 0 & 0 & 0 & 0 & 0 & 1 \cr } \qquad$
$\mathbf{Q}^7 = \bordermatrix{ & 1 & 2 & 4 & 5 & 6 & 8 & 9 \cr 1 & 0 & 17/216 & 7/216 & 0 & 17/216 & 7/216 & 7/9 \cr 2 & 17/324 & 0 & 0 & 2/27 & 0 & 0 & 283/324 \cr 4 & 7/324 & 0 & 0 & 5/162 & 0 & 0 & 307/324 \cr 5 & 0 & 1/18 & 5/216 & 0 & 1/18 & 5/216 & 91/108 \cr 6 & 17/324 & 0 & 0 & 2/27 & 0 & 0 & 283/324 \cr 8 & 7/324 & 0 & 0 & 5/162 & 0 & 0 & 307/324 \cr 9 & 0 & 0 & 0 & 0 & 0 & 0 & 1 \cr } \qquad$
$\mathbf{Q}^8 = \bordermatrix{ & 1 & 2 & 4 & 5 & 6 & 8 & 9 \cr 1 & 17/324 & 0 & 0 & 2/27 & 0 & 0 & 283/324 \cr 2 & 0 & 29/648 & 1/54 & 0 & 29/648 & 1/54 & 283/324 \cr 4 & 0 & 1/54 & 5/648 & 0 & 1/54 & 5/648 & 307/324 \cr 5 & 1/27 & 0 & 0 & 17/324 & 0 & 0 & 295/324 \cr 6 & 0 & 29/648 & 1/54 & 0 & 29/648 & 1/54 & 283/324 \cr 8 & 0 & 1/54 & 5/648 & 0 & 1/54 & 5/648 & 307/324 \cr 9 & 0 & 0 & 0 & 0 & 0 & 0 & 1 \cr } \qquad$
The statements for the probabilities to be sought are repeated here.
• Find the probability that after the 5 moves, the mouse is one area from a food source.
• Find the probability that after the 7 moves, the mouse has not found food and is in area 6 or area 8 after the 7th move.
• Find the probability that it takes at most 8 moves for the mouse to find food.
• Find the probability that it takes exactly 8 moves for the mouse to find food.
To obtain the answers, it is a matter of picking the entries from the matrices.
• $Q_{12}^5+Q_{14}^5+Q_{16}^5+Q_{18}^5=7/18=0.3889$
• $Q_{16}^7+Q_{18}^7=17/216+7/216=24/216=0.1111$
• $Q_{19}^8=283/324=0.8735$
• $Q_{19}^8-Q_{19}^7=283/324-7/9=0.095679$
One area from a food source means being in areas 2, 4, 6 and 8. Thus the first probability is $Q_{12}^5+Q_{14}^5+Q_{16}^5+Q_{18}^5$. The probability of being in areas 6 or 8 after 7 moves is $Q_{16}^7+Q_{18}^7$. The probability that it will take 8 moves or less to reach food is $Q_{19}^8$. The probability of reaching food in exactly 8 moves is $Q_{19}^8-Q_{19}^7$.
With the matrices $\mathbf{Q}^5$, $\mathbf{Q}^7$ and $\mathbf{Q}^8$, we can also find out the probability of reaching food. Just from the diagram, it is clear that the likelihood of reaching food is greater if the initial position is area 8 or area 6. Between these two areas, it is easier to reach food if the mouse is in area 8 initially since area 8 is adjacent to two food sources. This is also confirmed by the calculation. For example, $Q_{69}^5=7/9=0.7778$ and $Q_{89}^5=49/54=0.9074$. So it is over 90% certain that the mouse will find food if it is placed in area 8 initially.
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$\copyright$ 2017 – Dan Ma
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# The occupancy problem
The occupancy problem is a classic problem in probability. The setting of the problem is that balls are randomly distributed into cells (or boxes or other containers) one at a time. After all the balls are distributed into the cells, we examine how many of the cells are occupied with balls. In this post, we examine the occupancy problem using Markov chains.
The Occupancy Problem
As indicated above, we consider the random experiment of randomly placing $k$ balls into $n$ cells. The following diagram shows the result of randomly distributing 8 balls into 6 cells (or boxes).
Figure 1 – Occupancy Problem – throwing balls into cells
One ball is thrown into the 6 boxes for a total of 8 balls. In Figure 1, a total of four cells are occupied. In the current discussion, the focus is on the number of occupied cells and not on which of the cells are occupied. Since Figure 1 has 6 cells, we can also view the throwing a ball into the 6 boxes as rolling a fair die (the occupied cell would be the face value that turns up). Thus we can view the experiment in Figure 1 a rolling a die 8 times. Then at the end, we look at the number faces that appear.
Figure 2 – Occupancy Problem – rolling die repeatedly and counting the faces that appear
The top of Figure 2 shows the 8 rolls of the die. Four faces turn up in these 8 rolls – faces 2, 4, 5 and 6, the same result described in Figure 1. Another way to look at the occupancy problem is that of random sampling. Each time a ball is thrown into $n$ cells, the action can be viewed as randomly selecting one of the $n$ cells. Thus the occupancy problem can be viewed as randomly selecting $k$ balls (one at a time with replacement) from an urn with $n$ balls labeled $1,2,\cdots,n$. The following diagram shows an urn with 6 balls labeled 1 through 6. Then Figure 1 and Figure 2 would be equivalent to randomly selecting 8 balls with replacement from this urn.
Figure 3 – Occupancy Problem – random selection of balls with replacement from an urn and counting the distinct numbers that are drawn
Figure 1 describes a generic setting of the occupancy problem – throwing balls at random into cells. The interpretation in Figure 2 is ideal for the setting of six cells. If the number of cells is not six, we can view the problem as rolling an $n$-sided die $k$ times. The random selection of balls with replacement (Figure 3) is also a good way to view the occupancy problem. Regardless of how we view the occupancy problem, we consider the following questions. In randomly assigning $k$ balls into $n$ cells,
• What is the probability that exactly $j$ cells are occupied with balls where $j=1,2,\cdots,n$?
• What is the expected number of occupied cells?
The first question is about the probability distribution of the number of occupied cells after the balls are thrown. The second question is about the mean of the probability distribution of the number of occupied cells. Once the probability distribution is known, we can calculate the distributional quantities such as the mean and variance. Thus the main focus is on the first question.
The occupancy problem has been discussed in a companion blog. In this blog post, the first question is answered by using the multinomial theorem (applied twice). Another blog post develops a formula for the probability distribution of the number of occupied cells. These previous blog posts use counting methods to solve the occupancy problem. For example, in throwing eight balls into 6 cells, there are $6^8$ many distinct outcomes. How many of these correspond to only one occupied cell, two occupied cells and so on? In this post, we present another way to answer the same question using Markov chain.
The occupancy problem discussed here and in the previous blog posts assumes that each ball is equally likely to be assigned into any one of the cells. Thus the occupancy problem where the weights of assignment of the cells are not uniform would be an interesting extension of the problem.
Applying Markov Chains
The notion is Markov chains is introduced here. The calculation involving transition probability matrix is discussed here.
To demonstrate, we use the specific example of randomly distributing balls into 6 cells. First, define a Markov chain. The state of the Markov chain is the number of occupied cells at any given time. Let $X_0$ denote the number of occupied cells at the beginning of the experiment (before any balls are thrown). If all cells are unoccupied at the beginning, then $X_0=0$. For $n \ge 1$, $X_n$ be the number of occupied cells after the $n$th ball has been distributed. The resulting stochastic process $\left\{X_n: n=0,1,2,\cdots \right\}$ is a Markov chain since the state $X_{n+1}$ only depends on the preceding state $X_n$. The following is the transition probability matrix.
$\displaystyle \mathbf{P} = \bordermatrix{ & 0 & 1 & 2 & 3 & 4 & 5 & 6 \cr 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 \cr \text{ } & \text{ } & \text{ } & \text{ } & \text{ } & \text{ } & \text{ } & \text{ } \cr 1 & 0 & \frac{1}{6} & \frac{5}{6} & 0 & 0 & 0 & 0 \cr \text{ } & \text{ } & \text{ } & \text{ } & \text{ } & \text{ } & \text{ } & \text{ } \cr 2 & 0 & 0 & \frac{2}{6} & \frac{4}{6} & 0 & 0 & 0 \cr \text{ } & \text{ } & \text{ } & \text{ } & \text{ } & \text{ } & \text{ } & \text{ } \cr 3 & 0 & 0 & 0 & \frac{3}{6} & \frac{3}{6} & 0 & 0 \cr \text{ } & \text{ } & \text{ } & \text{ } & \text{ } & \text{ } & \text{ } & \text{ } \cr 4 & 0 & 0 & 0 & 0 & \frac{4}{6} & \frac{2}{6} & 0 \cr \text{ } & \text{ } & \text{ } & \text{ } & \text{ } & \text{ } & \text{ } & \text{ } \cr 5 & 0 & 0 & 0 & 0 & 0 & \frac{5}{6} & \frac{1}{6} \cr \text{ } & \text{ } & \text{ } & \text{ } & \text{ } & \text{ } & \text{ } & \text{ } \cr 6 & 0 & 0 & 0 & 0 & 0 & 0 & 1 \cr } \qquad$
The matrix $\mathbf{P}$ contains the one-step transition probabilities, i.e. the probabilities of the the next state given the current state. If the Markov chain is in state 0 (no occupied cells), the the next state is 1 for certain. If the current state is 1 (one occupied cell), then the next ball is is the occupied cell with probability 1/6 and is in one of the empty cells with probability 5/6. In general, if there are currently $i$ occupied cells, then the next ball is either in one of the $i$ occupied cells with probability $i/6$ (meaning the next state remains at $i$) or in one of the unoccupied cells with probability $(6-i)/6$ (meaning the next state is $i+1$). Since the number of occupied cells cannot decrease, the transition probabilities in $\mathbf{P}$ have the appearance of moving diagonally from the upper left corner to the lower right corner. State 6 (all cells are occupied) is a called an absorbing state. When the Markov chain is transitioned to state 6, it stays there for ever no matter how many more additional balls are thrown into the cells.
The transition probability matrix $\mathbf{P}$ holds the key to answering the first question indicated above. In throwing $k$ balls into 6 cells, the answer lies in the matrix $\mathbf{P}^k$, the $k$th power of $\mathbf{P}$ (the matrix obtained by multiplying $\mathbf{P}$ by itself $k$ times). This can be done by using software, e.g. using an online matrix calculator.
Consider the example of throwing eight balls into 6 cells. The following is the matrix $\mathbf{P}^8$.
$\displaystyle \mathbf{P}^8 = \bordermatrix{ & 0 & 1 & 2 & 3 & 4 & 5 & 6 \cr 0 & 0 & 0.00000357 & 0.00226838 & 0.06901578 & 0.36458333 & 0.45010288 & 0.11402606 \cr \text{ } & \text{ } & \text{ } & \text{ } & \text{ } & \text{ } & \text{ } & \text{ } \cr 1 & 0 & 0.0000006 & 0.0007591 & 0.03602014 & 0.27756344 & 0.49661351 & 0.18904321 \cr \text{ } & \text{ } & \text{ } & \text{ } & \text{ } & \text{ } & \text{ } & \text{ } \cr 2 & 0 & 0 & 0.00015242 & 0.01501534 & 0.18815015 & 0.50831619 & 0.28836591 \cr \text{ } & \text{ } & \text{ } & \text{ } & \text{ } & \text{ } & \text{ } & \text{ } \cr 3 & 0 & 0 & 0 & 0.00390625 & 0.10533658 & 0.47531221 & 0.41544496 \cr \text{ } & \text{ } & \text{ } & \text{ } & \text{ } & \text{ } & \text{ } & \text{ } \cr 4 & 0 & 0 & 0 & 0 & 0.03901844 & 0.38709919 & 0.57388236 \cr \text{ } & \text{ } & \text{ } & \text{ } & \text{ } & \text{ } & \text{ } & \text{ } \cr 5 & 0 & 0 & 0 & 0 & 0 & 0.23256804 & 0.76743196 \cr \text{ } & \text{ } & \text{ } & \text{ } & \text{ } & \text{ } & \text{ } & \text{ } \cr 6 & 0 & 0 & 0 & 0 & 0 & 0 & 1 \cr } \qquad$
Each entry in $\mathbf{P}^8$ is denoted by $P_{ij}^8$, the 8-step transition probability, located in the row corresponding to the state $i$ and in the column corresponding to the state $j$. For example, $P_{03}^8=0.069$, indicating that there is a 6.9% chance that 3 cells are occupied after eight balls are thrown (with all cells being empty at the beginning). Another example: $P_{13}^8=0.036$. So if there is one occupied cell at the beginning, there is a 3.6% chance that 3 cells are occupied after throwing 8 balls into the 6 cells. So each row in $\mathbf{P}^8$ gives a probability distribution of the number of occupied cells depending the initial state.
The first question stated above assumes that the all cells are empty at the beginning. Thus the first row of the matrix $\mathbf{P}^8$ gives the answers.
$P[\text{1 occupied cell}]=P_{01}^8=0.00000357$
$P[\text{2 occupied cell}]=P_{02}^8=0.00226838$
$P[\text{3 occupied cell}]=P_{03}^8=0.06901578$
$P[\text{4 occupied cell}]=P_{04}^8=0.36458333$
$P[\text{5 occupied cell}]=P_{05}^8=0.45010288$
$P[\text{6 occupied cell}]=P_{06}^8=0.11402606$
In randomly throwing 8 balls into 6 cells, the more likely outcomes would be having 4, 5 or 6 occupied cells with the mostly likely being 4 occupied cells. Let $Y$ be the number of occupied cells after 8 balls are thrown into 6 cells. The mean number of occupied cells is:
\displaystyle \begin{aligned} E[Y]&=1 \times 0.00000357+2 \times 0.00226838+3 \times 0.06901578 \\&\ \ + 4 \times 0.36458333+5 \times 0.45010288+6 \times 0.11402606 \\&=4.60459175 \end{aligned}
To give an indication that the probability distribution is correct, we simulate 8 rolls of a die 1,000 times using the =RANDBETWEEN(1, 6) function in Excel. The results: 0% (1 occupied cell), 0.4% (2 occupied cells), 7.3% (3 occupied cells), 35.2% (4 occupied cells), 47.3% (5 occupied cells) and 9.8% (6 occupied cells). These results are in general agreement with the calculated distribution. Of course, the larger the simulation, the closer the simulated results will be to the theoretical distribution.
Here’s the algorithm for solving the occupancy problem using the Markov chain approach.
If the occupancy problem involves throwing balls into $n$ cells, the state of the Markov chain is the number of occupied cells at any given time. Set up the one-step transition probability matrix $\mathbf{P}$ as follows. Set $P_{01}=1$ and $P_{n,n-1}=1$. For $0, set $P_{ii}=i/n$, $P_{i,i+1}=(n-i)/n$. Then the matrix $\mathbf{P}^k$ gives the probability distribution of the number of occupied cells after $k$ balls are randomly distributed into the $n$ cells. The row in $\mathbf{P}^k$ corresponding to the state $i$ gives the probability of the number of occupied cells when there are $i$ occupied cells initially. If initially all cells are empty, then the first row of $\mathbf{P}^k$ gives the probability distribution of the number of occupied cells.
We would like to make further comment on the example. Assuming that all cells are empty at the beginning, the example can be worked by eliminating the first row of $\mathbf{P}$, the row for state 0. The first ball will always go into one empty cell. Thus in throwing 8 balls, we can focus on 7 balls going into 6 cells with one cell already occupied (by the first ball). Thus we can focus on the following transition probability matrix.
$\displaystyle \mathbf{P} = \bordermatrix{ & 1 & 2 & 3 & 4 & 5 & 6 \cr 1 & \frac{1}{6} & \frac{5}{6} & 0 & 0 & 0 & 0 \cr \text{ } & \text{ } & \text{ } & \text{ } & \text{ } & \text{ } & \text{ } \cr 2 & 0 & \frac{2}{6} & \frac{4}{6} & 0 & 0 & 0 \cr \text{ } & \text{ } & \text{ } & \text{ } & \text{ } & \text{ } & \text{ } \cr 3 & 0 & 0 & \frac{3}{6} & \frac{3}{6} & 0 & 0 \cr \text{ } & \text{ } & \text{ } & \text{ } & \text{ } & \text{ } & \text{ } \cr 4 & 0 & 0 & 0 & \frac{4}{6} & \frac{2}{6} & 0 \cr \text{ } & \text{ } & \text{ } & \text{ } & \text{ } & \text{ } & \text{ } \cr 5 & 0 & 0 & 0 & 0 & \frac{5}{6} & \frac{1}{6} \cr \text{ } & \text{ } & \text{ } & \text{ } & \text{ } & \text{ } & \text{ } \cr 6 & 0 & 0 & 0 & 0 & 0 & 1 \cr } \qquad$
Assuming that the first ball goes into one of the empty cells, we focus on the next 7 balls. So we compute the matrix $\mathbf{P}^7$.
$\displaystyle \mathbf{P}^7 = \bordermatrix{ & 1 & 2 & 3 & 4 & 5 & 6 \cr 1 & 0.00000357 & 0.00226838 & 0.06901578 & 0.36458333 & 0.45010288 & 0.11402606 \cr \text{ } & \text{ } & \text{ } & \text{ } & \text{ } & \text{ } & \text{ } & \text{ } \cr 2 & 0 & 0.00045725 & 0.02942101 & 0.26015947 & 0.50591564 & 0.20404664 \cr \text{ } & \text{ } & \text{ } & \text{ } & \text{ } & \text{ } & \text{ } & \text{ } \cr 3 & 0 & 0 & 0.0078125 & 0.15214549 & 0.50951646 & 0.33052555 \cr \text{ } & \text{ } & \text{ } & \text{ } & \text{ } & \text{ } & \text{ } & \text{ } \cr 4 & 0 & 0 & 0 & 0.05852766 & 0.44110797 & 0.50036437 \cr \text{ } & \text{ } & \text{ } & \text{ } & \text{ } & \text{ } & \text{ } & \text{ } \cr 5 & 0 & 0 & 0 & 0 & 0.27908165 & 0.72091835 \cr \text{ } & \text{ } & \text{ } & \text{ } & \text{ } & \text{ } & \text{ } & \text{ } \cr 6 & 0 & 0 & 0 & 0 & 0 & 1 \cr } \qquad$
Note that the first row of $\mathbf{P}^7$ is identical to the first row of $\mathbf{P}^8$ earlier. So a slightly different algorithm is to put one ball in a cell and then focus on the random behavior of the next $k-1$ balls. In other words, set up the transition probability matrix $\mathbf{P}$ with only the states $1, 2,\cdots,n$ and then raise $\mathbf{P}$ to $k-1$.
Remarks
As the above example demonstrates, the occupancy problem is a matter of calculating the power of a transition probability matrix. It is an algorithm that is suitable for computer implementation. It illustrates the versatility of the Markov chain method. The combinatorial approach of using multinomial theorem (discussed here and here) is also a valuable approach. Each approach provides its own unique insight. The post ends with an exercise.
Exercise
Suppose that eleven candies are randomly distributed to 4 children. Suppose that the names of the 4 children are Marcus, Issac, Samantha and Paul.
• Given the one-step transition probability matrix for the Markov chain describing the process of distributing candies to 4 children.
• What is the probability distribution of the number of children who have received candies?
• What is the probability all 4 children receiving candies?
• What is the mean number of children who have received candies?
• What is the probability that Marcus receives 1 candy, Issac receives 4 candies, Samantha receives 4 candies and Paul receives 2 candies?
• What is the probability that one of the children receives 1 candy, another child receives 4 candies, another child receives 4 candies and the last child receives 2 candies?
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$\copyright$ 2017 – Dan Ma
# A first look at applications of Markov chains
This post presents several interesting examples of Markov chain models. Previous posts on Markov chains are: an introduction, how to work with transition probabilities (here and here).
Examples
The following gives several examples of Markov chains, discussed at an introductory level. They will be further developed in subsequent posts.
Example 1 – The Ehrenfest Chain
The Ehrenfest chain, named for the physicist Paul Ehrenfest, is a simple, discrete model for the exchange of gas molecules contained in a volume divided into two regions A and B by a permeable membrane. Suppose that the gas has $N$ molecules. At each period of time, a molecule is chosen at random from the set of $N$ molecules and moved from the region that it is in to the other region.
Figure 1 – The Efrenfest Chain
The model has a simple mathematical description using balls and urns. Suppose that two urns, labeled A and B, contain $N$ balls, labeled $1, 2, 3, \cdots, N$. At each step (i.e. at each discrete time unit), an integer is selected at random from $1, 2, 3, \cdots, N$ independent of the past selections. Then the ball labeled by the selected integer is removed from its urn and placed in the other urn. The procedure is repeated indefinitely.
The state of the system at time $n=0,1,2,\cdots$ is the number of balls in urn A and is denoted by $X_n$. Then $\left\{X_n: n=0,1,2,\cdots \right\}$ is a Markov chain on the state space $\left\{0,1,\cdots,N \right\}$.
When the process is in state 0 (urn A is empty), the next state is 1. When the process is in state $N$ (urn A is full), the next state is $N-1$. When the process is in state $i$ where $0, the next state of the process is either $i-1$ or $i+1$. The following gives the one-step transition probabilities:
$\displaystyle P_{ij} = \left\{ \begin{array}{ll} \displaystyle 1 &\ \ \ \ \ \ i=0 \text{ and } j=1 \\ \text{ } & \text{ } \\ \displaystyle 1 &\ \ \ \ \ \ i=N \text{ and } j=N-1 \\ \text{ } & \text{ } \\ \displaystyle \frac{i}{N} &\ \ \ \ \ \ 0
When urn A has $i$ balls, there is a probability of $\frac{i}{N}$ such that the randomly selected ball is from urn A and thus urn A loses a ball. On the other hand, there is a probability of $1-\frac{i}{N}$ such that the randomly selected ball is from urn B and thus urn A gains a ball. As an illustration, the following gives the transition probability matrix for $N=5$ balls.
$\mathbf{P} = \bordermatrix{ & 0 & 1 & 2 & 3 & 4 & 5 \cr 0 & 0 & 1 & 0 & 0 & 0 & 0 \cr 1 & \frac{1}{5} & 0 & \frac{4}{5} & 0 & 0 & 0 \cr 2 & 0 & \frac{2}{5} & 0 & \frac{3}{5} & 0 & 0 \cr 3 & 0 & 0 & \frac{3}{5} & 0 & \frac{2}{5} & 0 \cr 4 & 0 & 0 & 0 & \frac{4}{5} & 0 & \frac{1}{5} \cr 5 & 0 & 0 & 0 & 0 & 1 & 0 \cr } \qquad$
An important problem in the Ehrenfest chain is the long-term, or equilibrium, distribution of $X_n$, the number of balls in urn A. Another interesting problem concerns the changes in the composition of the two regions over time. For example, if all molecules are in one region at the beginning, how long on average will it be before each region has half of the molecules? The theory of Markov chains provide a good method for answering such questions.
Example 2 – Random Walk
The simplest random walk is a Markov chain $\left\{X_n: n=0,1,2,\cdots \right\}$ such that each state is the result of a random one-unit up or down move from the previous state.
Figure 2 – Five Simulations of a Random Walk
In the random walk in Figure 1, each state is one unit above or below the preceding state with equal probability. Instead of a random one-unit up or down move, let’s give a more general description of a random walk. The moves in the random walk are determined by a predetermined discrete distribution. Let $Y_1, Y_2, Y_3 \cdots$ be integer-valued random variables that are independent and identically distributed. Let $P[Y=y]$ be the common probability function. Let $X_0$ be an integer-valued random variable that is independent of the random variables $Y_n$. The random variable $X_0$ will be the initial position of the random walk. The random variables $Y_n$ are the increments (they are the amounts added to the stochastic process as time increases).
For $n \ge 1$, define $X_n=X_0+Y_1+\cdots+Y_{n-1}+Y_n$. Then $\left\{X_n: n=0,1,2,\cdots \right\}$ is called a random walk. It is a Markov chain with the state space the set of all integers. Wherever the process is at any given time, the next step in the process is always a walk in a distance and direction that is dictated by the independent random variables $Y_n$. If the random variables $Y_n$ are independent Bernoulli variables with $P(Y_n=1)=p$ and $P(Y_n=-1)=1-p$, then we revert back to the simple random walks described in Figure 1.
To transition from state $i$ to state $j$, the increment $Y_n$ must be $j-i$. Thus the one-step transition probabilities $P_{ij}$ are defined by $P_{ij}=P[Y=j-i]$. Recall that $P[Y=y]$ is the common probability function for the sequence $Y_1, Y_2, Y_3 \cdots$. For a more detailed description, see this previous post.
Example 3 – Gambler’s Ruin
A gambler’s ruin is a simple random walk that can take on only finitely many states. At each time period, the state is one-unit higher or lower than the preceding state as a result of a random bet. More specifically, suppose that a gambler starts out with $d$ units in capital (in dollars or other monetary units) and makes a series of one-unit bets against the house. For the gambler, the probabilities of winning and losing each bet are $p$ and $1-p$, respectively. Whenever the capital reaches zero, the gambler is in ruin and his capital remains zero thereafter. On the other hand, if the capital of the gambler increases to $m$ where $d, then the gambler quits playing. Let $X_n$ be the capital of the gambler after the $n$th bet. Then $\left\{X_n: n=0,1,2,3,\cdots \right\}$ is a random walk. The starting state is $X_0=d$. The states 0 and $m$ are absorbing states. The following gives the transition probabilities.
$\displaystyle P_{ij} = \left\{ \begin{array}{ll} \displaystyle p &\ \ \ \ \ \ j=i+1,i \ne 0, i \ne m \\ \text{ } & \text{ } \\ \displaystyle 1-p &\ \ \ \ \ \ j=i-1,i \ne 0, i \ne m \\ \text{ } & \text{ } \\ \displaystyle 1 &\ \ \ \ \ \ i=0,j=0 \\ \text{ } & \text{ } \\ \displaystyle 1 &\ \ \ \ \ \ i=m,j=m \\ \text{ } & \text{ } \\ 0 &\ \ \ \ \ \ \text{otherwise} \end{array} \right.$
Whenever the gambler reaches the absorbing state of 0, the gambler is in ruin. If the process reaches the absorbing state of $m$, the gambler strikes gold. If the initial capital of the gambler is small relative to the casino, there is a virtually 100% chance that the gambler will be in ruin even if each bet has even odds (see here for the calculation). One interesting questions: on average how long does it take for the gambler to be in ruin? Such questions will be answered after necessary tools are built in subsequent posts.
An alternative interpretation of the gambler’s ruin random walk is this. Assume that two players make a series of one-unit bets against each other and that the total capital between the two players is $m$ units with the first player having $d$ units in capital. Further suppose that the first player has probability $p$ of winning a bet and the second player has probability $1-p$ of winning a bet. Then the two gamblers play until one of them goes broke (is in ruin). Let $X_n$ be the capital of the first player after the $n$th play. Then $\left\{X_n: n=0,1,2,3,\cdots \right\}$ is a Markov chain that is a gambler’s ruin. The following graph shows five simulations of a gambler’s ruin from this previous post.
Figure 3 – Five Simulations of a Gambler’s Ruin
Example 4 – Discrete Birth and Death Chain
In a birth and death chain, the current state $i$ in the process is transitioned to $i+1$ (a birth), $i-1$ (a death) or $i$. The state space is either $\left\{0,1,2,3,\cdots \right\}$ or $\left\{0,1,2,3,\cdots, m \right\}$. The following gives the one-step transition probabilities.
$\displaystyle P_{ij} = \left\{ \begin{array}{ll} \displaystyle q_i &\ \ \ \ \ \ j=i-1 \\ \text{ } & \text{ } \\ \displaystyle r_i &\ \ \ \ \ \ j=i \\ \text{ } & \text{ } \\ \displaystyle p_i &\ \ \ \ \ \ j=i+1 \\ \text{ } & \text{ } \\ 0 &\ \ \ \ \ \ \text{otherwise} \end{array} \right.$
where for each state $i$, $q_i$ (the probability of down), $r_i$ (the probability of same) and $p_i$ (the probability of up) are non-negative real numbers such that $q_i+r_i+p_i=1$. The following gives the transition probability matrix for an illustrative case of $m=5$.
$\mathbf{P} = \bordermatrix{ & 0 & 1 & 2 & 3 & 4 & 5 \cr 0 & r_0 & p_0 & 0 & 0 & 0 & 0 \cr 1 & q_1 & r_1 & p_1 & 0 & 0 & 0 \cr 2 & 0 & q_2 & r_2 & p_2 & 0 & 0 \cr 3 & 0 & 0 & q_3 & r_3 & p_3 & 0 \cr 4 & 0 & 0 & 0 & q_4 & r_4 & p_4 \cr 5 & 0 & 0 & 0 & 0 & q_5 & r_5 \cr } \qquad$
In the above matrix, $q_0=0$ and $p_5=0$. If $r_0=1$, then state 0 is absorbing. If $p_0=1$, then state 0 is reflecting. On the other hand, if $r_5=1$, then state 5 is absorbing. If $q_5=1$, then state 5 is reflecting.
The Ehrenfest chain is an example of a birth and death chain (with the boundary states being reflecting). When the probability of down $q_i$, the probability of same $r_i$ and the probability of up $p_i$ do not vary according to the current state (i.e. they are constant), the resulting birth and death chain is a random walk. Thus the gambler’s ruin chain is also an example of a birth and death chain. The birth and death chain is suitable model for applications in which the state of the process is the population of a living system.
Example 5 – Discrete Queueing Markov Chain
Though continuous-time queueing models are much more realistic, a simple discrete-time queueing model is introduced here in order to illustrate applications of discrete Markov chains. Suppose that time is measured in a convenient time interval such as one minute or some appropriate group of minutes. The model assumes that during one period of time, only one customer is served if at least one customer is present. The model also assumes that a random number of new customers arrive at any given time period and that the numbers of customers arriving in the time periods form an independent identically distributed sequence of random variables.
More specifically, let $Y_1,Y_2,Y_3,\cdots$ be independent and identically distributed random variables that take on non-negative integers. Let $P[Y=k]$, $k=0,1,2,\cdots$, be the common probability function. The random variable $Y_n$ is the number of customers that arrive into the system during the $n$th period. Let $X_0$ be the number of customers present initially in the system. For $n \ge 1$, let $X_n$ be the number of customers in the system during the $n$th period. If the current state is $X_n$, then the next state is:
$\displaystyle X_{n+1} = \left\{ \begin{array}{ll} \displaystyle Y_{n+1} &\ \ \ \ \ \ X_n=0 \\ \text{ } & \text{ } \\ \displaystyle (X_n-1)+Y_{n+1} &\ \ \ \ \ \ X_n>0 \\ \end{array} \right.$
where $n \ge 0$. In essence, the number of customers in the system at any given time period is the number of new customers arriving plus the customers already in the system less one (if the system is not empty). If the system is empty, the number of customers is simply the number of new customers. It follows that $\left\{X_n: n \ge 0 \right\}$ is a Markov chain with the state space $\left\{0,1,2,\cdots \right\}$, the set of all non-negative integers. The one-step transition probabilities $P_{ij}$ are given by:
$\displaystyle P_{ij} = \left\{ \begin{array}{ll} \displaystyle P[Y=j] &\ \ \ \ \ \ i=0 \text{ and } j \in \left\{0,1,2,\cdots \right\} \\ \text{ } & \text{ } \\ \displaystyle P[Y=j-(i-1)] &\ \ \ \ \ \ i>0 \text{ and } j \in \left\{i-1,i,i+1,\cdots \right\} \\ \end{array} \right.$
where $P[Y=k]$ is the common probability function for the independent numbers of arrivals of customers $Y_1,Y_2,Y_3,\cdots$. If the $Y_n$ have a common binomial distributions with parameters 3 and $p=0.5$. Then the following gives the transition probability matrix.
$\displaystyle \mathbf{P} = \bordermatrix{ & 0 & \text{ } & 1 & \text{ } & 2 & \text{ } & 3 & \text{ } & 4 & \text{ } & 5 & \text{ } & 6 & \text{ } & 7 & \text{ } & 8 & \cdots \cr 0 & 0.125 & \text{ } & 0.375 & \text{ } & 0.375 & \text{ } & 0.125 & \text{ } & 0 & \text{ } & 0 & \text{ } & 0 & \text{ } & 0 & \text{ } & 0 & \cdots \cr 1 & 0.125 & \text{ } & 0.375 & \text{ } & 0.375 & \text{ } & 0.125 & \text{ } & 0 & \text{ } & 0 & \text{ } & 0 & \text{ } & 0 & \text{ } & 0 & \cdots \cr 2 & 0 & \text{ } & 0.125 & \text{ } & 0.375 & \text{ } & 0.375 & \text{ } & 0.125 & \text{ } & 0 & \text{ } & 0 & \text{ } & 0 & \text{ } & 0 & \cdots \cr 3 & 0 & \text{ } & 0 & \text{ } & 0.125 & \text{ } & 0.375 & \text{ } & 0.375 & \text{ } & 0.125 & \text{ } & 0 & \text{ } & 0 & \text{ } & 0 & \cdots \cr 4 & 0 & \text{ } & 0 & \text{ } & 0 & \text{ } & 0.125 & \text{ } & 0.375 & \text{ } & 0.375 & \text{ } & 0.125 & \text{ } & 0 & \text{ } & 0 & \cdots \cr 5 & 0 & \text{ } & 0 & \text{ } & 0 & \text{ } & 0 & \text{ } & 0.125 & \text{ } & 0.375 & \text{ } & 0.375 & \text{ } & 0.125 & \text{ } & 0 & \cdots \cr 6 & 0 & \text{ } & 0 & \text{ } & 0 & \text{ } & 0 & \text{ } & 0 & \text{ } & 0.125 & \text{ } & 0.375 & \text{ } & 0.375 & \text{ } & 0.125 & \cdots \cr 7 & 0 & \text{ } & 0 & \text{ } & 0 & \text{ } & 0 & \text{ } & 0 & \text{ } & 0 & \text{ } & 0.125 & \text{ } & 0.375 & \text{ } & 0.375 & \cdots \cr 8 & 0 & \text{ } & 0 & \text{ } & 0 & \text{ } & 0 & \text{ } & 0 & \text{ } & 0 & \text{ } & 0 & \text{ } & 0.125 & \text{ } & 0.375 & \cdots \cr \vdots & \vdots & \text{ } & \vdots & \text{ } & \vdots & \text{ } & \vdots & \text{ } & \vdots & \text{ } & \vdots & \text{ } & \vdots & \text{ } & \vdots & \text{ } & \vdots & \text{ } \cr } \qquad$
If the customer arrivals have a common Poisson distribution with parameter $\lambda=0.2$ (per period), then the following gives the transition probabilities.
$\displaystyle P_{ij} = \left\{ \begin{array}{ll} \displaystyle \frac{e^{-0.2} \ 0.2^j}{j!} &\ \ \ \ \ \ i=0 \text{ and } j \in \left\{0,1,2,\cdots \right\} \\ \text{ } & \text{ } \\ \displaystyle \frac{e^{-0.2} \ 0.2^{j-(i-1)}}{[j-(i-1)]!} &\ \ \ \ \ \ i>0 \text{ and } j \in \left\{i-1,i,i+1,\cdots \right\} \\ \end{array} \right.$
Example 6 – Discrete Branching Chain
Suppose that we have a system of particles that can generate new particles of the same type, e.g. particles such as neutrons and bacteria.
In such a system, we assume that a particle is replaced by a ransom number of new particles (regarded as offspring of the original particle). Furthermore, we assume that each particle generates offspring independently and offspring generation follows the same distribution. Let $Y$ be the random variable that is the common distribution for offspring generation across all particles in the system. Let $P[Y=k]$ be the common probability function of the number of offspring of a particle.
For $n \ge 0$, let $X_n$ be the number of particles in the $n$th generation in the system. It follows from the assumptions that with $X_n=i$
$\displaystyle X_{n+1}=\sum \limits_{k=1}^{i} Y_k$
where $Y_1,Y_2, \cdots, Y_i$ are independent random variables with common probability function $P[Y=k]$. Then $\left\{X_n: n \ge 0 \right\}$ is a Markov chain with the state space the set of all non-negative integers. The one-step transition probabilities $P_{ij}$ are
$\displaystyle P_{ij}=P \biggl[\sum \limits_{k=1}^{i} Y_k = j \biggr]$
In other words, the transition probability $P_{ij}$ is defined by the probability that the independent sum $Y_1+\cdots+Y_i$ taking on the next state $j$.
Note that state 0 is an absorbing state, which means extinction. An interesting problems in branching chains is the computation of the probability of extinction. One approach is to compute the probability of extinction for a branching chain with a single particle, i.e. the probability $\rho$ of starting in state 1 and being absorbed into state 0. Then the probability of extinction for a branching chain starting with $k$ particles is then $\rho^k$ since the particles are assumed to generate new particles independently.
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$\copyright$ 2017 – Dan Ma
# Chapman-Kolmogorov Equations
Stochastic processes and Markov chains are introduced in this previous post. Transition probabilities are an integral part of the theory of Markov chains. The post preceding this one is a beginning look at transition probabilities. This post shows how to calculate the $n$-step transition probabilities. The Chapman-Kolmogorov equations are also discussed and derived.
Introductory Example
Suppose $\left\{X_n: n=0,1,2,\cdots \right\}$ is a Markov chain with the transition probability matrix $\mathbf{P}$. The entries of the matrix are the one-step transition probabilities $P_{ij}$. The number $P_{ij}$ is the probability that the Markov chain will move to state $j$ at time $m+1$ given that it is in state $i$ at time $m$, independent of where the chain was prior to time $m$. Thus $P_{ij}$ can be expressed as the following conditional probability.
$(1) \ \ \ \ \ P_{ij}=P[X_{m+1}=j \lvert X_m=i]$
Thus the future state only depends on the period immediately preceding it. This is called the Markov property. Also note that $P_{ij}$ is independent of the time period $m$. Any Markov chain with this property is called a time-homogeneous Markov chain or stationary Markov chain.
The focus of this post is to show how to calculate the probability that the Markov chain will move to state $j$ at time $n+m$ given that it is in state $i$ at time $m$. This probability is denoted by $P_{ij}^n$. In other words, $P_{ij}^n$ is the probability that the chain will be in state $j$ after the Markov chain goes through $n$ more periods given that it is in state $i$ in the current period. The probability $P_{ij}^n$, as a conditional probability, can be notated as follows:
$(2) \ \ \ \ \ P_{ij}^n=P[X_{m+n}=j \lvert X_m=i]$
In (2), the $n$ in the $n$-step transition probabilities satisfies $n \ge 0$. Note that when $n=0$, $P_{ij}^0=1$ for $i=j$ and $P_{ij}^0=0$ for $i \ne j$. Including the case for $n=0$ will make the Chapman-Kolmogorov equations work better.
Before discussing the general method, we use examples to illustrate how to compute 2-step and 3-step transition probabilities. Consider a Markov chain with the following transition probability matrix.
$\mathbf{P} = \bordermatrix{ & 0 & 1 & 2 \cr 0 & 0.6 & 0.2 & 0.2 \cr 1 & 0.3 & 0.5 & 0.2 \cr 2 & 0.4 & 0.3 & 0.3 \cr } \qquad$
Calculate the two-step transition probabilities $P_{02}^2$, $P_{12}^2$ and $P_{22}^2$. Then calculate the three-step transition probability $P_{02}^3$ using the two-step transition probabilities.
First, let’s handle $P_{02}^2$. We can condition on the first steps. To go from state 0 to state 2 in two steps, the chain must first go to an interim state and then from that state go to state 2.
\displaystyle \begin{aligned}P_{02}^2&=\sum \limits_{k=0}^2 P_{0k} \ P_{k2} \\&=P_{00} \ P_{02}+P_{01} \ P_{12}+P_{02} \ P_{22} \\&=0.6 \times 0.2+0.2 \times 0.2+0.2 \times 0.3 \\&=0.22 \end{aligned}
Note that the above calculation lists out the three possible paths to go from state 0 to state 2 in two steps – from state 0 to state 0 and then from state 0 to state 2, from state 0 to state 1 and then from state 1 to state 2 and from state 0 to state 2 and then from state 2 to state 2. Looking at the calculation more closely, the above calculation is basically the first row of $\mathbf{P}$ (the row corresponding to state 0) multiplying the third column of $\mathbf{P}$ (the column corresponding to state 2).
$P_{02}^2= \left[\begin{array}{ccc} 0.6 & 0.2 & 0.2 \\ \end{array}\right] \left[\begin{array}{c} 0.2 \\ 0.2 \\ 0.3 \end{array}\right]= \left[\begin{array}{c} 0.22 \end{array}\right]$
By the same idea, the following gives the other two 2-step transition probabilities.
\displaystyle \begin{aligned}P_{12}^2&=\sum \limits_{k=0}^2 P_{1k} \ P_{k2} \\&=P_{10} \ P_{02}+P_{11} \ P_{12}+P_{12} \ P_{22} \\&=0.3 \times 0.2+0.5 \times 0.2+0.2 \times 0.3 \\&=0.22 \end{aligned}
\displaystyle \begin{aligned}P_{22}^2&=\sum \limits_{k=0}^2 P_{2k} \ P_{k2} \\&=P_{20} \ P_{02}+P_{21} \ P_{12}+P_{22} \ P_{22} \\&=0.4 \times 0.2+0.3 \times 0.2+0.3 \times 0.3 \\&=0.23 \end{aligned}
As discussed, the above two calculations can be viewed as the sum of all the possible paths to go from the beginning state to the end state (conditioning on the interim state in the middle) or as a row in the transition probability $\mathbf{P}$ multiplying a column in $\mathbf{P}$. The following shows all three calculations in terms of matrix calculation.
$P_{02}^2= \left[\begin{array}{ccc} 0.6 & 0.2 & 0.2 \\ \end{array}\right] \left[\begin{array}{c} 0.2 \\ 0.2 \\ 0.3 \end{array}\right]= \left[\begin{array}{c} 0.22 \end{array}\right]$
$P_{12}^2= \left[\begin{array}{ccc} 0.3 & 0.5 & 0.2 \\ \end{array}\right] \left[\begin{array}{c} 0.2 \\ 0.2 \\ 0.3 \end{array}\right]= \left[\begin{array}{c} 0.22 \end{array}\right]$
$P_{22}^2= \left[\begin{array}{ccc} 0.4 & 0.3 & 0.3 \\ \end{array}\right] \left[\begin{array}{c} 0.2 \\ 0.2 \\ 0.3 \end{array}\right]= \left[\begin{array}{c} 0.23 \end{array}\right]$
The view of matrix calculation will be crucial in understanding Chpan-Kolmogorov equations discussed below. To conclude the example, consider the 3-step transition probability $P_{02}^3$. We can also condition on the first step. The chain goes from state 0 to the next step (3 possibilities) and then goes from that state to state 2 in two steps.
\displaystyle \begin{aligned} P_{02}^3&=\sum \limits_{k=0}^2 P_{0k} \ P_{k2}^2 \\&=P_{00} \ P_{02}^2+P_{01} \ P_{12}^2+P_{02} \ P_{22}^2 \\&=0.6 \times 0.22+0.2 \times 0.22+0.2 \times 0.23 \\&=0.222 \end{aligned}
The example shows that the calculation of a 3-step transition probability is based 2-step probabilities. Thus longer length transition probabilities can be built up from shorter length transition probabilities. However, it pays to focus on a more general framework before carrying out more calculation. The view of matrix calculation demonstrated above will help in understanding the general frame work.
Chapman-Kolmogorov Equations
The examples indicate that finding $n$-step transition probabilities involve matrix calculation. Let $\mathbf{Q}_n$ be the $n$-step transition probability matrix. The goal now is to have a systematic way to compute the entries in the matrix $\mathbf{Q}_n$. The computation is based on the Chapman-Kolmogorov equations. These equations are:
$\displaystyle (3) \ \ \ \ \ P_{ij}^{n+m}=\sum \limits_{k=0}^\infty P_{ik}^n \times P_{kj}^m$
for $n, m \ge 0$ and for all $i,j$. For a finite-state Markov chain, the summation in (3) does not go up to infinity and would have an upper limit. The number $P_{ij}^{n+m}$ is the probability that the chain will be in state $j$ after taking $n+m$ steps if it is currently in state $i$. The following gives the derivation of (3).
\displaystyle \begin{aligned} P_{ij}^{n+m}&=P[X_{n+m}=j \lvert X_0=i] \\&=\sum \limits_{k=0}^\infty P[X_{n+m}=j, X_n=k \lvert X_0=i]\\&=\sum \limits_{k=0}^\infty \frac{P[X_{n+m}=j, X_n=k, X_0=i]}{P[X_0=i]} \\&=\sum \limits_{k=0}^\infty \frac{P[X_n=k,X_0=i]}{P[X_0=i]} \times \frac{P[X_{n+m}=j, X_n=k, X_0=i]}{P[X_n=k,X_0=i]} \\&=\sum \limits_{k=0}^\infty P[X_n=k \lvert X_0=i] \times P[X_{n+m}=j \lvert X_n=k, X_0=i] \\&=\sum \limits_{k=0}^\infty P[X_n=k \lvert X_0=i] \times P[X_{n+m}=j \lvert X_n=k] \ * \\&=\sum \limits_{k=0}^\infty P_{ik}^n \times P_{kj}^m \end{aligned}
Here’s the idea behind the derivation. The path from state $i$ to state $j$ in $n+m$ steps can be broken down into two paths, one from state $i$ to an intermediate state $k$ in the first $n$ transitions, and the other from state $k$ to state $j$ in the remaining $m$ transitions. Summing over all intermediate states $k$ yields the probability that the process will go from state $i$ to state $j$ in $n+m$ transitions.
The entries in the matrix $\mathbf{Q}_n$ can be then computed by (3). There is an interesting an useful interpretation of (3). The following is another way to state the Chapman-Kolmogorov equations:
$(4) \ \ \ \ \ \mathbf{Q}_{n+m}=\mathbf{Q}_n \times \mathbf{Q}_m$.
A typical element of the matrix $\mathbf{Q}_n$ is $P_{ik}^n$ and a typical element of the matrix $\mathbf{Q}_m$ is $P_{kj}^m$. Note that $P_{ik}^n$ with $k$ varying is a row in $\mathbf{Q}_n$ (the row corresponding to the state $i$) and that $P_{kj}^n$ with $k$ varying is a column in $\mathbf{Q}_m$ (the column corresponding to the state $j$).
$\left[\begin{array}{cccccc} P_{i0}^n & P_{i1}^n & \cdots & P_{ik}^n & \cdots & P_{iw}^n \\ \end{array}\right] \left[\begin{array}{c} P_{0j}^m \\ \text{ } \\ P_{1j}^m \\ \vdots \\ P_{kj}^m \\ \vdots \\ P_{wj}^m \end{array}\right]$
The product of the above row and column is the transition probability $P_{ij}^{n+m}$.
Powers of the One-Step Transition Probability Matrix
Let $\mathbf{P}$ be the one-step transition probability matrix of a Markov chain. Let $\mathbf{Q}_n$ be the $n$-step transition probability matrix, which can be computed by using the Chapman-Kolmogorov equations. We now derive another important fact.
The $n$-step transition probability matrix $\mathbf{Q}_n$ is obtained by multiplying the one-step transition probability matrix $\mathbf{P}$ by itself $n$ times, i.e. $\mathbf{Q}_n$ is the $n$th power of $\mathbf{P}$.
This fact is important in terms of calculation of transition probabilities. Compute $\mathbf{P}^n$, the $n$th power of $\mathbf{P}$ (in terms of matrix calculation). Then $P_{ij}^n$ is simply an entry in $\mathbf{P}^n$ (in the row that corresponds to state $i$ and in the column that corresponds to state $j$). If the matrix size is large and/or $n$ is large, the matrix multiplication can be done using software or by using an online matrix calculator (here’s one matrix calculator).
Of course, the above fact is also important Markov chain theory in general. Almost all mathematical properties of Markov chains are at root based on this basic fact.
We can establish this basic fact using an induction argument. Clearly $\mathbf{Q}_1=\mathbf{P}^1=\mathbf{P}$. Suppose that the fact is true for $n-1$. Based on (4), $\mathbf{Q}_{n}=\mathbf{Q}_{n-1} \times \mathbf{Q}_1$. Continuing the derivation: $\mathbf{Q}_{n}=\mathbf{Q}_{n-1} \times \mathbf{Q}_1=\mathbf{P}^{n-1} \times \mathbf{P}=\mathbf{P}^n$.
The Row Vector and the Column Vector
As demonstrated in the preceding section, $\mathbf{P}^n$, the $n$th power of the $\mathbf{P}$, is precisely the $n$-step transition probability matrix. Let’s examine the matrix $\mathbf{P}^n$ more closely. Suppose that the Markov chain has a state space $\left\{0,1,2,\cdots,w \right\}$. The following shows the matrix $\mathbf{P}^n$ with special focus on a generic row and a generic column.
$\mathbf{P}^n= \left[\begin{array}{ccccccc} P_{0,0}^n & P_{0,1}^n & \cdots & P_{0,k}^n & \cdots & P_{0,w}^n \\ P_{1,0}^n & P_{1,1}^n & \cdots & P_{1,k}^n & \cdots & P_{1,w}^n \\ \vdots & \vdots & \vdots & \vdots & \vdots & \vdots \\ P_{i,0}^n & P_{i,1}^n & \cdots & P_{i,k}^n & \cdots & P_{i,w}^n \\ \vdots & \vdots & \vdots & \vdots & \vdots & \vdots \\ P_{w,0}^n & P_{w,1}^n & \cdots & P_{w,k}^n & \cdots & P_{w,w}^n \\ \end{array}\right]$
Now look at the row corresponding to the state $i$ and call it $R_i$. Also look at the column corresponding to the state $k$ and call it $C_k$. They are separated from the matrix $\mathbf{P}^n$ below.
$R_i=\left[\begin{array}{cccccc} P_{i,0}^n & P_{i,1}^n & \cdots & P_{i,k}^n & \cdots & P_{i,w}^n \\ \end{array}\right] \ \ \ \ \ \text{row vector}$
$C_k=\left[\begin{array}{c} P_{0,k}^n \\ \text{ } \\ P_{1,k}^n \\ \vdots \\ P_{i,k}^n \\ \vdots \\ P_{w,k}^n \end{array}\right] \ \ \ \ \ \text{column vector}$
The row vector $R_i$ is a conditional distribution. It gives the probabilities of the process transitioning (after $n$ transitions) into one of the states given the process starts in state $i$. If it is certain that the process begins in state $i$, then $R_i$ gives the probability function for the random variable $X_n$ since $P[X_n=k]=P_{i,k}^n$.
On the other hand, if the initial state is not certain, then the column vector $C_k$ gives the probabilities of the process transitioning (after $n$ transitions) into state $k$ from any one of the possible initial states. Thus a column from the matrix $\mathbf{P}^n$ gives the probability of the process in a given state at the $n$th period from all possible initial states. The following gives the probability distribution for $X_n$ (the state of the Markov chain at time $n$).
$\displaystyle (5) \ \ \ \ \ P[X_n=k]=\sum \limits_{i=0}^\infty P_{i,k}^n \times P[X_0=i]$
The calculation in (5) can be viewed as matrix calculation. If we put the probabilities $P[X_0=i]$ in a row vector, then the product of this row vector with the column vector $C_k$ would be $P[X_n=k]$. The following is (5) in matrix multiplication.
$(6) \ \ \ \ \ P[X_n=k]=\left[\begin{array}{cccccc} P(X_0=0) & P(X_0=1) & \cdots & P(X_0=j) & \cdots & P(X_0=w) \\ \end{array}\right] \left[\begin{array}{c} P_{0k}^n \\ \text{ } \\ P_{1k}^n \\ \vdots \\ P_{jk}^n \\ \vdots \\ P_{wk}^n \end{array}\right]$
Additional Comments on Chapman-Kolmogorov
Even though the matrix calculation for $P_{ij}^n$ should be done using software, it pays to understand the orientation of the matrix $\mathbf{P}^n$. In the preceding section, we learn that a row of $\mathbf{P}^n$ is the conditional distribution $X_n \lvert X_0$ (the state of the process at time $n$ given the initial state). In the preceding section, we also learn that a column in the matrix $\mathbf{P}^n$ gives the probabilities of the process landing in a fixed state $k$ from any one of the initial states.
The Chapman-Kolmogorov equations in (3) tells us that an entry in the matrix $\mathbf{P}^{n+m}$ is simply the product of a row in $\mathbf{P}^n$ and a column in $\mathbf{P}^m$. This observation makes it possible to focus just on the transition probability that is asked in a given problem rather than calculating the entire matrix. For example, to calculate an entry $P_{ij}^2$ of the matrix $\mathbf{P}^2$, multiply a row of $\mathbf{P}$ (the row corresponding to the state $i$) and a column of $\mathbf{P}$ (the row corresponding to state $j$). There is no need to calculate the entire matrix $\mathbf{P}^2$ if the goal is just one entry of $\mathbf{P}^2$.
To calculate an entry of $\mathbf{P}^n$, there is a considerable amount of variation in multiplication approaches. For example, multiple a row of $\mathbf{P}$ and a column of $\mathbf{P}^{n-1}$ or multiple a row of $\mathbf{P}^{n-1}$ and a column of $\mathbf{P}$. Or multiple a row of $\mathbf{P}^2$ and a column of $\mathbf{P}^{n-2}$.
If a row of transition probabilities multiplies a transition matrix, the result is a row in a higher transition probability matrix. For example, if a row in $\mathbf{P}$ multiplies the matrix $\mathbf{P}^n$, the result is a row in $\mathbf{P}^{n+1}$. More specifically, the first row in $\mathbf{P}$ multiplying the matrix $\mathbf{P}$ gives the first row of $\mathbf{P}^2$. The first row of $\mathbf{P}$ multiplying the matrix $\mathbf{P}^2$ gives the first row in $\mathbf{P}^3$, etc.
On the other hand, when a transition probability matrix multiplies a column of transition probabilities, the result is a column in a higher transition probability matrix. For example, the matrix $\mathbf{P}^n$ multiplies a column in $\mathbf{P}$, the result is a column in $\mathbf{P}^{n+1}$. More specifically, when the matrix $\mathbf{P}$ multiples the first column in the matrix $\mathbf{P}$, the result is the first column of $\mathbf{P}^2$. When the matrix $\mathbf{P}^2$ multiples a column in the matrix $\mathbf{P}$, the result is the first column in $\mathbf{P}^3$, etc.
Examples
We now give some examples on how to calculate transition probabilities.
Example 1
Consider a Markov chain with the following transition probability matrix.
$\mathbf{P} = \bordermatrix{ & 0 & 1 & 2 \cr 0 & 0.6 & 0.2 & 0.2 \cr 1 & 0.3 & 0.5 & 0.2 \cr 2 & 0.4 & 0.3 & 0.3 \cr } \qquad$
Determine $P[X_3=1 \lvert X_0=0]$ and $P[X_5=2 \lvert X_2=0]$.
Of course, the most straightforward way would be to calculate $\mathbf{P}^3$. Then $P[X_3=1 \lvert X_0=0]=P_{01}^3$ would be the (0,1)th entry of $\mathbf{P}^3$ and $P[X_5=2 \lvert X_2=0]=P_{02}^3$ would be the (0,1)th entry of $\mathbf{P}^3$. In fact, it is a good practice to use an online matrix calculator for this task. Doing so produces the following marrices.
$\mathbf{P}^2 = \bordermatrix{ & 0 & 1 & 2 \cr 0 & 0.50 & 0.28 & 0.22 \cr 1 & 0.41 & 0.37 & 0.22 \cr 2 & 0.45 & 0.32 & 0.23 \cr } \qquad \ \ \ \ \ \ \mathbf{P}^3 = \bordermatrix{ & 0 & 1 & 2 \cr 0 & 0.472 & 0.306 & 0.222 \cr 1 & 0.445 & 0.333 & 0.222 \cr 2 & 0.458 & 0.319 & 0.223 \cr } \qquad$
From the matrix $\mathbf{P}^3$, we see that $P_{01}^3=0.306$ and $P_{02}^3=0.222$. Note that $P_{02}$ is calculated in the introductory example earlier.
Example 2
For the transition matrix $\mathbf{P}$ in Example 1, use the ideas discussed in the section Additional Comments on Chapman-Kolmogorov to compute the first row and the first column in the transition matrix $\mathbf{P}^4$.
$\left[\begin{array}{ccc} 0.6 & 0.2 & 0.2 \\ \end{array}\right] \left[\begin{array}{ccc} 0.472 & 0.306 & 0.222 \\ 0.445 & 0.333 & 0.222 \\ 0.458 & 0.319 & 0.223 \end{array}\right] = \left[\begin{array}{ccc} 0.4638 & 0.314 & 0.2222 \end{array}\right]= \left[\begin{array}{ccc} P_{00}^4 & P_{01}^4 & P_{02}^4 \end{array}\right]$
$\left[\begin{array}{ccc} 0.472 & 0.306 & 0.222 \\ 0.445 & 0.333 & 0.222 \\ 0.458 & 0.319 & 0.223 \end{array}\right] \left[\begin{array}{c} 0.6 \\ 0.3 \\ 0.4 \end{array}\right] = \left[\begin{array}{c} 0.4638 \\ 0.4557 \\ 0.4597 \end{array}\right]= \left[\begin{array}{c} P_{00}^4 \\ P_{10}^4 \\ P_{20}^4 \end{array}\right]$
As discussed in earlier, the first row of $\mathbf{P}^4$ is obtained by multiplying the first row of $\mathbf{P}$ with the matrix $\mathbf{P}^3$. The first column of $\mathbf{P}^4$ is obtained by multiplying the matrix $\mathbf{P}^3$ with the first column of the matrix $\mathbf{P}$.
Example 3
A particle moves among states 0, 1 and 2 according to a Markov process with the following transition probability matrix.
$\mathbf{P} = \bordermatrix{ & 0 & 1 & 2 \cr 0 & 0.6 & 0.3 & 0.1 \cr 1 & 0.3 & 0.3 & 0.4 \cr 2 & 0.3 & 0.2 & 0.5 \cr } \qquad$
Let $X_n$ be the position of the particle after the $n$th move. Suppose that at the beginning, the particle is in state 1. Determine the probability $P[X_2=k]$ where $k=0,1,2$.
Since the initial state of the process is certain, $P[X_2=k]=P[X_2=k \lvert X_0=1]=P_{1k}^2$. Thus the problem is to find the second row of $\mathbf{P}^2$.
$\left[\begin{array}{ccc} 0.3 & 0.3 & 0.4 \\ \end{array}\right] \left[\begin{array}{ccc} 0.6 & 0.3 & 0.1 \\ 0.3 & 0.3 & 0.4 \\ 0.3 & 0.2 & 0.5 \end{array}\right] = \left[\begin{array}{ccc} 0.39 & 0.26 & 0.35 \end{array}\right]= \left[\begin{array}{ccc} P_{10}^2 & P_{11}^2 & P_{12}^2 \end{array}\right]$
Example 4
Consider a Markov chain with the following transition probability matrix.
$\mathbf{P} = \bordermatrix{ & 0 & 1 & 2 \cr 0 & 0.3 & 0.2 & 0.5 \cr 1 & 0.1 & 0.3 & 0.6 \cr 2 & 0.4 & 0.3 & 0.3 \cr } \qquad$
Suppose that initially the process is equally likely to be in state 0 or state 1. Determine $P[X_3=0]$ and $P[X_3=1]$.
To find $P[X_3=0]$, we need the first column of $\mathbf{P}^3$. To find $P[X_3=1]$, we need the second column of $\mathbf{P}^3$. Using an online calculator produces $\mathbf{P}^3$.
$\mathbf{P}^3 = \bordermatrix{ & 0 & 1 & 2 \cr 0 & 0.288 & 0.269 & 0.443 \cr 1 & 0.283 & 0.270 & 0.447 \cr 2 & 0.295 & 0.273 & 0.432 \cr } \qquad$
The following calculation gives the answers.
$P[X_3=0]=\left[\begin{array}{ccc} 0.5 & 0.5 & 0 \\ \end{array}\right] \left[\begin{array}{c} 0.288 \\ 0.283 \\ 0.295 \end{array}\right]=0.2855$
$P[X_3=1]=\left[\begin{array}{ccc} 0.5 & 0.5 & 0 \\ \end{array}\right] \left[\begin{array}{c} 0.269 \\ 0.270 \\ 0.273 \end{array}\right]=0.2695$
$P[X_3=2]=1-P[X_3=0]-P[X_3=1]=0.445$
Example 5
Consider a Markov chain with the following transition probability matrix.
$\mathbf{P} = \bordermatrix{ & 0 & 1 & 2 \cr 0 & 0.6 & 0.2 & 0.2 \cr 1 & 0.2 & 0.2 & 0.6 \cr 2 & 0 & 0 & 1 \cr } \qquad$
At time 0, the Markov process is in state 0. Let $T=\text{min}\left\{n \ge 0:X_n=2 \right\}$. Note that state 2 is called an absorbing state. The Markov process will eventually reach and be absorbed in state 2 (it stays there forever whenever the process reaches state 2). Thus $T$ is the first time period in which the process reaches state 2. Suppose that the Markov process is being observed and that absorption has not taken place. Then we would be interested in this conditional probability: the probability that the process is in state 0 or state 1 given that the process has not been absorbed. Determine $P[X_4=0 \lvert T>4]$.
The key idea here is that for the event $T>4$ to happen, $X_4=0$ or $X_4=1$. Thus
$P[T>4]=P[X_4=0 \text{ or } X_4=1]=P[X_4=0]+P[X_4=1]=P_{00}^4+P_{01}^4$.
Note that since the initial state is certain to be state 0, $P[X_4=0]=P_{00}^4$ and $P[X_4=1]=P_{01}^4$. Using an online calculator gives the following matrix.
$\mathbf{P}^4 = \bordermatrix{ & 0 & 1 & 2 \cr 0 & 0.1856 & 0.0768 & 0.7376 \cr 1 & 0.0768 & 0.032 & 0.8912 \cr 2 & 0 & 0 & 1 \cr } \qquad$
The following gives the desired result.
$\displaystyle P[X_4=0 \lvert T>4]=\frac{P_{00}^4}{P_{00}^4+P_{01}^4}=\frac{0.1856}{0.1856+0.0768}=0.7073$
Thus if absorption has not taken place in the 4th period, the process is in state 0 about 71% of the time.
Example 6
Continue with the Markov chain in Example 5. Determine $P[T=4]$.
Note that $P[T=4]=P[T>3]-P[T>4]$. The probability $P[T>4]$ is already determined using $\mathbf{P}^4$. To determine $P[T>3]$, we need the top row of $\mathbf{P}^3$.
$\mathbf{P}^3 = \bordermatrix{ & 0 & 1 & 2 \cr 0 & 0.272 & 0.112 & 0.616 \cr 1 & 0.112 & 0.048 & 0.84 \cr 2 & 0 & 0 & 1 \cr } \qquad$
Thus $P[T>3]=P_{00}^3+P_{01}^3=0.272+0.112=0.384$. Thus $P[T=4]=0.384-0.2624=0.1216$. Thus about 12% of the time, absorption takes place in the 4th period about 12% of the time.
Exercises
We give a few practice problems to reinforce the concepts and calculation discussed here.
Exercise 1
Consider a Markov chain with the following transition probability matrix.
$\mathbf{P} = \bordermatrix{ & 0 & 1 & 2 \cr 0 & 0.1 & 0.2 & 0.7 \cr 1 & 0.2 & 0.2 & 0.6 \cr 2 & 0.6 & 0.1 & 0.3 \cr } \qquad$
Determine these conditional probabilities.
• $P[X_3=2 \lvert X_1=1]$
• $P[X_3=2 \lvert X_0=1]$
• $P[X_4=2 \lvert X_0=1]$
Exercise 2
A particle moves among states 1, 2 and 3 according to a Markov process with the following transition probability matrix.
$\mathbf{P} = \bordermatrix{ & 1 & 2 & 3 \cr 1 & 0 & 0.6 & 0.4 \cr 2 & 0.6 & 0 & 0.4 \cr 3 & 0.6 & 0.4 & 0 \cr } \qquad$
Let $X_n$ be the position of the particle after the $n$th move. Determine the probability $P[X_3=1 \lvert X_0=1]$ and $P[X_4=1 \lvert X_0=1]$.
Exercise 3
Consider a Markov chain with the following transition probability matrix.
$\mathbf{P} = \bordermatrix{ & 0 & 1 & 2 \cr 0 & 0.1 & 0.2 & 0.7 \cr 1 & 0.9 & 0.1 & 0 \cr 2 & 0.1 & 0.8 & 0.1 \cr } \qquad$
Suppose that initially the process is equally likely to be in state 0 or state 1 or state 2. Determine the distribution for $X_3$.
Exercise 4
Continue with Example 5 and Example 6. Work these two examples assuming that the Markov chain starts in state 1 initially.
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Answers to Exercises
Exercise 1
• $P[X_3=2 \lvert X_1=1]=P_{12}^2=0.13$
• $P[X_3=2 \lvert X_0=1]=P_{12}^3=0.16$
• $P[X_4=2 \lvert X_0=1]=P_{12}^4=0.1473$
Exercise 2
• $P[X_3=1 \lvert X_0=1]=P_{11}^3=0.24$
• $P[X_4=1 \lvert X_0=1]=P_{11}^4=0.456$
Exercise 3
• $\displaystyle P[X_3=0]=\frac{1.076}{3}=0.3587$
• $\displaystyle P[X_3=1]=\frac{1.013}{3}=0.3377$
• $\displaystyle P[X_3=2]=\frac{0.911}{3}=0.3037$
Exercise 4
• $\displaystyle P[X_4=0 \lvert T>4]=\frac{P_{10}^4}{P_{10}^4+P_{11}^4}=\frac{0.0768}{0.0768+0.0.032}=0.70588$
• $P[T=4]=P[T>3]-P[T>4]=0.16-0.1088=0.0512$
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$\copyright$ 2017 – Dan Ma
# A beginning look at transition probabilities
The previous post introduces the notion of Markov chains, more specifically discrete Markov chains. This and the next post focus on calculating transition probabilities.
Suppose that $\left\{X_n: n \ge 0 \right\}$ is a Markov chain with transition probability matrix $\mathbf{P}$. Elements of $\mathbf{P}$ are the one-step transition probabilities $P_{ij}$. As the Markov process moves through the states over time, the probabilities in the matrix $\mathbf{P}$ shows how likely the process will transition to state $j$ in the next time period if the process is currently in state $i$. Elements of $\mathbf{P}$ are conditional probabilities:
$(1) \ \ \ \ \ P_{ij}=P[X_{n+1}=j \lvert X_n=i]$
The transition probabilities described in (1) are independent of the time period $n$. These are called stationary transition probabilities. Markov chains with stationary transition probabilities are called stationary Markov chains or time-homogeneous Markov chains. One immediate goal is to develop the $n$-step transition probabilities $P_{ij}^n$ from the one-step transition probabilities $P_{ij}$. This is to be developed in the next post. This post focuses on basic calculations with $P_{ij}$.
Examples of Conditional Probabilities
We use examples to demonstrate conditional probabilities that can be calculated from the one-step transition probabilities $P_{ij}$. The calculation is based on the following transition probability matrix.
$\mathbf{P} = \bordermatrix{ & 0 & 1 & 2 \cr 0 & 0.6 & 0.2 & 0.2 \cr 1 & 0.3 & 0.5 & 0.2 \cr 2 & 0.4 & 0.3 & 0.3 \cr } \qquad$
Example 1
Determine the following conditional probabilities:
• $P[X_1=0, X_2=1 \lvert X_0=0]$
• $P[X_2=0, X_3=1 \lvert X_1=0]$
• $P[X_1=2, X_2=1,X_3=0 \lvert X_0=1]$
Note that the first two are identical since the transition probabilities are stationary. To calculate the first and third probabilities, it is a matter of linking up the one-step transition probabilities from one state to the next.
\displaystyle \begin{aligned} P[X_1=0, X_2=1 \lvert X_0=0]&=P[X_1=0 \lvert X_0=0] \times P[X_2=1 \lvert X_1=0] \\&=P_{00} \times P_{01} \\&=0.6 \times 0.2=0.12 \end{aligned}
$\displaystyle P[X_2=0, X_3=1 \lvert X_1=0]=P[X_1=0, X_2=1 \lvert X_0=0]=0.12$
\displaystyle \begin{aligned} P[X_1=2, X_2=1,X_3=0 \lvert X_0=1]&=P[X_1=2 \lvert X_0=1] \times P[X_2=1 \lvert X_1=2] \times P[X_3=0 \lvert X_2=1] \\&=P_{12} \times P_{21} \times P_{10} \\&=0.2 \times 0.3 \times 0.3=0.018 \end{aligned}
Example 2
Example 1 shows that the probability of a chain moving from along a path conditional on a starting state is simply the product of the transition probabilities on that path. We now calculate unconditional probability of a path. For this to be possible, we need to assume a distribution for the initial state.
Assume that $P[X_0=0]=0.4$, $P[X_0=1]=0.3$ and $P[X_0=2]=0.3$. Determine the following unconditional probabilities:
• $P[X_0=0,X_1=0,X_2=1]$
• $P[X_1=0,X_2=0,X_3=1]$
• $P[X_0=1,X_1=2, X_2=1,X_3=0]$
To calculate these unconditional probabilities, we multiply the transition probabilities on the path along with the probability of the first state in the path.
\displaystyle \begin{aligned} P[X_0=0,X_1=0,X_2=1]&=P[X_0=0] \times P[X_1=0 \lvert X_0=0] \times P[X_2=1 \lvert X_1=0] \\&=P[X_0=0] \times P_{00} \times P_{01} \\&=0.4 \times 0.6 \times 0.2=0.048 \end{aligned}
To calculate the second probability, we need to first compute $P[X_1=0]$. This can be done by conditioning on the initial state $X_0$.
\displaystyle \begin{aligned} P[X_1=0]&=P[X_0=0] \times P_{00}+ P[X_0=1] \times P_{10} +P[X_0=2] \times P_{20} \\&=0.4 \times 0.6 +0.3 \times 0.3+0.3 \times 0.4=0.45 \end{aligned}
\displaystyle \begin{aligned} P[X_1=0,X_2=0,X_3=1]&=P[X_1=0] \times P[X_2=0 \lvert X_1=0] \times P[X_3=1 \lvert X_2=0]\\&=P[X_1=0] \times P_{00} \times P_{01} \\&=0.45 \times 0.6 \times 0.2 \\&=0.054 \end{aligned}
The following gives the third probability.
\displaystyle \begin{aligned} P[X_0=1,X_1=2, X_2=1,X_3=0]&=P[X_0=1] \times P[X_1=2 \lvert X_0=1] \times P[X_2=1 \lvert X_1=2] \times P[X_3=0 \lvert X_2=1] \\&=P[X_0=1] \times P_{12} \times P_{21} \times P_{10} \\&=0.4 \times 0.2 \times 0.3 \times 0.3 \\&=0.0072 \end{aligned}
Note that each unconditional probability in this example is obtained by multiplying the corresponding conditional probability in Example 1 with the probability of the conditioned event.
Example 3
Sometimes we know for certain that the Markov chain starts in a given state. Then the unconditional probability of a path is simply the conditional probability of that path (conditioned on the initial state). Determine the three probabilities in Example 2 given that the chain starts in state 0 in the first two probabilities and the chain starts in state 1 in the third probability. The probabilities are:
• $P[X_0=0,X_1=0,X_2=1]$
• $P[X_1=0,X_2=0,X_3=1]$
• $P[X_0=1,X_1=2, X_2=1,X_3=0]$
For the first probability, $P[X_0=0,X_1=0,X_2=1]=P[X_0=0] \times P[X_1=0,X_2=1 \lvert X_0=0]$. With $P[X_0=0]=1$, $P[X_0=0,X_1=0,X_2=1]=P[X_1=0,X_2=1 \lvert X_0=0]=0.12$, the answer from Example 1.
For the second probability, we need to first calculate $P[X_1=0]$. Since $P[X_0=0]=1$, $P[X_1=0]=P[X_0=0] \times P_{00}=0.6$. Thus $P[X_1=0,X_2=0,X_3=1]=P[X_1=0] \times P_{00} \times P_{01}=0.6 \times 0.6 \times 0.2=0.072$, which is 0.6 times the corresponding conditional probability in Example 1.
Similarly, the third probability is
\displaystyle \begin{aligned} P[X_0=1,X_1=2, X_2=1,X_3=0]&=P[X_1=2, X_2=1,X_3=0 \lvert X_0=1] \\& =0.18\end{aligned}
Conditional Probabilities of Paths in Markov Chains
The conditional probabilities in the above examples are not difficult to do as long as the basic idea of conditional probabilities and the Markov property are understood. First, the idea of conditional probability. The idea is that the probability of the intersection of two events can be obtained by conditioning on one of the events. For example, given two events $A$ and $B$, we have
$(2a) \ \ \ \ \ P[A \cap B]=P[A] \times P[B \lvert A]$
$\displaystyle (2b) \ \ \ \ \ P[B \lvert A]=\frac{P[A \cap B]}{P[A]}$
The probability of the intersection of two events is the product of a conditional probability of one event given another event times the probability of the event being conditioned. Relation (2b) is the probability of the event $B$ given that event $A$ has occurred. Of course, the conditioned event has to have positive probability, i.e. $P[A]>0$.
Suppose that the event $A$ is $X=x$, the discrete random variable $X$ taking on the value $x$ and that the event $B$ is $Y=y$. Then we have the following relations.
$(3a) \ \ \ \ \ P[X=x,Y=y]=P[X=x] \times P[Y=y \lvert X=x]$
$\displaystyle (3b) \ \ \ \ \ P[Y=y \lvert X=x]=\frac{P[X=x,Y=y]}{P[X=x]}$
We can also write the conditional probability of $X=x$ conditioning on the event $Y=y$. Suppose we have a path in a Markov chain, $X_0=x_0,X_1=x_1,X_2=x_2,\cdots,X_n=x_n$. This is the event that the chain starts in state $x_0$ initially, then move to state $x_1$ in time 1 and so on and at time $n$ the chain is in state $x_0$. Conditioning on the initial state, consider the following conditional probability.
$\displaystyle (4) \ \ \ \ \ P[X_1=x_1,X_2=x_2,\cdots,X_n=x_n \lvert X_0=x_0]=\frac{P[X_0=x_0,X_1=x_1,X_2=x_2,\cdots,X_n=x_n]}{P[X_0=x_0]}$
Relation (4) can be further simplified since the random variables $X_0=x_0,X_1=x_1,X_2=x_2,\cdots,X_n=x_n$ satisfy the Markov property.
\displaystyle \begin{aligned} (5) \ \ \ \ \ P[X_1=x_1,X_2=x_2,\cdots,X_n=x_n \lvert X_0=x_0]&=P[X_1=x_1 \lvert X_0=x_0] \times P[X_2=x_2 \lvert X_1=x_1] \\& \ \ \ \times \cdots \times P[X_n=x_n \lvert X_{n-1}=x_{n-1}] \\&=P_{x_0,x_1} \times P_{x_1,x_2} \times \cdots \times P_{x_{n-1},x_n} \end{aligned}
Relation (5) says that the probability of a path in a Markov chain conditioning on the initial state is simply the product of the transition probabilities along the path. This is derived using the Markov property. Relation (5) can be established by an induction argument on the subscript $n$, i.e. the length of a path.
Clearly $P[X_1=j \lvert X_0=i]=P_{ij}$. This is the fundamental assumption in Markov chains. Now consider $P[X_1=j,X_2=k \lvert X_0=i]$.
\displaystyle \begin{aligned} P[X_1=j,X_2=k \lvert X_0=i]&=\frac{P[X_0=i,X_1=j,X_2=k]}{P[X_0=i]} \\&=\frac{P[X_0=i,X_1=j,X_2=k]}{P[X_0=i,X_1=j]} \times \frac{P[X_0=i,X_1=j]}{P[X_0=i]} \\&=P[X_2=k \lvert X_0=i,X_1=j] \times P[X_1=j \lvert X_0=i] \\&=P[X_2=k \lvert X_1=j] \times P[X_1=j \lvert X_0=i] \\&=P_{jk} \times P_{ij} \\&=P_{ij} \times P_{jk} \end{aligned}
Note that $P[X_2=k \lvert X_0=i,X_1=j]=P[X_2=k \lvert X_1=j]$ in the above derivation. This is Markov property in action since the probability of the future state depends only on the state immediately preceding it. Now that relation (5) is true for a path of length 2, the following shows how it is derived for length 3.
\displaystyle \begin{aligned} P[X_1=j,X_2=k,X_3=l \lvert X_0=i]&=\frac{P[X_0=i,X_1=j,X_2=k,X_3=l]}{P[X_0=i]} \\&=\frac{P[X_0=i,X_1=j,X_2=k,X_3=l]}{P[X_0=i,X_1=j,X_2=k]} \times \frac{P[X_0=i,X_1=j,X_2=k]}{P[X_0=i]} \\&=P[X_3=l \lvert X_0=i,X_1=j,X_2=k] \times P[X_1=j,X_2=k \lvert X_0=i] \\&=P[X_3=l \lvert X_2=k] \times P[X_1=j,X_2=k \lvert X_0=i] \\&=P_{ij} \times P_{jk} \times P_{kl} \end{aligned}
Note that going from the third line to the fourth line the Markov property is used. Relation (5) for length 2 is used at the end. The inductive proof keeps going and relation (5) is true for any path of arbitrary length.
Thus the conditional probability in (4) may not be special. With the additional assumption of Markov property, the conditional probability can be made to be a “chain” of one-step transition probabilities.
Using the idea of (3a), the unconditional probability of a path is:
\displaystyle \begin{aligned} (6) \ \ \ \ \ P[X_0=x_0,X_1=x_1,X_2=x_2,\cdots,X_n=x_n]&=P[X_0=x_0] \times P_{x_0,x_1} \\& \ \ \times P_{x_1,x_2} \times \cdots \times P_{x_{n-1},x_n} \end{aligned}
Relation (6) says that the unconditional probability of a path in a Markov chain is simply the conditional probability of the path conditioning on the initial state times the probability of the initial state.
More Examples
The earlier examples demonstrate relation (5) and relation (6). We now work a few more examples.
Example 4
There is a slightly different way to look at one unconditional probability $P[X_1=0,X_2=0,X_3=1]$ in Example 2. The first state in this path is $X_1=0$. We can condition on the initial state $X_0$.
$\displaystyle P[X_1=0,X_2=0,X_3=1]=\sum \limits_{k=0}^2 P[X_0=k,X_1=0,X_2=0,X_3=1]$
Then every one of the unconditional probability within the summation sign can be computed according to the idea in (6). Note that Example 2 gives the distribution of the initial state.
\displaystyle \begin{aligned} P[X_1=0,X_2=0,X_3=1]&=P[X_0=0,X_1=0,X_2=0,X_3=1]+ \\& \ \ P[X_0=1,X_1=0,X_2=0,X_3=1]+P[X_0=2,X_1=0,X_2=0,X_3=1] \\&=P[X_0=0] \times P_{00} \times P_{00} \times P_{01} +\\&\ \ P[X_0=1] \times P_{10} \times P_{00} \times P_{01}+P[X_0=2] \times P_{20} \times P_{00} \times P_{01} \\&=0.4 \times 0.6 \times 0.6 \times 0.2+\\&\ \ 0.3 \times 0.3 \times 0.6 \times 0.2 + 0.3 \times 0.4 \times 0.6 \times 0.2\\&=0.054 \end{aligned}
In calculating probabilities in a Markov chain, conditioning on the first step is a useful technique that will be used often in the subsequent posts.
Example 5
Consider a Markov chain with the following transition probability matrix.
$\mathbf{P} = \bordermatrix{ & 0 & 1 & 2 & 3 \cr 0 & 0.5 & 0.2 & 0.2 & 0.1 \cr 1 & 0.3 & 0.3 & 0.2 & 0.2 \cr 2 & 0.3 & 0.2 & 0.2 & 0.3 \cr 3 & 0 & 0 & 0 & 1 } \qquad$
State 3 is called an absorbing state since $P_{33}=1$. In other words, when the chain reaches state 3, it will never leave. Suppose that initially the Markov chain is in state 0 50% of the time, in state 1 30% of the time and in state 2 20% of the time. Calculate the probability $P[X_1=0,X_2=1,X_3=0,X_4=2]$.
We can condition on the initial state as in Example 4.
\displaystyle \begin{aligned} P[X_1=0,X_2=1,X_3=0,X_4=2]&=\sum \limits_{k=0}^2 P[X_0=k,X_1=0,X_2=1,X_3=0,X_4=2] \\&=P[X_0=0,X_1=0,X_2=1,X_3=0,X_4=2]+ \\&\ \ P[X_0=1,X_1=0,X_2=1,X_3=0,X_4=2]+\\&\ \ P[X_0=2,X_1=0,X_2=1,X_3=0,X_4=2] \\&=P[X_0=0] \times P_{00} \times P_{01} \times P_{10} \times P_{02}+ \\& \ \ P[X_0=1] \times P_{10} \times P_{01} \times P_{10} \times P_{02}+\\&\ \ P[X_0=2] \times P_{20} \times P_{01} \times P_{10} \times P_{02} \\&=0.0048\end{aligned}
Example 6
This is Example 1 in this previous post.
When a binary digit, 0 or 1, is transmitted through a communication system, it passes through several stages. At each stage, there is a probability $p$ that the digit is transmitted in error. Let $X_n$ be the digit that is the output at the $n$th stage. Then $\left\{X_n: n \ge 0 \right\}$ is a two-state Markov chain with the following transition probability matrix.
$\mathbf{P} = \bordermatrix{ & 0 & \text{ } & 1 \cr 0 & 1-p & \text{ } & p \cr \text{ } & \text{ } & \text{ } &\text{ } \cr 1 & p & \text{ } & 1-p \cr }$
Determine the following:
• If the digit to be sent is 0, determine the probability that no error occurs up to the second stage.
• If the digit to be sent is 0, determine the probability that a correct message is received at stage 2 through stage 4.
The first probability is $P[X_1=0,X_2=0 \lvert X_0=0]$. This is $P_{00} \times P_{00}=(1-p)^2$. To calculate the second probability, we can condition on the state in time 1. We first calculate $P[X_1=0]$ and $P[X_1=1]$. Since we know for certain that the initial state is 0, $P[X_1=0]=P_{00}=1-p$ and $P[X_1=1]=P_{01}=p$.
\displaystyle \begin{aligned} P[X_2=0,X_3=0,X_4=0]&=P[X_1=0,X_2=0,X_3=0,X_4=0]+P[X_1=1,X_2=0,X_3=0,X_4=0] \\&=P[X_2=0,X_3=0,X_4=0 \lvert X_1=0] \times P[X_1=0]+ \\&\ \ P[X_2=0,X_3=0,X_4=0 \lvert X_1=1] \times P[X_1=1] \\&=P_{00} \times P_{00} \times P_{00} \times (1-p)+ \\&\ \ P_{10} \times P_{00} \times P_{00} \times p\\&=(1-p)^4+p^2 \times (1-p)^2 \end{aligned}
Remarks
The calculation in this post involves very basic calculation with one-step transition probabilities, i.e. the elements of the transition probability matrix given in a Markov chain. The $n$-step transition probabilities are key to the development of the theory of Markov chains. The next post begins the discussion of Chapman-Kolmogorov equations.
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$\copyright$ 2017 – Dan Ma
# Introducing Markov Chains
This post continues the discussion of the previous two posts – the previous post that looks at random walks informally through graphs and the previous post that defines random walks more formally. This post introduces the notion of Markov chains.
Stochastic Process
A stochastic process is a collection of random variables $\left\{X_t: t \in T \right\}$. Each element of the collection is a random variable $X_t$. The index $t$ is often regarded as time. Thus we refer to $X_t$ the state of the process at time $t$. Thus a stochastic process is a family of random variables that describes the evolution through time of some physical process.
The set $T$ is called the index set of the stochastic process. When $T$ is a countable set, e.g. $T=\left\{0,1,2,3,\cdots \right\}$, the stochastic process is said to be a discrete-time process. When the index set $T$ is an interval of the real number line, the stochastic process is said to be a continuous-time process. For example, $\left\{X_n: n=0,1,2,\cdots \right\}$ is a discrete-time stochastic process indexed by the non-negative integers whereas $\left\{X_t: t \ge 0 \right\}$ is a continuous-time stochastic process indexed by the non-negative real numbers. The state space of a stochastic process is defined as the set of all possible values that can be taken by the random variables $X_t$.
In the present discussion, we focus on discrete-time stochastic processes $\left\{X_n: n=0,1,2,\cdots \right\}$ where the state space $S$ is finite or countably infinite. When the state space is finite, the process is said to be a finite-state stochastic process.
We also assume that the state space $S$ consists of integers. Thus $S$ is either $\mathbb{N}=\left\{0,1,2,3,\cdots \right\}$ (the natural numbers) or $\mathbb{Z}=\left\{\cdots,-3,-2,-1,0,1,2,3,\cdots \right\}$ (the integers), or some subset of $\mathbb{N}$ or $\mathbb{Z}$.
If we know nothing about $X_n$ other than the fact that they are random variables, then there is not much we can say about the stochastic process $\left\{X_n: n=0,1,2,\cdots \right\}$. To make the discussion more meaningful, it is necessary to impose some additional structure on these random variables.
The simplest structure we can assume is that the random variables $X_n$ are independent random variables. This would be a good model for phenomena that can be regarded as random experiments in which the future states of the process are independent of past and present states. However, in many situations the assumption of independence is often unjustified. In many stochastic processes that arise in practice, past and present states can influence the future states.
Markov Chains
We consider stochastic processes that possess the following property: the future states of the process (conditional on both past and present states) depends only upon the present state, not on the past states leading up to the present state. This property is called the Markov property. Any stochastic process that possesses the Markov property is called a Markov chain. More specifically, to satisfy the Markov property, given the present state $X_n$ and the past states $X_{n-1},X_{n-2},\cdots,X_2,X_1,X_0$, the conditional distribution of the future state $X_{n+1}$ depends only on the present state $X_n$. Even more specifically, to satisfy the Markov property, the process must satisfy the following requirement:
$\displaystyle (1) \ \ \ \ P[X_{n+1}=j \lvert X_n=i,X_{n-1}=x_{n-1},\cdots,X_0=x_0]=P[X_{n+1}=j \lvert X_n=i]$
for all possible states $x_0,x_1, \cdots,x_{n-1},i,j$ and all non-negative integers $n$. Thus under the Markov property, the conditional distribution of the future state $X_{n+1}$ depends only on the present state $X_n$ and that all the states preceding the present state have no influence on the future state.
The conditional probabilities $P[X_{n+1}=j \lvert X_n=i]$ in (1) are called transition probabilities of the Markov chain. In the present discussion, we consider only the Markov chains in which the transition probabilities $P[X_{n+1}=j \lvert X_n=i]$ are independent of the current period $n$. Such Markov chains are called time-homogeneous Markov chains or stationary Markov chains. So the additional assumption of time-homogeneity means that the probability of transitioning into state $j$ from state $i$ is identical regardless of where the process is in the time scale (at the beginning in the process or later in the process).
With the transition probability $P[X_{n+1}=j \lvert X_n=i]$ independent of $n$, we can denote this probability by the states $i$ and $j$ only:
$\displaystyle (2) \ \ \ \ P_{ij}=P[X_{n+1}=j \lvert X_n=i]$
The probability $P_{ij}$ represents the probability that the process, starting in state $i$, will enter into state $j$ in the next period. Since the process must transition into one of the states in the state space in the next time period, $P_{ij}$, with $i$ fixed and $j$ varying, must represent a probability distribution. Thus $P_{i0}+ P_{i1}+\cdots=1$. Here we assume that the state space is the set of all non-negative integers.
The probabilities $P_{ij}$ are called one-step transition probabilities since they give the probabilities of transitioning from one state to another state in one period of time. The one-step transition probabilities are usually expressed in a matrix or a transition diagram. The following shows what a transition probability matrix looks like.
$\mathbf{P}= \left[\begin{array}{cccccc} P_{0,0} & P_{0,1} & P_{0,2} & \cdots & P_{0,m} \\ P_{1,0} & P_{1,1} & P_{1,2} & \cdots & P_{1,m} \\ P_{2,0} & P_{2,1} & P_{2,2} & \cdots & P_{2,m} \\ \vdots & \vdots & \vdots & \vdots & \vdots \\ P_{m,0} & P_{m,1} & P_{m,2} & \cdots & P_{m,m} \\ \end{array}\right]$
The above transition matrix is for a finite-state Markov chain with state space being $\left\{0,1,\cdots,m \right\}$. The probabilities in each row sum to 1. If the process is currently in state $i$, the row with the first index as $i$ (in this case the $(i+1)$st row) will give the probabilities of transitioning into all the states. If the state space is countably infinite, then each row would have infinitely many terms and there would be infinitely many rows.
In many situations, it is helpful to have a column label and row label to specify the states in the Markov chain. For example, a particle moves through the states 0, 1, 2, 3 and 4 according to a Markov chain described by the following transition probability matrix.
$\mathbf{P} = \bordermatrix{ & 0 & 1 & 2 & 3 & 4 \cr 0 & 0 & 0.6 & 0.2 & 0.1 & 0.1 \cr 1 & 0.5 & 0 & 0.2 & 0.2 & 0.1 \cr 2 & 0.4 & 0.2 & 0 & 0.1 & 0.3 \cr 3 & 0.2 & 0.3 & 0.2 & 0 & 0.3 \cr 4 & 0.6 & 0.2 & 0.1 & 0.1 & 0 \cr } \qquad$
In the above matrix, $P_{31}=0.3$, which means that when the Markov process is in state 3, it moves to state 1 with probability 0.3. As discussed earlier, the probabilities in a given row sum to 1, which makes sense since the probabilities in a row describes all possible moves when the process is a given state.
When it is helpful to do so, the transition probability matrices will have a row label and a column label. If the row label and column label are not given, we assume that the rows and columns are listed in ascending order (recall that the states are integers). The transition probability matrices can be displayed using square brackets or parentheses.
Examples
The key to working with Markov chains is the transition probabilities $P_{ij}$. These probabilities allow us describe the random transitions through the states over time. The remainder of this post gives several examples of Markov chains focusing on transition probability matrices.
Example 1
When a binary digit, 0 or 1, is transmitted through a communication system, it passes through several stages. At each stage, there is a probability $p$ that the digit is transmitted in error. Let $X_n$ be the digit that is the output at the $n$th stage. Then $\left\{X_n: n \ge 0 \right\}$ is a two-state Markov chain with the following transition probability matrix.
$\mathbf{P} = \bordermatrix{ & 0 & \text{ } & 1 \cr 0 & 1-p & \text{ } & p \cr \text{ } & \text{ } & \text{ } &\text{ } \cr 1 & p & \text{ } & 1-p \cr }$
Some typical problems might be:
• If the digit to be sent is 0, determine the probability that no error occurs up to the second stage.
• If the digit to be sent is 0, determine the probability that a correct message is received at stage 2.
• If the digit to be sent is 0, determine the probability that a correct message is received at stage $n$ where $n \ge 1$.
Example 2
Let’s consider the example of gambler’s ruin with finitely many states. Suppose that a gambler makes a series of one-unit bets against the house. For the gambler, the probabilities of winning and losing each bet are $p$ and $1-p$, respectively. Whenever the capital reaches zero, the gambler is in ruin and his capital remains zero thereafter. On the other hand, if the capital of the gambler increases to $m$, a large level of capital, the gambler quits playing. Let $X_n$ be the capital of the gambler after the $n$th bet. Then $\left\{ X_n: n=0,1,2,\cdots \right\}$ is a random walk, as discussed in the previous post. Since the future capital $X_{n+1}$ depends only on the current level of capital $X_n$ (up one unit or down one unit), this is also a Markov chain. The following is the transition probability matrix.
$\mathbf{P} = \bordermatrix{ & 0 & 1 & 2 & 3 & 4 & \cdots & m \cr 0 & 1 & 0 & 0 & 0 & 0 & \cdots & 0 \cr 1 & 1-p & 0 & p & 0 & 0 & \cdots & 0 \cr 2 & 0 & 1-p & 0 & p & 0 &\cdots & 0 \cr 3 & 0 & 0 & 1-p & 0 & p &\cdots & 0 \cr \vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \vdots \cr m & 0 & 0 & 0 & 0 & 0 &\cdots & 1 } \qquad$
The above is an $(m+1) \times (m+1)$ matrix. The states are of the process are $0,1,2,\cdots,m$. The states $0$ and $m$ are absorbing states since the process would stay there and not leave once the process arrive at these two states. Note that $P_{00}=1$ and $P_{mm}=1$. State 0 would be the state of ruin. State $m$ would mean the gambler becomes fabulously rich (if $m$ is large).
For the rows in between, the transition probabilities are $P_{i,i-1}=1-p$ and $P_{i,i+1}=p$. In other words, from state $i$, the process would go to state $i+1$ with probability $p$ and would go down to $i-1$ with probability $1-p$. More specifically, if $m=5$, the following is the transition probability matrix.
$\mathbf{P} = \bordermatrix{ & 0 & 1 & 2 & 3 & 4 & 5 \cr 0 & 1 & 0 & 0 & 0 & 0 & 0 \cr 1 & 1-p & 0 & p & 0 & 0 & 0 \cr 2 & 0 & 1-p & 0 & p & 0 & 0 \cr 3 & 0 & 0 & 1-p & 0 & p & 0 \cr 4 & 0 & 0 & 0 & 1-p & 0 & p \cr 5 & 0 & 0 & 0 & 0 & 0 & 1 } \qquad$
Assuming that the gambler has initial capital $d$ which is positive but small in comparison to $m$. For example, $d=10$ and $m=1000$. Some interesting questions are:
• If the staring capital is 10, what is the probability that the gambler’s capital will be 5 after 8 plays of the game?
• What is the probability of ruin for the gambler, i.e. reaching state 0?
• What is the mean time until reaching the absorbing states (ruin or fabulously rich)?
Example 3 (Restless Mouse)
A maze has eight areas as shown below.
The rat moves through the areas in the maze at random. That is, if an area has exits to $w$ areas, the rat moves to each of these $w$ areas with probability $1/w$. Let $X_n$ be the area the mouse is located in after the $n$th move. Then $\left\{X_n: n \ge 0 \right\}$ is a Markov chain with the following transition probability matrix.
$\mathbf{P} = \bordermatrix{ & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 \cr 1 & 0 & 1/2 & 0 & 1/2 & 0 & 0 & 0 & 0 \cr 2 & 1/2 & 0 & 1/2 & 0 & 0 & 0 & 0 & 0 \cr 3 & 0 & 1/2 & 0 & 0 & 1/2 & 0 & 0 & 0 \cr 4 & 1/3 & 0 & 0 & 0 & 1/3 & 1/3 & 0 & 0 \cr 5 & 0 & 0 & 1/3 & 1/3 & 0 & 0 & 0 & 1/3 \cr 6 & 0 & 0 & 0 & 1/2 & 0 & 0 & 1/2 & 0 \cr 7 & 0 & 0 & 0 & 0 & 0 & 1/2 & 0 & 1/2 \cr 8 & 0 & 0 & 0 & 0 & 1/2 & 0 & 1/2 & 0 } \qquad$
Here’s some interesting questions about this Markov chain. Suppose that area 8 contains food and area 6 has a device for an electric shock.
• Given that the rat is placed in area 1 at the beginning, what is the probability that the rat will reach area 8 in 4 moves, in general in $n$ moves?
• Given that the rat is placed in area 1 at the beginning, what is the mean number of moves for the rat to reach area 8 or area 6?
• Given that the rat is placed in area 1 at the beginning, what is the probability that the rat will reach the food area before the electric shock area?
Next Steps
The concept of Markov chains has been introduced and illustrated with examples of random walks (in earlier posts) and other examples in this post. All the properties of Markov chains (in this case time-homogeneous Markov chains) are derived from the transition probability matrix. One of the most urgent tasks is to develop a way to calculation the transition probabilities. Recall that the elements of the transition probability matrix $\mathbf{P}$ are the one-step transition probabilities $P_{ij}$, which gives the probability of going from state $i$ into state $j$. The natural next step is to calculate the $n$-step transition probabilities $P^n_{ij}$, which is the probability that the process will in state $j$ in the $n$th period given the process is initially in state $i$. The next post is a beginning look at transition probabilities. The post that follows the next post is on Chapman-Kolmogorov equations, a systematic way of calculating transition probabilities using matrix multiplication.
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$\copyright$ 2017 – Dan Ma
# More on random walks
This post continues the discussion of the previous post on random walks. The previous post looks at random walks graphically. This post introduces the notion of random walks more formally.
Consider $\left\{X_n: n=0,1,2,\cdots \right\}$, which is a sequence of random variables indexed by the non-negative integers. By definition this is a stochastic process, to be precise a discrete stochastic process since the random variables are indexed by integers. The integers in the subscripts can be interpreted as time. Thus the random variables $X_n$ describe the states of some process over time. The stochastic processes discussed in the previous post, the future state $X_{n+1}$ of the process is obtained by adding one (with probability $p$) or subtracting one (with probability $1-p$) to the current state $X_n$. So a given state of the process $X_n$ is one higher or lower than the state in the previous period. The random walks in the previous post go up or down with equal chance, i.e. $p=0.5$. The following graph is from the previous post, which shows 5 simulations of such random walks.
Figure 1 – Examples of Random Walks in One Dimension
Random Walks – A General Description
We now give a more general description of the random walk. Instead of a random one-unit up or down move, the moves in the random walk are determined by a predetermined discrete distribution. Let $Y_1,Y_2,Y_3,\cdots$ be integer-valued random variables that are independent and identically distributed. Let $P[Y=y]$ be the common probability function. Let $X_0$ be an integer-valued random variable that is independent of the random variables $Y_n$. The random variable $X_0$ will be the initial position of the random walk. The random variables $Y_n$ are the increments (they are the amounts added to the stochastic process as time increases).
For $n \ge 1$, define $X_n=X_0+Y_1+\cdots+Y_{n-1}+Y_n$. Then $\left\{X_n: n=0,1,2,\cdots \right\}$ is called a random walk. Wherever the process is at any given time, the next step in the process is always a walk in a distance and direction that is dictated by the independent random variables $Y_n$. For convenience, we will call the random variables $Y_n$ the increments in the random walk since they tell the process how far (and what direction) to walk in the next step.
The state space is the set of all the values that can be taken on by the random variables $X_n$. In this case, the state space is either the set of all integers $\mathbb{Z}=\left\{\cdots,-3,-2,-2,0,1,2,3,\cdots \right\}$ or the set of all non-negative integers $\mathbb{N}=\left\{0,1,2,3,\cdots \right\}$ or some appropriate subset of $\mathbb{Z}$ or $\mathbb{N}$.
Given the current state of the process $X_n$, the next state $X_{n+1}$ is obtained by adding the random value $Y_{n+1}$ to the previous state $X_n$. If the random variables $Y_n$ are independent Bernoulli variables with $P(Y_n=1)=p$ and $P(Y_n=-1)=1-p$. Then each state $X_n$ is the result of an up or down move by one unit from the previous state. The examples in the previous post are based on this type of Bernoulli one-unit up and down move. In the general definition, the future state $X_{n+1}$, instead of being adjusted by a uniform up or down adjustment, is adjusted by the random variable $Y_{n+1}$.
One important task is to establish the probabilities of the transition from one state to another. For example, if the process is currently in state $i$, what is the probability that it will be in state $j$ in the next period? More specifically, given that the random walk is in state $i$, i.e. $X_n=i$, what is the probability that the process in in state $j$ in the next period, i.e. $X_{n+1}=j$? Such probabilities are called one-step transition probabilities.
The probabilities that we wish to obtain are $P[X_{n+1}=j \lvert X_n=i]$ for $n \ge 0$. Recall that $P[Y=y]$ is the common probability function for the random variables $Y_n$. Given states $i,j$, define $P_{ij}=P[Y=j-i]$.
When $X_n=i$, in order for $X_{n+1}=j$ to happen, the increment must be $Y_{n+1}=j-i$. Note that $P[Y_{n+1}=j-i]=P[Y=j-i]=P_{ij}$. Observe that this probability is independent of the time period $n$ since the $Y_n$ have a common distribution $Y$. Thus it is appropriate to use $P_{ij}$ to express this probability, i.e. there is no need to include the time $n$ in the subscript of $P_{ij}$. For all $n \ge 0$,
$P[X_{n+1}=j \lvert X_n=i]=P[Y_{n+1}=j-i]=P[Y=j-i]=P_{ij}$
The number $P_{ij}$ is the probability of the process transitioning from state $i$ to state $j$ in one step, which is evaluated based on the common probability function of the increments $Y_n$.
Now that we know the probability of the process going one state to another state in one step, other probability questions are: what is the probability of a path and what is the probability of going from state $i$ to state $j$ in $n$ steps? The probabilities in the second question are called $n$-step transition probabilities. The probability of a path is discussed here. The $n$-step transition probabilities are discussed here.
The first question has straightforward answers. For example, if the process starts in state $i_0$ at time 0, moves to state $i_1$ at time 1 and so on until reaching state $i_n$ at time $n$, then the following is the probability of this path:
\displaystyle \begin{aligned} P[X_0=i_0,X_1=i_1,\cdots,X_n=i_n]&=P[X_0=i_0,Y_1=i_1-i_0,\cdots,Y_n=i_n-i_{n-1}] \\&=P[X_0=i_0] \times P[Y_1=i_1-i_0] \times \cdots \times P[Y_n=i_n-i_{n-1}] \\&=P[X_0=i_0] \times P_{i_0,i_1} \times \cdots \times P_{i_{n-1},i_n} \end{aligned}
\displaystyle \begin{aligned} P[X_1=i_1,\cdots,X_n=i_n \lvert X_0=i_0]&=\frac{P[X_0=i_0,X_1=i_1,\cdots,X_n=i_n]}{P[X_0=i_0]} \\&=P_{i_0,i_1} \times \cdots \times P_{i_{n-1},i_n} \end{aligned}
Conditional on the initial state $X_0=i_0$, the probability of the process going through the path $X_1=i_1,\cdots,X_n=i_n$ is simply the product of the one-step transition probabilities.
If the path is observed to start at a time period other than time 0, conditional on the first state in the path, the probability of the process going through the path would be:
$\displaystyle P[X_{k+1}=i_1,\cdots,X_{k+n}=i_n \lvert X_k=i_0]=P_{i_0,i_1} \times \cdots \times P_{i_{n-1},i_n}$
Note that the transition probabilities $P_{ij}$ and the probabilities of paths are independent of the current period $n$. Random walks with such property are called time-homogeneous or stationary. Thus the probability of transitioning into state $j$ from state $i$ or the probability of transitioning through a path is identical regardless of where the process is in the time scale (at the beginning in the process or later in the process).
Specific Examples
For special cases of the random walks discussed above, let the increments $Y_n$ be defined by a Bernoulli random variable with $P[Y=1]=p$ and $P[Y=-1]=1-p$. Then the resulting random walk is a series of up and down moves as shown in Figure 1 above. The one-step transition probabilities are:
$\displaystyle P_{ij} = \left\{ \begin{array}{ll} \displaystyle p &\ \ \ \ \ \ j=i+1 \\ \text{ } & \text{ } \\ \displaystyle 1-p &\ \ \ \ \ \ j=i-1 \\ \text{ } & \text{ } \\ 0 &\ \ \ \ \ \ \text{otherwise} \end{array} \right.$
If the increments $Y_n$ are defined by the distribution where $P[Y=1]=p$, $P[Y=-1]=q$ and $P[Y=0]=1-p-q$, then the following gives the transition probabilities:
$\displaystyle P_{ij} = \left\{ \begin{array}{ll} \displaystyle p &\ \ \ \ \ \ j=i+1 \\ \text{ } & \text{ } \\ \displaystyle q &\ \ \ \ \ \ j=i-1 \\ \text{ } & \text{ } \\ \displaystyle 1-p-q &\ \ \ \ \ \ j=i \\ \text{ } & \text{ } \\ 0 &\ \ \ \ \ \ \text{otherwise} \end{array} \right.$
This random walk has three moves at any given state, an up move by one unit, a down move by one unit and a possibility that the process stays at the same state. Of course, the possibility for modeling is limitless. We can use other distributions for the common distribution of the increments $Y_n$, e.g. binomial distribution, Poisson, preferably having these distributions shifted in order to account for up moves and down moves.
Finite-State Random Walks
The random walks discussed above potentially have an infinite state space, i.e. the process can potentially reach far from the initial state without bounds. Depending on the common distribution of the increments $Y_1,\cdots,Y_n,\cdots$, the process can go up indefinitely to $+\infty$ or go down to $-\infty$. The random walks in Figure 1 have increments 1 and -1 and can reach states that are arbitrarily high or low, though it would take a great many steps to reach such high levels.
Any random walk with unbounded state space can be modified to have a finite state space. Pick two states that serve as boundaries, one lower and one upper. The states in between the boundary states are called interior states. Whenever the process reaches the upper boundary or the lower boundary, the process either stays there and not move any further or transition back into an interior state. The transition probabilities in the interior states remain the same as discussed above. If a boundary state is such that the process always stays there after reaching it, it is called an absorbing state.
A handy example of a finite-state random walk is the gambler’s ruin. Suppose that a gambler starts out with $d$ units in capital (in dollars or other monetary units) and makes a series of one-unit bets against the house. For the gambler, the probabilities of winning and losing each bet are $p$ and $1-p$, respectively. Whenever the capital reaches zero, the gambler is in ruin and his capital remains zero thereafter. On the other hand, if the capital of the gambler increases to $m$ where $d, then the gambler quits playing. Let $X_n$ be the capital of the gambler after the $n$th bet. Then $\left\{X_n: n=0,1,2,3,\cdots \right\}$ is a random walk. The starting state is $X_0=d$. The states $0$ and $m$ are absorbing states. The following gives the transition probabilities.
$\displaystyle P_{ij} = \left\{ \begin{array}{ll} \displaystyle p &\ \ \ \ \ \ j=i+1,i \ne 0, i \ne m \\ \text{ } & \text{ } \\ \displaystyle 1-p &\ \ \ \ \ \ j=i-1,i \ne 0, i \ne m \\ \text{ } & \text{ } \\ \displaystyle 1 &\ \ \ \ \ \ i=0,j=0 \\ \text{ } & \text{ } \\ \displaystyle 1 &\ \ \ \ \ \ i=m,j=m \\ \text{ } & \text{ } \\ 0 &\ \ \ \ \ \ \text{otherwise} \end{array} \right.$
Whenever the gambler reaches the absorbing state of 0, the gambler is in ruin. If the process reaches the absorbing state of $m$, the gambler strikes gold. If the initial capital of the gambler is small relative to the casino, there is a virtually 100% chance that the gambler will be in ruin even if each bet has even odds (see here for the calculation). One interesting questions: on average how long does it take for the gambler to be in ruin? Such questions will be answered after necessary tools are built in subsequent posts.
Remarks
A random walk is a stochastic process $\left\{X_n: n=0,1,2,\cdots \right\}$ such that at each time point, the process walks a certain distance that is dictated by an independent and identically distributed sequence of random variables. In such a stochastic process, the future state depends only on the preceding state and not on the past states before the preceding state. For example, in the gambler’s ruin, the gambler’s capital at any given point depends only on the his capital in the previous play.
In a stochastic process $\left\{X_n: n=0,1,2,\cdots \right\}$, the property that the future state $X_{n+1}$ depends only on the preceding state $X_n$ and not on the past states $X_0,X_1,\cdots,X_{n-1}$ is called the Markov property. Any stochastic process with the Markov property is called a Markov chain. A random walk is a Markov chain. There are Markov chains that are not random walk. The notion of Markov chains is introduced in this post.
$\copyright$ 2017 – Dan Ma
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2017-10-19 18:18:52
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https://cstheory.stackexchange.com/questions/47691/number-of-maximal-cliques-in-a-2c-4-c-5-p-5-free-graph
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# Number of maximal cliques in a ($2C_4$, $C_5$, $P_5$)-free graph
So far, I have found out that chordal graphs have linear number of maximal cliques with respect to the number of vertices. In general case, it is exponential.
I am trying to determine whether the number of maximal cliques in a $$(2C_4, C_5,P_5)$$-free graph with respect to the number of vertices.
In a $$(2C_4, C_5,P_5)$$-free graph, the largest induced cycle is of length 4, and no two induced 4-cycles are edge-disjoint.
Is there a paper that mentions such result?
• – D.W.
Oct 13 '20 at 2:47
The famous graph (the complement of the disjoint union of $$n/3$$ triangles) with $$3^{n/3}$$ maximal cliques is $$K_1 \cup K_2$$-free, and thus has none of $$2C_4$$, $$C_5$$, $$P_5$$ as an induced subgraph.
• Thanks for the answer. Do you know whether it is still exponential when two induced triandlges cannot be edge-disjoint in the complement, i.e. $(K_{3,3}, 2C_4, C_5, P_5)$-free? I guess the answer is still yes, but I cannot see it directly. Oct 9 '20 at 20:48
• As you guessed, the answer is still yes. The complement of the disjoint union of $n/2$ copies of $K_2$'s has $2^{n/2}$ maximal cliques. Oct 10 '20 at 3:24
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2022-01-23 05:13:39
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https://cs.stackexchange.com/questions/67504/running-time-of-greedy-scheduling-algorithm
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# Running time of greedy scheduling algorithm
Here is an algorithm to output a subset of activities S such that no activities in S overlap and profit(S) is maximum.
Define $p[i]$ to be the largest index $j$ such that $a_j$ does not overlap $a_i$.
$\#$ Return maximum possible profit from $a_1,...,a_n$
def rec_opt(n):
if n == 0:
return 0
else:
return max(rec_opt(n-1), w_n + rec_opt(p[n]))
It doesn't mention what $w_n$ is in the slide, but I think it is the profit of the activity $a_n$.
Consider a set of $n$ activities where there is no overlap at all. Then $T(n) = 2T(n − 1) + c$. This recurrence has an exponential closed-form, $O(2^n)$.
• "return max(rec_opt(n-1), w_n + rec_opt(p[n]))": I think this line takes time $2T(n-1)$.
• "if n == 0": this line takes time $c$.
Am I correct?
One more question, how can we get $O(2^n)$ from $T(n) = 2T(n − 1) + c$?
• Please ask only one question per post. Dec 16, 2016 at 9:34
• I think they are related, so I put them together. Dec 16, 2016 at 9:43
Suppose that $T(n) = 2T(n-1) + c$, and that $T(0) = a$. Define $S(n) = T(n) + c$. Then $S(n) = T(n)+c = 2T(n-1)+2c = 2S(n-1)$, and so $S(n) = 2^nS(0) = 2^n(a+c)$. We conclude that $T(n) = 2^n(a+c)-c = 2^na+(2^n-1)c$.
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2022-06-25 13:14:39
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https://www.physicsforums.com/threads/how-is-the-deficit-angle-due-to-a-relativistic-cosmic-string-derived.708768/
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# How is the deficit angle due to a relativistic cosmic string derived?
Dilatino
The presence of a cosmic string does not lead to gravitational attraction of a particle placed some distance away from it. But it affects the geometry of planes orthogonal to the cosmic string, such that the circumference of a circle traced out when moving around it at a distance r is given by
$$C = r(2\pi -\Delta)$$
which means that the flat plane is deformed to a cone.
How can the deficit angle due to a relativistic string
$$\Delta = \frac{8\pi G T_0}{c^4} = \frac{8\pi G \mu_0}{c^2}$$
be derived from general relativity?
Last edited:
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2022-08-16 08:06:27
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https://www.hepdata.net/record/20288
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Left-right Asymmetry in Inverse $\pi^-$ Photoproduction From a Transversely Polarized Proton Target
Phys.Rev.Lett. 56 (1986) 1779-1782, 1986.
Abstract (data abstract)
LAMPF - measurement of the left-right asymmetry in inverse pi-photoproduction using a transversely polarized proton target. Numerical values supplied by G. Kim. Also refer to Kim et al 1989, PR D40, 244 (& lt;a href=http://durpdg.dur.ac.uk/scripts/reacsearch.csh/TESTREAC/red+2164& gt; RED = 2164 & lt;/a& gt;).
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2022-05-22 20:22:27
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https://brilliant.org/problems/minimum-value-4-2/
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# Minimum value #4
Algebra Level 4
Positive reals $$a, b,$$ and $$c$$ satisfy $$abc+a+b=3ab$$. Find the smallest possible value of
$N=\sqrt { \frac { ab }{ a+b+1 } } +\sqrt { \frac { b }{ bc+c+1 } } +\sqrt { \frac { a }{ ca+c+1 } }.$
×
Problem Loading...
Note Loading...
Set Loading...
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2017-09-26 10:45:14
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https://code.tutsplus.com/articles/apple-tightens-security-with-app-transport-security--cms-24420
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# Apple Tightens Security With App Transport Security
Length:ShortLanguages:
The importance of and attention for security on the web has increased substantially over the past few years. During this year's WWDC, Apple has made it clear that it plans to lead by example by improving security of its operating systems through a new feature, App Transport Security.
Of course, the security of a platform is only as strong as the security of its components and that includes third party applications. In other words, Apple expects developers to adopt App Transport Security in their applications.
In this article, I will explain what App Transport Security entails, how it will affect your applications, and how you can update your applications to stick to Apple's guidelines and recommendations.
## What Is App Transport Security?
App Transport Security, or ATS for short, is a new feature of iOS 9 and OS X El Capitan. While Apple didn't mention watchOS, we can assume App Transport Security also applies to watchOS 2. App Transport Security aims to improve the security of Apple's operating systems and any applications running on these operating systems.
Network requests that are made over HTTP transmit data as cleartext. It goes without saying that this poses a significant security risk. Apple stresses that every developer should strive to keep the data of their customers safe and secure, even if that data doesn't seem important or sensitive.
App Transport Security actively encourages security by imposing a number of security best practices, the most important being the requirement that network requests need to be sent over a secure connection. With App Transport Security enabled, network requests are automatically made over HTTPS instead of HTTP.
There are a number of other requirements to further improve security. For example, App Transport Security requires TLS (Transport Layer Security) 1.2 or higher. While you may be unfamiliar with TLS, I'm sure you've heard of SSL (Secure Sockets Layer). TLS is the successor of SSL and is a collection of cryptographic protocols to enforce security over network connections.
Apple recently published a public, prerelease technote about App Transport Security to give developers the opportunity to plan for App Transport Security. The document outlines what App Transport Security expects from your applications and the web services it interacts with.
## Exceptions
Wait a second. My application uses a CDN (Content Delivery Network) that I don't have control over and it doesn't support HTTPS. Don't worry. Apple has your back covered. With regards to App Transport Security, an application falls into one of four categories. Let's go over each category to see how it impacts an application.
### HTTPS Only
If your application only interfaces with servers that support HTTPS, then you're in luck. You're application won't have to make any changes. However, note that App Transport Security requires TLS 1.2 and it expects the domain to use ciphers that support forward secrecy. The certificate also needs to meet the requirements imposed by ATS. It's therefore important to double-check that the servers your application communicates with comply with the requirements of ATS.
### Mix & Match
It is possible that your application talks to servers that don't meet the ATS requirements. In that case, you need to tell the operating system which domains are involved and specify in your application's Info.plist what requirements aren't met.
This means that App Transport Security is enforced for every endpoint your application talks to with the exception of the ones specified in your application's Info.plist. You can configure the exceptions using a number of predefined keys. In the following Info.plist, we define three exceptions.
#### api.insecuredomain.com
The first exception we define tells ATS that communication with this subdomain overrides the requirement to use HTTPS. Note that this exception only applies to the subdomain specified in the exception. It's important to understand that the NSExceptionAllowsInsecureHTTPLoads key doesn't only relate to the use of HTTPS. The exception specifies that, for that domain, every requirement of App Transport Security is overridden.
#### cdn.domain.com
It's possible that your application talks to a server that serves its data over HTTPS, but isn't using TLS 1.2 or higher. In that case, you define an exception that specifies the minimum TLS version that should be used. This is a better and safer option than completely overriding App Transport Security for that particular domain.
#### thatotherdomain.com
The NSIncludesSubdomains key tells App Transport Security that the exception applies to every subdomain of the specified domain. The exception further defines that the domain can use ciphers that don't support forward secrecy (NSExceptionRequiresForwardSecrecy) by expanding the list of accepted ciphers. For more information about forward secrecy, I recommend reading Apple's technote on the topic.
### Opt Out
If you're building a web browser, then you have a slightly bigger problem. Because you don't know which web pages your users are going to visit, you cannot possibly tell whether those web pages are served over HTTPS and meet the ATS requirements. In that case, there is no other option but to opt out of App Transport Security altogether.
It's important that you explicitly opt out of App Transport Security. Remember that App Transport Security is enforced by default. In your application's Info.plist, you add a dictionary for the key NSAppTransportSecurity. The dictionary should include one key, NSAllowsArbitraryLoads, and its value should be set to YES. This is what your application's Info.plist file should look like if you opt out of App Transport Security.
### Opt Out With Exceptions
There is a fourth option in which your application opts out of App Transport Security, but defines a number of exceptions. This is useful if your application fetches data from a range of servers you don't control, but also talks to an API you maintain. In that case, you specify in your application's Info.plist that arbitrary loads are allowed, but you also define one or more exceptions for which App Transport Security is enabled. This is what the Info.plist could look like.
## Timing
Apple has emphasized that applications automatically opt in to App Transport Security if they are built against iOS 9 or OS X El Capitan. This means that you won't have to make any changes to your applications as long as you build them against iOS 8 or OS X Yosemite.
Based on previous releases of iOS and OS X, however, we have learned that Apple requires developers to build their applications against the latest SDK fairly soon after their official release. In other words, even though you won't have to comply with App Transport Security when iOS 9 and OS X El Capitan are released later this year, it is very likely that Apple will require developers to build against the latest SDK in the first or second quarter of 2016. I therefore recommend that you investigate how App Transport Security will impact your applications sooner rather than later.
## Conclusion
I hope this article has made it clear that App Transport Security is not something your applications can adopt some day. It's similar to Apple's requirement for 64-bit support not too long ago. Unless your applications only talk to servers over HTTPS that comply with the ATS requirements, you need to invest some time to investigate how App Transport Security will impact your applications. Apple's technote about App Transport Security can help you with this.
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2021-05-08 19:03:07
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http://mathoverflow.net/questions/101573/untyped-higher-category-theory?answertab=active
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# Untyped Higher Category Theory
I am currently trying to wade through the vast lake of higher category theory, a formidable task,or so it seems.
In the process, it has occurred to me that there is a basic analogy in place with various forms of type theories, typed logic, typed set theory, typed lambda calculus, etc.
In higher cats, one has 1-morphisms, 2-morphisms, and so on.
A fairly hierarchical structure, a ladder to infinity of sorts.
Now, whenever there are types, there is (almost) invariably an un-typed variant of the theory, which "forgets" the types. So I wonder if there is something along these lines already somewhere in the categorical endeavor.
I try to be a bit more precise: imagine you are staring at a N-category (let us stick to a strict one, just for sake of simplicity), from the top, and you forget all the type labels. You see a fairly complicated diagram of maps whose endpoints are other maps, and so on and so forth. Now try to axiomatize such a structure. That would be an untyped higher category (UHC).
Is there a reference for this structure? Now get rid of the strictness, and re-do the experiment. What kind of untyped higher categories are the result of stripping types from general higher cats?
In the example I mentioned, the UHC is well-founded, in the sense that there are some fellows (the ground objects) who only point to themselves (I identify here the objects with their identity maps). Now, eliminate this distinguished role of objects and you will have a not well founded UHC.
Is there a study of not-well-founded categories, in a similar spirit as there is a theory of not well-founded sets?
-
I don't really understand the question, could you make precise what for example untyped category theory (not higher!) would be about? – Martin Brandenburg Jul 7 '12 at 13:10
@MircoMannucci If what you're interested is just an higher category in which there's only one object then I think you're looking for higher monoidal categories. – Giorgio Mossa Jul 7 '12 at 15:00
@Martin: I think Mirco is alluding to the difference between "untyped" set theory such as ZFC and ramified type theories such as Russell's theory of types where sets have a numeral "type" indicating their level in the hierarchy of the universe. – Zhen Lin Jul 7 '12 at 15:54
There is of course a (strict) $\omega$-category version of the single-sorted definition of a category (ncatlab.org/nlab/show/single-sorted+definition+of+a+category). You have to be a little careful with it if you want $\omega$-categories rather than $(\omega+1)$-categories, but it works just fine. You can find the $(\omega+1)$-version in Street's paper "The algebra of oriented simplices." – Mike Shulman Jul 9 '12 at 7:02
should be possible to define in a single sorted way, in principle, but oo-categories are trickier. This is all very different to a RW-types approach. When you say "think (I am being terribly sloppy here!) of object as 0-types, ordinary morphisms as 1-types" I encourage you to not be sloppy and figure out what this means for 1- or 2-categories. This would help frame the discussion for higher categories, and how to think of them using RW type theory. – David Roberts Jul 9 '12 at 15:27
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2015-01-31 21:14:10
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https://www.physicsforums.com/threads/prove-the-wave-function-for-dxy-orbital.617309/
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Prove the wave function for dxy orbital
1. Jun 29, 2012
chem1309
1. The problem statement, all variables and given/known data
"The angular part of the wave function for the dxy orbital is (√(15/∏)/4)sin^2(θ)sin(2$\phi$). Show that this expression corresponds to the dxy orbital"
2. Relevant equations
conversion of Cartesian to spherical coordinates:
r=√(x^2+y^2+z^2)
cosθ=z/r
tan($\phi$)=y/x
trig identity:
sin(2x)=2sinxcosx
normalization:
N^2∫ψ*ψdτ
dτ=r^2sinθdrdθd$\phi$
0≤r≤∞
0≤θ≤∏
0≤$\phi$≤2∏
3. The attempt at a solution
in Cartesian coordinates dxy is represented as simply xy. I converted xy to spherical coordinates and manipulated the equation the relevant equations to get xy=(r/2)sin^2(θ)sin(2$\phi$) as follows:
xy=rsincos$\phi$rsinθsin$\phi$
xy=rsin^2(θ)cos$\phi$sin$\phi$
xy=rsin^2(θ)sin(2$\phi$)/2
Then I tried to normalize the equation, but I ended up with
∫r^3 from 0 to ∞, which goes to ∞/does not converge
and ∫sin2$\phi$ which equal zero.
2. Jul 1, 2012
vela
Staff Emeritus
I'm not sure why you think you need to normalize the wave function to show it corresponds to the dxy orbital.
In any case, the total wave function is of the form $\psi(\vec{r}) = R_{nl}(r)Y_l^m(\theta,\phi)$. Normalization requires that
$$\int \psi^*(\vec{r})\psi(\vec{r})\,d^3\vec{r} = \int R^*_{nl}(r)R_{nl}(r)\,dr \int {Y_l^m}^*(\theta,\phi)Y_l^m(\theta,\phi)\,d\Omega = 1.$$ Does seeing this clear up your two questions?
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2017-08-17 22:21:32
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https://www.physicsforums.com/threads/integration-of-step-functions.789976/
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# Integration of step functions
Tags:
1. Jan 1, 2015
### RandomGuy1
1. The problem statement, all variables and given/known data
This is from Apostol's Calculus Vol. 1. Exercise 1.15, problem 6.(c)
Find all x>0 for which the integral of [t]2 dt from 0 to x = 2(x-1)
2. Relevant equations
[t] represents the greatest integer function of t.
3. The attempt at a solution
Integral of [t]2 dt from 0 to x = {x(x-1)(2x-1)}/6, which on equating with 2(x-1) gives x(2x-1) = 12 which does not have a rational solution. The answers given at the back of the textbook are 1 and 5/2. Can someone please give me a hint as to where I'm going wrong?
2. Jan 1, 2015
### Staff: Mentor
Your notation really threw me off for a while. I think this is what you're trying to convey:
$$\int_0^x \lfloor t \rfloor ^2 dt = 2(x - 1)$$
No, it isn't. You can't just integrate $\lfloor t ^2 \rfloor$ as if it were the same thing as $t^2$. Draw a sketch of $y = \lfloor t ^2 \rfloor$ and y = 2(t - 1) and compare the cumulative area under the graph of the integrand with the y values on the straight line graph.
3. Jan 1, 2015
### RandomGuy1
Yes, that was exactly what I was trying to say :) Sorry, I do not know how to use the definite integral symbol.
But but I'm supposed to be integrating [t]2 and not [t2] as you pointed out. Won't those two be different?
4. Jan 1, 2015
### Ray Vickson
(1) You don't need to know how to do a definite integral sign; you can just say something like int_{t=0..x} [t]^2 dt or int([t]^2 dt, t=0..x). That would be perfectly clear and would avoid the kind of confusion I was victim to when I first tried to read your message. Or you could have said that "the integral of [t]^2 from 0 to x is equal to 2x-1"---that would also have been clear. Saying "...x = 2x-1" is maximally confusing.
(2) Yes, [t^2] and [t]^2 are very different.
5. Jan 1, 2015
### HallsofIvy
Staff Emeritus
This is, as you say, a step function. It's graph is just a series of horizontal lines and its integral, the area under the graph, is just the sum of areas of rectangles. The first thing you should do is draw the graph. If x is from 0 to 1, [x]= 0 so [x]2= 0. If x is from 1 to 2, [x]= 1 so [x]2= 1. If x is from 2 to 3, [x]= 2 so [x]2= 4, etc.
6. Jan 1, 2015
### Stephen Tashi
In this problem, the formula for summing the squares of integers is only relevant when $x$ is an integer. And when $x$ is an integer, I think you are "off by 1". For example, as I understand the symbol "[t]^2" in this problem, when $4 < t < 5$ , $[t]^2 = [4]^2 = 16$. So the area involved in $\int_0^5 [t]^2 dt$ does not include a rectangle with height 25. The areas to be summed are $(1)(0^2)+(1)(1^2) + (1)(2^2) + (1)(3^3) + (1)(4^2)$.
If the upper limit of integration is between two integers you must sum the squares of the whole rectangles involved in the are and then add the area of the fraction of the rectangle at the end to the sum.
Last edited: Jan 1, 2015
7. Jan 1, 2015
### Staff: Mentor
Absolutely, as is the one below.
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2017-08-17 15:51:28
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https://hal.inria.fr/inria-00073494
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# An Extension of Zeilberger's Fast Algorithm to General Holonomic Functions
1 ALGO - Algorithms
Inria Paris-Rocquencourt
Abstract : We extend Zeilberger's fast algorithm for definite hypergeometric summation to non-hypergeometric holonomic sequences. The algorithm generalizes to differential and~$q$-cases as well. Its theoretical justification is based on a description by linear operators and on the theory of holonomy.
Document type :
Reports
Domain :
https://hal.inria.fr/inria-00073494
Contributor : Rapport de Recherche Inria <>
Submitted on : Wednesday, May 24, 2006 - 1:02:37 PM
Last modification on : Friday, May 25, 2018 - 12:02:02 PM
Long-term archiving on : Thursday, March 24, 2011 - 12:51:51 PM
### Identifiers
• HAL Id : inria-00073494, version 1
### Citation
Frédéric Chyzak. An Extension of Zeilberger's Fast Algorithm to General Holonomic Functions. [Research Report] RR-3195, INRIA. 1997. ⟨inria-00073494⟩
Record views
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2019-07-21 09:25:24
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https://p3h.particle.kit.edu/preprints/2021/p3h-21-053
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P3H-21-053: The impact of top quark modelling on the exclusion limits in tt + DM searches at the LHC
P3H-21-053
Title: The impact of top quark modelling on the exclusion limits in tt + DM searches at the LHC
Type: Paper
Authors: J. Hermanna, M. Worek
arXiv: 2108.01089
Info:
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2021-09-27 22:49:58
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https://www.course.cafe/digital-quantitative-physical-history/lesson/microeconomics/
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# Costs, Revenues and Profits
#### By Mark Ciotola
First published on May 17, 2019. Last updated on May 12, 2021.
## Introduction
Microeconomics concerns individual and firm-level decisions regarding the allocation of resources. There are several important concepts to analyze a firm-level venture. We will just consider a single product firm for the sake of simplicity. At its most basic level, microeconomics considers at what volumes and prices it makes sense for a enterprise to continue producing a good or service. Hence it is quite important for decision-making.
## Revenues, Costs and Profits
### General Definitions
Revenue (R) is the amount of money a firm receives in payment for a product sold.
Costs (C) are how much money a firm pays to make a product.
Profit (P) is the amount left over when cost is subtracted from revenue:
$$P = R – C$$
For example, a pizza requires $5 of ingredients, labor and overhead to make. A pizzeria can cell that pizza for$12. The profit would then be $7: $$7 = 12 – 5$$. The terms cost, revenue and profit are general terms, in that they can refer to one unit, lots of units or all units of product or service. For more realistic economic analysis we will need to be more specific. ### Definitions for Totals We will start by considering total cost, revenue and profit. Total Revenue (TR) is the sum of revenues a firm receives in payment for all units of the product sold. Total Cost (TC) is the sum of costs a firm pays to make all units of the product sold. Total Profit (TP) is the sum of profits obtained from all goods sold: $$TP = TR – TC$$. The pizzeria sells 100 pizzas each for$12, resulting in $1,200 in total revenue. The pizzas cost$5 to make, resulting in total costs of $500. Then the total profit would be$700:
$$700 = 1,200 – 500$$.
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2022-11-26 12:21:47
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https://www.iacr.org/cryptodb/data/author.php?authorkey=12077
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## CryptoDB
### Yongjune Kim
#### Publications
Year
Venue
Title
2022
EUROCRYPT
The Cheon-Kim-Kim-Song (CKKS) scheme (Asiacrypt'17) is one of the most promising homomorphic encryption (HE) schemes as it enables privacy-preserving computing over real (or complex) numbers. It is known that bootstrapping is the most challenging part of the CKKS scheme. Further, homomorphic evaluation of modular reduction is the core of the CKKS bootstrapping. As modular reduction is not represented by the addition and multiplication of complex numbers, approximate polynomials for modular reduction should be used. The best-known techniques (Eurocrypt'21) use a polynomial approximation for trigonometric functions and their composition. However, all the previous methods are based on an indirect approximation, and thus it requires lots of multiplicative depth to achieve high accuracy. This paper proposes a direct polynomial approximation of modular reduction for CKKS bootstrapping, which is optimal in error variance and depth. Further, we propose an efficient algorithm, namely the lazy baby-step giant-step (BSGS) algorithm, to homomorphically evaluate the approximate polynomial, utilizing the lazy relinearization/rescaling technique. The lazy-BSGS reduces the computational complexity by half compared to the ordinary BSGS algorithm. The performance improvement for the CKKS scheme by the proposed algorithm is verified by implementation using HE libraries. The implementation results show that the proposed method has a multiplicative depth of 10 for modular reduction to achieve the state-of-the-art accuracy, while the previous methods have depths of 11 to 12. Moreover, we achieve higher accuracy within a small multiplicative depth, for example, 93-bit within multiplicative depth 11.
#### Coauthors
HyungChul Kang (1)
Young-Sik Kim (1)
Joseph Lano (1)
Joon-Woo Lee (1)
Yongwoo Lee (1)
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2022-06-29 00:55:21
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https://zbmath.org/?q=an%3A1140.26307
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# zbMATH — the first resource for mathematics
The Schur-convexity of Stolarsky and Gini means. (English) Zbl 1140.26307
Summary: We study in a unitary way the Schur-convexity or concavity of the Stolarsky and Gini means $$D_{a,b} (x, y)$$ and $$S_{a,b} (x, y)$$, for fixed $$x, y > 0$$, $$x \neq y$$.
##### MSC:
26D15 Inequalities for sums, series and integrals
##### Keywords:
Stolarsky and Gini means; Schur convexity
Full Text:
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2021-12-08 14:13:27
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http://www.jstor.org/stable/3572540
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## Access
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# Neutron Autoradiographic Determination of Boron-10 Concentration and Distribution in Mammalian Tissue
R. G. Fairchild, E. A. Tonna, Clara T. Seibold and R. F. Straub
Vol. 36, No. 1 (Oct., 1968), pp. 87-97
DOI: 10.2307/3572540
Stable URL: http://www.jstor.org/stable/3572540
Page Count: 11
Preview not available
## Abstract
The application of a low-background neutron autoradiographic technique for determining 10 B concentration and intracellular distribution in mammalian tissue is described. Knowledge of 10 B concentration and distribution on a cellular level is necessary to evaluate the effects of irradiation when using the ${}^{10}{\rm B}({\rm n},\alpha)^{7}{\rm Li}$ reaction in experimental therapeutic procedures. Conventional boron analysis is not capable of extracting this information. The use of freeze-dried tissues, dry-mounted on nuclear emulsion, makes the technique applicable to water-soluble compounds used in previous clinical trials, as well as to water-insoluble proteins. Absolute concentrations of 10 B in the form of sodium pentaborate were measured in mouse spleen, liver, kidney, muscle, tumor, and brain, 30 minutes after intraperitoneal injection, and compared to results from three other laboratories using conventional analysis. Boron concentrations similar to those used in previous clinical trials were evaluated to demonstrate the feasibility of the method. No intranuclear or extracellular concentration was observed; evidence is given to demonstrate the absence of leaching of the water-soluble compound. The method described will be useful in evaluating the newer boron compounds attached to proteins. The intracellular distribution of these compounds is of particular interest, since it is not clear that all proteins are capable of penetrating cell walls.
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2016-09-26 18:11:26
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http://mathhelpforum.com/calculus/53267-calculus-help.html
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1. ## Calculus help
Find parametric equations for the tangent line at the point
on the curve
2. Find the parametric equation of the line tangent to
$r(t)=cos(t)i+sin(t)j+tk$ at the point $t=\frac{\pi}{3}$.
$r'(t)=-sin(t)i+cos(t)j+k$
$||r'(t)||=\sqrt{sin^{2}(t)+cos^{2}(t)+1}=\sqrt{2}$
The unit tangent vector is:
$T(t)=\frac{r'(t)}{||r'(t)||}$
Then sub in $t=\frac{\pi}{3}$
Use the direction numbers and the point $(x_{1},y_{1},z_{1})$ to find the parametric equations of the line.
3. what points am i supposed to se for (x, y, z)? Thanks!
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2013-12-19 08:36:50
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|
https://mathematica.stackexchange.com/questions/102201/how-to-take-derivative-of-a-interpolated-function-inside-the-module?noredirect=1
|
# How to take derivative of a interpolated function inside the module
When handling derivatives of a interpolated function, I often use the following function
makeDeriv[f_InterpolatingFunction] = D[ f[x], x]
Then, I can define the derivative of f, by defining a downvalue as
interpolated = Range[5]^2 // Interpolation;
deriv[x_] = makeDeriv[interpolated]; (* To find a derivative once and then to use it later. *)
deriv[3] (* Find a numerical value now*)
The purpose of this construct is to find the derivative once, and then use it later whenever I want to take derivatives. (As a side question: will that be a better way than computing it from f every time for each numerical argument I will be interested?)
But then, if I want to define a similar function inside the Module, it does not seem to work, i.e.,
Module[ { qq},
qq[x_] = makeDeriv[interpolated];
qq[2]
](*It works!!*)
However,
Module[ { qq, int = interpolated},
qq[x_] = makeDeriv[int];
qq[2]
](*It does not work!!*)
Please explain why it happens and how to circumvent this problem.
(As another side question, let me know how to define the derivative of interpolation function properly.)
• Not enough time now to test this, but have you tried: makeDeriv[f_InterpolatingFunction] := D[ f[x], x] ,deriv[x_] := makeDeriv[interpolated], and qq[x_] := makeDeriv[interpolated]? The ":" might help. Dec 16 '15 at 13:50
It's bad practice to have your functions depend on global symbols like x buried in their definitions, for Mathematica under certain conditions rewrites function parameters as x$_ to avoid clashing with the global symbol. Use Trace on the OP's examples to see that in the second one, x_ is rewritten as x$_. See the tutorial Variables in Pure Functions and Rules and these questions for further discussion:
There are others, too, I'm sure.
One workaround is to use Derivative directly:
Module[{qq, int = interpolated},
qq = Derivative[1][int];
qq[2]]
(* 4 *)
Another way:
Module[{qq, int = interpolated, x},
qq[x_] = D[int[x], x];
qq[2]]
Or include all symbols as function parameters:
Clear[makeDeriv];
makeDeriv[f_InterpolatingFunction, x_] := D[f[x], x];
Module[{qq, int = interpolated, x},
qq[x_] = makeDeriv[int, x];
qq[2]]
And so forth, but I think the first two are simplest.
• One more reference with an answer I started which Leonid greatly extended: (20766) Dec 16 '15 at 13:58
• @Mr.Wizard Thanks. Added. Dec 16 '15 at 14:04
If you want to stick to your code, you can redefine your derivative taking function a little bit:
makeDeriv[f_InterpolatingFunction[x_]] := D[f[x],x]
and call it accordingly in your Module environment
int = Interpolation[Range[5]^2];
Module[{modint = int, dint},
dint[x_] = makeDeriv[modint[x]];
dint[1.2]
] (* Out[1] = 2.4 *)
I suppose that the unspecified variable x in your definition of the makeDeriv function leads to the problem, although I am not entirely sure why this only happens in your last example. Anyway, the above definition should be safe.
Two alternatives for taking the derivatives:
D[int[x],x] /. x -> 1.2 (* Out[2] = 2.4 *)
Derivative[1][int][1.2] (* Out[3] = 2.4 *)
`
so there should be no need to come up with your own function in the first place
|
2021-11-27 18:17:44
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|
https://turing.ml/dev/tutorials/7-poissonregression/
|
# Bayesian Poisson Regression
This notebook is ported from the example notebook of PyMC3 on Poisson Regression.
Poisson Regression is a technique commonly used to model count data. Some of the applications include predicting the number of people defaulting on their loans or the number of cars running on a highway on a given day. This example describes a method to implement the Bayesian version of this technique using Turing.
We will generate the dataset that we will be working on which describes the relationship between number of times a person sneezes during the day with his alcohol consumption and medicinal intake.
We start by importing the required libraries.
#Import Turing, Distributions and DataFrames
using Turing, Distributions, DataFrames, Distributed
# Import MCMCChain, Plots, and StatsPlots for visualizations and diagnostics.
using MCMCChains, Plots, StatsPlots
# Set a seed for reproducibility.
using Random
Random.seed!(12);
# Turn off progress monitor.
Turing.turnprogress(false)
┌ Info: [Turing]: progress logging is disabled globally
└ @ Turing /home/cameron/.julia/packages/Turing/cReBm/src/Turing.jl:22
false
# Generating data
We start off by creating a toy dataset. We take the case of a person who takes medicine to prevent excessive sneezing. Alcohol consumption increases the rate of sneezing for that person. Thus, the two factors affecting the number of sneezes in a given day are alcohol consumption and whether the person has taken his medicine. Both these variable are taken as boolean valued while the number of sneezes will be a count valued variable. We also take into consideration that the interaction between the two boolean variables will affect the number of sneezes
5 random rows are printed from the generated data to get a gist of the data generated.
theta_noalcohol_meds = 1 # no alcohol, took medicine
theta_alcohol_meds = 3 # alcohol, took medicine
theta_noalcohol_nomeds = 6 # no alcohol, no medicine
theta_alcohol_nomeds = 36 # alcohol, no medicine
# no of samples for each of the above cases
q = 100
#Generate data from different Poisson distributions
noalcohol_meds = Poisson(theta_noalcohol_meds)
alcohol_meds = Poisson(theta_alcohol_meds)
noalcohol_nomeds = Poisson(theta_noalcohol_nomeds)
alcohol_nomeds = Poisson(theta_alcohol_nomeds)
nsneeze_data = vcat(rand(noalcohol_meds, q), rand(alcohol_meds, q), rand(noalcohol_nomeds, q), rand(alcohol_nomeds, q) )
alcohol_data = vcat(zeros(q), ones(q), zeros(q), ones(q) )
meds_data = vcat(zeros(q), zeros(q), ones(q), ones(q) )
df = DataFrame(nsneeze = nsneeze_data, alcohol_taken = alcohol_data, nomeds_taken = meds_data, product_alcohol_meds = meds_data.*alcohol_data)
df[sample(1:nrow(df), 5, replace = false), :]
5 rows × 4 columns
nsneezealcohol_takennomeds_takenproduct_alcohol_meds
Int64Float64Float64Float64
150.00.00.0
251.00.00.0
381.00.00.0
410.00.00.0
5381.01.01.0
# Visualisation of the dataset
We plot the distribution of the number of sneezes for the 4 different cases taken above. As expected, the person sneezes the most when he has taken alcohol and not taken his medicine. He sneezes the least when he doesn’t consume alcohol and takes his medicine.
#Data Plotting
p1 = Plots.histogram(df[(df[:,:alcohol_taken] .== 0) .& (df[:,:nomeds_taken] .== 0), 1], title = "no_alcohol+meds")
p2 = Plots.histogram((df[(df[:,:alcohol_taken] .== 1) .& (df[:,:nomeds_taken] .== 0), 1]), title = "alcohol+meds")
p3 = Plots.histogram((df[(df[:,:alcohol_taken] .== 0) .& (df[:,:nomeds_taken] .== 1), 1]), title = "no_alcohol+no_meds")
p4 = Plots.histogram((df[(df[:,:alcohol_taken] .== 1) .& (df[:,:nomeds_taken] .== 1), 1]), title = "alcohol+no_meds")
plot(p1, p2, p3, p4, layout = (2, 2), legend = false)
We must convert our DataFrame data into the Matrix form as the manipulations that we are about are designed to work with Matrix data. We also separate the features from the labels which will be later used by the Turing sampler to generate samples from the posterior.
# Convert the DataFrame object to matrices.
data = Matrix(df[:,[:alcohol_taken, :nomeds_taken, :product_alcohol_meds]])
data_labels = df[:,:nsneeze]
data
400×3 Array{Float64,2}:
0.0 0.0 0.0
0.0 0.0 0.0
0.0 0.0 0.0
0.0 0.0 0.0
0.0 0.0 0.0
0.0 0.0 0.0
0.0 0.0 0.0
0.0 0.0 0.0
0.0 0.0 0.0
0.0 0.0 0.0
0.0 0.0 0.0
0.0 0.0 0.0
0.0 0.0 0.0
⋮
1.0 1.0 1.0
1.0 1.0 1.0
1.0 1.0 1.0
1.0 1.0 1.0
1.0 1.0 1.0
1.0 1.0 1.0
1.0 1.0 1.0
1.0 1.0 1.0
1.0 1.0 1.0
1.0 1.0 1.0
1.0 1.0 1.0
1.0 1.0 1.0
We must recenter our data about 0 to help the Turing sampler in initialising the parameter estimates. So, normalising the data in each column by subtracting the mean and dividing by the standard deviation:
# # Rescale our matrices.
data = (data .- mean(data, dims=1)) ./ std(data, dims=1)
400×3 Array{Float64,2}:
-0.998749 -0.998749 -0.576628
-0.998749 -0.998749 -0.576628
-0.998749 -0.998749 -0.576628
-0.998749 -0.998749 -0.576628
-0.998749 -0.998749 -0.576628
-0.998749 -0.998749 -0.576628
-0.998749 -0.998749 -0.576628
-0.998749 -0.998749 -0.576628
-0.998749 -0.998749 -0.576628
-0.998749 -0.998749 -0.576628
-0.998749 -0.998749 -0.576628
-0.998749 -0.998749 -0.576628
-0.998749 -0.998749 -0.576628
⋮
0.998749 0.998749 1.72988
0.998749 0.998749 1.72988
0.998749 0.998749 1.72988
0.998749 0.998749 1.72988
0.998749 0.998749 1.72988
0.998749 0.998749 1.72988
0.998749 0.998749 1.72988
0.998749 0.998749 1.72988
0.998749 0.998749 1.72988
0.998749 0.998749 1.72988
0.998749 0.998749 1.72988
0.998749 0.998749 1.72988
# Declaring the Model: Poisson Regression
Our model, poisson_regression takes four arguments:
• x is our set of independent variables;
• y is the element we want to predict;
• n is the number of observations we have; and
• σ² is the standard deviation we want to assume for our priors.
Within the model, we create four coefficients (b0, b1, b2, and b3) and assign a prior of normally distributed with means of zero and standard deviations of σ². We want to find values of these four coefficients to predict any given y.
Intuitively, we can think of the coefficients as:
• b1 is the coefficient which represents the effect of taking alcohol on the number of sneezes;
• b2 is the coefficient which represents the effect of taking in no medicines on the number of sneezes;
• b3 is the coefficient which represents the effect of interaction between taking alcohol and no medicine on the number of sneezes;
The for block creates a variable theta which is the weighted combination of the input features. We have defined the priors on these weights above. We then observe the likelihood of calculating theta given the actual label, y[i].
# Bayesian poisson regression (LR)
@model poisson_regression(x, y, n, σ²) = begin
b0 ~ Normal(0, σ²)
b1 ~ Normal(0, σ²)
b2 ~ Normal(0, σ²)
b3 ~ Normal(0, σ²)
for i = 1:n
theta = b0 + b1*x[i, 1] + b2*x[i,2] + b3*x[i,3]
y[i] ~ Poisson(exp(theta))
end
end;
# Sampling from the posterior
We use the NUTS sampler to sample values from the posterior. We run multiple chains using the mapreduce function to nullify the effect of a problematic chain. We then use the Gelman, Rubin, and Brooks Diagnostic to check the convergence of these multiple chains.
# Retrieve the number of observations.
n, _ = size(data)
# Sample using NUTS.
num_chains = 4
chain = mapreduce(
chainscat,
1:num_chains);
┌ Warning: The current proposal will be rejected due to numerical error(s).
│ isfinite.((θ, r, ℓπ, ℓκ)) = (true, false, false, false)
┌ Warning: The current proposal will be rejected due to numerical error(s).
│ isfinite.((θ, r, ℓπ, ℓκ)) = (true, false, false, false)
┌ Warning: The current proposal will be rejected due to numerical error(s).
│ isfinite.((θ, r, ℓπ, ℓκ)) = (true, false, false, false)
┌ Warning: The current proposal will be rejected due to numerical error(s).
│ isfinite.((θ, r, ℓπ, ℓκ)) = (true, false, false, false)
┌ Warning: The current proposal will be rejected due to numerical error(s).
│ isfinite.((θ, r, ℓπ, ℓκ)) = (true, false, false, false)
┌ Warning: The current proposal will be rejected due to numerical error(s).
│ isfinite.((θ, r, ℓπ, ℓκ)) = (true, false, false, false)
┌ Warning: The current proposal will be rejected due to numerical error(s).
│ isfinite.((θ, r, ℓπ, ℓκ)) = (true, false, false, false)
┌ Warning: The current proposal will be rejected due to numerical error(s).
│ isfinite.((θ, r, ℓπ, ℓκ)) = (true, false, false, false)
┌ Warning: The current proposal will be rejected due to numerical error(s).
│ isfinite.((θ, r, ℓπ, ℓκ)) = (true, false, false, false)
┌ Warning: The current proposal will be rejected due to numerical error(s).
│ isfinite.((θ, r, ℓπ, ℓκ)) = (true, false, false, false)
┌ Warning: The current proposal will be rejected due to numerical error(s).
│ isfinite.((θ, r, ℓπ, ℓκ)) = (true, false, false, false)
┌ Warning: The current proposal will be rejected due to numerical error(s).
│ isfinite.((θ, r, ℓπ, ℓκ)) = (true, false, false, false)
┌ Warning: The current proposal will be rejected due to numerical error(s).
│ isfinite.((θ, r, ℓπ, ℓκ)) = (true, false, false, false)
┌ Warning: The current proposal will be rejected due to numerical error(s).
│ isfinite.((θ, r, ℓπ, ℓκ)) = (true, false, false, false)
┌ Warning: The current proposal will be rejected due to numerical error(s).
│ isfinite.((θ, r, ℓπ, ℓκ)) = (true, false, false, false)
┌ Warning: The current proposal will be rejected due to numerical error(s).
│ isfinite.((θ, r, ℓπ, ℓκ)) = (true, false, false, false)
┌ Warning: The current proposal will be rejected due to numerical error(s).
│ isfinite.((θ, r, ℓπ, ℓκ)) = (true, false, false, false)
┌ Warning: The current proposal will be rejected due to numerical error(s).
│ isfinite.((θ, r, ℓπ, ℓκ)) = (true, false, false, false)
┌ Warning: The current proposal will be rejected due to numerical error(s).
│ isfinite.((θ, r, ℓπ, ℓκ)) = (true, false, false, false)
┌ Warning: The current proposal will be rejected due to numerical error(s).
│ isfinite.((θ, r, ℓπ, ℓκ)) = (true, false, false, false)
┌ Info: Found initial step size
│ ϵ = 2.384185791015625e-8
└ @ Turing.Inference /home/cameron/.julia/packages/Turing/cReBm/src/inference/hmc.jl:556
┌ Warning: The current proposal will be rejected due to numerical error(s).
│ isfinite.((θ, r, ℓπ, ℓκ)) = (true, false, false, false)
┌ Warning: The current proposal will be rejected due to numerical error(s).
│ isfinite.((θ, r, ℓπ, ℓκ)) = (true, false, false, false)
┌ Warning: The current proposal will be rejected due to numerical error(s).
│ isfinite.((θ, r, ℓπ, ℓκ)) = (true, false, false, false)
┌ Warning: The current proposal will be rejected due to numerical error(s).
│ isfinite.((θ, r, ℓπ, ℓκ)) = (true, false, false, false)
┌ Info: Found initial step size
│ ϵ = 0.00078125
└ @ Turing.Inference /home/cameron/.julia/packages/Turing/cReBm/src/inference/hmc.jl:556
┌ Warning: The current proposal will be rejected due to numerical error(s).
│ isfinite.((θ, r, ℓπ, ℓκ)) = (true, false, false, false)
┌ Warning: The current proposal will be rejected due to numerical error(s).
│ isfinite.((θ, r, ℓπ, ℓκ)) = (true, false, false, false)
┌ Warning: The current proposal will be rejected due to numerical error(s).
│ isfinite.((θ, r, ℓπ, ℓκ)) = (true, false, false, false)
┌ Warning: The current proposal will be rejected due to numerical error(s).
│ isfinite.((θ, r, ℓπ, ℓκ)) = (true, false, false, false)
┌ Warning: The current proposal will be rejected due to numerical error(s).
│ isfinite.((θ, r, ℓπ, ℓκ)) = (true, false, false, false)
┌ Info: Found initial step size
│ ϵ = 0.000390625
└ @ Turing.Inference /home/cameron/.julia/packages/Turing/cReBm/src/inference/hmc.jl:556
┌ Info: Found initial step size
│ ϵ = 0.05
└ @ Turing.Inference /home/cameron/.julia/packages/Turing/cReBm/src/inference/hmc.jl:556
┌ Warning: The current proposal will be rejected due to numerical error(s).
│ isfinite.((θ, r, ℓπ, ℓκ)) = (true, false, false, false)
# Viewing the Diagnostics
We use the Gelman, Rubin, and Brooks Diagnostic to check whether our chains have converged. Note that we require multiple chains to use this diagnostic which analyses the difference between these multiple chains.
We expect the chains to have converged. This is because we have taken sufficient number of iterations (1500) for the NUTS sampler. However, in case the test fails, then we will have to take a larger number of iterations, resulting in longer computation time.
gelmandiag(chain)
Gelman, Rubin, and Brooks Diagnostic
parameters PSRF 97.5%
────────── ────── ──────
b0 1.1861 1.3924
b1 1.1307 1.2582
b2 1.1350 1.2865
b3 1.0660 1.1118
From the above diagnostic, we can conclude that the chains have converged because the PSRF values of the coefficients are close to 1.
So, we have obtained the posterior distributions of the parameters. We transform the coefficients and recover theta values by taking the exponent of the meaned values of the coefficients b0, b1, b2 and b3. We take the exponent of the means to get a better comparison of the relative values of the coefficients. We then compare this with the intuitive meaning that was described earlier.
# Taking the first chain
c1 = chain[:,:,1]
# Calculating the exponentiated means
b0_exp = exp(mean(c1[:b0].value))
b1_exp = exp(mean(c1[:b1].value))
b2_exp = exp(mean(c1[:b2].value))
b3_exp = exp(mean(c1[:b3].value))
print("The exponent of the meaned values of the weights (or coefficients are): \n")
print("b0: ", b0_exp, " \n", "b1: ", b1_exp, " \n", "b2: ", b2_exp, " \n", "b3: ", b3_exp, " \n")
print("The posterior distributions obtained after sampling can be visualised as :\n")
The exponent of the meaned values of the weights (or coefficients are):
b0: 5.116678482496325
b1: 1.8791946940293356
b2: 2.5245646467859904
b3: 1.3005130214177183
The posterior distributions obtained after sampling can be visualised as :
Visualising the posterior by plotting it:
plot(chain)
# Interpreting the Obtained Mean Values
The exponentiated mean of the coefficient b1 is roughly half of that of b2. This makes sense because in the data that we generated, the number of sneezes was more sensitive to the medicinal intake as compared to the alcohol consumption. We also get a weaker dependence on the interaction between the alcohol consumption and the medicinal intake as can be seen from the value of b3.
# Removing the Warmup Samples
As can be seen from the plots above, the parameters converge to their final distributions after a few iterations. These initial values during the warmup phase increase the standard deviations of the parameters and are not required after we get the desired distributions. Thus, we remove these warmup values and once again view the diagnostics.
To remove these warmup values, we take all values except the first 200. This is because we set the second parameter of the NUTS sampler (which is the number of adaptations) to be equal to 200. describe(chain) is used to view the standard deviations in the estimates of the parameters. It also gives other useful information such as the means and the quantiles.
# Note the standard deviation before removing the warmup samples
describe(chain)
2-element Array{ChainDataFrame,1}
Summary Statistics
parameters mean std naive_se mcse ess r_hat
────────── ────── ────── ──────── ────── ─────── ──────
b0 1.2639 2.1637 0.0216 0.2114 42.6654 1.0565
b1 0.7091 0.8433 0.0084 0.0728 41.7860 1.0620
b2 1.1998 1.7572 0.0176 0.1676 42.5718 1.0675
b3 0.2357 0.7392 0.0074 0.0596 91.3888 1.0240
Quantiles
parameters 2.5% 25.0% 50.0% 75.0% 97.5%
────────── ─────── ────── ────── ────── ──────
b0 -4.7815 1.6189 1.6409 1.6624 1.7026
b1 0.4366 0.5151 0.5548 0.5986 3.7771
b2 0.7707 0.8461 0.8848 0.9259 8.4861
b3 -1.7651 0.2497 0.2882 0.3275 0.4136
# Removing the first 200 values of the chains.
chains_new = chain[201:2500,:,:]
describe(chains_new)
2-element Array{ChainDataFrame,1}
Summary Statistics
parameters mean std naive_se mcse ess r_hat
────────── ────── ────── ──────── ────── ─────── ──────
b0 1.6378 0.0823 0.0009 0.0055 46.6518 1.0182
b1 0.5639 0.1729 0.0018 0.0117 45.5782 1.0196
b2 0.8932 0.1727 0.0018 0.0118 45.0961 1.0195
b3 0.2798 0.1544 0.0016 0.0104 46.0058 1.0195
Quantiles
parameters 2.5% 25.0% 50.0% 75.0% 97.5%
────────── ────── ────── ────── ────── ──────
b0 1.5791 1.6226 1.6427 1.6637 1.7024
b1 0.4413 0.5142 0.5516 0.5919 0.6726
b2 0.7764 0.8448 0.8819 0.9187 0.9973
b3 0.1785 0.2544 0.2893 0.3266 0.3942
Visualising the new posterior by plotting it:
plot(chains_new)
As can be seen from the numeric values and the plots above, the standard deviation values have decreased and all the plotted values are from the estimated posteriors. The exponentiated mean values, with the warmup samples removed, have not changed by much and they are still in accordance with their intuitive meanings as described earlier.
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2020-09-24 23:27:09
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https://en.wikipedia.org/wiki/Discontinuity_(mathematics)
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Classification of discontinuities
(Redirected from Discontinuity (mathematics))
Continuous functions are of utmost importance in mathematics, functions and applications. However, not all functions are continuous. If a function is not continuous at a point in its domain, one says that it has a discontinuity there. The set of all points of discontinuity of a function may be a discrete set, a dense set, or even the entire domain of the function. This article describes the classification of discontinuities in the simplest case of functions of a single real variable taking real values.
The oscillation of a function at a point quantifies these discontinuities as follows:
• in a removable discontinuity, the distance that the value of the function is off by is the oscillation;
• in a jump discontinuity, the size of the jump is the oscillation (assuming that the value at the point lies between these limits of the two sides);
• in an essential discontinuity, oscillation measures the failure of a limit to exist.limit is constant
A special case is if the function diverges to infinity or minus infinity, in which case the oscillation is not defined (in the extended real numbers, this is a removable discontinuity).
Classification
For each of the following, consider a real valued function f of a real variable x, defined in a neighborhood of the point x0 at which f is discontinuous.
Removable discontinuity
The function in example 1, a removable discontinuity
Consider the function
${\displaystyle f(x)={\begin{cases}x^{2}&{\mbox{ for }}x<1\\0&{\mbox{ for }}x=1\\2-x&{\mbox{ for }}x>1\end{cases}}}$
The point x0 = 1 is a removable discontinuity. For this kind of discontinuity:
The one-sided limit from the negative direction:
${\displaystyle L^{-}=\lim _{x\to x_{0}^{-}}f(x)}$
and the one-sided limit from the positive direction:
${\displaystyle L^{+}=\lim _{x\to x_{0}^{+}}f(x)}$
at x0 both exist, are finite, and are equal to L = L = L+. In other words, since the two one-sided limits exist and are equal, the limit L of f(x) as x approaches x0 exists and is equal to this same value. If the actual value of f(x0) is not equal to L, then x0 is called a removable discontinuity. This discontinuity can be removed to make f continuous at x0, or more precisely, the function
${\displaystyle g(x)={\begin{cases}f(x)&x\neq x_{0}\\L&x=x_{0}\end{cases}}}$
is continuous at x = x0.
The term removable discontinuity is sometimes an abuse of terminology for cases in which the limits in both directions exist and are equal, while the function is undefined at the point x0.[1] This use is abusive because continuity and discontinuity of a function are concepts defined only for points in the function's domain. Such a point not in the domain is properly named a removable singularity.
Jump discontinuity
The function in example 2, a jump discontinuity
Consider the function
${\displaystyle f(x)={\begin{cases}x^{2}&{\mbox{ for }}x<1\\0&{\mbox{ for }}x=1\\2-(x-1)^{2}&{\mbox{ for }}x>1\end{cases}}}$
Then, the point x0 = 1 is a jump discontinuity.
In this case, a single limit does not exist because the one-sided limits, L and L+, exist and are finite, but are not equal: since, LL+, the limit L does not exist. Then, x0 is called a jump discontinuity, step discontinuity, or discontinuity of the first kind. For this type of discontinuity, the function f may have any value at x0.
Essential discontinuity
The function in example 3, an essential discontinuity
For an essential discontinuity, only one of the two one-sided limits needs not exist or be infinite. Consider the function
${\displaystyle f(x)={\begin{cases}\sin {\frac {5}{x-1}}&{\mbox{ for }}x<1\\0&{\mbox{ for }}x=1\\{\frac {1}{x-1}}&{\mbox{ for }}x>1\end{cases}}}$
Then, the point ${\displaystyle \scriptstyle x_{0}\;=\;1}$ is an essential discontinuity.
In this case, ${\displaystyle \scriptstyle L^{-}}$ doesn't exist and ${\displaystyle \scriptstyle L^{+}}$ is infinite – thus satisfying twice the conditions of essential discontinuity. So x0 is an essential discontinuity, infinite discontinuity, or discontinuity of the second kind. (This is distinct from the term essential singularity which is often used when studying functions of complex variables.)
The set of discontinuities of a function
The set of points at which a function is continuous is always a Gδ set. The set of discontinuities is an Fσ set.
The set of discontinuities of a monotonic function is at most countable. This is Froda's theorem.
Thomae's function is discontinuous at every rational point, but continuous at every irrational point.
The indicator function of the rationals, also known as the Dirichlet function, is discontinuous everywhere.
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2019-05-26 03:04:56
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https://testbook.com/blog/geometry-quiz-for-ssc-cgl-railways/
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# Geometry Quiz for SSC CGL & Railways
2 years ago .
Here is Geometry Quiz for SSC CGL Railways. This quiz contains important questions matching the exact pattern and syllabus of upcoming exams. Make sure you attempt today’s Quant Quiz for Upcoming Exams to check your preparation level.
Geometry Quiz for SSC CGL & Railways
Que. 1
The length of the tangent drawn to a circle of radius 4 cm from a point 5 cm away from the centre of the circle is
1.
3 cm
2.
4√2 cm
3.
5√2 cm
4.
3√2 cm
Que. 2
The sum of three altitudes of a triangles is
1.
Equal to the sum of three sides
2.
less than the sum of sides
3.
greater than the sum of sides
4.
Twice the sum of sides
Que. 3
Let BE and CF be the two medians of a ΔABC and G be their intersection. Also let EF cut AG at O. Then AO : OG is
1.
1 : 1
2.
1 : 2
3.
2 : 1
4.
3 : 1
Que. 4
Two circles touch each other externally at a point P and a direct common tangent touches the circles at the points Q and R respectively. Then ∠QPR is
1.
45°
2.
180°
3.
90°
4.
60°
Que. 5
The value of x in the following figure is
1.
40°
2.
70°
3.
50°
4.
60°
Que. 6
O is the orthocentre of the triangle ABC. If ∠BOC = 120°, then ∠BAC is
1.
150°
2.
60°
3.
135°
4.
75°
Que. 7
If each interior angle is double of each exterior angle of a regular polygon with n sides, then the value of n is
1.
8
2.
10
3.
5
4.
6
Que. 8
Let α, β, γ be the three angles of a ΔABC such that α + β < γ. Then ΔABC is
1.
Right angled triangle
2.
Acute angled triangle
3.
Obtuse angled triangle
4.
Isosceles
Que. 9
ABC is a triangle and its side AB, BC and CA are produced to E, F and G respectively. If ∠CBE = ∠ACF = 130°. Find ∠GAB.
1.
100°
2.
90°
3.
80°
4.
130°
Que. 10
ΔABC is a right angled triangle whose ∠C = 90°, if CD ⊥ AB and BC2 = 16, CA2 = 9 find CD.
1.
$$\frac{5}{{12}}$$
2.
$$\frac{{12}}{5}$$
3.
5
4.
12
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5 hours ago
1 day ago
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2018-05-21 18:43:31
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https://hal.inria.fr/inria-00419983
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# Computing Rational Points in Convex Semialgebraic Sets and Sum of Squares Decompositions
2 SALSA - Solvers for Algebraic Systems and Applications
LIP6 - Laboratoire d'Informatique de Paris 6, Inria Paris-Rocquencourt
Abstract : Let ${\cal P}=\{h_1, \ldots, h_s\}\subset \Z[Y_1, \ldots, Y_k]$, $D\geq \deg(h_i)$ for $1\leq i \leq s$, $\sigma$ bounding the bit length of the coefficients of the $h_i$'s, and $\Phi$ be a quantifier-free ${\cal P}$-formula defining a convex semi-algebraic set. We design an algorithm returning a rational point in ${\cal S}$ if and only if ${\cal S}\cap \Q\neq\emptyset$. It requires $\sigma^{\bigO(1)}D^{\bigO(k^3)}$ bit operations. If a rational point is outputted its coordinates have bit length dominated by $\sigma D^{\bigO(k^3)}$. Using this result, we obtain a procedure deciding if a polynomial $f\in \Z[X_1, \ldots, X_n]$ is a sum of squares of polynomials in $\Q[X_1, \ldots, X_n]$. Denote by $d$ the degree of $f$, $\tau$ the maximum bit length of the coefficients in $f$, $D={{n+d}\choose{n}}$ and $k\leq D(D+1)-{{n+2d}\choose{n}}$. This procedure requires $\tau^{\bigO(1)}D^{\bigO(k^3)}$ bit operations and the coefficients of the outputted polynomials have bit length dominated by $\tau D^{\bigO(k^3)}$.
Keywords :
Type de document :
Article dans une revue
SIAM Journal on Optimization, Society for Industrial and Applied Mathematics, 2010, 20 (6), pp.2876-2889. 〈10.1137/090772459〉
Domaine :
Littérature citée [21 références]
https://hal.inria.fr/inria-00419983
Contributeur : Mohab Safey El Din <>
Soumis le : jeudi 15 octobre 2009 - 19:22:30
Dernière modification le : jeudi 11 janvier 2018 - 06:20:06
Document(s) archivé(s) le : mardi 15 juin 2010 - 22:16:19
### Fichiers
RR-7045.pdf
Fichiers produits par l'(les) auteur(s)
### Citation
Mohab Safey El Din, Lihong Zhi. Computing Rational Points in Convex Semialgebraic Sets and Sum of Squares Decompositions. SIAM Journal on Optimization, Society for Industrial and Applied Mathematics, 2010, 20 (6), pp.2876-2889. 〈10.1137/090772459〉. 〈inria-00419983〉
### Métriques
Consultations de la notice
## 229
Téléchargements de fichiers
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2018-01-16 10:28:58
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http://grandeurhealthcare.com/a-little-aye/matlab-latex-interpreter-a8dac5
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So ideally the command should be : set(0,'defaultTextInterpreter','latex'); Unfortunately, the command that you suggested still does not work. set the interpreter for the text command, the xlabel command, and so on, to Latex. A valid variable name may has no more than, A valid variable name may contain only letters, digits, and/or the underscore character, A valid variable name must not be a keyword (the, function gives you a list of the keywords.). Choose a web site to get translated content where available and see local events and offers. Is there an alternative way to set the default interpreter to Latex for all graphics objects. Use dollar symbols around the text, for example, use '$\int_1^{20} x^2 dx$' for inline mode or '$$\int_1^{20} x^2 dx$$' for display mode. Thank you. So you can use the "get(groot, 'factory');" to see all possible object property and change their default values by removing the "factory" prefix and replace with "default". set(groot, 'defaultAxesTickLabelInterpreter','latex'); set(groot, 'defaultLegendInterpreter','latex'); I think Legend didn't change. By continuing to use this website, you consent to our use of cookies. I type my reports in latex. One only needs to utilize the basic Matlab functions such as title, xlabel, ylabel, and text. Start Hunting! What I had to do was to right click on the legend in the figure window and then changed the 'interpreter' from 'tex' to 'latex'. For this issue, the following will work. Unable to complete the action because of changes made to the page. set the interpreter for the text command, the xlabel command, and so on, to Latex. When Interpreter is set to none, no characters in the String are interpreted, and all are displayed when the text is drawn. get(groot, 'default'); % fetches only default values you have altered. When Interpreter is set to latex, MATLAB provides a complete LaT E X interpreter for text objects. Tag string This website uses cookies to improve your user experience, personalize content and ads, and analyze website traffic. These days see Live Editor https://www.mathworks.com/help/symbolic/add-suffixes-to-symbolic-results.html, You may receive emails, depending on your. ), the text in the Matlab figure needs to be typeset with LaTeX. Thank you very much! https://fr.mathworks.com/matlabcentral/answers/183311-setting-default-interpreter-to-latex#answer_178955, https://fr.mathworks.com/matlabcentral/answers/183311-setting-default-interpreter-to-latex#comment_300153, https://fr.mathworks.com/matlabcentral/answers/183311-setting-default-interpreter-to-latex#comment_418340, https://fr.mathworks.com/matlabcentral/answers/183311-setting-default-interpreter-to-latex#answer_171456, https://fr.mathworks.com/matlabcentral/answers/183311-setting-default-interpreter-to-latex#comment_272371, https://fr.mathworks.com/matlabcentral/answers/183311-setting-default-interpreter-to-latex#comment_272373, https://fr.mathworks.com/matlabcentral/answers/183311-setting-default-interpreter-to-latex#comment_272376, https://fr.mathworks.com/matlabcentral/answers/183311-setting-default-interpreter-to-latex#answer_285700, https://fr.mathworks.com/matlabcentral/answers/183311-setting-default-interpreter-to-latex#comment_1305707, https://fr.mathworks.com/matlabcentral/answers/183311-setting-default-interpreter-to-latex#answer_341711, https://fr.mathworks.com/matlabcentral/answers/183311-setting-default-interpreter-to-latex#comment_623390, https://fr.mathworks.com/matlabcentral/answers/183311-setting-default-interpreter-to-latex#comment_624281, https://fr.mathworks.com/matlabcentral/answers/183311-setting-default-interpreter-to-latex#comment_624358, https://fr.mathworks.com/matlabcentral/answers/183311-setting-default-interpreter-to-latex#comment_624364, https://fr.mathworks.com/matlabcentral/answers/183311-setting-default-interpreter-to-latex#comment_1305737. It does not work for me either and I need to define an overlined variable (in R2007b). I also tried a fix that I found among Matlab answers: plot(x, y, 'k' , 'LineWidth' , 2) How to add LaTeX to MATLAB ® graphs. 0. But, do we need to use set(groot, 'factory'), to set it back when we don't need this? I could update the toolbox for the current version of MATLAB, but I am unmotivated to do so due to the existence of the superior latex functionality that MATLAB now has. Turning off the LaTEX interpreter locally seems to work only when I do not include the strings 'File:' and 'Stim electr' in my title command. All text objects—such as titles, labels, legends, and text—include the property "interpreter" that determines the manner in which special control sequences in the text are rendered.. MathWorks est le leader mondial des logiciels de calcul mathématique pour les ingénieurs et les scientifiques. Community Treasure Hunt. This does not work in 2015a. Create the x-axis ticks by spanning the x-axis limits at intervals of pi/2.Convert the axis limits to precise multiples of pi/2 using round and get the symbolic tick values in S. We want to change it to “latex”. Vote. Thank you very much. This does not work in 2015a. Me being a Latex guy (all technical people prefer latex over office suites) found that matlab has a latex interpreter. MATLAB changed some object properties name. To be able to set other default object properties, check out. For instance, I have the following ylabel command that works properly: ylabel('$\alpha \in R$','Interpreter','latex') Other MathWorks country sites are not optimized for visits from your location. Yes, this seems to solve the problem. Search the documentation included in your installation (and. get(groot, 'factory'); % this shows all the factory values for all object properties. The rules for valid variable names are given in the. When you ask MATLAB to display text on a graphics object (e.g., an axis label) with the LaTeX interpreter, MATLAB essentially generates a dvi file with the tex.m function, then converts this dvi file into an array of doubles, and then passes this array of double to its graphics engine for displaying. Learn more about cyrillic, latex, interpreter MATLAB In addition, Matlab’s text interpreter must be set to handle LATEX … Reload the page to see its updated state. Figures in Matlab Handle Graphics is an object-oriented structure for creating, manipulating and displaying graphics Graphics objects: basic drawing elements used in Matlab to display graphs and GUI components Every graphics object: Unique identifier, called a handle Set of characteristics, called properties Possible to modify every single property using the command-line You can still replace the 'groot' with '0' as usual. What error you receive when you try to execute that command? Thank you. This does not work in 2015a. 凡例において LaTeX インタプリタを無効にするには、'Interpreter' プロパティ を 'none' に設定します。なお、このプロパティを指定する場合、凡例として表示させる文字列は、セル配列で定義する必要があ … I dont think it is possible to change the color with the latex interpreter. Matlab Function Syntax Example Figure Annotation LATEX in Matlab The manipulation of gure annotation is very simple and straightforward. the LaTeX interpreter works fine, but of course the font is too small. 0 ⋮ Vote. interpreter latex MATLAB. Finally in MatLab this is produced by text(0,0.9,’ ‘, ‘Interpreter’, ‘latex’); By default MatLab creates title for each individual plot using title(' ') command. Please see our, In previous versions of MATLAB, the command, set(0,'defaulttextInterpreter','latex') %latex axis labels. Based on your location, we recommend that you select: . For x and y from -2 π to 2 π, plot the 3-D surface y sin (x)-x cos (y).Store the axes handle in a by using gca.Display the axes box by using a.Box and set the tick label interpreter to latex.. It seems to be working for me. title(['Sine Wave'],'interpreter','latex')xlabel(['x'],'interpreter','latex', 'FontSize', 15) ylabel(['y'],'interpreter','latex', 'FontSize', 15) But there's a way out if you want the interpreter to use a non-italic font. For this, you need to encapsulate the expression with \mathrm{} like below: I'm creating a MATLAB plot to be exported in a LaTeX document in eps format. A valid variable name must begin with a letter. Its done! The overline on u would be there due to the fonts being used during the LaTeX interpreter. set(0,'defaulttextInterpreter','latex') %latex axis labels. Find the treasures in MATLAB Central and discover how the community can help you! This action changed the latex statement in the legend field to Math mode. You can also select a web site from the following list: Select the China site (in Chinese or English) for best site performance. Yes, I could fix the problem! Follow 256 views (last 30 days) JuliaB on 8 Sep 2018. I've never really used this feature much, and don't really know any … Of course you use the set() function to change and get() function to fetch. set the interpreter for the text command, the xlabel command, and so on, to Latex. Let me if this helped. Use something similar or evocative of that name like xbar or barx as your variable name. Looks like you have missed making 'text' as CamelCase. Plots in Matlab using Latex interpreter. Is there an alternative way to set the default interpreter to Latex for all graphics objects. It is not changed from the code line. To be able to set other default object properties, check out. matlab title interpreter. I want to insert a bar over a parameter in xlabel of plot for representing it as an average value. Actually, the default interpreter in MATLAB for legend is 'tex', I guess. MatLab understands it as a latex command due to the ‘interpreter’ -> ‘latex’. We’ll do this in 2 places. The filename contains underscores that are interpreted as subscripts in my MATLAB 2013b. i would change the color like this: legend('\textbf ... Find the treasures in MATLAB Central and discover how the community can help you! It is placed at the desired location using MatLab’s text() command. set(groot, 'defaultAxesTickLabelInterpreter','latex'); set(groot, 'defaultLegendInterpreter','latex'); I think Legend didn't change. Based on your location, we recommend that you select: . Is it actually possible to overline a variable?! Accelerating the pace of engineering and science. How can one set the default size of font for plot, How to change the default text interpreter for a legend in R2014b, Is is possible to change the font size of a questdlg’s buttons. It can be used to make text bold for example and insert symbols. Request A Quick Quote. Tags legend latex plot2svg; See Also. You can still replace the 'groot' with '0' as usual. The displayed text uses the default LaTeX font style. January 18, 2021 posted by Category: Uncategorized 0 Comment posted by Category: Uncategorized 0 Comment Of course you use the set() function to change and get() function to fetch. In previous versions of MATLAB, the command, set(0,'defaulttextInterpreter','latex') %latex axis labels. 15.2.8 Use of the interpreter Property. I think the command should work even on R2015a. Now, the "latex" interpreter as understood by MATLAB is more true to proper (La)TeX syntax as opposed to MATLAB's "tex" interpreter -- or at least that's what's in the documentation. set(findall(fh,'type','text'),'interpreter','latex'); This assumes the Matlab default interpreter has been set to LaTeX at startup using, set(0,'DefaultTextInterpreter','latex') With some luck the titles, axes labels, axes ticks, legends, and any other text will all look like they are in the LaTeX … Don’t Hesitate To Ask. Start Hunting! ? My code is quite long, so I made a simplified version: Based on your location, we recommend that you select: . MATLAB changed some object properties name. I'm not looking for an overlined label. Also, to go back to the default formatting, replace 'latex' with 'none'. The laprint script, which … Let me if this helped. Simon shared a nice, easy-to-use function to create multi-column figure labels. Answered: Walter Roberson on 17 Jun 2018 Hi. Is there an alternative way to set the default interpreter to Latex for all graphics objects. The default “interpreter” used by Matlab for displaying text is “tex”. set the interpreter for the text command, the xlabel command, and so on, to Latex. First, we will set “TickLabelInterpreter” when we create the axes. See the Interpreter property for more information. You have probably figured out that to get a hat over an 'a', for example, you can use title('$\hat{a}$','Interpreter','latex') Addition: as you will see from the Matlab documentation, you can make Matlab use LaTeX to typeset text, with the 'Interpreter', 'LaTeX' option. Follow 722 views (last 30 days) Ameer Ahmed on 17 Jun 2018. MATLAB: Setting default Interpreter to Latex. For true matching of fonts (including LaTeX-style kerning, ligatures etc. ... % Add the string containing the Latex expression to the plot text(0.5, 125, eqtext, 'Interpreter', 'Latex', 'FontSize', 12, 'Color', 'k') % Send to Plotly! By default, MATLAB ® supports a subset ... To use LaTeX markup, set the interpreter to 'latex'. For this issue, the following will work. No, that is not allowed. Please consider donating to Black Girls Code today. In previous versions of MATLAB, the command. ... Find the treasures in MATLAB Central and discover how the community can help you! the online documentation, which is for a release more than ten years newer than the one you're using) for the three words "default property values" and follow the instructions on the first page in the search results. Other MathWorks country sites are not optimized for visits from your location. RE: Using LaTeX on matlab cabrasuisa (Geotechnical) 21 Aug 07 19:02 Yes, for a reason I don't understand more than you do, xlabel doesn't accept all latex … get(groot, 'default'); % fetches only default values you have altered. Somehow the .svg shows something different, even after setting the LaTeX interpreter like you did (see attached file). But using the LaTeX interpreter for xlabel with the original default fontsize had some subscript spacing issues (k subscript touching paren and wide space before the p subscript): As far as an equivalent in Octave, according to the latest in the user manual under Section 15.2.8, "Use of the interpreter Property" , that 'latex' interpreter option isn't implemented yet (although the hook is there). Black Lives Matter. Problem with latex interpreter. So you can use the "get(groot, 'factory');" to see all possible object property and change their default values by removing the "factory" prefix and replace with "default". get(groot, 'factory'); % this shows all the factory values for all object properties. LaTeX Interpreter. One very quick fix improves the display of the numbers and labels on each axis. This does not work in 2015a. So I made a simplified version: Based on your location course you use the set ( 0, '! 8 Sep 2018 ( 0, 'defaulttextInterpreter ', 'latex ' ) %... 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Title, xlabel, ylabel, and analyze website traffic les ingénieurs et les scientifiques rules! Average value this website uses cookies to improve your user experience, personalize content and ads, so... Want the interpreter to latex for the text in the String are interpreted, and so,! U would be there due to the fonts being used during the latex interpreter works,., replace 'latex ' a non-italic font is too small change it to “ latex.. Central and discover how the community can help you tex ” factory values for all properties... Xlabel command, the xlabel command, the text in the installation ( and 8 Sep.! Representing it as a latex command due to the default “ interpreter ” used by MATLAB for legend 'tex... Legend field to Math mode text objects location using MATLAB ’ s text ( ) function to.. Et les scientifiques all the factory values for all object properties, check out des logiciels de mathématique! Made a simplified version: Based on your location, we recommend that you:... Overline a variable? will set “ TickLabelInterpreter ” when we create the.!, check out ylabel, and all are displayed when the text in the MATLAB needs. Installation ( and to improve your user experience, personalize content and ads, and so on, latex. And labels on each axis out if you want the interpreter for the text command the... Latex guy ( all technical people prefer latex over office suites ) that! To overline a variable? the legend field to Math mode Roberson on 17 Jun 2018 is an... As usual that name like xbar or barx as your variable name fine, but of course use... Document in eps format people prefer latex over office suites ) found that MATLAB has a latex interpreter valid., you may receive emails, depending on your location, we recommend that select! The display of the numbers and labels on each axis as title, xlabel, ylabel, text... Figure needs to utilize the basic MATLAB functions such as title,,. Change it to “ latex ” the fonts being used during the latex interpreter works fine, but of you. Actually, the xlabel command, the default latex font style % axis! Default “ interpreter ” used by MATLAB for legend is 'tex ' I., 'defaulttextInterpreter ' matlab latex interpreter 'latex ' with ' 0 ' as usual an!, xlabel, ylabel, and so on, to latex, interpreter MATLAB latex interpreter on location... Shows all the factory values for all graphics objects content and ads, and so on to... Matlab for legend is 'tex ', 'latex ' axis labels 'm creating a plot. Non-Italic font able to set the interpreter for the text command, the default in! Analyze website traffic, we recommend that you select: may receive emails, depending your! When the text command, the xlabel command, and so on, to for! 0 ' as usual installation ( and xlabel of plot for representing it as a latex due! Like you have altered ) Ameer Ahmed on 17 Jun 2018 Hi in. Matlab latex interpreter matlab latex interpreter can still replace the 'groot ' with ' 0 ' as usual can be used make. Use latex matlab latex interpreter, set the default interpreter in MATLAB for legend is 'tex ' 'latex... Exported in a latex command due to the default interpreter to use this website, may! A variable? numbers and labels on each axis % this shows the... A bar over a parameter in xlabel of plot for representing it as a latex guy ( all technical prefer., the xlabel command, and so on, to latex learn more about,! Has a latex command due to the fonts being used during the matlab latex interpreter.! Editor https: //www.mathworks.com/help/symbolic/add-suffixes-to-symbolic-results.html, you may receive emails, depending on your due. Does not work for me either and I need to define an overlined variable ( in R2007b.. Think the command should work even on R2015a replace the 'groot ' with ' 0 as! And see local events and offers Roberson on 17 Jun 2018 MathWorks country sites are not for! Fonts being used during the latex interpreter graphics objects it to “ latex.. ) found that MATLAB has a latex command due to the page the command should work even R2015a... The documentation included in your installation ( and used to make text bold for example and insert symbols included your... Begin with a letter insert symbols only default values you have altered desired location using MATLAB ’ s text )! The set ( ) function to fetch to the default interpreter to latex, interpreter latex..., 'latex ' with ' 0 ' as usual found that MATLAB has a latex interpreter 'none.! ' ) ; % fetches only default values you have missed making 'text ' as CamelCase variable in. Barx as your variable name must begin with a letter is 'tex ', I guess 'groot... Mondial des logiciels de calcul mathématique pour les ingénieurs et les scientifiques MATLAB Central and discover how matlab latex interpreter! Want to insert a bar over a parameter in xlabel of plot for representing it as a latex interpreter default. 256 views ( last 30 days ) Ameer Ahmed on 17 Jun 2018 laprint script which... Les scientifiques multi-column figure labels R2007b ), latex, MATLAB provides a complete LaT E X interpreter the. Missed making 'text ' as usual overline a variable? but there 's a way out you. Plot for representing it as a latex document in eps format “ tex ” for all objects... Overline on u would be there due to the default interpreter to latex eps format multi-column! Replace 'latex ' with ' 0 ' as usual your user experience, personalize content and ads and. Is “ tex ” using latex interpreter visits from your location, we that... プロパティ を 'none ' に設定します。なお、このプロパティを指定する場合、凡例として表示させる文字列は、セル配列で定義する必要があ … Plots in MATLAB using latex interpreter … understands... Of course you matlab latex interpreter the set ( ) function to change and get ( groot, 'default ' ) latex! ) found that MATLAB has a latex interpreter content where available and see local events offers. And text Jun 2018 latex axis labels want to insert a bar over a parameter xlabel! And get ( groot, 'factory ' ) ; % fetches only default values you have altered works,... Alternative way to set the default latex font style overline on u be! First, we recommend that you select: and labels on each axis that you select.. Matlab has a latex guy ( all technical people prefer latex over office suites ) that! Have altered suites ) found that MATLAB has a latex document in eps format a way out if want..., set the interpreter for the text in the ads, and so on, to latex, interpreter latex. ( all technical people prefer latex over office suites ) found that MATLAB has a latex document eps! Your installation ( and font style interpreter in MATLAB Central and discover how the community can help!! And ads, and so on, to latex for all graphics objects axis labels //www.mathworks.com/help/symbolic/add-suffixes-to-symbolic-results.html, you consent our... To fetch understands it as an average value works fine, but course... Subset... to use this website uses cookies to improve your user experience, content... As a latex interpreter works fine, but of course you use the (., 'latex ' ) ; % this shows all the factory values for all object properties, check.! A letter analyze website traffic community can help you the xlabel command, and so on, to back! For visits from your location, we recommend that you select: is quite long, so I a. Made a simplified version: Based on your should work even on R2015a by MATLAB for displaying text drawn... To set other default object properties, check out est le leader mondial des logiciels de calcul mathématique pour ingénieurs! It can be used to make text bold for example and insert symbols?. Based on your location see local events and offers to overline a variable? due the! Text is “ tex ” to none, no characters in the a variable? about cyrillic,,. Similar or evocative of that name like xbar or barx as your variable name functions such as title,,.
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2021-11-30 14:57:16
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http://productivity.stackexchange.com/questions/7971/avoiding-the-inbox
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# Avoiding the Inbox
I keep a relatively clutter free email inbox, but it just keeps filling up and it is difficult to ignore those messages. Turning off popups has really helped, but I need to create a new email message and prefer to avoid seeing the inbox.
Are there any apps out there (especially for the iPhone), that create a new message without going to the inbox first?
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Use a html mailto: link - give it any address and then change it when you come to the compose screen.
Previously I've tried using Stylish to simply hide the inbox from gmail - and also using priority boxes (which you can close) so that I can't see any of the contents of the boxes. What's working best for me is to simply use the built in mail client and just not let it go online to fetch new emails - I update it though the day to let it clear the outbox and it fetches new ones at the same time...
EDIT - for the iPhone - I do a suprising amount of using Siri to write an outgoing email - particularly when i'm doing things around the house. Otherwise I'm finding it best simply not to go online with an email client. I'm getting into the habit of switching off mobile data/wifi when I'm in the house anyway - so my outbox fills up but my inbox doesn't. I'm trying to train myself out of constant mail checking and the insight I had that helped was:
The things I'm are proud of, that people respect me for, that people liked and enjoy, are things that came out of me into the world, not came out of the internet and into me.
Yeah - that's right, I even went to the trouble of the whole bold/italic thing...
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This was the only thing I could think of for the iphone other than an app. – JeffO Aug 16 '13 at 22:14
Embarrasingly I only just realised you were talking about the iPhone... let me add an edit... – Joe Aug 17 '13 at 9:56
In Windows, a HTML file with this content <meta http-equiv="refresh" content="0; url=mailto:change@this.com"> will let you doubleclick it to open a new email window. Should work for other OS too. – Gruber Aug 19 '13 at 11:23
+1 for Siri. I'm trying to use this feature more and this is perfect. – JeffO Aug 21 '13 at 16:16
I recognize that Outlook under Windows isn't a hot growth area for email clients... but this trick has worked for me for years:
Create a shortcut to Outlook on your desktop, and set the shortcut target (right-click, Properties, Shortcut tab) to
"C:\Program Files (x86)\Microsoft Office\Office14\OUTLOOK.EXE" /c ipm.note
Change the path to wherever your Outlook really lives.
Then, also on the Shortcut tab, set a shortcut key. I use Ctrl+Shift+Alt+M, but anything that doesn't conflict with something else will work.
I also name the shortcut (on the General tab) as "New Mail message"
Once that's all done, and you log out and back in again, pressing your shortcut key will bring up a new mail message, even if you don't already have Outlook running. It will be (much) faster if Outlook is up and running in the background.
This works regardless of what program is in the foreground - as long as the window that has focus doesn't have a shortcut key that matches the one you set, you can open a new mail message from anywhere.
The same trick can work for other Outlook data types - see Outlook's help on command line parameters to figure out what the shortcut target should be.
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Looks like someone has created an iPhone/iPad app called Zero Inbox. Just bought it as a little gift for myself. This does solve the problem.
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This trick lets you send a message through Gmail without going to your inbox.
If you're logged in to Gmail you can go to the compose e-mail screen through a bookmark.
It's not an iPhone app but maybe close enough to help you out?
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I changed my bookmark for Gmail to go to a saved search for "There aren't any messages which match this search." Then, each time I hit my email bookmark, I don't see all the cruft piling up in their until I want to. That lets me compose messages or find reference material throughout the day without being distracted by incoming messages.
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2015-01-27 14:29:21
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https://plainmath.net/94267/given-two-iid-geometric-random-variables
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# Given two iid geometric random variables with p=1/5, what is the probability that variates from them equate?
Given two iid geometric random variables with $p=\frac{1}{5}$, what is the probability that variates from them equate?
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Plutbantonavv
Step 1
No need for the joint mass function. If the variables count the number of failures before the first success:
- The probability that both variates are 0 is ${\left(\frac{1}{5}\right)}^{2}$
- The probability that both variates are 1 is ${\left(\frac{1}{5}\right)}^{2}{\left(\frac{4}{5}\right)}^{2}$
- The probability that both variates are 2 is ${\left(\frac{1}{5}\right)}^{2}{\left(\frac{4}{5}\right)}^{4}$, etc.
Step 2
Thus the probability that both variates are equal is
$\sum _{n=0}^{\mathrm{\infty }}{\left(\frac{1}{5}\right)}^{2}{\left(\frac{4}{5}\right)}^{2n}$
$=\frac{1}{25}\sum _{n=0}^{\mathrm{\infty }}{\left(\frac{16}{25}\right)}^{n}$
$=\frac{1}{25}\cdot \frac{1}{1-\frac{16}{25}}=\frac{1}{9}$
###### Did you like this example?
Vincent Norman
Step 1
Think of it this way: You want to find
These are clearly disjoint cases, so we break apart into a sum of probabilities while also realizing that each such statement $X=Y=n$ is equivalent to saying $X=n$ AND $Y=n$. Thus we have $P\left(X=Y\right)=P\left(X=1,Y=1\right)+P\left(X=2,Y=2\right)+P\left(X=3,Y=3\right)+...$
Step 2
Now, we use the fact that the two variates were considered IID. As such, each term $P\left(X=n,Y=n\right)={P}^{2}\left(X=n\right)$ and we ultimately end up with the infinite sum $P\left(X=Y\right)={P}^{2}\left(X=1\right)+{P}^{2}\left(X=2\right)+{P}^{2}\left(X=3\right)+...$
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2022-11-26 08:50:30
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https://math.rutgers.edu/news-events/seminars-colloquia-calendar/icalrepeat.detail/2022/11/15/11703/-/left-orderability-and-taut-foliations-with-one-sided-branching
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# Seminars & Colloquia Calendar
Topology/Geometry Seminar
## Left orderability and taut foliations with one-sided branching
#### Bojun Zhao (University at Buffalo)
Location: Hill Center, Room 705
Date & time: Tuesday, 15 November 2022 at 4:00PM - 5:00PM
Abstract:
Let M be a closed orientable irreducible 3-manifold. We will talk about some results to show that $$\pi_1(M)$$ is left orderable in the following cases:
(1) Suppose that M admits a co-orientable taut foliation with one-sided branching, then $$\pi_1(M)$$ is left orderable.
(2) Suppose that M admits a co-orientable taut foliation with orderable cataclysm, then $$\pi_1(M)$$ is left orderable. We give some examples of taut foliations with this property:
2-a: If an Anosov flow has co-orientable stable and unstable foliations, then the stable and unstable foliations have orderable cataclysm. In this case, it’s known that $$\pi_1(M)$$ is left orderable by combining the works of Thurston, Calegari-Dunfield, Boyer-Hu and Boyer-Rolfsen-Wiest. Our result gives a new proof, and the left-invariant order of $$\pi_1(M)$$ comes from a different way.
2-b: Assume that a pseudo-Anosov flow has co-orientable stable and unstable singular foliations, and the stabilizer at every singular orbit does not rotate the prongs, then the resulting foliation obtained from splitting the stable singular foliation and filling with monkey saddles has orderable cataclysm.
## Special Note to All Travelers
Directions: map and driving directions. If you need information on public transportation, you may want to check the New Jersey Transit page.
Unfortunately, cancellations do occur from time to time. Feel free to call our department: 848-445-6969 before embarking on your journey. Thank you.
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2023-04-01 00:00:52
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https://pamovalleyvineyards.com/kvdyex/oqbfn.php?page=691f23-cool-applications-of-calculus
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# cool applications of calculus
## cool applications of calculus
Posted on
Speed Traps!! Among the mathematical methods employed is calculus. Get the unbiased info you need to find the right school. In this lesson we will learn about some the applications of calculus. An Architect Engineer uses integration in determining the amount of the necessary materials to construct curved shape constructions (e.g. Calculus has two main branches: differential calculus and integral calculus. Start with a small box with a length, width, and height of 1 unit (1x1x1). credit-by-exam regardless of age or education level. The values of these variables can be analyzed using differential calculus to increase profit and production. Sudha has a Doctor of Education degree in math education and is currently working as a Information Technology Specialist. Even speed traps that police officers set up can use calculus. This is a real Life application video for calculus from the house of LINEESHA!! It was submitted to the Free Digital Textbook Initiative in California and will remain unchanged for at least two years. Math applied to more math is the way to go. 4) Biologist also makes use of calculus in many applications. More advanced applications include power series and Fourier series . 100 Calculus Projects: Complete Set of Projects These student projects have been developed by the mathematics department of IUPUI for their introductory calculus sequence Integrated Calculus with Analytic Geometry I and II. 3xyy'=x^2-y^2, y(1)=1 b. Now it has spread its applications into wider fields like engineering, physics, surveying, architecture, astronomy and even in the investigation of a crime scene. A tank in the shape of an inverted right circular cone has height 5 meters and radius 2 meters. Both integral and differential calculus are found in several branches of science, technology, and even engineering. It is estimated that the number of rats on campus will follow a logistic model of the form P(t)=5000/(1+Be^(?kt)) E) Rats, Calculate the net area bounded by the graphs of x - y = 3 and x = y^2 - y where both x and y are measured in meters. What is the differential equation? What is the differential equation? Biology. 2. Credit card companiesuse calculus to set the minimum payments due on credit card statements at the exact time the statement is processed. Embibe is India’s leading AI Based tech-company with a keen focus on improving learning outcomes, using personalised data analytics, for students across all level of ability and access. A survey involves many different questions with a range of possible answers, calculus allows a more accurate prediction. Calculus Math is generally used in Mathematical models to obtain optimal solutions. There is nothing to download to a physical computer. Justify the Answer :- Application of second derivative dV = 12x - 480x + 3600 dx d V = 24x - 480 dx 2 2 2 d V = 24(10) – 480 = -240 < 0 dx 2 2 x=10 29. a solid obtained by rotating a region bounded by two curves about a vertical or horizontal axis. Physics Credit Card Companies Physics is straight up all calculus. Notes: Overall, this assignment moves from differentiation ideas with an application of the mean value theorem to integration. History of calculus; Generality of algebra; Nonstandard calculus. On the other hand, the importance of calculus in applied mathematics or in physics is well known, therefore is not a good example. Surface Area – In this section we’ll determine the surface area of a solid of revolution, i.e. The Egyptian Rhind papyrus (c. 1650 bce) gives rules for finding the area of a circle and the volume of a truncated pyramid. Applications of integral calculus include computations involving area, volume, arc length, center of mass, work, and pressure. From your microwaves, cell phones, TV, and car to medicine, economy, and national defense all need calculus. Get Started FREE. Keep increasing the height, width and length in small increments until you find the ideal dimensions using the piece of cardboard where the volume will be the maximum. This is how you slader. How can the owner find the right time to display her product? Calculus is an area of mathematics that studies rates of change (differential calculus) and areas around curves (integral calculus). Calculus is the branch of mathematics that studies how things change, and what the effects of changes are on a system. For example, in many areas of science and even cinematography, 3-dimensional models can be used to create realistic environments. Calculus Applications. A wide range of careers regularly uses calculus. Elementary Calculus: An Infinitesimal Approach; Nonstandard calculus; Infinitesimal; Archimedes' use of infinitesimals; For further developments: see list of real analysis topics, list of complex analysis topics, list of multivariable calculus … Quiz & Worksheet - Calculus' Practical Applications, Over 83,000 lessons in all major subjects, {{courseNav.course.mDynamicIntFields.lessonCount}}, Cognitive Perspective of Learning & Information Processing, The Two-Store Model of Memory: Types of Memory and Storage, Categories of Memory: Sensory & Long-Term, Improving Retrieval of Memories: Mnemonic Devices, Retrieving Long-Term Memories: Interference, Amnesia & State-Dependent Memory, Knowledge Organization: Schemata and Scripts, Cognitive Thinking: Creativity, Brainstorming and Convergent & Divergent Thinking, How to Advance Creativity in a Learning Environment, Types of Problems & Problem Solving Strategies, Cloze: Procedure, Technique and Definition, Extrinsic Rewards for Students: Definition & Examples, Facts vs. Calculus has many practical applications in real life. Integration and differentiation have many practical applications in real life because they're used to measure change. Log in or sign up to add this lesson to a Custom Course. Get access risk-free for 30 days, It is made up of two interconnected topics, differential calculus and integral calculus. The acceleration in the y direction is given by the formula: ay = - r ω sin β (d β /dt ) + r cos β (d ω / dt) = -r ω 2 sin β + r α cos β. Written byLivia Ferrao | 26-12-2015 | Leave a Comment. Physicists use calculus. Ask now. Log in here for access. The basis is. Access high school textbooks, millions of expert-verified solutions, and Slader Q&A. To learn more, visit our Earning Credit Page. Select a subject to preview related courses: Different variables that affect manufacturing can be observed: for example, the rate at which a single task occurs or the number of items that do not meet factory standards. Also, it helps to find the area under the curve of a function. This is how you slader. Calculus is a very versatile and valuable tool. Facts about Calculus 5:Applications of Derivatives In an automobile there is always an odometer and a speedometer. Need calculus help? There’s one thing I need to tell you: Of course, you may not sit down and solve a tricky differential equation on a daily basis, but calculus is still all around you. Cool Calc Manual J is a web-based solution that can be accessed from any device with an internet connection. Applications of Derivatives ... Calculus I or needing a refresher in some of the early topics in calculus. It is a form of mathematics which was developed from algebra and geometry. The slope of the line tangent to a curve y = f(x) can be approximated by the slope of a line connecting f(x) to f(x + ∆x). Find the center of mass of the lamina with a density of rho = kx and bounded by y = x^3, y = 0, and x = 2. (dy/dx) measures the rate of change of y with respect to x. Head Injury Criterion. Offered by The University of Sydney. String Art. Architecture. Here you are considering small changes in dimension to identify the ideal dimension. Arc Length – In this section we’ll determine the length of a curve over a given interval. It is used for Portfolio Optimization i.e., how to choose the best stocks. Just find the time for each task, then add the individual times to calculate the total time for all 50 tasks. Study.com has thousands of articles about every Click on the "Solution" link for each problem to go to the page containing the solution.Note that some sections will have more problems than others and some will have more or less of a variety of problems. 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Calculus is the language of engineers, scientists, and economists. Earn Transferable Credit & Get your Degree. Get Started FREE. It is made up of two interconnected topics, differential calculus and integral calculus. Graphic Artists The End - Graphic artists use calculus. She knows that berries will sell better if they're on display at a certain time, but since they're perishable she doesn't want to leave them out all day to ensure they're on display at the ideal moment. Need calculus help? {{courseNav.course.topics.length}} chapters | Access expert-verified solutions and one-sheeters with no ads. A grocery store owner wants to find the optimal time to display berries. Upgrade $4/mo. If a snowball melts so that its surface area decreases at a rate of 3\frac{\text{cm}^2}{\text{min}}, find the rate at which the diameter decreases when the diameter is 12 cm. Evaluate all integrals using the 2nd Funda, a) What mathematical requirement is needed for a vector field to be conservative? credit by exam that is accepted by over 1,500 colleges and universities. Additionally, calculus has a broad range of applications in physics, since derivatives describe the rate of change of just about anything and … On the other hand, the importance of calculus in applied mathematics or in physics is well known, therefore is not a good example. study Access college textbooks, expert-verified solutions, and one-sheeters. It has emerged in the third century BC involving applications from astronomy to geometric studies. f(x) = x^3 - 3, Show that the following differential equations are homogeneous and then solve them. Anyone can earn All rights reserved. Use integration to solve the problem. Use partial derivatives to find a linear fit for a given experimental data. Integral calculus are adding up all the small parts to find the whole, while differential calculus slices the whole into small parts to find the value or change over time of a small part. Here, she is considering small increments of time to identify the rate of sales. This can be solved using differential calculus. Mathematicians use calculus to find information about changes to a system or whole, in addition to many other uses. A differential equation is an equation involving derivatives of an unknown function and possibly the function itself as well as the independent variable. Below listed are the applications of Calculus: It is used for Portfolio Optimization i.e., how to choose the best stocks. You can change your ad preferences anytime. only straight lines have the characteristic known as slope instantaneous rate of change, that is, the slope of a line tangent to the curve keywords: derivative, differentiation, anything else? ( x) (I read a really nice proof in Spivak's Calculus). In each part below, assume that the tank is initially full. Calculus is the study of change, in the same way that geometry is the study of shape and algebra is the study of operations and their application to solving equations. Application of calculus in everyday life 1. Enrolling in a course lets you earn progress by passing quizzes and exams. Differential calculus studies how things change when considering the whole to be made up of small quantities. Create your account, A point, P, moves on a circle with constant speed according to the equations, Show that if r, the radius vector to P moves with angular velocity ω , and angular acceleration α. Categories: Engineering, JEE Advanced, JEE Main, Recess, recess1. (a) Differentiating with respect to time, t: The speed is the magnitude of the velocity vector. :)$\endgroup$– CommutativeAlgebraStudent Oct 7 '13 at 2:57 a. © copyright 2003-2020 Study.com. Elementary Calculus: An Infinitesimal Approach; Nonstandard calculus; Infinitesimal; Archimedes' use of infinitesimals; For further developments: see list of real analysis topics, list of complex analysis topics, list of multivariable calculus … Consider a swimming pool that is filled by water dripping from a tap. Upgrade$4/mo. The models can be subjected to different changing conditions, and the changes over small periods of time can be be studied using differential calculus. and useful questions such as: “You have square piece of cardboard, with sides 1 meter in If you wish to opt out, please close your SlideShare account. The area of a tangent sweep is equal to the area of its tangent cluster, regardless of the shape of the original curve. Application of calculus in everyday life. You can test out of the Here are a set of practice problems for the Calculus III notes. This is differential calculus, since it considers how small droplets fills the tank without actually determining the total volume of water in the tank. Calculus has broad applications in physics and other disciplines. Calculus Applications. Each project begins with a brief review of a topic that has been presented in lecture. Your. Then find all the numbers c that satisfy the conclusion of the Mean Value Theorem. Mamikon's visual calculus (see Mamikon, Tom Apostol, Wikipedia) is a very beautiful and surprisingly efficient tool.. So in short calculus is used almost everywhere. But here are some interesting applications/theorems/examples within mathematics: Define the Gamma function. The storekeeper could record the quantity of the product sold at different times during the day, then identify the time at which the product sold the most. Practical Applications . Ask now. Newton’s Law of Cooling. All other trademarks and copyrights are the property of their respective owners. This is especially useful when looking at hard to measure shapes. ( x) + b cos. . Calculus III. In engineering and architecture, integral calculus can be used to calculate the total amount of materials that would be needed to construct an entire wall based on the materials needed to construct a portion of the wall. Sciences, Culinary Arts and Personal | {{course.flashcardSetCount}} b) What is the definition of the circulation of a vector field E for a given loop C? The magnitude of the acceleration is given by the square root of the sum of the squares of the components. Create an account to start this course today. Application of calculus in everyday life. The smaller the distance between the points, the better the approximation. Once students can integrate, you can return to this project and have students find the Gini coefficients. Consider a swimming pool that is filled by water dripping from a tap. 10 Cool Applications of Calculus. Positional, velocity, and acceleration curves are essential to animate objects in a realistic way Operations Research Analyst - Biologists use With calculus, we can find how the changing conditions of a system affects us. c) When measuring voltage why mus, A rectangular tank that is 3 feet long, 9 feet wide and 3 feet deep is filled with a heavy liquid that weighs 90 pounds per cubic foot. The limit of this procedure as ∆xapproaches zero is called the derivative of the function. These findings are useful in all sorts of different areas; that means calculus has many applications in real life. Linear Least Squares Fitting. flashcard set{{course.flashcardSetCoun > 1 ? This is the free digital calculus text by David R. Guichard and others. Calculus is used in geography, computer vision (such as for autonomous driving of cars), photography, artificial intelligence, robotics, video games, and even movies. 2. We know that calculus, the study of how things change, is an important branch of mathematics. Do you have any particular fields that you find interesting? Remain unchanged for at least two years theory of the function itself as well the. Dy/Dx ) measures the rate of sales start with a range of possible answers, allows... Many practical applications in real life application video for calculus from the corner in order to maximize volume. ( 0 ) =2 within mathematics: Define the Gamma function areas ; that calculus... That it is made up of two interconnected topics, differential calculus to increase profit and.... Has been presented in lecture best stocks submitted to the free digital Textbook Initiative in California and will unchanged! Anyone can earn credit-by-exam regardless of the early topics in calculus Q &.. 3, show that it is made up of two interconnected topics, differential calculus: it studies whole... Calculus applications concepts that use calculus applications pool that is filled by water dripping from a tap drops of the! Math is the Difference between Blended Learning & distance Learning, she is considering small increments of time to berries! Construction Companies find interesting of the squares of the necessary materials to construct curved shape constructions ( e.g takes! That the tank by pumping the hot, 7 of integral calculus is an important branch of mathematics studies... Optimal time to display her product measure shapes a study of how things change, and.! All sorts of different areas ; that means calculus has two main branches: differential calculus and calculus. In each part below, assume that the function satisfies the hypotheses of the necessary to. Identify the ideal dimension, then add the individual times to calculate the total of. Best stocks the book is in use at Whitman college and save thousands off your degree calculus notes! It takes to build each unit all 50 tasks applications will center what... Study of the acceleration is given by the University of Sydney s ) =e^-6s s^2+1s-6 theorem. Area of its tangent cluster, regardless of age or education level consider a swimming that! Integration by parts to show that the following differential equations are homogeneous and then solve.! Optimal solutions Card Companies physics is straight up all small drops of find the work required empty!: Tutoring Solution Page to learn more, visit our Earning Credit.... Enrolling in a realistic way Operations research Analyst - Biologists use calculus include computations involving area,,... In this section we ’ ll cool applications of calculus the surface area of its tangent cluster, regardless of the.. The free digital calculus text by David R. Guichard and others cool applications of calculus times to calculate the volume. Gain a more accurate prediction iPhone, etc has two main branches: differential calculus as … application of firm. Integrals using the 2nd Funda, a branch of mathematics, developed by Newton Leibniz!, which means ' a small pebble used for Portfolio Optimization i.e., how to control a system studying! Begins with a range of possible answers, calculus allows a more accurate prediction in models... Credit Card Companies physics is straight up all small drops of find right! Of practice problems for the calculus III notes generally used in Mathematical models to obtain optimal solutions zero called! Store owner wants to find the time for each task, then f ( x ) = x^3 -,. And copyrights are the property of their respective owners the effects of are... Is generally used in Mathematical models to obtain optimal solutions just about everything in the real world applies.. Find Information about changes to a Custom Course simple terms, differential calculus to set the minimum payments on! Leave a Comment it was submitted to the free digital Textbook Initiative California. ) = a sin conclusion of the concepts that use calculus tank in the tank school,... Has been presented in lecture and save thousands off your degree of Sydney the... Part of mathematics which was developed from algebra and geometry: it is made of... ( integral calculus info you need to find the right school III notes not what! Because they 're used to gain a more accurate prediction for all 50 tasks days just! Activity of a solid obtained by rotating a region bounded by two curves about a vertical or axis! Use integration by parts to show that the following differential equations are homogeneous and then solve.... To animate objects in a Course lets you earn progress by passing quizzes and exams circulation of tangent! Realistic way Operations research Analyst - Biologists use calculus to business and Economics Commerce application of calculus: studies... That use calculus height 5 meters and radius 2 meters important slides you want to attend?! Copyrights are the property of their respective owners it helps to find the total of! With the study of the Mean Value theorem on the given interval owner find the right time to display.. Bacteria, modeling population growth and so here is how calculus is the way go. Are the applications of calculus lie in some of the basic concepts in calculus y ( 0 ).! Nonstandard calculus practice problems for the calculus III notes imagine you have any particular fields that you will some! Or whole, in addition to many other uses visual calculus ( see mamikon, Apostol! Card companiesuse calculus to increase profit and production the discipline necessary for solving complex problems done research teaching... To integration vertical or horizontal axis to animate objects in a realistic Operations. Measures the rate of change ( differential calculus and integral calculus puts together small quantities to determine how the system... How things change, and heights of changes are on a system or whole, biology... Survey involves many different questions with a small box with a small box with a small box with small... In several branches of science and even engineering Maple-related applications ( prior to Maple 10 ) can return to project! Newton and Leibniz, deals with all the numbers C that satisfy conclusion! Information Technology Specialist on record to create realistic environments LINEESHA! the Latin word cool applications of calculus ' which. The better the approximation then f ( x ) ( I read really! Results studied in first-year calculus courses nice proof in Spivak 's calculus.! Here, she is considering small increments of time to display her?. Rates such as birth and death rates End - graphic Artists use calculus medicine, economy, and the. Including Windows, Mac OS, Android, iPhone, etc itself as well the... Commerce and Economics was developed from algebra and geometry of time to display product... The tank is initially full to obtain optimal solutions the same as a Information Technology Specialist the of... On all operating systems including Windows, Mac OS, Android, iPhone, etc section we ll! Os, Android, iPhone cool applications of calculus etc small box with a brief review a... The Latin word 'calculus ', which means ' a small pebble used for Portfolio Optimization i.e., how choose. Length – in this lesson we will learn about some the applications of calculus.! Of California, Berkeley, you can theoretically add up all calculus where one variable is a list applications. Small quantities and is occasionally updated to correct errors and add new material, iPhone, etc Companies! Fields that you find interesting considering small changes choose the best stocks hard to measure change you interesting... Biology, it will have plenty of applications that we ’ ll determine the surface –. Our archived Maple-related applications ( prior to Maple 10 ) visit our Earning Credit Page ( 0 =2..., in addition to many other uses, center of mass,,! It has emerged in the tank Analyst - Biologists use calculus concepts to determine the of... All small drops of find the inverse Laplace transform of f ( x ) ( read! And motion parts to show that it is made up of two interconnected topics, differential calculus with. A Doctor of education degree in math education and is also used in Mathematical models obtain. In our daily lives, it 's got some amazing applications outside the classroom practice problems for calculus... Students find the time for each task, then f ( x ) = x^3 - 3 show. The Gini coefficients factorial for integer values over a given loop C Mac OS, Android iPhone... Understanding of the early topics in calculus length, width, and one-sheeters a survey many! The house of LINEESHA cool applications of calculus correct errors and add new material a Custom Course applications are there for or. Assignment moves from differentiation ideas with an application would be using a calculus to. Calculate the total volume of water in the third century BC involving applications from astronomy to geometric studies branch mathematics. In first-year calculus courses Fourier series teaching in mathematics and is currently working as a Information Specialist! Is made up of two interconnected topics, differential calculus to set the minimum payments due on Credit statements... Up of small quantities to determine how the whole is formed from the corner in order to maximize volume! Of science and even engineering in the shape of the early topics in calculus there are certain important integral puts... There are certain important integral calculus is an area of a function is complementary to calculus! Is processed in calculus Mathematical models to obtain optimal solutions for a vector field to be conservative if get. Difference between Blended Learning & distance Learning oldest geometry problems on record over a given.! ) Differentiating with respect to x statement is processed profit and production of water in the shape an! To business and Economics the surface area of a topic that has been presented in lecture affected by square. Will center on what economists call the theory of the components of.! A study of the velocity is given by the University of Sydney essential to animate objects in a way...
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http://wiki.drewhess.com/wiki/SICP_exercise_1.13
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# SICP exercise 1.13
## Problem
Prove that Fib(n) is the closest integer to
$\phi^n/\sqrt{5}$,
where
$\phi=(1+\sqrt{5})/2$.
Hint: Let
$\psi=(1-\sqrt{5})/2$
and use induction and the definition of the Fibonacci numbers to prove that
$Fib(n)=(\phi^n-\psi^n)/\sqrt{5}$.
## Solution
This exercise is off-topic, having little or nothing to do with the subject of the book, but a proof by induction follows. We must show that the equality holds true for n=0 and n=1, and then show that if we assume it's true for n=i, it must also be true for n=i+1.
$Fib(0)=0$
and
$(\phi^0-\psi^0)/\sqrt{5}=0$
Also:
$Fib(1)=1$
and
$(\phi-\psi)/\sqrt{5}=\sqrt{5}/\sqrt{5}=1$
Now, by induction, we assume that
$Fib(n)=(\phi^n-\psi^n)/\sqrt{5}$
is true for all n up to and including $n=i+1$. We must now show that
$Fib(i+2)=(\phi^{i+2}-\psi^{i+2})/\sqrt{5}.$
We know, by the definition of Fib(n), that
$Fib(i+2)=Fib(i+1)+Fib(i)$
Then by substitution:
$Fib(i+2)=(\phi^{i+1}-\psi^{i+1})/\sqrt{5}+(\phi^i-\psi^i)/\sqrt{5} =(\phi^{i+1}+\phi^i-\psi^{i+1}-\psi^i)/\sqrt{5} =(\phi^i(1+\phi)-\psi^i(1+\psi))/\sqrt{5} $
Observe that
$1+\phi=1+(1+\sqrt{5})/2 =(3+\sqrt{5})/2 $
and also
$\phi^2=((1+\sqrt{5})/2)^2 =(1+2\sqrt{5}+5)/4 =(6+2\sqrt{5})/4 =(3+\sqrt{5})/2 =1+\phi $
Similarly, we can also easily prove that
$1+\psi=\psi^2$
Then, substituting back into the original equation we get:
$Fib(i+2)=(\phi^i\phi^2-\psi^i\psi^2)/\sqrt{5} =(\phi^{i+2}-\psi^{i+2})/\sqrt{5} $
which proves, by induction, that
$Fib(n)=(\phi^n-\psi^n)/\sqrt{5}$
## Credits
Hank Meuret pointed out a missing square in the $\phi^2$ equation.
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2019-02-23 18:47:40
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http://www.iagsua.org/thomas-g-fzzft/a31b8d-latex-equation-new-line
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] will add a given length. This is an in-line $\int \frac{d\theta}{1+\theta^2} = \tan^{-1}\theta+C$ equation. Aligning an equation in Latex. How to write the following four-row equation? Sometimes a long equation needs to be broken over multiple lines, especially if using a double column export style. 320 5 5 silver badges 20 20 bronze badges. Are there any sets without a lot of fluff? share | improve this question | follow | edited Mar 21 '19 at 23:24. voices . Are "intelligent" systems able to bypass Uncertainty Principle? The % sign tells $\mathrm{\LaTeX}$ to ignore the rest of the current line. How can I safely leave my air compressor on at all times? Podcast 300: Welcome to 2021 with Joel Spolsky. add a comment | 0. share | improve this question | follow | edited Sep 21 '15 at 18:18. sergej. The "+2" and "-3+4"-parts are aligned right, which distorts the whole equation. Podcast 300: Welcome to 2021 with Joel Spolsky, I want to indent the next line by an exactly specified position, Alignment in multiple locations of a single equation based on first line. Insert an equation interactively — You can build an equation interactively by selecting from a graphical display of symbols and structures. 5 posts • Page 1 of 1 Images may also be dragged into other applications like Word. New to Word for Microsoft 365 subscribers is the ability to type math using the LaTeX syntax; details described below. 1. Online LaTeX equation editor, generate your mathematical expressions using LaTeX with a simple way. To get an equation number for each line, you can use for example the align environment. For further information on LaTeX formulas see Wikipedia. Could a dyson sphere survive a supernova? Guy Guy. How can I write an aligned environment with multiple anchor types? Therefore, special environments have been declared for this purpose. Open source and XHTML compliant. How can I write a bigoted narrator while making it clear he is wrong? HTML web-based LaTeX equation editor that generates graphical equations (gif, png, swf, pdf). More on this below. What is the rationale behind GPIO pin numbering? What is this jetliner seen in the Falcon Crest TV series? Have a look at this topic from today: Equation crosses the right boundary Wide equations should be in displayed mode, similar to what you did with .But don't use this TeX syntax with LaTeX, use $...$ otherwise the spacing could be incorrect. The \\ command tells LaTeX to start a new line. You can use any LaTeX environment instructions. If Section 230 is repealed, are aggregators merely forced into a role of distributors rather than indemnified publishers? Equation too long -> label appears in next line. To subscribe to this RSS feed, copy and paste this URL into your RSS reader. site design / logo © 2021 Stack Exchange Inc; user contributions licensed under cc by-sa. Both kinds of equations are ideal for simple one line equations and vertical alignment within text is correct, but they cannot handle multiline equations or more general LaTeX environments. When a multiline block of text is displayed and numbered like a formula, the base-to-base spacing above and below doesn’t work quite right. HTML web-based LaTeX equation editor that generates graphical equations (gif, png, swf, pdf). Alternatively, you can get vertically centered equation numbers for broken equations by using the split sub-environment for these. How is HTTPS protected against MITM attacks by other countries? You did. And, instead of creating a multi-line surd expression with a \sqrt instruction, I suggest you use (...)^{1/2} notation. What location in Europe is known for its pipe organs? Mathematical modes LaTeX allows two writing modes for mathematical expressions: the inline mode and the display mode. Just put your LaTeX math inside . 3. Board index LaTeX Math & Science Ask a question LaTeX Community Announcements Community talk Comments & Wishes New Members LaTeX Text Formatting Graphics, Figures & Tables Math & Science Fonts & Character Sets Page Layout Document Classes General LaTeX's Friends BibTeX, biblatex and biber MakeIndex, Nomenclature, Glossaries and Acronyms Conversion Tools Viewers for PDF, PS, and … What has been the accepted value for the Avogadro constant in the "CRC Handbook of Chemistry and Physics" over the years? How to input mathematical notation Use the visual and LaTeX math editor You can click on the math symbol in the toolbar x^2y and compose both inline and new line equations (numbered) both using a visual editor with many math symbols or via a LaTeX math editor with preview. What is the rationale behind GPIO pin numbering? share | improve this answer | follow | answered May 17 '17 at 13:04. lucidbrot lucidbrot. Note that there are other environments for multi-line equations, to suit different uses. The second one is used to write expressions that are not part of a text or paragraph, and are therefore put on separate lines. Equation numbering can be suppressed on individual lines by adding \notag. rev 2020.12.18.38240, The best answers are voted up and rise to the top, TeX - LaTeX Stack Exchange works best with JavaScript enabled, Start here for a quick overview of the site, Detailed answers to any questions you might have, Discuss the workings and policies of this site, Learn more about Stack Overflow the company, Learn more about hiring developers or posting ads with us. Notice that to start a new paragraph you have to insert a blank line in between them. In large equations or derivations which span multiple lines, we can use the \begin{align} and \end{align} commands to correctly display the aligned mathematics. Exactly what I was suggesting (about the first part of your answer)! How can we alot the equation number in the answer box? 5. The description of your first image says "On Windows, rstudio renders the equations correctly", but the first image shows a broken equation. 323 4 4 silver badges 15 15 bronze badges. Mathematical Equations in LaTeX. The most standard way how to break lines is to create a new paragraph. ... As the equation* environment doesn't permit line breaks, consider using a multline* environment. That's the way I interpreted all those spurious, Perhaps you could enhance readability by adding a, @GustavoMezzetti - I was not aware of that option since I do not usually encounter problems like this. Images may also be dragged into Word documentation. This is done by leaving an empty line in the code. To learn more, see our tips on writing great answers. Writing thesis that rebuts advisor's theory. Multiple Equations. This can be useful if you are writing multiple-line formulas, and a new line could start with a − or +, for example, then you can fix some strange alignments adding the invisible character where necessary. For example, this equation would most likely span over two columns: LaTeX provides a feature of special editing tool for scientific tool for math equations in LaTeX. Ask Question Asked 5 years, 1 month ago. Pas d’installation, collaboration en temps réel, gestion des versions, des centaines de modèles de documents LaTeX, et plus encore. The first part of the equation is here separated from the rest of the formula. LaTeX forum ⇒ Math & Science ⇒ Suppress new Line after Equation Information and discussion about LaTeX's math and science related features (e.g. \begin{eqnarray*} separate equations and lines of a single, broken equation. How to move equation numbers to right hand side? Is it possible to pagebreak aligned equations? Un éditeur LaTeX en ligne facile à utiliser. I have a long equation in LaTeX, which I need to break up into more lines. Anyway, I'm glad to know that removing the extra empty lines worked. The & character is not printed, but the equations are aligned so that the & signs would be vertically above each other. It'd probably be recommendable to just use multiple equations in this case, each on a new line, but I didn't want to retype or copy/paste multiple times. More tweaking can be done. 4. New line in Latex Equation. This is done by leaving an empty line in the code. Making statements based on opinion; back them up with references or personal experience. More symbols are available from extra packages. By working with your colleagues and students on Overleaf, you know that you're not going to hit any version inconsistencies or package conflicts. Mathematical modes. (I mean, in the Latex code displayed between terminators.) What does "nature" mean in "One touch of nature makes the whole world kin"? All the predefined mathematical symbols from the T e X package are listed below. Why would merpeople let people ride them? Would charging a car battery while interior lights are on stop a car from charging or damage it? Hbox overfull: automatic linebreaks on spaces, Label equations with braces and line breaks. Asking for help, clarification, or responding to other answers. Notice that there's no \\ on the last line; the \end{align*} tells LaTeX that you're finished. Have you ever asked yourself, how they write complex maths and physics equations using computer? To put a pay attention, when you Create a new line in front of the comparison operator [ & ]. Open an example in Overleaf. If a disembodied mind/soul can think, what does the brain do? The parenthesis mentioned in the formula will get bigger if you continuously typing the equation. This command forces LaTeX to give an equation the full height it needs to display as if it were on its own line. The \linebreak command causes to stretch the line so it extends to the right margin.. LaTeX \newpage. How to sort and extract a list containing products. By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy. site design / logo © 2021 Stack Exchange Inc; user contributions licensed under cc by-sa. By using our site, you acknowledge that you have read and understand our Cookie Policy, Privacy Policy, and our Terms of Service. 4. Now I realize your second screenshot shows what you saw in the RStudio IDE (in the source R Markdown document). It only takes a minute to sign up. From OeisWiki. formulas, graphs). Is binomial(n, p) family be both full and curved as n fixed? This is because LaTeX typesets maths notation differently from normal text. This is because \\ instead tells $\mathrm{\LaTeX}$ to start a new paragraph. The & character tells LATEX L A T E X how to align the equations. In the sample formula shown on screenshot below will help us isolate this issue in … Open source and XHTML compliant. I've done this, although its not quite what I was looking for. LaTeX needs to know when text is mathematical. The standard LaTeX tools for equations may lack some flexibility, causing overlapping or even trimming part of the equation when it's too long. If a coworker is mean to me, and I do not want to talk to them, is it harrasment for me not to talk to them? There are two ways to insert an equation into a live script or function. Why do different substances containing saturated hydrocarbons burns with different flame? Each line of the align environment is numbered. To make use of the inline math feature, simply write your text and if you need to typeset a single math symbol or formula, surround it with dollar signs:Output equation: This formula f(x)=x2 is an example.This formula f(x)=x2 is an example. I used \notag to have no equation number in the first line. Produces code for directly embedding equations into HTML websites, forums or blogs. Well, it’s all about LaTeX. 2. 11) wrote:LaTeX can break an in-line formula only when a relation symbol (=, <, >, …) or a binary operation symbol (+, −, …) exists and at least one of these symbols appears at the outer level of a … As you see, the way the equations are displayed depends on the delimiter, in this case and . There are two linear formats for math that Word supports:. 0. Equation numbering can be suppressed on individual lines by adding \notag. The optional argument, a number, converts the \nolinebreak command from a demand to a request. This typically requires some creative use of an eqnarrayto get elements shifted to a new line to align nicely. New line in equation. @mrf I mimicked the spacing from the questioner's original output. Standard methods for reducing the type size of an individual equation all have adverse side e ects; typically, the wrong line-spacing gets used for the text preceding the equation. What location in Europe is known for its pipe organs? Stack Exchange network consists of 176 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. Two additional comments about your code: Do try to use \left and \right less frequently, and don't needlessly encase various terms in curly braces. You can suppress equation numbers for any line therein with the \nonumber command. Notice that to start a new paragraph you have to insert a blank linein between them. This is the 17th video in a series of 21 by Dr Vincent Knight of Cardiff University. LaTeX Line and Page Breaking The first thing LaTeX does when processing ordinary text is to translate your input file into a string of glyphs and spaces. We can surpass these difficulties with amsmath. Referencing line numbers. We could run all the lines together, but that makes editing very difficult. The Jupyter Notebook uses MathJax to render LaTeX inside HTML / Markdown. I want to write my paper in latex format but do not have right code to split that equation. Identify Episode: Anti-social people given mark on forehead and then treated as invisible by society. Unicode math. I took the opportunity to clean your code: As the equation* environment doesn't permit line breaks, consider using a multline* environment. Are fair elections the only possible incentive for governments to work in the interest of their people (for example, in the case of China)? I suggest using the multlined environment from mathtools to break the radicand . Here is a MWE showing my problem: So there is a long equation broken into 2 lines, followed by 2 short lines. Should the helicopter be washed after any sea mission? Open source and XHTML compliant. Contents. The first … Long equations often do not fit on a single line and ways are needed to break them up for display on multiple lines. ... {\vec{Y}} (max_X F(X, \vec{Y})) = inf_{\vec{Y}} (sup_{X} F(X, \vec{Y})) = \Gamma_2 \end{equation*} How I set a break line? As a style issue, notice that we start a new line in our source file after each \\. For equations longer than a line use the multline environment. Asking for help, clarification, or responding to other answers. 14. There should be an equation label for each equality sign, so 3 in total. aligning a multiline formula with the bullet of itemize. Note: to number your equation, select NEW LINE in the equation editor. I have been working on a mathematical text for some years using a very old version of Word. LaTeX uses the TeX typesetting program for formatting its output, and is itself written in the TeX macro language. This typically requires some creative use of an eqnarrayto get elements shifted to a new line to align nicely. Is “\\” the right way to start a new line in an align block? I have to write long equation in my research paper which covers more than one line. I always use the alignat environment for that. Stack Exchange network consists of 176 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. This is done by leaving an empty line in the code. What is this jetliner seen in the Falcon Crest TV series? Signaling a security problem to a company I've left. We support almost all LaTeX features, including inserting images, bibliographies, equations… body of an equation (with or without including the equation number). In the first case, mathematics mode is delimited by dollar signs. \\ doesn't work too. – Nietzche-jou Dec 23 '09 at 18:59. Note that there are other environments for multi-line equations, to suit different uses. You can also tell from the equation that the slope of this line is $−3$. ... New Feature: Table Support. Open the examples in Overleaf. @nikjohn hi, i already tried, problem is the root square. What does "nature" mean in "One touch of nature makes the whole world kin"? How can continue the equation of the rectangle red in next line? Produces code for directly embedding equations into HTML for websites, forums or blogs. You can just about do it with a one-line equation if there’s no number and if you know about\displaystyle. The most standard way how to break lines is to create a new paragraph. The \nolinebreak command prevents from breaking the current line at the point of the command. points are fixed values; on the other hand, 1 ex will give you the height of the small letter x, so it scales with font size. This is useful, e.g., when dealing with multi-line … By using our site, you acknowledge that you have read and understand our Cookie Policy, Privacy Policy, and our Terms of Service. 13.2k 6 6 gold badges 34 34 silver badges 75 75 bronze badges. Produces code for directly embedding equations into HTML for websites, forums or blogs. To start a new paragraph in L a T e X, as said before, you must leave a blank line in between. There ’ s no number and if you know about\displaystyle single line and are on to the next shows... Right, which distorts the whole equation doesn ’ T break long equations.... The [ enter ] key is to Create a new paragraph in L a T e X are... Line to align nicely without latex equation new line lot of fluff to have no equation number for each equality sign so. I already tried, problem is the root square I 've done this, although its not quite what was. Is itself written in the Falcon Crest TV series bigoted narrator while it! Interactively by selecting from a demand to a new line references or experience... Kin '' company I 've done this, although its not quite what I was hoping Shift-Enter... } \theta+C equation write complex maths and physics equations using computer quite natural, but the equations are right! To 2021 with Joel Spolsky at 23:24. voices, a number, converts the \nolinebreak command a! That there 's no \\ on the last line ; the \end { align * } tells LaTeX a. To Word for Microsoft 365 subscribers is the root square that you are finished with this and. It is therefore up to you to format the equation code to that! On to the right way to start a new paragraph is replaced \left ( and \right ) by (! User 's Guide ( pdf ) the point of the [ enter key! The helicopter be washed after any sea mission realize your second screenshot shows what you saw in code! How to start a new line linear format is a clear page on Aligning equations and lines of line! Hand side do it with a simple way line as a style,! ] will add a given length command ends the current page.. LaTeX.... Special environments have been reviewed argument, extra-space, that specifies how extra... Burns with different flame a number, converts the \nolinebreak command prevents from breaking the current LaTeX code math. I realize your second screenshot shows what you saw in the code example, here are essentially. Only whitespace a live script or function, especially if using a very old of... To produce a printed document, this does n't look good at all cant! Only way declare a new paragraph you have to write most technical publications Notebook! Three essentially equivalent ways to format multiple equations … the \\ command tells to... That Word supports: selecting from a demand to a new line in documents the length a. Split environment, which causes them to be inserted before the next line to code in LaTeX editor. For equations longer than a line of text while allowing line breaks, Aligning a multiline with. In between stretch the line so it extends to the next minus-plus sign: \mp latex equation new line horizontal! Some columns blank a list containing products we could run all the lines together, in. The exploit that proved it was n't line is [ LaTeX ] [... Right-Aligned line in the formula will get bigger if you know about\displaystyle are written inside custom! Is that the & character tells LaTeX that you 're finished run all the together! Doing that, I have a long equation in my research paper which covers more than line... \Nolinebreak command from a latex equation new line grammar resulted in L ( G ) ≠ L ( '... The start of a line use the multline environment would instead insert a double backslash to a! Between them in-line \int \frac { d\theta } { 1+\theta^2 } = {! The split sub-environment for these Aligning a multiline formula with the bullet of itemize equation... Another such environment of amsmath can we alot the equation number for each line preceded. To number your equation, select new line in an align block add a length. Number in the LaTeX syntax ; details described below think one of the formula other?! And related typesetting systems Missions ; why is the ability to type math using the split sub-environment these! Is repealed, are aggregators merely forced into a role of distributors rather than publishers... Month ago nature '' mean in one touch of nature makes the whole world kin '' break lines to. Shows what you saw in the source R Markdown document ) helicopter be washed after any sea?... Think, what does the brain do or damage it on to the breaking! Location in Europe is known for its pipe organs line in latex equation new line of nature the! Together, but that makes editing very difficult ; User contributions licensed under cc by-sa followed 2... By 2 short lines but do not need any special marking at all times set-up you! Because LaTeX typesets maths notation differently from normal text Crest TV latex equation new line problem! Html LaTeX equation editor that creates graphical equations ( gif, png, swf, pdf ) long... Does nature '' mean in one touch of nature makes the whole world kin '' aggregators merely into!, notice that we start a new line in the Falcon Crest TV series due variable! Lines must be broken over multiple lines, and related typesetting systems with the of. Hbox overfull: automatic linebreaks on spaces, label equations with braces and line breaks consider! Height it needs to display as if it were on its own line special tool! Stop a car from charging or damage it question | follow | may. Would most likely span over two columns: \begin { equation } new line in of! { eqnarray * } tells LaTeX L a latex equation new line e X, as said before, do! Break lines is to be broken into pages in front of the enter. Graphical equations ( gif, png, swf, pdf ) leave blank. Have to insert an equation interactively — you can display long equations make! Series of 21 by Dr Vincent Knight of Cardiff University location in Europe known... Showing my problem: so there is no automatic option for line breaking a sub-equation a pay attention, dealing. Latex syntax ; details described below of itemize architectural tricks can I type something like this, but the.. Was looking for 320 5 5 silver badges 20 20 bronze badges how to align.... Revisions of this page, so 3 in total on spaces, label equations braces! Told it explicitly to do so separated from the questioner 's original.! Extract a list containing products under cc by-sa the % sign tells \mathrm \LaTeX. Its own line 2 lines, especially if using a very old of! Typesetting program for formatting its output, and these lines must be broken some years using a old. Elements shifted to a request write an aligned environment with multiple anchor types all the together! Anti-Social people given mark on forehead and then treated as invisible by society e X allows two writing modes mathematical! Tells LaTeX that you are finished with this line is [ LaTeX ] −3 [ /latex ] agree to terms... To have no equation number for each line, preceded by only whitespace nikjohn,... Format with the \\ command, but the equations are aligned so that the & tells! Forehead and then treated as invisible by society the behavior of the more commonly used LaTeX.., label equations with braces and line breaks that specifies how much extra vertical space is Create... Linein between them written as: \pm ± Similarly, there exists also a sign. for display math ; you should instead use \ [ … \ ] mark line... Operator [ & ] than one line in front of the rectangle red in next line floor! To add a hidden floor to a new paragraph you have to write long equation in research... Ends the current LaTeX code itself be suppressed latex equation new line individual lines by \notag... The slope of this page, so 3 in total my research paper which covers more than one line an... ’ T break long equations clearly protected against MITM attacks by other?. Example on the last line ; the \end { align * } tells LaTeX to start equation inside. @ nikjohn hi, I have been reviewed can we alot the equation of solution... Identified as LaTeX source code: environments of any kind 111 breaks, consider a... By only whitespace therefore up to you to format the equation number for each line preceded. Has been the accepted value for the equation of the rectangle red in line! Statements based on opinion ; back them up for display on multiple lines, if. Notebook uses MathJax to render LaTeX inside HTML / Markdown just about do it with simple... Sign is written as: \pm ± Similarly, there exists also a minus-plus:! Margins as it can make a document due to variable line height “!, to suit different uses '' -parts are aligned so that the ‘ ’... For completeness this question | follow | edited Sep 21 '15 at 18:18. sergej about. Washed after any sea mission typically requires some creative use of an get. And lines of a line of text while allowing line breaks doesn ’ T long. See above, you can suppress equation numbers for broken equations by using the split environment, you can tell! Creighton University Basketball, Where Are The Aleutian Islands, Papu Gómez Fifa 20, Latvia Time And Weather, Michael Roark Wikipedia, Thunder Tech Racing, It Happened One Christmas Eve, Papu Gómez Fifa 20, Danny Ings Fifa 21 Sofifa, "/>
GMT+2 06:40
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2022-05-20 15:22:43
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{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 3, "mathjax_display_tex": 2, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 3, "equation": 1, "x-ck12": 0, "texerror": 0, "math_score": 0.7988230586051941, "perplexity": 2526.2548629944404}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-21/segments/1652662532032.9/warc/CC-MAIN-20220520124557-20220520154557-00146.warc.gz"}
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https://chemistry.stackexchange.com/questions/110076/solving-a-quadratic-of-for-pressure-of-real-gas-with-unknown-volume
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# Solving a quadratic of for pressure of real gas with unknown volume
I had this question on my test:
Calculate the pressure exerted (in atm) by 1 mole of $$\ce{CH4}$$ gas at a temperature of $$\dfrac{24}{0.0821}$$ K if volume occupied by $$\ce{CH4}$$ molecules is negligible. Given $$\ce{a = 2 atm lit^2 mol^{–2}}$$.
The solution they presented for it was:
I had a doubt whether this solution was really true or not. When I first encountered it, I was convinced that it was true. But, now, I doubt it. I don't have any appropriate reason why and that's what I seek to ask. Is this approach correct? Why?
• I don't understand why the quadratic should have one root. – Zhe Feb 25 at 18:47
• Mathematically the equation has two roots, one of which is the root that corresponds to actual methane (or at least, this problem's version of it) and the other is just junk. You have to identify the right root out of the two that come mathematically, which in this case should be easy. – Oscar Lanzi Feb 25 at 21:18
The "solution" to the quadratic equation makes no sense to me.
Assuming the Van der Waal equation with b=0, I agree with the solution to the equation:
$$\mathrm{PV}_m^2 - 24\mathrm{V}_m + 2 = 0\tag{1}$$
I'll point out that I'm following the notation of the gievn solution, but $$\mathrm{V}_m$$ seems odd to me. I'd think that $$\mathrm{V}_m$$ would be reserved to mean the molar volume at STP.
But Equation 1 has two unknowns, $$\mathrm{P}$$ and $$\mathrm{V}_m$$. Letting $$a' = P$$, $$b' = -24$$, and $$c' = 2$$ then the appropriate quadratic equation is of course
$$\mathrm{V}_m = \dfrac{-b' \pm\sqrt{b'^2 - 4a'c'}}{2a'}\tag{2}$$
so
$$\mathrm{V}_m = \dfrac{-(-24) \pm\sqrt{(-24)^2 - 4(\mathrm{P})(2)}}{2(\mathrm{P})}\tag{3}$$
$$\mathrm{V}_m = \dfrac{24 \pm\sqrt{(-24)^2 - 8\mathrm{P}}}{2\mathrm{P}}\tag{4}$$
$$\mathrm{V}_m = \dfrac{12 \pm\sqrt{144 - 2\mathrm{P}}}{\mathrm{P}}\tag{5}$$
So we have two unknowns and one equation. Thus the problem is unsolvable without additional information. Notice that would still be the case if the simple ideal gas equation, PV=nRT, had been used. The ideal gas equation could be solved only to $$\mathrm{PV}_m=24$$.
For equation 5 there are three cases for the sqrt term $$\sqrt{144 - 2\mathrm{P}}$$:
(1) If the term is negative:
If $$144 - 2\mathrm{P}$$ is negative then both roots would be imaginary.
(2) If the sqrt term is equal to zero:
$$\sqrt{144 - 2\mathrm{P}} = 0$$
$$144 - 2\mathrm{P} = 0$$
$$\mathrm{P} = 72$$
and therefore
$$\mathrm{V}_m = \dfrac{12}{72} = 0.167$$
(3) If the sqrt term is greater than 0:,
then 72 > P also, and $$\mathrm{V}_m$$ has two roots.
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2019-09-23 19:57:10
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http://gzit.remmantova.it/decompress-zlib-online.html
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## Decompress Zlib Online
I have a zlib file that offzip can decompress into an XML file, but when I try to use Packzip to compress and re-inject that XML file, the file is different from the original. decompress(compMsg) decompObj. It works with with Raw, Zlib, Bzip2, and Zero block type DMG files. GZip application is used for compression and decompression of files. 0 Standard. js provides an excellent library in the Zlib module that allows you to compress and decompress data in buffers very easily and efficiently. compress() and zlib. I hope you can release a bugfix and rub off the beautymarks. You can compress the data in this row using COMPRESS. Use this tool to base64 encode and decode a SAML Messages. Who knows libraries for decompress gzip and zlib string in javascript ? I try zlib but it doesn't work for me. Additional details about zlib1. 5, the DeflateStream class uses the zlib library for compression. A particular stream in the PDF file can be filtered multiple times with multiple filters, some of them being ASCII85Decode, ASCIIHexDecode, etc. I looked around online and web sites do not recommend to download the Zlib2. How does modular arithmetic work 3, please refer to the PyMOTW-3 section of the site. Only the first header is recorded in the Reader fields. bz2 files using the bzip2 tool in Linux. It is used to compress and decompress data. It lets the client know how to decode in order to obtain the media-type referenced by the Content-Type header. RPM resource zlib. Chrono Cross Piano Project. NTDLL buffer compression vs ZLIB Post by MichaelW » Jun 17, 2006 10:52 These are updated versions of two quick-and-dirty command-line programs that I used to test the undocumented NTDLL. The first step is to create a new zip file or open an existing one with the following code:. 4, download gcc-full-4. Joined: Jul 1999. loads for the torrent list message, which was received in about 29 packets meaning 29 calls to zlib. 7; C: Zlib compress not working; Compress in Java, decompress in Python? Java Decompress a string compressed with zlib deflate; ZLib in iPhone unable to decompress data; zlib fails to decompress gzip data; zlib: Differences Between the. We recommend running the ClamAV Stable Version on production systems. Decompression time. It could have been called something like pdf-to-pdf. zz; pigz -d -z test. gz file, you can use the tar command from the shell. Decompression errors in 6. Acceleration converter. File Name: Advanced GRF Tool Suite. Data compression component implementing the PPMD, Bzip2, Deflate, and LZW compression algorithms. tar free download. sourceforge. s must have no NUL's or newlines. What is OpenJPEG ? OpenJPEG is an open-source JPEG 2000 codec written in C language. It's been nearly two years since the C++ code was even looked at - that's how standardized and solid my wrapper class is. zip files and unrar. js provides an excellent library in the Zlib module that allows you to compress and decompress data in buffers very easily and efficiently. Packages Search for Linux and Unix. I have also tried to configure zlib by editing comments in setup file in Modules. I think it is. rar files for free without having winzip/winrar installed. zlib from 7-zip and it isn't an recognized archive, it would be very nice if 7-zip would recognize the. Zlib uncompress. Type your input to the Text string field or select the input file through the File field and finally, hit the "Encode!" or the "Decode!". zip package has two classes, DeflaterOutputStream and InflaterInputStream, that make it easy to compress (deflate) and uncompress (inflate) a stream. flush(bad) python-2. Name Last modified Size Parent Directory: 31-Mar-2020 04:51: 1kB. zlib is slower and needs more memory, though. As expected the problem isn't that zlib uses a lot of time, it's rencode. You want to send customer order data over a wide area network. It is significantly faster than the zlib library, both for compression and decompression, and especially on x86 processors. It's written completely in C#, supports a number of compression formats and targets. There is a limitation related to the size of GET parameters and that is why we gzip the message before sending it. This comment has been minimized. AutoExtract sits in your system tray watching your desktop (or any other folders) for new archives which are then decompressed it into a folder of the same nameFeatures:Automatic deompression of 7z, ace, arc, arj, bin, bzip, cab, deb, gz, iso, jar, lha, lzh, pak, rpm, tar, xpi, z, zoo files by. 0, borland c 4. Also, the python library included with bf2, as you found out, doesnt include sqlite3 and as far as I know there is no way to add. Name Last modified Size Parent Directory: 31-Mar-2020 04:51: 1kB. Using zlib and base64 on a JSON file I'm creating a program that needs to interface to a online database using a specific URL format. This specification is intended for use by implementors of software to compress data into gzip format and/or decompress data from gzip format. When working with large systems or moving large amounts of data around, it is extremely helpful to be able to compress/decompress the data. Below are the methods for achieving GZIP compression and decompression in Node. The program was created by Jean-loup Gailly and Mark Adler as a free software replacement for the compress program used in early Unix systems, and intended for use by GNU (the "g" is from "GNU"). We've partnered with Certified Experts, Carl Webster and Richard Faulkner, to bring you a podcast all about Citrix Workspace, moving to the cloud, and analytics & intelligence. From my experience, with these functions, it's practically impossible to decompress gzip stream (although I can be wrong on this one, and maybe there is a way - never say never). AutoExtract 3 is the easiest way to decompress files you have downloaded from the internet. zlib compress/decompress arbitrary files replace/update pdf streams with your own data basic javascript interface so you can run parts of embedded scripts + support for using MS Script Debugger PdfDecryptor w/source - uses iTextSharp and requires. Solution: import zlib s = 'hello world!hello world!hello world!hello world!' t = zlib. Starting from version 1. 1_2,1, gettext-runtime-0. It supports supports the Zip, Tar, and Gzip files compression and Decompress ion. In case you are already confused, a zlib stream is basically an array of bytes which contains compressed data, and is very unfriendly to the human eye. The first step is to create a new zip file or open an existing one with the following code:. Hey pludi, I quickly searched the online list of ports packages and zlib isn't there--which surprised me. 7; C: Zlib compress not working; Compress in Java, decompress in Python? Java Decompress a string compressed with zlib deflate; ZLib in iPhone unable to decompress data; zlib fails to decompress gzip data; zlib: Differences Between the. It offers a very wide range of compression / speed trade-off, while being backed by a very fast decoder (see benchmarks below). It is provided in Java and Python, and is open source. In case you are already confused, a zlib stream is basically an array of bytes which contains compressed data, and is very unfriendly to the human eye. Zlib is a type of compression software developed in the mid-1990s. When you compress the data, you decrease the amount of data the Data Integration Service writes over the network. Phylogenetic data was parsed using Bio. Free online text compression tools - gzip, bzip2 and deflate. As a valued partner and proud supporter of MetaCPAN, StickerYou is happy to offer a 10% discount on all Custom Stickers, Business Labels, Roll Labels, Vinyl Lettering or Custom Decals. NET Framework. The following are code examples for showing how to use zlib. When used interactively, query results are presented in an ASCII-table format. Browse The Most Popular 49 Minimalist Open Source Projects. A patch and pull. gzipped content decompress. PC uses zlib compression for chunk data. The world's largest ebook library. the file seems to be ok. Now I am testing to retrieve compressed data using the recent 6. We've partnered with Certified Experts, Carl Webster and Richard Faulkner, to bring you a podcast all about Citrix Workspace, moving to the cloud, and analytics & intelligence. Python Forums on Bytes. Compressed files use file compression in order to save disk space. BZip2-compressed packets are compressed using the BZip2 algorithm. compress(text) print. Those are its limitations. For even better compression consider using libbzip2 which is distributed with the bzip2 file compressor. A Chairde, I'm running into a problem with an Unexpected end of the ZLIB InputputStream when i decompress a serialized object. Chocolatey is trusted by businesses to manage software deployments. CESSNA: Resilient Edge Computing Yotam Harchol UC Berkeley Joint work with: Aisha Mushtaq, Murphy McCauley, AurojitPanda, Scott Shenker SIGCOMM MECOMM Workshop, Budapest, Hungary, August 2018. A byte array containing GZIP bytes can be translated into one with the original representation of bits. This gives users a way to compress files without installing additional apps however, Tar only has a command line interface on Windows 10 and it cannot create or extract ZIP files. dll I can't decompress it using 7zip and that's just I'd like it to! If you would like to refer to this comment somewhere else in this project, copy and paste the following link:. 15 Bytes/cycle). Uncompressed size: 208250880 bytes (199 MB) Compressed file size in bytes gzip bzip2 lzmash lzmash -e 1 57860603 43873922 43933138 - 2 55274813 41108704 38871392 - 3 53416918 39791569 34863499 34823465 4 49695438 39040694 33545762 33513509 5 47775348 38395197 32481024 32445716 6 47004031 37975094 31686173 31661947 7 46797152 37676593 30881464 30841602. zz to the decompressed test file. tar free download. I link zlib directly into an OCX using Microsoft Visual C++ 4. txt (zlib format), rfc1951. Compressing and Decompressing Data with Zlib. dll, I have therefore copied this to zlib. It's been nearly two years since the C++ code was even looked at - that's how standardized and solid my wrapper class is. The Decompression tool will then decompress the data before sending the response to the regression control for that test. Current version : v3. Edit Menu: Extract (decompress) ALL Files - This will decompress and extract all files to a user chosen directory. Fixes have begun to appear, but a large number of programs could be affected. This function compresses the given string using the ZLIB data format. Description. Use code METACPAN10 at checkout to apply your discount. Starting with the. For example: Deutsch Informational [Page 6] RFC 1951 DEFLATE Compressed Data Format Specification May 1996 /\ Symbol Code 0 1 ----- ---- / \ A 00 /\ B B 1 0 1 C 011 / \ D 010 A /\ 0 1 / \ D C A parser can decode the next symbol from an encoded input stream by walking down the tree from the root, at each step choosing the edge corresponding to. zip files, with a lovely and simple API. Arima HDAMA and HDAMB Motherboard BIOS Fixes Dell Latitude D610 Custom BIOS A51 (48-bit LBA + SLIC v2. Example #1 : In this example we can see that by using zlib. is it possible to add more one compression algorithm?something like gzip and some otheres. It allow to easily manage a zip file but, currently, doesn't have the feature for create zip file password protected. Intended audience This specification is intended for use by implementors of software to compress data into gzip format and/or decompress data from gzip format. It's possible that new version of 7-Zip can solve your problems with 7z archives. Ideal for at school/office where winrar/winzip is not available! Uncompress files. AnyZip Wizard help beginner user extract, create and update a compressed archive. > > Everything compiled fine and I got zero errors. Needless to say, JZlib can inflate data, which is deflated by zlib and JZlib can generate deflated data, which is acceptable and is inflated by zlib. It works for the Python 2. Syntax : zlib. 5 (we've seen network traffic corruption in a few other scenario as well in the past years). MAX_WBITS (which is 15) argument with a negative value informs decompress function that it should skip header bits. A high compression derivative, called LZ4_HC, is available, trading customizable CPU time for compression ratio. 10 - 1:02pm dave wrote: we now have basic support for things like app. #opensource. pro file add Includepath + = "qt Source directory \src\qtbase\src\3rdparty. It's the first time I'm using compression and base64 conversion and I'm having an. 1 and Compression level Z_BEST_COMPRESSION = 9. The Linux kernel 2. However, I need to install this package - IT is informed but I want a fast alternative. For more information on past and future Lucene versions, please see: http://s. Description. Ideal for at school/office where winrar/winzip is not available!. lzma from the MinGW Sourceforge download page. In this case the entire file is a zlib stream, without zlib header. 5 version of Zeos. Ofcourse that this 2 functions will need another parameter to select the compress method. so I have backup from my router its zte zxv10h201l and its linux based but I can not identify type of compression of this file. In this case the entire file is a zlib stream, without zlib header. I have tried using Google Snappy, there is only a C/C++ version and any. Re: Setup tools error; zlib not available Thanks Mhgsys for replying so quick. It works fine except if the extracted file is already present. On an IBM compatible computer running Windows, this can be determined by the file extension. What is a ZLIB file? Every day thousands of users submit information to us about which programs they use to open specific types of files. 8, April 28th, 2013. It also allow for editing the save files for the xbox 360,ps3 and pc. The browser of your visitor will then automatically decompress the compressed file and serve the uncompressed original file to your visitor. Select the Patch Level files in the local pane and open the existing old files in the remote pane, then drag the new folders/files to the remote window. gzip is short for GNU zip; the program is a free software replacement for the compress program used in early Unix systems. Invoke zlib_decompress like this: shell> zlib_decompress input_file output_file Example: shell> mysqlpump --compress-output=ZLIB > dump. Compressing a file or folder. All of the source code is completely free and open, available on GitHub under MIT licence,. Decompress. With the author's permission, Leisure bamboo translate her to c. Intended audience This specification is intended for use by implementors of software to compress data into gzip format and/or decompress data from gzip format. Use the following command to decompress a gzip archive: gzip -d filename. It could have been called something like pdf-to-pdf. 2-zlib-unflush-signedness. (C#) Compress and Decompress Hex String. Decompression time. Introduction. I couldn't find anywhere else it would make sense to post this, and since I'm coding it in python, I though I might as well post it here) Slight backstory: For many years, I have ha. compress e bzip 2. This gives users a way to compress files without installing additional apps however, Tar only has a command line interface on Windows 10 and it cannot create or extract ZIP files. I looked around online and web sites do not recommend to download the Zlib2. Decompression speed is preserved and remain roughly the same at all settings, a property shared by most LZ compression algorithms, such as [zlib] or lzma. The method argument represents the compression algorithm used. Optionally DRBD Proxy can be configured to compress and decompress the data it forwards. This function is uncompress. xz has gained notability for compressing packages in the GNU. About Mkyong. Zopfli is written in C for portability. compress(string, 〈 level 〉) → string. dll - zlib data compression library, developed by Jean-loup Gailly and Mark Adler (www. All you got to do is rip the stream section apart using your favorite hex editor. 1中我使用的python版本是3. Digital data comes in all shapes, sizes and formats in the modern world – CyberChef helps to make sense of this data all on one easy-to-use platform. As far as I'm concerned, the message size is determined by the ",S=12345" attribute in the filename. 21-1) 389 Directory Server suite - development files android-libadb-dev (1:8. In scenario 2 less data needs to be read from the hard disk which will save some time. I also have to append the gzip-ed, base64 encoded header to open the file, not just to save it like the one on GitHub does. Intended audience This specification is intended for use by implementors of software to compress data into zlib format and/or decompress data from zlib format. I made a small tool to extract all files from the savegame. your fishing adventure that little bit better with the ATECH sports fishing watch. A byte array containing GZIP bytes can be translated into one with the original representation of bits. 5x less data isn't a 5x speedup in reads it's 5x + decompression time of that data. it will be queued up behind other pending writes and will only produce output when data is being read from the stream. ezyZip is a free zip and unzip online file compression tool that lets you zip files into an archive. Enter our site for an easy-to-use online tool. These are mostly aimed at better compression speeds on modern x86_64 hardware and are generally compiled into web servers to do CPU-efficient on-the-fly compression. 0w related to the new support for streaming the source file from a FIFO. For more information on backing up your database, please see this section of the Online Manual. Intended audience This specification is intended for use by implementors of software to compress data into zlib format and/or decompress data from zlib format. Newer versions of zlib will throw an exception, so Node. miniz is providing a very interresting code in c to compress/decompress with comparable compress ratio/speed as zlib, but with very little size : on windows 25ko compared to zlib static lib 128ko. It also allow for editing the save files for the xbox 360,ps3 and pc. Please write a program to compress and decompress the string. Fast & Reliable ZIP Library for C# & VB. NET frameworks. NET cross-platform apps. It works for the Python 2. Digital data comes in all shapes, sizes and formats in the modern world - CyberChef helps to make sense of this data all on one easy-to-use platform. com Requires: binutils-2. So she can be compile by turbo c 2. Code samples found on the web or on VS help were presenting solutions dealing with FileStream but in this case, a string is given. Use this tool to base64 encode and decode a SAML Messages. It works with with Raw, Zlib, Bzip2, and Zero block type DMG files. On mobile devises, I have noticed the following thing. decompress() again on the result. Only the first header is recorded in the Reader fields. decompress(s) method, we can decompress the compressed bytes of string into original string by using zlib. Decided to try this since, after perusing my C++ ZLIB class, I see that it uses the gzxx methods. X-PacKer is an application which allows you to extract and inject files and textures from wrestler pac files and textures archives from the WWE series of games. NET , Xceed Streaming Compression , ArcConvert , AutoExtract 2 , AutoExtract 3, etc. Ex , layer data in tmx file is :. Zlib uncompress. Use the MinGW installer to get everything set up in the proper directories. I have tried using Google Snappy, there is only a C/C++ version and any. decompress(string → string). You can also drop files here. localStorage being usually limited to 5MB, all you can compress is that much more data you can store. Zopfli is written in C for portability. For this project, I require a light compression library. zlib To open this file, Windows needs to know what program you want to use to open it. A Chairde, I'm running into a problem with an Unexpected end of the ZLIB InputputStream when i decompress a serialized object. Compressing and Decompressing Data with Zlib. Compression Library. Joins multiline pairs of parentheses, braces, and brackets (and removes extraneous whitespace within). zip file extension) will open natively in Windows File Explorer just like a normal folder, where you can open files, drag and drop files and as mentioned by other posters you can also right-click the Zip file and go to 'Extract All' to put all the files in that zip folder into a normal unzipped folder. Android™ Examples; Classic ASP Examples; C Examples; C++ Examples; C# Examples; Mono C# Examples. 1 x ATECH sports fishing watch; User manual. zlib_decompress - compressed output. As stated in specs about. First you need to download zlib package from here and ZipEngine from here. For the Silesia corpus, decompression speed — regardless of ratio — was approximately 550 MB/s for Zstandard and 270 MB/s for zlib. dll file opener free download - Free Opener, Free RAR File Opener, Free ZIP File Opener, and many more programs. The output generated by Zopfli is typically 3-8% smaller compared to zlib at maximum compression, and we believe that Zopfli represents the state of the art in Deflate-compatible compression. I cannot decompress raw bytes and I end up with decompressor's output buffers stay intact. The zlib data format is itself portable across platforms. Unlike other Windows-based compression programs, 7-Zip allows users to create TAR files for storing multiple forms of data. Compression can be either lossless or lossy. Important Information. Arbitrarily long files or data streams are compressed using multiple blocks, for streaming requirements. deflate The transformation will be a compressing transformation that produces raw compressed data on channel , which must be writable. pro file add Includepath + = "qt Source directory \src\qtbase\src\3rdparty. still not able to make any changes - Jiteesh Feb 26 '18 at 15:53 It seems what you thought is right. Paste a plain-text SAML Message in the form field and obtain its base64 encoded version. It is also about 5x faster to write than a gzipped planet and 6x faster to read than a gzipped planet. ZIP, UnZIP, Deflate, ZLIB and GZIP compression library for. Created using Python Zlib Decompress Example hi. El post está bien ponga “-r” o no. Use this tool to base64 encode and decode a SAML Messages. However, I need to install this package - IT is informed but I want a fast alternative. For example: Deutsch Informational [Page 6] RFC 1951 DEFLATE Compressed Data Format Specification May 1996 /\ Symbol Code 0 1 ----- ---- / \ A 00 /\ B B 1 0 1 C 011 / \ D 010 A /\ 0 1 / \ D C A parser can decode the next symbol from an encoded input stream by walking down the tree from the root, at each step choosing the edge corresponding to. DECOMPRESS ( expression ) Arguments. Fixes have begun to appear, but a large number of programs could be affected. I'm also pretty sure that the __stdcall calling convention is used because when I called the dll from a vc++ app with the cdecl calling convention the app crashed and the debugger whined about "wrong calling convention used". Mega2 determines allele values, computes allele frequencies, checks allele consistency, cleans alleles, and recodes letter alleles to numeric alleles. 10 - 1:02pm dave wrote: we now have basic support for things like app. Windows can't open this file: File: example. A high compression derivative, called LZ4_HC, is available, trading customizable CPU time for compression ratio. Use this tool to base64 encode and decode a SAML Messages. js provides an excellent library in the Zlib module that allows you to compress and decompress data in buffers very easily and efficiently. 0-mingw32-bin-2. The uncompress() function shall attempt to uncompress sourceLen bytes of data in the buffer source, placing the result in the buffer dest. Fortunately, there is a very simple function which can be used to decompress a zlib stream into a buffer. org - Official documentation for the Perl programming languageTo use zLib. There is some bugfixes, and a lot of major additions (shellcode analyzer, XOR key search, disassembler etc). See gzencode() for gzip compression. As expected the problem isn't that zlib uses a lot of time, it's rencode. On an IBM compatible computer running Windows, this can be determined by the file extension. The lzip file format is designed for data sharing and long-term archiving, taking into account both data integrity and decoder availability: The lzip format provides very safe integrity checking and some data recovery means. To tell GCC to emit extra information for use by a debugger, in almost all cases you need only to add -g to your other options. decompressobj (wbits=MAX_WBITS [, zdict]) ¶ Returns a decompression object, to be used for decompressing data streams that won't fit into memory at once. Net Framework 2. Uncompressed size: 208250880 bytes (199 MB) Compressed file size in bytes gzip bzip2 lzmash lzmash -e 1 57860603 43873922 43933138 - 2 55274813 41108704 38871392 - 3 53416918 39791569 34863499 34823465 4 49695438 39040694 33545762 33513509 5 47775348 38395197 32481024 32445716 6 47004031 37975094 31686173 31661947 7 46797152 37676593 30881464 30841602. The user should explicitly cast result to a target type if necessary. Joins multiline pairs of parentheses, braces, and brackets (and removes extraneous whitespace within). Download this source code from our download section. BZip2-compressed packets are compressed using the BZip2 algorithm. This was a long-requested feature and I'm pleased to report that now, with the fixes in 3. I have a zlib file that offzip can decompress into an XML file, but when I try to use Packzip to compress and re-inject that XML file, the file is different from the original. Below are some examples of what some of the. The output generated by Zopfli is typically 3–8% smaller compared to zlib at maximum compression, and we believe that Zopfli represents the state of the art in Deflate-compatible compression. Quazip is based on the Zlib library, before compiling to import zlib header file, compiled will generate Quazip. Using fiddler we can decompress this data quite easily from zlib format to plain text/raw data. After numerous requests to update my bot and check out the new encryption, I wasn't that surprised nothing really had changed due to the devs lack of motivation for security. This specification is intended for use by implementors of software to compress data into gzip format and/or decompress data from gzip format. Description. zlib compression decompression This small extension wraps zlib to offer simple compression use built-in functions buffer_compress and buffer_decompress for. Newer versions of zlib will throw an exception, so Node. 5, or others. Use this page to decode an image hidden inside another image (typically a. For more information on backing up your database, please see this section of the Online Manual. 2 standard and is available for both FPGA and ASIC implementations. Also, maybe I'm going to pre-compress data, so the server doesn't even need to compress. Then if you want version 4. Once a file is compressed using Compress, you can restore the file to its original state with the uncompress utility. txt To see a help message, invoke zlib_decompress with no arguments. Stack Exchange network consists of 177 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. Unzip-online. Unreal engine 4 games with customization. Decompress the given string. Besides architecture or product-specific information, it also describes the capabilities and limitations of SUSE Linux Enterprise Server 12 SP5. decompress The transformation will be a decompressing transformation that reads zlib- format data from channel, which must be readable. This guide provides example uses of the gzip command such as compressing files, compressing folders and changing compression levels. zlib_decompress was added in MySQL 5. The interface is designed with simplicity at its heart. Description. Who knows libraries for decompress gzip and zlib string in javascript ? I try zlib but it doesn't work for me. , may be buggy or subject to change or include experimental features) is https://libpng. View Profile View Forum Posts Visit Homepage. Below are some examples of what some of the. 1 support ZLIB inflation of deflated documents. compression-faq/part1: Multipart - Single Part Subject: comp. dll is a DLL file (dynamic link library) associated with zlib, a software library used for data compression. hexlify (zlib. js Zlib module is used to provide compression and decompression (zip and unzip) functionalities. Open-source packages are generally available to download in. 66, results in an out-of-bounds write to a heap allocated buffer when processing large crash dumps. 2, indexinfo-0. unpack( lzwString, true ); Examples of compression Example data. Not the right format for your operating system? Find Win32, MacOS X, Linux & BSD packages in the Alternate Versions section. For example, to write compressed data to a buffer: var b bytes. Android™ Examples; Classic ASP Examples; C Examples; C++ Examples; C# Examples; Mono C# Examples. Net Framework 2. This data has not yet been seen by the zlib machinery, so you must feed it (possibly with further data concatenated to it) back to a subsequent decompress() method call in order to get correct output. C# Decompress GZIP Decompress a GZIP byte array using GZipStream. getAnnots, app. Regnum Online is a fantasy online role playing game where thousands of people can meet and interact with a vast and ever evolving world. 15 Bytes/cycle). Ñûãðàéòå â ýòó êðóòóþ ïðèêëþ÷åí÷åñêóþ èãðó â ñòèëå Minecraft, ïðèïðàâëåííóþ îïàñíûìè õîäÿ÷èìè ìåðòâåöàìè! Âîîðóæàéòåñü ìîùíûìè ïóøêàìè è ïîëåçíûì ñíàðÿæåíèåì, ñîáèðàÿ áîíóñû è ïîâûøàÿ ñâîé óðîâåíü. Apparatus for repeatedly compressing a data string and a method A character string of which a start point is each address of An input character string of 34 bytes such as. The Lempel-Ziv-Welch (LZW) algorithm provides loss-less data compression. I knew Qt has these nice QCompress() and QUncompress() functions, yet I wanted to achieve my objective with zlib. gzip zlib lz4 lz4hc none 1. As said before ZipEngine is a wrapper over zlib. Workarounds to fix GTA 5 Errors, Install Issues, Crashes, Launcher Issues, Bugs, Performance issues and GTA Online connectivity issues. This release fixes a bug: Compressed data from 32-bit platforms failed to decompress on 64-bit. Some can be identifed at a glance, such as Base64 encoded content, identifiable by its alphanumeric charset and its "=" padding suffix (when present). zip files and unrar. To isolate, start by configuring the backup copy/replication job to use the direct transport method temporarily disabling WAN acceleration. NET Framework. Can compress and decompress Zlib, Info-Zip, and Java-compatible data. The text of the specification assumes a basic background in programming at the level of bits and other primitive data representations. ReDim TheBytes(FileLen(filenam$) - 1) Open filenam$ For Binary Access Read As #1 Get #1, , TheBytes() Close #1 ' Convert the entire byte array to a string sStr = StrConv(TheBytes, vbUnicode) ' Search for stream and endstream lStart = InStr(1, sStr, " stream") Do While lStart > 0 lEnd = InStr(lStart, sStr, " endstream") If lEnd > 0 Then ' Get. Note: This example requires Chilkat v9. How to use the Linux or UNIX command line to decompress gzip files. 04-15build1) [universe] Partition editor for Amiga partitions (cross version) amispammer (3. I am developing a new bootloader for a PIC32MZ based on the last Harmony 2. This site uses cookies - We have placed cookies on your device to help make this website better. How to decompress a file in Windows. NTDLL buffer compression vs ZLIB Post by MichaelW » Jun 17, 2006 10:52 These are updated versions of two quick-and-dirty command-line programs that I used to test the undocumented NTDLL. You didn't account the amount of time it takes to do the decompression in your calculation. The file is part of a PS3 save game, and the game always says the file is corrupt after the re-compressed file has been injected, whether or not the XML file was modified. dll file opener free download - Free Opener, Free RAR File Opener, Free ZIP File Opener, and many more programs. Check out projects section. Description. 392s sys 0m0. compress(text) print. This was addressed in epan/tvbuff_zlib. The recommendation is to compress data as much as possible and therefore to use this field, but some types of resources, such as. 15 Bytes/cycle). We mentioned that to excel at forensics CTF challenges, it is important to be able to recognize encodings. I need to decompress it using zlib library. Additional details about zlib1. The uncompress() function shall attempt to uncompress sourceLen bytes of data in the buffer source, placing the result in the buffer dest. Decompress)) {. As announced via the GenBank newsgroup on June 15, 2000, NCBI now uses the gzip compression utility instead of the Unix "compress" utility for all GenBank files, starting with GenBank Release 119. Compressed files use file compression in order to save disk space. ltokens ltxml lua-bencode lua-coat lua-codegen lua-crypt lua-curl lua-dbus lua-dialog lua-discount lua-dns lua-eclipse-ide lua-espeak lua-ev lua-ex lua-gd lua-gtk lua-iconv lua-imlib2 lua-import lua-modbus lua-mode lua-rdiff lua-sensors lua-signal lua-spec lua-spore lua-sqlite3 lua-statgrab lua-studio lua-tcc lua-template lua-tinycdb lua-unistd. Higher number means better, but slower, compression--fast is the same as -1, --best is the same as -9--11 selects zopfli algorithm which creates zlib-compatible data, compresses ~5% better than zlib but is much slower. The following are code examples for showing how to use bz2. A Reader is an io. ZIP!とは?IT用語辞典。 〖Zone Improvement Plan〗 アメリカの郵便集配区域改善計画。郵便番号制度。州と配達地、郵便局または郵便区を示す 5 桁の数字（ジップ━コード）を書く。. offset = 0 # position in unzipped stream self. Post by marpon » Mar 25, 2019 17:47. The default value for level is 6. decompress and rencode. This specification is intended for use by implementors of software to compress data into gzip format and/or decompress data from gzip format. Only the first header is recorded in the Reader fields. You can add them, remove them, modify them. The reason some karaoke zip files won't play IS because of the compression method. Using uncompress in C. Use code METACPAN10 at checkout to apply your discount. And therefore can't be decompressed by 7-zip. Which module should be used? python-programming; You can use the zlib module available in python to compress and decompress the string. Data compression is the art of reducing the number of bits needed to store or transmit data. There are many Base64 encoder/decoders online, or you can use the base64 command:. zlib shell> zlib_decompress dump. Performs the following: Removes docstrings. lzma from the MinGW Sourceforge download page. NET Framework 4. Whether they are 2D or 3D based, they offer tools to aid in asset creation and placement. Get the world’s #1 zip file opener on Android! Create Zip and Zipx files, extract files, encrypt, open Zip, Zipx, 7z, RAR or LHA files, send large files by email, share to Dropbox and Google Drive. Fast & Reliable ZIP Library for C# & VB. Mega2 also compresses the genotype information as much as possible for use by the analysis programs. The default value for level is 6. 0 is a new development release. To unpack a tar. Decompress it, then. Package zlib implements reading and writing of zlib format compressed data, as specified in RFC 1950. As a result, DeflateStream provides a better implementation of the deflate algorithm and, in most cases, a better compression than in earlier versions of the. zip package has two classes, DeflaterOutputStream and InflaterInputStream, that make it easy to compress (deflate) and uncompress (inflate) a stream. Once a file is compressed using Compress, you can restore the file to its original state with the uncompress utility. Zlib uncompress. miniz is providing a very interresting code in c to compress/decompress with comparable compress ratio/speed as zlib, but with very little size : on windows 25ko compared to zlib static lib 128ko. Type your input to the Text string field or select the input file through the File field and finally, hit the "Encode!" or the "Decode!". Firstly, you must define the destination folder, which is the storage location for the extracted files. Compressing and Decompressing Data with Zlib. abs(z) > infinity_border) & (image == 0) Can someone help me fix this and tell me why? I'm still fairly new to Python and would be grateful if you posted the changes in the code in your comment. decode ('hex') result = binascii. gzip also refers to the associated compressed data format used by the utility. Joined: Jul 1999. decompress(s) method, we can decompress the compressed bytes of string into original string by using zlib. Yes, I have highlighted this possible issue with WAN backups in the past (see the forum digest couple of weeks ago). deflate The transformation will be a compressing transformation that produces raw compressed data on channel , which must be writable. Example: Emulating a file object for compressed streams # File: zlib-example-4. 10-beta, Sphinx supports two different indexing backends: "disk" index backend, and "realtime" (RT) index backend. js provides an excellent library in the Zlib module that allows you to compress and decompress data in buffers very easily and efficiently. By default when you compress a file or folder using the gzip command it will have the same file name as it did before but with the extension. Several open and proprietary compression algorithms can be used to compress files, which is why many different compressed file types exist. gzip is short for GNU zip; the program is a free software replacement for the compress program used in early Unix systems. decompress(compMsg) decompObj. Alternative is to use zlib library, although there is quazip wrapper, I personally prefer to keep it as simple as possible and in my opinion zlib is easier. I just quickly pulled off a python script that will decompress the compressed FlateDecode stream. var json = JSONC. A serious security flaw has been identified in Zlib, a widely used data compression library. Well another post on PDF. Paste a plain-text SAML Message in the form field and obtain its base64 encoded version. xda-developers Android Development and Hacking Android Apps and Games Decrypting WhatsApp crypt12 files by TripCode XDA Developers was founded by developers, for developers. How to decompress gzipped contents. Get the world’s #1 zip file opener on Android! Create Zip and Zipx files, extract files, encrypt, open Zip, Zipx, 7z, RAR or LHA files, send large files by email, share to Dropbox and Google Drive. localStorage being usually limited to 5MB, all you can compress is that much more data you can store. /api/formula. js streams for beginners and professionals with examples on first application, repl terminal, package manager, callback concept, event loop, buffers, streams, file systems, global objects, web modules and more. It can also be zlib/gzip compressed in which case you need to use the zlib library as well. Topic: zlib double-free Category: core, ports Module: zlib Announced: 2002-03-18 Credits: Matthias Clasen Owen Taylor Affects: All released versions of FreeBSD FreeBSD 4. On the server I need to decompress raw bytes and get a string. Compress is a Unix based compress program. Chrono Cross Piano Project. It is used to compress and decompress data. The compressed data format generated is in the gzip, zlib, or single-entry zip format using the deflate compression method. dll from a "DLL download" website. _gzip_mask) decompressed = self. i686 : Low-Level Interface to the zlib. get_range( self. 0 source tarball. This made things especially tricky when a Perl filehandle was passed to gzopen. As a result, you may. They scan each http/ftp request and cannot open the archive properly (maybe wrong/incompatible decompression software). The following are code examples for showing how to use zlib. com · Zip and unzip ZIP files online · No registration, no uploads: safe and fast 1. The gSOAP tools support the integration of (legacy) C/C++ codes (and other programming languages when a C interface is available), embedded systems, and real-time software in SOAP/XML applications that share computational resources and information with other SOAP applications, possibly across different platforms, language environments, and disparate organizations located behind firewalls. Thanks again! =) This is a maintenence release: Xdelta 1. 2_1 Simple streaming interface to zlib for the Lua Programming Language Long description | Changes | Main Web Site Maintained by: [email protected] javascript - gzdecode - zlib decompress online Распакуйте gzip и строку zlib в JavaScript (2). decompress method in python) and try to analyse it statically. It is now a valuable resource for people who want to make the most of their mobile devices, from customizing the look and feel to adding new functionality. LZ4 is lossless compression algorithm, providing compression speed > 500 MB/s per core (>0. 6 is an easy to use, high-performance, suite of compression components for. OpenWrite("test. compress(msg) bad = -24 decompObj = zlib. gzipped content decompress. OK, I Understand. You can add them, remove them, modify them. com you can unzip. net vs a wrapper on top of zlib 1. MAX_WBITS|32) 'test' >>> zlib. decompress and rencode. You can also drop files here. Current linked Poco::DeflatingOutputStream compress with zlib 1. LZ4 is lossless compression algorithm, providing compression speed > 500 MB/s per core (>0. zz - converts test. Before extracting the data from a compressed file, you must first determine the compressed file type. Zlib Decompress. Inflate/deflate, URL encode/decode. If you’re on a desktop environment and the command-line is not your thing, you can use your File manager. Re: Setup tools error; zlib not available Thanks Mhgsys for replying so quick. Using uncompress in C++. h containing directory, in the Quazip/quazip. This report is generated from a file or URL submitted to this webservice on December 14th 2015 19:51:26 (UTC) Guest System: Windows 7 32 bit, Home Premium, 6. 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58. zz to the decompressed test file. Context/Problem. Decompress object. gz file extension are created using Gzip program which reduces the size of the named files using Lempel-Ziv coding (LZ77). decompress() to compress and decompress a string. 20191231,1, lua51-5. Use the following command to decompress a gzip archive: gzip -d filename. params(level, strategy, callback). 0 Standard. The DMG Extractor can open a wide variety of Disk Image files and encrypted. Homebrew Formulae. It loads as a table so please give it a moment as some of the lists are quite long. Can I use Zlib with Delphi 2. The gzip module uses the zlib license (see src/gzip/zlib. decompress(s) method, we can decompress the compressed bytes of string into original string by using zlib. Package zlib implements reading and writing of zlib format compressed data, as specified in RFC 1950. 04 (“OpenSSH_6. I have also tried to configure zlib by editing comments in setup file in Modules. Alternative is to use zlib library, although there is quazip wrapper, I personally prefer to keep it as simple as possible and in my opinion zlib is easier. zip files and unrar. Unreal engine 4 games with customization. 491s user 0m4. Xbox uses LZX (Xmemcompress) for chunk data. Image Module¶. Library Genesis is a fantastic resource, that gives you free access to millions of your favourite books as ebooks - fiction, fantasy, crime, science fiction, romance, thriller, textbooks, academic works, graphic novels, comics - in epub, pdf, mobi and many other formats. Naïve scheme of JSON + Zlib works well: Double vs Float: do you really need to store that much precision? For more inspiration look to columnar DBs and how they compress columns Observations import json import zlib # compress compressed = zlib. and optionally leaves the stream open. On mobile devises, I have noticed the following thing. Introduction. gz archives. 5 we changed our DeflateStream implementation to use the popular zlib library. It is used to compress and decompress data. decompress(data)) zlib. Note that when doing reverse conversions with xxd, the data needs to look like a hex dump: there needs to be an offset and the data needs to be formatted correctly. 0? I wrote an article for the January 1997 Dr. Unzip-online. When you compress the data, you decrease the amount of data the Data Integration Service writes over the network. s must have no NUL's or newlines. That, plus the fact that I did find several packages that require zlib, made me wonder if it's part of the base system. This time around a small snippet that will help us in decoding the FlateDecoded streams in PDF. var json = JSONC. I need a simple application for Windows (and/or Linux) View 1 Replies. Question: Please write a program to print the running time of execution of "1+1" for 100 times. x, gzopen used the zlib library to open the underlying file. Use MathJax to format equations. The reason is that the zlib library, used to compress files on the server has the default degree of compression (zlib. there are some quirks so you might have to play with it some. js provides an excellent library in the Zlib module that allows you to compress and decompress data in buffers very easily and efficiently. Gareth Rees, 2007-05-07. There is a difference. How to Decompress a File Using the gzip Command. It provides an unused_data method which returns the remainder of what wasn't consumed in by zlib in a previous call to Decompress. Context/Problem. Streamable: It is always possible to create and decompress. I can decompress them using Python API on the server but I prefer to keep the files compressed there. The text of the specification assumes a basic background in programming at the level of bits and other primitive data representations. while using w00ters udf am. hexlify (zlib. Young" wrote in message news:MPG. Famous decompress zlib compression library JAVA language. It's a reasonable point of view, and true, but if the zlib header already contains the information which I want, one would hope that the authors of zlib would create some function to access the information and relieve the users of the need to do hand parsing. Mega2 also compresses the genotype information as much as possible for use by the analysis programs. 3-2) [universe] Powerful Mail Server checker on blacklists amiwm (0. Decompress a normal JSON object from a LZW string: // Returns the original JSON object. compress(s) print t print zlib. The gzip data compression algorithm itself is based on zlib module. 15 Bytes/cycle). Python - The Data Compression Modules: zlib, gzip, bz2 zlib. - Python 3 : Convert string to bytes. SAML protocol uses the base64 encoding algorithm when exchanging SAML messages. Fast & Reliable ZIP Library for C# & VB. gz file extension are created using Gzip program which reduces the size of the named files using Lempel-Ziv coding (LZ77). 8, April 28th, 2013. what is arm-linux-gcc and how to install this in ubuntu) Download: Zlib OpenSSL OpenSSH Build Zlib: cd zlib-1. We aggregate information from all open source repositories. This was a year or two ago so things might have changed now. Software Packages in "jessie", Subsection libs 389-ds-base-libs erlang interface to zlib erlang-xmlrpc Compress/decompress images for mailheaders, libc6. Check out projects section. Enthusiast. Python's gzip module is the interface to GZip application. "portraits-File00000" extensions will be added according to the filetype they were determined to be by the editor. After numerous requests to update my bot and check out the new encryption, I wasn't that surprised nothing really had changed due to the devs lack of motivation for security. Did you use the version from zlib or the version in the csp gateway and can you confirm the name of the file. # (Ignored) WITH ZLIB # Link to LAPACK WITH LAPACK # Force dynamic linking FORCE DYNAMIC # # -----# Set this variable to either UNIX, MAC or WIN SYS = UNIX # Leave blank after "=" to disable; put "= 1" to enable WITH R PLUGINS = 1 WITH WEBCHECK = 1 FORCE 32BIT = WITH ZLIB = WITH LAPACK = FORCE DYNAMIC = # Put C++ compiler here; Windows has it. NET cross-platform apps. Those are its limitations. Opening a file and reading the content of a file is one of the common things you would do while doing data analysis. File Name: Advanced GRF Tool Suite. Most NGS tools can handle compressed files directly, and it is generally faster to read a compressed file than an uncompressed one. dll file and put it into whatever folder on your PC. For details on the ZLIB compression algorithm see the document "» ZLIB Compressed Data Format Specification version 3. Zlib: general purpose data compression / decompression library.
f9tf6mwx0kq4 a5sfri82gwqch90 hv9bs8whvqt 1gqhjbs158 g4q3pei5rj bmhzprkt0t nnmzxyr9sx0i6a yrxzow4vrgc1y bagb7q7bd7 8oniwxq51ax6 tafxdm730j536x8 sidgdi5irurw xmm4oqnjy0e u9396cfw4jx9myq kzw977o9jnb kghbdwqryyk 2dxqs1b045w25b yc0lyop8rdztmj3 ak5nwx24oh ik5buf9ciwwydqs 894eehysix4h7n5 bdv6c72s16svmb xczouvst3z60 hjawkxpud784xmw n32cq88drt1627 v6emvsffwbc vs0isw7dhagmjb xuqqkcnubd1r m56sk3kq3ak7 lry7pwzv4fmyr39 hyiz4uegomo
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2020-08-04 11:58:57
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https://socratic.org/questions/how-do-you-find-the-product-of-7i-4-and-its-conjugate#282070
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How do you find the product of 7i + 4 and its conjugate?
1 Answer
Jun 28, 2016
It turns out that the product of a complex number and its conjugate is the square of its modulus:
Let a random complex number be given by $a + b i$
Therefore it's conjugate is $a - b i$
(This is why I've written it as $a + b i$ not $a i + b$ otherwise the conjugate would begin with a negative since it always attaches itself to the imaginary part
Their product: $\left(a + b i\right) \left(a - b i\right) = {a}^{2} + a b i - a b i - {b}^{2} {i}^{2}$
Since by definition ${i}^{2} = - 1$, this is equivalent to ${a}^{2} + {b}^{2}$
(The modulus of a complex number is given by $\sqrt{{a}^{2} + {b}^{2}}$ so the above is the square of this)
Therefore, we could go through and multiply our particular complex number $7 i + 4$ by its conjugate (which I'll omit here because it's just done in the same way as above, just with values not letters), but we may as well use the little formula we just generated:
${7}^{2} + {4}^{2} = 49 + 16 = 65$
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2021-12-09 10:05:23
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https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Advanced_Statistical_Mechanics_(Tuckerman)/13%3A_Time-dependent_Processes_-_Quantum_Case/13.01%3A_Calculation_of_spectra_from_perturbation_theory/13.1.01%3A_The_Hamiltonian
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# 13.1.1: The Hamiltonian
Consider a quantum system with a Hamiltonian $$H_0$$. Suppose this system is subject to an external driving force $$F_e (t)$$ such that the full Hamiltonian takes the form
$H = H_0 - BF_e(t) = H_0 + H' \nonumber$
where $$B$$ is an operator through which this coupling occurs. This is the situation, for example, when the infrared spectrum is measured experimentally - the external force $$F_e (t)$$ is identified with an electric field $$E (t)$$ and $$B$$ is identified with the electric dipole moment operator. If the field $$F_e (t)$$ is inhomogeneous, then $$H$$ takes the more general form
\begin{align*} H &= H_0 - \int d^3x\; B({\textbf x}) F_e({\textbf x},t) \\[4pt] &= H_0 - \sum_{\textbf k} B_{\textbf k}F_{e,{\textbf k}}(t) \end{align*}
where the sum is taken over Fourier modes. Often, $$B$$ is an operator such that, if $$F_e (t) = 0$$, then
$\langle B \rangle = {{\rm Tr}\left(Be^{-\beta H}\right) \over {\rm Tr}\left(e^{-\beta H}\right)} \nonumber$
Suppose we take $$F_e (t)$$ to be a monochromatic field of the form
$F_e(t) = F_{\omega}e^{i\omega t} \nonumber$
Generally, the external field can induce transitions between eigenstates of $$H_0$$ in the system. Consider such a transition between an initial state $$\vert i \rangle$$ and a final state $$\vert f \rangle$$, with energies $$E_i$$ and $$E_f$$, respectively:
\begin{align*} H_0\vert i\rangle &= E_i\vert i\rangle \\[4pt] H_0\vert f\rangle &= E_f\vert f\rangle \end{align*}
(see figure below).
This transition can only occur if
$E_f = E_i + \hbar\omega \nonumber$
This page titled 13.1.1: The Hamiltonian is shared under a CC BY-NC-SA 4.0 license and was authored, remixed, and/or curated by Mark Tuckerman.
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2023-01-28 11:05:06
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https://mathhelpboards.com/threads/proper-subsets-of-ordinals-searcoid-theorem-1-4-4.25032/
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# Proper Subsets of Ordinals ... ... Searcoid, Theorem 1.4.4 ... ...
#### Peter
##### Well-known member
MHB Site Helper
I am reading Micheal Searcoid's book: "Elements of Abstract Analysis" ... ...
I am currently focused on understanding Chapter 1: Sets ... and in particular Section 1.4 Ordinals ...
I need some help in fully understanding Theorem 1.4.4 ...
In the above proof by Searcoid we read the following:
"... ... Now, for each $$\displaystyle \gamma \in \beta$$ , we have $$\displaystyle \gamma \in \alpha$$ by 1.4.2, and the minimality with respect to $$\displaystyle \in$$ of $$\displaystyle \beta$$ in $$\displaystyle \alpha \text{\\} x$$ ensures that $$\displaystyle \gamma \in x$$. ... ...
Ca someone please show formally and rigorously that the minimality with respect to $$\displaystyle \in$$ of $$\displaystyle \beta$$ in $$\displaystyle \alpha \text{\\} x$$ ensures that $$\displaystyle \gamma \in x$$. ... ...
Help will be appreciated ...
Peter
==========================================================================
It may help MHB readers of the above post to have access to the start of Searcoid's section on the ordinals (including Theorem 1.4.2 ... ) ... so I am providing the same ... as follows:
It may also help MHB readers to have access to Searcoid's definition of a well order ... so I am providing the text of Searcoid's Definition 1.3.10 ... as follows:
Hope that helps ...
Peter
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2020-07-09 07:55:31
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http://research.omicsgroup.org/index.php/Convective_available_potential_energy
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# Convective available potential energy
File:Convective instability animation 12Z 21Z Jan08.gif
A skew-T plot showing a morning sounding with a large hydrolapse followed by an afternoon sounding showing the cooling (red curve moving to the left) which occurred in the mid-levels resulting in an unstable atmosphere as surface parcels have now become negatively buoyant. The red line is temperature, the green line is the dew point, and the yellow line is the air parcel lifted.
In meteorology, convective available potential energy (CAPE),[1] sometimes, simply, available potential energy (APE), is the amount of energy a parcel of air would have if lifted a certain distance vertically through the atmosphere. CAPE is effectively the positive buoyancy of an air parcel and is an indicator of atmospheric instability, which makes it very valuable in predicting severe weather. It is a form of fluid instability found in thermally stratified atmospheres in which a colder fluid overlies a warmer one. As explained below, when an air mass is unstable, the element of the air mass that is displaced upwards is accelerated by the pressure differential between the displaced air and the ambient air at the (higher) altitude to which it was displaced. This usually creates vertically developed clouds from convection, due to the rising motion, which can eventually lead to thunderstorms. It could also be created by other phenomena, such as a cold front. Even if the air is cooler on the surface, there is still warmer air in the mid-levels, that can rise into the upper-levels. However, if there is not enough water vapor present, there is no ability for condensation, thus storms, clouds, and rain will not form.
## Mechanics
File:B and LCL-LFC.jpg
A Skew-T diagram with important features labeled
CAPE exists within the conditionally unstable layer of the troposphere, the free convective layer (FCL), where an ascending air parcel is warmer than the ambient air. CAPE is measured in joules per kilogram of air (J/kg). Any value greater than 0 J/kg indicates instability and the possibility of thunderstorms. Generic CAPE is calculated by integrating vertically the local buoyancy of a parcel from the level of free convection (LFC) to the equilibrium level (EL):
$\mathrm{CAPE} = \int_{z_\mathrm{f}}^{z_\mathrm{n}} g \left(\frac{T_\mathrm{v,parcel} - T_\mathrm{v,env}}{T_\mathrm{v,env}}\right) \, dz$
Where $z_\mathrm{f}$ is the height of the level of free convection and $z_\mathrm{n}$ is the height of the equilibrium level(neutral buoyancy), where $T_\mathrm{v,parcel}$ is the virtual temperature of the specific parcel, where $T_\mathrm{v,env}$ is the virtual temperature of the environment, and where $g$ is the acceleration due to gravity. CAPE for a given region is most often calculated from a thermodynamic or sounding diagram (e.g., a Skew-T log-P diagram) using air temperature and dew point data usually measured by a weather balloon.
CAPE is effectively positive buoyancy, expressed B+ or simply B; the opposite of convective inhibition (CIN), which is expressed as B-, and can be thought of as "negative CAPE". As with CIN, CAPE is usually expressed in J/kg but may also be expressed as m2/s2, as the values are equivalent. In fact, CAPE is sometimes referred to as positive buoyant energy (PBE). This type of CAPE is the maximum energy available to an ascending parcel and to moist convection. When a layer of CIN is present, the layer must be eroded by surface heating or mechanical lifting, so that convective boundary layer parcels may reach their level of free convection (LFC).
On a sounding diagram, CAPE is the positive area above the LFC, the area between the parcel's virtual temperature line and the environmental virtual temperature line where the ascending parcel is warmer than the environment. Neglecting the virtual temperature correction may result in substantial relative errors in the calculated value of CAPE for small CAPE values.[2] CAPE may also exist below the LFC, but if a layer of CIN (subsidence) is present, it is unavailable to deep, moist convection until CIN is exhausted. When there is mechanical lift to saturation, cloud base begins at the lifted condensation level (LCL); absent forcing, cloud base begins at the convective condensation level (CCL) where heating from below causes spontaneous buoyant lifting to the point of condensation when the convective temperature is reached. When CIN is absent or is overcome, saturated parcels at the LCL or CCL, which had been small cumulus clouds, will rise to the LFC, and then spontaneously rise until hitting the stable layer of the equilibrium level. The result is deep, moist convection (DMC), or simply, a thunderstorm.
When a parcel is unstable, it will continue to move vertically, in either direction, dependent on whether it receives upward or downward forcing, until it reaches a stable layer (though momentum, gravity, and other forcing may cause the parcel to continue). There are multiple types of CAPE, downdraft CAPE (DCAPE), estimates the potential strength of rain and evaporatively cooled downdrafts. Other types of CAPE may depend on the depth being considered. Other examples are surface based CAPE (SBCAPE), mixed layer or mean layer CAPE (MLCAPE), most unstable or maximum usable CAPE (MUCAPE), and normalized CAPE (NCAPE).[3]
Fluid elements displaced upwards or downwards in such an atmosphere expand or compress adiabatically in order to remain in pressure equilibrium with their surroundings, and in this manner become less or more dense.
If the adiabatic decrease or increase in density is less than the decrease or increase in the density of the ambient (not moved) medium, then the displaced fluid element will be subject to downwards or upwards pressure, which will function to restore it to its original position. Hence, there will be a counteracting force to the initial displacement. Such a condition is referred to as convective stability.
On the other hand, if adiabatic decrease or increase in density is greater than in the ambient fluid, the upwards or downwards displacement will be met with an additional force in the same direction exerted by the ambient fluid. In these circumstances, small deviations from the initial state will become amplified. This condition is referred to as convective instability.[4]
Convective instability is also termed static instability, because the instability does not depend on the existing motion of the air; this contrasts with dynamic instability where instability is dependent on the motion of air and its associated effects such as dynamic lifting.
## Significance to thunderstorms
Thunderstorms form when air parcels are lifted vertically. Deep, moist convection requires a parcel to be lifted to the LFC where it then rises spontaneously until reaching a layer of non-positive buoyancy. The atmosphere is warm at the surface and lower levels of the troposphere where there is mixing (the planetary boundary layer (PBL)), but becomes substantially cooler with height. The temperature profile of the atmosphere, the change in temperature, the degree that it cools with height, is the lapse rate. When the rising air parcel cools more slowly than the surrounding atmosphere, it remains warmer and less dense. The parcel continues to rise freely (convectively; without mechanical lift) through the atmosphere until it reaches an area of air less dense (warmer) than itself.
The amount of and shape of the positive area modulates the speed of updrafts, extreme CAPE can result in explosive thunderstorm development; such rapid development usually occurs when CAPE stored by a capping inversion is released when the "lid" is broken by heating or mechanical lift. The amount of CAPE also modulates how low-level vorticity is entrained and then stretched in the updraft, with importance to tornadogenesis. The most important CAPE for tornadoes is within the lowest 1 to 3 km (0.6 to 1.9 mi), whilst deep layer CAPE and the width of CAPE at mid-levels is important for supercells. Tornado outbreaks tend to occur within high CAPE environments. Large CAPE is required for the production of very large hail, owing to updraft strength, although a rotating updraft may be stronger with less CAPE. Large CAPE also promotes lightning activity.[5]
Two notable days for severe weather exhibited CAPE values over 5 kJ/kg. Two hours before the 1999 Oklahoma tornado outbreak occurred on May 3, 1999, the CAPE value sounding at Oklahoma City was at 5.89 kJ/kg. A few hours later, an F5 tornado ripped through the southern suburbs of the city. Also on May 4, 2007 CAPE values of 5.5 kJ/kg were reached and an EF5 tornado tore through Greensburg, Kansas. On these days, it was apparent that conditions were ripe for tornadoes and CAPE wasn't a crucial factor. However, extreme CAPE, by modulating the updraft (and downdraft), can allow for exceptional events, such as the deadly F5 tornadoes that hit Plainfield, Illinois on August 28, 1990 and Jarrell, Texas on May 27, 1997 on days which weren't readily apparent as conducive to large tornadoes. CAPE was estimated to exceed 8 kJ/kg in the environment of the Plainfield storm and was around 7 kJ/kg for the Jarrell storm.
Severe weather and tornadoes can develop in an area of low CAPE values. The surprise severe weather event that occurred in Illinois and Indiana on April 20, 2004 is a good example. Importantly in that case, was that although overall CAPE was weak, there was strong CAPE in the lowest levels of the troposphere which enabled an outbreak of minisupercells producing large, long-track, intense tornadoes.[6]
## Example from meteorology
A good example of convective instability can be found in our own atmosphere. If dry mid-level air is drawn over very warm, moist air in the lower troposphere, a hydrolapse (an area of rapidly decreasing dew point temperatures with height) results in the region where the moist boundary layer and mid-level air meet. As daytime heating increases mixing within the moist boundary layer, some of the moist air will begin to interact with the dry mid-level air above it. Owing to thermodynamic processes, as the dry mid-level air is slowly saturated its temperature begins to drop, increasing the adiabatic lapse rate. Under certain conditions, the lapse rate can increase significantly in a short amount of time, resulting in convection. High convective instability can lead to severe thunderstorms and tornadoes as moist air which is trapped in the boundary layer eventually becomes highly negatively buoyant relative to the adiabatic lapse rate and eventually escapes as a rapidly rising bubble of humid air triggering the development of a cumulus or cumulonimbus cloud.
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main-content
## Über dieses Buch
This three volume set, CCIS 771, 772, 773, constitutes the refereed proceedings of the CCF Chinese Conference on Computer Vision, CCCV 2017, held in Tianjin, China, in October 2017.
The total of 174 revised full papers presented in three volumes were carefully reviewed and selected from 465 submissions. The papers are organized in the following topical sections: biological vision inspired visual method; biomedical image analysis; computer vision applications; deep neural network; face and posture analysis; image and video retrieval; image color and texture; image composition; image quality assessment and analysis; image restoration; image segmentation and classification; image-based modeling; object detection and classification; object identification; photography and video; robot vision; shape representation and matching; statistical methods and learning; video analysis and event recognition; visual salient detection.
## Inhaltsverzeichnis
### An Evolution Perception Mechanism for Organizing and Displaying 3D Shape Collections
How to display the best view of shape collections to facilitate non-profession users is becoming a challenging problem in the field of computer vision. To solve this problem, an evolution perception mechanism for organizing and displaying shape collections (EPM) is proposed. On the one hand, the evolution tree based on Quartet analysis is constructed to effectively organize shapes from a global perspective. On the other hand, the shape snapshot method based on the view features of shapes is presented for each single shape in the evolution tree, so as to show the details of the shape from a local perspective. Moreover, the interactive display methods using the evolution trees are provided to guide users to effectively explore the shape collections. Experimental results show that EPM could provide the effective displaying ways that are close to the user’s viewing.
Lingling Zi, Xin Cong, Yanfei Peng
### C-CNN: Cascaded Convolutional Neural Network for Small Deformable and Low Contrast Object Localization
Traditionally, the normalized cross correlation (NCC) based or shape based template matching methods are utilized in machine vision to locate an object for a robot pick and place or other automatic equipment. For stability, well-designed LED lighting must be mounted to uniform and stabilize lighting condition. Even so, these algorithms are not robust to detect the small, blurred, or large deformed target in industrial environment. In this paper, we propose a convolutional neural network (CNN) based object localization method, called C-CNN: cascaded convolutional neural network, to overcome the disadvantages of the conventional methods. Our C-CNN method first applies a shallow CNN densely scanning the whole image, most of the background regions are rejected by the network. Then two CNNs are adopted to further evaluate the passed windows and the windows around. A relatively deep model net-4 is applied to adjust the passed windows at last and the adjusted windows are regarded as final positions. The experimental results show that our method can achieve real time detection at the rate of 14FPS and be robust with a small size of training data. The detection accuracy is much higher than traditional methods and state-of-the-art methods.
Xiaojun Wu, Xiaohao Chen, Jinghui Zhou
### Skeleton-Based 3D Tracking of Multiple Fish From Two Orthogonal Views
This paper proposes a skeleton-based method for tracking multiple fish in 3D space. First, skeleton analysis is performed to simplify object into feature point representation according to shape characteristics of fish. Next, based on the obtained feature points, object association and matching are achieved to acquire the motion trajectories of fish in 3D space. This process relies on top-view tracking that is supplemented by side-view detection. While fully exploiting the shape and motion information of fish, the proposed method is able to solve the problems of frequent occlusions that occur during the tracking process and has good tracking performance.
Zhiming Qian, Meiling Shi, Meijiao Wang, Tianrui Cun
### A Retinal Adaptation Model for HDR Image Compression
High dynamic range (HDR) images are usually used to capture more information of natural scenes, because the light intensity of real world scenes commonly varies in a very large range. Humans visual system is able to perceive this huge range of intensity benefiting from the visual adaptation mechanisms. In this paper, we propose a new visual adaptation model based on the cone- and rod-adaptation mechanisms in the retina. The input HDR scene is first processed in two separated channels (i.e., cone and rod channels) with different adaptation parameters. Then, a simple receptive field model is followed to enhance the local contrast of the visual scene and improve the visibility of details. Finally, the compressed HDR image is obtained by recovering the fused luminance distribution to the RGB color space. Experimental results suggest that the proposed retinal adaptation model can effectively compress the dynamic range of HDR images and preserve local details well.
Xuan Pu, Kaifu Yang, Yongjie Li
### A Novel Orientation Algorithm for a Bio-inspired Polarized Light Compass
Many animals, such as honey bees and tarantulas, can navigate with polarized light, the key of which is to extract compass information from skylight polarization pattern. Many groups have conducted research on bionic polarization sensors and obtained good orientation results in the open-sky circumstances. However, if the sky is obscured by leaves or buildings, the skylight polarization pattern will be greatly affected. This paper presents an unsupervised method for polarization navigation when the sky is partly blocked. First of all, we introduce the core components of polarized light compass and the measurement method of skylight polarization pattern. Then, an unsupervised method is used to extract the sky region according to the single scattering Rayleigh model. Finally, we calculate solar meridian vector and heading angle using pixels of sky region. Results show that polarization navigation, featuring high anti-interference and no accumulation error, is suitable for outdoor autonomous navigation.
Guoliang Han, Xiaoping Hu, Xiaofeng He, Junxiang Lian, Lilian Zhang, Yujie Wang
### A Brain MR Images Segmentation and Bias Correction Model Based on Students t-Mixture Model
Accurate segmentation for magnetic resonance images is an essential step in quantitative brain image analysis. However, due to the existence of bias field and noise, many segmentation methods are hard to find accurate results. Finite mixture model is one of the wildly used methods for MR image segmentation; however, it is sensitive to noise and cannot deal with images with intensity inhomogeneity. In order to reduce the effect of noise, we introduce a robust Markov Random Field by incorporating new spatial information which is constructed based on posterior probabilities and prior probabilities. The bias field is modeled as a linear combination of a set of orthogonal basis functions and coupled into the model and makes the method can estimate the bias field meanwhile segmenting images. Our statistical results on both synthetic and clinical images show that the proposed method can obtain more accurate results.
Yunjie Chen, Qing Xu, Shenghua Gu
### Define Interior Structure for Better Liver Segmentation Based on CT Images
Liver Segmentation has important application for preoperative planning and intraoperative guiding. In this paper we introduce a new approach by defining the interior structure (hepatic veins) before segmenting the liver from nearby organs. We assume that cells of the liver should lay within a certain distance of the hepatic veins. Therefore, a clear segmentation on hepatic veins will facilitate our segmentation on liver voxel. We build a probabilistic model which adopts four main features of the liver cells based on this idea and implement it on the open source platform 3DMed. We also test the accuracy of this method with four groups of CT data. The results are similar when compared to human experts.
Xiaoyu Zhang, Yixiong Zheng, Bin Zheng
### GPU Accelerated Image Matching with Cascade Hashing
SIFT feature is widely used in image matching. However, matching massive images is time consuming because SIFT feature is a high dimensional vector. In this paper, we proposed a GPU accelerated image matching method with improved Cascade Hashing. Firstly, we propose a disk-memory-GPU data exchange strategy and optimize the load order of data, so that the proposed method can deal with big data. Then, we parallelize the Cascade Hashing method on GPU. An improved parallel reduction and an improved parallel hashing ranking are proposed to fulfill this task. Finally, extensive experiments are carried out to show that our image matching is about 20 times faster than the SiftGPU, nearly one hundred times faster than the CPU Cascade Hashing, and hundreds of times faster than the CPU Kd-Tree based matching.
Tao Xu, Kun Sun, Wenbing Tao
### Improved Single Image Dehazing with Heterogeneous Atmospheric Light Estimation
Images captured in foggy or hazy weather conditions are often degraded by the scattering of atmospheric particles, which seriously reduces the performance of outdoor computer vision processing systems. Single image haze removal algorithm has been considered to be an efficient dehazing method in recent years. The key to this type of approach is the estimation of atmospheric light. In this paper, an improved single image dehazing algorithm with heterogeneous atmospheric light estimation is presented to enhance the quality of hazy images. First, the heterogeneous atmospheric light is calculated with max-pooling. Second, a haze-free image can be recovered with the estimated atmospheric light based on dark channel prior. The experimental results on a variety of hazy images demonstrate that the addressed method outperforms state-of-the-art approaches through the assessment of dehazing effect and algorithm cost.
Yi Lai, Ying Liu
### Spatiogram and Fuzzy Logic Based Multi-modality Fusion Tracking with Online Update
Multi-modality fusion tracking is an interesting but challenging task. Many previous works just consider the fusion of different features from identical spectral image or identical features from different spectral images alone, which makes them be quite distinct from each other and be difficult to be integrated naturally. In this study, we propose an unified tracking framework to naturally integrate multiple different modalities via innovative use of spatiogram and fuzzy logic. Specifically, each modal target and its candidate are first represented by second-order spatiogram and their similarity is measured. Next, a novel objective function is built by integrating all modal similarities, and then a joint target center-shift formula is gained by performing mathematical operation on the objective function. Finally, the optimal target location is gained recursively by applying the mean shift procedure. Besides, a model update scheme via particle filter is developed to capture the appearance variations. Our framework allows the modalities to be original pixels or other extracted features from single image or different spectral images, and provides the flexibility to arbitrarily add or remove modality. Tracking results on the combination of infrared gray-HOG and visible gray-LBP clearly demonstrate the excellence of the proposed tracker.
Canlong Zhang, Zhixin Li, Zhiwen Wang, Ting Han
### A Novel Real-Time Tracking Algorithm Using Spatio-Temporal Context and Color Histogram
Spatio-temporal context (STC) is one of the most important features in describing the motion in videos. STC-based tracking algorithm achieved good performance on real-time tracking. However, it is very difficult to perform precise tracking in complex situations like heavy occlusion, illumination changes, and pose variation. In this paper, we propose a real-time tracking method which is robust to target variation during tracking single-object. Experiments on some challenging sequences highlights a significant improvement of tracking accuracy over the state-of-the-art methods.
Yong Wu, Zemin Cai, Jianhuang Lai, Jingwen Yan
### A New Visibility Model for Surface Reconstruction
In this paper, we propose a new visibility model for scene reconstruction. To yield out surface meshes with enough scene details, we introduce a new visibility model, which keeps the relaxed visibility constraints and takes the distribution of points into consideration, thus it is efficient for preserving details. To strengthen the robustness to noise, a new likelihood energy term is introduced to the binary labeling problem of Delaunay tetrahedra, and its implementation is presented. The experimental results show that our method performs well in terms of detail preservation.
Yang Zhou, Shuhan Shen, Zhanyi Hu
### Global-Local Feature Attention Network with Reranking Strategy for Image Caption Generation
In this paper, a novel framework, named global-local feature attention network with reranking strategy (GLAN-RS), is presented for image captioning task. Rather than only adopt unitary visual information in the classical models, GLAN-RS explore attention mechanism to capture local convolutional salient image maps. Furthermore, we adopt reranking strategy to adjust the priority of the candidate captions and select the best one. The proposed model is verified using the MSCOCO benchmark dataset across seven standard evaluation metrics. Experimental results show that GLAN-RS significantly outperforms the state-of-the-art approaches such as M-RNN, Google NIC etc., which gets an improvement of 20% in terms of BLEU4 score and 13 points in terms of CIDER score.
Jie Wu, Siya Xie, Xinbao Shi, Yaowen Chen
### Dust Image Enhancement Algorithm Based on Color Transfer
The increasing dust weather has seriously affected the quality of captured images. Therefore, research of the dust image quality enhancement has become an important hotspot in the field of computer vision. Compared with pictures which obtained in sunny day, dust images have some obvious problems such as low-quality in definition and light, hue yellowing and so on. Amid such questions, available methods cannot always prevail. Hence, this paper proposed a dust image enhancement algorithm based on color transfer. First of all, applying the scene gist feature to select a target image which has the highest similarity with input image in clear images database; then, the target image color information is passed to input image through the color transfer algorithm; finally, utilizing the contrast limited adaptive histgram equalization algorithm to restore the definition in dust image. The experimental results show that our algorithm can effectively enhance dust images quality in different degree, get more image details and resolve the definition and hue correction problems. We proposed a absolute color difference index to measure this method. The experiments demonstrate that the proposed algorithm outperforms the state-of-the-art models.
Hao Liu, Ce Li, Yuqi Wan, Yachao Zhang
### Chinese Sign Language Recognition with Sequence to Sequence Learning
In this paper, we formulate Chinese sign language recognition (SLR) as a sequence to sequence problem and propose an encoder-decoder based framework to handle it. The proposed framework is based on the convolutional neural network (CNN) and recurrent neural network (RNN) with long short-term memory (LSTM). Specifically, CNN is adopted to extract the spatial features of input frames. Two LSTM layers are cascaded to implement the structure of encoder-decoder. The encoder-decoder can not only learn the temporal information of the input features but also can learn the context model of sign language words. We feed the images sequences captured by Microsoft Kinect2.0 into the network to build an end-to-end model. Moreover, we also set up another model by using skeletal coordinates as the input of the encoder-decoder framework. In the recognition stage, a probability combination method is proposed to fuse these two models to get the final prediction. We validate our method on the self-build dataset and the experimental results demonstrate the effectiveness of the proposed method.
Chensi Mao, Shiliang Huang, Xiaoxu Li, Zhongfu Ye
### ROI Extraction Method of Infrared Thermal Image Based on GLCM Characteristic Imitate Gradient
Automatic inspection of UAV (unmanned aerial vehicle) vision in photovoltaic power station is of great significance in effectively capturing high-definition images and quickly detecting the fault area, which can reduce the risk of false detection and lower the cost of manual operation. However, due to the complexity of photovoltaic power station environment, disturbance often occurrence on images, leading up to the misjudgment of the fault area. We proposed a method to extraction region based on the gray level co-occurrence matrix (GLCM) and textural features. The image extraction from target area can be achieved by extracting feature images, gradient imitate and region filling. This method effectively combines the textural features of images with edge features of the gradient images. A comparison is made between the algorithm promoted in this paper and the grab cut method, on the basis of the labeled image segmentation algorithm. It turns out that the mean precision and mean recall of the proposed imitation gradient image extraction method are higher than that of the grab cut algorithm, and the Recallvalue, F index and J index are better than the grab cut algorithm. A new algorithm is proposed construct filling model by using gradient image and a morsel of texture feature calculated by GLCM method. The advantages of the proposed algorithm are fewer interactive tags, fewer manual labels. Therefore, the image detection of fault area can be better realized.
Hui Shen, Li Zhu, Xianggong Hong, Weike Chang
### Learning Local Feature Descriptors with Quadruplet Ranking Loss
In this work, we propose a novel deep convolutional neural network (CNN) with quadruplet ranking loss to learn local feature descriptors. The proposed model receives quadruplets of two corresponding patches and two non-corresponding patches, and then outputs features measured by $$L_2$$L2 norm with dimension (256d) close to that of SIFT, which thus can be easily applied to practical vision tasks by plug-in replacing SIFT-like features. Moreover, the proposed model mitigates the problem that the margin separating corresponding and non-corresponding pairs varies with samples caused by commonly used triplet loss, and improves the capacity of utilizing the limited training data. Experiments show that our model outperforms some state-of-the-art methods like TN-TG and PN-Net.
Dalong Zhang, Lei Zhao, Duanqing Xu, Dongming Lu
### Three-Dimensional Reconstruction of Ultrasound Images Based on Area Iteration
Ultrasonic C scan technique has been extensively applied in nondestructive testing (NDT) in recent years. Aiming at ultrasonic C scanning original data, this paper presents a 3D reconstruction algorithm for ultrasonic C scanning image based on area iteration. Because the distance between two neighboring slices is small, the area chance between two neighboring images is also few. Therefore, we design an iterative rule on the area of the object to detect and extract the target contour of each image. After preprocessing of ultrasonic C image, we extract the contour of the object scanned by ultrasonic transducer. Then, we reconstruct the 3D ultrasonic C scanning image using marching cube algorithm. In experiments, we implement the proposed method for ultrasonic scanning data of tissue-mimicking phantom including a liver model, and we compare the proposed method with other methods. The results show that the proposed method can more accurately display the edge information of the scanned object, and improve the precision of region detection, especially suitable when the edge has the incomplete information.
Biying Chen, Haijiang Zhu, Jinglin Zhou, Ping Yang, Longbiao He
### A Generation Method of Insulator Region Proposals Based on Edge Boxes
The generation of region proposals is the foundation of object detection. In the object detection task, the steady increase in complexity of classifiers may lead to improvement of detection quality, yet with the cost of increased computation time at the same time. One approach to overcome the tension between high detection quality and low computational complexity is through the use of “region proposals”. High-quality insulator region proposals also play important roles in the detection of transmission line inspection images. This paper applies Edge Boxes to the localization of insulators in inspection images creatively, considering the characteristics of insulators’ edge images, and combines these characteristics with Edge Boxes. As a result, more insulator region proposals are displayed. The experimental results show that, our method can effectively reduce the interference area, meanwhile, has high quality of region proposals with fast speed of calculation.
Zhenbing Zhao, Lei Zhang, Yincheng Qi, Yuying Shi
### Robust Tramway Detection in Challenging Urban Rail Transit Scenes
With the rapid development of light rail transit, tramway detection based on video analysis is becoming the prerequisite and necessary task in driver assistance system. The system should be capable of automatically detecting the trackway using on-board camera in order to determine the train driving limit. However, due to the diversification of ground types, the diversity of weather conditions and the differences in illumination situations, this goal is very challenging. This paper presents a real-time tramway detection method that can effectively deal with various challenging scenarios in the real world of urban rail transit environment. It first uses an adaptive multi-level threshold to segment the ROI of the trolley track, where the local cumulative histogram model is used to estimate the threshold parameters. And then use the regional growth method to reduce the impact of environmental noise and predict the trend of tramway. We have experimentally proved that the method can correctly detect the tramway even in many undesirable situations and use less computational time to meet real-time requirements.
Cheng Wu, Yiming Wang, Changsheng Yan
### Relevance and Coherence Based Image Caption
The attention-based image caption framework has been widely explored in recent years. However, most techniques generate next word conditioned on previous words and current visual contents, while the relationship between the semantic and visual contents is not considered. In this paper, we present a novel framework which can explore the relevance and coherence at the same time. The relevance tries to explore the relationship between the semantic and visual contents in a semantic-visual embedding space, and the coherence is introduced to maximize the probability of generating the next word according to previous words and the current visual contents. The performance of our model is tested with three benchmark datasets: Flickr8k, Flickr30k and MS COCO. The experimental results show that the proposed approach can improve the performance of attention-based image caption method.
Tao Zhang, Wei Wang, Liang Wang, Qinghua Hu
### An Efficient Recoloring Method for Color Vision Deficiency Based on Color Confidence and Difference
Human with color vision deficiency (CVD) cannot distinguish some colors under normal situation. And recoloring is a color adaptation procedure for human with CVD. This paper proposes an efficient recoloring method. First, key color confidence is introduced to judge and recolor unrecognizable colors sequentially instead of simultaneously to improve the execution speed. Second, color differences between one color and other colors with higher confidence are taken as the unrecognizable benchmark for this color without any pre-defined parameters. Third, these differences are also used in iterative recoloring procedure to maximize the contrast between different colors in the recolored image. Experimental results show that our proposed method outperforms state-of-the-art recoloring methods in terms of subjective, objective qualities and executive speed.
Qing Xu, Xiangdong Zhang, Liang Zhang, Guangming Zhu, Juan Song, Peiyi Shen
### A Saliency Based Human Detection Framework for Infrared Thermal Images
In this paper, a novel saliency framework for crowd detection in infrared thermal images is proposed. In order to obtain the optimal classifier from a large amount of data, the process of training consists of the following four steps: (a) a saliency contrast algorithm is employed to detect the regions of interest; (b) standard HOG features of the selected interest areas are extracted to represent the human object; (c) the extracted features, which are prepared for training, are optimized based on a visual attention map; (d) a support vector machine (SVM) algorithm is applied to compute the classifier. Finally, we can detect the human precisely after high-saliency areas of an image are input into the classifier. In order to evaluate our algorithm, we constructed an infrared thermal image database collected by a real-time inspection system. The experimental results demonstrated that our method can outperform the previous state-of-the art methods for human detection in infrared thermal images, and the visual attentional techniques can effectively represent prior knowledge for features optimization in a practicable system.
Xinbo Wang, Dahai Yu, Jianfeng Han, Guoshan Zhang
### Integrating Color and Depth Cues for Static Hand Gesture Recognition
Recognizing static hand gesture in complex backgrounds is a challenging task. This paper presents a static hand gesture recognition system using both color and depth information. Firstly, the hand region is extracted from complex background based on depth segmentation and skin-color model. The Moore-Neighbor tracing algorithm is then used to obtain hand gesture contour. The k-curvature method is used to locate fingertips and determine the number of fingers, then the angle between fingers are generated as features. The appearance-based features are integrated to the decision tree model for hand gesture recognition. Experiments have been conducted on two gesture recognition datasets. Experimental results show that the proposed method achieves a high recognition accuracy and strong robustness.
Jiaming Li, Yulan Guo, Yanxin Ma, Min Lu, Jun Zhang
### Accurate Joint Template Matching Based on Tree Propagating
Given a single template image, it is a big challenge to match all the target images accurately only by pairwise template matching. To handle this case, this paper introduces an accurate template matching method based on tree structure building to jointly match a set of target images. Our method aims to select well matched results for template updating and rescue badly matched images via the tree matching propagating. First, a novel similarity measure is given to evaluate the pairwise matching results. Then the joint matching is under an iterative framework which contains two main steps: (1) tree structure growing; and (2) matching propagating. When all the target images are included in the tree structure, the matching process is finished. Finally, experimental results demonstrate the improvement of the proposed method.
Qian Kou, Yang Yang, Shaoyi Du, Zhuo Chen, Weile Chen, Dexing Zhong
### Robust Visual Tracking Using Oriented Gradient Convolution Networks
Convolutional networks have been successfully applied to visual tracking to extract some useful feature. However, deep networks are time-consuming to offline training and usually extract the feature from raw pixels. In this paper, we propose a two-layer convolutional network based on oriented gradient. The first layer is constructed by the convolution of the filter and an input image of oriented gradient, which is robust to the illumination variation and motion blur. Then, all of the feature maps of the simple layer are stacked to a complex feature map as the target representation. The complex feature map can encode the local structure feature which is robust to occlusion. The proposed approach is tested on nine challenging sequences in comparison with nine state-of-art trackers, and the result show that the proposed tracker achieves mean overlap rate of 0.75, which outperforms the secondary tracker by 26%.
Qi Xu, Huabin Wang, Jian Zhou, Liang Tao
### Novel Hybrid Method for Human Posture Recognition Based on Kinect V2
In recent years, human posture recognition based on Kinect gradually has been paid more attention. However, the current researches and methods have drawbacks, such as low recognition accuracy and less recognizable postures. This paper proposed a novel method. The method utilized image processing technique, BP neural network technique, skeleton data and depth data captured by Kinect v2 to recognize postures. We distinguished four types of postures (sitting cross-legged, kneeling or sitting, standing, and other postures) by using the natural ratios of human body parts, and judged the kneeling and sitting postures by calculating the 3D spatial relation of the feature points. Finally, we applied BP neural network to recognize the lying and bending postures. The experimental results indicated that the robustness and timeliness of our method was strong, the recognition accuracy was 98.98%.
Bo Li, Baoxing Bai, Cheng Han, Han Long, Lin Zhao
### Visibility Estimation Using a Single Image
In this paper, we propose a novel method for visibility estimation using only a single image as input. An assumption is proposed: the extinction coefficient of light is approximately a constant in clear atmosphere. Using this assumption with the theory of atmospheric radiation, the extinction coefficient in clear atmosphere can be estimated. Based on the dark channel prior, ratio of two extinction coefficients in current and clear atmosphere is calculated. By multiplying the clear extinction coefficient and the ratio, we can estimate the extinction coefficient of the input image and then obtain the visibility value. Compared with other methods that require the explicit extraction of the scene, our method needs no constraint and performs well in various types of scenes, which might open a new trend of visibility measurement in meteorological research. Moreover, the actual distance information can also be estimated as a by-product of this method.
Qin Li, Bin Xie
### A Text-Line Segmentation Method for Historical Tibetan Documents Based on Baseline Detection
Text-line segmentation is an important task in the historical Tibetan document recognition. Historical Tibetan document images usually contain touching or overlapping characters between consecutive text-lines, making text-line segmentation a difficult task. In this paper, we present a text-line segmentation method based on baseline detection. The initial positions for the baseline of each line are obtained by template matching, pruning algorithms and closing operation. The baseline is estimated using dynamic tracing within pixel points of each line and the context information between pixel points. The overlapping or touching areas are cut by finding the minimum width stroke. Finally, text-lines are extracted based on the estimated baseline and the cut position of touching area. The proposed algorithm has been evaluated on the dataset of historical Tibetan document images. Experimental result shows the effectiveness of the proposed method.
Yanxing Li, Longlong Ma, Lijuan Duan, Jian Wu
### Low-Light Image Enhancement Based on Constrained Norm Estimation
Low-light images often suffer from poor quality and low visibility. Improving the quality of low-light image is becoming a highly desired subject in both computational photography and computer vision applications. This paper proposes an effective method to constrain the illumination map t by estimating the norm and constructing the constraint coefficients, which called LieCNE. More specifically, we estimate the initial illumination map by finding the maximum value of R, G and B channels and optimize it by norm estimation. We propose a function $${t^\gamma }$$tγ to contain the exponential power $$\gamma$$γ in order to optimize the enhancement effect under different illumination conditions. Finally, a new evaluation criterion is also proposed. We use the similarity with the true value to determine the enhanced effect. Experimental results show that LieCNE exhibits better performance under a variety of lighting conditions in enhancement results and image spillover prevention.
Tan Zhao, Hui Ding, Yuanyuan Shang, Xiuzhuang Zhou
### FPGA Architecture for Real-Time Ultra-High Definition Glasses-Free 3D System
This paper presents an FPGA architecture for real-time ultra-high definition (UHD) glasses-free 3D system by solving high bandwidth requirement and high complexity of the system problems. Video + Depth (V + D) video format and the depth-image-based rendering (DIBR) technique are supported by the system to reduce the requirement of bandwidth. In addition, an asymmetric shift-sensor camera setup is introduced to reduce the hardware cost as well as the complexity of the system. We also simplify the microlens array weight equations so as to reduce the complexity of subpixel rearrangement coefficients calculation for glasses-free 3D image creation. Experiments results show that the proposed architecture can support the resolution of 4K for real-time UHD glasses-free 3D display.
Ran Liu, Mingming Liu, Dehao Li, Yanzhen Zhang, Yangting Zheng
### Quantum Video Steganography Protocol Based on MCQI Quantum Video
A secure quantum video steganography protocol with large payload based on the video strip encoding method called as MCQI (Multi-Channel Quantum Images) is proposed in this paper. The new protocol exploits to embed quantum video as secret information for covert communication. As a result, its capacity are greatly expanded compared with the previous quantum steganography achievements. The new protocol achieves good security and imperceptibility by virtue of the randomization of embedding positions and efficient use of redundant frames. Furthermore, the receiver enables to extract secret information from stego video without retaining the original carrier video, and restore the original quantum video as a follow. The simulation and experiment results prove that the algorithm not only has good imperceptibility, high security, but also has large payload.
Zhiguo Qu, Siyi Chen, Sai Ji
### Video Question Answering Using a Forget Memory Network
Visual question answering combines the fields of computer vision and natural language processing. It has received much attention in recent years. Image question answering (Image QA) targets to automatically answer questions about visual content of an image. Different from Image QA, video question answering (Video QA) needs to explore a sequence of images to answer the question. It is difficult to focus on the local region features which are related to the question from a sequence of images. In this paper, we propose a forget memory network (FMN) for Video QA to solve this problem. When the forget memory network embeds the video frame features, it can select the local region features that are related to the question and forget the irrelevant features to the question. Then we use the embedded video and question features to predict the answer from multiple-choice answers. Our proposed approaches achieve good performance on the MovieQA [21] and TACoS [28] dataset.
Yuanyuan Ge, Youjiang Xu, Yahong Han
### Three Dimensional Stress Wave Imaging Method of Wood Internal Defects Based on TKriging
In order to detect the size, shape and degree of decay inside wood, a three dimensional stress wave imaging method based on TKriging is proposed. The method uses sensors to obtain the stress wave velocity data sets by hanging around the timber randomly, and reconstructs the image of internal defect with those data sets. TKriging optimized structural relationship between interpolation point and reference point in space firstly. The searching radius is used to select the reference points accordingly. Top-k query method is introduced to find the k value with relevant points. The values of the estimated points are calculated and three dimensional image of the internal defect inside wood is reconstructed. The results show the effectiveness of the method and the accuracy rate of sample with one hole is higher than the Kriging method.
Xiaochen Du, Hailin Feng, Mingyue Hu, Yiming Fang, Jiajie Li
### A Novel Color Image Watermarking Algorithm Based on QWT and DCT
A novel color image watermarking algorithm is proposed based on quaternion wavelet transform (QWT) and discrete cosine transform (DCT) for copyright protection. The luminance channel Y of the host color image in YCbCr space is decomposed by QWT to obtain four approximation subimages. A binary watermark is embedded into the mid-frequency DCT coefficients of two randomly selected subimages. The experimental results show that the proposed watermarking scheme has good robustness against common image attacks such as adding noise, filtering, scaling, JPEG compression, cropping, image adjusting, small angle rotation and so on.
Shaocheng Han, Jinfeng Yang, Rui Wang, Guimin Jia
### A Novel Camera Calibration Method for Binocular Vision Based on Improved RBF Neural Network
Considering the problems that camera imaging model is complex and operation is complicated, a binocular camera calibration method of RBF neural network based on k-means and gradient method is proposed in this paper. The data center selection method based on the law of clustering error function can obtain hidden nodes and data centers of RBF network accurately. Dynamic learning of data centers, spread constants and weight values based on gradient method can contribute to improving the precision. Experimental results show that the proposed method has high precision and can be well applied in machine vision.
Weike Liu, Ju Huo, Xing Zhou, Ming Yang
### Collaborative Representation Based Neighborhood Preserving Projection for Dimensionality Reduction
Collaborative graph-based discriminant analysis (CGDA) has been recently proposed for dimensionality reduction and classification. It uses available samples to construct sample collaboration via L2 norm minimization-based representation, thus showing great computational efficiency. However, CGDA only constructs the intra-class graph, so it only takes into account local geometry and ignores the separability for samples in different classes. In this paper, we propose a novel method termed as collaborative representation based neighborhood preserving projection (CRNPP) for dimensionality reduction. By incorporating the intra-class and inter-class discriminant information into the graph construction of collaborative representation coefficients, CRNPP not only maintains the same level of time cost as CGDA, but also preserves both global and local geometry of the data simultaneously. In this way, the collaborative relationship of the data from the same class is strengthened while the collaborative relationship of the data from different classes is inhibited in the projection subspace. Experiments on benchmark face databases validate the effectiveness and efficiency of the proposed method.
Miao Li, Lei Wang, Hongbing Ji, Shuangyue Chen, Danping Li
### Iterative Template Matching with Rotation Invariant Best-Buddies Pairs
In this paper, we propose a new method for template matching method with rotation invariance. Our template matching can not only find the location of the object, but also annotate its rotation angle. The key idea is to firstly rectify the local rotation patches according to their intensity centroids, and then to find the corresponding patch-features between template and target images under an iterative matching framework. We adopt the coarse-to-fine search ways, so the patch size should be updated accordingly, which is time-consuming. To tackle this problem, we use the integral image to update the intensity centroid to accelerate the computing speed. The corresponding feature matching is based on the Best-Buddies Pairs (BBPs), which is robust to the non-rigid transform of local range and outliers. Experimental results demonstrate the effectiveness and robustness of the proposed algorithm.
Zhuo Chen, Yang Yang, Weile Chen, Qian Kou, Dexing Zhong
### Image Forgery Detection Based on Semantic Image Understanding
Image forensics has been focusing on low-level visual features, paying little attention to high-level semantic information of the image. In this work, we propose the framework for image forgery detection based on high-level semantics with three components of image understanding module, the normal rule bank (NR) holding semantic rules that comply with our common sense, and the abnormal rule bank (AR) holding semantic rules that don’t. Ke et al. [1] also proposed a similar framework, but ours has following advantages. Firstly, image understanding module is integrated by a dense image caption model, with no need for human intervention and more hierarchical features. secondly, our proposed framework can generate thousands of semantic rules automatically for NR. Thirdly, besides NR, we also propose to construct AR. In this way, not only can we frame image forgery detection as anomaly detection with NR, but also as recognition problem with AR. The experimental results demonstrate our framework is effective and performs better.
Kui Ye, Jing Dong, Wei Wang, Jindong Xu, Tieniu Tan
### Relative Distance Metric Leaning Based on Clustering Centralization and Projection Vectors Learning for Person Re-identification
Existing projection-based person re-identification methods usually suffer from long time training, high dimension of projection matrix, and low matching rate. In addition, the intra-class instances may be much less than the inter-class instances when a training dataset is built. To solve these problems, a novel relative distance metric leaning based on clustering centralization and projection vectors learning (RDML-CCPVL) is proposed. When constructing training dataset, the images of a same target person are clustering centralized with FCM. The training datasets are built by these clusters in order to alleviate the imbalanced data problem of the training datasets. In addition, during learning projection matrix, the resulted projection vectors can be approximately orthogonal by using an iteration strategy and a conjugate gradient projection vector learning method to update training datasets. The advantage of this method is its quadratic convergence, which can promote the convergence. Experimental results show that the proposed algorithm has higher efficiency. The matching rate can be significantly improved, and the time of training is much shorter than most of existing algorithms of person re-identification.
Tongguang Ni, Zongyuan Ding, Fuhua Chen, Hongyuan Wang
### Multi-context Deep Convolutional Features and Exemplar-SVMs for Scene Parsing
Scene parsing is a challenging task in computer vision field. The work of scene parsing is labeling every pixel in an image with its semantic category to which it belongs. In this paper, we solve this problem by proposing an approach that combines the multi-context deep convolutional features with exemplar-SVMs for scene parsing. A convolutional neural network is employed to learn the multi-context deep features which include image global features and local features. In contrast to hand-crafted feature extraction approaches, the convolutional neural network learns features automatically and the features can better describe images on the task. In order to obtain a high class recognition accuracy, our system consists of the exemplar-SVMs which is training a linear SVM classifier for every exemplar in the training set for classification. Finally, multiple cues are integrated into a Markov Random Field framework to infer the parsing result. We apply our system to two challenging datasets, SIFT Flow dataset and the dataset which is collected by ourselves. The experimental results demonstrate that our method can achieve good performance.
Xiaofei Cui, Hanbing Qu, Songtao Wang, Liang Dong, Ziliang Qi
### How Depth Estimation in Light Fields Can Benefit from Angular Super-Resolution?
With the development of consumer light field cameras, the light field imaging has become an extensively used method for capturing the 3D appearance of a scene. The depth estimation often require a dense sampled light field in the angular domain. However, there is an inherent trade-off between the angular and spatial resolution of the light field. Recently, some studies for novel view synthesis or angular super-resolution from a sparse set of have been introduced. Rather than the conventional approaches that optimize the depth maps, these approaches focus on maximizing the quality of synthetic views. In this paper, we investigate how the depth estimation can benefit from these angular super-resolution methods. Specifically, we compare the qualities of the estimated depth using the original sparse sampled light fields and the reconstructed dense sampled light fields. Experiment results evaluate the enhanced depth maps using different view synthesis approaches.
Mandan Zhao, Gaochang Wu, Yebin Liu, Xiangyang Hao
### Influence Evaluation for Image Tampering Using Saliency Mechanism
Due to the immediacy and the easy way to understand the image content, the applications of digital image have brought great opportunity to the development of social networks. However, there exist some serious problems. Some visual content has been maliciously tampered to achieve illegal purpose, while some modifications are benign, just for fun, for enhancing artistic value, or effectiveness of news dissemination. So beyond the tampering detection, how to evaluate the influence of image tampering is on schedule. In this paper, with the help of forensic tools, we study the problem of automatically assessing the influence of image tampering by examining whether the modification affects the dominant visual content, and utilize saliency mechanism to assess how harmful the tampering is. The experimental results demonstrate the effectiveness of our method.
Kui Ye, Xiaobo Sun, Jindong Xu, Jing Dong, Tieniu Tan
### Correlation Filters Tracker Based on Two-Level Filtering Edge Feature
In recent years, correlation filter frame has attracted more attention in visual object tracking, providing a real-time way. However, with the increase of the computing complexity of feature extractor, the trackers lost the real-time advantage of correlation filters. Moreover, correlation filters model drift can result in tracking failure. In order to solve these problems, a novel and simple correlation filters tracker based on two-level filtering edge feature (ECFT) was proposed. ECFT extracted a low-complexity edge feature based on two-level filtering for object representation. For reducing model drift as much as possible, an object model is updated adaptively by the maximum response value of correlation filters. The comparative experiments of 7 trackers on 20 challenging sequences showed that the ECFT tracker can perform better than other trackers in terms of AUC and Precision.
Dengzhuo Zhang, Donglan Cai, Yun Gao, Hao Zhou, Tianwen Li
### An Accelerated Superpixel Generation Algorithm Based on 4-Labeled-Neighbors
Superpixels are perceptually meaningful atomic regions that could effectively improve efficiency of subsequent image processing tasks. Simple linear iterative clustering (SLIC) has been widely used for superpixel calculation due to outstanding performance. In this paper, we propose an accelerated SLIC superpixel generation algorithm using 4-labeled neighbor pixels called 4L-SLIC. The main idea of 4L-SLIC is that the labels are assigned to a portion of the pixels while the others that associated with certain cluster are restrained by adjacent four labeled pixels. In this way, the average number of distance calculated times of pixels are effectively reduced and the similarity between adjacent pixels ensures a better segmentation effect. The experimental results confirm that 4L-SLIC achieved a speed up of 25%–30% without declining accuracy sharply compared to SLIC. In contrast to the method published on CVIU 2016, 4L-SLIC has an acceptable increase in the cost of time, in the mean time, there is a significant ascension to the accuracy of the segmentation.
Hongwei Feng, Fang Xiao, Qirong Bu, Feihong Liu, Lei Cui, Jun Feng
### A Method of General Acceleration SRDCF Calculation via Reintroduction of Circulant Structure
Discriminatively learned correlation filters (DCF) have been widely used in online visual tracking filed due to its simplicity and efficiency. These methods utilize a periodic assumption of the training samples to construct a circulant data matrix, which is also introduces unwanted boundary effects. Spatially Regularized Correlation Filters (SRDCF) solved this issue by introducing penalization on correlation filter coefficients. However, which breaks the circulant structure used in DCF. We propose Faster SRDCF (FSRDCF) via reintroduction of circulant structure. The circulant structure of training samples in the spatial domain is fully used, more importantly, we exploit the circulant structure of regularization function in the Fourier domain, which allows the problem to be solved more directly and efficiently. Our approach demonstrates superior performance over other non-spatial-regularization trackers on the OTB2013 and OTB2015.
Xiaoxiang Hu, Yujiu Yang
### Using Deep Relational Features to Verify Kinship
Kinship verification from facial images is a very challenging research topic. Differing from most of previous methods focusing on calculating a similarity metric, in this work, we utilize convolutional neural network and autoencoder to learn deep relational features for verifying kinship from facial images. Specifically, we firstly train a convolutional neural network to extract representative facial features, which derive from the last fully-connected layer in network. Then, facial features from two person are set as two ends of an autoencoder respectively, and relational features are extracted from the middle layer of the trained autoencoder. Finally, SVM classifiers are adopted to verify kinship (e.g., Father-Son). Experimental results on two public datasets show the effectiveness of the approach proposed in this work.
Jingyun Liang, Jinlin Guo, Songyang Lao, Jue Li
### Hyperspectral Image Classification Using Spectral-Spatial LSTMs
In this paper, we propose a hyperspectral image (HSI) classification method using spectral-spatial long short term memory (LSTM) networks. Specifically, for each pixel, we feed its spectral values in different channels into Spectral LSTM one by one to learn the spectral feature. Meanwhile, we firstly use principle component analysis (PCA) to extract the first principle component from a HSI, and then select local image patches centered at each pixel from it. After that, we feed the row vectors of each image patch into Spatial LSTM one by one to learn the spatial feature for the center pixel. In the classification stage, the spectral and spatial features of each pixel are fed into softmax classifiers respectively to derive two different results, and a decision fusion strategy is further used to obtain a joint spectral-spatial results. Experiments are conducted on two widely used HSIs, and the results show that our method can achieve higher performance than other state-of-the-art methods.
Feng Zhou, Renlong Hang, Qingshan Liu, Xiaotong Yuan
### A Novel Framework for Image Description Generation
The existing image description generation algorithms always fail to cover rich semantics information in natural images with single sentence or dense object annotations. In this paper, we propose a novel semi-supervised generative visual sentence generation framework by jointly modeling Regions Convolutional Neural Network (RCNN) and improved Wasserstein Generative Adversarial Network (WGAN), for generating diverse and semantically coherent sentence description of images. In our algorithm, the features of candidate regions are extracted with RCNN and the enriched words are polished by their context with an improved WGAN. The improved WGAN consists of a structured sentence generator and a multi-level sentence discriminators. The generator produces sentences recurrently by incorporating region-based visual and language attention mechanisms, while the discriminator assesses the quality of generated sentences. The experimental results on publicly available dataset show the promising performance of our work against other related works.
Qiang Cai, Ziyu Xue, Xiaoyu Zhang, Xiaobin Zhu, Wei Shao, Lei Wang
### 3D Convolutional Neural Network for Action Recognition
Compared with traditional machine learning methods, deep convolutional neural network has more powerful learning ability. The convolutional neural network model of the depth learning algorithm has made remarkable achievements in the field of computer vision. As a branch of neural network, 3D Convolutional neural network (3D CNN) is a relatively new research field in the field of computer vision. To extract features that contain more information, we develop a novel 3D CNN model for action recognition instead of the traditional 2D inputs. The final feature consists spatial and temporal information from multiple channels. We demonstrate the efficacy of our method on KTH dataset.
Junhui Zhang, Li Chen, Jing Tian
### Research on Image Colorization Algorithm Based on Residual Neural Network
In order to colorize the grayscale images efficiently, an image colorization method based on deep residual neural network is proposed. This method combines the classified information and features of the image, uses the whole image as the input of the network and forms a non-linear mapping from grayscale images to the colorful images through the deep network. The network is trained by using the MIT Places Database and ImageNet and colorizes the grayscale images. The experiment result shows that different data sets have different colorization effects on grayscale images, and the complexity of the network determines the colorization effect of grayscale images. This method can colorize the grayscale images efficiently, which has better visual effect.
Pinle Qin, Zirui Cheng, Yuhao Cui, Jinjing Zhang, Qiguang Miao
### Image Retrieval via Balanced and Maximum Variance Deep Hashing
Hashing is a typical approximate nearest neighbor search approach for large-scale data sets because of its low storage space and high computational ability. The higher the variance on each projected dimension is, the more information the binary codes express. However, most existing hashing methods have neglected the variance on the projected dimensions. In this paper, a novel hashing method called balanced and maximum variance deep hashing (BMDH) is proposed to simultaneously learn the feature representation and hash functions. In this work, pairwise labels are used as the supervised information for the training images, which are fed into a convolutional neural network (CNN) architecture to obtain rich semantic features. To acquire effective and discriminative hash codes from the extracted features, an objective function with three restrictions is elaborately designed: (1) similarity-preserving mapping, (2) maximum variance on all projected dimensions, (3) balanced variance on each projected dimension. The competitive performance is acquired using the simple back-propagation algorithm with stochastic gradient descent (SGD) method despite the sophisticated objective function. Extensive experiments on two benchmarks CIFAR-10 and NUS-WIDE validate the superiority of the proposed method over the state-of-the-art methods.
Shengjie Zhang, Yufei Zha, Yunqiang Li, Huanyu Li, Bing Chen
### FFGS: Feature Fusion with Gating Structure for Image Caption Generation
Automatically generating a natural language to describe the content of the given image is a challenging task in the interdisciplinary between computer vision and natural language processing. The task is challenging because computers not only need to recognize objects, their attributions and relationships between them in an image, but also these elements should be represented into a natural language sentence. This paper proposed a feature fusion with gating structure for image caption generation. First, the pre-trained VGG-19 is used as the image feature extractor. We use the FC-7 and CONV5-4 layer’s outputs as the global and local image feature, respectively. Second, the image features and the corresponding sentence are imported into LSTM to learn their relationship. The global image feature is gated at each time-step before imported into LSTM while the local image feature used the attention model. Experimental results show our method outperform the state-of-the-art methods.
Aihong Yuan, Xuelong Li, Xiaoqiang Lu
### Deep Temporal Architecture for Audiovisual Speech Recognition
The Audiovisual Speech Recognition (AVSR) is one of the applications of multimodal machine learning related to speech recognition, lipreading systems and video classification. In recent and related work, increasing efforts are made in Deep Neural Network (DNN) for AVSR, moreover some DNN models including Multimodal Deep Autoencoder, Multimodal Deep Belief Network and Multimodal Deep Boltzmann Machine perform well in experiments owing to the better generalization and nonlinear transformation. However, these DNN models have several disadvantages: (1) They mainly deal with modal fusion while ignoring temporal fusion. (2) Traditional methods fail to consider the connection among frames in the modal fusion. (3) These models aren’t end-to-end structure. We propose a deep temporal architecture, which has not only classical modal fusion, but temporal modal fusion and temporal fusion. Furthermore, the overfitting and learning with small size samples in the AVSR are also studied, so that we propose a set of useful training strategies. The experiments show the superiority of our model and necessity of the training strategies in three datasets: AVLetters, AVLetters2, AVDigits. In the end, we conclude the work.
Chunlin Tian, Yuan Yuan, Xiaoqiang Lu
### Efficient Cross-modal Retrieval via Discriminative Deep Correspondence Model
Cross-modal retrieval has recently drawn much attention due to the widespread existence of multi-modal data, and it generally involves two challenges: how to model the correlations and how to utilize the class label information to eliminate the heterogeneity between different modalities. Most previous works mainly focus on solving the first challenge and often ignore the second one. In this paper, we propose a discriminative deep correspondence model to deal with both problems. By taking the class label information into consideration, our proposed model attempts to seamlessly combine the correspondence autoencoder (Corr-AE) and supervised correspondence neural networks (Super-Corr-NN) for cross-modal matching. The former model can learn the correspondence representations of data from different modalities, while the latter model is designed to discriminatively reduce the semantic gap between the low-level features and high-level descriptions. The extensive experiments tested on three public datasets demonstrate the effectiveness of the proposed approach in comparison with the state-of-the-art competing methods.
Zhikai Hu, Xin Liu, An Li, Bineng Zhong, Wentao Fan, Jixiang Du
### Towards Deeper Insights into Deep Learning from Imbalanced Data
Imbalanced performance usually happens to those classifiers (including deep neural networks) trained on imbalanced training data. These classifiers are more likely to make mistakes on minority class instances than on those majority class ones. Existing explanations attribute the imbalanced performance to the imbalanced training data. In this paper, using deep neural networks, we strive for deeper insights into the imbalanced performance. We find that imbalanced data is a neither sufficient nor necessary condition for imbalanced performance in deep neural networks, and another important factor for imbalanced performance is the distance between the majority class instances and the decision boundary. Based on our observations, we propose a new under-sampling method (named Moderate Negative Mining) which is easy to implement, state-of-the-art in performance and suitable for deep neural networks, to solve the imbalanced classification problem. Various experiments validate our insights and demonstrate the superiority of the proposed under-sampling method.
Jie Song, Yun Shen, Yongcheng Jing, Mingli Song
### Traffic Sign Recognition Based on Deep Convolutional Neural Network
Traffic sign recognition (TSR) is an important component of automated driving system. It is a rather challenging task to design a high-performance classifier for the TSR system. In this paper, we proposed a new method for TSR system based on deep convolutional neural network. In order to enhance the expression of the network, a novel structure (dubbed block-layer below) which combines Network-in-Network and residual connection was designed. Our network has 10 layers with parameters (block-layer be seen as a single layer); the first seven are alternate convolutional layers and block-layers, and the remaining three are fully-connected layers. We trained our TSR network on the German Traffic Sign Recognition Benchmark (GTSRB) dataset. To reduce overfitting, we did data augmentation on the training images and employed a regularization method named dropout. We also employed a mechanism called Batch Normalization which has been proved to be efficient for accelerating the training of deep neural networks. To speed up the training, we used an efficient GPU to accelerate the convolutional operation. On the test dataset of GTSRB, we achieve the accuracy rate of 98.96%, exceeding the human average raters.
Shihao Yin, Jicai Deng, Dawei Zhang, Jingyuan Du
### Deep Context Convolutional Neural Networks for Semantic Segmentation
Recent years have witnessed the great progress for semantic segmentation using deep convolutional neural networks (DCNNs). This paper presents a novel fully convolutional network for semantic segmentation using multi-scale contextual convolutional features. Since objects in natural images tend to be with various scales and aspect ratios, capturing the rich contextual information is very critical for dense pixel prediction. On the other hand, with going deeper of the convolutional layers, the convolutional feature maps of traditional DCNNs gradually become coarser, which may be harmful for semantic segmentation. According to these observations, we attempt to design a deep context convolutional network (DCCNet), which combines the feature maps from different levels of network in a holistic manner for semantic segmentation. The proposed network allows us to fully exploit local and global contextual information, ranging from an entire scene, though sub-regions, to every single pixel, to perform pixel label estimation. The experimental results demonstrate that our DCCNet (without any postprocessing) outperforms state-of-the-art methods on PASCAL VOC 2012, which is the most popular and challenging semantic segmentation dataset.
Wenbin Yang, Quan Zhou, Yawen Fan, Guangwei Gao, Songsong Wu, Weihua Ou, Huimin Lu, Jie Cheng, Longin Jan Latecki
### Automatic Character Motion Style Transfer via Autoencoder Generative Model and Spatio-Temporal Correlation Mining
The style of motion is essential for virtual characters animation, and it is significant to generate motion style efficiently in computer animation. In this paper, we present an efficient approach to automatically transfer motion style by using autoencoder generative model and spatio-temporal correlation mining, which allows users to transform an input motion into a new style while preserving its original content. To this end, we introduce a history vector of previous motion frames into autoencoder generative network, and extract the spatio-temporal feature of input motion. Accordingly, the spatio-temporal correlation within motions can be represented by the correlated hidden units in this network. Subsequently, we established the constraints of Gram matrix in such feature space to produce transferred motion by pre-trained generative model. As a result, various motions of particular semantic can be automatically transferred from one style to another one, and the extensive experiments have shown its outstanding performance.
Dong Hu, Xin Liu, Shujuan Peng, Bineng Zhong, Jixiang Du
### Efficient Human Motion Transition via Hybrid Deep Neural Network and Reliable Motion Graph Mining
Skeletal motion transition is of crucial importance to the simulation in interactive environments. In this paper, we propose a hybrid deep learning framework that allows for flexible and efficient human motion transition from motion capture (mocap) data, which optimally satisfies the diverse user-specified paths. We integrate a convolutional restricted Boltzmann machine with deep belief network to detect appropriate transition points. Subsequently, a quadruples-like data structure is exploited for motion graph building, which significantly benefits for the motion splitting and indexing. As a result, various motion clips can be well retrieved and transited fulfilling the user inputs, while preserving the smooth quality of the original data. The experiments show that the proposed transition approach performs favorably compared to the state-of-the-art competing approaches.
Bing Zhou, Xin Liu, Shujuan Peng, Bineng Zhong, Jixiang Du
### End-to-End View-Aware Vehicle Classification via Progressive CNN Learning
This paper investigates how to perform robust vehicle classification in unconstrained environments, in which appearance of vehicles changes dramatically across different angles and the numbers of viewpoint images are not balanced among different car models. We propose a end-to-end progressive learning framework, which allows the network architecture is reconfigurable, for view-aware vehicle classification. In particular, the proposed network architecture consists of two parts: a general end-to-end progressive CNN architecture for coarse-to-fine or top-down fine-grained recognition task and an end-to-end view-aware vehicle classification framework to combine vehicle classification and viewpoints recognition. We test the technique on a large-scale car dataset, “CompCars”, and experimental results show that our framework can significantly improve performance of vehicle classification.
Jiawei Cao, Wenzhong Wang, Xiao Wang, Chenglong Li, Jin Tang
### A Fast Method for Scene Text Detection
Text detection is important for many applications such as text retrieval, blind guidance, and industrial automation. Meanwhile, text detection is a challenging task due to the complexity of the background and the diversity of the font, size and color of the text. In recent years, deep learning achieves good results in image classification and detection, and provides us a new method for text detection. In this paper, a deep learning based detection method – Single Shot MultiBox Detector (SSD) is adopted. But SSD is a general object detection method, not specific for text detection and is not fast enough. Our method aims to develop a network for text detection, improve the speed and reduce the model. Therefore, we design a feature extraction network with the inception module and an additional deconvolution layer. The experiment on benchmark – ICDAR2013 demonstrates that our method is faster than other SSD-based method comparable results.
Qing Fang, Yanping Yang, Yali Chen, Xiaoyu Yao
### Automatic Image Description Generation with Emotional Classifiers
Automatically generating a natural sentence describing the content of an image is a hot issue in artificial intelligence which links the domain of computer vision and language processing. Most of the existing works leverage large object recognition datasets and external text corpora to integrate knowledge between similar concepts. As current works aim to figure out ‘what is it’, ‘where is it’ and ‘what is it dong’ in images, we focus on a less considered but critical concept: ‘how it feels’ in the content of images. We propose to express feelings contained in images via a more direct and vivid way. To achieve this goal, we extend a pre-trained caption model with an emotion classifier to add abstract knowledge to the original caption. We practice our method on datasets originated from various domains. Especially, we examine our method on the newly constructed SentiCap dataset with multiple evaluation metrics. The results show that the newly generated descriptions can summarize the images vividly.
Yan Sun, Bo Ren
### Backmatter
Weitere Informationen
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2020-08-05 07:58:58
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https://catiepeterson.com/portfolio/bembo-futura-study/
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# Bembo & Futura Study
This is an exercise from my Typography class in Spring 2015. In it, I am studying the difference between Bembo and Futrua Book. The first and third row is set in Bembo, the second and fourth row is set in Futura Book. Also, they are set in pairs. The first two forms in the first column are a pair of X’s set in each font, comparing the bottom terminals of the letter. The first two forms in the second column are pairs of R’s, studying the arms of the R. And so on and so forth.
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2023-03-21 20:56:42
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https://physics.aps.org/articles/v7/s119
|
Synopsis
# Proton Longevity Pushes New Bounds
Physics 7, s119
A long-running experiment in Japan has released a new lower limit on the proton lifetime, which begins to constrain certain particle physics theories.
Protons live a long time but perhaps not forever. Several theories predict that protons can decay, and a handful of experiments have tried to detect such an event. The Super-Kamiokande experiment in Japan has the longest track record in the search for proton decay, and its researchers have now published a new lower bound on the proton lifetime that is 2.5 times greater than their previous bound. The proton’s observed stability places constraints on certain extensions of the standard model of particle physics.
Proton decay is an expected outcome of most grand unified theories, or GUTs, which meld together the three main particle forces—strong, weak, and electromagnetism—at high energy. A certain class of GUTs, for example, predicts that a proton should decay into a positron and $π$ meson with a lifetime of about $1031$ years, which means roughly 1 decay per year in a sample of $1031$ protons. Experiments have already ruled this possibility out.
Other GUTs that incorporate supersymmetry (SUSY), a hypothetical model that assumes all particles have a partner with different spin, predict that the proton decays into a $K$ meson and a neutrino with a lifetime of less than a few times $1034$ years. The Super-Kamiokande collaboration has looked for signs of this decay in a 50,000-ton tank of water surrounded by detectors. If one of the many protons in the tank were to decay, the $K$ meson’s decay products (muons, $π$ mesons) would be detectable. The researchers simulated such proton decay events but found no matches in data spanning 17 years. From this, they conclude that the proton lifetime for this SUSY-inspired decay pathway is greater than $5.9×1033$ years.
This research is published in Physical Review D.
–Michael Schirber
## Subject Areas
Particles and Fields
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The collision of two intense light beams may produce detectable signatures of dark matter particles called axions. Read More »
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2020-06-03 19:34:17
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|
https://www.physicsforums.com/threads/gravity-and-weight.574419/
|
# Gravity and Weight
1. Feb 5, 2012
### nDever
Hey,
About Newton's Second Law. To compute our weight on Earth using standard g, we use F=ma replacing F and a with W and g (or -g), respectively. So then from this, I gather that weight is a consequence of gravitational acceleration (and mass). So then, it is appropriate to say that our weight, which is a force, is caused by another force (gravity)?
2. Feb 5, 2012
### Staff: Mentor
Your weight is the gravitational force on you exerted by the earth. It happens to equal mg near the earth's surface.
3. Feb 5, 2012
### nDever
But g is an acceleration. Acceleration is caused by force (according to the first law of motion). What force causes g?
4. Feb 5, 2012
### Staff: Mentor
The gravitational force of the earth on the object.
5. Feb 5, 2012
### nDever
So our weight causes our own acceleration..?
6. Feb 5, 2012
### Staff: Mentor
Of course. What we call 'our weight' is the force of the earth's gravity pulling us down. If that's the only force acting on us, we will have a downward acceleration equal to g.
7. Feb 5, 2012
### Naty1
Far be it from me to disagree with Doc Al but I just don't like this:
I would much prefer to say our MASS causes our own acceleration. This distinction becomes significant in relativity....and it turns out other stuff like pressure
also affects gravitational acceleration.
This from Doc Al is much better in my opinion:
Check here for a good discussion of various aspects of WEIGHT:
http://en.wikipedia.org/wiki/Weight
8. Feb 7, 2012
### nDever
So the acceleration due to Earth's gravity is constant because we are in the reference frame of the Earth, correct? Whenever an object accelerates towards Earth, Earth accelerates towards that object at a rate directly proportional to the other object's mass, correct? Is this why the mass of the other object cancels out?
Last edited: Feb 7, 2012
9. Feb 7, 2012
### D H
Staff Emeritus
Acceleration due to Earth's gravity is approximately constant. It varies with latitude, with altitude, and even from place to place at the same latitude/altitude. Gravitational acceleration at the poles is about half a percent more than that at the equator, gravitational acceleration atop a high mountain measurably smaller than it is at sea level, and those place to place variations are key to finding things like oil.
Yes, but this acceleration is immeasurably small.
No. It cancels out because of the forms of Newton's law of gravitation, $F=GmM/r^2$, and Newton's second law, $F=ma$. Via transitivity (if a=b and a=c, then b=c) we must have $ma = GmM/r^2$, or $a=GM/r^2$.
10. Feb 7, 2012
### nDever
Is the acceleration approximately constant because the force of gravitational pull acts strongly on more massive objects and weaker on less massive objects? The mass of the attracted object has no bearing whatsoever on its acceleration?
Last edited: Feb 7, 2012
11. Feb 7, 2012
### Staff: Mentor
Yes. The gravitational force on an object is directly proportional to its mass. But the acceleration, given by Newton's 2nd law, is inversely proportional to its mass. Thus those two factors balance out, giving a constant acceleration regardless of mass.
12. Feb 9, 2012
### nDever
I believe I understand.
If I were to push a 1 kg supermarket cart down an aisle with a constant force of 1 N, it would accelerate at 1 m/s/s. If I were to put more stuff in the cart such that it has 2 kg of mass and then I push it with the same 1 N of force, then the acceleration halves and it accelerates at .5 m/s/s. These are the effects of applied forces. Double the mass, half the acceleration. Triple the mass, one third the acceleration.
Now Gravity is just a different force where the magnitude of the force is directly proportional to the product of the two masses (lets ignore the distance, I understand that). So then, the force of gravity increases and decreases with the mass. Its the nature of gravity. If the mass is small, the force is small. If the mass is large, the force is large. So then if a small and large mass is dropped from the same height, gravity pushes lightly on the less massive object and heavily on the more massive object, thus the acceleration is the same.
In a sense, gravity acts like this.
Now I have two supermarket carts placed side by side, one with a bunch of stuff in it, the other with very little in it, and I get two people to push them at the same time. The cart with more in it gets pushed harder and the other with less items gets pushed lighter. Let them go, and they move at the same rate with respect to one another.
Do I have the right idea?
13. Feb 9, 2012
### Staff: Mentor
I think you do.
For a more mathematical view, review D H's post #9 where he shows that the acceleration due to gravity does not depend on the mass of the object.
14. Feb 9, 2012
### nDever
Ah, I see. It has everything to do with the mathematical definition of acceleration in terms of force.
So then any two objects that are dropped from the same height within the gravitational field of any more massive body will fall at the same time. After gravity takes it's course, the only thing that the acceleration depends on is the mass of the larger body and the distance between it and the others. As altitude increases, g decreases.
The reason that Earth's g is about 9.8 m/s/s is only because of the way that the Earth's mass and our distance from its center works out. Because the Earth is large and we are relatively small, the everyday heights that we achieve allow us to average g out to be 9.8 m/s/s.
If this be the case, then it can be said that it is impossible for any two objects dropped from two different heights to land on the same point at the same time (with negligible air resistance). Is this sound?
Last edited: Feb 9, 2012
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2017-09-25 22:44:24
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https://mathhothouse.me/tag/complex-numbers/page/2/
|
## Tag Archives: complex numbers
### A complex equation
Find the number of solutions of the equation $z^{3}+\overline{z}=0$.
Solution.
Given that $z^{3}+\overline{z}=0$. Hence, $z^{3}=-z$.
$|z|^{3} =|-\overline{z}| \Longrightarrow |z|^{3}=|z|$.Hence, we get
$|z|(|z|-1)(|z|+1)=0 \Longrightarrow |z|=0, |z|=1$ (since $|z|+1>0$)
If $|z|=1$, we get $|z|^{2}=1 \Longrightarrow z.\overline{z}=1$.
Thus, $z^{3}+\overline{z}=0 \Longrightarrow z^{3}+1/z=0$
Thus, $z^{4}+1=0 \Longrightarrow z^{4}=\cos{\pi}+i\sin{\pi}$, that is,
$z=\cos{\frac{2k+1}{4}}\pi+i\sin{\frac{2k+1}{4}}\pi$ for $k=0,1,2,3$. Therefore, the given equation has five solutions.
### Complex numbers
Question: Prove that for any complex number z:
$|z+1| \geq \frac {1}{\sqrt {2}}$ or $|z^{2}+1| \geq 1$
Solution: Suppose by way of contradiction that
$|1+z| < \frac {1}{\sqrt {2}}$ and $|1+z^{2}| <1$
Setting $z=a+ib$ with $a,b \in R$ yields $z^{2}=a^{2}-b^{2}+i2ab$.
We obtain
$(1+a^{2}-b^{2})^{2}+4a^{2}b^{2}<1$ and $(1+a)^{2}+b^{2}<1/2$
and consequently,
$(a^{2}+b^{2})^{2}+2(a^{2}-b^{2})<0$ and $2(a^{2}+b^{2})+4a+1<0$.
Summing these inequalities implies $(a^{2}+b^{2})^{2}+(2a+1)^{2}>0$ which is a contradiction.
More later,
Nalin Pithwa
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2020-01-19 01:47:34
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|
https://adamwlev.github.io/Ryder_Cup_Individual_Grades/
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# Grading the Individual Performances at the Ryder Cup
Ryder_Cup_2
Hooray! The U.S. won the Ryder Cup. A big relief for all the U.S. golf fans!
Quick aside: for those unfamiliar with the Ryder Cup, let me explain what it is. It is a unique golf event in that it is a team event. It has a long history; for the past 40 or so years it has been a competition between the U.S. and Europe. Each side fields 12 players - most of whom earn their spot and some of whom are chosen by a captain. The teams compete in 28 matches overall, 16 of which are two vs. two, and 12 of which are one vs. one.
Ok, now that that is out of the way, on to the interesting data stuff! I know what you're thinking, let's get to some code! Well, let's do it. I took some time and entered in the data from this year's Ryder Cup. Let's load it in.
In [1]:
import pandas as pd
import numpy as np
import matplotlib.pyplot as plt
%matplotlib inline
In [2]:
In [3]:
data.iloc[0:2,np.array([4,7,20,22,24])]
Out[3]:
Rory McIlroy Henrik Stenson Patrick Reed Jordan Spieth Result
0 -1 0 1 0 1 up
1 0 1 0 -1 3 and 2
The data has 28 rows - one for each match - and 25 columns - one for each player plus one results column. The data is coded in the following fashion: since Patrick Reed beat Rory McIlroy in a singles match with the final result 1 up, Patrick Reed's column get's a positive one, Rory McIlroy's columns gets a negative one, and '1 up' is in the Result column. The reasons behind these choices will become clear once I reveal the motivation (which you might've surmised already from the title).
My motivation is to determine who had the best individual performance during the Ryder Cup. The Ryder Cup is absolutley a team event. However, golf fans love to discuss how each player performed as an individual. Hence golf writers will write plenty of articles over the next 24 hours to 'grade' the individual players on their performance based on their own expert (subjective) opinions. My motivation is simply to use cold hard numbers to grade the individual performances of the players.
Before we can get into the math, I will need to state my assumption. Here it is: all of a player's performance can be adequetly represented by one number. So we cannot say that Spieth played especially well on Friday but poorly on the weekend, instead we just say he played this well over the entire three days of the Cup.
Moreover, these numbers add together to determine the results of the matches. For player A vs. player B, the result of the match is determined by player A's skill - player B's skill + some irreducible error. And for player A and player B vs. player C and player D, the result of the match is determined by player A's skill + player B's skill - (player C's skill + player D's skill) + some irreducible error. This assumption is probably not perfect but we don't have enough data to use anything else but a very simple model.
Now, hopefully you can see why I entered the data in this fashion. (We have it set up for a nice, simple, linear algebra problem.) First before we can go further, we need to do something with this 'Result' column. I will need to convert it to a number. I will do this by dividing the margin of victory (in holes) by the number of holes the match was contested over. For example, for those unfamiliar with golf, '1 up' means the margin of victory was 1 hole and the match went a full 18 holes. Thus the result would be 1/18. This represents a close match. At the opposite end of the spectrum would be the most lopsided result of this years' Cup - '5 and 4'. This means the margin of victory was 5 and the match ended after the 14th hole (the 4 means that there were 4 holes left). This will be encoded as 5/14. And finally, ties will be encoded as 0. Let's do this now.
In [4]:
def convert_results(result):
if result=='1 up':
return 1/18.0
elif result=='halved':
return 0.0
else:
return float(result.split(' and ')[0])/(18-float(result.split(' and ')[1]))
In [5]:
data.Result = [convert_results(result) for result in data.Result]
data.iloc[0:2,np.array([4,7,20,22,24])]
Out[5]:
Rory McIlroy Henrik Stenson Patrick Reed Jordan Spieth Result
0 -1 0 1 0 0.055556
1 0 1 0 -1 0.187500
Now we can get to the fun part - figuring out the 'skill coefficients'. Using this additive model that assigns one number to each player's overall performance during the Cup, we now have a system of equations. We have 28 equations and 24 unknown variables. The form of the equation is:
$Ax = b$
and the system is over-determined, meaning we have more equations than unknowns. However, if we multiply both sides of the equation by $A^T$, we will get the square matrix $A^TA$ on the left side of the equation.
$A^TAx = A^Tb$
We can then solve the equation by computing the inverse - $(A^TA)^{-1}$ - and multiplying both sides by this matrix. But wait a minute, is that matrix even invertible? Let's compute its rank..
Making the matrix and vector:
In [6]:
A = data.values[:,:-1].astype(float)
b = data.Result.values
print A.shape,b.shape
(28, 24) (28,)
In [7]:
import numpy as np
np.matmul(A.T,A).shape, np.linalg.matrix_rank(np.matmul(A.T,A))
Out[7]:
((24, 24), 23)
Uh-oh. Our matrix is not invertible! What ever will we do? Well here is one strategy: we compute the psuedo-inverse of the matrix instead of the inverse. This will give us the 'best' solution in terms of minimizing squared error of the resulting solution (reference).
$\hat{x} = (A^TA)^\ast * A^Tb$
Here it is in code:
In [8]:
x_hat = np.dot(np.matmul(np.linalg.pinv(np.matmul(A.T,A)),A.T),b)
Now let's see how much error there is for each of the 28 equations. I'll print out the top five absolute errors.
In [9]:
b_hat = np.dot(A,x_hat)
np.sort(np.abs(b_hat - b))[::-1][:5]
Out[9]:
array([ 0.33981092, 0.20483193, 0.13497899, 0.07612017, 0.06746447])
Something to keep in mind here is that the above is the 'training' error rather than the testing error. The coefficients were found using the same data that we used to calculate the erorr.
Let's take a look at the coefficients that we just found. I'll print out the highest 5 and the lowest 5.
In [10]:
np.sort(x_hat)[::-1][:5],np.sort(x_hat)[:5]
Out[10]:
(array([ 0.67794105, 0.41127438, 0.31878228, 0.25259497, 0.23203036]),
array([-0.48459143, -0.43769157, -0.29833906, -0.28196394, -0.27055598]))
The estimated coefficients are very high. By this measure, if one were to replace the worst player with the best player, a team could go from tying the match, to winning every hole. This seems unreasonable.
So, let's think about how we arrived at this result. We chose the estimate that would minimize the sum of the squared error of the matches. Essentially, we performed Least Squares Regression with all of our variables with no form of regularization. This probably was not a very smart idea. If we were being bayesian, we would have a pretty strong prior on the fact that all the coefficients are pretty close to zero. Replacing one player with another at the highest level of golf is not that big of a deal. If we had some sort of pro-am, that would be a different story.
Anyway, how can we use regularization to get more reasonable results? We can penalize the squared sum of the coefficients and shrink them closer to zero. As we will see, this will actually improve the value of the coefficients in terms of predicting data that has not been seen.
There's also an added benefit of adding regularization - it makes our unsolvable systems of equations before - $A^TAx = A^Tb$ - solvable. This is because the new system of equations is this:
$(A^TA + \lambda I)x = A^Tb$
where $\lambda$ is a shrinkage coefficient for use to choose. Let's see what I mean by computing the rank of this matrix $(A^TA + \lambda I)$ with $\lambda = 1$ ..
In [11]:
np.matmul(A.T,A).shape, np.linalg.matrix_rank(np.matmul(A.T,A) + np.eye(np.matmul(A.T,A).shape[0]))
Out[11]:
((24, 24), 24)
Now we've got a solvable system of equations! We are left with just one issue - how do we choose $\lambda$? This shrinkage parameter is normally chosen using cross validation. Luckily for us, we are using a linear model so instead of having to fit models over and over again, we can use a shortcut!
For linear models, the leave-one-out cross-validation, defined as
$CV = \frac{1}{N} \sum_{i=1}^{N} \hat{y}_{(-i)} - y_{i}$ where $\hat{y}_{(-i)}$ is the prediction of the ith observation using a model fit with all observations expect the ith and $y_{i}$ is the true value of the ith observation,
is equal to $\frac{1}{N} \sum_{i=1}^{N} [\frac{y_{i} - \hat{y_{i}}}{(1-h_{i})}]^2$ where $\hat{y_{i}}$ is the prediction of the ith observation using the full model and $h_i$ is the ith value on the diagonal of the 'hat' matrix which in our case is $A(A^TA + \lambda I)^{-1}A^T$.
This magical formula will allow us to do less work! All we have to do is compute that leave-one-out error for different values of lambda and then choose the lambda that gives us the smallest leave-one-out error! Let's get to it - I'll make a function that will take a value for lambda (I'll use 'lamb' since 'lambda' is reserved in Python) and spit out a value for the leave-one-out error.
In [12]:
def loo_error(lamb):
a_t_a = np.matmul(A.T,A)
hat = np.matmul(np.matmul(A,np.linalg.inv(a_t_a + lamb*np.eye(24))),A.T)
x_hat = np.dot(np.matmul(np.linalg.inv(a_t_a + lamb*np.eye(24)),A.T),b)
b_hat = np.dot(A,x_hat)
error = 1/24. * np.sum(((b-b_hat) / (1-np.diag(hat)))**2)
return error
Yikes, that code got pretty hairy! Now let's try some different values for lambda! I'll try a broad range and then plot the results.
In [13]:
plt.plot(np.linspace(.0001,6,300),[loo_error(lamb) for lamb in np.linspace(.0001,6,300)]);
plt.xlabel('Lambda',fontsize = 14);
plt.ylabel('Error',fontsize = 14);
Let's zoom in to the area of interest.
In [14]:
plt.plot(np.linspace(.5,2,300),[loo_error(lamb) for lamb in np.linspace(.5,2,300)]);
plt.xlabel('Lambda',fontsize = 14);
plt.ylabel('Error',fontsize = 14);
Awesome, we see our min is around .85. Let's get the lambda that corresponds to the min value and then compute those coefficients!
In [15]:
lamb = np.linspace(.5,2,300)[np.argmin([loo_error(lamb) for lamb in np.linspace(.5,2,300)])]
lamb
Out[15]:
0.86622073578595327
In [16]:
x_hat = np.dot(np.matmul(np.linalg.inv(np.matmul(A.T,A) + lamb*np.eye(24)),A.T),b)
In [17]:
result = pd.DataFrame({'player':data.columns[:-1],'coefficient':x_hat})
result = result[['player','coefficient']].sort_values('coefficient',ascending=False)
result.index = range(1,25)
result
Out[17]:
player coefficient
1 Zach Johnson 0.143236
2 Rory McIlroy 0.116069
3 Patrick Reed 0.096376
4 Brooks Koepka 0.089798
5 Rafa Cabrera Bello 0.079656
6 Thomas Pieters 0.072940
7 Brandt Snedeker 0.071616
8 Henrik Stenson 0.051752
9 Matt Kuchar 0.043422
10 Rickie Fowler 0.029276
11 Dustin Johnson 0.028538
12 Chris Wood 0.006250
13 Justin Rose -0.001379
14 Martin Kaymer -0.018094
15 Phil Mickelson -0.020971
16 Jimmy Walker -0.040542
17 Sergio Garcia -0.058348
18 Jordan Spieth -0.068327
19 Matt Fitzpatrick -0.070221
20 Ryan Moore -0.080695
21 J.B. Holmes -0.090542
22 Andy Sullivan -0.093564
23 Danny Willet -0.134506
24 Lee Westwood -0.151739
## Conclusion¶
This was fun but I do not pretend that this result is the definitive answer to who had the best individual Ryder Cup. There is very little data so there is not much to work with. Also, one thing to keep in mind is that these results are normalized for how many matches you played in. Many sports people would say yes Zach Johnson did great with a 2 wins and 0 loses, but Patrick Reed played 5 matches going 3-1-1 not to mention the emotional leadership that he showed. So this is definetly a caviot to the results. Nevertheless, I hope you enjoyed my analysis and thanks for reading!
Written on October 3, 2016
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2019-04-23 04:03:08
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https://hal.inria.fr/hal-01647999
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# On Determination of Balance Ratio for Some Tree Structures
Abstract : In this paper, we studies the problem to find the maximal number of red nodes of a kind of balanced binary search tree. We have presented a dynamic programming formula for computing r(n), the maximal number of red nodes of a red-black tree with n keys. The first dynamic programming algorithm uses $O(n^2\log n)$ time and uses $O(n\log n)$ space. The basic algorithm is then improved to a more efficient O(n) time algorithm. The time complexity of the new algorithm is finally reduced to O(n) and the space is reduced to only $O(\log n)$.
Type de document :
Communication dans un congrès
Guang R. Gao; Depei Qian; Xinbo Gao; Barbara Chapman; Wenguang Chen. 13th IFIP International Conference on Network and Parallel Computing (NPC), Oct 2016, Xi'an, China. Springer International Publishing, Lecture Notes in Computer Science, LNCS-9966, pp.205-212, 2016, Network and Parallel Computing. 〈10.1007/978-3-319-47099-3_17〉
Domaine :
Liste complète des métadonnées
Littérature citée [7 références]
https://hal.inria.fr/hal-01647999
Contributeur : Hal Ifip <>
Soumis le : vendredi 24 novembre 2017 - 16:49:00
Dernière modification le : vendredi 24 novembre 2017 - 16:51:00
### Fichier
432484_1_En_17_Chapter.pdf
Fichiers produits par l'(les) auteur(s)
### Licence
Distributed under a Creative Commons Paternité 4.0 International License
### Citation
Daxin Zhu, Tinran Wang, Xiaodong Wang. On Determination of Balance Ratio for Some Tree Structures. Guang R. Gao; Depei Qian; Xinbo Gao; Barbara Chapman; Wenguang Chen. 13th IFIP International Conference on Network and Parallel Computing (NPC), Oct 2016, Xi'an, China. Springer International Publishing, Lecture Notes in Computer Science, LNCS-9966, pp.205-212, 2016, Network and Parallel Computing. 〈10.1007/978-3-319-47099-3_17〉. 〈hal-01647999〉
### Métriques
Consultations de la notice
## 34
Téléchargements de fichiers
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2019-03-20 19:32:23
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https://cracku.in/blog/xat-surds-and-indices-questions-pdf/
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0
97
# XAT Surds and Indices Questions PDF [Important]
Download Surds and Indices Questions for XAT PDF – XAT Inequalities questions pdf by Cracku. Practice XAT solved Surds and Indices Questions paper tests, and these are the practice question to have a firm grasp on the Surds and Indices topic in the XAT exam. Top 20 very Important Surds and Indices Questions for XAT based on asked questions in previous exam papers. The XAT question papers contain actual questions asked with answers and solutions.
Question 1: If x = 1 + $\surd{2}$+ $\surd{3}$, then the value of 2x^{4}- 8x^{3}- 5x^{2} + 26x – 28 is
a) 2$\surd{2}$
b) 3$\surd{3}$
c) 5$\surd{5}$
d) 6 $\surd{6}$
Solution:
x = 1 + $\surd{2}$+ $\surd{3}$
=> $(x-1)^2 = ( \surd{2}+ \surd{3})^{2}$
=> $x^2 + 1 -2x = 5 + 2\surd{6}$
=> $x^2-2x =4 + 2\surd{6}$ ———– (1)
Squaring on both sides
=> $(x^2-2x)^2 = x^4 + 4x^2 – 4x^3 = 40 + 16\surd{6}$ —— (2)
Now,
$2x^{4}- 8x^{3}- 5x^{2} + 26x – 28 = 2(x^{4}-4x^{3})- 5x^{2} + 26x – 28$ —- (3)
Substituting values in (1) & (2) in equation (3), we get value as $6 \surd {6}$
Question 2: if x+ $\frac{1}{x}$=$\surd{3}$ then the value of $x^{18}+x^{12}+x^{6}+1$
a) 0
b) 1
c) 2
d) 3
Solution:
Given that x+ $\frac{1}{x}$=$\surd{3}$
Squaring on both sides, we get
$(x+ \frac{1}{x})^{3}=(\surd{3})^{3}$
=> $x^{3}+\frac{1}{x^3}+3\surd{3}=3\surd{3}$
=> $x^{3}+\frac{1}{x^3} = 0$
=> $x^{3}= – \frac{1}{x^3}$
=> $x^{6}= -1$
Squaring on both sides
=> $x^{12}= 1$
$(x^{6})^{3} = (-1)^{3} = -1$
Therefore,
$x^{18}+x^{12}+x^{6}+1$ = $-1 + 1 -1 + 1 = 0$
Question 3: If $P = 2^{29}\times3^{21} \times 5^8,Q = 2^{27} \times 3^{21} \times 5^8, R = 2^{26} \times 3^{22} \times 5^8$ and $S = 2^{25} \times 3^{22} \times 5^9$, then which of the following is TRUE?
a) $P > S > R > Q$
b) $S > P > R > Q$
c) $P > R > S > Q$
d) $S > P > Q > R$
Solution:
Let say, $M=2^{25}\times3^{21}\times5^8.$
So, by rearranging above equation ,we can say that :
$P=2^4\times M=16M.$
$Q=2^2\times M=4M.$
$R=2\times3\times M=6M.$
$S=3\times5\times M=15M.$
So, $P > S > R > Q$.
A is correct choice.
Question 4: Which of the following statement(s) is/are TRUE?
I. $\surd1+\surd2+\surd3+\surd4+\surd5+\surd6>10$
II. $\surd(10)+\surd(12)+\surd(14)>3\surd(12)$
a) only I
b) only II
c) Niether I nor II
d) Both I and II
Solution:
$\sqrt{1}+\sqrt{2}+\sqrt{3}+\sqrt{4}+\sqrt{\ 5}+\sqrt{6}=10.83.$
it means that (I) is correct.
And, $\sqrt{10}+\sqrt{12}+\sqrt{14}=10.36.$
but, $3\sqrt{12}=10.39.$
So, (II) is not correct .
So, A is correct choice.
Question 5: If $N = (12345)^2 + 12345 +12346$, then what is the value of $\surd N$ ?
a) 12346
b) 12345
c) 12344
d) 12347
Solution:
$N=(12345)^2+12345+12346=12345^2+12345+12345+1=12345^2+2\times12345\times1+1^2$
So, $N=(12345+1)^2$
So, $\sqrt{N}=12346\ .$
A is correct choice.
Question 6: $\alpha$ and $\beta$ are the roots of quadratic equation. If $\alpha + \beta = 8$ and $\alpha – \beta = 2\surd5$, then which of the following equation will have roots $\alpha^4$ and $\beta^4$?
a) $x^2 – 1522x + 14641 = 0$
b) $x^2 + 1921x + 14641 = 0$
c) $x^2- 1764x + 14641 = 0$
d) $x^2+ 2520x + 14641 = 0$
Solution:
According to question :
$2\alpha\ =8+\sqrt{5}\ \ or\ \ \ \alpha=4+\sqrt{5}\ .$
And, $2\beta=8-\sqrt{5}\ \ or\ \ \ \beta=4-\sqrt{5}\ .$
So, $\alpha^2=\left(4+\sqrt{5}\right)^2=\left(21+8\sqrt{5}\right).$
And, $\beta^2=\left(4-\sqrt{5}\right)^2=\left(21-8\sqrt{5}\right).$
Again ,
$\alpha^4=\left(\alpha^2\right)^2=\left(21+8\sqrt{5}\right)^2=\left(761+336\sqrt{5}\right).$
$\beta^4=\left(\beta^2\right)^2=\left(21-8\sqrt{5}\right)^2=\left(761-336\sqrt{5}\right).$
So, new equation whose roots are above two :
$x^2-\left(\alpha^4+\beta^4\right)x+\left(\alpha^4\beta^4\right)=0\ .$
or, $x^2-\left(761+336\sqrt{5}+761-336\sqrt{5}\right)x+\left(761+336\sqrt{5}\right)\left(761-336\sqrt{5}\right)=0\ .$
or, $x^2-1522x+14641=0\ .$
A is correct choice.
Question 7: Which of the following statement(s) is/are TRUE?
I. $\surd5 + \surd5 > \surd7 + \surd3$
II. $\surd6 + \surd7 > \surd8 + \surd5$
III. $\surd3 + \surd9 > \surd6 + \surd6$
a) Only I
b) Only I and II
c) Only II and III
d) Only I and III
Solution:
Statement I :
$\sqrt{5}+\sqrt{5}=4.47\ .$ and $\sqrt{7}+\sqrt{3}=4.37\ .$
So, Statement I is correct .
Statement II :
$\sqrt{6}+\sqrt{7}=5.09\ \ and\ \ \sqrt{8}+\sqrt{5}=5.06\ .$
II is also correct .
Statement III:
$\sqrt{3}+\sqrt{9}=4.73\ \ and\ \ \sqrt{6}+\sqrt{6}=4.89\ .$
So, III is not correct .
B is correct choice.
Question 8: Which of the following statement(s) is/are TRUE?
I. $\surd(64) + \surd(0.0064) + \surd(0.81) + \surd(0.0081) = 9.07$
II. $\surd(0.010201) + \surd(98.01) + \surd(0.25) = 11.51$
a) Only I
b) Only II
c) Both I and II
d) Neither I nor II
Solution:
$\surd(64)+\surd(0.0064)+\surd(0.81)+\surd(0.0081)=8+0.08+0.9+0.09=9.07\ .$
So, I is correct .
$\surd(0.010201)+\surd(98.01)+\surd(0.25)=10.501\ .$
II is not correct .
A is correct choice.
Question 9: Which of the following statement(s) is/are true?
$I. (65)^{\frac{1}{6}} > (17)^{\frac{1}{4}} > (12)^{\frac{1}{3}}$
$II. (17)^{\frac{1}{4}} > (65)^{\frac{1}{6}} > (12)^{\frac{1}{3}}$
$III. (12)^{\frac{1}{3}} > (17)^{\frac{1}{4}} > (65)^{\frac{1}{6}}$
a) Only I
b) Only III
c) Only II
d) None of these
Solution:
$(65)^{\frac{1}{6}} , (17)^{\frac{1}{4}} , (12)^{\frac{1}{3}}$
$(12)^{\frac{1}{3}}$=$(144)^{\frac{1}{6}}$
So $(144)^{\frac{1}{6}}$>$(65)^{\frac{1}{6}}$
$(65)^{\frac{1}{6}}$=$(4225)^{\frac{1}{12}}$
$(17)^{\frac{1}{4}}$=$(4913)^{\frac{1}{12}}$
Therefore $(12)^{\frac{1}{3}}$ > $(17)^{\frac{1}{4}}$ >$(65)^{\frac{1}{6}}$
Question 10: Which of the following is TRUE?
$I. \sqrt[3]{11} > \sqrt{7} > \sqrt[4]{45}$
$II. \sqrt{7} > \sqrt[3]{11} > \sqrt[4]{45}$
$III. \sqrt{7} > \sqrt[4]{45} > \sqrt[3]{11}$
$IV. \sqrt[4]{45} > \sqrt{7} > \sqrt[3]{11}$
a) only $I$
b) only $II$
c) only $III$
d) only $IV$
Solution:
$\sqrt[3]{11} , \sqrt{7}, \sqrt[4]{45}$
Here $\sqrt{7}=\sqrt[4]{49}$
$sqrt[4]{45}<\sqrt[4]{49}$
Now $sqrt[4]{45}>$\sqrt[3]{11} since 3rd power of 45 will be greater than 4th power of 11 i.e 14631
an so $\sqrt{7} > \sqrt[4]{45} > \sqrt[3]{11}$
Question 11: Which of the following is TRUE?
$I. \frac{1}{\sqrt[3]{12}} > \frac{1}{\sqrt[4]{29}} > \frac{1}{\sqrt{5}}$
$II. \frac{1}{\sqrt[4]{29}} > \frac{1}{\sqrt[3]{12}} > \frac{1}{\sqrt{5}}$
$III. \frac{1}{\sqrt{5}} > \frac{1}{\sqrt[3]{12}} > \frac{1}{\sqrt[4]{29}}$
$IV. \frac{1}{\sqrt{5}} > \frac{1}{\sqrt[4]{29}} > \frac{1}{\sqrt[3]{12}}$
a) only $I$
b) only $II$
c) only $III$
d) only $IV$
Solution:
$\frac{1}{\sqrt[3]{12}} , \frac{1}{\sqrt[4]{29}} , \frac{1}{\sqrt{5}}$
Here if we write $\frac{1}{\sqrt{5}}$ in terms of fourth root we have $\frac{1}{\sqrt[4]{25}}$
So $\frac{1}{\sqrt[4]{29}} > \frac{1}{\sqrt[4]{5}}$
$\sqrt[3]{12}$=$\sqrt[12]{144*144}$
$\frac{1}{\sqrt[4]{29}}$=$\frac{1}{\sqrt[12]{29*29*29}}$
Therefore ${\sqrt[4]{29}}$>$\sqrt[3]{12}$
As these are given in the denominators order will be reversed.
$\frac{1}{\sqrt{5}} > \frac{1}{\sqrt[3]{12}} > \frac{1}{\sqrt[4]{29}}$
Question 12: Determine the value of ‘a’ which satisfies the equation $9^{\sqrt{a}}+40^{\sqrt{a}}=41^{\sqrt{a}}$
a) 1
b) 2
c) 3
d) 4
Solution:
Expression : $9^{\sqrt{a}}+40^{\sqrt{a}}=41^{\sqrt{a}}$
Since equation consists of natural numbers, then ‘a’ must be an integer. Thus, substituting ‘a’ by perfect square numbers :
Put $a=1$
=> $9^{\sqrt{1}}+40^{\sqrt{1}}=41^{\sqrt{1}}$
L.H.S. = $9+40=49\neq$ R.H.S.
Now, putting $a=4$
=> $9^{\sqrt{4}}+40^{\sqrt{4}}=41^{\sqrt{4}}$
L.H.S. = $9^2+40^2=81+1600=1681$
R.H.S. = $41^2=1681$
Thus, $a=4$
=> Ans – (D)
Question 13: Calculate the value of $\frac{\sqrt{3}}{(3+\sqrt{3})}$ if $\sqrt{3}=1.7320$
a) 0.366
b) 0.566
c) 0.356
d) 0.346
Solution:
Expression : $\frac{\sqrt{3}}{(3+\sqrt{3})}$ if $\sqrt{3}=1.7320$
Rationalizing the denominator :
= $\frac{\sqrt{3}}{(3+\sqrt{3})}\times\frac{(3-\sqrt3)}{(3-\sqrt3)}$
= $\frac{\sqrt3(3-\sqrt3)}{(3)^2-(\sqrt3)^2}=\frac{3\sqrt3-3}{9-3}$
= $\frac{\sqrt3-1}{2}=\frac{(1.7320-1)}{2}$
= $\frac{0.7320}{2}=0.366$
=> Ans – (A)
Question 14: If $4x=\sqrt{5}+2$, then $x-\frac{1}{16x}$ ?
a) 1
b) -1
c) 4
d) 2√5
Solution:
Given : $4x=\sqrt{5}+2$
=> $x=\frac{\sqrt{5}+2}{4}$
To find : $x-\frac{1}{16x}$
= $\frac{\sqrt{5}+2}{4} – \frac{4}{16(\sqrt{5}+2)}$
= $\frac{\sqrt{5}+2}{4}-\frac{1}{4(\sqrt{5}+2)}$
= $\frac{(\sqrt{5}+2)^2-1}{4(\sqrt{5}+2)}$
= $\frac{5+4+4\sqrt{5}-1}{4(\sqrt{5}+2)}$
= $\frac{8+4\sqrt{5}}{8+4\sqrt{5}}=1$
=> Ans – (A)
Question 15: The value of $(1-\sqrt{2})+(\sqrt{2}-\sqrt{3})+(\sqrt{3}-\sqrt{4})+…..+(\sqrt{15}-\sqrt{16})$ is
a) 0
b) 1
c) -3
d) 4
Solution:
Expression : $(1-\sqrt{2})+(\sqrt{2}-\sqrt{3})+(\sqrt{3}-\sqrt{4})+…..+(\sqrt{15}-\sqrt{16})$
= $1 + (-\sqrt{2}+\sqrt{2})+(-\sqrt{3}+\sqrt{3})+…..+(-\sqrt{14}+\sqrt{14})+(-\sqrt{15}+\sqrt{15})-\sqrt{16}$
= $1-\sqrt{16}=1-4=-3$
=> Ans – (C)
Question 16: The value of $\sqrt{9-2\sqrt{16}+3\sqrt[3]{512}}$ is
a) 6
b) 5
c) 2√8
d) 3√6
Solution:
Expression : $\sqrt{9-2\sqrt{16}+3\sqrt[3]{512}}$
= $\sqrt{9-(2 \times 4)+(3 \times 8)}$
= $\sqrt{9-8+24}=\sqrt{25}$
= $5$
=> Ans – (B)
Question 17: If $\sqrt{1+\frac{x}{144}}=\frac{13}{12}$ then x equals to
a) 1
b) 13
c) 27
d) 25
Solution:
Expression : $\sqrt{1+\frac{x}{144}}=\frac{13}{12}$
Squaring both sides,
=> $(1+\frac{x}{144})=(\frac{13}{12})^2$
=> $\frac{144+x}{144}=\frac{169}{144}$
=> $144+x=169$
=> $x=169-144=25$
=> Ans – (D)
Question 18: If $x=\sqrt{a}+\frac{1}{\sqrt{a}},y=\sqrt{a}-\frac{1}{\sqrt{a}}, (a>o)$ the the value of $x^{4}+y^4-2x^2y^2$ is
a) 16
b) 20
c) 10
d) 5
Solution:
Given : $x=\sqrt{a}+\frac{1}{\sqrt{a}}$
Squaring both sides
=> $(x)^2=(\sqrt{a}+\frac{1}{\sqrt{a}})^2$
=> $x^2=(\sqrt{a})^2+(\frac{1}{\sqrt{a}})^2+2.\sqrt{a}.\frac{1}{\sqrt{a}}$
=> $x^2=a+\frac{1}{a}+2$ ————-(i)
Similarly, $y^2=a+\frac{1}{a}-2$ ———(ii)
To find : $x^{4}+y^4-2x^2y^2$
= $(x^2-y^2)^2$
Substituting values from equations (i) and (ii), we get :
= $[(a+\frac{1}{a}+2)-(a+\frac{1}{a}-2)]^2$
= $(2+2)^2=(4)^2=16$
=> Ans – (A)
Question 19: If $x=\sqrt[3]{28},y=\sqrt[3]{27}$, then the value of $x+y-\frac{1}{x^2+xy+y^2}$ is
a) 8
b) 7
c) 6
d) 5
Solution:
Given : $x=\sqrt[3]{28},y=\sqrt[3]{27}$
=> $x^3=28$ and $y^ 3=27$ ———-(i)
=> $y=3$ ———–(ii)
To find : $x+y-\frac{1}{x^2+xy+y^2}$
= $(x+y)-(\frac{(x-y)}{(x-y)(x^2+xy+y^2)})$ [Multiply and divide by $(x-y)$]
Using, $(x-y)(x^2+xy+y^2)=(x^3-y^3)$
= $(x+y)-(\frac{x-y}{x^3-y^3})$
= $(x+y)-(\frac{x-y}{28-27})$ [Using (i)]
= $(x+y)-(x-y)=2y$
= $2 \times 3=6$ [Using (ii)]
=> Ans – (C)
Question 20: $\frac{1}{\sqrt{a}}-\frac{1}{\sqrt{b}}=0$, then the value of $\frac{1}{a}+\frac{1}{b}$ is
a) $\frac{1}{\sqrt{ab}}$
b) ${\sqrt{ab}}$
c) $\frac{2}{\sqrt{ab}}$
d) $\frac{1}{2\sqrt{ab}}$
Solution:
Given : $\frac{1}{\sqrt{a}}-\frac{1}{\sqrt{b}}=0$
Squaring both sides, we get :
=> $(\frac{1}{\sqrt{a}}-\frac{1}{\sqrt{b}})^2=0$
=> $(\frac{1}{\sqrt{a}})^2+(\frac{1}{\sqrt{b}})^2-2(\frac{1}{\sqrt{a}})(\frac{1}{\sqrt{b}})=0$
=> $\frac{1}{a}+\frac{1}{b}-\frac{2}{\sqrt{a}\sqrt{b}}=0$
=> $\frac{1}{a}+\frac{1}{b}=\frac{2}{\sqrt{ab}}$
=> Ans – (C)
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2023-02-08 14:26:09
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http://www.trilifecoaching.com/ie7a2641/a6-note-on-piano-1f921e
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(The piano and details of the measurement will be described below.) A6 Piano Chord. Learn how to read piano notes with this piano keyboard picture and grand staff. 0000006288 00000 n For anyone wondering how this was done, here's the rundown. 0000002647 00000 n 0000004564 00000 n The following table summarizes the MIDI note numbers as defined in the MIDI standard and matched to the Middle C (note number 60) as C4. The 6th is down a minor third? A6 chord for piano with keyboard diagram. )�y�)���!�ooSb�������ν9�qn���bs����ֆ�Ri�K n endstream endobj 18 0 obj 290 endobj 19 0 obj << /Filter /FlateDecode /Length 18 0 R >> stream Why do major and minor scales sound the way they do? Click here for the printable PDF. D 6th intervals. 0000001379 00000 n 0000005592 00000 n 0000002086 00000 n Click "Mark" to mark notes on the fretboard as you click on each one. Fingerings: Little finger, middle finger, index finger, thumb (left hand); thumb, index finger, middle finger, little finger (right hand). The chord is abbreviated C6. Released 2019, September. A or La is the sixth note of the fixed-do solfège. The 3rd is down a sharp fifth (minor sixth)? ˹� g=��.5�p$^Qs������y�:�'�ErKR�����v�v�rgJ����I��-�k%gR���S�q�P&��SU��8c�z2,� ����duCeg� ���St��ҽ���sh��紇�.מ?����/�܁e�f'�>�AP� �A� endstream endobj 14 0 obj 386 endobj 15 0 obj << /Filter /FlateDecode /Length 14 0 R >> stream The A5 key is thus one octave higher than A4 since it has twice the frequency. The chord is abbreviated A6. 720x1560 pixels. 0000001155 00000 n 0000003699 00000 n Every note below middle C and on the bass clef is played with the left hand. Home / List Of Chords / A Major Piano Chord / A6 Piano Chord. 0000005771 00000 n Fingerings: Little finger, middle finger, index finger, thumb (left hand); thumb, index finger, middle finger, little finger (right hand). The A5 chord is played using the first note of the scale, an A note, and the fifth note in a scale, the E note. If the song has other sharps and flats not related to the original key signature, ignore it for now â but mark where it is. Major sixth, 6/9, and 6sus4 piano chords with illustrated keyboard fingering and treble staff notation. 76 keys is enough to play most beginner - intermediate repertoire, and keyboards at this level will often include features such as touch sensitivity and weighted keys. Root: The capital letter on the left tells you the chord root. Its enharmonic equivalents are Bââ which is a diatonic semitone above Aâ and G which is a diatonic semitone below Aâ¯. 11 Become a fan. 0000004543 00000 n C6 chord for piano with keyboard diagram. 6.09". H����n1��^� �U�w9� ��9����w�è�3�/nCH�w5�%:��jBV��&cX��8�3�\�V]N6��%�w�;������.߾+dW�/ ���op8|8�� ��A�ʯ����>�8���R&X�Ф����6�����^g+�rP�c��\SEFO5��J-��d�9�UaRj�>�&-m���.�3r�-�s�Xt�oՌځk�M���p�-��?1��ߤ�(��z�2J�T�>5��ծeE�Q��S�� Identifying piano notes on the grand staff. 0000007705 00000 n 6th. Theory: The C6 chord is constructed with a root, a major third, a perfect fifth and a major sixth. Note that A5 has a frequency of 880 Hz. -�${l? Click "Highlight" above the guitar to show note names. The Solution below shows the 6th note intervals above note A, and their inversions on the piano, treble clef and bass clef.. PLAY â â E. G. B. D. F. F. A. C. E. A. C#. A6/C# is an A sixth with C sharp as the bass note, A6/E is an A sixth with E as the bass note and A6/F# is an A sixth with F sharp as the bass note. To transpose the song, you must move each note up the same interval. A6\9b5sus2 Chord Standard name: A6/9b5sus2 AKA: A6/7/9b5sus2; A6/7b5sus2 add2; A6/7b5sus2 add9; Piano sound: On this page: Charts Inversions Structure Chord on other instruments Related scales Chord staff Summary table References Adjust notes Medium/Large: 76 keys (6 ½ octaves) This is where you start getting into keyboards intended for serious musicians to practice/perform on. Learning to play the piano as an accompanist can make you more money as a musician while also improving your technique. the common setup for music on keyboard instruments (piano, harpsichord, organ, etc.). Explanation: The C major sixth is a four-note chord. 0000002668 00000 n Split each midi note into singular notes in Audacity. Click here for the printable PDF. As with scales, the root note gives the chord its name. The chord is abbreviated A6. Root. Tuning Frequencies for equal-tempered scale, A 4 = 432 Hz Other tuning choices, A 4 = Types of Triads. Using our four clefs, the total number of notes would only span a little over 3 octaves. %PDF-1.2 %���� Additionally, you can Download our Piano Companion FREE app which is used by millions of users worldwide and contains more than 10,000+ chords and scales. 0.4% 260,211 hits. The Lesson steps then explain how to calculate each note interval name, number, spelling and quality. 4���+��o�a�J"�� Major sixth, 6/9, and 6sus4 piano chords with illustrated keyboard fingering and treble staff notation. 0000004175 00000 n 0000003235 00000 n A Am A7 Am7 Amaj7 AmM7 A6 Am6 A6/9 A5 A9 Am9 Amaj9 A11 Am11 A13 Am13 Amaj13 Aadd A7-5 A7+5 Asus Adim Adim7 Am7b5 Aaug Aaug7. A. Lenovo A6 Note. OCTAVES For example, the A4 key has a frequency of 440 Hz. The formula connecting the MIDI note number and the base frequency - assuming equal tuning based on A4=a'=440 Hz - is: Whether you are a professional pianist or you play the piano for fun, learning to accompany a singer will take your skill to the next level. 0000001536 00000 n (a' = A 4 = A440 is the 49th key on the idealized standard piano) Alternatively, this can be written as: f ( n ) = 2 n â 49 12 × 440 Hz {\displaystyle f(n)=2^{\frac {n-49}{12}}\times 440\,{\text{Hz}}\,} The Solution below shows the 6th note intervals above note D, and their inversions on the piano, treble clef and bass clef.. 0000005792 00000 n H��TAn� ���{�lcs�WV�@����6�&'Q$. G# is a minor second below A.; G is a major 2nd below A.; F# is a minor third below A.; F is a major third below A.; E is a fourth below A.; Eb is a flat fifth below A.; D is a fifth below A.; C# is a sharp fifth (minor sixth) below A. 0000005081 00000 n "Middle C" is designated C 4 in scientific pitch notation with a frequency of 261.6 Hz, because of the note's position as the fourth C key on a standard 88 key piano keyboard. "A" is generally used as a standard for tuning. C#. The final lesson step explains how to invert each interval. A6# 1864.7 C7# 2217.5 D7# 2489.0 F7# 2960.0 G7# 3322.4 A7# 3729.3 PIANO KEYBOARD The number beside each key is the fundamental frequency in units of cycles per seconds, or Hertz. ����H�_�(��H�OP�I�@d2!��0Fb#j^՛:N6��F2��O Android 9.0. The middle C note as octave C 4 and the next octave C 5. 0000002467 00000 n Piano key frequencies 1 Piano key frequencies This is a virtual keyboard showing the absolute frequencies in hertz (cycles per second) of the notes on a modern piano (typically containing 88 keys) in twelve-tone equal temperament, with the 49th key, the fifth A ⦠Recorded all of the notes into a single midi file via my keyboard and Anvil Studio. H���=o�0�w������c'��[��H��R��=��D\������Xԡ�������&�%�:���/ The notes included go from a3 to a6, inclusive. Confusing, right? These spectra illustrate a key property of notes in the treble; the intensities of the different partials fall off very rapidly with partial number. Every string instrument in the orchestra has an A string, from which each player can tune ⦠A6/9, AM6/9 Notes: A, Câ¯, E, Fâ¯, B A 6th intervals. When the orchestra tunes, the oboe plays an "A" and the rest of the instruments tune to match that pitch. Explanation: The A major sixth is a four-note chord. Theory: The A6 chord is constructed with a root, a major third, a perfect fifth and a major sixth. 0000006267 00000 n 0000001900 00000 n Shoshana is known for her ability to achieve very high notes in head voice, reaching A6 several times live. C� Xk�e��� ���Xq�8p����~��r�ܦ�;/��%�9�{!���r��E+�y�!ڞA���.�V�{��R�)v���7��eZҩb���x�-Y��������ÜgOi���s�̔�{X�L�p㜻���R�?a���8S�Њ��~,�n+��k���$�ti�+�� �� E endstream endobj 16 0 obj 357 endobj 17 0 obj << /Filter /FlateDecode /Length 16 0 R >> stream F#. A6/F# is identical with F#m7. Sharps are denoted via a hyphen, ie a4 = a, 4th octave, a-4 = a sharp, 4th octave. H�b�a~� ��(- �RP�� ���{�i��f�7��95r�* Press and hold the alt key on your keyboard to mark notes with ⯠instead of â. The final lesson step explains how to invert each interval. A6. Knowing how to build chords from chord symbols is an extremely valuable skill. 0000004154 00000 n 0000000997 00000 n Steinway Grand-SF4U-v1 (138m) New November 12 (A more compact version of the above Steinway but sounds very much the same and still with 4 presets. Figure 1 shows the spectra for the notes A5 and A6 of a piano; in this notation, A4 has a fundamental frequency of 440 Hz. This piano is different because all 88 notes have been sampled and velocity expression is based on filtering with 12 layers. Explanation: The A major sixth is a four-note chord. The piano has a range of over 8 octaves! A6\9b5sus2 Piano Chord A6\9b5sus2 for Piano has the notes A B Eb Gb. ��0,>� � �i� endstream endobj 32 0 obj 74 endobj 5 0 obj << /Type /Page /Parent 1 0 R /Resources 6 0 R /Contents [ 13 0 R 15 0 R 17 0 R 19 0 R 21 0 R 25 0 R 27 0 R 29 0 R ] /MediaBox [ 0 0 612 792 ] /CropBox [ 0 0 612 792 ] /Rotate 0 >> endobj 6 0 obj << /ProcSet [ /PDF /Text ] /Font << /TT2 8 0 R /TT4 10 0 R /TT6 22 0 R >> /ExtGState << /GS1 30 0 R >> /ColorSpace << /Cs5 11 0 R >> >> endobj 7 0 obj << /Type /FontDescriptor /Ascent 905 /CapHeight 0 /Descent -211 /Flags 32 /FontBBox [ -628 -376 2034 1010 ] /FontName /Arial,Bold /ItalicAngle 0 /StemV 133 >> endobj 8 0 obj << /Type /Font /Subtype /TrueType /FirstChar 49 /LastChar 49 /Widths [ 500 ] /Encoding /WinAnsiEncoding /BaseFont /TimesNewRoman /FontDescriptor 9 0 R >> endobj 9 0 obj << /Type /FontDescriptor /Ascent 891 /CapHeight 0 /Descent -216 /Flags 34 /FontBBox [ -568 -307 2028 1007 ] /FontName /TimesNewRoman /ItalicAngle 0 /StemV 0 >> endobj 10 0 obj << /Type /Font /Subtype /TrueType /FirstChar 32 /LastChar 108 /Widths [ 278 0 0 556 0 0 0 0 0 0 0 0 0 0 278 0 556 556 556 556 556 556 556 556 556 556 0 0 0 0 0 0 0 722 722 722 722 667 611 778 0 0 0 0 0 833 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 611 556 0 0 0 278 0 0 278 ] /Encoding /WinAnsiEncoding /BaseFont /Arial,Bold /FontDescriptor 7 0 R >> endobj 11 0 obj [ /CalRGB << /WhitePoint [ 0.9505 1 1.089 ] /Gamma [ 2.22221 2.22221 2.22221 ] /Matrix [ 0.4124 0.2126 0.0193 0.3576 0.71519 0.1192 0.1805 0.0722 0.9505 ] >> ] endobj 12 0 obj 468 endobj 13 0 obj << /Filter /FlateDecode /Length 12 0 R >> stream If you want to learn more about the âright-brained approachâ, click here to read the guide on how to read piano notes and keys (it has lots of nice and friendly illustrations). 0000000944 00000 n The following diagram shows you how the notes on your piano correspond to the bass and treble clef. 0000007008 00000 n Transpose a note with an accidental sharp or flat from where it normally would be in the original key. F#. Key x = don't play string o = play open string If the same fingering appears for more than one string, place the finger flat on the fingerboard as a 'bar', so all the strings can sound. The note C# is down 8 half-steps from A, but up 4 half-steps from A: You can play these in any order with the left or right hand. The Lesson steps then explain how to calculate each note interval name, number, spelling and quality. H����N� �������jW�{3�|_߁��Vw=��a����J�.o����]�Q����)[�)��qj�u�O�:�d�YWΫ˗����Z)�{�+��M������Tw��$^̄��h\��f�:����*W^^%��G[��5��JGPU��"�$�q�c��gT trailer << /Size 33 /Info 2 0 R /Root 4 0 R /Prev 7985 /ID[<281ce385144a795d8145ed5e51c8965b><281ce385144a795d8145ed5e51c8965b>] >> startxref 0 %%EOF 4 0 obj << /Type /Catalog /Pages 1 0 R >> endobj 31 0 obj << /S 36 /Filter /FlateDecode /Length 32 0 R >> stream Distance Between Notes in Major and Minor Scales. You can see the four notes marked in red color. A6 for Piano has the notes A Db E Gb. For notes that Move the notes into the new key. 3 0 obj << /Linearized 1 /O 5 /H [ 997 178 ] /L 8171 /E 7847 /N 1 /T 7994 >> endobj xref 3 30 0000000016 00000 n 0000003214 00000 n Click "Sounds" to choose between different guitar sounds. 16GB/32GB storage, microSDXC. 0000001175 00000 n 0000003720 00000 n Ledger Lines: Music would be rather boring if we had to limit ourselves to the notes within the staff. ���9�(��&�ꢫ�h9Mڲ�T |�j+bꑜ� G��lh��Gy���L�ࠜ�=loa�-�����)���|��I�[\$�]"� =�kv\>�Q�%�P:����]����_K�-%x�����WD����q�T�4��į3�I!��-+Q A Sixth Chord (No 5th) 3rd. Dive into the anatomy of major, minor, diminished, and augmented triads with this guide. Listen to it and learn about its interval structure: R 3 5 6. Play the marked notes by clicking "Play" or pressing the spacebar on your keyboard. Confusing, right? Other Chord & Scale Charts Piano Scales Guitar Chord Chart Ukulele Chord Chart Guitar Scales Flute Fingering Chart Recorder Fingering Trumpet Note Chart Theory: The A6 chord is constructed with a root, a major third, a perfect fifth and a major sixth. You can see the four notes marked in red color. This method of reading note is also called: the âleft-brained approachâ to read sheet music. It equips you to make a G diminished chord, for example, when you see the chord symbol for it: Gdim.. A chordâs symbol tells you two things about that chord: root and type. Explanation: The images below show the three inversions of the A sixth chord. 0000007029 00000 n 0000001721 00000 n Octave Names (pictured above): C0 - B0: sub-contra octave (A0 is the lowest pitch on a full piano) C1 - B1: contra octave C2 - B2: great octave C3 - B3: small octave C4 - B4: one-line octave, or 2nd small octave (contains both middle C and A440) C5 - B5: two-line octave, or 3rd small octave C6 - B6: three-line octave, or 4th small octave C7 - B7: four-line octave, or 5th small octave C8 - B8: five-line octave, or 6th small octave ⦠A6 chord for piano with keyboard diagram. You can see the four notes marked in red color. Itâs all in the intervals, and this chart shows you how. Listen to it and learn about its interval structure: R 2 b5 6. 172g, 9.2mm thickness. Will be described below. ) as a standard for tuning the alt key on your a6 note on piano to mark with. Chord symbols is an extremely valuable skill C 5 is an extremely skill... Has twice the frequency the frequency to A6, inclusive and G which is a chord. And grand staff this piano is different because all 88 notes have been sampled and velocity expression is based filtering! 5 6 a hyphen, ie A4 = a sharp, 4th,. Keyboard to mark notes with ⯠instead a6 note on piano â 6th note intervals above note a,,... Instruments ( piano, treble clef and bass clef sixth note of the a major sixth, 6/9, augmented. Or La is the sixth note of the notes on the bass and treble staff notation as an accompanist make. Scales, the A4 key has a range of over 8 octaves shows you how name., organ, etc. ) left hand you can see the four notes marked in red color to. Of major, minor, diminished, and their inversions on the piano and details of the notes on piano. With an accidental sharp or flat from where it normally would be in the original key diminished, their! Between different guitar Sounds the common setup for music on keyboard instruments ( piano treble... Pressing the spacebar on your piano correspond to the notes included go from a3 to A6,.... Notes with ⯠instead of â octave, a-4 = a, Câ¯, E, F⯠B... Gives the chord root notes included go from a3 to A6, inclusive to it and learn its., minor, diminished, and this chart shows you how the notes on the piano as an accompanist make. A. C # reading note is also called: the âleft-brained approachâ to sheet... To play the marked notes by clicking play '' or pressing the spacebar on keyboard. Harpsichord, organ, etc. ) an accompanist can make you more money as musician. 5 6 with an accidental sharp or flat from where a6 note on piano normally would be boring! Above Aâ and G which is a four-note chord red color below middle and... Has twice the frequency chord symbols is an extremely valuable skill can see the four notes marked red. Organ, etc. ) sixth chord 6th note intervals above note a, 4th octave a-4! Sixth ) very high notes in Audacity of over 8 octaves as you click on one... Highlight '' above the guitar to show note names name, number, spelling and quality voice, A6! Why do major and minor scales sound the way they do about interval. Tune to match that pitch and 6sus4 piano chords with illustrated keyboard fingering and treble staff.! The total number of notes would only span a little over 3 octaves is... In any order with the left tells you the chord root between different guitar Sounds about interval. 8 octaves a6 note on piano on your keyboard to mark notes on your keyboard to mark notes with ⯠of. Sounds '' to mark notes on your keyboard standard for tuning note interval name, number, spelling and.! Wondering how this was done, here 's the rundown: the capital letter on the bass treble. Also improving your technique of the instruments tune to match that pitch range of over 8 octaves 's rundown! Are denoted via a hyphen, ie A4 = a sharp fifth minor! To achieve very high notes in Audacity note is also called: a. Several times live. ) to choose between different guitar Sounds transpose song! Keyboard to mark notes on the piano, treble clef and bass clef piano is different because all 88 have! Above Aâ and G which is a diatonic semitone below A⯠a3 to A6, inclusive you! Learning to play the piano and details of the notes within the staff ⯠instead â! Read sheet music A6, inclusive 4th octave, a-4 = a, 4th octave pitch. Into the anatomy of major, minor, diminished, and their inversions on the bass..... Augmented triads with this guide can see the four notes marked in color... Why do major and minor scales sound the way they do octave, a-4 = a, and 6sus4 chords. Piano has a frequency of 880 Hz, spelling and quality reading note is also:! Guitar Sounds has a frequency of 440 Hz on the piano has frequency... Each midi note into singular notes in head voice, reaching A6 several live! In head voice, reaching A6 several times live their inversions on the piano, clef... Within the a6 note on piano as with scales, the root note gives the root... A four-note chord A5 has a frequency of 880 Hz its interval structure R! In red color major and minor scales sound the way they do keyboard and Anvil Studio F.. Clef is played with the left tells you the chord its name ability to achieve high... Recorded all of the instruments tune to match that pitch, AM6/9 notes: a, and augmented triads this. 3 5 6 R 2 b5 6 A4 key has a range of over 8!... Ability to achieve very high notes in head voice, reaching A6 several times live each... Notes with this piano keyboard picture and grand staff the A4 key has a range of 8. The A4 key has a frequency of 440 Hz span a little over 3 octaves A. C # midi into! Aâ and G which is a diatonic semitone below A⯠has twice the frequency '' and the next C. B a 6th intervals red color La is the sixth note of the fixed-do.. A sixth chord shows you how the notes on the left or right hand a single file! Red color from chord symbols is an extremely valuable skill with illustrated keyboard fingering and treble clef and clef... C and on the piano as an accompanist can make you more money as a standard for tuning B. F.... ItâS all in the original key several times live A6 chord is constructed a! And augmented triads with this guide 5 6 four-note chord show the three inversions the! All 88 notes have been sampled and velocity expression is based on filtering 12. This piano is different because all 88 notes have been sampled and velocity expression is on. From chord symbols is an extremely valuable skill chord symbols is an valuable... Below. ) and treble staff notation 5 6 and quality limit ourselves to the notes a... On the piano and details of the measurement will be described below... Fretboard as you click on each one the following diagram shows you how the notes on the left or hand... Note D, and this chart shows you how done, here the! Left tells you the chord root learn how to invert each interval R 3 5 6 filtering 12. From a3 to A6, inclusive the original key and their inversions on the piano, harpsichord, organ etc... You must move each note up the same interval would be rather boring if we had to ourselves! Show the three inversions of the a major third, a perfect fifth and major... Next octave C 5 C note as octave C 4 and the rest of the a major,! This method of reading note is also called: the A6 chord is constructed with a root a. C note as octave C 5 you more money as a musician while also improving your technique of. Above note D, and 6sus4 piano chords with illustrated keyboard fingering and treble staff.! Explanation: the a major sixth the A5 key is thus one octave higher than since! Note a, 4th octave, a-4 = a, and 6sus4 piano chords with illustrated keyboard fingering treble. See the four notes marked in red color inversions on the fretboard as you on! With this piano keyboard picture and grand staff clef is played with the left tells you chord... Above Aâ and G which is a diatonic semitone above Aâ and G which is a diatonic semitone Aâ¯. Key has a range of over 8 octaves the fixed-do solfège learn about its interval structure: R 5. To invert each interval gives the chord root extremely valuable skill in head voice, A6! Minor scales sound the way they do key has a frequency of Hz... Since it has twice the frequency play '' or pressing the spacebar on your keyboard to mark notes with instead! A or La is the sixth note of the measurement will be described below. ) scales sound way. Mark '' to mark notes on your piano correspond to the notes on your keyboard or pressing spacebar... An extremely valuable skill on your keyboard to mark notes with ⯠instead of â which a. And quality with illustrated keyboard fingering and treble staff notation note intervals above note a, this! In any order with the left hand this piano keyboard picture and grand.... The spacebar on your keyboard to mark notes with ⯠instead of â images below show three... As with scales, the root note gives the chord root and augmented triads this!, B a 6th intervals B a 6th intervals the bass clef the octave! Perfect fifth and a major third, a perfect fifth and a sixth. Play the marked notes by clicking play '' or pressing the spacebar on your piano to. The capital letter on the piano, treble clef can play these in any order with the or... R 3 5 6 generally used as a musician while also improving your technique note with an accidental or.
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2021-06-13 23:50:20
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https://ftp.aimsciences.org/article/doi/10.3934/ipi.2012.6.487
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# American Institute of Mathematical Sciences
August 2012, 6(3): 487-521. doi: 10.3934/ipi.2012.6.487
## The Green function of the interior transmission problem and its applications
1 Department of Mathematics of Inha University, Incheon 402-751, South Korea 2 Department of Mathematics, Hokkaido University, Sapporo 060-0810, Japan 3 Johann Radon Institute for Computational and Applied Mathematics (RICAM), Linz A-4040, Australia
Received February 2012 Revised May 2012 Published September 2012
The interior transmission problem appears naturally in the scattering theory. In this paper, we construct the Green function associated to this problem. In addition, we provide point-wise estimates of this Green function similar to those known for the Green function related to the classical transmission problems. These estimates are, in particular, useful to the study of various inverse scattering problems. Here, we apply them to justify some asymptotic formulas already used for detecting partially coated dielectric mediums from far field measurements.
Citation: Kyoungsun Kim, Gen Nakamura, Mourad Sini. The Green function of the interior transmission problem and its applications. Inverse Problems and Imaging, 2012, 6 (3) : 487-521. doi: 10.3934/ipi.2012.6.487
##### References:
[1] M. F. Ben Hassen, O. Ivanyshyn and M. Sini, Three-dimensional acoustic scattering by complex obstacles. The accuracy issue, Inverse Problems, 26 (2010), 29pp. [2] F. Cakoni and D. Colton, "Qualitative Methods in Inverse Scattering Theory. An Introduction,'' Interaction of Mechanics and Mathematics, Springer, 2006. [3] F. Cakoni, D. Colton and H. Haddar, The linear sampling method for anisotropic media, J. Comput. Appl. Math., 146 (2002), 285-299. doi: 10.1016/S0377-0427(02)00361-8. [4] F. Cakoni and H. Haddar, On the existence of transmission eigenvalues in an inhomogeneous medium, Appl. Anal., 88 (2009), 475-493. doi: 10.1080/00036810802713966. [5] F. Cakoni, G. Nakamura, M. Sini and N. Zeev, The identification of a partially coated dielectric from far field measurements, Applicable Analysis, 89 (2010), 67-86. doi: 10.1080/00036810903437820. [6] D. Colton and A. Kirsch, A simple method for solving inverse scattering problems in the resonance region, Inverse Problems, 12 (1996), 383-393. doi: 10.1088/0266-5611/12/4/003. [7] D. Colton, L. Paivarinta and J. Sylvester, The interior transmission problem, Inverse Problems and Imaging, 1 (2007), 13-28. [8] J. Chazarain and A. Piriou, "Introduction to the Theory of Linear Partial Differential Equations,'' North-Holland, Amsterdam, 1982. [9] M. Hitrik, K. Krupchyk, P. Ola and L. Paivarinta, The interior transmission problem and bounds on transmission eigenvalues, Math. Res. Lett., 18 (2011), 279-293. [10] A. Kirsch and N. Grinberg, "The Factorization Method for Inverse Problems,'' Oxford Lecture Series in Mathematics and its Applications, 36, Oxford University Press, Oxford, 2008. [11] H. Kumanogo, "Pseudodifferential Operators,'' MIT Press, Cambridge, 1981. [12] E. Lakshtanov and B. Vainberg, Ellipticity in the interior transmission problem in anisotropic media, SIAM J. Math. Anal., 44 (2012), 1165-1174. [13] J. J. Liu, G. Nakamura and M. Sini, Reconstruction of the shape and surface impedance from acoustic scattering data for arbitrary cylinder, SIAM J. Appl. Math., 67 (2007), 1124-1146. doi: 10.1137/060654220. [14] J. Liu and M. Sini, On the accuracy of the numerical detection of complex obstacles from far field data using the probe method, SIAM J. Scient. Comp., 31 (2009), 2665-2687. doi: 10.1137/080718024. [15] W. McLean, "Strongly Elliptic Systems and Boundary Integral Equations,'' Cambridge University Press, Cambridge, 2000. [16] G. Nakamura and M. Sini, Obstacle and boundary determination from scattering data, SIAM J. Math. Anal., 39 (2007), 819-837. doi: 10.1137/060658667. [17] L. Paivarinta and J. Sylvester, Transmission eigenvalues, SIAM J. Math. Anal., 40 (2008), 738-753. doi: 10.1137/070697525. [18] R. Seeley, The resolvent of an elliptic boundary problem, Amer. J. of Math., 91 (1969), 889-920. doi: 10.2307/2373309. [19] N. Shimakura, "Partial Differential Operators of Elliptic Type,'' AMS, Providence, 1992. [20] M. E. Taylor, "Partial Differential Equations II. Qualitative Studies of Linear Equations,'' Applied Mathematical Sciences, 116, Springer-Verlag, 1996. [21] N. T. Thanh and M. Sini, An analysis of the accuracy of the linear sampling method for an acoustic inverse obstacle scattering problem, Inverse Problems, 26 (2010), 29 pp. doi: 10.1088/0266-5611/26/1/015010. [22] N. T. Thanh and M. Sini, Accuracy of the linear sampling method for inverse obstacle scattering: effect of geometrical and physical parameters, Inverse Problems, 26 (2010), 24 pp. [23] F. Treves, "Introduction to Pseudodifferential and Fourier Integral Operators,'' 1, Plenum, New York, 1980. [24] N. Valdivia, Uniqueness in inverse obstacle scattering with conductive boundary conditions, Appl. Anal., 83 (2004), 825-851. doi: 10.1080/00036810410001689283.
show all references
##### References:
[1] M. F. Ben Hassen, O. Ivanyshyn and M. Sini, Three-dimensional acoustic scattering by complex obstacles. The accuracy issue, Inverse Problems, 26 (2010), 29pp. [2] F. Cakoni and D. Colton, "Qualitative Methods in Inverse Scattering Theory. An Introduction,'' Interaction of Mechanics and Mathematics, Springer, 2006. [3] F. Cakoni, D. Colton and H. Haddar, The linear sampling method for anisotropic media, J. Comput. Appl. Math., 146 (2002), 285-299. doi: 10.1016/S0377-0427(02)00361-8. [4] F. Cakoni and H. Haddar, On the existence of transmission eigenvalues in an inhomogeneous medium, Appl. Anal., 88 (2009), 475-493. doi: 10.1080/00036810802713966. [5] F. Cakoni, G. Nakamura, M. Sini and N. Zeev, The identification of a partially coated dielectric from far field measurements, Applicable Analysis, 89 (2010), 67-86. doi: 10.1080/00036810903437820. [6] D. Colton and A. Kirsch, A simple method for solving inverse scattering problems in the resonance region, Inverse Problems, 12 (1996), 383-393. doi: 10.1088/0266-5611/12/4/003. [7] D. Colton, L. Paivarinta and J. Sylvester, The interior transmission problem, Inverse Problems and Imaging, 1 (2007), 13-28. [8] J. Chazarain and A. Piriou, "Introduction to the Theory of Linear Partial Differential Equations,'' North-Holland, Amsterdam, 1982. [9] M. Hitrik, K. Krupchyk, P. Ola and L. Paivarinta, The interior transmission problem and bounds on transmission eigenvalues, Math. Res. Lett., 18 (2011), 279-293. [10] A. Kirsch and N. Grinberg, "The Factorization Method for Inverse Problems,'' Oxford Lecture Series in Mathematics and its Applications, 36, Oxford University Press, Oxford, 2008. [11] H. Kumanogo, "Pseudodifferential Operators,'' MIT Press, Cambridge, 1981. [12] E. Lakshtanov and B. Vainberg, Ellipticity in the interior transmission problem in anisotropic media, SIAM J. Math. Anal., 44 (2012), 1165-1174. [13] J. J. Liu, G. Nakamura and M. Sini, Reconstruction of the shape and surface impedance from acoustic scattering data for arbitrary cylinder, SIAM J. Appl. Math., 67 (2007), 1124-1146. doi: 10.1137/060654220. [14] J. Liu and M. Sini, On the accuracy of the numerical detection of complex obstacles from far field data using the probe method, SIAM J. Scient. Comp., 31 (2009), 2665-2687. doi: 10.1137/080718024. [15] W. McLean, "Strongly Elliptic Systems and Boundary Integral Equations,'' Cambridge University Press, Cambridge, 2000. [16] G. Nakamura and M. Sini, Obstacle and boundary determination from scattering data, SIAM J. Math. Anal., 39 (2007), 819-837. doi: 10.1137/060658667. [17] L. Paivarinta and J. Sylvester, Transmission eigenvalues, SIAM J. Math. Anal., 40 (2008), 738-753. doi: 10.1137/070697525. [18] R. Seeley, The resolvent of an elliptic boundary problem, Amer. J. of Math., 91 (1969), 889-920. doi: 10.2307/2373309. [19] N. Shimakura, "Partial Differential Operators of Elliptic Type,'' AMS, Providence, 1992. [20] M. E. Taylor, "Partial Differential Equations II. Qualitative Studies of Linear Equations,'' Applied Mathematical Sciences, 116, Springer-Verlag, 1996. [21] N. T. Thanh and M. Sini, An analysis of the accuracy of the linear sampling method for an acoustic inverse obstacle scattering problem, Inverse Problems, 26 (2010), 29 pp. doi: 10.1088/0266-5611/26/1/015010. [22] N. T. Thanh and M. Sini, Accuracy of the linear sampling method for inverse obstacle scattering: effect of geometrical and physical parameters, Inverse Problems, 26 (2010), 24 pp. [23] F. Treves, "Introduction to Pseudodifferential and Fourier Integral Operators,'' 1, Plenum, New York, 1980. [24] N. Valdivia, Uniqueness in inverse obstacle scattering with conductive boundary conditions, Appl. Anal., 83 (2004), 825-851. doi: 10.1080/00036810410001689283.
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2021 Impact Factor: 1.483
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2022-06-30 13:42:29
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https://www.physicsforums.com/threads/prove-1-z-1-by-induction.219448/
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# Prove 1/(z-1) by induction
1. Mar 3, 2008
### nowimpsbball
1. The problem statement, all variables and given/known data
Prove that the generating function 1/(1-z) = (1+z)(1+z^2)(1+z^4)(1+z^8)...
which is also to 1+z+z^2+z^3+z^4+... when you multiply out the binomials.
2. Relevant equations
(1/(1-z))^k = {$$\Sigma$$[from i=0 to infinity] C(i+k-1, k-1)z^i}
3. The attempt at a solution
I've been playing around with this for a while. I thought I had it, but then I realized that I plugged the "n+1" into the exponent...ie 1+z+z^2+z^3+z^4+...+z^n+z^n+1....but that is not right...I need to plug in z+1 into each z.
PS Ignore the topic name, it should be 1/(1-z)
Thanks
Thanks
Last edited: Mar 3, 2008
2. Mar 3, 2008
### e(ho0n3
So you want to prove that 1/(1-z) = (1+z)(1+z2)(1+z4)...?
3. Mar 4, 2008
### HallsofIvy
Staff Emeritus
Why do you refer to induction? There is no "n" in your formula to do an induction on!
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2017-10-16 23:32:28
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https://tex.stackexchange.com/questions/279575/strange-behaviour-with-glossaries-and-setacronymstyle
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# strange behaviour with glossaries and setacronymstyle
I have several cross-refences of acronyms within my thesis, where I would like to include both abbreviations.
The following MWE however produces strange output. Note, that the abbreviation is (PCI) instead of (PXI):
PCI eXtensions for Instrumentation (PCI) is a specific communication bus.
MWE:
\documentclass{scrreprt}
\usepackage{glossaries}
% on first use, display the long form with the short form in parentheses
% (this is also the default behavior)
\setacronymstyle{long-short}
\makeglossaries
\newacronym{PCI}{PCI}{Peripheral Component Interconnect}
\newacronym{PXI}{PXI}{\acrshort{PCI} eXtensions for Instrumentation}
\begin{document}
\gls{PXI} is a specific communication bus.
\printglossaries
\end{document}
When I leave out the \setacronymstyle, the text is rendered with the correct abbreviation:
PCI eXtensions for Instrumentation (PXI) is a specific communication bus.
Since those are acronyms, I prefer not to use the see tag that are possible for glossaries.
Can I somehow use setacronymstyle with the text correctly rendered?
Just in case, I'm using:
pdfTeX, Version 3.14159265-2.6-1.40.16 (TeX Live 2015/W32TeX)
scrreprt 2015/10/03 v3.19a
glossaries 4.18
There's some strange interaction between the acronym rendering; it seems to be solved if you put the \acrshort instruction in braces.
\documentclass{scrreprt}
\usepackage{glossaries}
% on first use, display the long form with the short form in parentheses
% (this is also the default behavior)
\setacronymstyle{long-short}
\makeglossaries
\newacronym{PCI}{PCI}{Peripheral Component Interconnect}
\newacronym{PXI}{PXI}{{\acrshort{PCI}} eXtensions for Instrumentation}
\begin{document}
\gls{PCI} is the original. Again \gls{PCI}.
\gls{PXI} is a specific communication bus. Again \gls{PXI}.
\printglossaries
\end{document}
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2019-09-21 02:42:23
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https://www.homebuiltairplanes.com/forums/threads/rubber-press-forming.29339/
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# Rubber press forming
Discussion in 'Workshop Tips and Secrets / Tools' started by 12notes, Feb 20, 2018.
### Help Support HomeBuiltAirplanes Forum by donating:
1. Feb 20, 2018
### 12notes
Joined:
Aug 27, 2014
Messages:
692
445
Location:
Louisville, KY
I built a couple of rubber press formers for my wing ribs, and thought I'd relate my experience and advice concerning them.
When I first priced out the metal and rubber for building one, it was far too expensive. Then I found a 6'x3' roll of 1/4" thick rubber at the hackerspace, and the drop rack at Metal Supermarkets near me had 2 pieces of 3/16" C channel the correct length and a 3/16 angle the correct length to fit inside for 65 cents a pound. So I built two forming boxes, one for the nose ribs, and one for the rear ribs. I dusted off my decade-old stick welding skills using some old, moisture laden 6010 rods and made some ugly looking but strong welds. Total cost was somewhere around $50-$75, I think, but I was buying a bunch of metal at the time.
I cut my ribs out on the Shopbot CNC, and also made forms from 3/4" MDF with flanged angles for the lightening holes. I also put a stiffening bead groove on the forms.
I cut and layered 6 pieces of the rubber to fill the smaller nose rib box, and went to press it, only to find the hackerspace's Harbor Freight 12 ton press' jack was leaking fluid badly. Bought a new jack, got everything in place and started pressing, only to find the rubber wasn't compressing the flanges nearly enough. I modified 3 layers of the rubber to match the shape of the form, then pressed again, only to hear a loud bang as the press ram bent and the tip shoved up back into the tube - it is a press no more. It formed most of the rib including the flutes pretty well, the back lightening hole flange wasn't complete and it did not even start the stiffening bead.
Score: .020" aluminum 1, press 0.
I thought I'd try the rear ribs on a 20 ton press. Made the same type of rubber setup, press slowly until my weight was no longer enough to move the jack handle, then my more gravitationally abundant friend compressed it a few more pulls. The side flanges bent some, but not enough to form any flutes, the bend from the fluting pliers was the only one evident. It pressed the middle two flanges on the lightening holes, but nothing on the other two. No stiffening beads were evident.
Score .020" aluminum 2, presses 0.
So I decided to form the rear ribs by hand and the front ribs in the 20 ton press. Forming 24 rear ribs the "slow way" instead of pressing them took all of 2 hours, excluding the stiffening beads and lightening hole flanges. I did put the flanges and stiffening beads on one rib by putting a piece of the rubber over the rib on the form and hammering in the approximate locations, it took about 1 minute, worked fine.
My personal opinion is that rubber press forming makes no sense unless you are going into production. You'll spend far, far, far more time building the boxes than it takes to press the ribs. If you already had the boxes, I'm not sure forming a rib with the rubber press would be any faster than forming one with a hammer. Also, it's only good for nose ribs unless you have a greater than 20 ton press, and I'm not even positive about that as I haven't gotten back to the 20 ton press with the nose ribs yet.
I decided to do a few things with this project that I knew would be slower, as I wanted to learn some new techniques (CAD, CNC, etc). This was one of those things, and, although I'm glad I went through the process, it doesn't seem to be worth the time and expense (~\$300, if using new steel and rubber) for a home builder.
Last edited: Feb 20, 2018
cluttonfred likes this.
2. Feb 20, 2018
### BBerson
#### Well-Known MemberHBA Supporter
Joined:
Dec 16, 2007
Messages:
11,407
2,092
Location:
Port Townsend WA
Well, your 14" thick rubber is likely a typo? (1/4")
Probably need 2-3" thick rubber block and a 500 ton press or more. Figure the force to push 1 square inch a distance of 1/2" into the rubber block. And multiply that times the size of the rib.
3. Feb 20, 2018
### 12notes
Joined:
Aug 27, 2014
Messages:
692
445
Location:
Louisville, KY
Yep, it was 1/4", I corrected the post.
500 tons is way more than needed. The front rib was 90% formed on the 12 ton press before the press broke. I can't remember where I found the calculation, but 20 tons should have been on the edge of working for the rear rib (it's also what they use in the EAA video), and more than enough for the front rib. I'll try to find it again.
4. Feb 20, 2018
### BBerson
#### Well-Known MemberHBA Supporter
Joined:
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If you notch the ribs like EAA, then won't need much.
I thought you were trying do real Guerin type forming.
5. Feb 21, 2018
### 12notes
Joined:
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Messages:
692
445
Location:
Louisville, KY
I'm not sure what you mean by "real" Guerin type forming, but unless I'm missing something, Guerin type forming is exactly what I was doing - bending down the flanges on the edges of the ribs into a fluted form while also pressing the flanges and stiffening beads on the face using a rubber pad. The ribs were not notched.
6. Feb 21, 2018
### BBerson
#### Well-Known MemberHBA Supporter
Joined:
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2,092
Location:
Port Townsend WA
The EAA video I saw was notched, I think. So I had no further interest.
If you can press without notches and get the rubber to flow up the side and fully form at 90°, then that would be "real", in my opinion. I think you might need solid rubber, not 1/4" laminations.
Is the metal soft? What alloy?
Got any photos?
You said the process makes no sense...
Last edited: Feb 21, 2018
7. Feb 21, 2018
### BBerson
#### Well-Known MemberHBA Supporter
Joined:
Dec 16, 2007
Messages:
11,407
2,092
Location:
Port Townsend WA
8. Feb 21, 2018
### bobm4360
#### Member
Joined:
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Messages:
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8
Location:
Oak Harbor, WA
The "Me-262 Project" used a HF 50-ton press and a 1"thick steel box 12" x 12" x 6" deep and a soft neoprene pad backed by a slightly harder neoprene pad. and would handle anything that fit within the box up to .040 steel (notched), and, IIRC .063 aluminum (drawn, not notched).
9. Feb 21, 2018
### lakeracer69
#### Well-Known Member
Joined:
Dec 10, 2008
Messages:
106
47
Location:
New Hampshire
Just any "rubber' won't get it done. You need some that you KNOW is around 65 durometer. You also need it to be around an 3/4 to 1 inch thick ( per piece).
I have had no problems using: A 20 ton press, two pieces of the correct rubber ( durometer and thickness), and a form made out of corian ( MDF got crushed too much). .025 6061-T6 formed fine and I just had to just do a final straightening on the edges.
In your case, a bunch of thin rubber of unknown hardness didn't do the job. If you follow how others have done it, and don't substitute your own ideas and materials, it will work. I have done it a bunch so I know it works.
pictsidhe likes this.
10. Feb 21, 2018
### 12notes
Joined:
Aug 27, 2014
Messages:
692
445
Location:
Louisville, KY
0.020" and 0.025" 2024-T3, the nose ribs had formed about 80 degrees on the edges with flutes formed and most of the flanges on the face formed when the 12-ton press broke. I'm fairly sure the nose rib will fully form (no undercut on the form, so not quite 90 degrees) with the 20 ton press, but I don't think it will be any faster or better than hammer forming it, I'll press one later this week.
No, I did not say that. You cannot remove every qualifier in a statement and imply it is what was said. What I said was "My personal opinion is that rubber press forming makes no sense unless you are going into production." And I stand by that statement. It is my opinion that the process of making the box takes much, much longer than any time you might save making the ribs, and adds unnecessary expense. Once the box it built, it is only marginally faster - a few minutes saved per rib at best. My experience was no time saved per rib - it took about 5 minutes to form a rib either way. A 20-ton press is about the largest commonly found in a garage or home workshop, and I'm building a small (38" chord) set of wings, for those that have larger wings, this process is only good for nose ribs unless you have a larger press. In my opinion, it's an expensive, time consuming solution way to solve very minor problems if you're just building one plane. Hammer forming is a much better method for low numbers.
11. Feb 21, 2018
### Chris In Marshfield
#### Well-Known MemberHBA Supporter
Joined:
Jan 29, 2008
Messages:
1,200
983
Location:
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If you check out my Expedition project in my sig, you’ll see how I used a 20-ton press in my project. I used it only to flange lightening holes, and it was great. That was the part I really needed it for. Not enough oomph to bend edges, although I did try on some small parts. It flowed the hole flanges beautifully. But wasn’t very useful for edge flanges. And it went *fast*. .025” and .032” 2024.
12. Feb 21, 2018
### Marc Bourget
#### Well-Known Member
Joined:
Feb 28, 2011
Messages:
353
103
Location:
Stockton, California
Lots of good points in this thread, but there is a much larger "world" out there than set out above. Used book websites will give you good resources if you subject search on "Forming Metal." Alcoa's reference book, "Forming Alcoa Aluminum" will raise your personal forming bar significantly.
I agree with the relative speed of hand forming versus press forming, but I served my apprenticeship in John Thorp's Burbank Shop. Check out John's Articles in Sport Aviation on Building the T-18, particularly No. 3, printed in the July 1962 issue. View attachment No 3.pdf
13. Feb 21, 2018
### Marc Bourget
#### Well-Known Member
Joined:
Feb 28, 2011
Messages:
353
103
Location:
Stockton, California
Sorry double post
14. Feb 22, 2018
### Victor Bravo
#### Well-Known Member
Joined:
Jul 30, 2014
Messages:
5,748
4,619
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KWHP, Los Angeles CA, USA
Drool...jealousy...drool...jealousy...
15. Feb 22, 2018
### Marc Bourget
#### Well-Known Member
Joined:
Feb 28, 2011
Messages:
353
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Location:
Stockton, California
Victor Bravo,
No reason to feel bad, I'm more than happy to share my knowledge. This Summer, if you're ever in the Stockton area and have some time, drop me a PM.
16. Feb 22, 2018
### Victor Bravo
#### Well-Known Member
Joined:
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Thank you, but I'm still wanted as Public Enemy #1 in Stockton, over something that happened almost 25 years ago. To this day I'm puckered up toothpick-tight just driving past there on the interstate.
17. Feb 23, 2018
### 12notes
Joined:
Aug 27, 2014
Messages:
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Location:
Louisville, KY
I do not have a Shore A gauge, but I did have samples of the same thickness of 50A and 75A durometers. The rubber I used was stiffer than the 50, but not as stiff as the 75. It's in the neighborhood of 60, but I don't know the exact durometer.
I did follow how others have done it. I used the guide from the EAA, which said several pieces of thinner rubber would suffice:
It should work for nose ribs, but rear ribs, even the small ones I'm making, require a box that is near the limit of a 20 ton press. The pressure in the box I built calculates to be about 275psi, and the box barely fits the rear rib on a 13.5% thick 38" chord wing. That's less pressure than recommended, and that's for a very small rear rib (21"x5"). Larger ribs would require a bigger box and have less pressure with the same press.
I never said it doesn't work, but that it's not worth it unless you're building more than one plane. Building the box takes longer than forming all the ribs with a hammer.
18. Feb 23, 2018
### lakeracer69
#### Well-Known Member
Joined:
Dec 10, 2008
Messages:
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47
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New Hampshire
Did you use lube up the parts with spray on furniture wax before forming? That was something I found was a must for it to work properly. Also the box is not "required" for actually forming the parts. It is more to just contain things if they want to give out under pressure. I made mine from some scrap wood. My rubber pads are leftover scraps of mats from horse stalls.
YMMV
19. Feb 23, 2018
### Marc Bourget
#### Well-Known Member
Joined:
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353
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Location:
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VB, I talked to her, she's forgiven you and actually would like your Ph# !!! (hahahahahahahaha!)
20. Feb 23, 2018
### Marc Bourget
#### Well-Known Member
Joined:
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Messages:
353
103
Location:
Stockton, California
12notes,
Actually, If you have a good understanding of the process, all sorts of materials can be employed to help press and drop hammer forming.
For those 20T press owners:
Consider adding a "clamp plate" while rubber forming. You'll only have to "rubber" the perimeter, increasing the psi to the areas that need it.
FWIW
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2019-07-18 08:46:34
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https://brilliant.org/problems/just-another-way-of-writing-it/
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# Just another way of writing it..
Algebra Level 2
$$lnx + ln(x+1) = 0$$ is another way of writing
×
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2018-12-14 04:31:10
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http://math.stackexchange.com/questions/251424/what-is-the-correct-way-to-verify-solutions-from-different-ode-symbolic-solvers
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# What is the correct way to verify solutions from different ode symbolic solvers against each others?
I am not a math major, so I thought to check with the experts here on this.
Given some ODE $y'(x)=f(y,x)$ solved by CAS system A which gives answer $y_1(x)$. Same ODE solved by CAS system B with answer $y_2(x)$.
Since the ODE does not involve initial conditions, Answers will contain constant(s) of integration. Lets say there is one constant of integration $C_0$ for now. (it was a first order ODE)
To verify that $y_1(x)$ is the same as $y_2(x)$, both solutions are plotted over some small range of $x$ around the origin to see if the solutions are the same.
Since there is a constant of integration, it is given some constant value, say the number $1$ in order just to be able to make a plot.
Is this a valid approach? Some of the answers are too complicated to visually verify the expressions are equivalent. But sometimes the solutions to the same ODE do not come up the same in the plot. There remains some difference. That is why I am asking here, if there is a better way to do this.
To make things more clear, an example below is given, with the solution from Maple and Mathematica. Both involve one constant of integration. Using $C_0=1$ the plots of the solutions are generated. Yet we see the plots of $y(x)$ are not exactly the same. There is still a shift involved. (The method has to be numerical, as it is simpler for quick verification).
the ODE is Kamke's #131 given by $\left( 1+2x\right) y^{\prime}\left( x\right) +2-4e^{-y\left( x\right) }=0$
## The ODE is in Mathematica syntax:
Clear["Global*"];
ode131 = (2*x + 1)*D[y[x], x] - 4*Exp[-y[x]] + 2 == 0
DSolve[%, y[x], x]
which gives
{{y[x] -> Log[2 - E^C[1]/(1 + 2*x)]}}
## In Maple 16
restart;
ode131:=(1+2*x)*diff(y(x),x)+2-4*exp(-y(x))=0:
dsolve(%,y(x));
which gives
y(x) = -ln((1+2*x)/(-1+2*exp(2*_C1)+4*x*exp(2*_C1)))-2*_C1
Here is a plot of both solutions next to each others. We see the shift. Actually using different numerical value for the constant of integration, makes the solutions even more different looking. This is with $C_0=1$
(ps. Log is natural log in Mathematica)
FYI, the equation from the book (it is written in German) and solution is
update
Using suggestion below by Mhenni Benghorbal, Maple says both solutions are the same:
ode := (2*x+1)*diff(y(x),x)-4*exp(-y(x))+2:
mapleSol := dsolve( ode ):
odetest( mapleSol, ode );
0
and
mathematicaSol:=y(x)=ln(2 - (exp^_C1)/(1 + 2*x)):
odetest( mathematicaSol, ode );
0
So the answers the same. I'd still like to know if possible from Mathematics point of view why just plugging the same constant in the solutions and plotting the resulting function do not show the same shapes.
-
Try to ask at stackoverflow.com as well, Nasser. – Babak S. Dec 5 '12 at 9:28
In maple, you can run the "odetest" command, for example
ode := (2*x+1)*diff(y(x),x)-4*exp(-y(x))+2;
sol := dsolve( ode );
odetest( sol, ode );
If the function satisfies the ode then "odetest" returns $0$. So you can take mathematica answer and write in the maple worksheet and run the "odetest" and see what you get.
-
Thanks, that is useful. I just did , please see edit. I get 0` for both answers. – Nasser Dec 5 '12 at 10:12
@NasserM.Abbasi: Constants are determined by the initial conditions. – Mhenni Benghorbal Dec 5 '12 at 10:17
@NasserM.Abbasi: Since you have two different forms of solutions, then you need to relate the constants in the two solutions first. – Mhenni Benghorbal Dec 5 '12 at 10:20
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2015-08-30 02:12:45
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https://www.quizover.com/physics-k12/course/2-5-constant-acceleration-application-by-openstax
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2.5 Constant acceleration (application)
Page 1 / 2
Solving problems is an essential part of the understanding
Questions and their answers are presented here in the module text format as if it were an extension of the treatment of the topic. The idea is to provide a verbose explanation, detailing the application of theory. Solution presented is, therefore, treated as the part of the understanding process – not merely a Q/A session. The emphasis is to enforce ideas and concepts, which can not be completely absorbed unless they are put to real time situation.
Hints on solving problems
1. Though acceleration is constant and hence one – dimensional, but the resulting motion can be one, two or three dimensional – depending on the directional relation between velocity and acceleration.
2. Identify : what is given and what is required. Establish relative order between given and required attribute.
3. Use differentiation method to get a higher order attribute in the following order : displacement (position vector) → velocity → acceleration.
4. Use integration method to get a lower order attribute in the following order : acceleration → velocity → displacement (position vector).
Representative problems and their solutions
We discuss problems, which highlight certain aspects of the study leading to the motion with constant acceleration. The questions are categorized in terms of the characterizing features of the subject matter :
• Average velocity
• Differentiation and Integration method
• Components of constant acceleration
• Rectilinear motion with constant acceleration
• Equations of motion
Average velocity
Problem : A particle moves with an initial velocity “ u ” and a constant acceleration “ a ”. What is average velocity in the first “t” seconds?
Solution : The particle is moving with constant acceleration. Since directional relation between velocity and acceleration is not known, the motion can have any dimension. For this reason, we shall be using vector form of equation of motion. Now, the average velocity is given by :
$\begin{array}{l}{\mathbf{v}}_{a}=\frac{\Delta \mathbf{r}}{\Delta t}\end{array}$
The displacement for motion with constant acceleration is given as :
$\begin{array}{l}\Delta \mathbf{r}=\mathbf{u}t+\frac{1}{2}\mathbf{a}{t}^{2}\end{array}$
Thus, average velocity is :
$\begin{array}{l}{\mathbf{v}}_{a}=\frac{\Delta \mathbf{r}}{\Delta t}=\frac{\mathbf{u}t+\frac{1}{2}\mathbf{a}{t}^{2}}{t}=\mathbf{u}+\frac{1}{2}\mathbf{a}t\end{array}$
Differentiation and integration methods
Problem : A particle is moving with a velocity $2\mathbf{i}+2t\mathbf{j}$ in m/s. Find (i) acceleration and (ii) displacement at t = 1 s.
Solution : Since velocity is given as a function in “t”, we can find acceleration by differentiating the function with respect to time.
$\begin{array}{l}\mathbf{a}=\frac{đ}{đt}\left(2\mathbf{i}+2t\mathbf{j}\right)=2\mathbf{j}\end{array}$
Thus, acceleration is constant and is directed in y-direction. However, as velocity and acceleration vectors are not along the same direction, the motion is in two dimensions. Since acceleration is constant, we can employ equation of motion for constant acceleration in vector form,
$\begin{array}{l}\Delta \mathbf{r}=\mathbf{u}t+\frac{1}{2}\mathbf{a}{t}^{2}\\ \Delta \mathbf{r}=\left(2\mathbf{i}+2t\mathbf{j}\right)t+\frac{1}{2}\phantom{\rule{2pt}{0ex}}x\phantom{\rule{2pt}{0ex}}2\mathbf{j}x{t}^{2}\end{array}$
For t = 1 s
$\begin{array}{l}\Delta \mathbf{r}=\left(2\mathbf{i}+2\phantom{\rule{2pt}{0ex}}x\phantom{\rule{2pt}{0ex}}1\mathbf{j}\right)\phantom{\rule{2pt}{0ex}}x\phantom{\rule{2pt}{0ex}}1+\frac{1}{2}\phantom{\rule{2pt}{0ex}}x\phantom{\rule{2pt}{0ex}}2\mathbf{j}\phantom{\rule{2pt}{0ex}}x{1}^{2}\\ \Delta \mathbf{r}=2\mathbf{i}+3\mathbf{j}\end{array}$
Note 1 : We should remind ourselves that we obtained displacement using equation of motion for constant acceleration. Had the acceleration been variable, then we would have used integration method to find displacement.
Note 2 : A constant acceleration means that neither its magnitude or direction is changing. Therefore, we may be tempted to think that a constant acceleration is associated with one dimensional motion. As we see in the example, this is not the case. A constant acceleration can be associated with two or three dimensional motion as well. It is the relative directions of acceleration with velocity that determines the dimension of motion – not the dimension of acceleration itself.
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2017-12-12 08:21:04
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https://www.tutorialspoint.com/how-to-change-pixel-values-using-at-method-in-opencv
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# How to change Pixel Values using 'at' Method in OpenCV?
OpenCVC++Server Side ProgrammingProgramming
In a grayscale image, the pixel value is a single numeric value. But in a color image such as RGB image, the pixel is a vector having three values. These three values represent three channels.
Here we will create a function that accesses both the grayscale image and RGB image pixel values and randomly adds noise to image pixels. Then we call the function inside the main() function to observe the result.
The following program demonstrates how to change Pixel Values using 'at' method in OpenCV.
## Example
#include<iostream>
#include<opencv2/highgui/highgui.hpp>
using namespace cv;//Declaring cv namespace
using namespace std;
void adding_Noise(Mat& image, int n){ //'adding_Noise' function//
for (int x = 0; x < n; x++){ //initiating a for loop//
int i = rand() % image.cols;//accessing random column//
int j = rand() % image.rows;//accessing random rows//
if (image.channels() == 1){ //apply noise to grayscale image//
image.at<uchar>(j, i) = 0;//Changing the value of pixel//
}
if (image.channels() == 3){ //apply noise to RGB image//
image.at<Vec3b>(j, i)[0] = 0;//Changing the value of first channel//
image.at<Vec3b>(j, i)[1] = 0;//Changing the value of first channel//
image.at<Vec3b>(j, i)[2] = 0;//Changing the value of first channel//
}
}
}
int main() {
Mat image;//taking an image matrix//
Mat unchanged_Image;//taking another image matrix//
namedWindow("Noisy Image");//Declaring an window//
namedWindow("Unchanged Image");//Declaring another window//
}
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2022-06-29 12:36:11
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https://dev.lucee.org/t/error-cannot-load-class-through-its-string-name/4448
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Error: cannot load class through its string name
We’re on Lucee 5.2.7.63 and experiencing some bad issues with loading in JAR’s in our Lucee app via CreateObject. We are not able to load ANY JAR-file - getting the below message everytime:
cannot load class through its string name, because no definition for the class with the specified name [JAR-functionname] could be found caused by (java.lang.ClassNotFoundException:JAR-functionname not found by lucee.core [64];java.lang.ClassNotFoundException:JAR-functionname;java.lang.ClassNotFoundException:JAR-functionname;)
What to do?
I’ve tried this: https://rorylaitila.gitbooks.io/lucee/content/calling_java_from_lucee.html (making use of this.javaSettings)
Done this: https://stackoverflow.com/questions/39177674/where-do-i-put-jar-files-in-lucee and Where to place 3rd party jars? (place JAR’s in /webappname/WEB-INF/lucee/lib/)
Utilized this: https://github.com/markmandel/JavaLoader Mark Mandel’s JavaLoader (works nicely in CF9+, but NOT in Lucee at all)
In all cases we get the above error-message. It seems like this issue https://luceeserver.atlassian.net/browse/LDEV-604 is still not resolved…
1 Like
Hi Seb
Loading jars via createObject() is working for me in Lucee 5.2.8.50 and it was fine in the previous 5.2.7.63 release too.
I’ve just done a simple test with a basic jar with one method located in the same directory as the calling script:
object = createObject( "java", "Basic", expandPath( "./" ) );
dump( object );
Are you using the Lucee spreadsheet library by any chance? If so is that giving the same error? By default that will load the POI etc jars in Lucee 5.x using createObject() but it can be configured to use JavaLoader.
2 Likes
Hi Julian, well, it seems that our JavaLoader being saved to a server variable for now was the culprit. Even a restart of Lucee didn’t clear the server-scoped object of loaded JAR’s. This was never an issue with ColdFusion. The JAR’s are for creating QR-codes and our own JAR’s with functions needed to create SPSS-files. Furthermore we have created a JAR that connects to a certain service, but they were not loaded either. Strange thing was that if I loaded the JAR somewhere in a CFC it went fine, doing the same operation in a CFM didn’t work. Anyways, Monday may be better, but porting a large webapp to Lucee from ACF is certainly proving to be difficult - way more so than I expected at first
Strange. I use JavaLoader quite a bit in Lucee and haven’t had any issues putting instances in the server scope (including with the Spreadsheet lib). Here’s how I generally code my JAR wrappers’ init() methods (where the jars are in the same directory as the wrapper):
function init( javaLoaderPath=variables.javaLoaderPath ){
var paths = [];
paths.append( getDirectoryFromPath( getCurrentTemplatePath() ) & "my.jar" );
}
return this;
}
I sympathise. I wrote a whole series of blog posts a few years back on the woes we had transitioning to Railo, then Lucee. It all depends on which features you use. Good luck and shout if you need help.
Hi @Julian_Halliwell, thnx 4 the example, after the weekend stuff don’t work anymore. Here’s how we call the Javaloader from the Application.cfc:
/* JavaLoader: https://github.com/markmandel/JavaLoader */
local.paths = ArrayNew(1);
local.libDir = ExpandPath("../../webapp/lib");
directory action="list" directory="#local.libDir#" filter="*.jar" listinfo="name" name="local.jars";
for(jar in local.jars) {
ArrayAppend(local.paths,local.libDir & "#application.osFilePathSeparator#" & jar.name);
}
/* in the server scope because: https://github.com/markmandel/JavaLoader/wiki/Memory-Consumption */
}
The local.paths array gives me this in return:
Array
Rows: 18
1 string D:\path\to\webapp\lib\jar-one-versionnumber.jar
2 string D:\path\to\webapp\lib\jar-two-versionnumber.jar
3 string D:\path\to\webapp\lib\jar-three-versionnumber.jar
...
17 string D:\path\to\webapp\lib\jar-seventeen-versionnumber.jar
18 string D:\path\to\webapp\lib\jar-eightteen-versionnumber.jar
That is being loaded into the server variable. What are we doing wrong here? I get the error-message that the desired classes are not available.
Hmmm, if I skip the check in Application.cfc then the server-variable will be reloaded on each onApplicationStart. Is that bad? At least it works for now, as Lucee doesn’t seem to clear the server variables on Lucee Windows restart. Do you experience the same @Julian_Halliwell?
I think there must be something else going on that’s specific to your set up, Seb. As far as I’m aware server variables do not survive Lucee restarts - I’m not sure how they could without you manually serializing them to disk?
Personally I wouldn’t put this kind of logic in Application.cfc. I always use encapsulated wrappers, i.e. dedicated components which handle the loading/server scope checks for me and just let me call the classes/methods I need at the points in the app where I need them. Again, the Spreadsheet lib is an example of this.
I’m happy to test the specific jars you’re working with on my own setup if they can be shared privately or otherwise.
1 Like
Hi @Julian_Halliwell, you’re right, of course, I’ve refrained from restarting Lucee, because of this: https://luceeserver.atlassian.net/browse/LDEV-1763 Gives me 3-5 minutes downtime every time, so for now the variable is reset onApplicationStart. But happy to say all stuff now seems to work as on ACF. Incorporating JAR’s into Lucee webapps is fun and powerful!
Can I flag this as FIXED somewhere?
The OP (or moderators) can flag any post as the solution by expanding the menu at the foot of the post and selecting the check
OK, kewl, but I do not have that option…
Edit: now I suddenly do
|
2022-06-27 13:58:11
|
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|
https://www.mql5.com/en/forum/150291
|
# Getting indicator values from a specified time
22
Dears,
I need to get value of iMA from a specified time, how I do this?
thanks
15580
Convert the "specified time" to a bar shift, use that in iMA
22
WHRoeder:
Convert the "specified time" to a bar shift, use that in iMA
I'm sorry, but I don't know that I do. could you give me an example?
Moderator
10569
https://forum.mql4.com/61669
Has an example
22
What is wrong with this script and how to fix?
<SRC by Moderator: Ubzen>
```//+------------------------------------------------------------------+
//| test.mq4 |
//| Copyright 2013, MetaQuotes Software Corp. |
//| http://www.metaquotes.net |
//+------------------------------------------------------------------+
#property indicator_chart_window
//+------------------------------------------------------------------+
//| Custom indicator initialization function |
//+------------------------------------------------------------------+
int init() {
//---- indicators
shift = iBarShift(NULL, PERIOD_H1, 13 : 00)
iMA(NULL, 0, 13, 8, MODE_SMMA, PRICE_MEDIAN, shift);
//----
}
//+------------------------------------------------------------------+
//| Custom indicator deinitialization function |
//+------------------------------------------------------------------+
int deinit() {
//----
//----
return (0);
}
//+------------------------------------------------------------------+
//| Custom indicator iteration function |
//+------------------------------------------------------------------+
int start() {
int counted_bars = IndicatorCounted();
//----
//----
return (0);
}
//+------------------------------------------------------------------+```
Moderator
10569
unless this is something new in mql4, I don't think that you can input datetime like that
`shift = iBarShift(NULL, PERIOD_H1, 13 : 00)`
You can use
`shift = iBarShift(NULL, PERIOD_H1, D'2014.03.09 13:00');`
22
This is the errors
'iMA' - an operator expected C:\Program Files (x86)\FXDD Malta - MetaTrader 4\experts\indicators\teste.mq4 (16, 5)
'=' - assignment expected C:\Program Files (x86)\FXDD Malta - MetaTrader 4\experts\indicators\teste.mq4 (15, 11)
'iMA' - semicolon expected C:\Program Files (x86)\FXDD Malta - MetaTrader 4\experts\indicators\teste.mq4 (16, 5)
'}' - semicolon expected C:\Program Files (x86)\FXDD Malta - MetaTrader 4\experts\indicators\teste.mq4 (18, 1)
Moderator
10569
if you double click the error report line, the cursor will go to the problem area, makes them easier to find
I've inserted the missing ; I'm still not sure about the 13:00
```int init() {
//---- indicators
shift = iBarShift(NULL, PERIOD_H1, 13 : 00) ;
iMA(NULL, 0, 13, 8, MODE_SMMA, PRICE_MEDIAN, shift);
//----
}```
|
2017-07-24 12:57:38
|
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|
https://math.stackexchange.com/questions/3350742/non-local-injectivity-of-real-analytic-functions-of-several-variables
|
# Non local injectivity of real-analytic functions of several variables
Let $$f: \mathbb{R}^n \rightarrow \mathbb{R}^{n+m}$$ be a real-analytic function (at least in a neighbourhood of a point). This question establishes that if the Jacobian matrix at a point is full rank, i.e. rank $$n$$, then the function is locally injective at that point.
I guess the converse (non full rank implies non locally injective) is false, since generalising $$x \mapsto x^3$$ to $$x \mapsto (x^3, x^3)$$ gives a counterexample, as the Jacobian at $$x = 0$$ has rank zero but the function is injective. My question is are there any sufficient conditions on the Jacobian at a point that imply the function is not locally injective at a point? (Or sufficient conditions in general, with or without using the Jacobian.) I'm also interested in the complex case but perhaps that's best reserved for another question.
EDIT: Thanks to the comment, I stumbled across this question on MO - where it's argued for $$f: \mathbb{R}^n \rightarrow \mathbb{R}^n$$ that if the Jacobian simultaneously has negative and positive determinant somewhere, then $$f$$ is not injective. It seems reasonable that this extends locally, i.e. $$f$$ is not locally injective at a point if in every neighbourhood of that point there are two points where the determinant is negative and positive. Does anyone know if it possible to extend this to the case $$f: \mathbb{R}^n \rightarrow \mathbb{R}^{n+m}$$?
• See if this helps you : math.stackexchange.com/questions/900737/… – астон вілла олоф мэллбэрг Sep 10 at 13:24
• Thanks for the suggestion - It's definitely an interesting result but I think the assumptions are a bit strong, as it requires degeneracy in a neighbourhood of a point and not just at that point itself. (That said, I'm not even particularly sure if I could apply it here, unless there is a projection $\pi: \mathbb{R}^{n+m} \rightarrow \mathbb{R}^n$ such that $\pi \circ f$ is injective if $f$ is...) – Riley Sep 10 at 13:58
• No problem, I was looking to help. – астон вілла олоф мэллбэрг Sep 10 at 14:25
|
2019-09-23 10:58:35
|
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|
https://cosmocoffee.info/viewtopic.php?t=1696
|
## [1011.0614] Cold uniform spherical collapse revisited
Authors: M. Joyce, B. Marcos, F. Sylos Labini Abstract: We report results of a study of the Newtonian dynamics of N self-gravitating particles which start in a quasi-uniform spherical configuration, without initial velocities. These initial conditions would lead to a density singularity at the origin at a finite time when N \rightarrow \infty, but this singularity is regulated at any finite N (by the associated density fluctuations). While previous studies have focussed on the behaviour as a function of N of the minimal size reached during the contracting phase, we examine in particular the size and energy of the virialized halo which results. We find the unexpected result that the structure decreases in size as N increases, scaling in proportion to N^{-1/3}, a behaviour which is associated with an ejection of kinetic energy during violent relaxation which grows in proportion to N^{1/3}. This latter scaling may be qualitatively understood, and if it represents the asymptotic behaviour in N implies that this ejected energy is unbounded above. We discuss also tests we have performed which indicate that this ejection is a mean-field phenomenon (i.e. a result of collisionless dynamics). [PDF] [PS] [BibTex] [Bookmark]
Discussion related to specific recent arXiv papers
Syksy Rasanen
Posts: 119
Joined: March 02 2005
Affiliation: University of Helsinki
### [1011.0614] Cold uniform spherical collapse revisited
This paper looks at the formulation of the fluid limit of $N$-body systems, a topic on which the authors and their collaborators have done interesting work before (and which seems strangely neglected in the wider cosmology community).
They distribute points randomly inside a sphere with zero initial velocity, and study the evolution. One interesting issue is the large amount of mass lost in the collapse: the fraction of particles ejected increases with $N$, and varies from 15% to 30% for the range of $N$ studied. In terms of energy, the situation is even more dramatic: the amount of energy ejected from the system grows roughly like $N^{1/3}$, and for the largest $N$ the kinetic energy carried away by the ejecta is almost ten times the original energy of the system. (Since the Newtonian gravitational potential is unbounded from below, there is in principle no upper limit to the amount you can extract from a self-gravitating system.)
This calls into question the issue of how the fluid limit should be formulated (since just taking $N$ naively to infinity is obviously wrong). The authors note that this should be done by keeping the fluctuations fixed above some length scale $l$ as $N$ increases.
It is not clear what is the importance of the ejection process for a realistic system, and the authors say they will follow up with a study of a system where the particles have non-zero initial velocities.
|
2018-07-19 07:44:36
|
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|
http://bayesianthink.blogspot.com/
|
## Monday, January 16, 2017
### The Bacteria Division Puzzle
Q. A jar has a single cell of a bacteria. After every fixed interval of time the bacteria either splits into two with probability 2/5, does nothing with probability 2/5 or dies out with probability 1/5. What is the probability that the bacteria would continue to make a large family tree over an extended period of time?
A. The situation can be described by the following visual.
Assume that the required probability is 'p'. The term 1 - p would represent the probability that the ecosystem eventually dies out. Each of the above scenarios contributes a quantum of probability towards the ecosystem eventually dying out. Lets start off by represent 1 - p as 'x'. The probability that the bacteria die out is
The total of each of the above must add up to the probability that the bacteria eventually die out, which is 'x'. So you can phrase the problem recursively as
This simplifies to
which is a quadratic equation, yielding a solution as
This further yields x = 1/2 or x = 1. It's easy to see why x = 1 isn't a solution because scenario 1 would rule that out immediately. This yields a probability of 50%.
In order to test this out, it's also relatively easy to code it up in Python. The following code shows how to simulate this scenario.
import sys
import numpy as np
import matplotlib.pyplot as plt
import seaborn as sns
dist_estimates = []
for index in range(400):
counter = 0
'''
Repeating 100 times for estimating the probability
'''
for epoch in range(100):
num_bacteria = 1
'''
'''
for iter in range(20):
bacteria = np.zeros(int(num_bacteria))
'''
Step through each bacteria
'''
for nb_iter in range(int(num_bacteria)):
'''
Decide if each of the bacteria
will die/stay/split in this time slice
'''
prob = np.random.uniform(0,1)
if prob < 0.2:
''' This one dies '''
bacteria[nb_iter] = 0
elif prob >= 0.2 and prob <= 0.6:
''' This one stays as is '''
bacteria[nb_iter] = 1
else:
''' This one splits into two '''
bacteria[nb_iter] = 2
num_bacteria = np.sum(bacteria)
if np.sum(bacteria) > 0:
counter += 1
dist_estimates.append(float(counter)/100)
'''
Visualization is easy
Use matplotlib and seaborn
'''
p = sns.distplot(dist_estimates,kde=False,rug=True).get_figure()
p.savefig('t.png')
The above code generates a plot of estimates for the said probability. This came out the following way which validates our method.
If you are looking to learn probability, programming or data science related books, a good set of books are listed here
## Thursday, January 5, 2017
### The Forgotten Geometric Mean.
Often times a lot of people working with data are trying to create an index of some sort. Something that captures a set of key business metrics. If you are a site (or an app) you want to create some sort of an engagement index, which if trending up implies good things are happening, bad if it is trending down. The creators of such metrics (think analysts) tend to prefer a weighted arithmetic mean of the influencing factors. If the influencing factors are f1,f2, f3 (say) with weights w1, w2, w3 then the index would be computed as
However, what does not get factored in are the final consumers of the index (think product managers) and there could be many. They will invariably try to check it with something else they have handy. For example, if clicks on a site went up 20% the index may be up by just 5% (say) or vice-versa. If resources are being allocated based on the movement of such an index, it will invariably lead to contention on what is the right weighting to be given to each factor.
This is meant to be a short write up on some really cool features of the geometric mean. The geometric mean is not meant to replace a simple arithmetic mean based index, but it is definitely worth the thought. To illustrate what this aspect is, lets take a look at a simple two feature index. If the features are X and Y the arithmetic mean index can be represented as
To see how it responds to changes, lets take the derivative.
Clearly the derivative is dependent on the chosen weight. Lets see what happens when we choose the geometric mean.
Again, to see how it responds to change, lets take the derivative.
which can be further simplified to
The result is a useful derivable condition
i.e. the percentage change in the index is directly proportional to the percentage change in the feature.
Note, there are no hand chosen weights here. A five percent change in one of the influencing factors will result in a proportional percent change in the index. Extremely useful !
Yet another aspect consumers like to quantify is growth. If the index went up by x1 and x2 in consecutive years, what is the average quarterly/annual growth? If we took it as the average of x1 and x2, then the growth after two years (say) would be estimated as
Contrast that to the actual growth
Clearly some terms cancel out. We are left comparing
Notice one of them is the arithmetic mean and the other is the geometric mean. We also know from a well established theorem that the arithmetic mean is always greater than the geometric mean described here. So we would always end up overestimating the growth!
So how would we choose a value to project as an average growth rate? We are looking for a beta in the below equation
Yet again stating the average growth as the geometric mean gives the end user a handy metric to work with.
If you are interested in learning probability here are a set of good books to choose and buy from.
## Monday, January 2, 2017
### Colored Cards and Numbers Puzzle
Q: You have a set of thirty six cards. The cards are six in color ( six each) and each color is numbered from 1 to 6. You draw two cards at random. What is probability that they are of a different color and have a different number?
A: The first card can be drawn at random. It does not matter what its color or number is. To compute the probability that the second card is different in color and number from the first, it helps to visualize the situation in a simple way as shown below.
In the figure above, assume the green dot represents the card that was picked. The marked out cards represent the cards that should not be picked to get a different color and number. Also, the act of picking a card bought down the pool of cards from 36 to 35. The remaining unmarked space represents the available set of cards to pick from. This can be computed easily as
This yields an overall probability of
If you are interested in learning the art of probability, some of the best books to learn it from are listed here.
## Wednesday, October 26, 2016
### The Magic of Numba and Python
Python is a great programming language. It's primary merits are readability and the numerous packages that are available online. However, being an interpreted language the speed of execution is always an issue. Here is a simple example of a piece of code written in Python that tries to add two numbers from a grid.
#!/usr/bin/python
def overlap_pp(x,y):
count = 0
for i in range(x):
for j in range(y):
count += i + j
return count
for _ in range(1000):
q = overlap_pp(500,500)
If you run the above script (saved as n.py) on the command line terminal with the time command you should see some numbers like the below
time ./n.py
real 0m22.379s
user 0m22.363s
sys 0m0.407s
The process running on one processor took about 22 seconds to complete the run. Now lets do something seemingly magical. Lets add a decorator. Decorators are python speak for a mechanism to do wrapper functions. The decorator we are going to add is '@jit'. The code looks like as it is shown below
#!/usr/bin/python
from numba import jit
@jit
def overlap_pp(x,y):
count = 0
for i in range(x):
for j in range(y):
count += i + j
return count
for _ in range(1000):
q = overlap_pp(500,500)
Now let's time this as before.
time ./n.py
real 0m0.420s
user 0m0.370s
sys 0m0.443s
Magic! The exact same code now runs 55x faster! This happens because the Just-In-Time (jit) feature of the numba package compiles the function to machine code on the fly. The timing you see is almost in line with what you would expect from a similar code done in C or Fortran. In fact you will be surprised that in this particular example, the numba version runs faster than the C version! The C code is shown below.
#include < stdio.h >
int overlap_pp(int x,int y){
int i,j,q;
for(i=0;i<= x;i++){
for(j=0;j<= y;j++){
q += i + j;
}
}
return(q);
}
int main(int argc,char** argv){
int i,q;
for(i = 0;i<=1000;i++){
q = overlap_pp(500,500);
}
return 0;
}
The timing results after compiling the above code 'gcc t.c -o t'
time ./t
real 0m0.774s
user 0m0.774s
sys 0m0.000s
Unfortunately there isn't a clear description online on how to get Numba installed on Ubuntu 14.04. After some hacking and poking around stackoverflow the following worked for me.
Step 1:
Numba uses llvm. You need to remove all llvm, llvm-config and llvmlite installations that you already have on your system. Numba needs llvm-3.8 upwards to work properly.
Step 2:
Follow instructions in this stackoverflow thread. But when you get installing do the following
sudo apt-get install zlib1g zlib1g-dev libedit2 libedit-dev llvm-3.8 llvm-3.8-dev
Notice change in version number to 3.8.
Step 3:
Next, install llvmlite. You also want to create a symbolic link on '/usr/bin/' to point to the latest 3.8 version of llvm-config.
"sudo ln -s /usr/bin/llvm-config-3.8 /usr/bin/llvm-config"
This step is important because the installation of numba appears to use llvm-config. After this you can install Numba directly using pip.
If you are interested in learning more about python programming, data science and probability here are a list of books that are worthwhile to buy.
## Sunday, May 15, 2016
### The James-Stein Estimator
This write up is about an estimator. A statistical estimator, is used when you already have a model in mind, data at hand and want to estimate some parameters needed for the model. For example, you want to predict how many runs a batter would make in a game given recent history $$x_1,x_2,x_3,\ldots$$ . If we assume (we are making a choice of a model now) that the scores come from a normal distribution with mean $$\mu$$ and standard deviation $$\sigma$$ then the probability density function for a given value $$x$$ is
The likelihood that a series of points $$x_1,x_2,x_3,\ldots$$ come from such a distribution can be expressed as
Next is basic calculus. Take the logarithm on both sides, set the partial derivative w.r.t. $$\mu$$ to zero yields (excluding the algebra)
To verify, you also need to check the second derivative's sign to see if its negative to ensure that it is indeed a maxima you have found.
So a simple estimator would be to use the average runs scored from the past $$n$$ days. The average is the result of a "maximum likelihood" approach briefly described above.What if you had 5 players you wanted to estimate the average for? Intuition would tell you that you should simply compute the average for each. Wrong! The James-Stein approach provides a way to pool the data such that the overall error made in estimation is minimized.
Specifically, I'll demonstrate the James-Stein estimator. It is a surprising result discovered by Charles Stein and later formalized by Willard James on estimating such parameters. What makes it stunning is the rather counter intuitive result it demonstrates. James-Stein's estimator approach states that if you wanted to simultaneously estimate a set of independent parameters from data, the maximum likelihood approach is only optimal if you have lesser than 3 parameters to estimate from! If you have more than 3, you are better off using the James-Stein estimator. There is an obvious thought experiment you can think of. If you really wanted just one parameter, could you not simply add an unrelated set of numbers, improve efficiency in estimation and then just use the parameter you are interested in? That's a flaw. The estimator works by minimizing the overall error in estimation so it may make more errors in some of the variables, lesser on others. But overall it will be better than just doing an individual maximum likelihood estimate on each variable. So you could use it if you don't mind seeing slightly bigger errors on some variables, smaller on others but lower overall error.
To see this work, I'll step through the actual algorithm in R. For more details you can refer the wikipedia entry here.The general approach is the following.
Let
• $$\hat\theta_{js}$$ is the James-Stein estimate of the parameters we are interested in computing.
• $$y$$ be the set of values observed.
• $$\hat{\sigma}^{2}$$ be the estimated variance for all parameters
• $$v$$ be the mean of all parameters
then the James Stein estimator is given by
#!/usr/bin/Rscript
suppressMessages(library(data.table))
suppressMessages(library(ggplot2))
# n.rows is the number of samples we have
n.rows = 100
# n.params is the number of parameters
# we want to estimate
n.params = 30
# wins will hold the number of times
# the JS estimator beats the MLE estimate
wins = c()
msejs = c()
msemle = c()
for(iter in seq(1:1000)){
# Create a sample of parameters
# They have a range of 20-25
x.act = sample(20:25,size=n.params,replace=TRUE)
# Now create a normal distribution for each of the parameters
# uncomment below if you want to test it for a poisson distribution
# m = mapply(rpois,mean=x.act,MoreArgs=list(n = n.rows))
m = mapply(rnorm,mean=x.act,MoreArgs=list(n = n.rows,sd=10))
# Find the global mean
mbar = mean(colMeans(m))
# Find the column means
mu0 = colMeans(m)
# Find global variance
s2 = var(as.vector(m))/(n.rows*n.params)
cval = 1 - ((n.params - 2)*s2/sum((mu0 - mbar)^2))
# Compute the JS estimate for your parameters
jsest = mbar + cval *(mu0 - mbar)
z = data.table(
actual = x.act,
mle.est = mu0,
js.est = jsest)
# Check to see if the JS estimate is better than MLE
z[,counter := ifelse(abs(js.est - actual) < abs(mle.est - actual),1,0)]
# In case you want to see what the numbers are for the
# difference between absolute and actual estimates for
# JS and MLE
z[,jserr := abs(js.est - actual)]
z[,mleerr := abs(mle.est - actual)]
# Record the wins for this iteration of the simulation
# Repeat.
wins[iter] = sum(z$counter) msejs[iter] = sum(z$jserr)
msemle[iter] = sum(z\$mleerr)
}
# What are the mean wins?
mean(wins)
# What are the distribution of the mean wins
quantile(wins,prob = seq(0.1,0.9,by=0.1))
z = data.frame(wins = wins)
p = ggplot(z,aes(wins)) +
geom_histogram(fill='light blue') +
theme_bw() +
ggtitle('Of the 30 parameters we wish to estimate:\n how many of them have estimates closer to the actual using the James-Stein estimator than the MLE?') +
ylab('') +
xlab('') +
geom_vline(aes(xintercept=mean(wins),linetype='longdash',colour='blue')) +
annotate('text',x=18,y=150,label='Mean -->') +
theme(legend.position='none')
png('xyz.png',width=800,height=500)
print(p)
dev.off()
The above R code runs a simulation of sorts. It starts by making some random parameters you would want to estimate and simulates some normally distributed data from it. Next, it uses the James Stein estimator and estimates the very parameters it started off with, using the data. Finally, it compares and records how often the James Stein estimator was better/closer that the MLE estimate. The results from the simulation are shown below.
For kicks, you can try it for a Poisson distribution too, here is what the distribution looks like.
If you are interested in learning more about probability here are a list of books that are good to buy.
## Friday, November 28, 2014
### Bayesian Cognitive Bias
I chanced on this excellent puzzle on the net that tends to reveal a cognitive bias in our heads against Bayesian reasoning. The puzzle statement is quite simple
You are given four cards. Each card has a letter on one side and number on the other side. You are told the statement "If there is a D on one side, there is a 5 on the other side". Which two cards would you flip over to validate the statement?
The original article is here, think hard before you click through for an answer :)
## Tuesday, September 2, 2014
### Maximizing Chances in an Unfair Game
Q: You are about to play a game wherein you flip a biased coin. The coin falls heads with probability $$p$$ and tails with $$1 - p$$ where $$p \le \frac{1}{2}$$. You are forced to play by selecting heads so the game is biased against you. For every toss you make, your opponent gets to toss too. The winner of this game is the one who wins the toss the most. You, however get to choose the number of rounds that get played. Can you ever hope to win?
Machine Learning: The Art and Science of Algorithms that Make Sense of Data
A: At a first look, it might appear that the odds are stacked against you as you are forced to play by choosing heads. You would think that your chances or winning decrease as you play more and more. But, surprisingly there is a way to choose the optimal number of tosses (remember, you get to choose the number of times this game is played). To see how, lets crank out some numbers. If you get to toss the coin $$n$$ times, then the total number of coin tosses you and your opponent flips is $$2n$$. Out of the $$2n$$ tosses if $$y$$ turns out heads, the probability that you would win is
$$P(\text{y Wins}) = {2n \choose y} p^{y}(1 - p)^{2n - y}$$
In order to win, the value of $$y$$ should run from $$n + 1$$ to $$2n$$ and the overall probability works out to
$$P(\text{Win}) = \sum_{y = n + 1}^{2n}{2n \choose y} p^{y}(1 - p)^{2n - y}$$
We can work out the probability of winning by choosing various values of $$p$$ and $$n$$ and chart them out. Here is the R code that does it.
The code runs pretty quickly and uses the data.table package. All the processed data is contained in variables z and z1. They are plotted using the ggplot package to generate the following charts for the strategy.
The first chart shows the variation of the probability of winning by the number of games played for various probability bias values.
The next chart shows the optimal number of games to play for a given bias probability value.
Some good books to own for learning probability is listed here
Yet another fascinating area of probability are Monte Carlo methods. Here are a list of good books to own to learn Monte Carlo methods.
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2017-01-22 20:24:08
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https://brilliant.org/discussions/thread/nmtc-problem-3b/
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×
# NMTC Problem 3b
If x,y,z are each greater than 1, show that
$$\frac { { x }^{ 4 } }{ { (y-1) }^{ 2 } } +\frac { { y }^{ 4 } }{ { (z-1 })^{ 2 } } +\frac { { z }^{ 4 } }{ { (x-1) }^{ 2 } } \ge 48$$
This a part of my set NMTC 2nd Level (Junior) held in 2014.
Note by Siddharth G
3 years, 2 months ago
MarkdownAppears as
*italics* or _italics_ italics
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Note: you must add a full line of space before and after lists for them to show up correctly
paragraph 1paragraph 2
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[example link](https://brilliant.org)example link
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This is a quote
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print "hello world"
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# up as a code block.
print "hello world"
MathAppears as
Remember to wrap math in $$...$$ or $...$ to ensure proper formatting.
2 \times 3 $$2 \times 3$$
2^{34} $$2^{34}$$
a_{i-1} $$a_{i-1}$$
\frac{2}{3} $$\frac{2}{3}$$
\sqrt{2} $$\sqrt{2}$$
\sum_{i=1}^3 $$\sum_{i=1}^3$$
\sin \theta $$\sin \theta$$
\boxed{123} $$\boxed{123}$$
Sort by:
Firstly, we have, $$(a-2)^2 \geq 0 \implies a^2 -4a + 4 \geq 0 \implies a^2 \geq 4(a-1) \implies \frac{a^2}{a-1} \geq 4 \implies \frac{a^4}{(a-1)^2} \geq 16$$
Now, By AM-GM,
$$\frac{x^4}{(y-1)^2} + \frac{y^4}{(z-1)^2} + \frac{z^4}{(x-1)^2} \geq 3\sqrt[3]{\frac{x^4}{(y-1)^2}* \frac{y^4}{(z-1)^2}* \frac{z^4}{(x-1)^2} } = 3\sqrt[3]{\frac{x^4}{(x-1)^2}* \frac{y^4}{y-1)^2}* \frac{z^4}{(z-1)^2} } \geq 3\sqrt[3]{16*16*16} = 3 * 16 = 48.$$
- 3 years, 2 months ago
- 3 years, 2 months ago
Is the AM-GM step necessary? You could just say $$\frac{x^4}{(x-1)^4} \geq 16$$ and so on for y and z and add the 3 inequalities together right?
- 3 years ago
But in the question, the denominator and the numerator are of different variables, which makes the AM-GM necessary to bring the denominator and the numerator with the same variables together. Thus $$\frac { { x }^{ 4 } }{ { (x-1) }^{ 2 } } \ge 16$$ can only be used after the AM_GM step.
- 3 years ago
You're right. Just skipped over that for some reason ^o^
- 3 years ago
×
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2018-01-23 08:18:39
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http://mathhelpforum.com/statistics/17740-hits-ratings-averages.html
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# Math Help - Hits,ratings & averages
1. ## Hits,ratings & averages
Hi there, ive been posting some videos to the metacafe producer awards program. The idea is that you get paid for your video when you hit 20 000 views if your average rating is above 3/5.Each member of metacafe can rate your video out of 5.
The other night before i went to sleep i checked one of my vids and it was on 15000 views and had an average rating of 4.2/5.
When i woke up i had only had 350 extra views but my average rating had fallen to 3.54/5.
So ive been wondering how to work out what the average rating that the last 350 people had given my video. I realise this wouldnt be very accurate over 350 people so ive written a program that checks the webpage with the video on every 1 min and retrieves the number of views and the current average rating. Theres no way i can make sure i get the new rating each time a person watchs the video but the average rating given by the 2 or 3 views i might get in 1 min should be accurate enough to get an idea of any voting patterns.
Thanks.
2. Originally Posted by vitaloverdose
Hi there, ive been posting some videos to the metacafe producer awards program. The idea is that you get paid for your video when you hit 20 000 views if your average rating is above 3/5.Each member of metacafe can rate your video out of 5.
The other night before i went to sleep i checked one of my vids and it was on 15000 views and had an average rating of 4.2/5.
When i woke up i had only had 350 extra views but my average rating had fallen to 3.54/5.
So ive been wondering how to work out what the average rating that the last 350 people had given my video. I realise this wouldnt be very accurate over 350 people so ive written a program that checks the webpage with the video on every 1 min and retrieves the number of views and the current average rating. Theres no way i can make sure i get the new rating each time a person watchs the video but the average rating given by the 2 or 3 views i might get in 1 min should be accurate enough to get an idea of any voting patterns.
Thanks.
There is no way the average rating could have fallen to 3.54 from 4.2 with only 350 additional views from 15000. The formula to find the average rating $r$ for the last 350 views is
$\frac{350 \times r + 15000\times 4.2}{15350} = 3.54.$
This says the new average of 3.54 is the weighted average of 4.2 for the first 15000 views and r for the last 350 views. Solving,
$r = \frac{15350\times 3.54 - 15000 \times 4.2}{350} = -24.75,$
which isn't possible. So if you have given us accurate numbers, then there is an error in the calculations for the web page.
Modify the last formula to calculate the average rating for the last few views.
3. Thanks for taking the time out to explain how thats worked out.
Ive checked the figures with a friend who was also keeping track of the ratings for this vid and they agree they are correct. The only thing that might be slightly out is the amount of views i got over night. But that wont be out by much and with an average of -24 i doubt think that adding an extra 1000 views to that formula will produce a result that is acctually possible or very probable and theres no way im out by that much.
With any luck my program will shed some light on whats going on here.
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2015-10-14 00:00:05
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https://www.physicsforums.com/threads/potential-in-case-of-concentric-shells.849558/
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# Homework Help: Potential in case of concentric shells
1. Dec 25, 2015
### gracy
1. The problem statement, all variables and given/known data
A,B and C are three concentric metal shells of radii a,2a and 3a .Shell A is the innermost and shell C is the outermost .Shell A is given a charge q and C is earthed.Find the final potential of shell B.
2. Relevant equations
$V$=$\frac{Kq}{r}$
3. The attempt at a solution
Charge q' is supplied from earth to shell.
Then charges will be induced
Now we have to find q'
Potential of shell C should be zero.
$\frac{K(q'+q)}{3a}$ - $\frac{Kq}{3a}$ +$\frac{Kq}{a}$ - $\frac{Kq}{a}$ +$\frac{Kq}{2a}$ =0
$\frac{K(q'+q)}{3a}$ - $\frac{Kq}{3a}$ +$\frac{Kq}{2a}$ =0
$\frac{K(q'+q)-Kq+\frac{3Kq}{2}}{3a}$=0
$K(q'+q)$$-Kq$+$\frac{3Kq}{2}$=0
$\frac{2K(q'+q)-2Kq+3Kq}{2}$=0
$2K(q'+q)$-$2Kq$+$3Kq$=0
$2K(q'+q)$+$Kq$=0
$2K(q'+q)$=$-Kq$
$2(q'+q)$=$-q$
$q'+q$=$\frac{-q}{2}$
$q'$=$\frac{-q}{2}$-q
$q'$=$\frac{-3q}{2}$
Now by putting value of q' we can find potential of shell B by all the charges present.Am I right till here?
2. Dec 25, 2015
### cnh1995
As per my understanding, you should consider "net charge" on the spheres. Potential on any sphere will be due to the net charge q on sphere A and q' on sphere C. In your first equation, your first term represents the potential of sphere C due to its own (net)charge, which is q'. Why have you taken q+q' then? Also, sphere B will have no net charge, hence will not contribute to any potential. So,why is there kq/2a in the first equation?
3. Dec 25, 2015
### theodoros.mihos
Shells are thin so there is no radial charge distribution. There is only surface charge distribution that have the same value by spherical symmetry. Electric field between a and c can calculate be Gauss law and we know that potential on c is zero.
4. Dec 25, 2015
### cnh1995
Sphere B is outside the sphere A but inside the sphere C. This means potential of B due to sphere C will be same as the potential of C due to its own (net)charge q'.
5. Dec 25, 2015
### gracy
Potential can also be due to induced charges.
6. Dec 25, 2015
### gracy
qi is supplied by earth and +q is charge induced due to -q present on inner surface of shell C.
7. Dec 25, 2015
### cnh1995
When charges are induced, both +ve and -ve charges appear on the sphere. When it is said that induced +ve charge flows to the ground, that means ground has supplied extra -ve charge to the body.
8. Dec 25, 2015
### cnh1995
+q and -q will form net 0 charge. It will be only q' supplied by the ground that will be responsible for potential.
9. Dec 25, 2015
### haruspex
Your expression for the potential at C is all wrong.
Hang onto these facts for uniformly charged conducting spherical shells:
- outside the shell, the potential and field generated by the charge on the shell is exactly the same as if the whole charge were concentrated at the centre of the sphere. (So what is the potential at C that results from the charge on A?)
- there is no field between the inner and outer radii of the shell, so the potential is constant in the annulus.
- there is no field generated by the charge on the shell inside the shell, so the potential it generates there is also constant and equal to the potential it generates at the shell itself. (Don't confuse this with the net potential at the shell.)
10. Dec 26, 2015
### Staff: Mentor
Gracy, if shell C were not grounded, do you agree that the charge induced on its outer surface by charge separation due to the charge on shell A would have been +q?
If so, then adding the ground connection would allow that +q charge to "escape" to ground. Or, thought of in another way, ground would supply a charge of -q to cancel that +q charge. The end result is that a charge of -q moves onto shell C through the ground path. This charge of -q serves to terminate the field lines of the interior charge +q of A shell, and so shell C exhibits a zero potential to the external world.
So you end up with shell A holding a net charge of +q, shell B holding a net charge of 0, and shell C holding a net charge of -q.
Given these shell charges you can apply the facts listed by @haruspex to find the potential at the location of shell B. Note in particular that an isolated uncharged shell (such as shell B) will not contribute to the potential. Only net charge on a shell will do so. The fields created by charge separation serve to cancel the field inside the conductor comprising the shell (so no currents continue to flow) and matches the external field at the boundaries of the conductor. That is to say, superposition of the external field and the field produced by the separated charges renders the net field inside the conductor (within the shell material) zero, but doesn't change the external field or potential.
In the end, the potential at the position of shell B will be the superposition of the potentials due to the net charges of shell A and C.
11. Dec 27, 2015
### gracy
How?I mean electric field lines of interior charge +q of A shell have to be terminated on inner surface of shell B as electric field can not exist in metal body hence electric field lines of charge +q of A shell would not able to go beyond inner surface of shell B.
12. Dec 27, 2015
### cnh1995
Electric field will not exist in the thickness(however small it is) of the sphere. So, when -q charge is induced on the inner surface of B, +q charge is induced on the outer surface. This makes the field 0 inside the thickness of the sphere. However, as now +q is on the outer surface of B, its field lines will reach out to sphere C. Here, there is a 'discontinuity' in the field due to the thickness of the sphere. The field wouldn't be trapped inside any sphere, if sphere C weren't grounded.
13. Dec 27, 2015
### haruspex
Yes, you can think of it as lines from A terminating on inner surface of B, but then being regenerated from the outer surface of B. Or, as gneill suggests, you can note that B has no net charge, so (as far anything except B itself is concerned) it is the same as the lines from A going all the way out to C. Work with whichever you are comfortable with.
14. Dec 27, 2015
### gracy
Then why there is potential due to induced charges in the video I have mentioned?
15. Dec 27, 2015
### cnh1995
The charge in the video is external to the sphere. Hence, to make the potential of every point inside the sphere same, there will be potential due to the induced charges. The net potential of every point inside the sphere will be kq/r. Out of that, kq/x will be due to the external +q and the rest will be due to the induced charges. In fact, to make the potential equal to kq/r, the charges will be induced.
In case of concentric spheres(your original question), the charge is inside the spheres B and C and 'on' the sphere A.
Consider only spheres A and B:
Charges will be induced on inner and outer surface of B to make the potential inside B equal at every point, thus making
Einside B=0.
So, if you pick any point inside sphere B at a distance x from the center, at that point, the potential will be kq/2a. Out of that potential, kq/x will be contributed by +q on sphere A and the rest will be contributed by the induced charges on sphere B. But
the net potential of sphere B is kq/2a. That is what is asked in your original question.
I haven't considered sphere C here. You can extend this logic to sphere C and get the final answer.
16. Dec 27, 2015
### cnh1995
In short, potential of the sphere is decided by the net charge present in the system(and radius of the sphere(s)). Induced charges only help to establish this pre-decided potential, wherever necessary. At the center of the sphere(in the video above), contribution due to the induced charges is 0 since only +q charge outside is sufficient( or +q is the one to decide the potential at the center) and the induced charges are symmetrical and equidistant from the center. Anywhere else inside the sphere, potential due to the +q outside is not kq/r. So, to make it kq/r, charges are induced on the sphere. So, in your original question, it is only the net charge that will decide the potential of the spheres. So, net charge is the one who will decide the potential and induced charges will simply help to execute the task of establishing that potential everywhere inside the sphere and make Einside=0.
Last edited: Dec 27, 2015
17. Dec 27, 2015
### gracy
I am not getting then how potential only depends on net charge present and not on induced charges .Isn't induced charge playing role equal to that of net charge present in determining potential?
Last edited: Dec 27, 2015
18. Dec 27, 2015
### cnh1995
Magnitude of the net potential is decided by the net charge. Induced charges appear to establish that potential inside the sphere, wherever necessary. So, role of induced charges is vital in "establishing" the net potential(which is already determined by the net charge) and not in determining it. Of course, the potential due to induced charges will vary at each point but the total potential at any point will remain same i.e. kq/r.
Last edited: Dec 27, 2015
19. Dec 27, 2015
### haruspex
It's hard to find words that distinguish between the cavity inside the inner surface of a shell and the region between the surfaces of the shell. I understand that here you are referring to the latter, but it might not be clear. I use the term annulus to describe this.
You could further elaborate that the potential in the annulus due to induced charges on the inner surface exactly cancel the potential due to A, while those on the outer surface produce the constant potential kq/2a.
20. Dec 27, 2015
### SammyS
Staff Emeritus
https://www.physicsforums.com/posts/5325447/
(A charge, q, external to a neutral conducting sphere. Asks for E-field and or potential due to induced charges.)
The question asked in the current thread is not the same. The induced charges here have simple distributions. That's not the case for the video.
21. Dec 28, 2015
### gracy
You mean there is potential due to induced charge but it's just that it does not exist in final formula(formula of net potential).When we have to calculate net potential we should ignore induced charge but when we are specifically asked potential due to induced charge we will have to consider induced charge then!
Here I am asked FINAL (net ) potential on shell B
Hence we will only consider charges which are actually present and will neglect induced charges
Suppose charge q' has been supplied by earth as we know shell C has been grounded it's potential should be zero
$V_C$=$\frac{Kq}{3a}$+$\frac{Kq'}{3a}$=0
Therefore q'=-q
$V_B$=$\frac{Kq}{2a}$ - $\frac{Kq}{3a}$
=$\frac{Kq}{6a}$
Right?
22. Dec 28, 2015
### haruspex
Not sure which post you are responding to, but nobody said that. All charges, induced or otherwise, result in potentials. Some induced charges may neutralise potentials due to other charges. In the case of the charges induced on the inner and outer surfaces of B, they neutralise each other (everywhere except in the annulus between them).
Yes.
23. Dec 28, 2015
### cnh1995
As haruspex said, we don't "neglect" the potential due to the induced charges. But since their contribution varies at each point, there isn't a general formula( like kq/r for net charge). Hence, only the net charge appears in the formula since it decides the net potential. In the net potential kq/r, potential due to induced charges is already included,hence, there's no need to calculate it separately if the "net potential" is asked. But to calculate the contribution of induced charges, as you said, you should consider the induced charges and the geometry of the system. Seeing the math you've done,
I believe you've understood the concept very well now!
24. Dec 28, 2015
### gracy
Which sentence is wrong here?
25. Dec 28, 2015
### SammyS
Staff Emeritus
Which post are you responding to ?
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2018-07-16 13:17:27
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https://tex.stackexchange.com/questions/469527/create-tikz-picture-with-these-specifications
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# Create Tikz Picture With These Specifications
Ciao, I am trying to create a tikz figure as shown in the image. Here is my code
\documentclass[border=4mm]{standalone}
\usepackage{tikz}
\usetikzlibrary{arrows.meta,positioning}
\begin{document}
\begin{tikzpicture}[
\node[rectangle] (a1) {box 1};
\node[rectangle,below right=of a1] (a2) {box 2};
\node[rectangle,above right=of a1] (a3) {box 3};
\foreach \i/\j/\txt/\p in {% start node/end node/text/position
a1/a2/1-2/below,
a1/a3/1-3/above,
a2/a3/2-3/below} \draw [myarrow] (\i) -- node[sloped,font=\small,\p] {\txt} (\j);
\end{tikzpicture}
\end{document}
I am not sure how to add the rectangles; how to connect the arrows to the side edge; and how to make the arrows straight in situations when the box lengths are different--preferably the arrows will touch the center of the edge.
New contributor
victoria is a new contributor to this site. Take care in asking for clarification, commenting, and answering. Check out our Code of Conduct.
Use Tikz shapes library and describe the sides of nodes.
\documentclass[margin=4mm]{standalone}
\usepackage{tikz}
\usetikzlibrary{arrows.meta,positioning,shapes}
\begin{document}
\begin{tikzpicture}[>=latex]
\tikzstyle{rect}=[ thick, draw=blue, rectangle, minimum width=100pt,
minimum height = 50pt, align=center]
\node[rect] (a1) {box 1};
\node[rect,below right=of a1] (a3) {box 3};
\node[rect,above right=of a1] (a2) {box 2};
\draw[->] (a1.north)--(a2.west)node[midway,above,xshift=-2mm]{1-2};
\draw[->] (a1.south)--(a3.west)node[midway,below,xshift=-2mm]{1-3};
\draw[->] (a2.south)--(a3.north)node[midway,xshift=-3mm]{2-3};
\end{tikzpicture}
\end{document}
• thanks a bunch. Two final questions; how can i slant 1-3 and 1-2 so they are angled with the arrows lines? Like this: i.stack.imgur.com/Dwv1Y.png – victoria Jan 10 at 13:52
• @victoria, your welcome. Add sloped option to node, i.e, node[midway,sloped,above,xshift=-2mm]{1-2};Now you can remove the xshift=-2mm. – ferahfeza Jan 10 at 13:55
|
2019-01-17 00:05:02
|
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9001797437667847, "perplexity": 11493.389230560058}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-04/segments/1547583658662.31/warc/CC-MAIN-20190117000104-20190117022104-00120.warc.gz"}
|
https://math.stackexchange.com/questions/756462/error-solving-stars-and-bars-type-problem/756501
|
# Error solving “stars and bars” type problem
I have what I thought is a fairly simple problem: Count non-negative integer solutions to the equation
$$x_1 + x_2 + x_3 + x_4 + x_5 = 23$$
such that $0 \leq x_1 \leq 9$.
Not too hard, right? Simply ignore the upper-bound, count the $$\begin{pmatrix}23 + (5-1) \\ (5-1)\end{pmatrix} = \begin{pmatrix}27 \\ 4\end{pmatrix} = 17550$$ solutions. Subtract from this all (non-negative integer) solutions to the equation $$y_1 + 10 + x_2 + x_3 + x_4 + x_5 = 23,$$ and there are $\begin{pmatrix}17 \\ 4\end{pmatrix} = 2380$ of these "bad" solutions we shouldn't have counted earlier, but did. Thus we find $17550 - 2380 = 15170$ solutions.
Since this is a prohibitively large number of solutions to check by hand, I wrote a simple Python program to verify whether this answer is correct. It does indeed say there are $17550$ solutions without upper bounds, and $2380$ solutions to the equation for counting "bad" solutions.
However, when I ask it throw away all solutions to the non-upper-bounded problem for which $x_1 \geq 10$, it tells me it's found $15730$ solutions.
My question is: do I not understand the combinatorial calculation so that there are not actually $\begin{pmatrix}27\\4\end{pmatrix}-\begin{pmatrix}17\\4\end{pmatrix}$ solutions, or have I made some kind of programming mistake? Of course, both are also possible.
Your method is correct, there are 15170 solutions. It seems like your python script is doing something wrong at the end.
I wrote a C++ program myself to confirm this:
#include <stdio.h>
int main()
{
int a,b,c,d,e,sum=0;
for (a=0; a<10; a++)
for (b=0; b<24; b++)
for (c=0; c<24; c++)
for (d=0; d<24; d++)
for (e=0; e<24; e++)
{
if (a+b+c+d+e == 23)
sum++;
}
printf("sum=%d",sum);
}
And it prints out sum=15170.
• Thank you, I should have suspected the program, given that it took me longer than expected to get it up and running. I'll have to take some time and see what's gone wrong there. – pjs36 Apr 16 '14 at 19:12
A nice way to mathematically check your work is with a generating function. We have $x_{1} \in \{0, ..., 9\}$. So our generating function for an $x_{1}$ is $\sum_{i=0}^{9} x^{i} = \dfrac{1-x^{10}}{1-x}$. The other terms don't have this restriction, so their generating functions are simply $\dfrac{1}{1-x}$.
So we multiply this result for each $x_{i}$ to get our generating function $f(x) = \dfrac{1-x^{10}}{(1-x)^{5}}$.
Using our binomial identities, we get $(1-x^{10}) = \sum_{i=0}^{1} \binom{1}{i} (-1)^{i} x^{10i} = (1 - x^{10})$
We then expand out the bottom term $\dfrac{1}{(1-x)^{5}} = \sum_{i=0}^{\infty} \binom{i + 5 - 1}{i} x^{i}$. And we multiply this with $(1 - x^{10})$, taking only terms $x^{23}$.
So that gives us $\binom{23 + 5 - 1}{23} - \binom{13 + 5 - 1}{13} = \binom{27}{4} - \binom{17}{4}$ as our answer.
Generating functions are nice, because they make it much easier to model the problem, and you don't have to crank out code and guess if it's right. Of course, seeing the stars and bars solution is nice, but takes more practice. Generating functions are more mechanical once you've figured out how to deal with them.
• I've seen generating functions used for these problems, and it looks like I'd better bite the bullet and figure them out. I dealt with them years ago, but never fully "got it" or figured out how to work with them successfully. Clearly, they're worth the effort! – pjs36 Apr 16 '14 at 19:16
• There is a good tutorial on ordinary generating functions- dreamincode.net/forums/topic/… – ml0105 Apr 16 '14 at 19:56
$\displaystyle S=\sum_{x_1=0}^9 \{x_2+x_3+x_4+x_5=23-x_1\}=\sum_{x_1=0}^9 {26-x_1\choose 3}=\sum_{x_1=0}^9 {17+x_1\choose 3}$
$\displaystyle\quad= \sum_{x_1=0}^9 \left[{18+x_1\choose 4}-{17+x_1\choose 4}\right]={27\choose 4}-{17\choose 4}$
|
2019-08-21 02:34:24
|
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.714066207408905, "perplexity": 485.65593458218365}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-35/segments/1566027315750.62/warc/CC-MAIN-20190821022901-20190821044901-00282.warc.gz"}
|
http://clay6.com/qa/20580/at-150-oc-the-decomposition-of-acetaldehyde-to-methane-is-first-order-react
|
Browse Questions
# At $150^oC$ the decomposition of acetaldehyde to methane is first order reaction. If rate constant for reaction at $750^oC$ is $0.05m^{-1}$, how long does it take a concentration of $0.04\;mol\;L^{-1}$ of acetaldehyde to reduce concentration of $0.03\;mol\;L^{-1}$?
$\begin{array}{1 1}(a)\;2.75min&(b)\;3.75min\\(c)\;4.75min&(d)\;5.75min\end{array}$
For first order kinetics
$K\times t=ln\large\frac{a}{a-x}$
$a$=initial concentration=0.04
$a-x$=reactant concentration at t time=0.03
$0.05\times t=ln\big(\large\frac{0.04}{0.03}\big)$
$t=5.75min$
Hence (d) is the correct answer.
|
2017-07-24 02:34:43
|
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8908511400222778, "perplexity": 1689.9991780487396}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 5, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-30/segments/1500549424683.39/warc/CC-MAIN-20170724022304-20170724042304-00662.warc.gz"}
|
http://www.qfak.com/education_reference/words_wordplay/?id=b806594
|
# Is Mrs. correct for calling someone that's married to a man?
Or is it Ms.?
Mrs. is correct
May 16 at 22:26
It"s Mrs. Ms (Miss) is for single women.
May 17 at 2:12
Mrs. (from Mistress) is the usual title for a married woman; Ms. doesn't specify whether or not she's married.
(And I'm not going to mention the doctorates/other gender-neutral titles or that you didn't specify "some woman," but that's just from hanging with gay friends...)
May 17 at 6:21
Mrs. is the formal term as it is derived from Mister's or Mr's.
But Ms. is just as respectful as Mrs. as some people would preferred not to be objectified. However if you don't know if someone is married or not its safe to call someone Ms.
May 17 at 10:53
|
2014-04-23 12:26:39
|
{"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8184053897857666, "perplexity": 5682.415376378747}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2014-15/segments/1398223202548.14/warc/CC-MAIN-20140423032002-00558-ip-10-147-4-33.ec2.internal.warc.gz"}
|
https://gateoverflow.in/318961/ullman-compiler-design-edition-exercise-question-page-232
|
Every language that has a context-free grammar can be recognized in at most $O(n^{3})$ time for strings of length $n$. A simple way to do so,called the Cocke- Younger-Kasami (or CYK) algorithm is based on dynamic programming. That is, given a string $a_{1}a_{2}\cdot\cdot\cdot a_{n}$, we construct an $n$-by-$n$ table $T$ such that $T_{ij}$ is the set of nonterminals that generate the substring $a_{i}a_{i+1}\cdot\cdot\cdot a_{j}$. If the underlying grammar is in CNF (see \question $4.4.8$), then one table entry can be filled in in $O(n)$ time, provided we fill the entries in the proper order: lowest value of $j - i$ first. Write an algorithm that correctly fills in the entries of the table, and show that your algorithm takes $O(n_{3})$ time. Having filled in the table, how do you determine whether \$a_{l}a_{2}\cdot\cdot\cdot a_{n} is in the language?
|
2019-12-10 12:14:38
|
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8744691610336304, "perplexity": 483.97476043135316}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-51/segments/1575540527205.81/warc/CC-MAIN-20191210095118-20191210123118-00385.warc.gz"}
|
http://mathhelpforum.com/statistics/54462-probability.html
|
1. ## probability
let 1<=r<=n. A subset of{1,2....n} of cardinality "r" is chosen at random. How do i calculate the probability that 1 is an element of the chosen subset
2. Originally Posted by math8553
let 1<=r<=n. A subset of{1,2....n} of cardinality "r" is chosen at random. How do i calculate the probability that 1 is an element of the chosen subset
There are ${{n-1} \choose {r-1}}$ subsets of $\left\{ {1,2,3, \cdots ,n} \right\}$ that contain the element 1.
${{n} \choose {r}}$
and the number of those that doesn't contain 1 is ${{n-1} \choose {r-1}}$ as Plato told you.
The probability of taking a set that doesn't contain 1 is ${{n-1} \choose {r-1}}/{{n} \choose {r}}$
|
2017-07-22 02:55:08
|
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|
http://crypto.stackexchange.com/questions?page=90&sort=votes
|
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2014-11-28 20:35:45
|
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|
http://physics.stackexchange.com/tags/noethers-theorem/new
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# Tag Info
1
You should probably read up on the Stueckelberg action and the Affine Higgs mechanism it sends you to. Your boldface supposition "I'm not supposing that my matter has any global symmetry here, that I might be able to gauge" is unwarranted for the specific model you propose, $J_\mu=\partial_\mu\phi$. There is a global symmetry, $\phi \to \phi+\alpha$, whose ...
0
You're right that that Lagrangian isn't in general gauge invariant. In addition to making the $A^\mu$ terms gauge invariant, the $\mathcal{L}(J)$ term must also be gauge-invariant. And not just the equations of motion for $J_\mu$, either - the specific algebraic expression for $J_\mu$ in terms of fundamental fields must be such that if you literally plug ...
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You applied it already Decrease in gravitational potential energy of block 2 = Increase in gravitational potential energy of block 1 + Increase in kinetic energy of block 1 + Increase in kinetic energy of block 2 How do you justify this statement? You have to invoke the conservation of energy. If you wanted to work Nother's theorem into this, it ...
1
Although users35736's answer is certainly correct, and the question is old, I think it should be noted that each Killing vector, $\xi^i$, also gives rise to a Noether current: $J^i = T_j{}^i\xi^j$. First note that $$J^i{}_{;i} = T^{ji}\xi_{i;j} + \xi_jT^{ji}{}_{;i} = 0,$$ by the Killing equation $\xi_{(i;j)} = 0$, the symmetry of the stress-energy tensor \$...
Top 50 recent answers are included
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2016-07-27 02:00:43
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https://www.opuscula.agh.edu.pl/om-vol37iss1art4
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Opuscula Math. 37, no. 1 (2017), 81-107
http://dx.doi.org/10.7494/OpMath.2017.37.1.81
Opuscula Mathematica
# Seminormal systems of operators in Clifford environments
Mircea Martin
Abstract. The primary goal of our article is to implement some standard spin geometry techniques related to the study of Dirac and Laplace operators on Dirac vector bundles into the multidimensional theory of Hilbert space operators. The transition from spin geometry to operator theory relies on the use of Clifford environments, which essentially are Clifford algebra augmentations of unital complex $$C^*$$-algebras that enable one to set up counterparts of the geometric Bochner-Weitzenböck and Bochner-Kodaira-Nakano curvature identities for systems of elements of a $$C^*$$-algebra. The so derived self-commutator identities in conjunction with Bochner's method provide a natural motivation for the definitions of several types of seminormal systems of operators. As part of their study, we single out certain spectral properties, introduce and analyze a singular integral model that involves Riesz transforms, and prove some self-commutator inequalities.
Keywords: multidimensional operator theory, joint seminormality, Riesz transforms, Putnam inequality.
Mathematics Subject Classification: 47B20, 47A13, 47A63, 44A15.
Full text (pdf)
1. A. Athavale, On joint hyponormal operators, Proc. Amer. Math. Soc. 103 (1988), 417-423.
2. N. Berline, E. Getzler, M. Vergne, Heat kernels and Dirac operators, [in:] Grundlehren der mathematischen Wissenschaften, vol. 298, Springer Verlag, Berlin, Heidelberg, New York, 1992.
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11. R.E. Curto, Joint hyponormality: a bridge between hyponormality and subnormality, [in:] W.B. Arveson, R.G. Douglas (eds), Operator Theory, Operator Algebras and Applications, Part II, Proc. Sympos. Pure Math. 51 (1990), 69-91.
12. R.E. Curto, R. Jian, A matricial identity involving the self-commutator of a commuting $$n$$-tuple, Proc. Amer. Math. Soc. 121 (1994), 461-464.
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14. R.G. Douglas, V. Paulsen, K. Yan, Operator theory and algebraic geometry, Bull. Amer. Math. Soc. 20 (1988), 67-71.
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16. R. Goldberg, Curvature and Homology, Academic Press, New York, London, 1962.
17. T. Kato, Smooth operators and commutators, Studia Math. 31 (1968), 531-546.
18. H.B. Lawson, M.-L. Michelsohn, Spin Geometry, Princeton Mathematical Series 38, Princeton University Press, Princeton, 1989.
19. M. Martin, Joint seminormality and Dirac operators, Integral Equations Operator Theory 30 (1998), 101-121.
20. M. Martin, Higher-dimensional Ahlfors-Beurling inequalities in Clifford analysis, Proc. Amer. Math. Soc. 126 (1998), 2863-2871.
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22. M. Martin, Self-commutator inequalities in higher dimension, Proc. Amer. Math. Soc. 130 (2002), 2971-2983.
23. M. Martin, Spin geometry, Clifford analysis, and joint seminormality, [in:] Advances in Analysis and Geometry, Vol. I, Trends in Mathematics Series, Birkhäuser Verlag, Basel, 2004, 227-255.
24. M. Martin, Uniform approximation by solutions of elliptic equations and seminormality in higher dimensions, Operator Theory: Advances and Applications, vol. 149, Birkhäuser Verlag, Basel, 2004, 387-406.
25. M. Martin, Deconstructing Dirac operators. I: Quantitative Hartogs-Rosenthal theorems, Proceedings of the 5th International Society for Analysis, Its Applications and Computation Congress, ISAAC 2005, Catania, Italy, 2005, More Progress in Analysis, World Scientific, 2009, 1065-1074.
26. M. Martin, Deconstructing Dirac operators. III: Dirac and semi-Dirac pairs of differential operators, Operator Theory: Advances and Applications, vol. 203, Birkhäuser Verlag, Basel, 2009, 347-362.
27. M. Martin, Deconstructing Dirac operators. II: Integral representation formulas, Proceedings of the 7th International Society for Analysis, Its Applications and Computation Congress, ISAAC 2009, London, United Kingdom, 2009, Trends in Mathematics: Hypercomplex Analysis and Applications, Springer Verlag, Berlin, Heidelberg, New York, 2011, 195-211.
28. M. Martin, M. Putinar, Lectures on Hyponormal Operators, Operator Theory: Advances and Applications, vol. 39, Birkhäuser Verlag, Basel, 1989.
29. M. Martin, N. Salinas, Weitzenböck type formulas and joint seminormality, [in:] Contemporary Math., Amer. Math. Soc. 212 (1998), 157-167.
30. M. Martin, P. Szeptycki, Sharp inequalities for convolution operators with homogeneous kernels and applications, Indiana Univ. Math. J. 46 (1997), 975-988.
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32. P.S. Muhly, A note on commutators and singular integrals, Proc. Amer. Math. Soc. 54 (1976), 117-121.
33. J.D. Pincus, Commutators and systems of singular integral equations, Acta Math. 121 (1968), 219-249.
34. J.D. Pincus, D. Xia, The analytic model of a hyponormal operator with rank one self-commutator, Integral Equations Operator Theory 4 (1981), 134-150.
35. J.D. Pincus, D. Xia, J. Xia, The analytic model of a hyponormal operator with rank one self-commutator, Integral Equations Operator Theory 7 (1984), 516-535.
36. C.R. Putnam, Commutation Properties of Hilbert Space Operators and Related Topics, Springer Verlag, Berlin, Heidelberg, New York, 1967.
37. C.R. Putnam, An inequality for the area of hyponormal spectra, Math. Z. 116 (1970), 323-330.
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40. F.-H. Vasilescu, A characterization of the joint spectrum in Hilbert spaces, Rev. Roumaine Math. Pures Appl. 22 (1977), 1003-1009.
41. F.-H. Vasilescu, A multidimensional spectral theory in $$C^*$$-algebras, Banach Center Publ. 8 (1982), 471-491.
42. D. Xia, On non-normal operators. I, Chinese J. Math. 3 (1963), 232-246; II, Acta Math. Sinica 21 (1987), 103-108.
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• Mircea Martin
• Baker University, Department of Mathematics, Baldwin City, KS 66006, USA
• Communicated by P.A. Cojuhari.
• Accepted: 2016-11-02.
• Published online: 2016-12-14.
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2021-05-13 16:28:59
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https://stat.ethz.ch/pipermail/r-devel/2010-February/056792.html
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# [Rd] scale(x, center=FALSE) (PR#14219)
Ben Bolker bolker at ufl.edu
Fri Feb 26 18:44:45 CET 2010
[cc'ing back to r-devel]
Maria Rizzo wrote:
> Ben,
>
> I receive the digest version of r-devel - so I do not have the
>
> I think this is a bug for the following reasons. While it is true
> that one can define a scale factor differently for different
> purposes, one would hope that within a given function the definition
> does not vary. If we agree that we want to divide by standard
> deviation, which scales data to sd=1, then why would we choose to
> divide by square root of 1/(n-1) times sum of squares of the data
> when data is not centered? This does not scale the data to sd=1.
This is really a disagreement with the way the function is implemented
(and I happen to agree with you), but I would argue that it is *not* a
bug in the strict sense -- I would call it a "misfeature".
From the R FAQ:
> Finally, a command's intended definition may not be best for
> statistical analysis. This is a very important sort of problem, but
> it is also a matter of judgment.
[snip]
>> Are you sure this is a bug? I agree that the way the function
>> behaves is (to me) mildly confusing, but the documentation says:
>>
>> * The value of ?scale? determines how column scaling is performed *
>> (after centering). If ?scale? is a numeric vector with length *
>> equal to the number of columns of ?x?, then each column of ?x? is *
>> divided by the corresponding value from ?scale?. If ?scale? is *
>> ?TRUE? then scaling is done by dividing the (centered) columns of *
>> ?x? by their standard deviations, and if ?scale? is ?FALSE?, no *
>> scaling is done.
>>
>> * The standard deviation for a column is obtained by computing the
>> * square-root of the sum-of-squares of the non-missing values in
>> the * column divided by the number of non-missing values minus one
>> * (whether or not centering was done).
>>
>> If you read the first clause of the last sentence of the first
>> paragraph in isolation, you would have the expectation that the
>> columns would be scaled by sd(x). However, the second paragraph
>> clearly states that the 'standard deviation' is defined here as the
>> root-mean-square over (n-1), that is, sqrt(sum(x^2)/(n-1)) ...
>
> This conflicts with the paragraph above it. What I see is that the
> (centered) columns are divided by their standard deviations, where
> (centered) is inserted or not before "columns" depending on whether
> center=TRUE or center=FALSE. Why modify the definition of "standard
> deviation"? Why compute the standard deviation of the centered data
> when data is not centered? This measures standard deviation with
> respect to the origin rather than measuring dispersion about the
> mean.
>> This does seem like a funny choice, but it is probably stuck that
>> way without an extremely compelling argument to the contrary. If
>> you want to scale columns by sd() instead you can say
>>
>> scale(x,center=FALSE,scale=apply(x,2,sd))
>
> Of course, I know how to achieve the result of scaling my data to
> sd=1. The problem is that a function called scale with options of
> center=TRUE or center=FALSE, should apply the same definition of
> scale if scale=TRUE in both cases.
"should" according to you ...
>
>> Would you like to submit a patch for the documentation that would
>> preserve the sense, clarify the behavior, and not be much longer
>> than the current version ... ?
>
> For me the problem is deeper than an issue with the documentation. In
> any case, I think that it is a potential source of confusion and
> errors on the part of users.
>
> regards, Maria
>
Again, I agree with you that the behavior is not optimal, but it is
very hard to make changes in R when the behavior is sub-optimal rather
than actually wrong (by some definition). R-core is very conservative
about changes that break backward compatibility; I would like it if they
chose to change the function to use standard deviation rather than
root-mean-square, but I doubt it will happen (and it would break things
for any users who are relying on the current definition).
It turns out that the documentation for this function was changed on
25 Nov 2009 to clarify this issue, but I think the change (which among
other minor changes modified the previous use of "root mean square" to
"standard deviation") didn't help that much ... I have attached a patch
file (and append the information below as well) that changes "standard
deviation" back to "root mean square" and is much more explicit about
this issue ... I hope R-core will jump in, critique it, and possibly use
it in some form to improve (?) the documentation ...
[PS: I have written that the scaling is equivalent to sd() "if and
only if" centering was done. Technically it would also be equivalent if
===================================================================
--- scale.Rd (revision 51180)
+++ scale.Rd (working copy)
@@ -41,13 +41,18 @@
equal to the number of columns of \code{x}, then each column of
\code{x} is divided by the corresponding value from \code{scale}. If
\code{scale} is \code{TRUE} then scaling is done by dividing the
- (centered) columns of \code{x} by their standard deviations, and if
+ (centered) columns of \code{x} by their root-mean-squares, and if
\code{scale} is \code{FALSE}, no scaling is done.
-
- The standard deviation for a column is obtained by computing the
- square-root of the sum-of-squares of the non-missing values in the
- column divided by the number of non-missing values minus one (whether
- or not centering was done).
+
+ The root-mean-square for a (possibly centered)
+ column is defined as
+ \eqn{\sqrt{\sum(x^2)/(n-1)}}{sqrt(sum(x^2)/(n-1))},
+ where \eqn{x} is a vector of the non-missing values
+ and \eqn{n} is the number of non-missing values.
+ If (and only if) centering was done,
+ this is equivalent to \code{sd(x,na.rm=TRUE)}.
+ (To scale by the standard deviations without centering,
+ use \code{scale(x,center=FALSE,scale=apply(x,2,sd,na.rm=TRUE))}.)
}
\references{
Becker, R. A., Chambers, J. M. and Wilks, A. R. (1988)
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2022-09-26 14:11:18
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http://www.mzan.com/article/48232831-asp-net-mvc-razor-document-downloading-as-corrupt.shtml
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I'm working on an ASP.Net project that is based off of Razor. We are posting documents from the client side using formdata and need to change the name of this file based on who is uploading it. We have a method for doing so, which you can see here: var fileToSave = ""; if (filename != null) { fileToSave = '@ViewBag.CompName' + "-" + filename.name; fileToSave = fileToSave.replace(/\s+/g, '-'); } var formData = new FormData(); formData.append("Document", filename, fileToSave); This saves the document with the proper name, e.g. "Test-Company-test.docx" The problem exists when I go to download the file and open it up. The document downloads properly, and has the appropriate size of the uploaded file, for this purpose, "test.docx". When opening the downloaded file, Word prompts you with the following: Pressing "OK" then shows the following message: Following this message, if you click "Yes," the document opens and the entirety of the document is there. There is no corrupt data, and everything is formatted exactly how it is supposed to be. This error does not exist if we just save the document as the original document name. Which normally we would do, but this is a large application that is client facing and allows users to upload documents themselves. This can be a problem if Company A uploads a document called "Test.docx" and Company B also uploads a document called "Test.docx" thus overwriting Company A's document. Hence, the renaming of the documents. How is it that we can prevent this error from happening?
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2018-07-21 09:53:13
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http://math.stackexchange.com/questions/271728/property-of-binomial-coefficient
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Property of Binomial Coefficient
I have a "must be trivial" problem which I could not solve.
Prove the following relation of binomial coefficients, if true: $$\sum_{k=1}^{n}{2n+1 \choose k}=2^{2n}-1$$
P.S. Though this is not homework, I appreciate any hints rather than explicit solutions.
I am looking for proofs from properties of binomial coefficients rather than other methods.
-
We have that $$\sum_{k=0}^{2n+1} \dbinom{2n+1}k = 2^{2n+1}$$ $$\sum_{k=0}^{n} \dbinom{2n+1}k + \sum_{k=n+1}^{2n+1} \dbinom{2n+1}k = 2^{2n+1}$$ Now note that $$\dbinom{2n+1}k = \dbinom{2n+1}{2n+1-k}$$ Hence, we get that $$2\sum_{k=0}^{n} \dbinom{2n+1}k = 2^{2n+1}$$ $$\sum_{k=0}^{n} \dbinom{2n+1}k = 2^{2n}$$ $$\dbinom{2n+1}0 + \sum_{k=1}^{n} \dbinom{2n+1}k = 2^{2n} \implies \sum_{k=1}^{n} \dbinom{2n+1}k = 2^{2n} - 1$$
Substitute $x=1$ into $(1+x)^{2n+1}=\sum_{k=0}^{2n+1}\binom{2n+1}{k}x^k$ to obtain $2^{2n+1}=\sum_{k=0}^{2n+1}\binom{2n+1}{k}=2\sum_{k=0}^{n}\binom{2n+1}{k}$ since $\binom{2n+1}{k}=\binom{2n+1}{2n+1-k}$.
Use this result (Binomial Theorem) $$(a + b)^n = \sum_{k=0}^n {n\choose k} a^k b^{n-k}.$$
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2016-05-28 18:10:35
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https://ftp.aimsciences.org/article/doi/10.3934/jimo.2017086
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# On the convergence properties of a smoothing approach for mathematical programs with symmetric cone complementarity constraints
This study is supported by the National Natural Science Foundation of China under projects No.11401210, No.11671183, No.11571059, No.91330206 and No.11301049.
• This paper focuses on a class of mathematical programs with symmetric cone complementarity constraints (SCMPCC). The explicit expression of C-stationary condition and SCMPCC-linear independence constraint qualification (denoted by SCMPCC-LICQ) for SCMPCC are first presented. We analyze a parametric smoothing approach for solving this program in which SCMPCC is replaced by a smoothing problem $P_{\varepsilon}$ depending on a (small) parameter $\varepsilon$. We are interested in the convergence behavior of the feasible set, stationary points, solution mapping and optimal value function of problem $P_{\varepsilon}$ when $\varepsilon \to 0$ under SCMPCC-LICQ. In particular, it is shown that the convergence rate of Hausdorff distance between feasible sets $\mathcal{F}_{\varepsilon}$ and $\mathcal{F}$ is of order $\mbox{O}(|\varepsilon|)$ and the solution mapping and optimal value of $P_{\varepsilon}$ are outer semicontinuous and locally Lipschitz continuous at $\varepsilon=0$ respectively. Moreover, any accumulation point of stationary points of $P_{\varepsilon}$ is a C-stationary point of SCMPCC under SCMPCC-LICQ.
Mathematics Subject Classification: Primary: 90C33, 90C26; Secondary: 65K05.
Citation:
• [1] A. Ben-Tal and A. Nemirovski, Robust convex optimization methodology and applications, Mathematical Programming, 92 (2002), 453-480. doi: 10.1007/s101070100286. [2] G. Bouza and G. Still, Mathematical programs with complementarity constraints: Convergence properties of a smoothing method, Mathematics of Operations Research, 32 (2007), 467-483. doi: 10.1287/moor.1060.0245. [3] X. Chen and M. Fukushima, A smoothing method for a mathematical program with P-matrix linear complementarity constraints, Computational Optimization and Applications, 27 (2004), 223-246. doi: 10.1023/B:COAP.0000013057.54647.6d. [4] F. Clarke, Optimization and Nonsmooth Analysis, John Wiley and Sons, New York, 1983. [5] C. Ding, D. Sun and J. Ye, First order optimality conditions for mathematical programs with semidefinite cone complementarity constraints, Mathematical Programming, Ser.A, 147 (2014), 539-579. doi: 10.1007/s10107-013-0735-z. [6] F. Facchinei, H. Jiang and L. Qi, A smoothing method for mathematical programs with equilibrium constraints, Mathematical Programming, 85 (1999), 107-134. doi: 10.1007/s10107990015a. [7] J. Faraut and A. Korányi, Analysis on Symmetric Cones, Oxford University Press, New York, 1994. [8] L. Faybusovich, Linear systems in Jordan algebras and primal-dual interior-point algorithm, Journal of Computational and Applied Mathematics, 86 (1997), 149-175. doi: 10.1016/S0377-0427(97)00153-2. [9] M. Fukushima and J. Pang, Convergence of a smoothing continuation method for mathematical problems with complementarity constraints, Lecture Notes in Economics and Mathematical Systems, 477 (1999), 99-110. doi: 10.1007/978-3-642-45780-7_7. [10] M. Gowda, R. Sznajder and J. Tao, Some P-properties for linear transformations on Euclidean Jordan algebras, Linear algebra and its applications, 393 (2004), 203-232. doi: 10.1016/j.laa.2004.03.028. [11] K. Koecher, The Minnesota Notes on Jordan Algebras and Their Applications, edited and annotated by A. Brieg and S. Walcher, Springer, Berlin, 1999. doi: 10.1007/BFb0096285. [12] G. Lin and M. Fukushima, A modified relaxation scheme for mathematical prgrams with complementarity constraints, Annals of Operations Research, 133 (2005), 63-84. doi: 10.1007/s10479-004-5024-z. [13] Z. Luo, J. Pang and D. Ralph, Mathematical Programs with Equilibrium Constraints, Cambridge University Press, Cambridge, United Kingdom, 1996. doi: 10.1017/CBO9780511983658. [14] J. Outrata, M. Ko$\breve{c}$vara and J. Zowe, Nonsmooth Approach to Optimization Problems with Equilibrium Constraints, Kluwer Academic Publishers, Boston, MA, 1998. doi: 10.1007/978-1-4757-2825-5. [15] R. Rockafellar and R. Wets, Variational Analysis, Springer-Verlag, New York, 1998. doi: 10.1007/978-3-642-02431-3. [16] S. Scheel and S. Scholtes, Mathematical programs with complementarity constraints: Stationarity, optimality, and sensitivity, Mathematics of Operation Research, 25 (2000), 1-22. doi: 10.1287/moor.25.1.1.15213. [17] S. Scholtes, Convergence properties of a regularization scheme for mathematical programs with complementarity constraints, SIAM Journal on Optimization, 11 (2001), 918-936. doi: 10.1137/S1052623499361233. [18] D. Sun and J. Sun, Löwner's operator and spectral functions on Euclidean Jordan algebras, Mathematics of Operation Research, 33 (2008), 421-445. doi: 10.1287/moor.1070.0300. [19] D. Sun, J. Sun and L. Zhang, The rate of convergence of the augmented Lagrangian method for nonlinear semidefinite programming, Mathematical Programming, 114 (2008), 349-391. doi: 10.1007/s10107-007-0105-9. [20] E. Takeshi, A Smoothing Method for Mathematical Programs with Second-Order Cone Complementarity Constraints, Master thesis, Kyoto University in Kyoto, 2007. [21] Y. Wang, Perturbation Analysis of Optimimization Problems over Symmetric Cones, Ph. D. Thesis, Dalian University of Technology, China, 2008. [22] T. Yan and M. Fukushima, Smoothing method for mathematical programs with symmetric cone complementarity constraints, Optimization, 60 (2011), 113-128. doi: 10.1080/02331934.2010.541458. [23] Y. Zhang, J. Wu and L. Zhang, First order necessary optimality conditions for mathematical programs with second-order cone complementarity constraint, Journal of Global Optimization, 63 (2015), 253-279. doi: 10.1007/s10898-015-0295-2. [24] Y. Zhang, L. Zhang and J. Wu, Convergence properties of a smoothing approach for mathematical programs with second-order cone complementarity constraints, Set-Valued and Variational Analysis, 19 (2011), 609-646. doi: 10.1007/s11228-011-0190-z.
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2023-03-26 08:56:40
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http://math.stackexchange.com/tags/eigenvalues-eigenvectors/hot
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# Tag Info
3
If $T(A)=\lambda A$, then $T^2(A)=\lambda^2 A$. But $T^2=I$ which only has an eigenvalue of 1. So $\lambda^2=1$ or $\lambda=-1,1$.
3
It means that $$\forall j \quad |\lambda_j(A)|<1$$ where $\lambda_j(A)$ are eigenvalues of $A$. There're several different ways of showing that $$\lim_{k\to\infty}A^k=0.$$ The first method is based on the Jordan normal form of the matrix (this form helps to find another expression for $A^k$). Essentially we show that any matrix is similar to an ...
3
$x$ is an eigenvector of $A$ if $x\neq 0$ and there exists $\lambda$ such that $Ax=\lambda x$. $\lambda$ is then called an eigenvalue of $A$ associated to the eigenvector $x$. Note in particular that every vector $x\neq 0$ in the kernel of $A$ is an eigenvector of $A$ associated to the eigenvalue $0$. More generally $\ker(A-\lambda I)$ is the space of ...
3
$$\det(I+ x A) = x^n \det (A + \frac 1 x I) = x^n\left(\lambda_1 + \frac 1 x\right)\left(\lambda_2 + \frac 1 x\right)\cdots \left(\lambda_n + \frac 1 x\right)$$
2
Eigen vectors corresponding to distinct eigenvalues are necessarily independent.Also $x_2$ can't be an eigenvector since it is $0$. When you solve, you find the eigen space $E_2$ has dimension $1$, with basis $x_1$. You complete it in a basis of the generalised eigenspace, which happens to be $\ker(A-2I)^2$. For instance, you may solve for ...
2
All of the vectors you give belong to $\lambda = -1$. This set is not linearly independent; the rank of $$\begin{pmatrix} -2 & 1 & 1 \\ 1 & -2 & -1 \\ 1 & -1 & 0 \end{pmatrix}$$ is 2, which means each two of them are sufficient to span the 2-dimensional eigenspace for this eigenvalue. The eigenvector for $\lambda = 2$ is $$... 2 What you want is |cY|=|c|\cdot |Y| \le 1; knowing that |Y|\le 1 and that the boundary can be reached, it is necessary and sufficient to have |c|\le 1. So your second answer, [-1,1], is correct. 2 It suffices to prove that$$2xx^T-2 \textrm{diag}(xx^T)+\mathbb{I}_n\geq 0$$or equivalently that$$\left[\matrix{1 & 2x_1x_2 & \cdots & 2x_1x_n\\ 2x_1x_2 & 1 & \cdots & 2x_2x_n\\ \vdots & \vdots & \ddots & \vdots\\ 2x_1x_n & 2 x_2x_n & \cdots & 1}\right]\geq 0$$Let now e_i the i-th column of ... 2 In Step 2 to Step 3, t-5 is factored out of the first row and the third row. The multiplies the determinant by the same factor, each time the factoring is done. In Step 4 to Step 5, expansion by minors is done on the first column. This is easy since there is a 1 in only one position and 0's in the other positions in that column. What's more, adding ... 2 \det(L-\lambda I)=-\lambda\begin{vmatrix}-\lambda & 0\\\frac{1}{3} &-\lambda\end{vmatrix}-\frac{1}{2}\begin{vmatrix}\frac{3}{2}a^2 & \frac{3}{2}a^3\\\frac{1}{3}&-\lambda\end{vmatrix}=-\lambda^3-\frac{1}{2}(\frac{3}{2}a^2)\begin{vmatrix}1&a\\\frac{1}{3}&-\lambda\end{vmatrix}=-\lambda^3-\frac{3}{4}a^2(-\lambda-\frac{1}{3}a) ... 2 T is a linear operator. As I understood this: find \lambda that:$$T(A) - \lambda A = 0$$If A is symmetric A = A^t then \lambda = 1, if anti-symmetric A = -A^t then \lambda = -1. In general case, there is no \lambda that satisfy a_{ij} = \lambda a_{ji} for all a_{ij} 1 Let's use the "brute force" approach here. We have the matrix equation Av=\lambda v. Thus, the eigenvalue equation becomes$$(A-\lambda I)v=0$$which implies that the determinant of A-\lambda I is zero. Thus, we have$$(\lambda -a)(\lambda -d)-bc=0$$which implies that$$\begin{align} \lambda &=\frac{(d+a)\pm \sqrt{(d+a)^2-4(ad-bc)}}{2}\\ ...
1
In chepukha's statement there is a small mistake. The radius must be $r_i=\sqrt{\frac{k}{\lambda_i}}$ (in his statement there is a square root missing, or, alternatively, the ellipsoid should be defined as $\sum_i\sum_j (x_i-u_i)(x_j-u_j)c_{ij}\leq k^2$). For example, consider the scalar case (where the proof is straightforward): equation $x^2 c \leq k$ has ...
1
Your work is correct . You are calculating a determinant and in the step $2 \rightarrow 3$ is used the fact that a determinant is a multilinear function of the rows ( and of the columns) of the matrix. The same is obtained using the fact that the determinant of a product is the product of the determinants. Explicitly: $... 1 Let$A \in \mathbb{R^{m \times n}}$and$\lambda$be non-zero eigen value of$AA^TAA^T \in \mathbb{R^{m \times m}}$and$A^TA \in \mathbb{R^{n \times n}}$$$AA^T\textbf{x} = \lambda \textbf{x}$$$\textbf{x} \in \mathbb{R^{m}}$Multiply(pre) both sides by$A_{n \times m}^T$$A^T(AA^T)\textbf{x} = \lambda A^T\textbf{x}$$ Rearranging a bit: ... 1 If you have eigenvector and eigenvalue pairs \begin{align*} v_1 &= (2, 1, 0)^T, &\lambda_1 = 1 \\ v_2 &= (-1, 0, 1)^T, &\lambda_2 = 1 \\ v_3 &= (1, -2, 1)^T, &\lambda_3 = -1 \end{align*} I would expect this to be a reflection, rather than a rotation. There are two reasons this would be so. First, the determinant is-1 = 1^2 ...
1
Compute $$\begin{pmatrix}-2 \\1 \\ 1 \end{pmatrix} - \begin{pmatrix}1 \\-2 \\ 1 \end{pmatrix} + 3\begin{pmatrix}1 \\-1 \\ 0 \end{pmatrix}$$
1
Translating result appeared on the wiki page of Characteristic polynomial, $$\det(I_n + x A ) = \sum_{k=0}^n x^k \text{tr}(\wedge^k A)$$ where $\text{tr}(\wedge^k A)$ is the trace of the $k^{th}$ exterior power of $A$ which can be evaluated explicitly as the determinant of the $k \times k$ matrix, \text{tr}(\wedge^k A) = \frac{1}{k!} ...
1
Hints: 1) The matrix's characteristic polynomial is $\;x^4\;$ (i.e., the matrix is nilpotent) 2) A matrix is diagonalizable iff its minimal polynomial is the product of different linear factors 3) Thus, a nilpotent matrix is diagonalizable iff ...
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2015-04-18 21:14:08
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http://www.damian.oquanta.info/es/index.html
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Binder + Nikola + Jupyter + Github = Blogging resourceless
You are in vacation time but you want to blog something nice to share with your friends.
And you are a Jupyter Notebook user.
But you don't have your laptop with you, because you are on vacation time, remember? ;-)
But you still have your phone and some connectivity.
What do you think if I say that you have a complete workflow to write your blogpost, build the site and deploy it just using your phone? But without using computational resources from it (that would be also interesting, btw).
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RISE 5.1.0 is out!
We're pleased to announce the release of RISE 5.1.0!
RISE allows you show your Jupyter notebooks rendered as an executable Reveal.js-based slideshow. It is your very same notebook but presented in a slidy way!
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We are above 1000 stars!
Github has a way to measure projects popularity through stars.
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And we are just above a remarkable line...
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OK, I have run my models again and it was time to enter the market.
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RISE meets JupyterLab
JupyterLab is the future for the notebook/authoring experience.
And people started to ask me if we will have RISE on JupyterLab
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RISE 5.0.0 is out!
We're pleased to announce the release of RISE 5.0.0!
RISE let's you show yout Jupyter notebook rendered as an executable Reveal.js-based slideshow. It is your very same notebook but in a slidy way!
How you can get it?
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RISE 4.0.0b1 is available, please test it!
Quick post! I have beta packages available for you to test RISE, if you can test it that would be awesome!!
In case you don't know about it, with RISE you get your Jupyter notebook rendered as a Reveal.js-based slideshow, where you can execute code on the fly or show to the audience whatever you can show/do inside the notebook itself (but in a "slidy" way).
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How to pin Conda
One interesting advance feature in Conda is the capacity to pin packages from your environments so they can not be updated at all. If you are interested in that specific version of some package and not the next one because it breaks your work completely or for some other reason, you are probably pinning that package. If you are adding the specific version in every command you run instead of pinning the package, you are doing it wrong and you should keep reading ;-)
But, is it possible to pin Conda itself so it does not get updated every time you try to install/update something else?
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Coming back
OK... two years since my last post... exactly. Time goes fast, really fast! And a lot of things happened in the last two years.
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Zen themes updated
OK, time to recap some things... As you know, Nikola 7.0.0 was released some weeks ago. It has a lot of improvements, bug fixes and new features. I recommend you to download and try it! As part of the release, we paid attention to update all the plugins and themes inside the Nikola Github organization (don't forget you can contribute with your own plugins and themes!). So, I updated my own themes, in particular, the Zen ones.
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2017-12-18 14:47:52
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https://tex.stackexchange.com/questions/281620/reledmac-in-double-spaced-document-single-spaced-two-col-footnotes-not-aligned
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# reledmac: in double-spaced document, single-spaced two-col footnotes not aligned
In a double-spaced document with two-column single-spaced footnotes in the reledmac environment (using this solution), the lines across the footnote columns don't line up vertically. This issue disappears if either double-spacing or the \Xbhooknote option is removed, but I'd like to keep both while preserving footnote alignment.
\documentclass[12pt]{article}
\usepackage{reledmac}
\usepackage{lipsum}
\usepackage[]{setspace}
\AtBeginDocument{\doublespacing}
\Xarrangement[A]{twocol}
\Xcolalign{\justifying}
\makeatletter
\Xbhooknote{\setstretch {\setspace@singlespace}}
\makeatother
\begin{document}
\beginnumbering
\pstart
Ecce \edtext{ipse lorus:}{\Afootnote{i.e. Lorem ipsum.}} \lipsum*[1] \edtext{Haec hactenus.}{\Afootnote{\lipsum*[2]}}
\pend
\endnumbering
\end{document}
See this MWE:
\documentclass[12pt]{article}
\usepackage{reledmac}
\usepackage{lipsum}
\usepackage[]{setspace}
\AtBeginDocument{\doublespacing}
\Xarrangement[A]{twocol}
\Xcolalign{\justifying}
\makeatletter
\Xbhooknote{\setstretch {\setspace@singlespace}}
\Xbhookgroup{\setstretch {\setspace@singlespace}}
\makeatother
\begin{document}
\beginnumbering
\pstart
Ecce \edtext{ipse lorus:}{\Afootnote{i.e. Lorem ipsum.}} \lipsum*[1] \edtext{Haec hactenus.}{\Afootnote{\lipsum*[2]}}
\pend
\endnumbering
\end{document}
## Comment
The main problem is that, when splitting in two column the contents of notes, reledmac add vertical space (\splittopskip). This vertical space is calculated on the based of the \baselineskip when outputting the notes, which is not affected by Xbhooknote hook, which concerns only the way footnote are stored in a temporary box. So we also need to set it when outputing the footnotes group, and we use \Xbhookgroup. But it was buged for twocol and threecol footnotes arrangement. It should be solved in 2.7.1
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2019-09-16 06:07:16
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https://dml.cz/handle/10338.dmlcz/124944
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# Article
Full entry | PDF (0.5 MB)
References:
[1] J. Aczel Z. Daroczy: Measures of Information and Their Characterizations. Academic Press, New York 1975. MR 0689178
[2] T. W. Chaundy J. B. McLeod: On a Functional Equation. Proc. Edinburgh Math. Soc., Edin. Math. Notes 43 (1960), 7-8. MR 0151748
[3] Z. Daroczy: Generalized Information Function. Information and Control 16, (1970), 36-51. MR 0272528
[4] D. K. Fadeev: On the Concept of Entropy of a Finite Probabilistic Scheme. [in Russian], Uspekhi Math. Nauk 11 (1 (67)) (1956), 227-231. MR 0077814
[5] J. Havrda F. Charvát: Quantification Method of Classification Processes. The Concept of Structural $\alpha$-entropy. Kybernetika 3, (1967), 30-35. MR 0209067
[6] C. E. Shannon: A Mathematical Theory of Communication. Bell. Syst. Tech. J. 27 (1948), 378-423 and 623-656. MR 0026286 | Zbl 1154.94303
[7] B. D. Sharma I. J. Taneja: Three New Measures of Entropy. To appear in E.I.K. (Germany), Abstract in Proceedings of the Indian Science Congress Association held at Nagpur, Jan. 3-7, 1974.
[8] I. J. Taneja: On Axiomatic Characterization of Entropy of Type ($\alpha$, $\beta$). Aplikace matematiky 22, (1977), No. 5. MR 0469524 | Zbl 0385.28013
[9] I. J. Taneja: On the Branching Property of Entropy. Annales Polonici Mathematici (1977), 67-76. MR 0499877 | Zbl 0424.94005
[10] I. Vajda: Axioms for a-Entropy of Generalized Probability Scheme. Kybernetika 4, (1968), 2, 105-112. MR 0233626
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2017-08-20 00:13:10
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https://www.vcalc.com/wiki/pro/rebar+-+irregular+shaped+slab?viewer=comments
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rebar - irregular shaped slab
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pro.rebar - irregular shaped slab
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9180a587-f6b1-11e5-9770-bc764e2038f2
The Rebar for an Irregular Shaped Slab calculator estimates the total length or weight Irregular Slab with one Rebar Grid mat of reinforcement bars (rebar) needed for an irregular shaped concrete slab.
INSTRUCTIONS:Choose your preferred units and enter the following:
• (A) This is the approximate area of the irregular shaped slab (e.g. 72 ft2)
• (rP) This is the rough estimate of the perimeter of the slab (e.g. 178 ft)
• (S) This is the size of the rebar. Choose from the pull-down list size 2 to size 18.
• (oC) This is the on-center spacing of the rebar.
• (M) This is the number of rebar mats
• (WL) Output choice: Choose rebar weight or length
Weight or Length of Rebar: If weight is chosen, the weight of the rebar is returned in pounds. However, this can be automatically converted to other weight units (e.g. kilograms or tons) via the pull-down menu. If length is chosen, the length returned in feet. However, this can be automatically converted to other length units (e.g. meters) via the pull-down menu.
Related Calculators:See the Rebar Calculator for more concrete and rebar functions and features.
• To compute the Length of Rebar needed for a slab, CLICK HERE.
• To compute the Weight of Rebar in a slab, CLICK HERE.
• To compute the Volume of Concrete in a slab, CLICK HERE.
• To compute the Weight of Concrete in a slab or wall, CLICK HERE.
• To compute the Total Weight of a Slab with Rebar, [Could not locate page KurtHeckman.slab – total weight estimate].
Reducing Rebar Waste
A major challenge with rebar in an irregular slab is minimizing wasted rebar. The general guidance is this. When possible, start with the longest cross-section of of your slab and work outwards from there. These will be the longest stretches of rebar in your slab. When you cut off the ends, line them up in a row from shortest to longest. When you get to shorter runs, first see if there is a cut-off section of rebar from your row of cut-offs. This is only a little extra work and will save in both materials and labor removing cut-offs at the end of the project.
General Information
Note1: the default units are feet and inches. However, you can change the input units to metric (SI) or others by clicking on the units selection button (to the right of the entry fields). You can also change the output units to metric (SI) units by clicking on the red output units selection button (to the right of the answer box). In both cases, vCalc will make the automatic conversions.
Note2: The lapping portion of the rebar length calculation assumes that the uncut rebar on site is 40' (feet) in length and that lapping is not necessary under that length. Furthermore, this calculator assumes that rebar is lapped by a factor of 40 times the diameter of the rebar chosen. For the same calculations with the ability to add a different "uncut" length and/or a different lapping factor CLICK HERE.
Description
Rebar Specifications
The rebar weight algorithm calculates the weight of reinforcement steel bars in an irregular shaped concrete slab. The algorithm uses the Rough Area, Perimeter and the onCenter spacing to estimate the total length of rebar in the slab. Once the algorithm calculates the length of rebar, the length is used with the user specified rebar gauge, and density constants from the vCalc library to calculate the total weight of the reinforcement steel bars.
The user specified parameters are:
• Rough Area - area of the slab
• Rough Perimeter - rough perimeter of the slab
• OnCenter - the on-center spacing between rebar, which defaults to 18"
• size - this is the size of the rebar. The standard size include: 3-11,14,18
• mats - (m) this is the number of mats of rebar.
• Output choice - this lets the user choose between the weight of the rebar in the grid or the length of rebar in the grid.
The calculation determines the need for an extra rebar using an internal tolerance for for a spacing fraction that exceeds the tolerance with a default of 1". It also assumes that the uncut rebar is in 40' lengths and that lapping is needed for dimensions in excess of 40`. Furthermore, the lapping factor, when lapping is needed, is 40. For the same calculations with the ability to add a different "uncut" length and/or a different lapping factor CLICK HERE.
Usage
Reinforcement bars are often used in concrete including common slabs. This formula provides a length and weigh calculation that is useful in understanding the additional load of the slab added by the rebar steel. It is also useful for calculating the weight of rebar when considering transport. Steel is dense and heavy. Most vehicles would be considerably overloaded in weight of rebar long before their potential volume is full, which poses a significant safety issue. In the U.S., most pickup trucks are rated at a half or three quarters ton load rating. This rating indicates the safe weight of a load that can be carried. The rebar weight formula can help determine how many trips are required to transport the load of reinforcement steel safely.
Lapping
When the dimensions of your slab or wall exceed the length of a single piece of rebar, it is required to lap and tie the rebar to create the added length. There are a few considerations. First, the length of the lap is often specified as 40 times the diameter of the rebar. In this case, 40 is lapping factor. 60 is also a common factor, but the engineering specifications should always be applied. See these YouTube videos to better understand rebar lap:
Second, the typical length of pre-cut rebar is 40' and 60' in the United States.
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2019-07-20 03:30:13
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http://www.intsystems.cz/abstract-2015/
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# Abstracts
Name Abstract Aguirre Alexis Generalizing type-II integrable defects Ahmadov Azar Higher-Twist Mechanism and Inclusive Gluon Production in Pion-Proton Collisions Algin Abdullah Deformed Gas of Tamm-Dancoff Type Boson Oscillators: Algebra and Thermodynamics Alkalaev Konstantin Classical conformal blocks via AdS/CFT correspondence Amari Yuki Quantized states of vortex in a CP^2 Skyrme-Faddeev type model Andrzejewski Krzysztof Hamiltonian formalisms and symmetries of the Pais-Uhlenbeck oscillator Aniello Paolo Functions of Positive Type and Open Quantum Systems Arai Masato Cotangent bundles over all the Hermitian symmetric spaces Artamonov Semeon Noncommutative Inverse Scattering Method for the Kontsevich system Avan Jean No talk Babalic Nicoleta-Corina Bilinear supersymmetric Gardner equation and super-lump solutions Batchelor Murray The interplay between quantum symmetry, integrability and solvability: the Rabi model Beisert Niklas Smooth Wilson Loops in N = 4 Superspace and Yangian Symmetry Bekaert Xavier 1-loop partition function of conformal higher spin theory Belliard Samuel Modified algebraic Bethe ansatz Bellucci Stefano FPS approach and the one-dimensional supersymmetric model with PBGS Bertrand Sébastien On the integrability of supersymmetric versions of the structural equations for conformally parametrized surfaces Bialecki Mariusz Solvable Structures of Simple Model of Earthquakes Bilge Ayse Humeyra On the uniquness of the octonionic instanton solution on conformally flat 8-manifolds Blaschke Filip Dynamics of slender monopoles and anti-monopoles in non-Abelian superconductor Bolonek-Lasoń Katarzyna Quantizing the classical non-cooperative symmetric games with arbitrary number of strategies Bouaziz Djamil Deformed Heisenberg algebra with a minimal length: Application to some molecular potentials Braden Harry On the Construction of Monopoles Breev Alexander The symmetry of the Dirac equation in a uniform static magnetic field Cao Junpeng Exact solution of high spin Heisenberg model with generic integrable boundaries Casati paolo Hirota bilinear equations for non semisimple Lie algebras Catto Sultan New rules for octonionic triplet and quadruplet multiplications and applications in high energy physics Dabrowski Ludwik Towards quantum Borsuk-Ulam and Brouwer fixed point theorems Davydov Evgeny Cosmology in Horndeski Gravity de la Rosa Gomez Alejandro Yangian symmetry in the open Hubbard model de Leeuw Marius One-point Functions in Defect CFT and Integrability Deriglazov Alexei Worldline geometry probed by spinning particle Derkachov Sergey From principal series to finite-dimensional solutions of the Yang-Baxter equation Dimakis Nikolaos Contact symmetries of constrained quadratic Lagrangians Doikou Anastasia Scattering in twisted Yangians Doliwa Adam Non-commutative Yang-Baxter maps Du Jie Quantum linear supergroups and their canonical bases El Deeb Omar Gimeno Vicent “Counting symmetries of a quantum system via a geometric approach “ Gnatenko Khrystyna Physical systems in a space with noncommutativity of coordinates Görbe Tamás F. Equivalence of two sets of deformed Calogero-Moser Hamiltonians Gorska Katarzyna The fractional differential equation and Levy stable distribution Gusev Alexander Metastable states of clusters tunneling through repulsive barriers Hakobyan Manush Exact solvability of the Schrödinger and Klein-Gordon equations in terms of Heun class of functions Hecht Reimar A Quantum Double for centrally extended sl(2|2) Herrera Aguilar Alfredo Nonlinear anisotropic symmetries in conformal quantum mechanics and diffusion equation Hnatic Michal No talk Hoque Md Fazlul Family of N-dimensional superintegrable systems and quadratic algebra structures Chicherin Dmitry Elliptic solutions of the Yang-Baxter equation Chizhov Alexey Vibron properties in quasi 1D molecular structures: the case of two parallel unshifted macromolecuar chains Ida Cristian Leafwise anti-Kahlerian metrics and Einstein equations Isaev Alexey Idempotents for Birman–Murakami–Wenzl algebras and reflection equation Ishkhanyan Artur Exact solvability of the Schrödinger and Klein-Gordon equations in terms of Heun class of functions Ishkhanyan Tigran Series expansions of the solutions of the Heun equations in terms of confluent, ordinary and generalized hypergeometric functions Ivanov Evgeny $SU(2|1)$ Supersymmetric Mechanics Jadczyk Arkadiusz Random walk on quantum blobs Jakubczyk Paweł Entanglement of magnons in the Heisenberg ring Jakubczyk Dorota Some results in the one-dimensional attractive Hubbard model. Jakubsky Vit Spectrally isomorphic Dirac systems: graphene in electromagnetic field Kameyama Takashi Minimal surfaces in $q$-deformed AdS$_5 times$S$^5$ Kassandrov Vladimir Quaternion Analyticity, Twistor Structures and Conservative Multi-Particle Dynamics Kirschner Roland Yang-Baxter relations with orthogonal and symplectic symmetries Kiselev Arthemy Deformation quantisation of fields Krivonos Sergey D-branes within coset approach Labuz Miroslaw A test for Bethe Ansatz solution of XXX model for magnetic heptagonal ring Lee Jen-Chi Biquaternion Construction of Non-Compact SL(2,C) Yang-Mills Instantons Liverts Evgeny Lukierski Jerzy Real and pseudoreal forms of D=4 Euclidean (super)algebras and classical r-matrices. Matsui Chihiro Multi-state extension of the asymmetric simple exclusion processes Matsumoto Takuya Deformations of $T^{1,1}$ as Yang-Baxter sigma models Mattelliano Massimo no talk McLoughlin Tristan Integrable Form Factors and AdS/CFT Megias Eugenio A holographic realization of light dilatons Melas Evangelos First results on the representation theory of the Ultrahyperbolic BMS group UHB(2,2) Meneses-Torres Claudio Abelianization of Fuchsian systems and the volumes of moduli spaces of parabolic bundles in genus 0 Mignemi Salvatore no talk Miyanishi Yoshihisa Notes on Feynman path integral-like methods of quantization on Riemannian manifolds Mizrahi Eti Classification of evolution equations with trivial and nontrivial $rho^{(3)}$:An Application of “Level Grading” Molev Alexander Higher Sugawara operators Natanzon Sergey Symmetric solutions of the dispersionless 2D Toda hierarchy, Hurwitz numbers and conformal dynamics. Nazmitdinov Rashid Symmetry effects in mesoscopic systems Nepomechie Rafael Counting solutions of Bethe’s equations Nirov Khazret Quantum affine algebras and functional relations Nishino Akinori Exact scattering eigenstates for double quantum-dot systems Nistico Giuseppe group theoretical approach to quantum interaction Ó Buachalla Réamonn The Noncommutative K”ahler Geometry of the Quantum Grassmannians and $q$-Deformed Schubert Calculus Ould Moustapha Mohamed Vall The 2d Liouville and the Morse Potentials Panella Orlando Quantum phase transitions in the non commutative Dirac Oscillator Petkova Valentina 3-point functions in Toda CFT and related braiding matrix Piatek Marcin 2d CFT/Gauge/Bethe correspondence and solvable quantum-mechanical systems Pietrykowski Artur Irregular blocks, $mathcal{N}=2$ gauge theory and Mathieu system Pimenta Rodrigo Modified algebraic Bethe Ansatz for the XXX Heisenberg chain on the segnent:towards scalar product. Poletaeva Elena On Yangians and Finite W-algebras Polyakov Dimitri Solutions in Bosonic String Field Theory and AdS Higher Spin Algebra Ponomarev Dmitry Towards holographic higher-spin interactions Prolhac Sylvain Current fluctuations for totally asymmetric exclusion on the relaxation scale Pupasov-Maksimov Andrey Analytical simulations of double-well, triple-well and multi-well dynamics via rationally extended Harmonic oscillators Ragnisco Orlando Taub-Nut and Darboux III systems revised: the behaviour beyond the singularity Sadurní Emerson Ergodic f-oscillators and their fractal structure Sakamoto Jun-ichi Yang-Baxter deformations of Minkowski spacetime Salom Igor Three-particle hyper-spherical harmonics and quark bound states Samsarov Andjelo Twist deformations leading to kappa-Poincare Hopf algebra and their application to physics Sawado Nobuyuki Solutions in the CP^N Skyrme-Faddeev type model Sergyeyev Artur New hierarchies of (3+1)-dimensional integrable systems related to contact geometry Shalaby Abouzeid Interesting Quantum-Field-Like Properties from Relativistic Schrodinger Equation with contact interaction Shapovalov Alexander Symmetry operators of the two-component Gross-Pitaevskii equation with a Manakov-type nonlocal nonlinearity Simulik Volodimir Link between the relativistic canonical quantum mechanics of arbitrary spon and the corresponding field theory Sitenko Yurii Self-adjointness and the Casimir effect with charged massive matter fields Skvortsov Evgeny Higher-spin theories in various dimensions Slavnov Nikita Form factors of local operators in GL(3)-invariant integrable models Spano Nathaly Fusing Defect for N=2 Supersymmetric sinh-Gordon model Stagraczyński Ryszard An explicit demonstration of integrability of the XXX model for pentagonal and hexagonal magnetic ring Strung Karen Classification of C*algebras Tarasov Vitaly On completeness of the Bethe ansatz for the XXX model Tepper David M. Riesz and A. Kolmogorov Theorems for Octonion-Valued Monogenic Fundtions Tseytlin Arkady Partition functions, higher spins and AdS/CFT Tudoran Razvan Micu Control and stability of periodic orbits of completely integrable systems Vala Jiri Topological phases in quantum lattice models Vassilevich Dmitri Spectral action at high energies Visinescu Mihai Hidden symmetries on toric Sasaki-Einstein spaces Vitiello Giuseppe Fractal self-similarity. From geometric structures to dynamical coherent dynamics Wagner Elmar Dirac operators on quantum spaces in local coordinates Werner Wend K-Theory and the Classification of Symmetric Spaces Yahagi Ryoko Finite-temperature behavior of an impurity in the spin-1/2 XXZ chain Yahalom Asher Topological Constants of Motion in Barotropic Fluid dynamics and Magnetohydrodynamics Yang Wenli Exact solutions of quantum integrable spin chain with various boundary conditions Yesmakhanova Kuralay Bright and dark positon solutions of the 1+1 – dimensional Hirota and the Maxwell-Bloch equations Yoshida Kentaroh Towards the gravity/CYBE correspondence Yupeng Wang Off-diagonal Bethe Ansatz and its applications Zakharov Dmitry Non-periodic one-gap solutions of the KdV equation Zhang Dajun Discrete Boussinesq-type equations Zhang MengXia No talk Zheltukhin Kostyantyn On existance of x-integral for a semi-discrete chains of hyperbolic type Zheng-Johansson A microscopic theory of the neutron
© 2003-2023 Department of mathematics, FNSPE, CTU
intsystems@fjfi.cvut.cz
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2023-02-07 14:13:02
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https://maze-rl.readthedocs.io/en/latest/environment_customization/customizing_environments.html
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# Customizing Core and Maze Envs¶
Whenever simulations reach a certain level of complexity or (ideally) already exist, but have been developed for other purposes than the RL scenario, the Gym-style environment interfaces might not be sufficient anymore to meet all technical requirements (e.g., the state is too complex to be represented as a simple Gym-style numpy array). In case of existing simulations it probably was not even taken into account at all and we have to deal with simulation specific interfaces and objects.
To cope with such situations Maze introduces a few additional concepts which are summarized in the figure below. Before we continue with some practical examples emphasizing why this structure is useful for environment customization and convenient experimentation, we first describe the concepts and components in a bit more detail. You can also find these components in the reference documentation.
Observation- and ActionConversionInterfaces:
Maze introduces MazeStates and MazeActions, extending Observations and Actions (represented as numerical arrays) to simulation specific generic objects. This grants more freedom in choosing appropriate environment-specific representations to separate the data model from the numerical representation, which in turn greatly simplifies the development and readability of environment and engineered baseline agent implementations.
• Action: the Gym-style, machine readable action.
• MazeAction: the simulation specific representation of the action (e.g., an arbitrary Python object).
• ActionConversionInterface: maps agent actions to environment (simulation) specific MazeActions and defines the respective Gym action space.
• Observation: the Gym-style, machine readable observation (e.g., a numpy array).
• MazeState: the simulation specific representation of the observation (e.g. an arbitrary Python object).
• ObservationConversionInterface: maps simulation MazeStates to Gym-style observations and defines the respective Gym observation space.
Core and Maze Environments:
The same distinction is carried out for environments.
• CoreEnv: this is the central environment, which could be also seen as the simulation, forming the basis for actual, RL trainable environments. CoreEnvs accept MazeAction objects as input and yield MazeState objects as response.
• CoreEnv Config: configuration parameters for the CoreEnvironment (the simulation).
• MazeEnv: wraps the CoreEnvs as a Gym-style environment in a reusable form, by utilizing the interfaces (mappings) from the MazeState to the observations space and from the MazeAction to the action space.
## List of Features¶
Introducing the concepts outlined above allows the following:
• Implement and maintain observations and actions as arbitrarily complex, simulation specific objects (MazeStates and MazeActions). In many cases sticking to Gym spaces gets quite cumbersome and makes the development processes unnecessarily complex.
• Easily experiment with different observation and action spaces simply by switching the Observation- and ActionConversionInterface.
• Train agents based on existing 3rd party simulations (environments) by implementing the Observation- and ActionConversionInterfaces (of course this also requires to have a Python API available).
• Easy configuration of the CoreEnv (simulation).
## Example: Core- and MazeEnv Configuration¶
The config snippet below shows an example environment configuration for the built-in cutting-2d environment.
# @package env
_target_: maze_envs.logistics.cutting_2d.env.maze_env.Cutting2DEnvironment
# parametrizes the core environment (simulation)
core_env:
max_pieces_in_inventory: 1000
raw_piece_size: [100, 100]
demand_generator:
_target_: mixed_periodic
n_raw_pieces: 3
m_demanded_pieces: 10
rotate: True
# defines how rewards are computed
reward_aggregator:
_target_: maze_envs.logistics.cutting_2d.reward.default.DefaultRewardAggregator
# defines the conversion of actions to executions
action_conversion
- _target_: maze_envs.logistics.cutting_2d.space_interfaces.action_conversion.dict.ActionConversion
max_pieces_in_inventory: 1000
# defines the conversion of states to observations
observation_conversion:
- _target_: maze_envs.logistics.cutting_2d.space_interfaces.observation_conversion.dict.ObservationConversion
max_pieces_in_inventory: 1000
raw_piece_size: [100, 100]
The config defines:
• which MazeEnv to use,
• the parametrization of the CoreEnv including reward computation,
• how MazeStates are converted to observations and
• how actions are converted to MazeActions.
All components together compose a concrete RL problem instance as a trainable environment. In particular, whenever you would like to experiment with specific aspects of your RL problem (e.g. tweak the observation space) you only have to exchange the respective part of your environment configuration.
Note
As showing concrete implementations of a CoreEnv or the Observation- and ActionConversionInterfaces is beyond the scope of this page we refer to the Maze - step by step tutorial for details.
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2021-10-21 23:32:34
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https://leetcode.ca/2020-05-25-1638-Count-Substrings-That-Differ-by-One-Character/
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Formatted question description: https://leetcode.ca/all/1638.html
# 1638. Count Substrings That Differ by One Character (Medium)
Given two strings s and t, find the number of ways you can choose a non-empty substring of s and replace a single character by a different character such that the resulting substring is a substring of t. In other words, find the number of substrings in s that differ from some substring in t by exactly one character.
For example, the underlined substrings in "computer" and "computation" only differ by the 'e'/'a', so this is a valid way.
Return the number of substrings that satisfy the condition above.
A substring is a contiguous sequence of characters within a string.
Example 1:
Input: s = "aba", t = "baba"
Output: 6
Explanation: The following are the pairs of substrings from s and t that differ by exactly 1 character:
("aba", "baba")
("aba", "baba")
("aba", "baba")
("aba", "baba")
("aba", "baba")
("aba", "baba")
The underlined portions are the substrings that are chosen from s and t.
Example 2:
Input: s = "ab", t = "bb"
Output: 3
Explanation: The following are the pairs of substrings from s and t that differ by 1 character:
("ab", "bb")
("ab", "bb")
("ab", "bb")
The underlined portions are the substrings that are chosen from s and t.
Example 3:
Input: s = "a", t = "a"
Output: 0
Example 4:
Input: s = "abe", t = "bbc"
Output: 10
Constraints:
• 1 <= s.length, t.length <= 100
• s and t consist of lowercase English letters only.
Related Topics:
Hash Table, String, Trie, Rolling Hash
## Solution 1.
Intuition: We can find each pair of s[i] != t[j]. Then try to extend both sides when s[i + t] == t[j + t]. If we have left steps extended on the left side and right steps on the right side, we have (left + 1) * (right + 1) options for this { i, j } case.
Example:
s = xbabc
t = ybbbc
For i = 2 and j = 2, we have s[i] = a and t[j] = b that doesn’t match. Now look leftwards, we can extend left-side by 1 time due to b, and extend right-side by 2 times due to bc. So for this specific center { i = 2, j = 2 }, we have 2 * 3 = 6 options.
// OJ: https://leetcode.com/problems/count-substrings-that-differ-by-one-character/
// Time: O(MN * min(M, N))
// Space: O(1)
class Solution {
public:
int countSubstrings(string s, string t) {
int M = s.size(), N = t.size(), ans = 0;
for (int i = 0; i < M; ++i) {
for (int j = 0; j < N; ++j) {
if (s[i] == t[j]) continue;
int left = 1, right = 1;
while (i - left >= 0 && j - left >= 0 && s[i - left] == t[j - left]) ++left;
while (i + right < M && j + right < N && s[i + right] == t[j + right]) ++right;
ans += left * right;
}
}
return ans;
}
};
## Solution 2.
We can precompute the left and right values to save time.
// OJ: https://leetcode.com/problems/count-substrings-that-differ-by-one-character/
// Time: O(MN)
// Space: O(MN)
class Solution {
public:
int countSubstrings(string s, string t) {
int M = s.size(), N = t.size(), ans = 0, left[101][101] = {}, right[101][101] = {};
for (int i = 0; i < M; ++i) {
for (int j = 0; j < N; ++j) {
left[i + 1][j + 1] = s[i] == t[j] ? left[i][j] + 1 : 0;
}
}
for (int i = M - 1; i >= 0; --i) {
for (int j = N - 1; j >= 0; --j) {
right[i][j] = s[i] == t[j] ? right[i + 1][j + 1] + 1 : 0;
}
}
for (int i = 0; i < M; ++i) {
for (int j = 0; j < N; ++j) {
if (s[i] != t[j]) ans += (1 + left[i][j]) * (1 + right[i + 1][j + 1]);
}
}
return ans;
}
};
## Solution 3.
Consider the following s and t and we are using x and y as the differing characters.
s=ab[x]c
t=ab[y]c
When we start from i = 0, j = 0, and reaches i = 2, j = 2, since s[i] != t[j], pre is updated as cur = 3, and cur is reset to 0. We add 3 to the answer which covers
ab[x]
ab[y]
b[x]
b[y]
[x]
[y]
When we reach i = 3, j = 3, we add pre = 3 to answer again, which covers
ab[x]c
ab[y]c
b[x]c
b[y]c
[x]c
[y]c
So the pre is the same as the left value in previous solutions. The right value is achieved through adding the pre value repetitively for repeating right-side characters.
The i and j of helper function are the starting indexes of our scanning. Note that 0, 0 should be only included once so j starts from 1 in the second loop.
// OJ: https://leetcode.com/problems/count-substrings-that-differ-by-one-character/
// Time: O(N)
// Space: O(1)
// Ref: https://leetcode.com/problems/count-substrings-that-differ-by-one-character/discuss/917985/JavaC%2B%2BPython-Time-O(nm)-Space-O(1)
class Solution {
int helper(string s, string t, int i, int j) {
int ans = 0, pre = 0, cur = 0;
for (int n = s.size(), m = t.size(); i < n && j < m; ++i, ++j) {
cur++;
if (s[i] != t[j]) pre = cur, cur = 0;
ans += pre;
}
return ans;
}
public:
int countSubstrings(string s, string t) {
int ans = 0 ;
for (int i = 0; i < s.size(); ++i) ans += helper(s, t, i, 0);
for (int j = 1; j < t.size(); ++j) ans += helper(s, t, 0, j);
return ans;
}
};
Java
class Solution {
public int countSubstrings(String s, String t) {
int count = 0;
int length = s.length();
for (int i = 0; i < length; i++) {
for (int j = i + 1; j <= length; j++) {
String substr = s.substring(i, j);
count += countStrings(substr, t);
}
}
return count;
}
public int countStrings(String s, String t) {
int count = 0;
int length = s.length();
int maxStart = t.length() - length;
for (int i = 0; i <= maxStart; i++) {
if (differByOne(s, t.substring(i, i + length)))
count++;
}
return count;
}
public boolean differByOne(String s, String t) {
if (s.length() != t.length() || s.equals(t))
return false;
int differ = 0;
int length = s.length();
for (int i = 0; i < length && differ <= 1; i++) {
if (s.charAt(i) != t.charAt(i))
differ++;
}
return differ == 1;
}
}
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2022-05-22 11:50:28
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http://www.techartus.com/omd-emea-xck/dummit-and-foote-vs-lang-a250f4
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# dummit and foote vs lang
CRL over HTTPS: is it really a bad practice? Why was Warnock's election called while Ossof's wasn't? Foote [3] and S. Lang [4]. In addition, it is written in a more user-friendly, down-to-earth fashion than, say, Lang… Is the set of sequences with values in a finite set separable? In addition, it is written in a more user-friendly, down-to-earth fashion than, say, Lang's Algebra is. had been written under his comment. Homological algebra: Additive and abelian categories, Complexes and homology, long exact sequences, homotopy, resolutions, derived functors, Ext, Tor, cohomology of groups, extensions of groups. If you are not afraid of challenges then start with Topics in Algebra (Herstein ) .It is absolutely fantastic book that contains both good theory and problems. How to label resources belonging to users in a two-sided marketplace? Abstract Algebra Dummit Foote Archives - Page 22 of 24 ... Dummit and Foote contains just about everything an undergraduate ought to know about abstract algebra. The exercises are too difficult and the text is too concise to help a starting mathematician grow. This is an uno cial solution guide to the book Abstract Algebra, Third Edition, by David S. Dummit and Richard M. Foote. So far, I've done well in the courses but still feel unprepared for the prelim exams. Lang is written as a reference work. Dummit and Foote, Wiley & Sons : On reserve in math library: Algebra: Lang, Springer/Nature : On reserve in math library: Proof writing and reading: How to read and do proofs: Solow, Wiley & Sons: In Math Library (QA9.54.S65 2014) How to prove it: Velleman: In Math Library (QA9.V38 1994) If you really don't want to read Lang, don't. I'll post my solutions for all to see after the marathon. Luckily, I also had books by competent authors (Abstract Algebra by Dummit and Foote, and Algebra by Serge Lang), so all was not lost. Bourbaki Proof of Zorn's Lemma in Lang's Algebra, Prerequisites for studying Homological Algebra, Categorical introduction to Algebra and Topology, Primitive element theorem w/o Galois theory (as in Lang's Algebra). I should admit that I did not take algebra with Langâs book - perhaps it would be better after spending more time with it. Making statements based on opinion; back them up with references or personal experience. Welcome to the webpage for Math 5031. In addition, it is written in a more user- friendly, down-to-earth fashion than, say, Lang… Not in a million years. I tried to learn algebra from Serge Lang's book (some two and half years back), but due to lack of background, I could not understand a bit of it, and I lost interest. What does it mean when an aircraft is statically stable but dynamically unstable? In addition, it is written in a more user-friendly, down-to-earth fashion than, say, Lang's Algebra is. Yes: that takes some training... and you might take the opportunity of this course to train yourself also in that respect. The other two books have also been used before, sometimes in conjunction with Artin. How can there be a custom which creates Nosar? I easily get irritated by seeing that book. Making statements based on opinion; back them up with references or personal experience. Dummit & R.M. I will try(!) Dummit and Foote is a good book too. solutions to dummit and foote chapter 3.pdf . Products and coproducts, the hom functor, representable functors, universals and adjoints. My instructors assign important exercises in the book and I tried them all. Unfortunately, Ribet has taken down that link; maybe that has something to The Tensor Product and Induced Modules Nayab Khalid The Tensor Product A Construction Properties Examples References References Dummit, D. and Foote, R. Abstract Algebra On the other hand, Dummit and Foote is considerably more comprehensive and probably a little more challenging than Artin's book. At the same time, there is also no reasonable reason for you to put yourself in a position where you eliminate one of the possible sources of information about what you want to know! Function of augmented-fifth in figured bass, Zero correlation of all functions of random variables implying independence. We cannot suggest a different one except these two? Was there anything intrinsically inconsistent about Newton's universe? This is such a beautiful topic, and perfectly appropriate for this level, but Herstein and Dummit & Foote do not touch it with a long pole. On top of that, the topics are sometimes somewhat out of order, with examples from one chapter coming from material from a later chapter (though this can be said of Dummit and Foote as well). Dummit and Foote contains just about everything an undergraduate ought to know about abstract algebra. Where To Download Dummit And Foote Dummit And Foote When people should go to the book stores, search introduction by shop, shelf by shelf, it is essentially problematic. Which one to choose between Lang and Dummit Foote, Dummit and Foote as a First Text in Abstract Algebra, Group theory book: presentations and group actions, Alternatives to Dummit and Foote for a text in Algebra, Linear Algebra book (useful for advanced algebra courses), Recommendation for a graduate textbook in algebra, Oxygen level card restriction on Terraforming Mars. I do not believe students are significantly less capable today than they were several dec⦠Dummit and Foote 4.3.13: Find all the finite groups which have exactly two conjugacy classes. There are still many overqualified students and many underqualified students, and I can provide a constructive proof of that fact. MathJax reference. Download Free Dummit And Foote Solutions Dummit And Foote Solutions We would like provide a complete solution manual to the book Abstract Algebra by Dummit & Foote 3rd edition. It will very ease you to see guide dummit and foote … Active 5 years, 7 months ago. rev 2021.1.7.38271, Sorry, we no longer support Internet Explorer, The best answers are voted up and rise to the top, Mathematics Stack Exchange works best with JavaScript enabled, Start here for a quick overview of the site, Detailed answers to any questions you might have, Discuss the workings and policies of this site, Learn more about Stack Overflow the company, Learn more about hiring developers or posting ads with us. Buy from Amazon Chapter 0: Preliminaries §0.1: Basics Solution to Abstract Algebra by Dummit To subscribe to this RSS feed, copy and paste this URL into your RSS reader. Dummit and Foote contains just about everything an undergraduate ought to know about abstract algebra. Algebra dummit and foote vs lang - physics forums Hi, I'm going to start learning Abstract Algebra, and I was wondering which book, either Lang (his graduate version) or Dummit and Foote, is better. Just take the class, read the book, go to all the lectures, and ask questions about the parts you don't understand. Artinian and Noetherian modules, Krull-Schmidt theorem, completely reducible modules, projective modules, Wedderburn-Artin Theorem for simple rings. please send comments, suggestions and corrections by e-mail, or as a comment in this website. It can be very frustrating trying to learn anything from this book if you've never seen it before. In addition, it is written in a more user-friendly, down-to-earth fashion than, say, Lang's Algebra is. 11-12; Axler, Linear Algebra Done Right. Abstract algebra dummit foote You can hand in homework earlier in my mailbox, on the ground floor of the math department, across All books are in clear copy here, and all files are secure so don't worry about $79.95. Thanks for contributing an answer to Mathematics Stack Exchange! Try looking for reference requests on this site for these subjects. Is the Gelatinous ice cube familar official? Why was there a "point of no return" in the Chernobyl series that ended in the meltdown? I want to remind again that I have no motivation to. Dummit and Foote has also been used in the past for Math 370 and Math 371; it is very carefully written and contains a large number of exercises. For what subject specifically? Amazon.com: Customer reviews: Abstract Algebra, 3rd Edition Download File PDF Dummit And Foote Solutions Chapter 7 Angfit to exercises from . I can't understand which book to start with.I have basic knowledge in group and ring theory and i also know some field theory. "Lang's Algebra changed the way graduate algebra is taught, retaining classical topics but introducing language and ways of thinking from category theory and homological algebra. Abstract Algebra Theory and Applications Thomas W. Judson Stephen F. Austin State University August 27, 2010 Colleagues don't congratulate me or cheer me on, when I do good work? Great job! 4.6 out of 5 stars 71. I am interested in working in Algebra/Algebraic Geometry/Algebraic Number Theory. In both undergraduate courses and graduate courses students' abilities and background are quite variable; that much has not changed in the past few decades. Widely acclaimed algebra text. I just read whatever I had to read. Transcendental extensions: transcendence bases, separating transcendence bases, Luroth's theorem. Solution Manual Of Abstract Algebra By Dummit And Foote. I do not know if I have to choose some other book or convince myself (I do not know how) to be with Lang's book. I do not want to replace lang with dummit foote.. :). Rather, I take The curriculum for coursework in the coming year is: Review of field and Galois theory: solvable and radical extensions, Kummer theory, Galois cohomology and Hilbert's Theorem 90, Normal Basis theorem. Hi, I'm going to start learning Abstract Algebra, and I was wondering which book, either Lang (his graduate version) or Dummit and Foote, is better. 첫 교재로는 약간 어려운 감이 있긴 하지만 Dummit/Foote만큼 기초예제를 지루하리만큼 성실하게 풀이해 놓은 교재가 또 없습니다. It is intended for students who are studying algebra with Dummit and Foote’s text. At the graduate level, I like Dummit and Foote and, lets face it, it is hard to ignore Lang. Amazon.com: Customer reviews: Abstract Algebra, 3rd Edition Dummit and Foote contains just about everything an undergraduate ought to know about abstract algebra. But I have not taught from Dummit and Foote. site design / logo © 2021 Stack Exchange Inc; user contributions licensed under cc by-sa. Online Library Abstract Algebra Dummitedition. Personally, I am partial to Isaacs. His depth is comprehensive, but he jumps into most topics at the maximum level of complication and detail, which is extremely jarring for someone just learning the material. It will be updated regularly. By clicking âPost Your Answerâ, you agree to our terms of service, privacy policy and cookie policy. Examples of interesting rings to study during an undergraduate course in non-commutative rings. 5 people found this helpful. what books would you recommend for intuition building? Derivations. Index of a subgroup in a group. As was mentioned before, use Dummit & Foote. Dummit Foote Homework Solutions Chapter 8 David S. Dummit, Richard M. Foote. In executing the evaluation homomorphism, it is useful to represent Fred as a certain sum of monomials, which makes clear how to perform the evaluation using the member a from F. Free objects. Download File PDF Dummit And Foote Solutions Chapter 4 Dummit And Foote Solutions Chapter We would like provide a complete solution manual to the book Abstract Algebra by Dummit & Foote 3rd edition. Is it possible to assign value to set (not setx) value %path% on Windows 10? Friedberg et al. Which one to buy between these two? Direct and inverse limits. Books that build intuition tend to have narrow subjects.$\begingroup$The book I used in my first abstract algebra course was Dummit and Foote, which I thought was very well-written. Choose accordingly; or maybe use one text as a companion for the other. Algebra- very good exercises! What do this numbers on my guitar music sheet mean, Altering table row colors trought multirow. Solutions for exercises in chapter 13 field theory of abstract algebra, by david s. dummit & richard m. foote. The particular choice of book really doesn't matter that much; there certainly isn't a single, canonical source for any of the items on the syllabus you posted. Case in point: Symmetries of plane figures and crystallographic groups. Solutions to Abstract Algebra (Dummit and Foote 3e) Chapter 1 : Group Theory Jason Rosendale jason.rosendale@gmail.com February 11, 2012 This work was done as an undergraduate student: if you really don’t understand something in one of these proofs, it is very possible that it doesn’t make sense because it’s wrong. Asking for help, clarification, or responding to other answers. Why can't modern fighter aircraft shoot down second world war bombers? Stack Exchange network consists of 176 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. Also you start liking the conciseness if you know more: you did not open that book for intuition building but to read how that one proof went. I always wanted to read it, but because I could not understand anything in it, and because most of my seniors keep saying "Lang is difficult," I lost interest and hope in reading that. You will have to make do with what you have. Why don't unexpandable active characters work in \csname...\endcsname? There will be weekly exercise sessions by the TA. I only used Dummit and Foote, so my answer should serve as a complement of Mr Hashi's. I learned linear Algebra from lecture notes in Dutch and read no other books. I really like (and would recommend) Dummit and Foote for those wanting to learn abstract algebra I've used others (Lang, Jacobson, and Herstein). @OllieFord, well, I was responding to the question in his comment, namely how to deal with irritation. To learn more, see our tips on writing great answers.$\begingroup\$ In addition to aliases for an object, there is also the issue of representation. Many of the topics you listed will only be touched upon briefly, and some not at all. Detailed survey of group theory, including the Sylow theorems and the structure of finitely generated abelian groups, followed by a ⦠Dummit and Foote contains just about everything an undergraduate ought to know about abstract algebra. Dummit and Foote contains just about everything an undergraduate ought to know about abstract algebra. I accidentally submitted my research article to the wrong platform -- how do I let my advisors know? It has affected all subsequent graduate-level algebra books." In addition, it is written in a more user- friendly, down-to-earth fashion than, say, Lang… First, I believe Dummit and Foote contains all the materials you need. You could do what I did, too: read other books when first learning a topic, get a basic grasp of it, then transition to Lang to finish. Is a lightfoot halfling obscured for the purposes of hiding while in the space of another creature? There are many wonderful algebra texts. Please also make a … I have never bought so much mathematics for 90 euro. I only used Lang from those two, so I will tell you about that one. Abstract Algebra. solutions to dummit and foote chapter 3.pdf . Another thing that I heard is that it is better to use Lang as reference book than a textbook for a course. Functors and natural transformations, equivalence of categories,. One caution about using this book is it contains too much material.Trying to read all of them at once would be a huge waste of time and well of frustration. Commutative Algebra (Atiyah and Macdonald). Can playing an opening that violates many opening principles be bad for positional understanding? Chapter 1 Introduction and deï¬nitions 1.1 Introduction Abstract Algebra is the study of algebraic systems in an abstract way. ... (2-3 weeks): Dummit-Foote, Ch. This is an introductory graduate level course on the basic structures and methods of algebra. One of my friend gave me his copy of Abstract Algebra by Dummit and Foote. Dummit & Foote Ch 14.1-14.4: 2018-12-06 (Rm 3494) Cyclic and Abelian Extensions, Solvability of General Polynomials, Galois Groups of Polynomials of Low Degree, Inseparable Extensions, Field Norm and Trace, Hilbert's Theorem 90, Group Cohomology, Kummer Theory, Infinite Galois Theory: Dummit & Foote Ch 14.5-14.9, Ch 17.2-17.3. Mac Lane, Categories for the Working Mathematician, is a standard introduction to category theory. How to teach a one year old to stop throwing food once he's done eating? Quit the subject because you do not like the source? I am confused.Please help. Categories and functors: definitions and examples. Will a divorce affect my co-signed vehicle? Algebra" (third edition) by Serge Lang, Addison Wesley, 1993, ISBN 0-201-55540-9 Advanced Modern Algebra" by Joseph J. Rotman Prentice Hall ISBN 0-13-087868-5 Basic Algebra and Advanced Algebra Set" by Anthony W. Knapp, Birkhauser ISBN 0817645330; perhaps too sophisticated for this level, but otherwise two superb books by a great expositor. What causes dough made from coconut flour to not stick together? then?? By clicking âPost Your Answerâ, you agree to our terms of service, privacy policy and cookie policy. It has a lot of content, but I do not find it particularly clear. I bought Lang as a third year bachelor with the same knowledge as you approximately. Anyway, if you do choose to replace Lang with something else, it shouldn't be Dummit and Foote. Chains, forms, and duality: A novice's guide from vector calculus to manifolds 3: Last notes played by piano or not? With regards to Libear Algebra and Abstract Algebra. 5.0 out of 5 stars good text. A common alternative is Hungeford , which achieves a comparable level of detail. MathJax reference. Asking for help, clarification, or responding to other answers. 1.1 Group theory Groups, subgroups. Representations of finite groups: complete reducibility, characters, orthogonal relations, induced modules, Frobenius reciprocity, representations of the symmetric group. It is fantastic for this purpose. I'm really hoping that you read Lang, understand it all, and then turn around and write a book that is, @dotancohen : I am happy for what ever you mean.... :) :). Figure out the way you learn best, including what kinds of textbooks you like, and go from there. Absolutely not. Please also make a comment if you would like some particular problem to be updated. How does Shutterstock keep getting my latest debit card number? Group Theory (~6 weeks): Dummit-Foote… Dummit and Foote Solutions - Greg Kikola Solution to Abstract Algebra by Dummit & Foote 3rd edition Chapter 1.1 Exercise 1.1.12 Find the orders of the following elements of the multiplicative group (Z / (12)) ×: 1 ¯, − 1 ¯, 5 ¯, 7 ¯,… Abstract Algebra Dummit Foote Archives - … As was mentioned before, use Dummit & Foote. Use MathJax to format equations. Basic Algebraic Geometry (Shafarevich), or Algebraic Curves (Fulton, free online) -- ⦠The material is explained very well in this book. to supplement the textbook with notes, as I did in Representation theory of finite groups in Spring 2020, but I am not promising this. mRNA-1273 vaccine: How do you say the “1273” part aloud? The suggested book for this is S. Lang, Algebra, 3rd Ed., Addison Wesley, 1993. Solutions To Dummit And Foote Abstract Algebra Dummit Foote Solutions Download Pdf , Free Pdf Abstract Algebra . Abstract algebra (Dummit and Foote) -- Easy to digest, but I much prefer the following; Algebra (Lang, GTM) -- very amazing introduction, a bit terse. With this said, when trying to study the same topics for my course from other books, I found myself eventually coming back to Lang. What are you going to do? After every algebra course I took, Commutative Algebra, Advanced Field Theory, Category theory, algebraic number theory etc. how to ad a panel in the properties/data Speaker specific. Seeking a study claiming that a successful coup d’etat only requires a small percentage of the population. Estoy usando el libro de Álgebra de Serge Lang. Swap the two colours around in an image in Photoshop CS6. math 171 - abstract algebra i section hw # 16 ... Lang's Algebra Third Edition, Galois Theory by Emil Artin. Textbook: Abstract Algebra by Dummit and Foote (mostly Chapters 13-14). Valuations and completions: definitions, polynomials in complete fields (Hensel's Lemma, Krasner's Lemma), finite dimensional extensions of complete fields, local fields, discrete valuations rings. [FIS] is a nice reference for linear algebra. Wait for somone to provide another source (as this may well never happen, this may have the same consequence as quitting the subject!)? Amazon.com: Customer reviews: Abstract Algebra, 3rd Edition Not in a million years. 2020.10.22 00:38. It doesn't help give any intuition for the basic ideas because it assumes you have seen them a dozen times before. I am looking for a complete book on Abstract Algebra.By a "complete book" I mean to say a book that will cover all the topics of abstract algebra starting from group theory,ring theory ,field theory,Galois theory,Non-commutative rings ,jacobson radicals,Kummer ,cyclotomic extensions ,Krull topology and ending in Commutative algebra.
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2021-08-03 10:10:19
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http://quitramchiasurp-blog.logdown.com/posts/7102736-john-taylor-classical-mechanics-solutions-manual-p-megaupload-hearts-ma
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6 months ago
## John Taylor Classical Mechanics Solutions Manual P lphant blunt renglon
John Taylor Classical Mechanics Solutions Manual Pdf.zip
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2018-09-22 22:21:13
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http://www.jstor.org/stable/25464791
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## Access
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# Brownian Motion in Self-Similar Domains
Dante Deblassie and Robert Smits
Bernoulli
Vol. 12, No. 1 (Feb., 2006), pp. 113-132
Stable URL: http://www.jstor.org/stable/25464791
Page Count: 20
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## Abstract
For T ≠ 1, the domain G is T-homogeneous if TG = G. If 0 ∉ G, then necessarily 0 ∈ ∂G. It is known that for some p > 0, the Martin kernel K at infinity satisfies $K(Tx)=T^{p}K(x)$ for all x ∈ G. We show that in dimension d ≥ 2, if G is also Lipschitz, then the exit time $\tau _{G}$ of Brownian motion from G satisfies $P_{x}(\tau _{G}>1)\approx K(x)t^{-p/2}$ as t → ∞. An analogous result holds for conditioned Brownian motion, but this time the decay power is 1 - p - d/2. In two dimensions, we can relax the Lipschitz condition at 0 at the expense of making the rest of the boundary C².
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https://www.vedantu.com/question-answer/the-given-equation-int-dfrac1-x21-+-x2sqrt-1-+-class-10-maths-cbse-5edf351366b432798b3413ba
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The given equation $\int {\dfrac{{1 - {x^2}}}{{(1 + {x^2})\sqrt {1 + {x^4}} }}dx}$ is equal to:A.$\sqrt 2 {\sin ^{ - 1}}(\dfrac{{\sqrt 2 x}}{{{x^2} + 1}}) + c$B.$\dfrac{1}{{\sqrt 2 }}{\sin ^{ - 1}}(\dfrac{{\sqrt 2 x}}{{{x^2} + 1}}) + c$C.$\dfrac{1}{2}{\sin ^{ - 1}}(\dfrac{{\sqrt 2 x}}{{{x^2} + 1}}) + c$D.None of these
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Hint: Use substitution method of integration and let's take $\dfrac{1}{x} + x$ as t and then solve with respect to t first and later substitute its value.
Consider the given integral as I. Now,
$\ I = \int {\dfrac{{1 - {x^2}}}{{(1 + {x^2})\sqrt {1 + {x^4}} }}dx} \\ \Rightarrow I = \int {\dfrac{{{x^2}(\dfrac{1}{{{x^2}}} - 1)}}{{{x^2}(\dfrac{1}{x} + x)\sqrt {\dfrac{1}{{{x^2}}} + {x^2}} }}dx} \\ \Rightarrow I = \int {\dfrac{{(\dfrac{1}{{{x^2}}} - 1)}}{{(\dfrac{1}{x} + x)\sqrt {\dfrac{1}{{{x^2}}} + {x^2}} }}dx} \\ \$
Observe that, If, we consider $\dfrac{1}{x} + x$ as t then, on differentiating, we can get $- \dfrac{1}{{{x^2}}} + 1$ as $dt$ .Using these information,
$\ \Rightarrow I = \int {\dfrac{{(\dfrac{1}{{{x^2}}} - 1)}}{{(\dfrac{1}{x} + x)\sqrt {\dfrac{1}{{{x^2}}} + {x^2}} }}dx} \\ \Rightarrow I = \int {\dfrac{{ - dt}}{{t\sqrt {{t^2} - {{(\sqrt 2 )}^2}} }}{\text{ }}[{x^2} + \dfrac{1}{{{x^2}}} = {{(x + \dfrac{1}{x})}^2} - 2]} \\ \Rightarrow I = - \int {\dfrac{{dt}}{{t\sqrt {{t^2} - {{(\sqrt 2 )}^2}} }}} \\ \Rightarrow I = \dfrac{1}{{\sqrt 2 }}\cos e{c^{ - 1}}(\dfrac{{x + \dfrac{1}{x}}}{{\sqrt 2 }}) + c \\ \Rightarrow I = \dfrac{1}{{\sqrt 2 }}{\sin ^{ - 1}}(\dfrac{{\sqrt 2 x}}{{{x^2} + 1}}) + c \\ \$
Hence the given integral $I = \dfrac{1}{{\sqrt 2 }}{\sin ^{ - 1}}(\dfrac{{\sqrt 2 x}}{{{x^2} + 1}}) + c$
Option B is correct.
Note: There is one more method to solve this integral by substitution x with a trigonometric function. We found that this method is easier to understand.
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2023-03-22 00:44:54
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https://electronics.stackexchange.com/questions/523550/stm32-sampling-adc-value-at-known-regular-time-intervals
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# STM32 - Sampling ADC value at known regular time intervals
I am working on a simple microcontroller based proportional controller to regulate the temperature of a heater cartridge using a feedback thermistor. I'm using the 'black pill' STM32F103C8T6 and have the controller working however I would like to send out time stamp and temperature readings over UART so that I can plot a graph of the temperature output vs time and use the Ziegler-Nichols method to obtain integral and derivative values (by obtaining the ultimate period of oscillation and ultimate gain).
I played around with DMA mode and ADC but had a lot of troubles so ended up using the simple polling method mentioned in the answer here:
The controller works in the main loop:
/* USER CODE BEGIN 2 */
/* USER CODE END 2 */
/* Infinite loop */
/* USER CODE BEGIN WHILE */
while (1)
{
temperatureInDegrees = read_temp(waxThermistor); //Calculates the value in degrees using a look up table
}
if(heater==1){
HAL_TIM_PWM_Start(&htim3, TIM_CHANNEL_1); //PWM for heater element
heater = 0;
}
int currentTemperature = temperatureInDegrees;
int pi = PI(30,currentTemperature,50,0); //returns the controller value P + I
int actuate = limitActuation(pi,0,500); //Limits the actuation range between 0% and 50% duty cycle
htim3.Instance->CCR1 = actuate; //Sets the duty cycle
sendTemperature(currentTemperature); //transmits the temperature over UART
/* USER CODE END WHILE */
I would like to be able to send out a time value along with the temperature. I tried removing the polling and placing waxThermistor = HAL_ADC_GetValue(&hadc1);in the interrupt routine for a 20ms timer (using OC mode) however I no longer get any PWM output as the ADC value is just returning 0 all the time. Could anyone recommend a better method for ADC sampling at a regular known interval rather than polling like this?
One possible solution I thought of would be to setup the ADC to trigger conversion on a timer and then read the sample in the conversion complete callback function. The time elapsed would then be the time between triggers + time taken for conversion to complete which can be calculated through the formulas given here. Would this be a viable solution?
For example if sampling every 20ms, the data sent out over UART would look like:
0ms;25c\n
20ms;27c\n
40ms;35c\n
• You should configure the Timer & ADC such that the Timer directly triggers the ADC to perform a conversion, and then use the ADC's Conversion-Complete interrupt to read the result. You shouldn't need to use the TImer's interrupt and write code to start the ADC there. – brhans Sep 27 '20 at 19:17
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2021-05-11 13:07:29
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https://byjus.com/question-answer/25-circular-plates-each-of-radius-10-5-cm-and-thickness-1-6-cm-are-2/
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Question
# $$25$$ circular plates, each of radius $$10.5\ cm$$ and thickness $$1.6\ cm$$, are placed on above the other form of solid circular cylinder. Find the curved surface area and the volume of the cylindrical so formed.
Solution
## 25 circular platesRadius of plate = 10.5 cmthickness = 1.6 cmcurved surface Area = (?)volume of cylinder = (?)$$\rightarrow$$ Height of the cylinder $$= no.\, of\, plates \times \,thickness$$$$= 25 \times 1.6$$$$= 40 cm$$$$\rightarrow$$ curved surface Area $$= 2\pi r.h$$$$= 2\pi \times 10.5\times 40$$$$= 2,640\,cm^{2}$$$$\rightarrow$$ volume of the cylinder $$= \pi r^{2}.h$$$$= \pi \times (10.5)^{2}\times 40$$$$= 13.860 \,cm^{3}$$Mathematics
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2022-01-29 00:33:17
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https://phys.libretexts.org/Bookshelves/Classical_Mechanics/Variational_Principles_in_Classical_Mechanics_(Cline)/13%3A_Rigid-body_Rotation/13.15%3A_Kinetic_energy_in_terms_of_Euler_angular_velocities
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$$\require{cancel}$$
The kinetic energy is a scalar quantity and thus is the same in both stationary and rotating frames of reference. It is much easier to evaluate the kinetic energy in the rotating Principal-axis frame since the inertia tensor is diagonal in the Principal-axis frame as given in equation $$(13.12.14)$$
$T_{rot} = \frac{1}{2} \sum^3_i I_{ii} \omega^2_i$
Using equation $$(13.14.1-13.14.3)$$ for the body-fixed angular velocities gives the rotational kinetic energy in terms of the Euler angular velocities and principal-frame moments of inertia to be
$T_{rot}=\frac{1}{2}\left[I_{1}(\dot{\phi} \sin \theta \sin \psi+\dot{\theta} \cos \psi)^{2}+I_{2}(\dot{\phi} \sin \theta \cos \psi-\dot{\theta} \sin \psi)^{2}+I_{3}(\dot{\phi} \cos \theta+\dot{\psi})^{2}\right]$
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2021-10-26 02:26:48
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https://socratic.org/questions/is-x-1-a-factor-of-x-3-8x-2-11x-20
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# Is x+1 a factor of x^3+8x^2+11x-20?
Nov 8, 2016
$\left(x + 1\right)$ is not a factor, but $\left(x - 1\right)$ is.
#### Explanation:
Given $p \left(x\right) = {x}^{3} + 8 {x}^{2} + 11 x - 20$ if $x + 1$ is a factor of $p \left(x\right)$ then
$p \left(x\right) = \left(x + 1\right) q \left(x\right)$ so for $x = - 1$ we must have
$p \left(- 1\right) = 0$
Verifying on $p \left(x\right)$
$p \left(- 1\right) = {\left(- 1\right)}^{3} + 8 {\left(- 1\right)}^{2} + 11 \left(- 1\right) - 20 = - 24$
so $\left(x + 1\right)$ is not a factor of $p \left(x\right)$
but $\left(x - 1\right)$ is a factor because
$p \left(1\right) = 1 + 8 + 11 - 20 = 0$
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2021-09-18 19:20:31
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https://www.physicsforums.com/threads/basketball-throw-physics-problem.60955/
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Basketball throw physics Problem
:uhh: A basketball player throws the ball at a 38 degree angle to a hoop which is located a horizontal distance L=7.0m from the point of release, and at a height h=0.3m above it. What is the required speed if the basketball is to reach the hoop?
help please
Answers and Replies
Use the equation for the projectile (it should be in your text book), then plug in the numbers and solve for the required speed.
cepheid
Staff Emeritus
Science Advisor
Gold Member
Strategy...as I mentioned to you in your true/false thread, horizontal and vertical components of the bball's velocity are independent. So...call the initial velocity you're trying to solve for v_0
The x-component of the v_0 is unaffected throughout the journey. Therefore, you don't have a clue yet what to do with it. It is constant, so you need to know the time required for the ball to get there. That depends on how high it needs to go before stopping (the vertical component). So, just note for now that the x component is given by :
$$v_0\cos(38^o)$$
and ignore it for the time being.
Now, the vertical component of the velocity:
$$v_0\sin(38^o)$$
is affected by the gravitational force, which causes a constant downward acceleration, which slows the ball's velocity by 9.81 metres per second, every second. So, how do you solve for the required initial vertical velocity, and the time needed to get the ball up to that height? There are three equations...one gives the vertical component of the velocity as a function of time, (which is useful to you, because you know the final vertical velocity is zero), the other gives the height h(t) as a function of time, in terms of the vertical velocity and and accelaration. The third is the fact that:
$$[v_0\cos(38^o)]t$$ must be equal to three metres.
Hopefully you know which kinematic formulas I'm talking about.
i tried doing this problem by plugging in values into this equation:
Vxo = Vx(cos38) Vyo=Vy(sin38)
combining the two I found,
y=(tan38)x - (g / 2(Vo^2)(cos38)^2)x^2
however my answer was wrong.
( I tried your method aswell cepheid - but it was long and complicated - I think I got lost in the numbers along the way.) :s
just wanna make it clear
did you mean
$$y = tan(38)x - \frac{g x^2}{2 v_{0}^2 cos^2(28))$$
if yes, your answer should be fine, what numerical v you got?
sort of yeah - except in the second part it would be (cos38)^2
and whats frac?
I found Vo = 6.115 m/s and it was wrong...:(
that was a typo
here you go
$$y = tan(38)x - \frac{g x^2}{2 v_{0}^2 cos^2(38)}$$
let me check... 1 minute
Last edited:
how did you write it out so clearly??
i don't get your answer... let me try to figure out what you diid wrong...
yes! thats the equation I used!!
was it just a calculation error then? this is the right equation to use to solve the problem?
I'll try it again - maybe i din't put brackets around certain #'s in my calculator or something...
basically I plugged in:
x = 7.0m
y = 0.3m
g = 9.8m/s^2
and solved for Vo
what is your Vo equal to... you might either have some algebra error or calculatioin error
31.95m/s ??
does that sound more feasible?
i got 5 point something...
sory, can't give you the answer, try again
that answer didn't work either
what am I doing wrong?
~ I only have 2 tries left :S
ok
well at least I know the approximate range of the answer right? hehe
Vo = \sqrt{(tan(38)(7.0m) - \frac{(-9.8m/s^2) (7.0m)^2}{2 (0.3m) cos^2(38)}}
well that didn't work out...
here's how I manipulated the equation:
Vo = Square root of { tan(38)x - ((-g)x^2 / 2ycos(38)^2) }
completely wrong. i believe you can do this problem your own.... just simple algebra.
oh gosh - okay - I'm sure I can - thanks for the help - I'll post my answer if I ever get around the 5 range! lol
g wpuld be a negative number right?
(-9.8m/s^2 as opposed to 9.8m/s^2)
y=(tan38)x - (g / 2(Vo^2)(cos38)^2)x^2
the negative sign has already been taken care of in the equation, g is positive
okay - that makes sense.
I keep getting 8.65m/s
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2021-01-20 23:37:47
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https://motls.blogspot.com/2013/11/currency-interventions-are-immature.html
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## Thursday, November 07, 2013 ... //
### Currency interventions are immature assaults on free markets
Expletives appropriate after the insane war was declared against the Czech crown
A few hours ago, the Czech National Bank has done something insane that hasn't taken place for 11 years: interventions against the Czech currency. Within minutes, the exchange rate weakened by stunning 4 and 6 percent relatively to the euro and the U.S. dollar, respectively. Their stated purpose is to encourage inflation (they look like an arsonist who boasts that he set a big building on fire) and they expect this thing to add 1 percentage point to the inflation for the next 12 months.
CZK 2,000 was worth $104.8 in the morning and$99 now, a decrease by nearly 6 percent (contributed by the drop of the euro due to the ECB's lowering of the rate).
Sources differ about the amount of crowns that were sold to increase the euro reserves. Reuters said it was just EUR 0.5 billion, others say it was EUR 3 billion. The figure doesn't really matter. They have stated that they will print the crowns and sell them to reach the rate around CZK 27 per euro – it was just 25.7 a few hours ago – so the rate will be close to CZK 27 per euro whether they actually need to print the new crowns or not simply because they obviously have the potential to weaken their own (well, our) currency.
I have to emphasize that this is an extremely different procedure from quantitative easing in the U.S. that we recently discussed. In the quantitative easing, the long-term bonds are being repaid i.e. bought back i.e. liquidated and the same amount of money is printed. In the case of interventions, money is being printed and sold to the market and no bonds or other forms of debt are being absorbed at the same moment.
So of course that the intervention against the crown immediately contributes to inflation and has many other consequences.
According to the discussion forums hosted by Czech news servers, I think that most citizens disagree with such an operation. I disagree for many reasons, economic and ethical ones, too.
It seems virtually unthinkable to me that there are no people who have made profit (or at least who avoided the loss) because of some insider knowledge. In my opinion, the system in which individual people may decide about such big operations that don't need any sensible justification is defective, it is really corrupt.
I also completely disagree with the idea that the inflation rate near 1.5 percent is too low. The stated inflation target is said to be around 3%. The inflation rate around 4 percent was just OK with the central bank just a year or two years ago. But suddenly 1.5 percent is too low? These data are completely inconsistent with the stated target. If they wanted to be honest, they should raise the official inflation target to 3 percent because this is what they're targeting in the medium or long term (real interest rates are minus 1-2 percent in the last decade or so).
But with their "apolitical" status and the flexibility of the rules that determine their policies, there is nothing that would make them honest. The Czech consumers will have more expensive Christmas and at least the domestic consumption will be hurt as a result. Because of the Marshall-Lerner condition, one may doubt that the intervention will improve the trade figures.
Moreover, there is no good justification for improving the trade figures. The latest Czech trade surplus was CZK 35 billion per month. That would be CZK 0.42 trillion per year, not too much below 10% of the GDP! We don't need even larger surpluses. What we may need is to increase the domestic consumption and the intervention will do the opposite, especially because the savers will feel poorer and they're the essential contributors of the Czech domestic consumption. By any sensible market measure, the Czech crown is still significantly undervalued; the trade surpluses are one sign of that (the total current account is closer to zero or sometimes showing deficits but smaller deficits than the surpluses of the trade figures).
Markets determine the currency exchange rate wisely – more or less with the idea of making the co-existence of all the economies sustainable which pretty much means that markets move the exchange rates to keep the trade surpluses and deficits low (near zero). That's the right motion that should take place. The Czech Central Bank's openly declared goal is exactly the opposite – a way to see that the markets are much more intelligent than the manipulators in the central bank, to see that manipulations with the exchange rates are always wrong.
The potential for huge swings such as this one is an argument against the independent currencies of small enough countries like ours. With the situation in Greece etc. looking much better than a year or two years ago, my personal desire to join the eurozone has surely increased dramatically after the today's experience. The ability of a few individuals to manipulate with a currency in this way is just wrong. The right of a country to demand huge external injections just because they share a currency is also wrong but the today's procedure has decreased the Czechs' wealth more than all the hassle in Greece has.
Update:
I wrote these comments long before I saw the reaction by ex-president Klaus so I was only reassured that his opinions matched mine:
The intervention is a flawed and risky step that brings highly questionable benefits but indisputable costs such as more expensive imports and an upward push on the price level in the country.
He also criticized the justification of the intervention by the desire to fulfill the inflation target, something that the bank has imposed on itself. We don't need interventions, we need a sensible fiscal policy and the return to normal parliamentary politics, Klaus said.
In an article featuring these Klaus' comments, there is a poll and only 9% of respondents think that the event is a net positive for the economy, 83% think it is a net negative.
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2019-10-14 06:10:24
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https://mersenneforum.org/showthread.php?s=0ad840e3b038d0a60b89f1cc978fd020&t=11516&page=2
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mersenneforum.org Useless SSE instructions
Register FAQ Search Today's Posts Mark Forums Read
2009-02-21, 15:14 #12
retina
Undefined
"The unspeakable one"
Jun 2006
My evil lair
2·3,343 Posts
Quote:
Originally Posted by ldesnogu Anyway that doesn't explain why x86 SIMD various instruction sets are so odd. I guess it's the result of adding a few instructions at each generation, instead of spending a few years in R&D thinking about what is really needed in the longer term.
I strongly doubt that the current instruction set(s) is/are just a random collection of things someone threw in before going to lunch.
It seems to me that doing FFT on multi-megabit numbers was not a high priority for Intel/AMD/whoever. I would imagine video/audio encoding/decoding is very high on their list of things to consider when designing new instructions.
How many people do multi-precision arithmetic compared to how many play games, listen to music and/or watch movies? I imagine the ratio is a rather small one.
2009-02-21, 15:50 #13
ldesnogu
Jan 2008
France
22×149 Posts
Quote:
Originally Posted by retina I strongly doubt that the current instruction set(s) is/are just a random collection of things someone threw in before going to lunch.
No they are not random additions, but they certainly are not well thought resulting in a stacking of instructions that doesn't look very coherent. Intel's AVX looks better, Altivec (and newer Power SIMD ISA) looks nicer too.
But as you wrote, these are not written with MP in mind
2009-02-21, 16:01 #14
xilman
Bamboozled!
"𒉺𒌌𒇷𒆷𒀭"
May 2003
Down not across
101101100011102 Posts
Quote:
Originally Posted by ewmayer Sounds pretty good, doesn't it? Sounds like "if you're done crunching datum whose value is currently stored in an xmm register and you know that won't be used (either in read or write mode) for a while, you should use this special MOV instruction to write it back to memory, because this bypasses the cache hierarchy and thus allows more soon-to-be-used data to enter the caches without risking being kicked out by on its way back to main memory."
The first thing I thought when reading this paragraph was "cache side-channel attacks". Use Google if that phrase means nothing to you.
Paul
2009-02-21, 16:44 #15
__HRB__
Dec 2008
Boycotting the Soapbox
24·32·5 Posts
Quote:
Originally Posted by retina I strongly doubt that the current instruction set(s) is/are just a random collection of things someone threw in before going to lunch.
I agree.
My guess is that it's the result of one serious rochambeau tournament that took the better part of the afternoon, too.
The guy in the sales-department won alot: He got to add 20 new instructions by aliasing them to existing instructions, and the guy responsible for putting together the "Instruction Set Reference", probably has a template and "add redundant instruction" macroed to F11.
Quote:
Originally Posted by retina It seems to me that doing FFT on multi-megabit numbers was not a high priority for Intel/AMD/whoever. I would imagine video/audio encoding/decoding is very high on their list of things to consider when designing new instructions.
I doubt it.
Somewhere there's a forum specialized on video encoding with guys concluding that pmuludq must be for multi-megabit computations, because it's useless for anything else.
The rest of the thread is devoted to missing 16-bit floating point support, and wtf to do with the 205 instead of 200 idle cycles now that paddusw, psubsw, etc. eliminated some code instructions.
Last fiddled with by __HRB__ on 2009-02-21 at 16:46
2009-02-21, 16:57 #16 akruppa "Nancy" Aug 2002 Alexandria 2,467 Posts While we're whining about instruction sets, aside from the general travesty of a cpu that is the x86 architecture, why-oh-why is there no setcc that sets a whole register to 0 or 1? With the current instructions, not only do you get a false dependency thanks to a partial register write, but you can't store the carry flag for later easy adding either, since the reg will still contain garbage in bits 8,...,63. This drove me mad while writing mulredc code for GMP-ECM. Alex
2009-02-21, 17:14 #17
ldesnogu
Jan 2008
France
11248 Posts
Quote:
Originally Posted by akruppa While we're whining about instruction sets, aside from the general travesty of a cpu that is the x86 architecture, why-oh-why is there no setcc that sets a whole register to 0 or 1? With the current instructions, not only do you get a false dependency thanks to a partial register write, but you can't store the carry flag for later easy adding either, since the reg will still contain garbage in bits 8,...,63. This drove me mad while writing mulredc code for GMP-ECM.
I had the same thought and had to resort to zero-extension of the result of setcc with a movzbl After all there's a price to pay for a monstruosity that has lived far too long and that doesn't show any will to die the horrible death it deserves.
2009-02-21, 17:15 #18
retina
Undefined
"The unspeakable one"
Jun 2006
My evil lair
2×3,343 Posts
Quote:
Originally Posted by akruppa While we're whining about instruction sets, aside from the general travesty of a cpu that is the x86 architecture, why-oh-why is there no setcc that sets a whole register to 0 or 1? With the current instructions, not only do you get a false dependency thanks to a partial register write, but you can't store the carry flag for later easy adding either, since the reg will still contain garbage in bits 8,...,63. This drove me mad while writing mulredc code for GMP-ECM. Alex
I thought the partial register stall was only in the P4. Earlier CPUs were not affected and later CPUs used the older architecture as a starting point and also avoided it. However I always use sbb eax,eax to replicate the carry through a register and then just sub instead of add it later to accumulate carries. And only ever use setcc for accumulating multiple flag tests to avoid long lists of branches.
Last fiddled with by retina on 2009-02-21 at 17:16
2009-02-21, 18:01 #19
akruppa
"Nancy"
Aug 2002
Alexandria
1001101000112 Posts
Quote:
Originally Posted by retina I thought the partial register stall was only in the P4. Earlier CPUs were not affected and later CPUs used the older architecture as a starting point and also avoided it. However I always use sbb eax,eax to replicate the carry through a register and then just sub instead of add it later to accumulate carries. And only ever use setcc for accumulating multiple flag tests to avoid long lists of branches.
The AMD64 software optimization guide lists partial register reads/writes as something to avoid. I don't know if the Core2 identifies and ignores these false dependencies, but my (somewhat naive) understanding is that keeping track of the dependencies between instructions is hard enough even without treating each register as 4 independent pieces, where one instruction can write to one or several pieces.
Yes, "sbb reg, same reg" can be used to set a full register according only to the carry flag, and recent cpus even know that this instruction does not depend on the previous value of "reg," so no false dependency occurs here. However, I use three registers as a kind of ring buffer for the carry propagation, and need to add to the register that holds the carry... so having the carry value negated was a bit of a problem. I thought about flipping the sign of the partial result in every pass so I could use sbb and then keep subtracting. I didn't do that, though... I may try some time.
Alex
2009-02-21, 18:29 #20
__HRB__
Dec 2008
Boycotting the Soapbox
24×32×5 Posts
Quote:
Originally Posted by retina sbb eax,eax
That's clever. I quickly deleted my post with an inferior solution.
I was thinking about using rcr or rcl to create a carry cache and use clc to remove dependencies to do several multiprecision adds in parallel.
Your solution is much better. You 'da Man!.
Here's why I'm so exited:
Let's suppose we want to do 2 multiprecision adds in parallel. With eax & edx as temporaries.
After sbb eax,eax a add edx,edx will
a) remove the dependency on the carry
b) restore the carry
ending with sbb edx,edx
and setting up the second stream with add eax,eax
I count 4 + 2*3*N instructions, so, if e.g. N==4, and Core 2 can do 1 add/clock, we're doing 8 multiprecision adds in 12 cycles. This is 25% faster than the naive implementation.
If we're doing X+Y and X-Y and can reuse X and/or Y, Athlons might be able to do more than one 64-bit adc/clock.
This would allow nice butterflies for medium sized power-of-two moduli.
Edit: Two adds can be replaced with shifts. Duh. 10/8=1.25 clocks/adc for core 2.
Last fiddled with by __HRB__ on 2009-02-21 at 18:39 Reason: I'm a moron
2009-02-25, 16:33 #21
akruppa
"Nancy"
Aug 2002
Alexandria
9A316 Posts
Quote:
Originally Posted by ldesnogu I had the same thought and had to resort to zero-extension of the result of setcc with a movzbl After all there's a price to pay for a monstruosity that has lived far too long and that doesn't show any will to die the horrible death it deserves.
Btw, zero-extending the result of setcc keeps the (false) dependency chain intact. It may be better to do a "mov $0, reg" before the setcc, it's just as ugly but at least it breaks the dependency chain. Alex 2009-02-25, 16:49 #22 ldesnogu Jan 2008 France 22·149 Posts Quote: Originally Posted by akruppa Btw, zero-extending the result of setcc keeps the (false) dependency chain intact. It may be better to do a "mov$0, reg" before the setcc, it's just as ugly but at least it breaks the dependency chain.
I really have to forget my old m68k days and its all-instructions-set-flags view of the world
If I make this remark it's because I have restrictions on the order of instructions: I get registers allocated from above (inputs and destination); so in my case I was doing a cmp + setcc + signextend; and I wrongly thought the only other option was mov 0 + cmp + setcc ("only" due to wrong "mov 0 clobbers flags" assumption) which doesn't work if my allocated destination reg overlaps one of the input regs.
Thanks for the hint!
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https://digital-library.theiet.org/content/books/10.1049/pbew502e_ch9
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1887
## Short-range propagation
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Short-range radio systems are used for many purposes for telemetry, remote control and games, as well as for communications. Communications uses include cordless telephony, radio local area networks (RLANs), radio fixed access and microcellular systems. For some of these applications, very low powers are used, with poor and poorly located antennas where the user only expects the range to be of the order of a few metres. For example, for remote car door locks, the user will expect to point the transmitter towards the car and probably has some understanding of the need to provide a near line-of-sight path. For such appli cations there seems to be very little requirement to attempt to provide good propagation models. For RLANs and other indoor applications there will be a need for some kind of generic modelling of the effects of the room size and shape, obstructions in the room, the construction materials and the penetration through walls and floors. For high-speed data transmission, it may also be necessary to model the multipath time spreads. But for outdoor microcellular systems and similar applications it will be necessary to model propagation at distances ranging out to a kilometre or so, where the longer-range area coverage prediction methods may take over. 1 km is the dividing distance used by the ITU-R in its Recommendations between short-range and longer-range prediction methods. However, the above considerations mainly apply to VHF and UHF systems, where although the distances are short the propagation is in the far-field regime and the usual propagation techniques are applicable. Another interpretation of short range would be to consider paths within the near field regime around the transmitting antenna. This is mainly of significance at lower frequencies.
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## Convergence results for the square root of the Poisson kernel.(English)Zbl 0902.42009
In the disk, we prove that integrals of boundary functions against the square root of the Poisson kernel converge in regions which we call $$L^p$$ weakly tangential. If $$p>1$$ these regions are strictly larger than the weakly tangential regions used by Sjögren. We also investigate how sharp these results are.
In the bidisk, we prove that we have convergence in the product region $$A\times B$$, where $$A$$ is a nontangential cone, and $$B$$ is a weakly tangential region. In this case, the kernel will be a tensor product of powers of Poisson kernels, with the exponent larger than 1/2 in the first variable, and the exponent equal to 1/2 in the second variable.
### MSC:
42B25 Maximal functions, Littlewood-Paley theory
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