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A145556 Denominators of partial sums of the alternating series of inverse central binomial coefficients. 3
2, 3, 60, 84, 126, 2772, 72072, 360360, 1225224, 23279256, 11639628, 503217, 13385572200, 2007835830, 465817912560, 2888071057872, 627841534320, 2489716429200, 534293145706320, 18423901576080, 4381203794791824, 941958815880242160, 261655226633400600 (list; graph; refs; listen; history; text; internal format)
OFFSET 1,1 COMMENTS See A145375 for the numerators, references and a W. Lang link. LINKS FORMULA a(n)=denominator(r(n)), with the rationals (in lowest terms) r(n):=sum(((-1)^(k+1))/binomial(2*k,k),k=1..n). EXAMPLE Rationals r(n) (in lowest terms): [1/2,1/3,23/60,31/84,47/126,1031/2772,26827/72072,...]. CROSSREFS Sequence in context: A299172 A154253 A097961 * A124083 A112098 A144545 Adjacent sequences: A145553 A145554 A145555 * A145557 A145558 A145559 KEYWORD nonn,frac,easy AUTHOR Wolfdieter Lang Oct 17 2008, Nov 17 2008, Nov 25 2008 STATUS approved
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https://chuntey.wordpress.com/2013/09/08/how-to-write-zx-spectrum-games-chapter-8/amp/
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# How To Write ZX Spectrum Games – Chapter 8
## Sprites
Note: This article was originally written by Jonathan Cauldwell and is reproduced here with permission.
Converting Pixel Positions to Screen Addresses
UDGs and character graphics are all very fine and dandy, but the better games usually use sprites and there are no handy ROM routines to help us here because Sir Clive didn’t design the Spectrum as a games machine. Sooner or later a games programmer has to confront the thorny issue of the Spectrum’s awkward screen layout. It’s a tricky business converting x and y pixel coordinates into a screen address but there are a couple of methods we might employ to do this.
Using a Screen Address Look-up Table
The first method is to use a pre-calculated address table containing the screen address for each of the Spectrum’s 192 lines such as this, or a similar variation:
``` xor a ; clear carry flag and accumulator.
ld d,a ; empty de high byte.
ld a,(xcoord) ; x position.
rla ; shift left to multiply by 2.
ld e,a ; place this in low byte of de pair.
rl d ; shift top bit into de high byte.
ld hl,addtab ; table of screen addresses.
add hl,de ; point to table entry.
ld e,(hl) ; low byte of screen address.
inc hl ; point to high byte.
ld d,(hl) ; high byte of screen address.
ld a,(ycoord) ; horizontal position.
rra ; divide by two.
rra ; and again for four.
rra ; shift again to divide by eight.
and 31 ; mask away rubbish shifted into rightmost bits.
add a,e ; add to address for start of line.
ld e,a ; new value of e register.
ret ; return with screen address in de.
.
.
defw 16640
defw 16896
.
.```
On the plus side this is very fast, but it does mean having to store each of the 192 line addresses in a table, taking up 384 bytes which might be better employed elsewhere.
The second method involves calculating the address ourselves and doesn’t require an address look-up table. In doing this we need to consider three things: Which third of the screen the point is in, the character line to which it is closest, and the pixel line upon which it falls within that cell. Judicious use of the and operand will help us to decide all three. It is a complicated business however, so please bear with me as I endeavour to explain how it works.
We can establish which of the three screen segments a point is situated in by taking the vertical coordinate and masking away the six least significant bits to leave a value of 0, 64 or 128 each of the segments being 64 pixels apart. As the high bytes of the 3 screen segment addresses are 64, 72 and 80 – a difference of 8 going from one segment to another – we take this masked value and divide it by 8 to give us a value of 0, 8 or 16. We then add 64 to give us the high byte of the screen segment.
Each segment is divided into 8 character cell positions which are 32 bytes apart, so to find that aspect of our address we take the vertical coordinate and mask away the two most significant bits we used to determine the segment along with the three least significant bits which determine the pixel position. The instruction and 56 will do nicely. This gives us the character cell position as a multiple of 8, and as the character lines are 32 bytes apart we multiply this by 4 and place our number in the low byte of the screen address.
Finally, character cells are further divided into pixel lines 256 bytes apart, so we again take our vertical coordinate, mask away everything except the bits which determine the line using and 7, and add the result to the high byte. That will give us our vertical screen address. From there we take our horizontal coordinate, divide it by 8 and add it to our address.
Here is a routine which returns a screen address for (xcoord, ycoord) in the de register pair. It could easily be modified to return the address in the hl or bc registers if desired.
```scadd ld a,(xcoord) ; fetch vertical coordinate.
ld e,a ; store that in e.
; Find line within cell.
and 7 ; line 0-7 within character square.
add a,64 ; 64 * 256 = 16384 = start of screen display.
ld d,a ; line * 256.
; Find which third of the screen we're in.
ld a,e ; restore the vertical.
and 192 ; segment 0, 1 or 2 multiplied by 64.
rrca ; divide this by 8.
rrca
rrca ; segment 0-2 multiplied by 8.
add a,d ; add to d give segment start address.
ld d,a
; Find character cell within segment.
ld a,e ; 8 character squares per segment.
rlca ; divide x by 8 and multiply by 32,
rlca ; net calculation: multiply by 4.
and 224 ; mask off bits we don't want.
ld e,a ; vertical coordinate calculation done.
; Add the horizontal element.
ld a,(ycoord) ; y coordinate.
rrca ; only need to divide by 8.
rrca
rrca
and 31 ; squares 0 - 31 across screen.
add a,e ; add to total so far.
ld e,a ; de = address of screen.
ret```
Shifting
Once the address has been established we need to consider how our graphics are shifted into position. The three lowest bit positions of the horizontal coordinate indicate how many pixel shifts are needed. A slow way to plot a pixel would be to call the scadd routine above, perform an and 7 on the horizontal coordinate, then right shift a pixel from zero to seven times depending on the result before dumping it to the screen.
A shifter sprite routine works in the same way. The graphic image is taken from memory one line at a time, shifted into position and then placed on the screen before moving to the next line down and repeating the process. We could write a sprite routine which calculated the screen address for every line drawn, and indeed the first sprite routine I ever wrote worked in such a way. Fortunately it is much simpler to determine whether we’re moving within a character cell, crossing character cell boundaries, or crossing a segment boundary with a couple of and instructions and to increment or decrement the screen address accordingly. Put simply, and 63 will return zero if the new vertical position is crossing a segment, and 7 will return zero if it is crossing a character cell boundary and anything else means the new line is within the same character cell as the previous line.
This is a shifter sprite routine which makes use of the earlier scadd routine. To use it simply set up the coordinates in dispx and dispy, point the bc register pair at the sprite graphic, and call sprite.
```sprit7 xor 7 ; complement last 3 bits.
inc a ; add one for luck!
sprit3 rl d ; rotate left...
rl c ; ...into middle byte...
rl e ; ...and finally into left character cell.
dec a ; count shifts we've done.
jr nz,sprit3 ; return until all shifts complete.
; Line of sprite image is now in e + c + d, we need it in form c + d + e.
ld a,e ; left edge of image is currently in e.
ld e,d ; put right edge there instead.
ld d,c ; middle bit goes in d.
ld c,a ; and the left edge back into c.
jr sprit0 ; we've done the switch so transfer to screen.
sprite ld a,(dispx) ; draws sprite (hl).
ld (tmp1),a ; store vertical.
ld a,16 ; height of sprite in pixels.
sprit1 ex af,af' ; store loop counter.
push de ; store screen address.
ld c,(hl) ; first sprite graphic.
inc hl ; increment pointer to sprite data.
ld d,(hl) ; next bit of sprite image.
inc hl ; point to next row of sprite data.
ld (tmp0),hl ; store in tmp0 for later.
ld e,0 ; blank right byte for now.
ld a,b ; b holds y position.
and 7 ; how are we straddling character cells?
jr z,sprit0 ; we're not straddling them, don't bother shifting.
cp 5 ; 5 or more right shifts needed?
jr nc,sprit7 ; yes, shift from left as it's quicker.
and a ; oops, carry flag is set so clear it.
sprit2 rr c ; rotate left byte right...
rr d ; ...through middle byte...
rr e ; ...into right byte.
dec a ; one less shift to do.
jr nz,sprit2 ; return until all shifts complete.
sprit0 pop hl ; pop screen address from stack.
ld a,(hl) ; what's there already.
xor c ; merge in image data.
ld (hl),a ; place onto screen.
inc l ; next character cell to right please.
ld a,(hl) ; what's there already.
xor d ; merge with middle bit of image.
ld (hl),a ; put back onto screen.
inc hl ; next bit of screen area.
ld a,(hl) ; what's already there.
xor e ; right edge of sprite image data.
ld (hl),a ; plonk it on screen.
ld a,(tmp1) ; temporary vertical coordinate.
inc a ; next line down.
ld (tmp1),a ; store new position.
and 63 ; are we moving to next third of screen?
jr z,sprit4 ; yes so find next segment.
and 7 ; moving into character cell below?
jr z,sprit5 ; yes, find next row.
dec hl ; left 2 bytes.
dec l ; not straddling 256-byte boundary here.
inc h ; next row of this character cell.
sprit6 ex de,hl ; screen address in de.
ld hl,(tmp0) ; restore graphic address.
ex af,af' ; restore loop counter.
dec a ; decrement it.
jp nz,sprit1 ; not reached bottom of sprite yet to repeat.
ret ; job done.
sprit4 ld de,30 ; next segment is 30 bytes on.
jp sprit6 ; repeat.
sprit5 ld de,63774 ; minus 1762.
add hl,de ; subtract 1762 from physical screen address.
jp sprit6 ; rejoin loop.```
As you can see, this routine utilises the xor instruction to merge the sprite onto the screen, which works in the same way that PRINT OVER 1 does in Sinclair BASIC. The sprite is merged with any graphics already present on screen which can look messy. To delete a sprite we just display it again and the image magically vanishes.
If we wanted to draw a sprite on top of something that is already on the screen we would need some extra routines, similar to the one above. One would be required to store the graphics on screen in a buffer so that that portion of the screen could be re-drawn when the sprite is deleted. The next routine would apply a sprite mask to remove the pixels around and behind the sprite using and or or, then the sprite could finally be applied over the top. Another routine would be needed to restore the relevant portion of screen to its former state should the sprite be deleted. However, this would take a lot of CPU time to achieve so my advice would be not to bother unless your game uses something called double buffering – otherwise known as the back screen technique, or you’re using a pre-shifted sprites, which we shall discuss shortly.
Another method you may wish to consider involves making sprites appear to pass behind background objects, a trick you may have seen in Haunted House or Egghead in Space. While this method is handy for reducing colour clash it requires a sizeable chunk of memory. In both games a 6K dummy mask screen was located at address 24576, and each byte of sprite data was anded with the data on the dummy screen before being xored onto the physical screen located at address 16384. Because the physical screen and the dummy mask screen were exactly 8K apart it was possible to flip between them by toggling bit 5 of the h register. To do this for the sprite routine above our sprit0 routine might look like this:
```sprit0 pop hl ; pop screen address from stack.
set 5,h ; address of dummy screen.
ld a,(hl) ; what's there already.
and c ; mask away parts behind the object.
res 5,h ; address of physical screen.
xor (hl) ; merge in image data.
ld (hl),a ; place onto screen.
inc l ; next character cell to right please.
set 5,h ; address of dummy screen.
ld a,(hl) ; what's there already.
and d ; mask with middle bit of image.
res 5,h ; address of physical screen.
xor (hl) ; merge in image data.
ld (hl),a ; put back onto screen.
inc hl ; next bit of screen area.
set 5,h ; address of dummy screen.
ld a,(hl) ; what's already there.
and e ; mask right edge of sprite image data.
res 5,h ; address of physical screen.
xor (hl) ; merge in image data.
ld (hl),a ; plonk it on screen.
ld a,(tmp1) ; temporary vertical coordinate.```
Pre-shifted Sprites
A shifter sprite routine has one major drawback: its lack of speed. Shifting all that graphic data into position takes time, and if your game needs a lot of sprites bouncing around the screen, you should consider using pre-shifted sprites instead. This requires eight separate copies of the sprite image, one in each of the shifted pixel positions. It is then simply a matter of calculating which sprite image to use based on the horizontal alignment of the sprite, calculating the screen address, and copying the sprite image to the screen. While this method is much faster it is fantastically expensive in memory terms. A shifter sprite routine requires 32 bytes for an unmasked 16×16 pixel sprite, a pre-shifted sprite requires 256 bytes for the same image. Writing a Spectrum game is a compromise between speed and available memory. In general I prefer to move my sprites 2 pixels per frame meaning the odd pixel alignments are not required. Even so, my pre-shifted sprites still require 128 bytes of precious RAM.
You may not necessarily want the same sprite image in each pre-shifted position. For example, by changing the position of a sprite’s legs in each of the pre-shifted positions a sprite can be animated to appear as if it is walking from left to right as it moves across the screen. Remember to match the character’s legs to the number of pixels it is moved each frame. If you are moving a sprite 2 pixels each frame it is important to make the legs move 2 pixels between frames. Less than this will make the sprite appear as if it is skating on ice, any more and it will appear to be struggling for grip. I’ll let you into a little secret here: believe it or not, this can actually affect the way a game feels so getting your animation right is important.
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https://www.askiitians.com/forums/Mechanics/10/56125/acceleration.htm
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# What is the acceleration exactly ?
34 Points
9 years ago
Acceleration in rate of change of velocity..... a=dv÷dt
Karteek Valluri
32 Points
9 years ago
Acceleration is nothing but it''s a change of speed of an object with respect to time.While you are moving in a bike you may observe that.For example if you are moving with constant speed say 60kmph means change in velocity zero so acceleration is zero.
The main thing you keep in mind that is accelaration positive or negtive.....Because If a person slowing down from 60kmph to 50kmph,here change in velocity is 10 so acceleration is 10 but it is negative(decreasing)..Same way he person again come to 60kmph from50kmph..obviosly his acceleration is 10.but it is in increasing so we say positive or Increasing acceleration....
lokesh soni
37 Points
9 years ago
accelaration is the rate of change of velocity.it shows how velocity is chaging with respect to time
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https://forum.quartertothree.com/t/steam-summer-sale-2020/148120/355
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# Steam Summer Sale 2020
Well for example the left island:
Solution
To start with, you can deduce that the two tiles on the left and right must contain mines:
because there can only be two mines around the centre tile, so the left and right 2s must each contribute one apiece to that, otherwise you end up with a contradiction. Knowing that, you can reveal the two above the centre, as we know neither mine can be up there.
Which reveals two 4s, so we can fill in the unknown tiles around these as containing mines, and finally we now know the two bottom unrevealed tiles must be empty.
Yeah that exact one had me baffled for a little bit, thinking ‘no guesswork? lies!’ until I went back and gave it another go and the logic clicked.
I think I’m missing some piece of logic. 1 of each color allowed. Nothing around the borders. If you use the pen to map out what would happen if the lower right pink squared was a mine, then it looks like everything should work out, but that’s not actually the correct solution.
I’m at a loss how I’m supposed to know what’s next.
I think my problem is I’m not getting (and therefor not using) the pen.
I’ve not got to this level yet, but I think there are multiple ways you could map out mines in this one consistent with the information there, so what you’re looking for is something which causes a contradiction, so you’re certain of the next move. For instance, if you assume the leftmost orange block is a mine, that leads to both pink blocks being empty, so you know that orange block must be empty. Hopefully revealing that one will give you more to go on.
Btw sorry everyone for derailing the thread a bit. If it helps, I’ve bought quite a few Steam games in the sale, including Desparados 3 (via CDkeys), Sunless Skies, and Cloudpunk, but nothing has so far gotten a look in apart from Tametsi and Bloody Rally Show.
Ah, good call. I was looking for positives, but I need to look for negatives. That might unstick me on another one or two.
Oh, uh… :shame: Back to your regularly scheduled show.
I still think this at least once per level, which is a testament to the variety (& difficulty) of puzzles the developer has come up with. But always, if you stare at it for long enough (or take a break and come back) the next move to make will be revealed.
Btw this is my favourite screenshot I’ve seen posted to Steam, sums it up well:
Summary
You have to start with what you know. Each one of those islands you can look at and determine without guessing what certain hexes must be. For example, the island on the left: there are two that must be clear. On the bottom: two that must be mines. On the right, three that must be clear
I have three people that want to get Tabletop Simulator, anybody want to fill out a 4-pack? \$7.50, you can paypal me. PM if interested!
I appreciate the Temetsi comments about guesswork. I finished over 30 levels, then quit because I was absolutely convinced guessing was required in some cases. I suspect that it’s possible to trap yourself by making incorrect moves which lead to “must guess” situations, but I don’t plan to revisit to test that. I had fun with it for a while anyway.
Thanks for all the Tametsi talk in this thread because I bought it and played the first 11 puzzles. Also glad it was made clear that there is no guessing needed because it caused me to really evaluate the puzzles until I didn’t need to guess.
Here is my haul so far. The big question is do I pick up Transport Fever 2. I want to play some American Truck / Euro Truck Simulator before I pick up any DLC.
Once again I held off on some that I’ve repeatedly considered buying, like Wildermyth, Star Traders Frontiers. Also, I forgot Circle Empires was in my cart and bought that before I fully decided to get that instead of Circle Empires Rivals while it is on sale from Indigala.
Both Wildermyth and Star traders Frontiers are EXCELLENT games, and well worth their time and dime spent on them :-)
No doubt :-)
I had to cut something. The main reason I cut Wildermyth is that it still is in EA. I fully expect to get it at some point. Even if it is in a good state already, I wouldn’t fully dive in until it’s done anyways.
Makes perfect sense! It will be well worth the wait, I am certain, and just more fun , the longer we wait!
Is wildermyth worth it, at this moment in EA? I’d say something like Hades is a good example of an EA product thats worth it in its current state, for reference.
On sale, I’d say so. I think I got it for 35-50% off, but could be wrong. It’s currently 20% and just above what I would consider ‘throw away’ price range. But it’s got good bones and enough muscle to warrant playing now. Unless you want to just power thru making it your primary game for a couple of weeks. In that case, I’d say there is not enough there. But again, the price isn’t too high regardless of your playstyle.
There is a lot of game there already - I enjoyed it immensely. But there is still quite a few systems that aren’t fully developed, or even there yet, so it kinda depends on how you feel about games not baked yet :-)
I bought it, but try to avoid playing it!
I’m not the type to milk a game a for all its worth. So based on that I think I can go ahead and get it >:) Thanks!
I need to search tonight for a game or DLC that is on sale for around \$1 as I have \$29.06 worth of stuff in my cart. =)
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#jsDisabledContent { display:none; } My Account | Register | Help
Axioms
Article Id: WHEBN0000420595
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Title: Axioms Author: World Heritage Encyclopedia Language: English Subject: Axiomatic design Collection: Publisher: World Heritage Encyclopedia Publication Date:
Axioms
"Postulation" redirects here. For the term in algebraic geometry, see Postulation (algebraic geometry).
An axiom, or postulate, is a premise or starting point of reasoning. As classically conceived, an axiom is a premise so evident as to be accepted as true without controversy.[1] The word comes from the Greek ἀξίωμα (āxīoma) 'that which is thought worthy or fit' or 'that which commends itself as evident.'[2][3] As used in modern logic, an axiom is simply a premise or starting point for reasoning.[4] Axioms define and delimit the realm of analysis; the relative truth of an axiom is taken for granted within the particular domain of analysis, and serves as a starting point for deducing and inferring other relative truths. No explicit view regarding the absolute truth of axioms is ever taken in the context of modern mathematics, as such a thing is considered to be an irrelevant and impossible contradiction in terms.
In mathematics, the term axiom is used in two related but distinguishable senses: "logical axioms" and "non-logical axioms". Logical axioms are usually statements that are taken to be true within the system of logic they define (e.g., (A and B) implies A), while non-logical axioms (e.g., a + b = b + a) are actually defining properties for the domain of a specific mathematical theory (such as arithmetic). When used in the latter sense, "axiom," "postulate", and "assumption" may be used interchangeably. In general, a non-logical axiom is not a self-evident truth, but rather a formal logical expression used in deduction to build a mathematical theory. As modern mathematics admits multiple, equally "true" systems of logic, precisely the same thing must be said for logical axioms - they both define and are specific to the particular system of logic that is being invoked. To axiomatize a system of knowledge is to show that its claims can be derived from a small, well-understood set of sentences (the axioms). There are typically multiple ways to axiomatize a given mathematical domain.
In both senses, an axiom is any mathematical statement that serves as a starting point from which other statements are logically derived. Within the system they define, axioms (unless redundant) cannot be derived by principles of deduction, nor are they demonstrable by mathematical proofs, simply because they are starting points; there is nothing else from which they logically follow otherwise they would be classified as theorems. However, an axiom in one system may be a theorem in another, and vice versa.
Etymology
The word "axiom" comes from the Greek word ἀξίωμα (axioma), a verbal noun from the verb ἀξιόειν (axioein), meaning "to deem worthy", but also "to require", which in turn comes from ἄξιος (axios), meaning "being in balance", and hence "having (the same) value (as)", "worthy", "proper". Among the ancient Greek philosophers an axiom was a claim which could be seen to be true without any need for proof.
The root meaning of the word 'postulate' is to 'demand'; for instance, Euclid demands of us that we agree that some things can be done, e.g. any two points can be joined by a straight line, etc.[5]
Ancient geometers maintained some distinction between axioms and postulates. While commenting Euclid's books Proclus remarks that "Geminus held that this [4th] Postulate should not be classed as a postulate but as an axiom, since it does not, like the first three Postulates, assert the possibility of some construction but expresses an essential property".[6] Boethius translated 'postulate' as petitio and called the axioms notiones communes but in later manuscripts this usage was not always strictly kept.
Historical development
Early Greeks
The logico-deductive method whereby conclusions (new knowledge) follow from premises (old knowledge) through the application of sound arguments (syllogisms, rules of inference), was developed by the ancient Greeks, and has become the core principle of modern mathematics. Tautologies excluded, nothing can be deduced if nothing is assumed. Axioms and postulates are the basic assumptions underlying a given body of deductive knowledge. They are accepted without demonstration. All other assertions (theorems, if we are talking about mathematics) must be proven with the aid of these basic assumptions. However, the interpretation of mathematical knowledge has changed from ancient times to the modern, and consequently the terms axiom and postulate hold a slightly different meaning for the present day mathematician, than they did for Aristotle and Euclid.
The ancient Greeks considered geometry as just one of several sciences, and held the theorems of geometry on par with scientific facts. As such, they developed and used the logico-deductive method as a means of avoiding error, and for structuring and communicating knowledge. Aristotle's posterior analytics is a definitive exposition of the classical view.
An "axiom", in classical terminology, referred to a self-evident assumption common to many branches of science. A good example would be the assertion that
When an equal amount is taken from equals, an equal amount results.
At the foundation of the various sciences lay certain additional hypotheses which were accepted without proof. Such a hypothesis was termed a postulate. While the axioms were common to many sciences, the postulates of each particular science were different. Their validity had to be established by means of real-world experience. Indeed, Aristotle warns that the content of a science cannot be successfully communicated, if the learner is in doubt about the truth of the postulates.[7]
The classical approach is well-illustrated by Euclid's Elements, where a list of postulates is given (common-sensical geometric facts drawn from our experience), followed by a list of "common notions" (very basic, self-evident assertions).
Postulates
1. It is possible to draw a straight line from any point to any other point.
2. It is possible to extend a line segment continuously in both directions.
3. It is possible to describe a circle with any center and any radius.
4. It is true that all right angles are equal to one another.
5. ("Parallel postulate") It is true that, if a straight line falling on two straight lines make the interior angles on the same side less than two right angles, the two straight lines, if produced indefinitely, intersect on that side on which are the angles less than the two right angles.
Common notions
1. Things which are equal to the same thing are also equal to one another.
2. If equals are added to equals, the wholes are equal.
3. If equals are subtracted from equals, the remainders are equal.
4. Things which coincide with one another are equal to one another.
5. The whole is greater than the part.
Modern development
A lesson learned by mathematics in the last 150 years is that it is useful to strip the meaning away from the mathematical assertions (axioms, postulates, propositions, theorems) and definitions. One must concede the need for primitive notions, or undefined terms or concepts, in any study. Such abstraction or formalization makes mathematical knowledge more general, capable of multiple different meanings, and therefore useful in multiple contexts. Alessandro Padoa, Mario Pieri, and Giuseppe Peano were pioneers in this movement.
Structuralist mathematics goes further, and develops theories and axioms (e.g. field theory, group theory, topology, vector spaces) without any particular application in mind. The distinction between an "axiom" and a "postulate" disappears. The postulates of Euclid are profitably motivated by saying that they lead to a great wealth of geometric facts. The truth of these complicated facts rests on the acceptance of the basic hypotheses. However, by throwing out Euclid's fifth postulate we get theories that have meaning in wider contexts, hyperbolic geometry for example. We must simply be prepared to use labels like "line" and "parallel" with greater flexibility. The development of hyperbolic geometry taught mathematicians that postulates should be regarded as purely formal statements, and not as facts based on experience.
When mathematicians employ the field axioms, the intentions are even more abstract. The propositions of field theory do not concern any one particular application; the mathematician now works in complete abstraction. There are many examples of fields; field theory gives correct knowledge about them all.
It is not correct to say that the axioms of field theory are "propositions that are regarded as true without proof." Rather, the field axioms are a set of constraints. If any given system of addition and multiplication satisfies these constraints, then one is in a position to instantly know a great deal of extra information about this system.
Modern mathematics formalizes its foundations to such an extent that mathematical theories can be regarded as mathematical objects, and logic itself can be regarded as a branch of mathematics. Frege, Russell, Poincaré, Hilbert, and Gödel are some of the key figures in this development.
In the modern understanding, a set of axioms is any collection of formally stated assertions from which other formally stated assertions follow by the application of certain well-defined rules. In this view, logic becomes just another formal system. A set of axioms should be consistent; it should be impossible to derive a contradiction from the axiom. A set of axioms should also be non-redundant; an assertion that can be deduced from other axioms need not be regarded as an axiom.
It was the early hope of modern logicians that various branches of mathematics, perhaps all of mathematics, could be derived from a consistent collection of basic axioms. An early success of the formalist program was Hilbert's formalization of Euclidean geometry, and the related demonstration of the consistency of those axioms.
In a wider context, there was an attempt to base all of mathematics on Cantor's set theory. Here the emergence of Russell's paradox, and similar antinomies of naïve set theory raised the possibility that any such system could turn out to be inconsistent.
The formalist project suffered a decisive setback, when in 1931 Gödel showed that it is possible, for any sufficiently large set of axioms (Peano's axioms, for example) to construct a statement whose truth is independent of that set of axioms. As a corollary, Gödel proved that the consistency of a theory like Peano arithmetic is an unprovable assertion within the scope of that theory.
It is reasonable to believe in the consistency of Peano arithmetic because it is satisfied by the system of natural numbers, an infinite but intuitively accessible formal system. However, at present, there is no known way of demonstrating the consistency of the modern Zermelo–Fraenkel axioms for set theory. The axiom of choice, a key hypothesis of this theory, remains a very controversial assumption. Furthermore, using techniques of forcing (Cohen) one can show that the continuum hypothesis (Cantor) is independent of the Zermelo–Fraenkel axioms. Thus, even this very general set of axioms cannot be regarded as the definitive foundation for mathematics.
Other sciences
Axioms play a key role not only in mathematics, but also in other sciences, notably in theoretical physics. In particular, the monumental work of Isaac Newton is essentially based on Euclid's axioms, augmented by a postulate on the non-relation of spacetime and the physics taking place in it at any moment.
In 1905, Newton's axioms were replaced by those of Albert Einstein's special relativity, and later on by those of general relativity.
Another paper of Albert Einstein and coworkers (see EPR paradox), almost immediately contradicted by Niels Bohr, concerned the interpretation of quantum mechanics. This was in 1935. According to Bohr, this new theory should be probabilistic, whereas according to Einstein it should be deterministic. Notably, the underlying quantum mechanical theory, i.e. the set of "theorems" derived by it, seemed to be identical. Einstein even assumed that it would be sufficient to add to quantum mechanics "hidden variables" to enforce determinism. However, thirty years later, in 1964, John Bell found a theorem, involving complicated optical correlations (see Bell inequalities), which yielded measurably different results using Einstein's axioms compared to using Bohr's axioms. And it took roughly another twenty years until an experiment of Alain Aspect got results in favour of Bohr's axioms, not Einstein's. (Bohr's axioms are simply: The theory should be probabilistic in the sense of the Copenhagen interpretation.)
As a consequence, it is not necessary to explicitly cite Einstein's axioms, the more so since they concern subtle points on the "reality" and "locality" of experiments.
Regardless, the role of axioms in mathematics and in the above-mentioned sciences is different. In mathematics one neither "proves" nor "disproves" an axiom for a set of theorems; the point is simply that in the conceptual realm identified by the axioms, the theorems logically follow. In contrast, in physics a comparison with experiments always makes sense, since a falsified physical theory needs modification.
Mathematical logic
In the field of mathematical logic, a clear distinction is made between two notions of axioms: logical and non-logical (somewhat similar to the ancient distinction between "axioms" and "postulates" respectively).
Logical axioms
These are certain formulas in a formal language that are universally valid, that is, formulas that are satisfied by every assignment of values. Usually one takes as logical axioms at least some minimal set of tautologies that is sufficient for proving all tautologies in the language; in the case of predicate logic more logical axioms than that are required, in order to prove logical truths that are not tautologies in the strict sense.
Examples
Propositional logic
In propositional logic it is common to take as logical axioms all formulae of the following forms, where $\phi$, $\chi$, and $\psi$ can be any formulae of the language and where the included primitive connectives are only "$\neg$" for negation of the immediately following proposition and "$\to\,$" for implication from antecedent to consequent propositions:
1. $\phi \to \left(\psi \to \phi\right)$
2. $\left(\phi \to \left(\psi \to \chi\right)\right) \to \left(\left(\phi \to \psi\right) \to \left(\phi \to \chi\right)\right)$
3. $\left(\lnot \phi \to \lnot \psi\right) \to \left(\psi \to \phi\right).$
Each of these patterns is an axiom schema, a rule for generating an infinite number of axioms. For example, if $A$, $B$, and $C$ are propositional variables, then $A \to \left(B \to A\right)$ and $\left(A \to \lnot B\right) \to \left(C \to \left(A \to \lnot B\right)\right)$ are both instances of axiom schema 1, and hence are axioms. It can be shown that with only these three axiom schemata and modus ponens, one can prove all tautologies of the propositional calculus. It can also be shown that no pair of these schemata is sufficient for proving all tautologies with modus ponens.
Other axiom schemas involving the same or different sets of primitive connectives can be alternatively constructed.[8]
These axiom schemata are also used in the predicate calculus, but additional logical axioms are needed to include a quantifier in the calculus.[9]
Mathematical logic
Axiom of Equality. Let $\mathfrak\left\{L\right\}\,$ be a first-order language. For each variable $x\,$, the formula
$x = x\,$
is universally valid.
This means that, for any variable symbol $x\,,$ the formula $x = x\,$ can be regarded as an axiom. Also, in this example, for this not to fall into vagueness and a never-ending series of "primitive notions", either a precise notion of what we mean by $x = x\,$ (or, for that matter, "to be equal") has to be well established first, or a purely formal and syntactical usage of the symbol $=\,$ has to be enforced, only regarding it as a string and only a string of symbols, and mathematical logic does indeed do that.
Another, more interesting example axiom scheme, is that which provides us with what is known as Universal Instantiation:
Axiom scheme for Universal Instantiation. Given a formula $\phi\,$ in a first-order language $\mathfrak\left\{L\right\}\,$, a variable $x\,$ and a term $t\,\!$ that is substitutable for $x\,$ in $\phi\,$, the formula
$\forall x \, \phi \to \phi^x_t$
is universally valid.
Where the symbol $\phi^x_t$ stands for the formula $\phi\,$ with the term $t\,\!$ substituted for $x\,$. (See Substitution of variables.) In informal terms, this example allows us to state that, if we know that a certain property $P\,$ holds for every $x\,$ and that $t\,\!$ stands for a particular object in our structure, then we should be able to claim $P\left(t\right)\,$. Again, we are claiming that the formula $\forall x \phi \to \phi^x_t$ is valid, that is, we must be able to give a "proof" of this fact, or more properly speaking, a metaproof. Actually, these examples are metatheorems of our theory of mathematical logic since we are dealing with the very concept of proof itself. Aside from this, we can also have Existential Generalization:
Axiom scheme for Existential Generalization. Given a formula $\phi\,$ in a first-order language $\mathfrak\left\{L\right\}\,$, a variable $x\,$ and a term $t\,\!$ that is substitutable for $x\,$ in $\phi\,$, the formula
$\phi^x_t \to \exists x \, \phi$
is universally valid.
Non-logical axioms
Non-logical axioms are formulas that play the role of theory-specific assumptions. Reasoning about two different structures, for example the natural numbers and the integers, may involve the same logical axioms; the non-logical axioms aim to capture what is special about a particular structure (or set of structures, such as groups). Thus non-logical axioms, unlike logical axioms, are not tautologies. Another name for a non-logical axiom is postulate.[10]
Almost every modern mathematical theory starts from a given set of non-logical axioms, and it was thought that in principle every theory could be axiomatized in this way and formalized down to the bare language of logical formulas.
Non-logical axioms are often simply referred to as axioms in mathematical discourse. This does not mean that it is claimed that they are true in some absolute sense. For example, in some groups, the group operation is commutative, and this can be asserted with the introduction of an additional axiom, but without this axiom we can do quite well developing (the more general) group theory, and we can even take its negation as an axiom for the study of non-commutative groups.
Thus, an axiom is an elementary basis for a formal logic system that together with the rules of inference define a deductive system.
Examples
This section gives examples of mathematical theories that are developed entirely from a set of non-logical axioms (axioms, henceforth). A rigorous treatment of any of these topics begins with a specification of these axioms.
Basic theories, such as arithmetic, real analysis and complex analysis are often introduced non-axiomatically, but implicitly or explicitly there is generally an assumption that the axioms being used are the axioms of Zermelo–Fraenkel set theory with choice, abbreviated ZFC, or some very similar system of axiomatic set theory like Von Neumann–Bernays–Gödel set theory, a conservative extension of ZFC. Sometimes slightly stronger theories such as Morse-Kelley set theory or set theory with a strongly inaccessible cardinal allowing the use of a Grothendieck universe are used, but in fact most mathematicians can actually prove all they need in systems weaker than ZFC, such as second-order arithmetic.
The study of topology in mathematics extends all over through point set topology, algebraic topology, differential topology, and all the related paraphernalia, such as homology theory, homotopy theory. The development of abstract algebra brought with itself group theory, rings and fields, Galois theory.
This list could be expanded to include most fields of mathematics, including measure theory, ergodic theory, probability, representation theory, and differential geometry.
Combinatorics is an example of a field of mathematics which does not, in general, follow the axiomatic method.
Arithmetic
The Peano axioms are the most widely used axiomatization of first-order arithmetic. They are a set of axioms strong enough to prove many important facts about number theory and they allowed Gödel to establish his famous second incompleteness theorem.[11]
We have a language $\mathfrak\left\{L\right\}_\left\{NT\right\} = \\left\{0, S\\right\}\,$ where $0\,$ is a constant symbol and $S\,$ is a unary function and the following axioms:
1. $\forall x. \lnot \left(Sx = 0\right)$
2. $\forall x. \forall y. \left(Sx = Sy \to x = y\right)$
3. $\left(\left(\phi\left(0\right) \land \forall x.\,\left(\phi\left(x\right) \to \phi\left(Sx\right)\right)\right) \to \forall x.\phi\left(x\right)$ for any $\mathfrak\left\{L\right\}_\left\{NT\right\}\,$ formula $\phi\$ with one free variable.
The standard structure is $\mathfrak\left\{N\right\} = \langle\N, 0, S\rangle\,$ where $\N\,$ is the set of natural numbers, $S\,$ is the successor function and $0\,$ is naturally interpreted as the number 0.
Euclidean geometry
Probably the oldest, and most famous, list of axioms are the 4 + 1 Euclid's postulates of plane geometry. The axioms are referred to as "4 + 1" because for nearly two millennia the fifth (parallel) postulate ("through a point outside a line there is exactly one parallel") was suspected of being derivable from the first four. Ultimately, the fifth postulate was found to be independent of the first four. Indeed, one can assume that exactly one parallel through a point outside a line exists, or that infinitely many exist. This choice gives us two alternative forms of geometry in which the interior angles of a triangle add up to exactly 180 degrees or less, respectively, and are known as Euclidean and hyperbolic geometries. If one also removes the second postulate ("a line can be extended indefinitely") then elliptic geometry arises, where there is no parallel through a point outside a line, and in which the interior angles of a triangle add up to more than 180 degrees.
Real analysis
The object of study is the real numbers. The real numbers are uniquely picked out (up to isomorphism) by the properties of a Dedekind complete ordered field, meaning that any nonempty set of real numbers with an upper bound has a least upper bound. However, expressing these properties as axioms requires use of second-order logic. The Löwenheim-Skolem theorems tell us that if we restrict ourselves to first-order logic, any axiom system for the reals admits other models, including both models that are smaller than the reals and models that are larger. Some of the latter are studied in non-standard analysis.
Role in mathematical logic
Deductive systems and completeness
A deductive system consists of a set $\Lambda\,$ of logical axioms, a set $\Sigma\,$ of non-logical axioms, and a set $\\left\{\left(\Gamma, \phi\right)\\right\}\,$ of rules of inference. A desirable property of a deductive system is that it be complete. A system is said to be complete if, for all formulas $\phi$,
$\text\left\{if \right\}\Sigma \models \phi\text\left\{ then \right\}\Sigma \vdash \phi$
that is, for any statement that is a logical consequence of $\Sigma\,$ there actually exists a deduction of the statement from $\Sigma\,$. This is sometimes expressed as "everything that is true is provable", but it must be understood that "true" here means "made true by the set of axioms", and not, for example, "true in the intended interpretation". Gödel's completeness theorem establishes the completeness of a certain commonly used type of deductive system.
Note that "completeness" has a different meaning here than it does in the context of Gödel's first incompleteness theorem, which states that no recursive, consistent set of non-logical axioms $\Sigma\,$ of the Theory of Arithmetic is complete, in the sense that there will always exist an arithmetic statement $\phi\,$ such that neither $\phi\,$ nor $\lnot\phi\,$ can be proved from the given set of axioms.
There is thus, on the one hand, the notion of completeness of a deductive system and on the other hand that of completeness of a set of non-logical axioms. The completeness theorem and the incompleteness theorem, despite their names, do not contradict one another.
Further discussion
Early mathematicians regarded axiomatic geometry as a model of physical space, and obviously there could only be one such model. The idea that alternative mathematical systems might exist was very troubling to mathematicians of the 19th century and the developers of systems such as Boolean algebra made elaborate efforts to derive them from traditional arithmetic. Galois showed just before his untimely death that these efforts were largely wasted. Ultimately, the abstract parallels between algebraic systems were seen to be more important than the details and modern algebra was born. In the modern view axioms may be any set of formulas, as long as they are not known to be inconsistent.
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# Cambridge CAT
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01 May 2006, 11:58
Hello,
Did anybody take cambridge CAT?
I just did. Unfortunatelly, they dont give you a score, but offer a rather nice review. I found quant pretty tricky and verbal very hard.
How the difficulty of cam.questions correlatted with those of the real test?
Regards,
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01 May 2006, 14:52
never used this, where can i get it...books, online??
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01 May 2006, 15:47
What is Cambridge CAT?
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01 May 2006, 22:17
I don't have a book. Only software. It has 1 diag.test, 1 untimed CAT, and 1 timed CAT.
it seemed to me, that its questions are on the same difficulty level as those of a Kaplan.
What do you think?
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07 May 2006, 08:29
This is related to Manhattan GMAT. If you have their package you would have the complete CD that has about 6-8 tests. Yes, they do not give absolute scores but ranges.
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07 May 2006, 08:29
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# Permutation Group
1. Apr 14, 2005
### gravenewworld
say you have the alternating group An for some permutation group Sn. If you are given An and then 1 odd permutation, must you be able to generate all of Sn? I tried it for S3 and I multiplied all the even perms in S3 by only 1 element that wasn't in A3 and was able to generate all of S3. Does this hold for any n?
Last edited: Apr 14, 2005
2. Apr 15, 2005
### Galileo
Yes, that is generally true.
Note that in any group multiplication on the left by an element in the group is a bijection.
$$ax=b \iff x=a^{-1}b$$
Use this to prove it for the general case $S_n ,n\geq 2$
The case n=1 is special, since the A1=S1.
3. Apr 15, 2005
### gravenewworld
Alright thanks a lot galileo. I just wanted to be sure of that fact before I brought it up in my presentation that I have to give.
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# Why RPMs Are Lower Than CPMs [Infographic]
One of the most common questions I get asked is why RPMs often turn out to be lower than CPMs. In many circles, RPM and CPM are used interchangeably so this has understandably led to abject horror when someone discovers that the RPM they are paid for hosting ads is less than the CPM that the advertiser paid to get ads onto their site.
No-one is being ripped off and nothing underhand is happening. The truth is that RPM and CPM are in fact different ideas. They are just used interchangeably because advertising people talk about CPM a lot and so forget that there is another side to the equation.
This has led to CPM being used to mean RPM, but that is incorrect. There in fact are lots of reasons why CPMs and RPMs can get knocked out of sync.
Before I explore them all, let’s take a quick step back to look at the differences between RPMs and CPMs. We can then celebrate our newfound understanding with an easily digestible infographic at the end, which will make everything as clear as can be.
## What Does CPM Mean?
CPM stands for Cost Per Thousand. The M is commonly thought to stand for mille which is the French word for 1,000 (although I have a bad feeling that is something I made up). It could also just be M – as that is the Roman numeral for 1,000. Weirdly, #7excludeGlossary knows for sure.
Either way, CPM means Cost Per Thousand. Per Thousand what you ask? Well, that depends.
In advertising CPM almost always refers to the cost per 1,000 ad impressions. It is an ad payment model, meaning that advertisers can purchase groups of 1,000 ad impressions for whatever the CPM price is.
The word cost is used because this ad pricing model is from the advertiser’s point of view. Eg it was the answer to #8excludeGlossary.
Click to enlarge
CPM = (Ad Revenue x 1000) ÷ Impressions
When advertisers had direct relationships with where the ads were going, it made sense that CPM was the universal term used. This direct relationship between advertisers and websites doesn’t generally exist anymore however, hence the rise of the use of RPM instead of CPM.
Before jumping into what RPM means, you should know that CPM is also a metric. The average CPM of multiple campaigns can be used to work out how much any set of 1,000 ad impressions cost. For example, if an advertiser had a CPC and a CPA campaign running, they might want to convert the two different ad models into one comparable price to see which is the most #24excludeGlossary.
For this comparison, CPM used to be the standard. Technically it would be the eCPM – Equivalent Cost Per Thousand or Effective Cost Per Thousand (again, weirdly #25excludeGlossary knows which is correct). Nowadays however it is more likely that CPA or ROAS is used to compare campaigns from an advertisers point of view.
While it is rarely used now, eCPM paved the way for RPM. This is because although advertisers don’t necessarily need to know how much impressions are costing them, websites do need to know how much page views are earning.
## What Does RPM Mean?
RPM stands for Revenue Per Thousand. Again the M is commonly thought to stand for mille (the French word for 1,000), or M as in the Roman numeral for 1,000. It just means how much a website is being paid for 1,000 of something. As a measure, it is intentionally #26excludeGlossary and is not a payment model at all.
RPM is also different from CPM in that it doesn’t exactly have a default meaning. CPM almost always implies Cost Per Thousand Ad Impressions. While RPM can also be used to mean Revenue Per Thousand Ad Impressions, it is also commonly used to mean Revenue Per Thousand Page Views, or even Revenue Per Thousand Sessions or Revenue Per Thousand Users. Because of this when people say RPM it is usually clarified with a modifier – eg Page RPM, Impression RPM, Session RPM etc.
Click to enlarge
RPM = (Ad Revenue x 1000) ÷ Impressions
Whether an ad network sells your ad space using an RTB auction or by using a #42excludeGlossary sales team, the one thing that is sure is that they won’t be selling all of your ad inventory on a flat rate CPM basis. They will be making a variety of deals at different price levels, getting as much for your ad inventory as possible.
What has this got to do with RPM? Well, this plethora of deals at different prices is confusing and impractical to report on. It makes life much easier for website owners if they are just told the amount of money they will receive overall. And it makes even more sense if revenue is presented in a way that is directly comparable to other time periods.
Page views and therefore ad inventory fluctuate after all, and so earning £10,000 from 500 page views isn’t the same as earning it from 10,000 page views.
With Page RPM it doesn’t matter how many page views you receive in a month. You can compare the amount you receive per page view (or more specifically, per 1,000 page views). This makes it simple to judge whether your ad network is doing better or worse for you on an ongoing basis.
## CPMs and RPMs Are Different
So a CPM is what an advertiser pays, and an RPM is what an advertiser receives. How can they be different? Oh let me count the ways:
### 1. They Can Be Talking About Different Things.
As mentioned above, CPMs are almost always talking about ad impressions, but RPMs can be talking about Page Views, Sessions, or Users. Which RPM is being used is generally specified, but when it’s not this can lead to confusion.
If an RPM is not an Impression RPM, you will see an RPM which is higher than the corresponding CPM. This is because there are many ad impressions per page view, session or user (in almost all cases). This would lead to an RPM being higher than a CPM, which is literally the opposite of what this article is about, but worth mentioning anyway.
Solution: Always check which RPM is being used if it isn’t specified.
### 2. Not All Impressions Are Being Counted
Ad Networks do a lot of shady stuff, but the one which annoys me most is when they try to create their own definition of a standard term (I’m looking at you Facebook Ads). A common stunt they try to pull is to not count all ad impressions for one reason or another, and then use a term like #56excludeGlossary or #57excludeGlossary in their T&Cs to get away with it.
Two common(ish) ways this happens are:
• Ad Networks give away free inventory (called Added Value) to make up for a mistake they made (or as an introductory offer to advertisers called a Test Budget). This is usually around 10% of the total inventory sold. So for example, the advertiser will pay £10 CPM on 100,000 impressions, and get 10,000 impressions free – with the deal costing £1,000 in total. The website will be paid £1,000 but has sold 110,000 impressions so their Impression RPM will be £9.09 (do the maths yourself).
• An ad network won’t have a 100% fill rate, meaning they don’t use up all your ad impressions. They will only (cheekily) report on the ad impressions they do sell, so it looks like they sold at a high CPM. However, you only received a low RPM as the revenue you receive is averaged out over all the impressions (including the unsold ones).
Solution: Always look at the Page RPM as your main measure and ignore any other metrics.
Ad Networks are businesses that want to encourage customer loyalty like any other business. The way they often do this is to offer a 15% discount to Ad Agencies. Ad Agencies represent many advertisers after all, so getting one to prefer booking with you is an easy way to earn lots of money.
An example of how it works is: an advertiser agrees to pay \$10 CPM to go on a website. The agency books that deal with the ad network. Because ad serving algorithms take price into account (and #64excludeGlossary ads get better ad placements), the deal is booked into the system as a \$10 CPM.
However, the Ad Network pays the Ad Agency 15% of the cost of that deal back. The Ad Network are unlikely to swallow that cost, so they generally pass that 15% cost back to the website. This turns a \$10 CPM deal into an \$8.50 Impression RPM deal (while also making actual \$10 CPM deals less successful as they have to compete with this one).
Solution: This only really happens with Ad Networks with a sales team who need to use discounts to make sales. This sort of Ad Network usually pays out much higher rates than an automated ad platform, so it should even out. However, if they aren’t paying out a page RPM that you are happy with, move to another ad network as they are profiting at your expense.
### 4. RPM Is Only An Estimate Until Payout
Advertisers (especially larger ones) will delay paying for ad space as long as possible. This is not only for accounting reasons but also because they need time to verify results. This sounds suspicious, but attribution is one of the biggest issues in marketing, and it still has no clear cut solution.
As a quick example – imagine someone clicks on an ad on your website but doesn’t make a purchase and closes the page. However, they see the same ad on another website and change their mind and so click and make a purchase. Because of the way conversion pixels work, both your website and the other website will record that the ad on their site made a sale. The advertiser, however, is not going to pay both for the same sale, and so they have to go through a process called #65excludeGlossary. After they have decided who gets paid they tell the ad networks, who up until this point will have been making best guesses at results.
This leads to this situation where an ad network wants to report revenue as soon as possible but doesn’t actually know what it will be paying out. This can lead to the average CPM being reported as high at the time an ad ran, but when the payout actually occurs the RPM is significantly lower.
Solution: If your ad network reports average CPMs, ignore it. Look at the revenue and RPM at the time of payout only.
Even if both the CPM and RPM are referring to the ad impressions, and there are no shenanigans or payment delays going on, RPM and CPM will still be different. This is because CPMs don’t only pay for advertising space.
When an advertiser buys ad space, they also have a whole set of other associated costs. I don’t even mean making and planning an ad, I mean from when the ads are ready to when they appear on a site there are bills to pay. These bills are paid by the advertiser, but reduce the amount the website receives.
These costs include:
• Costs an Ad Agency charges to manage the ads (which will include the 15% discount above when applied)
• Ad Serving Costs – ads need a specific type of server to run them, and these aren’t free. They often charge a CPM cost which is a fraction of a penny, but it still ads up. On top of this there might be multiple ad servers involved – usually at least one used by the ad agency, and a different one used by the ad network.
• Ad Agencies also often use third-party tracking, which will also have an associated cost. They might even use multiple ones at the same time for various reasons.
• There are also potentially other tools that might get used, such as verification tools which track where an ad is shown or fraud prevention tools. These will likely all need to get paid too.
• The Ad Network also needs to get paid – and this will be the largest chunk of the CPM that gets taken. Ad Networks will generally keep at least 50% of the CPM cost.
Whatever is leftover goes to the website and will count towards the RPM.
While the other reasons for CPM/RPM discrepancies listed above only occur some of the time, this reason occurs every time.
Advertisers need to know how much they will pay to run ads so they lump all the costs of running advertising together – and this is the CPM. Websites need to know how much they will be paid after everything out of their control is considered – and this is the RPM.
Solution: There isn’t one, this is just how it works.
## Why RPMs are lower than CPMs Infographic
Without further ado, here is the above information condensed into a helpful infographic. Feel free to share it.
Click to enlarge
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Cody
# Problem 512. Spot the rectangle
Solution 2227285
Submitted on 21 Apr 2020
This solution is locked. To view this solution, you need to provide a solution of the same size or smaller.
### Test Suite
Test Status Code Input and Output
1 Fail
a = [ 1 0 0 0 0 0 0 0 1 0 0 0 0 0 1 0 0 0 0 1 0 0 1 0 0 ]; tf = false; assert(isequal(has_rectangle(a),tf))
Unrecognized function or variable 'A'. Error in has_rectangle (line 2) [R,C] = size(A) Error in Test1 (line 7) assert(isequal(has_rectangle(a),tf))
2 Fail
a = [ 1 0 1 0 0 0 1 0 1 0 1 0 0 0 1 1 ]; tf = true; assert(isequal(has_rectangle(a),tf))
Unrecognized function or variable 'A'. Error in has_rectangle (line 2) [R,C] = size(A) Error in Test2 (line 6) assert(isequal(has_rectangle(a),tf))
3 Fail
a = [ 1 0 1 0 0 0 1 1 0 0 1 1 ]; tf = true; assert(isequal(has_rectangle(a),tf))
Unrecognized function or variable 'A'. Error in has_rectangle (line 2) [R,C] = size(A) Error in Test3 (line 5) assert(isequal(has_rectangle(a),tf))
4 Fail
a = [ 1 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 1 0 1 0 0 0 0 0 0 1 0 ]; tf = true; assert(isequal(has_rectangle(a),tf))
Unrecognized function or variable 'A'. Error in has_rectangle (line 2) [R,C] = size(A) Error in Test4 (line 5) assert(isequal(has_rectangle(a),tf))
5 Fail
a = zeros(20); tf = false; assert(isequal(has_rectangle(a),tf))
Unrecognized function or variable 'A'. Error in has_rectangle (line 2) [R,C] = size(A) Error in Test5 (line 3) assert(isequal(has_rectangle(a),tf))
6 Fail
a = ones(9); tf = true; assert(isequal(has_rectangle(a),tf))
Unrecognized function or variable 'A'. Error in has_rectangle (line 2) [R,C] = size(A) Error in Test6 (line 3) assert(isequal(has_rectangle(a),tf))
7 Fail
a = double(magic(6)<9); tf = false; assert(isequal(has_rectangle(a),tf))
Unrecognized function or variable 'A'. Error in has_rectangle (line 2) [R,C] = size(A) Error in Test7 (line 3) assert(isequal(has_rectangle(a),tf))
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# How many square feet will 2 gallons of epoxy cover?
When you have a project in mind, it’s important to have realistic expectations of what can be achieved. If you try to build a cathedral out of sandcastles, the results are going to be disastrous. With that in mind, let’s hone your expectations for how many square feet your two gallons of epoxy will cover.
## How much does 2 gallons of epoxy cover?
For the most part, it’s hard to tell how much you’ll need until you actually do a project and see how the epoxy works.
However, EpoxyMaster Epoxy, which is a standard industrial-quality epoxy for coating concrete floors and other surfaces, will give about 75 square feet of coverage at 1/8 inch thick. When you double that thickness to 1/4 inch, the coverage rate goes up to 50 square feet.
Knowing this will help you figure out how much material you need for your project. And if you’re hiring someone to do the job for you, this information could also help ensure that they’re not overcharging or using substandard materials.
## How many sq ft does a gallon of epoxy cover?
It’s important to keep in mind that the thicker the pour, the more epoxy you will need. How much 1 gallon of epoxy will cover depends on how thick it is poured, but as a general rule for our products, 1 gallon of Pro Marine Supplies table top epoxy will cover up to 13 square feet if poured at 1/8″ thickness.
This equates to approximately 12 sq ft per coat. So if you are pouring your epoxy at a thickness greater than 1/8″ inch and are interested in knowing how many coats you’ll need, simply divide the total number of square feet by 12. For example:
80 sq ft / 12 sq ft per coat = 6 coats
## How many square feet will 3 gallons of epoxy cover?
The amount of square footage 3 gallons of epoxy will cover is approximately 450.
## How do I calculate how much epoxy I need?
You’ll need to calculate the volume of the space you want to fill and the amount of epoxy needed.
• Volume in cubic inches = Length (in inches) x Width (in inches) x Thickness (in inches).
• Epoxy needed in cubic inches = Volume in cubic inches ÷ Coverage, where coverage equals 1/4″ thickness for every 8 oz. of liquid epoxy.
To convert the total volume needed to gallons, divide by 231 (1 gallon is 231 cubic inches).
## How much epoxy do I need for 400 square feet?
How much epoxy do you need? This can be a tricky question to answer, and it depends on a lot of different factors. To give an example, let’s look at the most commonly asked question: how much epoxy do I need for 400 square feet?
Now that you know how many gallons of epoxy you need, what’s next? Most people use liquid epoxy in two coats (with sanding in between). How long should you wait between coats of epoxy if you are going very deep into the concrete? The answer is important because if it’s too wet or too dry, your final product will not turn out as well.
It all depends on the brand of concrete and other factors like humidity. The manufacturer of your liquid epoxy probably has some recommendations for this number. Let’s assume the recommendation is 12 hours; we’ll go with that number for now.
## How long should you typically wait between coats of epoxy if you are pouring it very deep?
The exact amount of time you should wait between coats depends on a few factors:
• The type of epoxy you’re using
• The thickness of the pour
• How deep your project is. The deeper it is, the longer you’ll need to wait between pours because the epoxy will take longer to cure.
For most epoxies, it’s recommended that you wait at least 12-24 hours before applying another coat. If you want to create a deep pour and create multiple layers, however, you’ll need to give it more time than that—you should usually wait at least 48 hours between layers. Sometimes the label may specify how long you should wait for a given product; if so, follow these instructions instead of this general advice.
## How much is epoxy per gallon?
So, for example, if you purchased three gallons of epoxy, you’d have a total of 36 square feet. With this in mind, you can calculate the amount of epoxy you need by dividing your project’s area by 12 (the number of square feet one gallon covers).
For instance, let’s say we want to cover an area that measures 72 square feet. Divide 72 by 12 and we get 6—so we should buy six gallons. (Tip: You should always round up to the nearest whole number when purchasing anything to cover an area—in this case, six gallons instead of 5.67.)
## How much epoxy Do I need to make a river table?
The first question to answer is how much resin you’ll use in your project, and for this, you need to know the size of your surface area. Then you can apply the volume formula:
Area = length x width
Volume = Area x depth
If you have a small end table that is 24” long by 12” wide, then one half-gallon kit (that coats 10 square feet) will be sufficient. But if you decide to make a very large river table that’s 7 feet long and 2 feet wide, then at least two gallons will be needed per side (and possibly more depending on how thickly you want it poured).
And remember—you may need additional epoxy if your project is not perfectly flat or if it contains holes or voids. When in doubt, go with a little extra just in case.
## Conclusion
To recap: epoxy is a great material to use when coating a surface, and it comes in many different forms and variations.
With the proper tools and knowledge, you can make almost any surface look brand new. It is important to calculate how much epoxy you will need by measuring the square footage of your project area before buying.
One gallon of epoxy will cover 250-300 square feet, so two gallons will cover roughly 500-600 square feet.
Martin Flood
Martin Flood has been working in the construction industry for over 20 years as a general contractor with expertise in remodeling projects that are large or small. He has furthered his career by specializing in epoxy resin flooring, providing excellent service to both commercial and residential clients. Martin’s experience enables him to offer professional advice on how to choose the right type of project based on your needs and budget.
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Seven countries signed a treaty binding each of them to : GMAT Critical Reasoning (CR)
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# Seven countries signed a treaty binding each of them to
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20. Seven countries signed a treaty binding each of them to perform specified actions on a certain fixed date, with the actions of each conditional on simultaneous action taken by the other countries. Each country was also to notify the six other countries when it had completed its action.
The simultaneous-action provision of the treaty leaves open the possibility that
(A) the compliance date was subject to postponement, according to the terms of the treaty
(B) one of the countries might not be required to make any changes or take any steps in order to comply with the treaty, whereas all the other countries are so required
(C) each country might have a well-founded excuse, based on the provision, for its own lack of compliance
(D) the treaty specified that the signal for one of the countries to initiate action was notification by the other countries that they had completed action
(E) there was ambiguity with respect to the date after which all actions contemplated in the treaty are to be complete
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26 Apr 2005, 17:13
(C) each country might have a well-founded excuse, based on the provision, for its own lack of compliance
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26 Apr 2005, 18:33
Not too sure but will go with (C).
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26 Apr 2005, 23:59
I was toen apart between B & C and I think I will go with B.
My reasoning.. that the country that has to start the last .. need not comply with the Treaty but for the 7th to start it actions the 6 will have to comply with the treaty!
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27 Apr 2005, 01:07
christoph wrote:
MA wrote:
(C) each country might have a well-founded excuse, based on the provision, for its own lack of compliance
can you explain why C) ? thx
ahh i got it. its C)...because the actions of each are conditional on simultaneous action taken by the other countries, a country does not take action because it "waits" until another takes the action. thats why each country has a well-founded excuse based on the treaty to act simultaneously. in the end it it possible that no country starts the action.
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27 Apr 2005, 04:32
THe OA is C.
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27 Apr 2005, 05:50
christoph wrote:
christoph wrote:
MA wrote:
(C) each country might have a well-founded excuse, based on the provision, for its own lack of compliance
can you explain why C) ? thx
ahh i got it. its C)...because the actions of each are conditional on simultaneous action taken by the other countries, a country does not take action because it "waits" until another takes the action. thats why each country has a well-founded excuse based on the treaty to act simultaneously. in the end it it possible that no country starts the action.
thanks, it means no one would begin first.
Could you tell me what does the choice D talk about?
Re: CR: Treaty provisions [#permalink] 27 Apr 2005, 05:50
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, 02.07.2019 15:00 chaparoonie15
# Stephanie wants to take \$10,000 out of her bank savings account and put it in a certificate of deposit (cd) that charges a penalty if she withdraws the money before two years are over. what is the economic trade-off in this investment? 1.higher returns for more risk 2.higher returns for higher liquidity 3.higher returns for higher volatility 4.higher returns for lower liquidity
### Another question on Mathematics
Mathematics, 21.06.2019 20:30
W-16=-12 solve each one step equation plz
Mathematics, 21.06.2019 21:30
Iwill give brainliest. suppose tommy walks from his home at (0, 0) to the mall at (0, 5), and then walks to a movie theater at (6, 5). after leaving the theater tommy walks to the store at (6, 0) before returning home. if each grid square represents one block, how many blocks does he walk?
Mathematics, 21.06.2019 23:10
Which best describes the function on the graph? direct variation; k = −2 direct variation; k = -1/2 inverse variation; k = −2 inverse variation; k = -1/2
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Notes on Principle of Conversation of Linear Motion | Grade 11 > Physics > Laws of Motion | KULLABS.COM
Notes, Exercises, Videos, Tests and Things to Remember on Principle of Conversation of Linear Motion
Please scroll down to get to the study materials.
• Note
• Things to remember
• Videos
#### Linear momentum
Momentum is the motion contained in a body. The quantity of motion possessed by a body depends on upon both of its mass and velocity. So, the product of mass and velocity is the measure of the momentum,
$$\vec p = m\vec v$$
It is also called linear momentum. It is a vector quantity whose direction is in the direction of the velocity. Its unit is kg m/s in SI-units and dimension, [ML-1].
##### Principle of Conversation of Linear Motion
The law of conservation of linear momentum states that if no external forces act on the system of two colliding objects, then the vector sum of linear momentum of each body remains constant and is not affected by their mutual interaction.
Let us consider an isolated system of n particles having initial momentum p1, p2 …..pn. Due to the collision, let the momentum of the particles after collision be p1’, p2’ ….. pn’ respectively. Then according to the principle of conservation of linear momentum, in the absence of external force,
$$p_1 + p_2 + \dots p_n = p_1 ‘ + p_2 ‘ + \dots p_n ‘$$
For verification, we consider a collision between two spheres A and B having masses of m1 and m2 respectively. Let u1 and u2 be the velocities of the spheres before collision such that u1 > u2 and moving on the same straight line as shown in the figure. After collision, let their velocities be v1 and v2 on the same line. If they collide each other for short interval of time t, each sphere exerts a force on the other sphere and so, the force experienced by A is given as
\begin{align*} F_2 &= \frac {\text {change in momentum}}{\text {time}} = \frac {m_1v_1 – m_1 u_1}{t} \\ \text {Similarly, force experienced by B is } \\ F_1 &= \frac {\text {change in momentum}}{\text {time}} = \frac {m_2v_2 – m_2 u_2}{t}\\ \end{align*}
According to Newton’s Third law of motion, the forced experienced by A and B are equal and opposite \begin{align*} \\ F_1 &= -F_2 \\ \text {or,} \: \frac {m_1 (v_1 –u_1)} {t} &= -\frac {m_2 (v_2 – u_2) } {t} \\ \text {or,} \: m_1v_1 –m_1v_1 &= -m_2v_2 + m_2u_2 \\ \text {or,} \: m_1u_1 + m_2u_2 &= m_1v_1 + m_2v_2 \\ \end{align*}
This proves that total momentum before collision is equal to the total momentum after collision if no external forces at on them prove the principle of conservation of linear momentum.
Momentum is the motion contained in a body. Quantity of motion possessed by a body depends upon both of its mass and velocity. So, the product of mass and velocity is the measure of the momentum.
linear momentum is a vector quantity whose direction is in the direction of the velocity. Its unit is kg m/s in SI-units and dimension, [ML-1].
The law of conservation of linear momentum states that if no external forces act on the system of two colliding objects, then the vector sum of linear momentum of each body remains constant and is not affected by their mutual interaction.
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#### What is a Liquidity Ratio?
A liquidity ratio is a financial measure used to assess a company's ability to meet its short-term financial obligations. In other words, a liquidity ratio measures a company's ability to pay its bills on time. This metric is also known as a current ratio, and it generally measures the amount of cash or readily available cash relative to current liabilities.
Liquidity ratios are important financial ratios used to evaluate a company's financial health. A company's liquidity ratios help investors and analysts understand the company's financial position, including its ability to generate cash, pay its bills, and meet its financial obligations. Additionally, liquidity ratios are important measures to test a company’s solvency, in addition to its potential ability to handle economic shocks.
There are several types of liquidity ratios used in financial analysis. The most commonly used liquidity ratios include the current ratio, quick ratio, and cash ratio.
The current ratio is the most basic liquidity ratio and is calculated by dividing current assets by current liabilities. The current ratio measures a company's ability to pay off its short-term liabilities with its short-term assets. A current ratio of 1 or higher is considered good, indicating that the company has enough current assets to cover its current liabilities.
The quick ratio is a more stringent liquidity ratio that only considers a company's most liquid assets. The quick ratio is calculated by dividing the sum of cash, marketable securities, and accounts receivable by current liabilities. This ratio excludes inventory, which may take longer to convert to cash. A quick ratio of 1 or higher is considered good, indicating that the company can cover its short-term obligations with its most liquid assets.
The cash ratio is the most conservative liquidity ratio and only considers a company's cash and cash equivalents. The cash ratio is calculated by dividing cash and cash equivalents by current liabilities. A cash ratio of 1 or higher is considered good, indicating that the company has enough cash to cover its short-term liabilities.
Liquidity ratios are important because they indicate a company's ability to meet its short-term obligations. If a company has a low liquidity ratio, it may not be able to pay its bills on time, which could lead to financial problems and possibly even bankruptcy. Additionally, a low liquidity ratio may indicate that a company is struggling to generate cash, which could be a sign of poor management or a weak business model.
On the other hand, a high liquidity ratio may indicate that a company has excess cash or is not investing enough in its business. While having excess cash is not necessarily a bad thing, it may indicate that a company is not utilizing its resources efficiently. Additionally, a high liquidity ratio may indicate that a company is not taking advantage of growth opportunities, which could be a sign of poor management.
It's important to note that liquidity ratios are not the only financial ratios used in financial analysis. Investors and analysts also use profitability ratios, efficiency ratios, and leverage ratios to evaluate a company's financial health. However, liquidity ratios are an important component of financial analysis because they provide insight into a company's short-term financial health.
In addition to measuring a company's short-term financial health, liquidity ratios are also useful for comparing companies within the same industry. By comparing liquidity ratios across companies, investors and analysts can identify companies that are better positioned to weather economic downturns and other financial shocks.
For example, suppose two companies in the same industry have similar revenue and profit margins. In that case, the company with a higher liquidity ratio may be better positioned to weather economic shocks because it has more cash and liquid assets to meet its financial obligations.
A liquidity ratio is a financial measure used to assess a company's ability to meet its short-term financial obligations. The most commonly used liquidity ratios include the current ratio, quick ratio, and cash ratio. Liquidity ratios are important because they indicate
a company's ability to pay its bills on time, which is critical for maintaining financial health and avoiding bankruptcy. By measuring a company's liquidity, investors and analysts can gain insight into its short-term financial position and ability to handle economic shocks.
However, liquidity ratios should not be viewed in isolation, as they do not provide a complete picture of a company's financial health. It's important to consider other financial ratios, such as profitability ratios and efficiency ratios, to gain a more comprehensive understanding of a company's financial position.
Furthermore, liquidity ratios can be influenced by a company's industry and business model. For example, a company in a capital-intensive industry may have a lower liquidity ratio due to high levels of investment in long-term assets, such as property, plant, and equipment. In contrast, a company in a service-based industry may have a higher liquidity ratio due to lower levels of investment in fixed assets.
Investors and analysts should also be aware of potential limitations when using liquidity ratios to evaluate a company's financial health. For example, a company may have high levels of accounts receivable, which may artificially inflate its liquidity ratio. Similarly, a company may have large amounts of inventory that take longer to convert to cash, which may negatively impact its liquidity ratio.
In addition, liquidity ratios may not be as useful for companies with unpredictable cash flows or significant debt obligations. In these cases, investors and analysts may need to use other financial ratios, such as debt-to-equity ratios and interest coverage ratios, to gain a better understanding of the company's financial position.
Liquidity ratios are an important financial measure used to assess a company's short-term financial health and ability to meet its financial obligations. By measuring a company's liquidity, investors and analysts can gain insight into its financial position and ability to handle economic shocks. However, liquidity ratios should be viewed in conjunction with other financial ratios and should be considered in the context of a company's industry and business model. By taking a holistic approach to financial analysis, investors and analysts can make more informed decisions about potential investments and better understand the financial health of companies they are considering.
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# Revealed Preference Theory, Introduction and Graphical Representation
## Introduction
• Propounded by Paul Samuelson
• Ordinal utility analysis
• This theory analyses preference for a combination of goods on the basis of observed consumer behavior in the market.
• The demand theorem given by Samuelson, also known as the Fundamental theorem of consumption theory, states that a commodity that is known to have an increased demand when the income rises must have a decrease in demand when there is a rise in its price.
• In other words, when income elasticity of demand is positive, price elasticity of demand is negative.
## Choice Reveals Preference
• This theory of demand is based on the revealed preference hypothesis which states that choice reveals preference.
• According to this theorem, a consumer buys a combination of goods because of two reasons:
• Either the consumer likes the combination more than the other combinations even if it costs more,
• Or the combination is cheaper than the other combinations.
## Assumptions
• If a consumer chooses a combination, he reveals his preference for that combination.
• Tastes and preferences of the consumer are constant
• A consumer always prefers a combination with more goods than a combination with less goods.
• Only one combination is chosen at a given price-income line.
• Based on strong ordering.
• If A is preferred to B in one situation, combination B cannot be preferred to A in another situation. Therefore, the consumer is consistent in his behavior. This is called two-term consistency.
• If combination A is preferred to B, and B is preferred to C, then A is assumed to be preferred to C. This is known as transitivity or three-term consistency.
• The income elasticity of demand is positive, that is, when the income of the consumer rises, he demands more of the commodity and vice-versa.
## Demand Theorem in Case of Rise and Fall in Price
• We will now analyze the demand theorem in two cases:
• when price rises,
• and when price falls.
• We will assume that there are two commodities, X and Y.
## Mcqs
Q. 1. Who propounded the revealed preference theory?
1. Milton Friedman
2. Paul Krugman
3. Paul Samuelson
D. Kenneth Arrow
Q. 2. When income elasticity of demand is positive, the price elasticity of demand is:
A. Negative
B. Positive
C. Not correlated
D. None of the above
Q3. Which of the following is an assumption of the revealed preference theory?
A. A consumer will always choose a combination with less goods as opposed to a combination with more goods.
B. If a consumer chooses combination X in one situation, he may choose other combinations such as Y or Z in other situations.
C. If combination A is preferred to B and B is preferred to C, combination A is assumed to be preferred to C.
D. The income elasticity of demand is negative.
Q. 4. . Revealed preference theory assumes
A. Weak ordering
B. Strong ordering
C. Constant ordering
D. Multiple ordering
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# Rotor Power
Ok, here we go.
In a rotating mechanism (helicopter rotor), at a state of equilibrium, the rotor consumes a certain amount of energy from the shaft to maintain a constant angular velocity (since there is a measure of resistance present over the span of the rotor).
Lets suppose that the moment of inertia from all sources of drag acting on the rotor, transmitted to the shaft at a speed of 600RPM is measured to be 200Nm. If the rotor is made of two blades then each blade will have a moment of 100Nm?
Each rotor blade measures 3m from the shaft centre point and has a mass of 80N.
Second, assuming that the moment of inertia remains 200Nm for all angular velocities, how long will it take to slow the rotor to 0 RPM?
Sorry if its vague, Im engaged in a purely academic design of a helicopter but have become lost in the rotating physics!
"Moment of inertia" is essentially how hard it is to accelerate or decelerate the rotor. When you say "moment of inertia from drag is 200Nm" you're thinking of the torque from drag. That said if the total torque on the shaft is 200 Nm then indeed each blade will contribute 100 Nm.
In order to answer how long it will take to slow the rotor to 0 rpm, THIS is where the "moment of inertia" comes in. This depends on the shape of the rotor, but your rotor seems simple enough that you could estimate a reasonable value by some calculations (you need to google the formulas for these). Once you have torque and moment of inertia, how fast the rotational speed changes is a matter of another simple formula (again a google search away).
Wow, thats probly the most useless piece of advice anyone gives everyone nowadays.
The whole point of coming on this forum is NOT to just be redirected away to another search query. Perhaps you may not have thought of this but, maybe I would like to interact with a real person, not just some static page I cant ask questions to. Why dont we just all get our university degrees from google, the google university... preposterous.
HallsofIvy
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Lorentz Transformation of the Electromagnetic Field
Back to Contents
In Section 6 of his famous paper "On the Electrodynamics of Moving Bodies", Albert Einstein worked out the transformation of the electromagnetic field as measured by two observers moving relative to each other with some unvarying velocity. He obtained that result from applying the Lorentz Transformation for spatio-temporal coordinates to the third and fourth of Maxwell痴 Equations as one observer has them and then comparing the result to the same equations as the other observer has them, applying the basic relativistic criterion that the Lorentz Transformation applied to electric and magnetic fields must leave Maxwell痴 Equations unchanged; specifically, both observers must use the same mathematical form of the equations. However, a closer look at what Einstein did reveals that he already had the transformation equations for the electromagnetic field and merely substituted them into his equations to verify that they did, indeed, represent the electromagnetic field痴 analogue of the Lorentz Transformation of spatio-temporal coordinates, which he had deduced from his two postulates of Relativity.
So now we have before us the task of working out the Lorentz Transformation of the electromagnetic field from scratch. We want to start with equations that describe the relationships among the components of the electric and magnetic fields calculated with respect to the usual rectangular Cartesian grid of spatial components augmented by the temporal component (represented here with lower-case letters). We call the inertial frame occupied and marked by that grid the L-frame. We also have a U-frame, on whose grid observers use upper-case letters to represent measurements of distance and duration. Both of the grids of the L-frame and the U-frame share a common x-axis and the U-frame moves in the positive x-direction at the speed v relative to the L-frame. The Lorentz Transformation of L-frame coordinates into U-frame coordinates thus takes the form,
(Eq地s 1)
in which we have the Lorentz factor between the two frames as
(Eq地 2)
and β=v/c.
For convenience we can represent Equations 1 as a matrix multiplication,
(Eq地 3)
in which we have used the Einstein convention of summing over repeated indices and we have represented the Lorentz Transformation by the matrix,
(Eq地 4)
In order to use that matrix we need to represent the distance-duration 4-vector as xj={x,y,z,ict}.
We start by transforming the electromagnetic potential. For a charged particle immersed in an electromagnetic field we have {p,iE/c}={qA,iqφ/c}, which means that the potential, Φ={A,iφ/c}, represents a proper 4-vector. If the particle touches two events separated by a minuscule displacement δx={δx,icδt}, then the product ΦAδx remains invariant under a Lorentz Transformation. If the U-frame observers use primed variables to represent their measurements of the electromagnetic field and the L-frame observers use unprimed variables, then we have
(Eq地 5)
which gives us
(Eq地 6)
That equation tells us that, as a proper 4-vector, the electromagnetic potential transforms in accordance with the standard Lorentz Transformation. Thus we have the electromagnetic-potentials analogue of Equations 1,
(Eq地s 7)
Next we want to look at the forcefields that come from the potential fields. In preparation for taking that step we need to look at how the Lorentz Transformation applies to the process of partial differentiation.
For a differential change in displacement we have the transformation between two coordinate grids as
(Eq地 8)
Comparing that equation with Equation 3 tells us that
(Eq地 9)
Next we differentiate a scalar function, f(Xi)=f(xj), that remains invariant under a Lorentz Transformation and get
(Eq地 10)
Because we used a function not subject to the Lorentz Transformation, we can therefore infer the operator relation
(Eq地 11)
so now we know that we can represent the four-dimensional differentiation operator, {L, i/ct}, as a proper 4-vector.
From the appendix we get the tensor representation of the electromagnetic field, which we can expand into a explicit matrix,
(Eq地 12)
When we multiply that matrix by the column vector representing the current associated with a charged particle the first three rows yield the three components of the force exerted upon the particle and the fourth row yields a description of the rate at which the field alters the particle痴 energy.
We know how to apply the Lorentz Transformation matrix to a description of the 4-potential and to the 4-differentiation that converts it into a description of the electromagnetic field, so now we have the means to apply the Lorentz Transformation to the electromagnetic field directly. We have
(Eq地 13)
In going from the second line to the third line we exploit the fact that the differentiation operator does not affect the Lorentz Transformation matrix and the fact that commuting a matrix with a vector changes the matrix into its transposed version. We also made use of the fact that Λik=Λjm=Λ. So now we can rewrite Equation 13 in a more explicit form;
(Eq地 14)
In that equation I have assumed that the two inertial frames occupied by our observers move only in the x-direction at the speed β=Vx/c. In drawing out the transformed matrix I also made us of the fact that
(Eq地 15)
in the upper right and lower left corners of the matrix. If we now compare the electromagnetic field matrix in the first line, term by term, with the matrix in the third line, we get the Lorentz Transformation of the electromagnetic field:
(Eq地s 16)
And, of course, multiplying those components by the appropriate components of a 4-current gives us the electromagnetic 4-force acting on the particle carrying the electric charge of the 4-current. And thus we get the same equations that Einstein used in Section 6 of his first Relativity paper.
Appendix: Electromagnetic Force
Given the 4-potential, we want to derive an explicit description of the electromagnetic field and the force that it exerts upon an electrically charged particle. We begin the derivation by restating the principle of least action,
(Eq地 A-1)
which leads to the Euler-Lagrange Equations,
(Eq地 A-2)
For the purpose of applying those equations we devise the Lagrangian function by adding the negative of the relativistic Lagrangian of a free particle to the Lagrangian created from taking the inner product of the 4-potential and the 4-velocity,
(Eq地 A-3)
Substituting that expression into Equation A-2 then gives us
(Eq地 A-4)
Rearranging that equation and working through the vector identities transforms that equation into
(Eq地 A-5)
In that equation we see that the rate at which the particle痴 linear momentum changes (the force acting on the particle) equals the sum of the electrostatic force and the Lorentz force.
Next we want to put Equation A-2 into tensor form so that we can derive the tensor form of the electromagnetic field. In this derivation we have the 4-Lagrangian as
(Eq地 A-6)
in which we sum the subscripts for the values i=1, 2, 3, and 4.
Let痴 go back to the beginning, the principle of least action,
(Eq地 A-7)
In that equation we have T representing the proper time, the elapse of time measured in the inertial frame in which the endpoints of the measured action coincide in space. Because the processes of variation (a kind of differentiation) and integration commute with each other, we rewrite the left side of that equation as
(Eq地 A-8)
As usual, the middle term in the third line vanishes because it gives us the function in parentheses evaluated between the endpoints of the action, where the variation of location goes to zero. Because I left the inertial reaction term (the first term on the right side of Equation A-3) out of my Lagrangian, the expression in square brackets on the fourth line does not go to zero, but, rather, equals the rate at which the linear momentum of the forced particle changes; in other words, it represents the force exerted by the electromagnetic field.
It may seem strange, even illegitimate, to include a variation in time, δx4, in that derivation. But a proper relativistic treatment of the principle of least action demands such an inclusion. To understand such a variation imagine that the force acting on the particle varies with the elapse of time: the variation then represents the particle coming to a given value of the force sooner or later than it does on the true path.
We also have the total time derivative operator as
(Eq地 A-9)
Noting that the 4-velocity in that expression represents the motion of an inertial frame and using the Lagrangian function from Equation A-6, we write the Euler-Lagrange expression from Equaton A-8 as
(Eq地 A-10)
In that expression the lower case eff represents the force that the field exerts upon the particle. The derivatives of the 4-velocity go to zero, because the 4-velocity represents the motion of an inertial frame, which the particle occupies, however temporarily. In going from the second line to the third I interchanged the indices in the second term on the right so that I could factor out the 4-velocity and thereby represent the 4-force as the product of the 4-current (quj) associated with the particle and the electromagnetic force,
(Eq地 A-11)
which we use in the transformation above. In form that description looks like a four-dimensional analogue of the curl of a vector.
efefabefef
Back to Contents
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# Chapter 1
### Operations & Properties
TermDefinition
Estimate are close to the exact answer but are usually easier and faster to find.
Compatible Number are close to the numbers in the problem, and they can help do math mentally.
Underestimate an estimate that is less than the exact answer.
Overestimate an estimate that is greater than the exact answer.
Dividend the number to be divided into groups.
Divisor the number by which the dividend is divided.
Exponent a number by which tells how many times the base is used as a factor.
Base the number that is used as a factor.
Exponential Form is when a number is written with a base and an exponent.
Numerical Expression is a mathematical phrase that includes only numbers and operation symbols.
Simplify finding the value of a numerical expression.
Order of Operations a rule for evaluating expressions: Parenthesis, Exponents, Multiplication/Division (order from L→R), Addition/Subtraction (order from L→R).
Created by: Juchrin
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# Resistance In Series/Parallel circuits
1. Oct 27, 2014
### Mark Rice
1. The problem statement, all variables and given/known data
Hi, I need to find the total resistance of a circuit (attached file). I'm pretty sure it's really and straight forward but I'm a chemical engineer and this is coursework so just wanted to make sure I was doing it right.
2. Relevant equations
3. The attempt at a solution
Do I just simplify by adding the following to each other in series to give three resistor value (R2+R3) (R4+R5+R6) (R7+R8) then work it out using these in the parallel question? ie Rcircuit= R1 + R9 + [(1/Rtotal)=1/(R2+R3) + 1/(R4+R5+R6)] + 1/(R7+R8)]
I totally appreciate they way I have just typed this was not very clear so I'm happy to explain myself more!
*Edited*
#### Attached Files:
• ###### Resistor.png
File size:
6 KB
Views:
67
2. Oct 27, 2014
### haruspex
That cannot be right because R4 is not parallel to R7 and R8. But you have the right approach - collapse combinations in stages, starting with the simplest ones.
3. Oct 27, 2014
### Mark Rice
So do I add R1 + R4 + R9 in series. Then do the parallel calculation [(1/Rtotal)=1/(R2+R3) + 1/(R5+R6)] + 1/(R7+R8)]? Sorry not done physics in 3 years so I'm a bit rusty!
4. Oct 27, 2014
### Mark Rice
Is this what I do anyone?
5. Oct 27, 2014
### haruspex
Not R4. There's a connection with other resistors in between R1 and R4.
In general, you look for:
- two resistors in sequence with no other connection between them; if found, combine them, adding resistances
- two resistors joined to each other at both ends; if found, combine them using the parallel resistance rule
- repeat as necessary
Yes, you can combine 2 with 3, 5 with 6, and 7 with 8. What next?
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The purpose of the assignment is to develop students’ abilities in using datasets to apply the concepts of sampling distributions and confidence intervals to make management decisions.
Assignment Steps
Resources: Microsoft Excel®, The Payment Time Case Study, The Payment Time Case Data Set
Review the Payment Time Case Study and Data Set.
Develop a 700-word report including the following calculations and using the information to determine whether the new billing system has reduced the mean bill payment time:
• Assuming the standard deviation of the payment times for all payments is 4.2 days, construct a 95% confidence interval estimate to determine whether the new billing system was effective. State the interpretation of 95% confidence interval and state whether or not the billing system was effective.
• Using the 95% confidence interval, can we be 95% confident that µ ≤ 19.5 days?
• Using the 99% confidence interval, can we be 99% confident that µ ≤ 19.5 days?
• If the population mean payment time is 19.5 days, what is the probability of observing a sample mean payment time of 65 invoices less than or equal to 18.1077 days?
Latest completed orders:
Completed Orders
# Title Academic Level Subject Area # of Pages Paper Urgency
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https://www.bloombergprep.com/gmat/practice-question/1/1018/verbal-section-critical-reasoning-critical-reasoning-investigation-questions/
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We cover every section of the GMAT with in-depth lessons, 5000+ practice questions and realistic practice tests.
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# Critical Reasoning: Investigation Questions
A new polymer, recently developed as a material for minor car repairs, was imported to Country M after being sold successfully in Country B. Despite a high number of initial purchases following an effective marketing campaign, sales figures in Country M were unsatisfactory, and importation of the product was eventually discontinued. It can be deduced that consumers in Country M are more critical about car repair products than they are in Country B.
Which of the following would it be most useful to establish in order to evaluate the argument?
Incorrect.
[[snippet]]
This question would not assist us in evaluating the argument since it deals with an issue - health - that applies to consumers in both Country M and in Country B. Note that the author draws a conclusion about the difference between the consumers of those countries.
The crucial factor we are looking for should also point to a difference between the people in M and B. This difference should explain the discrepancy in sales and show us whether it's truly due to M's citizens' critical sense.
Incorrect.
[[snippet]]
Knowing who performed the repairs using the polymer will not help us establish whether M's citizens are more critical than people in B are. M's citizens could buy the polymer and use it themselves, or ask a professional to use this polymer. In both cases someone buys the product and increases sales, but we still don't know why they abandoned this product.
Incorrect.
[[snippet]]
This question is quite pointless since the second premise mentions that the consumers did buy the product at first. This means that they were open to trying the product despite its being new to the market. For some reason, however, they abandoned it later. Was this beause of their critical sense and high standards?
Incorrect.
[[snippet]]
The answer to this question would affect the consumers of both countries. However, the conclusion focuses on the differences in the product sales in two countries, so this answer choice cannot explain this discrepancy.
Excellent work!
[[snippet]]
This answer choice asks a relevant question that may explain the different sales figures in each country and, therefore, may prove or disprove the author's conclusion. If the polymer works better in Country B than it does in M, then this is very likely the reason for the low sales figures. If the polymer is equally effective in each country, then maybe the consumers are more picky when it comes to such products.
Whether the use of polymers without professional safety equipment such as gas masks can be harmful to the health of the average consumer
Whether consumers in Country M are prepared to perform repairs on their own cars instead of hiring paid professionals
Whether consumers in Country M are open-minded as far as the introduction of new products to the market is concerned
Whether the polymer's effectivity is influenced by climactic differences between regions
Whether repairs done using the new polymer can be painted with regular car paint
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Summer minicourses 2019
The summer minicourses are a chance for Michigan graduate students to teach each other interesting math in a friendly, informal setting.
Generally, courses meet once a day, in the afternoon, for a week. All are welcome to attend, but keep in mind the target audience is graduate students in the mathematics department.
The line-up this year will appear below, though additions and changes are likely! You can show or hide all the abstracts for printing purposes.
If you're giving one of the minicourses and you'd like to post notes, just send them to me and I'll post them!
Topic Speaker Dates Location and Time Abstract Notes
Motivic integration Devlin Mallory May 13–May 17 (M–F) EH 3088 1–2:30
Abstract. Motivic integration was introduced by Kontsevich in 1995 as a tool to prove that birational Calabi–Yau manifolds have the same Hodge numbers (generalizing previous results using p-adic integration). Since then, it's found numerous applications in algebraic and arithmetic geometry, particularly towards the study of cohomological invariants and singularities. In this course, we'll introduce arc schemes and the Grothendieck ring of varieties before developing the basic theory of motivic integration; we'll then apply the theory to prove Kontsevich's result on Hodge numbers. Additional time permitting, we'll give applications to the study of Igusa zeta functions and Hodge spectra. Throughout, we will focus on examples and motivation rather than technical proofs; the only prerequisite is basic algebraic geometry.
Homotopy theory and derived algebra via ∞-categories Montek Singh Gill May 20–May 23 (M–Th) EH 3088 1–3
Abstract. I will describe foundational aspects of homotopy theory and derived algebra, from the perspective of ∞-categories. First, I'll cover some basics of homotopy theory and that of ∞-categories as a context in which to do homotopy theory, including the ∞-category of spaces as the most important example. Next, I'll discuss ∞-(co)limits, and the corresponding notion of homotopy (co)limits for spaces. Another formal framework in which to do homotopy theory is that of model categories, and I will describe the relation between these two frameworks. Next, I'll describe the idea behind derived/homotopy coherent/higher algebra, and how this leads us to stable ∞-categories, as well as to the stable ∞-category of spectra, the most important example of such categories. Finally, I'll discuss how to do more of the derived algebra via operads in spaces and via ∞-operads. All throughout, there will be an emphasis on detail and precision.
Combinatorial commutative algebra Francesca Gandini May 28–May 31 (Tu–F) EH 3088 1–3
Abstract. This mini-course is a gentle introduction to a modern field of mathematics via examples. Starting in the seventies with the work of Stanley and Hochster, new results in commutative algebra have been proved by studying associated combinatorial objects. Ever since, new synergies between algebraic and combinatorial objects have been established. Each day I will introduce one of the main four families of examples: monomial ideals, binomial ideals, determinantal ideals, and linear ideals. The course will emphasize applications of the results from the literature to specific computations, so that we can understand via concrete examples what we can find out about these algebraic objects using the combinatorial perspective. The course will be a mix of lecture and problem solving as each day we work through a worksheet.
Hodge theory for combinatorial geometries Shelby Cox, Will Dana,
Sameer Kailasa, Harry Richman, and
Robert M. Walker
June 3–June 7 (M–F) EH 3088 1–2:30
Abstract. In 2015, Adiprasito, Huh, and Katz resolved a longstanding conjecture of Rota and Welsh stating that the coefficients of the characteristic polynomial of a matroid (for example, the chromatic polynomial of a graph) form a log-concave sequence. For matroids representable over a field, Huh and Katz had already proven this by associating an algebraic variety to a matroid and using inequalities from Hodge theory. Inspired by this, the general proof defines a "matroid Chow ring" and proves precise analogues of the Hodge-theoretic results without using algebraic geometry, creating an exciting new vantage point from which to study matroids.
This minicourse will outline Adiprasito, Huh, and Katz's argument, across 5 talks:
• The requisite background on matroids.
• The inequalities from Hodge theory which motivate analogous results for the matroid Chow ring.
• The "wonderful compactification" construction, which connects the Chow ring of a representable matroid to an actual Chow ring.
• How the matroid Chow ring allows us to show log-concavity for the characteristic polynomial.
• Sketching the proof of the Hodge-theoretic inequalities for the matroid Chow ring.
Master course on algebraic stacks Ruìān Chén June 10–June 14 (M–F) EH 3088 1–2:30
Abstract. This minicourse recaptures a part of “A master course on algebraic stacks” taught by Bertrand Toën in University of Toulouse in 2005 (hence the minicourse title). More specifically, we will take the perspective a homotopy theorist to try to understand stacks, or more specifically, stacks in groupoids.
Originally conceived by Grothendieck for construction of moduli “spaces” in descent theory, via the algebraic notions of “pseudofunctors” and “fibered categories”, the notion of stacks is a (higher) categorical device which takes into account of the automorphisms of the objects being parametrized. However, as many other higher categorical notions, the algebraic definition of a stack can sometimes be less motivating and always cumbersome to write down.
In this minicouse, we will see precisely, via the homotopical viewpoint, how a stack (resp. stack in groupoids) is nothing but a 2-sheaf (resp. (2,1)-sheaf), and how this viewpoint connects to Grothendieck’s algebraic definitions. We will discuss basic examples of stacks, such as stacks of sheaves, of geometric/algebraic spaces, and of quasi-coherent modules. We will also briefly demystify some of the key ideas like “stackification”, quotient stacks, classifying stacks, and algebraic (Artin, or Deligne–Mumford) stacks. Finally, in the last day of the minicourse, we will outline the construction of (∞,1)-stacks based on simiplicial (pre)sheaves, following a part of the lectures of Toën–Vezzosi on homotopical algebraic geometry.
Explicit class field theory for global function fields Angus Chung June 17–June 21 (M–F) EH 3088 2–3
Abstract. Class field theory is the study of abelian Galois extensions. A great example would be cyclotomic extensions of Q. In fact, the Kronecker-Weber theorem states that all finite abelian extensions of Q are contained in some cyclotomic extensions. So we can construct any abelian extensions of Q very explicitly. Another fantastic point of the theory is that we can obtain explicitly abelian extensions where only a certain set of primes ramify. Apart from Q, we only know how to construct all abelian extensions explicitly for imaginary quadratic field. This is by using theory of complex multiplication. The story is a mystery for other number fields.
As a lot have said, things are easier in function fields. Indeed, we have the explicit class field theory for not just the rational function field F_q(t), but in fact all global function field (that is, finite extensions of F_q(t)). This is developed by Hayes in 1980s, using the theory of Drinfeld module. In this course, our goal is to understand the construction. We will begin with introducing the theory of Drinfeld modules, and move towards the construction of 'cyclotomic fields' afterwards. We will be following David Goss' book Basic Structure of Function Field Arithmetic, in particular Chapter 3, 4 and 7.
Prerequisites: Galois theory, basic understanding of ramification theory of primes. No prerequisite is required for function field or algebraic geometry.
Algebraic K-theory Shubhankar Sahai June 24–June 28 (M–F) EH 3088 12:30–2
Abstract. Algebraic K-Theory has its roots in Grothendieck's proof of the celebrated Grothendieck-Riemann-Roch (GRR), a generalisation of earlier analytical results of Hirzebruch. In his proof, Grothendieck, among other things , developed the algebraic K^0-group of coherent sheaves on a scheme. Following his success, Atiyah and Hirzebruch developed topological K-theory, a generalised Eilenberg Steenrod cohomology theory, by applying the K^0-functor to topological vector bundles. Using results of Bott on the periodicity of certain homotopy groups, they were able to extend the topological K^0-functor to a sequence of functors K^i's which satisfied the Eilenberg Steenrod axioms. Coupled with the Atiyah-Hirzebruch spectral sequence (AHSS), this theory is extremely powerful and has far reaching applications to Index Theory, Stable Homotopy theory, etc.
However, despite attempts by Bass, Milnor etc, defining the higher K- group of Schemes remained a difficult problem until the work of Quillen in the late 60's and early 70's when he gave two constructions for higher K-groups - the 'Plus' construction for rings and the 'Q' construction for exact categories both of which are equivalent when appropriate. These constructions used hitherto unknown and novel techniques from homotopy theory such as building a homotopy theory for categories, and presented the K groups as the homotopy groups of a certain space.
Since then Algebraic K-Theory has absolved itself of its difficulty by exhibiting deep relationships with fields such as l-adic cohomology, Motivic Cohomology, Intersection Theory, Stable Homotopy Theory and so on. On the other hand computing the K-group of integers remains an open problem.
In this course we start by reviewing the GRR, topological K-theory and the AHSS as way of motivation. Then we move on to discuss the + construction and the Q construction. We then apply some of the main technical theorems of Quillen Q-construction to establish the relationship of K-Groups with the Chow Ring, and compute K-groups of various projective bundles on smooth schemes etc. Time remaining we may discuss Waldhausen K-Theory.
While the prerequisites are basic algebraic topology (singular cohomology, higher homotopy groups, basics of spectral sequences), basic algebraic geometry (passing familiarity with schemes and sheaf cohomology) and basic category theory (representable functors, adjunctions etc); we will try to be as self contained as possible and all are welcome to attend!
Elasticity and geometry Ian Tobasco June 24–June 28 (M–F) EH 3088 2–3
Abstract. This one-week mini-course is an introduction to elasticity theory --- the study of deformable bodies --- and geometry with a particular emphasis on the recent geometric rigidity theorem of Friesecke, James, and Muller. After introducing the basic concepts of strains and displacements and reviewing John’s counterexample to Linfty-Linfty rigidity, we prove L2-L2 rigidity following FJM. Time permitting, we explain the use of this rigidity theorem to derive Kirchhoff’s plate theory as a Gamma-limit and other recent developments along these lines. The course is aimed at graduate students having some basic familiarity with Sobolev spaces, though we will spend some time reviewing the basics in the first lecture. No knowledge of elasticity theory will be assumed.
The Atiyah–Singer index theorem Shubhankar Sahai July 1–July 5 (M–W, F) EH 3088 1–2:30
Abstract. The Atiyah–Singer Index Theorem is an immensely powerful theorem relating the analytical index (the difference of the dimension of the kernel and cokernel of the operator) to the topological index (the integral of the chern character and the Todd Class appropriately defined). In this course we will present the K-Theoretic proof of the same by roughly following Gregory Landweber's article 'K-Theory and Elliptic Operators', which itself is a distilled account of the original papers of Atiyah and Singer - 'The Index of Elliptic Operators I and III'. We would like to note that the proof will be presented from a topological perspective and analytical facts will be blackboxed. However we will introduce the requisite K-Theory and therefore the prerequisites for the course are just some familiarity with Algebraic Topology and Topological Vector bundles. Time remaining we may prove equivariant versions of the same.
Bruhat–Tits buildings (for GL(n) and SL(n)) Yiwang Chen July 8–July 12 (M–F) EH 4096 1–2
Abstract. We will discuss some topics about Bruhat-Tits building that are normally used in the study of p-adic groups and their representation theory. Rough schedule that I am now having in mind as follows: Monday, a brief discussion of p-adic numbers and some properties, then define the p-adic group we concerned throughout the minicourse and examine its structure. Tuesday, define parahoric subgroups, and how they reduce many problems in representation theory to the case of a finite group. Thursday, define the Bruhat–Tits building. Friday, construct an important class depth zero supercuspidal representations.
Prerequisites. Some familiarity on root systems and Weyl groups would be helpful.
The Fargues–Fontaine curve Shubhodip Mondal July 22–July 26 (M–F) EH 4096 4–5
Abstract. In this minicourse we will define the Fargues–Fontaine curve. We will discuss geometric structures such as vector bundles on this curve. We will discuss geometric interpretations of comparison theorems in p-adic cohomology theories in terms of the curve. Also, we will mention geometric interpretations of problems in p-adic galois representation theory in terms of this curve. Prerequisites on perfectoid rings will be discussed briefly when needed.
A K-theoretic approach to the representation theory of finite groups Attilio Castano July 29–August 2 (M–F) EH 4096 3–4
Abstract. Abstract: A celebrated theorem of Atiyah and Segal provides a comparison map from the ring of complex representations Rep(G) of a finite group, to the 0th complex K-theory KU^0 (BG) of a certain space BG. This comparison map is not an isomorphism, but rather it presents KU^0(BG) as the completion of Rep(G) with respect to a certain augmentation ideal. In this mini course, we will investigate a way in which one can “decomplete” KU^0(BG) in order for it to have all the information about the representation theory of G.
Our approach will be algebro-geometric, and we hope to provide a geometric explanation for the failure of this comparison map to be an isomorphism. This explanation will ultimately rely in the geometry of p-divisible groups, we will see that there is a natural p-divisible group associated to Rep(G), and that KU^0(G) only sees a connected component of this p-divisible group.
Time permitting, we will see what kind of advantages this tempered version of KU has, as it comes equipped with a great deal of favorable categorical properties, and discuss potential applications to the representation theory of the infinite symmetric group.
Abel's theorem on complex abelian integrals Jason Liang August 12–August 16 (M–F) EH 3088 9–10
Abstract.
Translation surfaces Mark Greenfield and
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### Problem Statement
You are given an infinite sequence P, where the i-th member is defined as follows:
``` P[i] = (A[i % A.size] ^ (B[i % B.size] + i / B.size)) % 1000000007
```
where % denotes modulo, ^ denotes exponentiation, / denotes integer division, and X.size represents the number of elements in X.
Alternatively, the following pseudocode can be used to calculate the elements of P successively:
```int i = 0;
loop forever
P[i] = (A[i % A.size] ^ B[i % B.size]) % 1000000007;
B[i % B.size] = B[i % B.size] + 1;
i = i + 1;
end of loop
```
You are given int[]s A and B, and a String N containing an integer. Calculate the sum of the first N elements of the sequence (i.e., P[0] + P[1] + ... + P[N - 1]), and return that sum modulo 1000000007.
### Definition
Class: StrangeArray Method: calculateSum Parameters: int[], int[], String Returns: int Method signature: int calculateSum(int[] A, int[] B, String N) (be sure your method is public)
### Constraints
-A and B will each contain between 1 and 50 elements, inclusive.
-Each element of A and B will be between 1 and 10^9, inclusive.
-N will represent an integer between 1 and 10^18, inclusive, with no leading zeros.
### Examples
0)
`{1,2,3}` `{3,4}` `"2"`
`Returns: 17`
1^3 + 2^4 = 17.
1)
`{2, 3, 4}` `{2, 3}` `"3"`
`Returns: 95`
2^2 + 3^3 + 4^(2 + 1) = 95.
2)
`{2, 3, 4}` `{2, 3}` `"5"`
`Returns: 192`
2^2 + 3^3 + 4^(2 + 1) + 2^(3 + 1) + 3^(2 + 2) = 192.
3)
`{1, 2, 3, 4}` `{1}` `"1000000000"`
`Returns: 607570807`
1^1 + 2^2 + 3^3 + 4^4 + 1^5 + 2^6 + ... + 4^1000000000 = 607570807 (mod 1000000007).
#### Problem url:
http://www.topcoder.com/stat?c=problem_statement&pm=8086
#### Problem stats url:
http://www.topcoder.com/tc?module=ProblemDetail&rd=11121&pm=8086
Relja
#### Testers:
PabloGilberto , Olexiy , marek.cygan , ivan_metelsky
Math, Recursion
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http://www.hometrainingtools.com/electricity-battery-science-experiments/a/1233/
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1-800-860-6272 | Customer Service | My Account | Cart (0 item, \$0.00)
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Electricity Experiments
Electron Attraction
One of the laws of electromagnetism is that like (similar) charges repel and opposite charges attract. The more charge something has, the more it attracts or repels. Also, the closer two charges are to each other, the greater the attraction or repulsion between them.
You can demonstrate how charges attract and repel with the following experiment. Hang two inflated balloons from a door frame or ceiling so that they are just touching. Take a sweater or wool sock and rub the sides of both balloons to negatively charge them - the balloons will pick up extra electrons from the sweater. What happens? The balloons will repel each other - the two negative charges push each other away. What do you think will happen if you stick your hand, which is positively charged, in between the two balloons? Because the balloons have a negative charge, they will be attracted and move toward the positive charge of your hand.
The Law of Change Conservation states that the net amount of electric charge is constant, so electrons can only move from one place to another - they don't disappear. This means that if one thing gains electrons and becomes negatively charged, something else has lost electrons and become positively charged. In your experiment, the balloons picked up the extra electrons and became negatively charged, while the wool lost electrons.
Demonstrate Static Electricity
Static electricity makes your hair stand up during a pillow fight or shocks your fingers when you touch a cold door handle. Static (unmoving) electricity occurs when insulating materials (ones that electric current can't flow through, such as plastic) get negatively or positively charged. Since the current caused by this force can't flow through the insulator, the static electricity just sort of sticks to the surface and builds up until it can exchange photons with an opposite charge.
Since lightning is an example of static electricity, it's not surprising that two results of static electricity are sparks and a crackling sound. You can observe this in a dark bathroom with a rubber or glass friction rod (a comb or a glass rod could also work) and a friction pad or piece of fur or wool.
1. Rub the friction rod against the pad for 15-20 seconds and then touch the rod to the sink faucet. You should hear a crackling sound and maybe even see a small blue spark as extra electrons flow from the rod to the positively-charged metal faucet.
2. Can you think of other ways to get a static electric spark or sound? Are there some materials that work better than others (e.g., a cotton sweater vs. wool, a plastic spatula vs. rubber comb, etc.)? Do more experiments to test your hypotheses!
You can make a simple battery with some pennies and circles of aluminum foil and wet paper towels (soaked in a salt water solution - try one teaspoon salt to 6 oz. tap water) that are the same size as the pennies. You will also need insulated wire. (Adult supervision recommended.)
1. Make your battery by stacking 12 "cells" against each other, each cell made up of one penny, one wet paper towel circle, and one foil circle in that order.
2. Wrap a 8-10" piece of insulated wire with stripped ends around the battery once and twist the ends together against the battery so that the wire holds the cells together.
3. Next, touch the bare ends of the wire to each end of the battery. If you're in a dark room, you might see a spark as your battery produces an electric current. Another way to test the battery is with a voltage meter or multimeter.
Note that U.S. pennies made before 1982 are 95% copper, but newer pennies only have a 2.5% copper coating. For further experimentation, compare the electric current when you make a battery using only older pennies and one using only newer pennies. You could also experiment with a stronger salt water solution or plain tap water.
Experiment with Electromagnets
Electromagnets are temporary magnets that only work as a magnet when electrical current is flowing through them. With some insulated wire and a large nail, you can make your own electromagnet. Loosely wrap a long piece of insulated wire around the nail. Connect the ends to a battery and see what happens.
Does your electromagnet attract any metal objects, such as paperclips? What happens if you set a magnetic compass close to the electromagnet? Experiment with coiling the wire more times around the nail; does that strengthen the magnetic field so that the electromagnet attracts heavier objects? If you have iron filings, you can test the magnetic field produced by the electromagnet. Sprinkle the iron filings on a paper plate, then hold the electromagnet directly underneath. The iron filings will arrange themselves in the pattern of the magnetic field. Electromagnets are used in electric bells, magnetic levitation (maglev) trains, and much more.
Interested in more projects like this? Sign up for our e-mail newsletters to get new science experiments, cool facts, and more each month!
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Enviro-Battery Kit \$12.95
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# Terminologies - lags, order in time series model
I am facing some difficulties with the terminologies - lag ($p$) and sequence length (number of data points) (N) used in time series model such as Moving average and Autoregressive model. Considering a Moving Average model described by the equation
$$y(n) = \sum_{i=1}^{p} h(i) x(i-n)$$ If the number of bits in the input is N=16 defined by $\mathbf{x} = \{1,0,1,1,0,1,0,1,1,0,1,1,0,0,1,1\}$ and the number of lags are p=10 then
Q1: should the model generate a time series of length 'N=16 i.e, would the output of the above model $\mathbf{y} = [y_1,y_2,\ldots,y_N]$ contain 16 elements where $n = 1,2,\ldots,16$?
I am writing out the full expression. I can write the MA process in vector form as $y_n = \mathbf{h}^T \mathbf{x}_n$ where $\mathbf{x}_n = [x_n,x_{n-1},\ldots,x_{n-p+1}]^T$ is the input signal vector and $\mathbf{h} = [h_0,h_1,\ldots,h_{p-1}]^T$ is the MA system coefficients. For example, in programming using Matlab the model would look like for
h = [1,0.2,0.3]; %channel coefficients
N=16;%number of data points
x = rand(1,N); %input data vector
p=2;
y(1) =0.0;
y(2) = 0.0;
for n = 3:N
y(n) = h(1)*x(n) +h(2)*x(n-1)+h(3)*x(n-2);
end
Since the order p=2 would the physical interpretation be that the channel can take in only 2 data values as input? Also, does p=2 mean that the channel has 2 paths through which the data input travels?
Q2: Is there a relationship between the number of input elements and the number of lags?
Please correct me if I have misinterpreted the concepts. Thank you.
• $$p \le N {}{}$$ Commented May 11, 2017 at 2:38
• For a filtering operation, you want $x(n-i)$ to describe a discrete convolution. What you have now is a discrete cross-correlation. Either way, if you assume the sequences to be zero padded the complete convolution results in $N+p-1$ output terms. Commented May 11, 2017 at 12:35
Q1: should the model generate a time series of length 'N=16 i.e, would the output of the above model $\mathbf{y} = [y_1,y_2,\ldots,y_N]$ contain 16 elements where $n = 1,2,\ldots,16$?
If one thinks about your question in terms of FIR filtering, then an input signal, $x$ of length $N$ filtered with an FIR filter of length $p$ ($p$ coefficients) will have an output of length $p + N - 1$.
Note: there are ways to define the output so that only $N$ outputs are obtained, but you need to understand that is what you are doing.
Since the order p=2 would the physical interpretation be that the channel can take in only 2 data values as input?
It means the channel does not change the input much at all. It's a two coefficient FIR filter.
Also does p=2 mean that the channel has 2 paths through which the data input travels?
One way to interpret the FIR (MA) nature of the filter you are using is that it involves $p$ paths with a time delay of $nT$ (where $T$ is the sample period).
Q2: Is there a relationship with the number of input elements and the number of lags?
In general, no, there will not be a relationship.
(1) Is this related to "multipath" often used in Rayleigh fading? If there are $p$ paths then would the transmitted signal travel in two directions?
The usual method of modeling multi-path is to say that the receiver receives: $$y^c(t) = \sum_{i=1}^{p} h^c_i x^c(t - d_i)$$ where this is the continuous-time version so the $d_i$ are not integers and $d_1 = 0$.
Moving to the discrete-time version means you need to sample everything using $t = nT$. So if you stick with $p$ components, you can only really have $d_i = (i-1)T$.
(2) What is the general way to set $nT$? Using Matlab, I did not do any thing in my code. Can you say what is $nT$ for my example?
$T$ is determined by your modeling. How did you relate the actual continuous-time problem to the discrete case?
• @SrishtiM See response in my edit.
– Peter K.
Commented May 11, 2017 at 23:17
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http://oeis.org/A173586
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This site is supported by donations to The OEIS Foundation.
Annual Appeal: Please make a donation to keep the OEIS running. In 2018 we replaced the server with a faster one, added 20000 new sequences, and reached 7000 citations (often saying "discovered thanks to the OEIS"). Other ways to donate
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A173586 Decimal values a(n) of the binary numbers b(n) obtained by starting from first prime number (2), sequentially concatenating the binary representation of all prime numbers till n-th prime, and after that, sequentially concatenating the binary representation of all prime numbers, from (n-1)th till the first prime. 0
2, 46, 1502, 96222, 12316638, 3153031134, 1614350348254, 1653094690025438, 1692768964130074590, 1733395419356639752158, 1774996909423485572837342, 3635193670499109531489365982 (list; graph; refs; listen; history; text; internal format)
OFFSET 1,1 LINKS FORMULA a(n) = binary_to_decimal(concatenate(10, 11, 101, ..., binary((n-2)th prime), binary((n-1)th prime), binary(n-th prime), binary((n-1)th prime), binary((n-2)th prime), ..., 101, 11, 10)) EXAMPLE a(1)=binary_to_decimal(10)=2, a(2)=binary_to_decimal(101110)=46, a(3)=binary_to_decimal(10111011110)=1502, a(4)=binary_to_decimal(10111011111011110)=96222 etc. CROSSREFS Cf. A066622. This sequence uses the term generation rule of A066622 (Concatenation of prime numbers in increasing order up to the n-th and then in decreasing order.), albeit with the binary base instead of the decimal base. Sequence in context: A012006 A279524 A264500 * A074041 A277554 A000191 Adjacent sequences: A173583 A173584 A173585 * A173587 A173588 A173589 KEYWORD base,nonn AUTHOR Umut Uludag, Feb 22 2010 STATUS approved
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Last modified December 18 14:22 EST 2018. Contains 318229 sequences. (Running on oeis4.)
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https://www.planetanalog.com/signal-chain-basics-45-is-high-speed-adc-clock-jitter-being-over-specified-for-communication-systems/
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# Signal Chain Basics #45: Is high-speed ADC clock jitter being over-specified for communication systems?
(Editor's note : go to https://m.eet.com/images/common/planetanalog/2010/02/SCBlist.pdf for a complete, linked list of all previous installments of the Signal Chain Basics series.)
There is a well-known relationship in analog-to-digital converters (ADCs) between the sample clock jitter and the resulting ADC signal-to-noise ratio (SNR) degradation (derived in reference 1), Equation 1 :
SNRjitter (dBc) = 20 * log10 (2 * p * fIN * tj )
where tj is the RMS jitter (typically in picoseconds or femtoseconds), fIN is the analog input frequency, and SNR jitter is the ADC SNR, if the only noise source is clock jitter. The total ADC SNR includes other noise sources such as thermal noise.
Here are some interesting points regarding Equation 1. First, there is no direct dependence on sample frequency. However, the integration of phase-noise to calculate jitter depends on the sample frequency. Also, since the RMS jitter value is the integrated phase-noise across frequency, the phase-noise dependence frequency dependence is lost in the analysis.
Using the RMS jitter effectively averages the phase-noise across the entire ADC output bandwidth, regardless of the actual phase-noise spectrum. Since the clock phase-noise typically decreases with increasing offset frequency, the noise due to clock jitter is highest near the large signal frequency.
This is the case when a band-pass filter is used on the clock signal, as described in Reference 2 , where a crystal filter removes the clock phase-noise above ~100 kHz. This is illustrated in Figure 1 , when the phase-noise is integrated in the wanted signal bandwidth, the noise estimated using jitter results in a higher estimate than integrating the in-band phase-noise.
Figure 1: ADC spectrum for a large blocker and small wanted signal.
(Click on image to enlarge)
How do you translate clock phase-noise to the ADC output noise? To demonstrate the relationship, we create a known level of phase-noise and measure the ADC output spectra. A 250-MHz clock with noise is generated using a high-speed DAC, such as the DAC5681 16-bit/1Gsps, and input as a clock for the ADC, using the ADS4149 14-bit/250Msps. The DAC pattern and capture size are set so the bin resolution in the DAC and ADC FFT’s are equal in size.
The DAC output pattern in Figure 2 consists of a 250-MHz tone and –60 dBc of random noise from 240 to 250-MHz. In a typical clock, the phase-noise is symmetric around the clock, but for clarity we use a single-sided noise.
Figure 2: 250-MHz clock with –60dB noise.
(Click on image to enlarge)
The ADC output using the DAC generated clock for input frequencies of 10 and 100MHz is shown in Figure 3 . The clock phase-noise energy is mixed in the sample process with the input tone and is symmetrical around the carrier. For the 100-MHz input tone, the noise due to the clock phase-noise is ~71dB across ±10 MHz from the tone. For the 10-MHz input tone, the noise due to the clock phase-noise is ~91dB (per FFT bin). This is consistent with the SNR jitter equation, which predicts a 20-dB change with 10 times the input frequency.
(Click on image to enlarge)
The ADC noise from the clock phase-noise can be described by Equation 2 :
ADCNoise (fOFFSET ) = – Phase-noise(fOFFSET ) – 20 * log10 (fIN /fCLK
where fOFFSET is the offset frequency, phase-noise is the one-sided phase-noise density, fIN is the input frequency, and fCLK is the clock frequency. Note that the units of phase-noise and ADCNoise are the same, i.e., dBc/Hz.
Returning to our example in Figure 3, the ADC noise at 100MHz of –71dB is 11dB lower than clock phase-noise of 60dB, –8dB is from the fIN /fCLK term in Equation 2, and 3dB is due to the clock phase-noise being on one side only, rather than symmetrical.
When used for specifying the required phase-noise for the ADC clock in communication systems, the ADCNoise should be integrated across the bandwidth of the wanted signal at the blocker offset to calculate the total that falls in the wanted signal (Figure 1).
The ADC clock phase-noise spectrum translates directly to noise in the ADC output with the same offset spectrum. Therefore, using jitter to calculate SNR is a simplification that often results in over specification of ADC clock phase-noise requirements.
Conclusion
Join us next month when we will discuss clock jitter specifications in high speed serial data links.
References
1. y Eduardo Bartolome, Vineet Mishra, Goutam Dutta, and David Smith, Texas Instruments, 1Q 2005.
2. Russell Hoppenstein and Firoj Kabir, Texas Instruments, December 2004.
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https://earlymath.erikson.edu/get-on-the-right-path-make-your-own-number-sense-activities/
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# Get on the Right Path: Make Your Own Number Sense Activities
Get on the Right Path: Make Your Own Number Sense Activities
Path games are fantastic ways for families to spend time together and have fun while doing math. Path games develop number sense, counting skills and, depending on children’s ages and the tools you use, computational fluency. These opportunities come about as players take turns rolling dice, counting spaces to move, and figuring out what they need to roll to win the game.
As we talk with children while playing a path game using a number path, we can highlight their ability to see small quantities without counting (subitizing) and build number relationships, such as: spatial sense, one/two more than and one/two less than a given number. Asking questions such as, “What do you hope to roll to win the game?” or “How many spaces do you need to get to the finish line?” also helps develop their mathematical skills. Kindergartners and primary grade children can develop addition and subtraction fluency by playing with two or more dice and finding the sums or differences to determine the amount of spaces to move. Children will want to play these games often because they have fun doing it, while games and number sense activities can provide a chance to connect with family and friends.
Best of all, there is no need to run out and buy these games! There are simple and creative ways to make path games and other number sense activities at home using materials from around the house. Making a path game at home allows you to tailor the game to the experience and age of your children. Also, it allows for the children to participate in making the game. The game-creation process adds to the math learning by incorporating measurement and geometry skills. Some, but not all of the elements that need to be considered are: How many spaces will the game be? Will you make your own dice? Will the game have a theme? Will you use different shapes to form the path? You can make as many different games as you like. Enjoy the journey.
#### Materials needed:
• Paper grocery bag
• Scissors
• Markers/Pen
• Dice (game dice or homemade) or other number generators such as a spinner or playing cards
• Playing pieces
#### Directions:
1. Cut paper bag
2. The bag should open up fully (see photo below) after you cut it
3. Cut off the extra piece from the bottom, so that the bag looks like the photo below
4. Draw out a path game board on the paper bag
Note: The path of the game can be modified to fit the needs of the child, varying the number sense activities like this require. You can make a short path game (5-10 spaces) or a longer path game (10-20 spaces) or even longer if you like.
#### Playing the Game
• Each player needs their own playing piece.
• Options, pieces from other games, different coins or any other ideas you have.
• You can use a dice to determine the number of spaces to move,
• One die for beginning players
• Two or more dice for more advanced players or boards with longer paths
• Cover up the faces of the dice with tape and write smaller numerals or quantities on the tape
• If you do not have dice, you can make a spinner from a piece of cardboard. Cut a circle. Divide the circle into 4 or 6 equal pieces and label with dots (and/or numerals). Use a pencil and paper clip to make the pointer of the spinner (see top photo)
• You can also use a deck of cards, to pull one or more cards to represent how many spaces the child will move on each turn.
• Pull out only some of the cards to use. For example, you can pull out the cards with quantities 1-3 or 1-5 and only use those as part of the game.
• If using playing cards, you can have children add the amounts on the cards to compare them to the number of spaces traveled on the path. They should be the same.
#### Modifications
• Develop the number relationship of +1/+2 or –1/–2 by covering the faces of one of the dice with 3 green sides and 3 red sides for adding or subtracting 1 or 2
• If a player rolls an even amount on the dice, they move one less space but if they roll an odd amount on the dice, they move one more space than on the dice.
• Create a path with spaces that include spaces marked +1/+2 or –1/–2.
• Make your own dice or a spinner to change the dot designs or number of dots on the faces.
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http://tech.bragboy.com/2010/02/coding-problem.html?showComment=1266080276532
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## 06 February 2010
### Progress software - interview coding problem
This is one of the problems that have been asked in a company called Progress Software, Hyderabad. I have posted the question as well as the working Java code for the same.
Question
Sports Associations in India
-------
The Sports Associations in India (SAI) wants to choose 2 teams of 4 people each
to send to Asain games. There are 13 people who want to be members of
the teams. The SAI tries grouping them in various ways to see which
athletes perform well together. Each grouping gets one test run on the
test track and their time is recorded. Your task is to help the SAI
choose two disjoint teams of 4 such that the sum of their practice times
is minimal.
Input
-----
There will be several input instances. The first line of each instance
gives the total number of practice runs. This will be followed by n lines.
Each of those lines will contain 5 numbers: p1 p2 p3 p4 t
t is the time taken (in milliseconds) by the team consisting of p1, p2, p3 and p4.
The time taken will not be more than 2 minutes.
The end of the input will be indicated by a line with n=0.
Output
------
Output the best total time for the two teams that you choose.
If it is impossible to choose two disjoint teams from the test runs given, output -1.
Sample Input
------------
6
1 2 3 4 30000
10 11 12 13 15000
5 6 7 8 37800
1 5 10 12 20000
5 6 9 11 43000
1 9 12 13 11000
3
1 4 7 9 10000
3 5 7 11 17890
6 7 12 13 20000
0
Sample Output
-------------
45000
-1
Along this you will get a file consisting of an input and they ask you to send the output. I am sharing the solution to this problem.
```package main;
import java.io.BufferedWriter;
import java.io.FileWriter;
import java.io.IOException;
import java.util.ArrayList;
public class Main {
/**
* @param args
*/
public static void main(String[] args) {
new Main().start();
}
private ArrayList<Event> events = null;
public static final String DIR = "D:/Question2/";
private String filename = DIR + "input.txt";
public void start() {
readInput();
printInput();
solve();
writeToFile();
}
private void printInput() {
Event eventRound = null;
for (int i = 0; i < events.size(); i++) {
eventRound = events.get(i);
System.out.println("Round Count: " + eventRound.roundCount);
System.out.println("Rounds : " + eventRound.toString());
}
}
public void readInput() {
FileRead read = new FileRead(filename);
events = read.getEvents();
}
public void solve() {
Event eventRound = null;
for (int i = 0; i < events.size(); i++) {
eventRound = events.get(i);
eventRound.findMinTime();
eventRound.printVal();
}
}
private void writeToFile() {
try {
BufferedWriter out = new BufferedWriter(new FileWriter(DIR
+ "output.txt"));
Event eventRound = null;
for (int i = 0; i < events.size(); i++) {
eventRound = events.get(i);
out.write(eventRound.minTime+System.getProperty("line.separator"));
}
out.close();
}catch(IOException e) {
e.printStackTrace();
}
}
}
```
```package main;
import java.util.StringTokenizer;
/**
* Class to reprsent a round. A round consist of an array of players and time taken
* by them on that particular round.
* @author Bragadeesh
*
*/
public class Round {
public int[] players;
public int time;
/**
* Represents the player count for the event
*/
public static final int PLAYER_COUNT = 4;
public Round(String line){
StringTokenizer tokens = new StringTokenizer(line);
players = new int[PLAYER_COUNT];
for(int i=0;i<PLAYER_COUNT;i++){
players[i] = Integer.parseInt(tokens.nextToken());
}
time = Integer.parseInt(tokens.nextToken());
}
/**
* Returns the formatized string for a particular Round
*/
public String toString(){
String op = "";
for(int i=0;i<players.length;i++){
op = op + players[i] + " ";
}
op = op + time;
return "[" + op + "]";
}
}
```
```package main;
/**
* Event is a class which is comprises an array of Rounds.
* @author Bragadeesh
*
*/
public class Event {
public Round[] rounds;
public int roundCount;
private static final int MAX_VAL = 99999999;
public int minTime = -1;
/**
* Variable used to show the the minimum round1 should there be any
*/
public Round minRound1;
/**
* Variable used to show the the minimum round2 should there be any
*/
public Round minRound2;
/**
* Returns the formatized string for a particular Event
*/
public String toString(){
String op = "";
for(int i=0;i<rounds.length;i++){
op+=rounds[i].toString();
}
return "{" + op + "}";
}
/**
* Method to calculate the minimum time
* @return
*/
public int findMinTime(){
int min = MAX_VAL;
Round roundA = null;
Round roundB = null;
for(int i=0;i<roundCount;i++){
for(int j=0;j<roundCount&&i!=j;j++){
roundA = rounds[i];
roundB = rounds[j];
if(!disjoint(roundA, roundB)){
if(roundA.time + roundB.time < min){
minRound1 = roundA;
minRound2 = roundB;
}
min = Math.min(roundA.time + roundB.time, min);
}
}
}
if(min != MAX_VAL){
minTime = min;
}
return minTime;
}
/**
* Method to print the Round values and the minimum time for a
* given event. Prints -1 if there is no minimum time set.
*/
public void printVal(){
if(minRound1!=null){
System.out.println(minRound1.toString());
System.out.println(minRound2.toString());
System.out.println(minTime);
}else{
System.out.println("-1");
}
}
/**
* Simple bubble sort that does the sorting of the events based on the round time
* @return void
*/
public void sortRounds() {
for(int i=0;i<roundCount;i++){
for(int j=i;j<roundCount;j++){
if(rounds[i].time > rounds[j].time){
Round temp = rounds[i];
rounds[i] = rounds[j];
rounds[j] = temp;
}
}
}
}
/**
* Given two arbitrary rounds A and B, returns whether they are disjoint or not
* @param roundA
* @param roundB
* @return boolean value whether given set is disjoint or not
*/
public boolean disjoint(Round roundA, Round roundB){
for(int i=0;i<Round.PLAYER_COUNT;i++){
for(int j=0;j<Round.PLAYER_COUNT;j++){
if(roundA.players[i] == roundB.players[j]){
return true;
}
}
}
return false;
}
}
```
```package main;
import java.io.BufferedReader;
import java.io.DataInputStream;
import java.io.FileInputStream;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.ArrayList;
public class FileRead {
private String filename = null;
private ArrayList<Event> events;
public FileRead(String filename){
this.filename = filename;
events = new ArrayList<Event>();
load();
}
private void load(){
FileInputStream fstream = null;
DataInputStream in = null;
BufferedReader br = null;
Event event = null;
try {
fstream = new FileInputStream(filename);
in = new DataInputStream(fstream);
br = new BufferedReader(new InputStreamReader(in));
for(;;){
event = new Event();
event.roundCount = Integer.parseInt(br.readLine());
if(event.roundCount == 0){
break;
}
event.rounds = new Round[event.roundCount];
for(int i=0;i<event.roundCount;i++){
event.rounds[i] = new Round(br.readLine());
}
event.sortRounds();
events.add(event);
}
} catch (Exception e) {
System.err.println("Error: " + e.getMessage());
} finally{
try {
in.close();
} catch (IOException e) {
e.printStackTrace();
}
}
}
public ArrayList<Event> getEvents(){
return events;
}
}
```
Input file : Input.txt
Output file : Output.txt
Cheers,
Bragaadeesh.
#### 4 comments:
JP Arun said...
adengappa! veriya aazha varuvae Braga! :)
ahmed said...
thanks for sharing the question.
is it necessary to implement it in java or can it be implemented in any programming language.
please post any other problems if u know for progress software
BragBoy said...
@ahmed - it is not necessary to do in Java. You can do it any language of your choice. All that matters is the end solution.
And I don't have any other problems from them.
Venky said...
import java.io.BufferedReader;
import java.io.FileInputStream;
import java.io.FileNotFoundException;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.StringTokenizer;
public class GetTeams {
void readInput()
{
try {
FileInputStream fis = new FileInputStream("input.txt");
BufferedReader breader= new BufferedReader(new InputStreamReader(fis));
String line1=breader.readLine();
while(line1!=null && !line1.trim().equals("0"))
{
line1=initialize(breader, line1);
line1=breader.readLine();
}
} catch (FileNotFoundException e) {
// TODO Auto-generated catch block
e.printStackTrace();
} catch (IOException e) {
// TODO Auto-generated catch block
e.printStackTrace();
}
}
/**
* @param breader
* @param line1
* @throws NumberFormatException
* @throws IOException
*/
public String initialize(BufferedReader breader, String line1)
throws NumberFormatException, IOException {
Integer initCount= Integer.parseInt(line1.trim());
int a [][]= new int [initCount][14];
int b[][]=new int [initCount][5];
int i=0;
while(i sum)
{
for(int x=0;x<13;x++)
{
if(a[m][x]+a[n][x]==2)
{
flag=false;
}
}
if(flag)
{
minTime=sum;
minTimeI=m;
minTImeJ=n;
}
}
}
}
System.out.println( minTime);
return line1;
}
public static void main(String[] args) {
GetTeams a = new GetTeams();
a.readInput();
}
}
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CC-MAIN-2024-26
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latest
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en
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|
https://math.stackexchange.com/questions/54915/is-there-general-formula-for-the-exponential-of-a-tridiagonal-matrix
| 1,653,682,520,000,000,000
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text/html
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| 450,288,642
| 66,344
|
# Is there general formula for the exponential of a tridiagonal matrix?
For an arbitrary tridiagonal matrix of the form
$$A = \begin{pmatrix} b_1 & c_1 & 0 & 0 & ... \\ a_2 & b_2 & c_2 & 0 & ... \\ 0 & a_3 & b_3 & c_3 & ... \\ \vdots &&\ddots&\ddots&\ddots\end{pmatrix}$$
is there a formula to calculate $\exp(A)$? Or at least for some special tridiagonal matrices?
The special case I am most interested in is a $(2n+1)^2$ matrix with $b_k = i(k-n-1)$ and $c_k = (a_{2n+2-k})^*$, i.e.
$$\begin{pmatrix} -in & c_1 & 0 & \\ c_{2n}^* & -i(n-1) & c_2 & \\ 0 & c_{2n-1}^* & -i(n-2) & \ddots \\ &&\ddots&\ddots \end{pmatrix}$$
• A closed form for that exponential would entail finding a closed form for the characteristic polynomial of the tridiagonal matrix, since the eigenvectors can be expressed in terms of derivatives of the characteristic polynomial evaluated at appropriate values... Aug 10, 2011 at 8:53
• Did you ever find a solution to your problem? Jan 10, 2013 at 16:55
• @JohnSalvatier I'm afraid not :-/ Jan 10, 2013 at 17:00
• I'm looking for a way to compute exp(At)*x_0 cheaply when A's a symmetric tridiagonal matrix. I think I may just have to eigen-decompose A and do it that way. Luckily I only have to decompose A once, and then it's O(n**2), which I guess is okay. Since you should be able to compute Ax_0 in O(n) steps since its tridiagonal, I was hoping for something better, but maybe that's not possible. Jan 10, 2013 at 20:26
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| 3.859375
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CC-MAIN-2022-21
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latest
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en
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https://blackestfest.com/how-do-you-get-30-of-80/
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crawl-data/CC-MAIN-2024-26/segments/1718198861916.26/warc/CC-MAIN-20240620074431-20240620104431-00448.warc.gz
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|
2021-10-25
## How do you get 30% of 80?
Percent means per one-hundred and hence 30% is 30/100. Thus 30% of 80 is (30/100) x 80 = 24.
28
## How do you find 70 percent of a number?
Example 1. Find 70% of 80. Following the shortcut, we write this as 0.7 × 80. Remember that in decimal multiplication, you multiply as if there were no decimal points, and the answer will have as many “decimal digits” to the right of the decimal point as the total number of decimal digits of all of the factors.
21
## How do you find percentages without a calculator?
If you need to find a percentage of a number, here’s what you do – for example, to find 35% of 240: Divide the number by 10 to find 10%. In this case, 10% is 24. Multiply this number by how many tens are in the percentage you’re looking for – in this case, that’s 3, so you work out 30% to be 24 x 3 = 72.
## How do you calculate percentages quickly?
To calculate 10 percent of a number, simply divide it by 10 or move the decimal point one place to the left. For example, 10 percent of 230 is 230 divided by 10, or 23. 5 percent is one half of 10 percent. To calculate 5 percent of a number, simply divide 10 percent of the number by 2.
52
14
32
## What number is 35% of 70?
Latest calculated numbers percentages
35% of 70 = 24.5 Mar UTC (GMT)
8.32% of 100,000 = 8,320 Mar UTC (GMT)
186% of 30.5 = 56.73 Mar UTC (GMT)
1.8% of 100 = 1.8 Mar UTC (GMT)
74% of 2,000 = 1,480 Mar UTC (GMT)
75%
42
30
15
8
7
24
## How do you find 40% 70?
Percentage Calculator: What is 40 percent of 70? = 28.
## How do you find 15% of a number?
15% is 10% + 5% (or 0.15 = 0.1 + 0.05, dividing each percent by 100). Thinking about it this way is useful for two reasons. First, it’s easy to multiply any number by 0.1; just move the decimal point left one digit. For example, 75.00 x 0.1 = 7.50, or 346.43 x 0.1 = 34.64 (close enough).
16
## How do you find 75%?
How much is 15 percent off?
1. Divide your original number by 20 (halve it then divide by 10).
2. Multiply this new number by 3.
3. Subtract the number from step 2 off of your original number.
4. You’ve just found your percentage off!
## What is the formula for calculating percentage?
To calculate the percentage, multiply this fraction by 100 and add a percent sign. 100 * numerator / denominator = percentage . In our example it’s 100 * 2/5 = 100 * 0.4 = 40 . Forty percent of the group are girls.
## How do I figure out percentages?
How to calculate percentage
1. Determine the whole or total amount of what you want to find a percentage for.
2. Divide the number that you wish to determine the percentage for.
3. Multiply the value from step two by 100.
20
50
## How do I calculate percentage on a test?
You can find your test score as a percentage by dividing your score by the total number of points, then multiplying by 100.
17
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## CLOSED** SETI/BOINC Milestones (tm) XVII **CLOSED
Message boards : Number crunching : CLOSED** SETI/BOINC Milestones (tm) XVII **CLOSED
Message board moderation
To post messages, you must log in. "Oldest first Newest first Highest rated posts first
Previous · 1 . . . 10 · 11 · 12 · 13 · 14 · 15 · 16 · Next
AuthorMessage
Sutaru Tsureku
Volunteer tester
Joined: 6 Apr 07
Posts: 7105
Credit: 147,663,825
RAC: 5
Message 906942 - Posted: 12 Jun 2009, 21:55:47 UTC
Congrats to all milestones!
BTW.
Is somewhere written at which steps are the milestones?
But ..also.. for some the 1,000 , 10,000 and 100,000 milestones are different far away..
Or.. how could be the calculation for the different RAC-milestones?
Or.. the milestones are all the time 1/4, 2/4, 3/4 and 4/4 ?
For example:
250 , 500 , 750 , 1,000 .
But then
1,250 , 1,500 , 1,750 , 2,000 ?
Or
2,500 , 5,000 , 7,500 , 10,000
25,000 , 50,000 , 75,000 , 100,000
250,000 , 500,000 , 750,000 , 1,000,000
2,500,000 , 5,000,000 , 7,500,000 , 10,000,000
25,000,000 , 50,000,000 , 75,000,000 , 100,000,000 ?
Ohh.. ;-)
ID: 906942 ·
Siran d'Vel'nahr
Volunteer tester
Joined: 23 May 99
Posts: 7379
Credit: 44,181,323
RAC: 238
Message 906955 - Posted: 12 Jun 2009, 22:10:11 UTC - in response to Message 906942.
Congrats to all milestones!
BTW.
Is somewhere written at which steps are the milestones?
But ..also.. for some the 1,000 , 10,000 and 100,000 milestones are different far away..
Or.. how could be the calculation for the different RAC-milestones?
Or.. the milestones are all the time 1/4, 2/4, 3/4 and 4/4 ?
For example:
250 , 500 , 750 , 1,000 .
But then
1,250 , 1,500 , 1,750 , 2,000 ?
Or
2,500 , 5,000 , 7,500 , 10,000
25,000 , 50,000 , 75,000 , 100,000
250,000 , 500,000 , 750,000 , 1,000,000
2,500,000 , 5,000,000 , 7,500,000 , 10,000,000
25,000,000 , 50,000,000 , 75,000,000 , 100,000,000 ?
Ohh.. ;-)
Greetings Sutaru,
There are no steps for milestones. Basically, just set a goal to reach and when you get there, announce it. Personal milestones is up to each individual. For instance, my last milestone was reaching my 10th year crunching for SETI@Home. My next one will be in July when I reach my 5th year with BOINC. After that, my next will be when I reach 750,000 credits. I have even seen some users declare milestones for the number of posts made on this forum. lol :D
Milestones is what you decide them to be. Have fun!
Keep on BOINCing!
CAPT Siran d'Vel'nahr - L L & P _\\//
Winders 11 OS? "What a piece of junk!" - L. Skywalker
"Logic is the cement of our civilization with which we ascend from chaos using reason as our guide." - T'Plana-hath
ID: 906955 ·
Fred W
Volunteer tester
Joined: 13 Jun 99
Posts: 2524
Credit: 11,954,210
RAC: 0
Message 906964 - Posted: 12 Jun 2009, 22:16:29 UTC
Your personal milestones are whatever you define them to be. Personally, I don't track cobblestones as milestones - machines have just got so much faster over the years that I can now add a million in a few weeks whereas such a number would have been unthinkable when I first started crunching SETI. In about 45 minutes, I hit a milestone that does float my boat - 10 years since I first crunched Classic (though I did take a few years furlough in the middle and missed that actual change-over to Boinc.
F.
ID: 906964 ·
Carlos
Volunteer tester
Joined: 9 Jun 99
Posts: 30182
Credit: 57,275,487
RAC: 157
Message 907108 - Posted: 13 Jun 2009, 2:46:59 UTC
Milestones are personal achivements. As already discussed they are goals you set. Number of years, number of units, number of posts, and of course there's misfit with number of friends.
ID: 907108 ·
kittyman
Volunteer tester
Joined: 9 Jul 00
Posts: 51470
Credit: 1,018,363,574
RAC: 1,004
Message 907126 - Posted: 13 Jun 2009, 4:17:22 UTC - in response to Message 906964.
Your personal milestones are whatever you define them to be. Personally, I don't track cobblestones as milestones - machines have just got so much faster over the years that I can now add a million in a few weeks whereas such a number would have been unthinkable when I first started crunching SETI. In about 45 minutes, I hit a milestone that does float my boat - 10 years since I first crunched Classic (though I did take a few years furlough in the middle and missed that actual change-over to Boinc.
F.
I luv the kibblestones...........
All 31 million of them and counting....
In a few weeks I will have been at it for 9 years.
I have risen to #31 in the WORLD for total Seti credits.......in about 6 days, that will be #30.
I am #4 in the SETI.USA team ranks for Seti....#117 in the world by RAC...
The Frozen Nehi still hangs on as the last non-server, non-cuda i7 in the top 20 computers on Seti...
And the Kitty City (Seti City), which I founded here, has risen to #13 in the USA.....
These are some things that float the kitties' boat....LOL.
"Time is simply the mechanism that keeps everything from happening all at once."
ID: 907126 ·
Fred W
Volunteer tester
Joined: 13 Jun 99
Posts: 2524
Credit: 11,954,210
RAC: 0
Message 907152 - Posted: 13 Jun 2009, 5:51:55 UTC - in response to Message 907126.
I luv the kibblestones...........
All 31 million of them and counting....
In a few weeks I will have been at it for 9 years.
I have risen to #31 in the WORLD for total Seti credits.......in about 6 days, that will be #30.
I am #4 in the SETI.USA team ranks for Seti....#117 in the world by RAC...
The Frozen Nehi still hangs on as the last non-server, non-cuda i7 in the top 20 computers on Seti...
And the Kitty City (Seti City), which I founded here, has risen to #13 in the USA.....
These are some things that float the kitties' boat....LOL.
You forgot your unique milestone:
Record number of time to return from banishment ;-P
Keep up the good work, good-buddy. My quaddie + CUDA was right on the Nehi's tail but I have now switched on my CPDN models again as they have been hanging around for a while and SETI is reluctant to rebuild my cache after a detach / re-attach to clean up the cache following the abortive attempt to upgrade to v6.6.34.
F.
ID: 907152 ·
kittyman
Volunteer tester
Joined: 9 Jul 00
Posts: 51470
Credit: 1,018,363,574
RAC: 1,004
Message 907154 - Posted: 13 Jun 2009, 5:57:16 UTC - in response to Message 907152.
I luv the kibblestones...........
All 31 million of them and counting....
In a few weeks I will have been at it for 9 years.
I have risen to #31 in the WORLD for total Seti credits.......in about 6 days, that will be #30.
I am #4 in the SETI.USA team ranks for Seti....#117 in the world by RAC...
The Frozen Nehi still hangs on as the last non-server, non-cuda i7 in the top 20 computers on Seti...
And the Kitty City (Seti City), which I founded here, has risen to #13 in the USA.....
These are some things that float the kitties' boat....LOL.
You forgot your unique milestone:
Record number of time to return from banishment ;-P
Keep up the good work, good-buddy. My quaddie + CUDA was right on the Nehi's tail but I have now switched on my CPDN models again as they have been hanging around for a while and SETI is reluctant to rebuild my cache after a detach / re-attach to clean up the cache following the abortive attempt to upgrade to v6.6.34.
F.
LOL....that's another stat I am working on....my MTBB (mean time between banishments)...longer is better...for everybody involved.
"Time is simply the mechanism that keeps everything from happening all at once."
ID: 907154 ·
Robert Barton
Joined: 16 Aug 99
Posts: 9
Credit: 3,264,524
RAC: 76
Message 907343 - Posted: 13 Jun 2009, 18:24:22 UTC
Looks like I will be one of the next ones to reach the 1 Million work unit mark !!
I never thought it would happen.
I am pleased, should happen today or tomorrow.
ID: 907343 ·
_heinz
Volunteer tester
Joined: 25 Feb 05
Posts: 744
Credit: 5,539,270
RAC: 0
Message 907374 - Posted: 13 Jun 2009, 21:22:54 UTC
Congrats to all your personal milestones :-)
every point counts
summer and vacations are comming soon, bring us most outages.
heinz
D5400XS V8-Xeon
ID: 907374 ·
Terror Australis
Volunteer tester
Joined: 14 Feb 04
Posts: 1817
Credit: 262,693,308
RAC: 44
Message 907476 - Posted: 14 Jun 2009, 3:14:32 UTC
Ticked over 7,500,000 Credits today
#12 in Oz
Will break into the world top 400 in about 10 days
Only one (very basic) CUDA machine.
Congrats to everyone else on their achievements !!
Brodo
ID: 907476 ·
Speedy
Volunteer tester
Joined: 26 Jun 04
Posts: 1643
Credit: 12,921,799
RAC: 89
Message 907690 - Posted: 14 Jun 2009, 20:57:25 UTC - in response to Message 907343.
Last modified: 14 Jun 2009, 21:06:05 UTC
Looks like I will be one of the next ones to reach the 1 Million work unit mark !!
Out of interest how do people keep track of how many work units they have completed, is it by writing down how many are returned each day or by a program? I'm close to reaching a milestone of sorts that is my remaining ap unit will be granted creait I returned it on 26 Apr 2009 5:09:10 UTC there's another 124,313 ap units to be returned. Imo it would be a milestone record to have all ap units returned be for new ap data comes online.
Congrats to everyone else on their achievements
ID: 907690 ·
Dorphas
Joined: 16 May 99
Posts: 118
Credit: 8,007,247
RAC: 0
Message 907813 - Posted: 15 Jun 2009, 12:49:10 UTC
just passed 2,000,000 in seti.
long time coming for me...... :)
ID: 907813 ·
KB4SH MentorSim
Volunteer tester
Joined: 17 May 99
Posts: 5
Credit: 39,807,916
RAC: 5
Message 907842 - Posted: 15 Jun 2009, 14:37:43 UTC
I passed 2,000,000 on May 26th. My results really accelerated after the Lunatics published the setup for installing a GPU properly and 6.6.20 came out. Thanks.
--
mentorsim
ID: 907842 ·
Charles Anspaugh
Volunteer tester
Joined: 11 Aug 00
Posts: 48
Credit: 22,715,083
RAC: 0
Message 907862 - Posted: 15 Jun 2009, 16:19:55 UTC
//
ID: 907862 ·
Bob Mahoney Design
Joined: 4 Apr 04
Posts: 178
Credit: 9,205,632
RAC: 0
Message 908014 - Posted: 16 Jun 2009, 0:44:05 UTC
I retired the Top Host that served so well for so long. It was a wonderful machine.
Check out my profile here for a picture of the computer, plus some technical data.
Bob
ID: 908014 ·
Byron Leigh Hatch @ team Carl Sagan
Volunteer tester
Joined: 5 Jul 99
Posts: 4548
Credit: 35,667,570
RAC: 4
Message 908034 - Posted: 16 Jun 2009, 1:48:58 UTC - in response to Message 908014.
I retired the Top Host that served so well for so long. It was a wonderful machine.
Check out my profile here for a picture of the computer, plus some technical data.
Bob
WOW! ... now that's SETI@home Crunching Machine!
Way To Go!
ID: 908034 ·
Ghery S. Pettit
Joined: 7 Nov 99
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Message 908041 - Posted: 16 Jun 2009, 2:35:33 UTC
5,000,000 total for BOINC. SETI in a few days.
ID: 908041 ·
Byron Leigh Hatch @ team Carl Sagan
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Message 908111 - Posted: 16 Jun 2009, 10:25:41 UTC - in response to Message 908014.
Last modified: 16 Jun 2009, 10:46:51 UTC
.
I retired the Top Host that served so well for so long. It was a wonderful machine.
Check out my profile here for a picture of the computer, plus some technical data.
Bob
Bob
Congratulation! on reaching the top Host spot!
the Computer you built is Awesome!
Ghery S. Pettit
Congratulation! on reaching your five (5) million Milestone WTG!
@ all
Congratulation! to all on reaching your own personal Milestones!
Byron
.
ID: 908111 ·
OzzFan
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Message 908212 - Posted: 16 Jun 2009, 22:13:40 UTC - in response to Message 908014.
I retired the Top Host that served so well for so long. It was a wonderful machine.
Check out my profile here for a picture of the computer, plus some technical data.
Bob
Congrats! That was an amazing host, and your next project seems very ambitious indeed!
Off topic, but is that the same ExecPC that started out as a BBS back in the 80s and 90s? ExecPC was one of my top favorite BBSes to dial into, and it always made me want to start my own BBS because it seemed so fun.
ID: 908212 ·
Fabe
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Message 908333 - Posted: 17 Jun 2009, 7:15:53 UTC
Just went pass 1,000,000 for Boinc total.
ID: 908333 ·
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Message boards : Number crunching : CLOSED** SETI/BOINC Milestones (tm) XVII **CLOSED
©2024 University of California
SETI@home and Astropulse are funded by grants from the National Science Foundation, NASA, and donations from SETI@home volunteers. AstroPulse is funded in part by the NSF through grant AST-0307956.
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sfortney - 1 year ago 106
Python Question
# Creating the Cartesian Product of a set of Vectors in Python?
Given the standard basis vectors
`(e_1,e_2,e_3)`
in 3 dimensions and letting the elements of
`(e_1,e_2,e_3)`
be restricted to, say
`(0,1,2,3,4)`
is there a simple pythonic way to create the cartesian product of all the vectors in this vector space?
For example, given [1,0,0],[0,1,0] and [0,0,1], I would like to get a list of all of the linear combinations (where the a_i's are restricted to the naturals between 0 and 4) of these vectors between [0,0,0] and [4,4,4].
I could program this up myself but before going to that trouble I thought I would ask if there is a simple pythonic way of doing it, maybe in numpy or something similar.
For the specific case of a space of natural numbers, you want `np.indices`:
``````>>> np.indices((4, 4)).reshape(2,-1).T
array([[0, 0],
[0, 1],
[0, 2],
[0, 3],
[1, 0],
[1, 1],
[1, 2],
[1, 3],
[2, 0],
[2, 1],
[2, 2],
[2, 3],
[3, 0],
[3, 1],
[3, 2],
[3, 3]])
``````
(numpy actually outputs these in a grid, but you wanted a 1-D list of points, hence the `.reshape`)
Otherwise, what you're describing is not a powerset but a cartesian product
``````itertools.product(range(4), repeat=3)
``````
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Thread: [SOLVED] Proving that a polynomial is 0
1. [SOLVED] Proving that a polynomial is 0
Hi
Challenging (more or less) question. I put it in this subforum as to avoid giving hints on the way to solve it.
My friend told me a method, but I'm curious to know if there are others.
So we have $\displaystyle P$ a polynomial such that its degree $\displaystyle \text{deg}(P)\leq n$, $\displaystyle n \in \mathbb{N}^*$
If $\displaystyle \int_0^1 P(t)t^k ~dt=0 \quad \forall k \in \mathbb{N} ~:~ 0\leq k\leq n$,
prove that $\displaystyle P(x)=0$.
2. Salut
Let $\displaystyle P(t)=\sum_{k=0}^{n}\:a_k\:t^k$
$\displaystyle \int_0^1 P(t)t^k ~dt=0 \quad \forall k \in \mathbb{N} ~:~ 0\leq k\leq n$
$\displaystyle \sum_{k=0}^{n}\:a_k \int_0^1 P(t)t^k ~dt=0$
$\displaystyle \int_0^1 P(t)\:\sum_{k=0}^{n}\:a_k t^k ~dt=0$
$\displaystyle \int_0^1 P^2(t) ~dt=0$
But of course $\displaystyle P^2(t) \geq 0$
3. Very neat solution
This is how a teacher told my friend :
Define the scalar product $\displaystyle \langle x,y\rangle=\int_0^1 x(t)y(t) ~dt$
$\displaystyle y(t)=t^k$ belongs to the basis of $\displaystyle P[X]$
So we have $\displaystyle \langle x,y\rangle=0$, for any y in the basis.
Hence x=0.
I don't know which theorem was used though
4. Originally Posted by Moo
Very neat solution
This is how a teacher told my friend :
Define the scalar product $\displaystyle \langle x,y\rangle=\int_0^1 x(t)y(t) ~dt$
$\displaystyle y(t)=t^k$ belongs to the basis of $\displaystyle P[X]$
So we have $\displaystyle \langle x,y\rangle=0$, for any y in the basis.
Hence x=0.
I don't know which theorem was used though
Because it implies it is orthogonal to itself. (exactly what running-gag showed)
Remember that $\displaystyle \left\langle {p_1,p_1} \right\rangle = 0 \Leftrightarrow p_1 = \bold 0$ (the 0 polynomial ) by the definition of inner-product
Okay, let $\displaystyle \mathcal{P}_n \left[ t \right]$ be the space of polynomials of degree less or equal than $\displaystyle n$ over R. (our field will be R)
We define: $\displaystyle \left\langle , \right\rangle :\mathcal{P}_n \left[ t \right] \times \mathcal{P}_n \left[ t \right] \to \mathbb{R}$ by: $\displaystyle \left\langle {a\left( t \right),b\left( t \right)} \right\rangle = \int_0^1 {a\left( t \right) \cdot b\left( t \right)dt}$
You can check that this is an inner-product.
Now $\displaystyle B = \left\{ {1,t,...,t^n } \right\}$ is a basis of $\displaystyle \mathcal{P}_n \left[ t \right]$
And $\displaystyle p\left( t \right)$ is orthogonal to each of them, hence it's orthogonal to all vectors of $\displaystyle \mathcal{P}_n \left[ t \right]$ just see that $\displaystyle \left\langle {p\left( t \right),b_0 + \sum\limits_{k = 1}^n {b_k \cdot t^k } } \right\rangle = b_0 \cdot \left\langle {p\left( t \right),1} \right\rangle + b_k \cdot \sum\limits_{k = 1}^n {\left\langle {p\left( t \right),t^k } \right\rangle }$. (in particular to itself and hence $\displaystyle p(t)\equiv 0$)
5. This problem has a nice generalization.
If $\displaystyle f:[0,1]\to \mathbb{R}$ is continous with $\displaystyle \int_0^1 f(x)x^t dx = 0$ for all $\displaystyle t=0,1,2,3,...$ then $\displaystyle f$ is identically zero.
6. Originally Posted by ThePerfectHacker
This problem has a nice generalization.
If $\displaystyle f:[0,1]\to \mathbb{R}$ is continous with $\displaystyle \int_0^1 f(x)x^t dx = 0$ for all $\displaystyle t=0,1,2,3,...$ then $\displaystyle f$ is identically zero.
it's a nice and simple application of the Weierstrass approximation theorem.
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# Converting Celsius to Kelvin Easy but slightly confusing
1. Nov 6, 2008
### Spartan Erik
1. The problem statement, all variables and given/known data
I'm doing calorimetry, and I have 34.73C and 63C. I need to subtract 63 from 34.73, and get the answer in Kelvin.
2. Relevant equations
3. The attempt at a solution
I know you add 273 to Celsius values to get Kelvin, but I am not sure which of these methods is correct:
1. (34.73C+273) - (63C+273) = -28.27K
2. 34.73C-63C = -28.27C. Then, -28.27C + 273 = 244.73K
Which one is the correct method? The first one seems unlikely just because the Kelvin's are essentially canceled out in the equation..
Last edited: Nov 6, 2008
2. Nov 6, 2008
### HallsofIvy
The first one. You are subtracting one temperature from another so the "base"- that is what temperature corresponds to 0- is irrelevant. You could as easily have just said that 34.73- 63= -28.27 degrees. Since 1 degree Celcius is the same "size" as 1 degree Kelvin, the difference is the same in both systems.
3. Nov 6, 2008
### Spartan Erik
Ah I just realized that I was calculating the DECREASE in temperature of hot water, so a negative decrease is a positive value. Thanks!
Last edited: Nov 6, 2008
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This blog is managed by Song Hock Chye, author of Improve Your Thinking Skills in Maths (P1-P3 series), which is published and distributed by EPH.
## Monday, September 28, 2009
### Rosyth Sch 2007 PSLE Math Prelim Q48
A bus was travelling at a constant speed from Town A to Town B. It passed a car travelling at a constant speed of 90 km/h in the opposite direction. 1 and 1/2 hours later, the bus reached Town B but the car was still 25 km away from Town A. If the bus took 4 hours to complete the whole journey, what is the distance between the two towns?
Solution
Anonymous said...
A fruit seller had some oranges, pears and durians. After selling some of them, there were 6 pears for every 5 oranges left and 7 oranges for every 3 durians left. He sold 168 oranges. He had 456 pears and oranges left. The number of pears sold was the same as the number of durians sold. a) How many oranges had he at first? (2m) b) If the number of pears was twice the number of durians at first, what was the total number of pears and durians sold? (3m)
Anonymous said...
Joe bought some chocolates and gave half of it to Ken. Ken bought some sweets and gave half of it to Joe. Joe ate 12 sweets and Ken ate 18 choclates.The ratio of sweets to chocolates Joe had became 1 : 7 and the ratio of sweets to chocolates Ken had became 1:4. How many sweets did Ken buy ?
Anonymous said...
?: Sally baked some cookies to sell. ¾ of them were chocolate cookies and the remaining were almond cookies. After she sold 210 chocolate cookies and 5/6 of the almond cookies, she had 1/5 of the cookies left. How many cookies did she sell? Any info wrong with this question?
Anonymous said...
For the cookie question, I thought Sally sold 210 almond cookies and 5/6 of the chocolate cookies. What is the solution?
Anonymous said...
lol omg .. psle question leaked? Oct 3 he got that question...
Anonymous said...
He got the qns on 8 oct.. that's the day of PSLE Mathematics.
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# CS 245 Assembly Language Programming Intro to Computer Math
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## Transcription
1 Binary-Hexadecimal Worksheet 1 CS 245 Assembly Language Programming Intro to Computer Math Text: Computer Organization and Design, 4 th Ed., D A Patterson, J L Hennessy Section 2.4 Objectives: The Student shall be able to: Convert numbers between decimal, binary, hexadecimal Add binary and hexadecimal numbers. Perform logical operations: AND and OR on binary or hexadecimal numbers Determine the range of possible numbers given a number of bits for storage. Form or translate a negative number from a positive number and vice versa. Class Time: Binary, Octal, Hexadecimal Signed and Unsigned numbers Exercise Total 1 hour 1 hour 1 hour 3 hours
2 Binary-Hexadecimal Worksheet 2 Hello Binary! Imagine a world of 1s and 0s no other numbers exist. Welcome to the world of the computer. This is how information and instructions are stored in the computer. Well, what happens when we add 1+1? We must get 10. What happens if we add 10+1? We get 11. What happens if we add 11+1? We get 100. Do you see the pattern? Try it for yourself below, by continually adding one to get the decimal number on the left: Notice: what is the value of each digit? For example, if we have a binary number: B11111 what does each binary number stand for? For example, in decimal the number would be: ,000. Using the same idea, what do the binary values: 1, 10, 100, 1000, translate into in decimal? Do you notice that each binary digit is basically a double of the digit to its right? That is very important to remember. Always remember that each place is multiplied by 2! Translate the following binary numbers to decimal using this rule: B = B = B = B =
3 Binary-Hexadecimal Worksheet 3 Addition using Binary EXERCISE: BINARY ADDITION Checking with Decimal B B 1010 = B = 15 B B 0011 = B = 15 B B 0011 = B = 12 Let s try something more complicated: B B = Carry: B B B Add the following numbers: Binary Check your work with the Decimal Equivalent
4 Binary-Hexadecimal Worksheet 4 EXERCISE: AND & OR Now let s play with AND and OR. AND: If both bits are set, set the result: & OR: If either bit is set, set the result: We can define truth tables for these operations. The bold italicized numbers IN the table are the answers. The column header and row header are the two numbers being operated on. AND & This table shows that: 0 & 0 = 0 1 & 0 = 0 0 & 1 = 0 1 & 1 = 1 OR This table shows that: 0 0 = = = = 1 I will show how these operations work with larger binary numbers: B B B B & B B AND B OR B B B B B Now you try some: B B B B AND B OR B & B B Below, show what binary value you would use to accomplish the operation. Then do the operation to verify that it works! Bits are ordered: Using ORs to turn on bits: Using ANDs to turn off bits: B B Turn on bits 0-3 Turn off bits 3-4 B Turn on bit 0 B Turn on bits 3-4 B Turn off bits 0-3 B Turn off bits 0-4
5 Binary-Hexadecimal Worksheet 5 Let s build a table for each numbering system. You fill in the blanks Decimal = Base 10 Binary = Base 2 Octal = Base 8 Hexadecimal = Base A B C D E F A E F There is something very special about Base 8 and Base 16 they are compatible with Base 2. So for example, let s take the binary number11000 = Notice that Base 8 operates basically modulo 8, whereas base 16 operates modulo 16. It is not easy to convert between decimal and binary, but it is easy to convert between binary and octal or hexadecimal. It is useful to know that the octal or base 8 number = (3 x 8 2 ) + (2 x 8) + 4 And the hexadecimal or base 16 number = (3 x 16 2 ) + (2 x 16) + 4
6 Binary-Hexadecimal Worksheet 6 Hello Octal! Binary is rather tedious isn t it? It is hard to keep track of all those 1s and 0s. So someone invented base 8 and base 16. These are also known as octal and hexadecimal systems, respectively. The octal (base 8) numbering system works as follows: Base 8: To convert between binary and octal: Step 1: group the binary digits by threes, similar to how we use commas with large numbers: B becomes B B becomes B Step 2: Add zeros to the left (most significant digits) to make all numbers 3 bit numbers: B becomes B Step 3: Now convert each three bit number into a octal number: 0..7 B becomes B becomes Likewise we can convert from Octal to Binary: = = Now you try! Binary -> Octal B = = Octal -> Binary B = = B = = B = = If we want to convert from octal to decimal, we do: = (8x8 2 ) + (9x8 1 ) + (3x8 0 ) = 8x64 + 9x8 + 3 = 587 Now you try! = = 64 8 = =
8 Binary-Hexadecimal Worksheet 8 Now convert from hexadecimal back to binary: Hexadecimal Binary Hexadecimal Binary 0x23 0x31 0x4a 0x18A 0x23F 0x xAB1 0xC0D 0x44 0xF x3C 0x58 0x28 0x49 Okay, now we can convert between binary and hexadecimal and we can do ANDs and ORs. Let s try doing these together. Let s AND hexadecimal numbers together: 0x254 AND 0x0f0 = 0010, 0101, 0100 & 0000, 1111, , 0101, 0000 = 0x 050 Notice that what we are doing is that we convert the hexadecimal to binary, and do the AND, and convert the resulting binary digits back to hexadecimal. Let s try an OR: 0x254 OR 0x0f0 = 0010, 0101, , 1111, , 1111, 0100 = 0x 2f4 ANDs and ORs are useful to turn on and off specific bits. Now you do some: 0x1a3 & 0x111 = 0x273 & 0x032 = 0x1a3 0x111= 0x273 0x032 =
9 Binary-Hexadecimal Worksheet 9 Conversions: Hexadecimal Binary Decimal There are two ways to convert between Base 16 or Base 8 and Decimal. Method 1: Convert to Binary, then Decimal: 0x1af = = = = x456 = = = = Method 2: Use division remainders: Convert from base 10 to base N (Example base 2): Number / 2 -> remainder is digit 0 -> quotient / 2 -> remainder is digit 1 -> quotient / 2 -> remainder is digit 2 Example 1: Convert into binary: Quotient/2 ->Remainder 36/2 ->0 18/2 ->0 9/2 ->1 4/2 ->0 2/2 ->0 1/2 -> = Example 2: Convert into base 16: 36/16 ->4 2/16 -> = Example 3: Convert 0x1af to base 10: 0x1af = 1 x 16 2 = = 256 a x 16 = 10 x 16 = 160 f = +15 Total = Now you try some conversions between base 16 and base 10 (Your choice of method!) 0x AC4= 0x C4A= = =
10 Binary-Hexadecimal Worksheet 10 Signed & Unsigned Numbers Assuming 1 byte: Binary Signed Unsigned Notice that you get HALF of the total positive numbers with signed integers!!! Example: Convert to a signed 8-bit integer: Converting to Decimal: Powers of Two The sign bit (bit 7) indicates both sign and value: If top N bit is 0, sign & all values are positive: top set value: 2 N If top N bit is 1, sign is negative: -2 N Remaining bits are calculated as positive values: = = = = = = 85 Changing Signs: Two s Compliment A positive number may be made negative and vice versa using this technique Method: Take the inverse of the original number and add 1. Original: = = -85 invert: add 1: sum: = = 85 First we determine what the range of numbers is for signed versus unsigned numbers: 4 bits 8 bits 12 bits Low High Low High Low High Unsigned Signed Total number of possible numbers Unsigned: Low = 0 High = 15 Set of 16 numbers
11 Binary-Hexadecimal Worksheet 11 Now you try some conversions between positive and negative numbers. Assume 8-bit signed numbers (and top bit is signed bit). Hexadecimal Actual Signed Decimal Value: Change sign: value: 0x = In binary: = -121 Invert: Add 1: = 0x 79 Translate: =121 0x ba 0x fc 0x 03 0x 81 0x 33
12 Binary-Hexadecimal Worksheet 12 Real-World Exercise: Conversion Let s do something USEFUL! Below is a table to show how IP headers are formatted. In yellow is shown the formatting for an ICMP header for a PING message Version HLenth Service Type Total Length Datagram Identification Flags Fragment Offset Time to Live Protocol Header Checksum Source IP Address Destination IP Address Type Code Checksum Identifier Sequence Number Data You are writing logic to decode this hexadecimal sequence and now you want to verify that the interpreted packet is correct you must convert it manually to verify! dc 039c b c0a c0a c a 6b6c 6d6e 6f What are the decimal values for the following fields: Word 1: Version: HLenth: Total Length: Word 2: Datagram Id: Fragment Offset: Word 2: A flag is a one-bit field. Flags include bits 16-18: Don t Fragment (Bit 17): More Fragment (Bit 18): Word 3: Time to Live: Protocol: For the two addresses below, convert each byte in word to decimal and separate by periods (e.g., ): Word 4: Source IP Address: Word 5: Destination IP Address: ICMP: Type: Code: Sequence Number:
### Useful Number Systems
Useful Number Systems Decimal Base = 10 Digit Set = {0, 1, 2, 3, 4, 5, 6, 7, 8, 9} Binary Base = 2 Digit Set = {0, 1} Octal Base = 8 = 2 3 Digit Set = {0, 1, 2, 3, 4, 5, 6, 7} Hexadecimal Base = 16 = 2
### Base Conversion written by Cathy Saxton
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### Binary Numbers. Bob Brown Information Technology Department Southern Polytechnic State University
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# Does GMAT ask Logs questions? Logx^5 = 32 kind of...? Thanks
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# 63991 (number)
63,991 (sixty-three thousand nine hundred ninety-one) is an odd five-digits composite number following 63990 and preceding 63992. In scientific notation, it is written as 6.3991 × 104. The sum of its digits is 28. It has a total of 2 prime factors and 4 positive divisors. There are 63,184 positive integers (up to 63991) that are relatively prime to 63991.
## Basic properties
• Is Prime? No
• Number parity Odd
• Number length 5
• Sum of Digits 28
• Digital Root 1
## Name
Short name 63 thousand 991 sixty-three thousand nine hundred ninety-one
## Notation
Scientific notation 6.3991 × 104 63.991 × 103
## Prime Factorization of 63991
Prime Factorization 89 × 719
Composite number
Distinct Factors Total Factors Radical ω(n) 2 Total number of distinct prime factors Ω(n) 2 Total number of prime factors rad(n) 63991 Product of the distinct prime numbers λ(n) 1 Returns the parity of Ω(n), such that λ(n) = (-1)Ω(n) μ(n) 1 Returns: 1, if n has an even number of prime factors (and is square free) −1, if n has an odd number of prime factors (and is square free) 0, if n has a squared prime factor Λ(n) 0 Returns log(p) if n is a power pk of any prime p (for any k >= 1), else returns 0
The prime factorization of 63,991 is 89 × 719. Since it has a total of 2 prime factors, 63,991 is a composite number.
## Divisors of 63991
1, 89, 719, 63991
4 divisors
Even divisors 0 4 2 2
Total Divisors Sum of Divisors Aliquot Sum τ(n) 4 Total number of the positive divisors of n σ(n) 64800 Sum of all the positive divisors of n s(n) 809 Sum of the proper positive divisors of n A(n) 16200 Returns the sum of divisors (σ(n)) divided by the total number of divisors (τ(n)) G(n) 252.964 Returns the nth root of the product of n divisors H(n) 3.95006 Returns the total number of divisors (τ(n)) divided by the sum of the reciprocal of each divisors
The number 63,991 can be divided by 4 positive divisors (out of which 0 are even, and 4 are odd). The sum of these divisors (counting 63,991) is 64,800, the average is 16,200.
## Other Arithmetic Functions (n = 63991)
1 φ(n) n
Euler Totient Carmichael Lambda Prime Pi φ(n) 63184 Total number of positive integers not greater than n that are coprime to n λ(n) 31592 Smallest positive number such that aλ(n) ≡ 1 (mod n) for all a coprime to n π(n) ≈ 6404 Total number of primes less than or equal to n r2(n) 0 The number of ways n can be represented as the sum of 2 squares
There are 63,184 positive integers (less than 63,991) that are coprime with 63,991. And there are approximately 6,404 prime numbers less than or equal to 63,991.
## Divisibility of 63991
m n mod m 2 3 4 5 6 7 8 9 1 1 3 1 1 4 7 1
63,991 is not divisible by any number less than or equal to 9.
## Classification of 63991
• Arithmetic
• Semiprime
• Deficient
• Polite
• Square Free
### Other numbers
• LucasCarmichael
## Base conversion (63991)
Base System Value
2 Binary 1111100111110111
3 Ternary 10020210001
4 Quaternary 33213313
5 Quinary 4021431
6 Senary 1212131
8 Octal 174767
10 Decimal 63991
12 Duodecimal 31047
20 Vigesimal 7jjb
36 Base36 1ddj
## Basic calculations (n = 63991)
### Multiplication
n×y
n×2 127982 191973 255964 319955
### Division
n÷y
n÷2 31995.5 21330.3 15997.8 12798.2
### Exponentiation
ny
n2 4094848081 262033423551271 16767780806469382561 1072987061586782259460951
### Nth Root
y√n
2√n 252.964 39.9981 15.9049 9.14584
## 63991 as geometric shapes
### Circle
Diameter 127982 402067 1.28643e+10
### Sphere
Volume 1.0976e+15 5.14574e+10 402067
### Square
Length = n
Perimeter 255964 4.09485e+09 90496.9
### Cube
Length = n
Surface area 2.45691e+10 2.62033e+14 110836
### Equilateral Triangle
Length = n
Perimeter 191973 1.77312e+09 55417.8
### Triangular Pyramid
Length = n
Surface area 7.09248e+09 3.08809e+13 52248.4
## Cryptographic Hash Functions
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# Continuity proof
• Dec 11th 2007, 07:52 PM
spoon737
Continuity proof
I need to show that x^2 is continuous on its domain using the epsilon-delta definition. So far, I've started with |(x^2) - (c^2)|< E, which is the same as |x+c||x-c|< E. However, I'm not quite how to get from here to |x-c|< D, where D is delta (which I'm pretty sure is supposed to depend on epsilon and c).
• Dec 11th 2007, 08:08 PM
ThePerfectHacker
Quote:
Originally Posted by spoon737
I need to show that x^2 is continuous on its domain using the epsilon-delta definition. So far, I've started with |(x^2) - (c^2)|< E, which is the same as |x+c||x-c|< E. However, I'm not quite how to get from here to |x-c|< D, where D is delta (which I'm pretty sure is supposed to depend on epsilon and c).
Let $x_0 \in \mathbb{R}$ we want to show $f(x) = x^2$ is continous at that point. We want to show $|f(x) - f(x_0)|< \epsilon$ can be made small, thus, $|x^2 - x_0^2| = |x - x_0||x+x_0|< \epsilon$. Now say $|x-x_0|<\delta$ then $|x| = |x-x_0+x_0|\leq |x-x_0| + |x_0| \implies |x|+|x_0| \leq |x-x_0|+2|x_0|$ but then $|x+x_0|\leq |x|+|x_0|\leq |x-x_0|+2|x_0| < \delta + 2|x_0|$. Thus, $|f(x)-f(x_0)| = |x-x_0||x+x_0| < \delta (\delta + 2|x_0|)$ choose $\delta \leq 1$ then $|\delta (\delta + 2|x_0|) \leq \delta (1+2|x_0|)$. Thus, we have that $|f(x)-f(x_0|< \delta (1+2|x_0|)$, this means we have to chose $\delta = \min \left\{ \frac{\epsilon}{1+2|x_0|}, 1\right\}$
• Dec 11th 2007, 08:13 PM
ThePerfectHacker
Quote:
Originally Posted by kalagota
so, if $\varepsilon > 0$, choose $\delta = \frac{\varepsilon}{1+2c}$
Careful what if $c=-\frac{1}{2}$ :eek:
• Dec 11th 2007, 08:17 PM
kalagota
Quote:
Originally Posted by ThePerfectHacker
Careful what if $c=-\frac{1}{2}$ :eek:
yeah, i should have edited it before i posted it.. anyways, you already had it.. thanks..
hey, can you help me with my problem i posted in the algeb section? thanks.
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| 10,845
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Velocity Reviews > Looks like the "conspiracy theories" really were true after all...
Looks like the "conspiracy theories" really were true after all...
schoenfeld.one@gmail.com
Guest
Posts: n/a
10-22-2007
Most people don't know that there were actually 3 buildings which came
crashing down on the day of 9/11.
The third building, WTC 7, can be seen here
There is no mention of this building in 911 Omission Report.
Can fire make a building come crashing down at free fall speed?
If you think it can, patent the idea and make billions in the
demolitions industry!
How do we know WTC 7 was demolished?
If WTC 7 collapsed in 6 seconds, and it takes 6 seconds to free fall
from the roof of WTC 7, then you got it - WTC 7 underwent a free fall.
This means as the each floor was falling straight to the ground it did
so without crashing into anything on the way. ONLY CONTROLLED
DEMOLITION CAN ACCOMPLISH THAT!
PROPOSITION 1:
It took a total of 6 seconds for the roof of WTC 7 to reach the
ground. This proposition is supported by the empirical,
Collapse start time: 17 seconds
Collapse end time: 23 seconds
Total collapse time: 23-17 = 6 seconds
PROPOSITION 2:
A free fall from a height equal to the roof of WTC 7 would take 6
seconds. This proposition derives trivially through (Galilean)
kinematical considerations alone:
Displacement = initial velocity * total time + 1/2 * acceleration *
total time^2
or
s = ut + 1/2at^2
where
s = 174 m (height of building)
u = 0 m/s (building was stationary prior to collapse)
a = 9.8 m/s^2 (since gravitational field strengh averages at
a constant)
Thus,
174 = 0 t + 1/2 9.8 t^2
Solving for t
t = sqrt( 2 * 174 / 9.
= 5.9590
~ 6 seconds
R. Mark Clayton
Guest
Posts: n/a
10-22-2007
<(E-Mail Removed)> wrote in message
news:(E-Mail Removed) oups.com...
> Most people don't know that there were actually 3 buildings which came
> crashing down on the day of 9/11.
At least one other building fell when hit by debris from one of the towers -
IIRC the Vista Hotel, which sat between the two towers, but several other
buildings were badly damaged and at least one was demolished shortly
afterwards as it was dangerous.
>
> The third building, WTC 7, can be seen here
>
>
> There is no mention of this building in 911 Omission Report.
>
> Can fire make a building come crashing down at free fall speed?
Probably, but being hit by thousands of tons of steel and masonary already
plunging downwards at high velocity can make it come down faster than free
fall speed.
>
> If you think it can, patent the idea and make billions in the
> demolitions industry!
>
> How do we know WTC 7 was demolished?
>
> If WTC 7 collapsed in 6 seconds, and it takes 6 seconds to free fall
> from the roof of WTC 7, then you got it - WTC 7 underwent a free fall.
6 seconds is free fall from 180m, so how tall was WTC7?
>
> This means as the each floor was falling straight to the ground it did
> so without crashing into anything on the way. ONLY CONTROLLED
> DEMOLITION CAN ACCOMPLISH THAT!
Er wrong again, the two towers pretty much collapsed like a deck of cards,
with only the top section tipping over.
But anyway, what are you actually trying to prove?
Dan
Guest
Posts: n/a
10-22-2007
<(E-Mail Removed)> wrote in message
news:(E-Mail Removed) oups.com...
>
plonk.
dan
Christopher Benson-Manica
Guest
Posts: n/a
10-22-2007
In comp.lang.c R. Mark Clayton <(E-Mail Removed)> wrote:
> some response to spam
Any chance you could take this charming discussion somewhere other
than comp.lang.c? I assure you that no one here is interested.
(F'ups set.)
--
C. Benson Manica | I appreciate all corrections, polite or otherwise.
cbmanica(at)gmail.com |
----------------------| I do not currently read any posts posted through
sdf.lonestar.org | Google groups, due to rampant unchecked spam.
Essex Laptops - Andy Usher
Guest
Posts: n/a
10-22-2007
<(E-Mail Removed)> wrote in message
news:(E-Mail Removed) oups.com...
> Most people don't know that there were actually 3 buildings which came
> crashing down on the day of 9/11.
Most people in the UK that I know of could not care less about your
obsession with the WTC, The only thing to have come out of all this is
people not being able to move as freely as they should be
Just A User
Guest
Posts: n/a
10-22-2007
http://www.velocityreviews.com/forums/(E-Mail Removed) wrote:
> Most people don't know that there were actually 3 buildings which came
> crashing down on the day of 9/11.
>
> The third building, WTC 7, can be seen here
>
>
> There is no mention of this building in 911 Omission Report.
>
> Can fire make a building come crashing down at free fall speed?
>
> If you think it can, patent the idea and make billions in the
> demolitions industry!
>
> How do we know WTC 7 was demolished?
>
> If WTC 7 collapsed in 6 seconds, and it takes 6 seconds to free fall
> from the roof of WTC 7, then you got it - WTC 7 underwent a free fall.
>
> This means as the each floor was falling straight to the ground it did
> so without crashing into anything on the way. ONLY CONTROLLED
> DEMOLITION CAN ACCOMPLISH THAT!
>
> PROPOSITION 1:
> It took a total of 6 seconds for the roof of WTC 7 to reach the
> ground. This proposition is supported by the empirical,
>
> Collapse start time: 17 seconds
> Collapse end time: 23 seconds
> Total collapse time: 23-17 = 6 seconds
>
> PROPOSITION 2:
> A free fall from a height equal to the roof of WTC 7 would take 6
> seconds. This proposition derives trivially through (Galilean)
> kinematical considerations alone:
>
> Displacement = initial velocity * total time + 1/2 * acceleration *
> total time^2
>
> or
>
> s = ut + 1/2at^2
> where
> s = 174 m (height of building)
> u = 0 m/s (building was stationary prior to collapse)
> a = 9.8 m/s^2 (since gravitational field strengh averages at
> a constant)
>
> Thus,
> 174 = 0 t + 1/2 9.8 t^2
>
> Solving for t
> t = sqrt( 2 * 174 / 9.
> = 5.9590
> ~ 6 seconds
>
And the explosives were put in the building by Elvis and the mystery
shooter on the grassy knoll. And they were of explosives were supplied
by the aliens that are still alive that crashed in Roswell in 1947.
Richard
Guest
Posts: n/a
10-22-2007
Christopher Benson-Manica <(E-Mail Removed)> writes:
> In comp.lang.c R. Mark Clayton <(E-Mail Removed)> wrote:
>
>> some response to spam
>
> Any chance you could take this charming discussion somewhere other
> than comp.lang.c? I assure you that no one here is interested.
>
> (F'ups set.)
Well done. You just woke the thread up for those of us with properly
configured spam filters and thread scoring who hadn't seen it. Well done
indeed.
prettybaby@softhome.net
Guest
Posts: n/a
10-22-2007
On Oct 22, 5:51 am, (E-Mail Removed) wrote:
> Most people don't know that there were actually 3 buildings which came
> crashing down on the day of 9/11.
>
> The third building, WTC 7, can be seen here
>
>
> There is no mention of this building in 911 Omission Report.
>
> Can fire make a building come crashing down at free fall speed?
>
> If you think it can, patent the idea and make billions in the
> demolitions industry!
>
> How do we know WTC 7 was demolished?
>
> If WTC 7 collapsed in 6 seconds, and it takes 6 seconds to free fall
> from the roof of WTC 7, then you got it - WTC 7 underwent a free fall.
>
> This means as the each floor was falling straight to the ground it did
> so without crashing into anything on the way. ONLY CONTROLLED
> DEMOLITION CAN ACCOMPLISH THAT!
>
> PROPOSITION 1:
> It took a total of 6 seconds for the roof of WTC 7 to reach the
> ground. This proposition is supported by the empirical,
>
> Collapse start time: 17 seconds
> Collapse end time: 23 seconds
> Total collapse time: 23-17 = 6 seconds
>
> PROPOSITION 2:
> A free fall from a height equal to the roof of WTC 7 would take 6
> seconds. This proposition derives trivially through (Galilean)
> kinematical considerations alone:
>
> Displacement = initial velocity * total time + 1/2 * acceleration *
> total time^2
>
> or
>
> s = ut + 1/2at^2
> where
> s = 174 m (height of building)
> u = 0 m/s (building was stationary prior to collapse)
> a = 9.8 m/s^2 (since gravitational field strengh averages at
> a constant)
>
> Thus,
> 174 = 0 t + 1/2 9.8 t^2
>
> Solving for t
> t = sqrt( 2 * 174 / 9.
> = 5.9590
> ~ 6 seconds
whatever you are trying to say, it is still a sign of the end of this
age.
prettybaby
http://spiritofart123.blogspot.com
Niel J Humphreys
Guest
Posts: n/a
10-22-2007
"Essex Laptops - Andy Usher" <(E-Mail Removed)> wrote in message
news:(E-Mail Removed)...
>
> <(E-Mail Removed)> wrote in message
> news:(E-Mail Removed) oups.com...
>> Most people don't know that there were actually 3 buildings which came
>> crashing down on the day of 9/11.
>
> Most people in the UK that I know of could not care less about your
> obsession with the WTC, The only thing to have come out of all this is
> people not being able to move as freely as they should be
You're joking. Everytime I get held up in a bloody queue at the airport it
makes me hate those ****ing Muslim extremists even more.
--
Niel H
Essex Laptops - Andy Usher
Guest
Posts: n/a
10-22-2007
"Richard" <(E-Mail Removed)> wrote in message
news:(E-Mail Removed)...
> Christopher Benson-Manica <(E-Mail Removed)> writes:
>
>> In comp.lang.c R. Mark Clayton <(E-Mail Removed)> wrote:
>>
>>> some response to spam
>>
>> Any chance you could take this charming discussion somewhere other
>> than comp.lang.c? I assure you that no one here is interested.
>>
>> (F'ups set.)
>
> Well done. You just woke the thread up for those of us with properly
> configured spam filters and thread scoring who hadn't seen it. Well done
> indeed.
tiy didnt have to open or reply to it, no one cares about the post and no
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# The pressure unit
Hi Dear
Is the unit of pressure atmosphere in real unit really? I’m very confused. I run a simulation for polymer chains to calculate constant elastic properties by deform command, but the obtained pressure are 100 times the real values, while the temperature and energy are logical. Before deform command, In the relaxation step in NPT fix command the fluctuations of pressure is following:
Hi Dear
Is the unit of pressure atmosphere in real unit really?
The documentation says so, why should it not be?
I’m very confused. I run a simulation for polymer chains to calculate constant elastic properties by deform command, but the obtained pressure are 100 times the real values, while the temperature and energy are logical.
The pressure fluctuations are just as logical. You already have been told that pressure fluctuations are large for small samples and dense objects like liquids and solids. If you want to get a better handle on this, please spend some more time with your favorite statistical mechanics text book and study the underlying relations.
Before deform command, In the relaxation step in NPT fix command the fluctuations of pressure is following:
Instantaneous values are pretty much useless to look at for this. You need to look at averages and trends. Most of MD analysis is based on averaging over time instead of volume, provided you have a system in equilibrium.
Axel
Hi Dear
The output in the fix deform command in my simulation is pressure/atom. there was no pressure fluctuations in fix deform command. With increasing deformation of cell the pressure/atom increase linearly. This is logical. The pressure/atom divided by volume to obtain stress in cell. But the values of stress are 100 times the real values. This is confusing. The output of fix deform is following:(fx is the summation of stress/atom)
Step Temp Press fx Lx Ly Lz Volume
160000 300.3934 -389.455 78215537 64.37766 64.37766 64.37766 266812.2
160100 300.3934 -389.312 91670384 64.3841 64.37444 64.37444 266812.2
160200 300.3934 -389.172 1.05E+08 64.39054 64.37123 64.37123 266812.2
160300 300.3934 -389.037 1.19E+08 64.39698 64.36801 64.36801 266812.2
160400 300.3934 -388.911 1.32E+08 64.40341 64.36479 64.36479 266812.2
160500 300.3934 -388.789 1.46E+08 64.40985 64.36157 64.36157 266812.2
160600 300.3934 -388.674 1.60E+08 64.41629 64.35836 64.35836 266812.2
160700 300.3934 -388.57 1.73E+08 64.42273 64.35514 64.35514 266812.2
160800 300.3934 -388.47 1.87E+08 64.42916 64.35193 64.35193 266812.2
160900 300.3934 -388.378 2.00E+08 64.4356 64.34871 64.34871 266812.2
161000 300.3934 -388.291 2.14E+08 64.44204 64.3455 64.3455 266812.2
161100 300.3934 -388.21 2.28E+08 64.44848 64.34228 64.34228 266812.2
161200 300.3934 -388.145 2.41E+08 64.45492 64.33907 64.33907 266812.2
161300 300.3934 -388.076 2.55E+08 64.46135 64.33586 64.33586 266812.2
161400 300.3934 -388.019 2.69E+08 64.46779 64.33265 64.33265 266812.2
161500 300.3934 -387.969 2.82E+08 64.47423 64.32943 64.32943 266812.2
161600 300.3934 -387.926 2.96E+08 64.48067 64.32622 64.32622 266812.2
161700 300.3934 -387.892 3.09E+08 64.4871 64.32301 64.32301 266812.2
161800 300.3934 -387.862 3.23E+08 64.49354 64.3198 64.3198 266812.2
161900 300.3934 -387.841 3.37E+08 64.49998 64.31659 64.31659 266812.2
162000 300.3934 -387.825 3.50E+08 64.50642 64.31338 64.31338 266812.2
162100 300.3934 -387.815 3.64E+08 64.51286 64.31017 64.31017 266812.2
162200 300.3934 -387.814 3.78E+08 64.51929 64.30696 64.30696 266812.2
162300 300.3934 -387.823 3.91E+08 64.52573 64.30376 64.30376 266812.2
162400 300.3934 -387.834 4.05E+08 64.53217 64.30055 64.30055 266812.2
162500 300.3934 -387.853 4.19E+08 64.53861 64.29734 64.29734 266812.2
162600 300.3934 -387.881 4.32E+08 64.54504 64.29413 64.29413 266812.2
162700 300.3934 -387.912 4.46E+08 64.55148 64.29093 64.29093 266812.2
162800 300.3934 -387.952 4.59E+08 64.55792 64.28772 64.28772 266812.2
162900 300.3934 -387.996 4.73E+08 64.56436 64.28452 64.28452 266812.2
163000 300.3934 -388.048 4.87E+08 64.5708 64.28131 64.28131 266812.2
163100 300.3934 -388.111 5.00E+08 64.57723 64.27811 64.27811 266812.2
163200 300.3934 -388.181 5.14E+08 64.58367 64.2749 64.2749 266812.2
163300 300.3934 -388.256 5.28E+08 64.59011 64.2717 64.2717 266812.2
163400 300.3934 -388.334 5.41E+08 64.59655 64.2685 64.2685 266812.2
163500 300.3934 -388.42 5.55E+08 64.60298 64.2653 64.2653 266812.2
163600 300.3934 -388.513 5.69E+08 64.60942 64.26209 64.26209 266812.2
163700 300.3934 -388.613 5.82E+08 64.61586 64.25889 64.25889 266812.2
163800 300.3934 -388.719 5.96E+08 64.6223 64.25569 64.25569 266812.2
163900 300.3934 -388.832 6.10E+08 64.62874 64.25249 64.25249 266812.2
164000 300.3934 -388.956 6.23E+08 64.63517 64.24929 64.24929 266812.2
164100 300.3934 -389.089 6.37E+08 64.64161 64.24609 64.24609 266812.2
164200 300.3934 -389.225 6.51E+08 64.64805 64.24289 64.24289 266812.2
164300 300.3934 -389.369 6.65E+08 64.65449 64.2397 64.2397 266812.2
164400 300.3934 -389.516 6.78E+08 64.66092 64.2365 64.2365 266812.2
164500 300.3934 -389.668 6.92E+08 64.66736 64.2333 64.2333 266812.2
164600 300.3934 -389.831 7.06E+08 64.6738 64.2301 64.2301 266812.2
164700 300.3934 -390.002 7.19E+08 64.68024 64.22691 64.22691 266812.2
164800 300.3934 -390.183 7.33E+08 64.68668 64.22371 64.22371 266812.2
164900 300.3934 -390.371 7.47E+08 64.69311 64.22051 64.22051 266812.2
165000 300.3934 -390.563 7.60E+08 64.69955 64.21732 64.21732 266812.2
165100 300.3934 -390.76 7.74E+08 64.70599 64.21412 64.21412 266812.2
165200 300.3934 -390.967 7.88E+08 64.71243 64.21093 64.21093 266812.2
165300 300.3934 -391.18 8.02E+08 64.71886 64.20774 64.20774 266812.2
165400 300.3934 -391.399 8.15E+08 64.7253 64.20454 64.20454 266812.2
165500 300.3934 -391.625 8.29E+08 64.73174 64.20135 64.20135 266812.2
165600 300.3934 -391.858 8.43E+08 64.73818 64.19816 64.19816 266812.2
165700 300.3934 -392.097 8.56E+08 64.74462 64.19497 64.19497 266812.2
165800 300.3934 -392.345 8.70E+08 64.75105 64.19178 64.19178 266812.2
165900 300.3934 -392.602 8.84E+08 64.75749 64.18858 64.18858 266812.2
166000 300.3934 -392.863 8.98E+08 64.76393 64.18539 64.18539 266812.2
166100 300.3934 -393.129 9.11E+08 64.77037 64.1822 64.1822 266812.2
166200 300.3934 -393.402 9.25E+08 64.7768 64.17901 64.17901 266812.2
166300 300.3934 -393.686 9.39E+08 64.78324 64.17583 64.17583 266812.2
166400 300.3934 -393.973 9.53E+08 64.78968 64.17264 64.17264 266812.2
166500 300.3934 -394.27 9.66E+08 64.79612 64.16945 64.16945 266812.2
166600 300.3934 -394.57 9.80E+08 64.80256 64.16626 64.16626 266812.2
166700 300.3934 -394.876 9.94E+08 64.80899 64.16308 64.16308 266812.2
166800 300.3934 -395.19 1.01E+09 64.81543 64.15989 64.15989 266812.2
166900 300.3934 -395.514 1.02E+09 64.82187 64.1567 64.1567 266812.2
167000 300.3934 -395.844 1.04E+09 64.82831 64.15352 64.15352 266812.2
167100 300.3934 -396.183 1.05E+09 64.83474 64.15033 64.15033 266812.2
167200 300.3934 -396.533 1.06E+09 64.84118 64.14715 64.14715 266812.2
167300 300.3934 -396.886 1.08E+09 64.84762 64.14396 64.14396 266812.2
167400 300.3934 -397.244 1.09E+09 64.85406 64.14078 64.14078 266812.2
167500 300.3934 -397.61 1.10E+09 64.86049 64.1376 64.1376 266812.2
167600 300.3934 -397.98 1.12E+09 64.86693 64.13441 64.13441 266812.2
167700 300.3934 -398.359 1.13E+09 64.87337 64.13123 64.13123 266812.2
167800 300.3934 -398.743 1.15E+09 64.87981 64.12805 64.12805 266812.2
167900 300.3934 -399.137 1.16E+09 64.88625 64.12487 64.12487 266812.2
168000 300.3934 -399.537 1.17E+09 64.89268 64.12169 64.12169 266812.2
168100 300.3934 -399.946 1.19E+09 64.89912 64.11851 64.11851 266812.2
168200 300.3934 -400.358 1.20E+09 64.90556 64.11533 64.11533 266812.2
168300 300.3934 -400.776 1.21E+09 64.912 64.11215 64.11215 266812.2
168400 300.3934 -401.201 1.23E+09 64.91843 64.10897 64.10897 266812.2
168500 300.3934 -401.633 1.24E+09 64.92487 64.10579 64.10579 266812.2
168600 300.3934 -402.074 1.26E+09 64.93131 64.10261 64.10261 266812.2
168700 300.3934 -402.523 1.27E+09 64.93775 64.09943 64.09943 266812.2
168800 300.3934 -402.978 1.28E+09 64.94419 64.09626 64.09626 266812.2
168900 300.3934 -403.442 1.30E+09 64.95062 64.09308 64.09308 266812.2
169000 300.3934 -403.911 1.31E+09 64.95706 64.0899 64.0899 266812.2
169100 300.3934 -404.387 1.32E+09 64.9635 64.08673 64.08673 266812.2
169200 300.3934 -404.868 1.34E+09 64.96994 64.08355 64.08355 266812.2
169300 300.3934 -405.359 1.35E+09 64.97637 64.08038 64.08038 266812.2
169400 300.3934 -405.858 1.37E+09 64.98281 64.0772 64.0772 266812.2
169500 300.3934 -406.363 1.38E+09 64.98925 64.07403 64.07403 266812.2
169600 300.3934 -406.875 1.39E+09 64.99569 64.07086 64.07086 266812.2
169700 300.3934 -407.394 1.41E+09 65.00213 64.06768 64.06768 266812.2
169800 300.3934 -407.919 1.42E+09 65.00856 64.06451 64.06451 266812.2
169900 300.3934 -408.454 1.44E+09 65.015 64.06134 64.06134 266812.2
170000 300.3934 -408.992 1.45E+09 65.02144 64.05817 64.05817 266812.2
Hi Dear
The output in the fix deform command in my simulation is pressure/atom. there was no pressure fluctuations in fix deform command. With increasing deformation of cell the pressure/atom increase linearly. This is logical. The pressure/atom divided by volume to obtain stress in cell. But the values of stress are 100 times the real values. This is confusing. The output of fix deform is following:(fx is the summation of stress/atom)
Well, there are a few things that don’t add up here, but it is difficult to make any statements based on so little information.
First off, there is no such thing as a per atom pressure. Then, fix deform does not return any property as far as i know. And there doesn’t seem to be any time integration going on in your simulation, which would render the entire exercise pointless.
Axel.
Hi Dear Axel
The input script for calculation of elastic constant of vinylester/TiO2 is following:(My problem is why the calculated stress in deform fix command is 100 times real value?)
# LAMMPS input for titan/vinylester nanocomposites elastic constants analysis
#Initialization
boundary p p p
units real
atom_style hybrid molecular charge
# create geometry and potentials-------------------------------------------
neighbor 0.4 bin
neigh_modify every 10 one 10000
bond_style harmonic
angle_style cosine/squared
dihedral_style charmm
read_data 3Vr10.txt
pair_style hybrid buck/coul/cut 10.5 6 lj/cut/coul/cut 10.5 6 lj/cut 10.5
pair_coeff * 7 lj/cut 0 3.00
pair_coeff 7 * lj/cut 0 3.00
pair_coeff 8 9 buck/coul/cut 391049.1 0.194 290.331
pair_coeff 8 8 buck/coul/cut 717647.4 0.154 121.067
pair_coeff 9 9 buck/coul/cut 271716.3 0.234 696.888
pair_coeff 1 1 lj/cut/coul/cut 0.1094 3.816
pair_coeff 2 2 lj/cut/coul/cut 0.0860 3.816
pair_coeff 3 3 lj/cut/coul/cut 0.2104 3.442
pair_coeff 4 4 lj/cut/coul/cut 0.2100 3.3224
pair_coeff 5 5 lj/cut/coul/cut 0.0157 2.974
pair_coeff 6 6 lj/cut/coul/cut 0.0150 2.918
pair_coeff 1 2 lj/cut/coul/cut 0.09699 3.816
pair_coeff 1 3 lj/cut/coul/cut 0.15171 3.629
pair_coeff 1 4 lj/cut/coul/cut 0.15157 3.5692
pair_coeff 1 5 lj/cut/coul/cut 0.04144 3.395
pair_coeff 1 6 lj/cut/coul/cut 0.04051 3.367
pair_coeff 1 8 lj/cut/coul/cut 0.04312 3.4955
pair_coeff 1 9 lj/cut/coul/cut 0.15157 3.5692
pair_coeff 2 3 lj/cut/coul/cut 0.1345 3.629
pair_coeff 2 4 lj/cut/coul/cut 0.1344 3.5692
pair_coeff 2 5 lj/cut/coul/cut 0.03675 3.395
pair_coeff 2 6 lj/cut/coul/cut 0.03591 3.367
pair_coeff 2 8 lj/cut/coul/cut 0.03823 3.4955
pair_coeff 2 9 lj/cut/coul/cut 0.1344 3.5692
pair_coeff 3 4 lj/cut/coul/cut 0.2102 3.3822
pair_coeff 3 5 lj/cut/coul/cut 0.0574 3.208
pair_coeff 3 6 lj/cut/coul/cut 0.05618 3.18
pair_coeff 3 8 lj/cut/coul/cut 0.05980 3.3085
pair_coeff 3 9 lj/cut/coul/cut 0.2102 3.3822
pair_coeff 4 5 lj/cut/coul/cut 0.0574 3.1482
pair_coeff 4 6 lj/cut/coul/cut 0.0561 3.1202
pair_coeff 4 8 lj/cut/coul/cut 0.05974 3.2487
pair_coeff 4 9 lj/cut/coul/cut 0.2100 3.3224
pair_coeff 5 6 lj/cut/coul/cut 0.01534 2.946
pair_coeff 5 8 lj/cut/coul/cut 0.01633 3.0745
pair_coeff 5 9 lj/cut/coul/cut 0.0574 3.1482
pair_coeff 6 8 lj/cut/coul/cut 0.01596 3.0465
pair_coeff 6 9 lj/cut/coul/cut 0.0561 3.1202
region nanoparticle sphere 25.84 25.84 25.84 10.5 side in units box
group nanoparticle region nanoparticle
region resin sphere 25.84 25.84 25.84 10.5 side out units box
group resin region resin
velocity all create 300 1231
# equilibration stage 1 (NVT dynamics at 300 k )-------------------------------------------------------------------------
fix 1 all nvt temp 300 300 50
thermo_style custom step temp epair etotal press lx ly lz
timestep 0.5
thermo 1000
run 20000
unfix 1
#unfix 2
# equilibration stage 2 (NPT dynamics from 300 k —> 300 K )-------------------------------------------------------------------------
fix 1 all npt temp 300 300 50 iso 10 10 1000 drag 2
fix 2 all momentum 1 linear 1 1 1
compute a resin group/group nanoparticle
thermo_style custom step temp epair etotal press lx ly lz
timestep 0.5
thermo 100
dump 1 nanoparticle xyz 5000 nano.xyz
run 50000
unfix 1
unfix 2
# Tensile’s modulus calculation---------------------------------------------------------------------------------------------------------
fix 1 resin deform 1 x scale 1.01 y volume z volume units box remap x
compute px resin stress/atom
compute fx resin reduce sum c_px[1]
thermo_style custom step temp press c_fx lx ly lz vol
thermo 100
timestep 1
run 10000
unfix 1
Hi Dear Axel
The input script for calculation of elastic constant of vinylester/TiO2 is
following:(My problem is why the calculated stress in deform fix command is
100 times real value?)
this is far too complex an input to quickly pinpoint any issues. does
this run well for regular MD? have you discussed this and the choice
of settings with somebody that has more experience in MD than you? if
not, do it. you should particularly discuss the issues of
equilibration and time scales.
i am not surprised that this gives bogus results since there are
several rather unusual and questionable choices. i strongly suggest to
first practice doing this kind of study with a much simpler to model
material and reproduce those results and then gradually increase the
complexity, making sure that you get correct results at every step.
building and debugging such a complex input for such a complex system
in one go is close to impossible and you will for certain not find
anybody on this mailing list do it for you.
axel.
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CC-MAIN-2024-26
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http://www.dummies.com/how-to/content/finding-the-unit-vector-of-a-vector.navId-611287.html
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crawl-data/CC-MAIN-2016-36/segments/1471982292675.36/warc/CC-MAIN-20160823195812-00160-ip-10-153-172-175.ec2.internal.warc.gz
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Every nonzero vector has a corresponding unit vector, which has the same direction as that vector but a magnitude of 1. To find the unit vector u of the vector
you divide that vector by its magnitude as follows:
Note that this formula uses scalar multiplication, because the numerator is a vector and the denominator is a scalar.
A scalar is just a fancy word for a real number. The name arises because a scalar scales a vector — that is, it changes the scale of a vector. For example, the real number 2 scales the vector v by a factor of 2 so that 2v is twice as long as v.
As you may guess from its name, the unit vector is a vector.
For example, to find the unit vector u of the vector
you first calculate its magnitude |q|:
Now use the previous formula to calculate the unit vector:
You can check that the magnitude of resulting vector u really is 1 as follows:
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https://ccssmathanswers.com/mcgraw-hill-my-math-kindergarten-chapter-7-check-my-progress-answer-key/
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# McGraw Hill My Math Kindergarten Chapter 7 Check My Progress Answer Key
All the solutions provided in McGraw Hill My Math Kindergarten Answer Key PDF Chapter 7 Check My Progress will give you a clear idea of the concepts.
## McGraw-Hill My Math Kindergarten Chapter 7 Check My Progress Answer Key
Check My Progress Page No. (461- 462)
Vocabulary Check
Question 1.
eleven 11
Given that there are 8 buttons.
For 11 buttons draw 3 more buttons.
Therefore the number of buttons are 11.
Question 2.
fifteen 15
Draw the 15 objects in the boxes.
In first box draw 10 objects and in second box draw 5 objects.
Therefore the total number of boxes are 10 + 5 = 15.
Concept Check
Question 3.
In first box there are 10 glasses.
In second box there are 4 glasses.
The total number of glasses are 10 + 4 = 14.
Directions: 1. Count the objects. Draw more to show 11. 2. Color the boxes in the ten-frames to show 15. 3. Count 10. Color the objects red to show 10. Write the number. Count how many more. Color the objects yellow to show how many more. Write the number. Trace the number made.
Question 4.
In first box there are 10 bells.
Circle the 10 bells.
In second box there are 3 bells.
The total number of glasses are 10 + 3 = 13.
Question 5.
In first box there are 10 inserts.
Circle the 10 inserts.
In second box there are 2 inserts.
The total number of glasses are 12 + 2 = 12.
Question 6.
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https://www.mathassignmentexperts.com/multiple-integrals-assignment-help
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crawl-data/CC-MAIN-2021-49/segments/1637964361064.69/warc/CC-MAIN-20211202024322-20211202054322-00621.warc.gz
| 921,276,217
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## Multiple Integrals Assignment Help
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https://byjus.com/question-answer/a-runner-jogs-along-a-straight-road-in-the-x-direction-for-30-min-travelling/
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1
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Question
# A runner jogs along a straight road (in the +x direction) for 30 min, travelling a distance of 6 km. She then turns around and walks back towards her starting point for 20 min, travelling 2 km during this time.The runners average velocity while jogging is 0.4 km min−1.
A
True
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B
False
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Solution
## The correct option is B FalseAverage Velocity (Jogging) =450=0.08Km min−1 (FALSE).
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https://ch.mathworks.com/matlabcentral/answers/2121761-training-validating-and-testing-model?s_tid=prof_contriblnk
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Training, validating, and testing model.
4 views (last 30 days)
NeedHelp55 on 23 May 2024
Answered: Star Strider on 23 May 2024
Hi,
I have written three linear regression models and would like to know how to train, validate, and thest them with a random 60 20 20 split. any help would be much appreciated. here is the code.
PTT = filtered_BP_Data(:, 1);
systolicBP = filtered_BP_Data(:, 2);
diastolicBP = filtered_BP_Data(:, 3);
avgBP = filtered_BP_Data(:, 4);
lm1 = fitlm(PTT, systolicBP);
disp(lm1);
% plot PTT vs systolic BP
figure;
scatter(PTT, systolicBP, 'x');
hold on;
regressionLine1 = lm1.Coefficients.Estimate(1) + lm1.Coefficients.Estimate(2) * PTT;
plot(PTT, regressionLine1, '-r');
xlabel('PTT (ms)');
ylabel('Systolic Blood Pressure (mmHg)');
title('Average PTT vs Systolic BP');
legend('Data Points', 'Location', 'best');
grid on;
hold off;
lm2 = fitlm(PTT, diastolicBP);
disp(lm2);
% plot PTT vs diastolic BP
figure;
scatter(PTT, diastolicBP, 'x');
hold on;
regressionLine2 = lm2.Coefficients.Estimate(1) + lm2.Coefficients.Estimate(2) * PTT;
plot(PTT, regressionLine2, '-r');
xlabel('PTT (ms)');
ylabel('Diastolic Blood Pressure (mmHg)');
title('Average PTT vs Diastolic BP');
legend('Data Points', 'Location', 'best');
grid on;
hold off;
lm3 = fitlm(PTT, avgBP);
disp(lm3);
% plot PTT vs avg BP
figure;
scatter(PTT, avgBP, 'x');
hold on;
regressionLine3 = lm3.Coefficients.Estimate(1) + lm3.Coefficients.Estimate(2) * PTT;
plot(PTT, regressionLine3, '-r');
xlabel('PTT (ms)');
ylabel('Avg Blood Pressure (mmHg)');
title('Average PTT vs Avg BP');
legend('Data Points', 'Location', 'best');
grid on;
hold off;
Star Strider on 23 May 2024
Perhaps I am missing something in your problem statement, however linear regressions such as that done by fitlm is a least-squares approach and entirely deterministic, not ‘training’. There is nothing heuristic about it. You should get the same result (within the limits of floating-point calculation precision) every time you run it using the same data. If you use subsets of the entire data set, you will of course get different regression coefficients depending on the data you use, and they may be different (I would expect them to be different) from the values using the entire data set. Also, if you use the predict function to calculate the ‘regressionLine’ values, it will also supply the confidence limits on the regression.
With respect to the systolic, diastolic, and mean blood pressure, note that the mean blood pressure (averaged over the entire cardiac cycle) is approximately one-third the difference of the systolic and diastolic pressures, not their arithmetic mean, so —
SBP = 120;
DBP = 80;
meanBP = (SBP - DBP)/3 + DBP
meanBP = 93.3333
not —
avgBP = mean([SBP DBP])
avgBP = 100
.
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# Set-16 General Science (English) Railway NTPC and Railway Group D 2019 Previous Year Questions | Study Virus
Dear Aspirants,
We are providing the most important General Science Previous year Questions for RRB NTPC 2019, RRB Group D 2019, SSC 2019 and all other competitive exams. These questions have very high chances to be asked in RRB NTPC 2019, RRB Group D 2019.
Set-16 General Science Previous Year Questions
Download our GK Gaming Application “GKPK” (30000+) for upcoming RRB NTPC 2019,RRB Group D 2019.
1.Instrument for measuring rainfall is called –
1. Lucimeter
2. Galactometer
3. Hyetometer
4. Hygrometer
Ans:3
2.Who established the foundations of the quantum theory?
1. Max Planck
2. Mark Nicholas
3. Albert Einstein
4. Alfred Hitchcock
Ans:1
3. If in a motion, the axis of the rotation passes through an object, the motion is called _________.
1. Orbital motion
2. Circulatory motion
3. Spin motion
4. Oscillatory motion
Ans:3
4.What is the angle between centripetal acceleration and tangential acceleration?
1. 90°
2. 45°
3. 0°
4. 180°
Ans:1
5. Rectifiers convert _____.
1. high voltage to low voltage
2. low voltage to high voltage
3. AC to DC
4. DC to AC
Ans:3
6. Name the first Indian who got Nobel Prize in physics.
1. C K Naidu
2. Rangnath Mishra
3. Amartya Sen
4. C V Raman
Ans:4
7.Which physical quantity is measured in ‘siemens’?
1. Electric potential
2. Electrical conductance
3. Magnetic flux
4. Refractive index
Ans:2
8.Who invented Universal Standard Time?
1. Enrico Fermi
3. Sandford Fleming
4. Benoit Fourneyron
Ans:3
9._____________________ states that the total current entering a junction is equal to the total current leaving the junction.
1. Lenz’s Law
2. Hooke’s Law
3. Ohm’s Law
4. Kirchhoff’s First Law
Ans:4
10.Acceleration due to gravity on a planet decreases with ______.
1. decrease in radius of the planet
2. increase in mass of the planet
3. decrease in mass of the body
4. increase in altitude from surface of the planet
Ans:4
11.Instrument for measuring wind velocity is called –
1. Coulombmeter
2. Anemometer
3. Cyanometer
4. Chronometer
Ans:2
12.Who was the first Indian astronaut to travel in space?
1. Rakesh Sharma
2. Ravish Malhotra
3. Kalpana Chawla
4. Sheikh Muszaphar Shukor
Ans:1
13.AV= constant, where A= area of cross-section and V= velocity of fluid. This equation is called _____.
1. Equation of discontinuity
2. Equation of continuity
3. Equation of sustenance
4. Equation of parallelism
Ans:2
14.A body in equilibrium _______.
1. can move with constant acceleration
2. is always at rest
3. can move with constant velocity
4. can move with variable acceleration
Ans:3
15.What is the viscosity of an ideal fluid?
1. Equal to its mass
2. Equal to its weight
3. Zero
4. One
Ans:3
16.A parachute descends slowly whereas a stone dropped from the same height falls rapidly, because –
1. Stone is heavier than parachute
2. Special mechanisms are present in parachute
3. A parachute has a larger surface area and air resistance is more
4. None of the above
Ans:3
17.Shaving mirror is –
1. Convex
2. Concave
3. Plane
4. Parabolic
Ans:2
18.The spoon dropped by an astronaut in a satellite will –
1. Fall to the floor
2. Remain stationary
3. Continue to follow the motion of the satellite
4. Move tangentially away
Ans:3
19.“Curie” is the unit of –
2. Temperature
3. Heat
4. Energy
Ans:1
20.The Source of the sun’s energy, is the process of –
1. Photoelectric emission
2. Nuclear fission
3. Nuclear fusion
4. Thermionic emission
Ans:3
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## Geometric Series
Let $a,r\in\mathbb{R}$ be constants. Let $\{a_n\}$ be a sequence of real numbers with the following recursive definition:
$a_n=\begin{cases}a & \text{ if }n=1\\ r\cdot a_{n-1}&\text{ if }n>1\end{cases}$.
Therefore the sequence is given by:
$a,ar,ar^2,ar^3,ar^4,\dots$
Such a sequence is called a geometric sequence with common ratio $r$.
When we add up the terms a sequence we have a geometric sum:
$S_n=a+ar+ar^2+ar^3+\cdots ar^{n-1}$.
Here $S_n$ is the sum of the first $n$ terms.
We can find a formula for $S_n$ using the following ‘trick’:
$r\cdot S_n=ar+ar^2+ar^3+\cdots ar^n$
$\Rightarrow a+r\cdot S_n-ar^n=S_n$
$\Rightarrow S_n(r-1)=a(r^n-1)$
$\displaystyle \Rightarrow S_n=\frac{a(r^n-1)}{r-1}$.
### Exercises
Assuming that $|r|<1$, find a formula for the geometric series
$\displaystyle S_{\infty}=\lim_{n\rightarrow \infty}S_n$.
## Binary Numbers
### Exercises
• Write the following as fractions:
$0.1_2,\,0.11_2,\,0.101_2$.
• Use infinite geometric series to show that:
• $0.111\dots_2=1$
• $0.0111\dots_2=\frac12$
• $0.101010\dots_2=\frac23$
## Doubling Mapping
The doubling mapping $D:[0,1)\rightarrow [0,1)$ is given by:
$\displaystyle D(x)=\begin{cases}2x & \text{ if }x<1/2 \\ 2x-1 & \text{ of }x\geq 1/2\end{cases}$.
### Exercises
• Find the first six iterates of the point $x_0=\frac17$ under $D$.
• Find the first four iterates of the point
$x_0=\frac{1}{2}+\frac{1}{2^2}+\frac{0}{2^3}+\frac{1}{2^4}=0.1101_2$.
• Where $x$ has the binary representation
$x = 0.a_1a_2a_3a_4a_5a_6a_7a_8\dots$ ,
write down expressions for $D(x)$ and $D^5(x)$.
• Hence find points $y, z \in [0, 1]$ such that $y$ and $z$ agree to 5 binary digits but $D^N(y)$ and $D^N(z)$ differ in the first binary digit for some $N \in \mathbb{N}$.
• Describe the period-5 points of $D$.
• Let $w \in [0, 1]$ have a binary representation beginning $w = 0.01001\dots$ . Find a period-5 point $\gamma$ of $D$ such that $w$ and $\gamma$ agree to five binary digits.
• Find a $\delta \in [0, 1]$ such that there are iterates of $\delta$, $D^{n_1}(\delta),D^{n_2}(\delta),D^{n_3}(\delta)$, with $n_1, n_2, n_3 \in \mathbb{N}$, that agree with 0.111 , 0.101, and 0.010, to three binary
digits.
## Sensitivity to Initial Conditions
### Exercise
Let $f(x)=4x\cdot (1-x)$. Where $[0,1]$ is the set of states, and $f:[0,1]\rightarrow [0,1]$ the iterator function, by looking at the first seven iterates of $x_0=0.8$ and $y_0=0.81$, show that this dynamical system displays sensitivity to initial conditions [HINT:4*ANS*(1-ANS)]
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# Thread: the expectation of the sum of yi minus ybar squared
1. ## the expectation of the sum of yi minus ybar squared
In general what is the expectation of the sum of yi minus ybar squared
$\displaystyle E(\sum( y_i - \bar{y})^2) = E(\sum( y_i^2 - 2\bar{y}y_i + \bar{y}^2)) =$
$\displaystyle E(\sum y_i^2 -\sum \bar{y}^2)) = \sum E(y_i^2) -\sum E(\bar{y}^2)$
is this the same as the sum of of VAR(Y)?
2. ## Re: the expectation of the sum of yi minus ybar squared
Hey kingsolomonsgrave.
What do you mean by sum of Var[Y]? Can you write it out in terms of sigma notation and expectation/variance of random variables?
3. ## Re: the expectation of the sum of yi minus ybar squared
var(Y) in terms of expectation is $\displaystyle E (Y_i - \bar{Y)}^2 = E(Y^2)-[E(y)]^2$
and
$\displaystyle E(\sum (Y_i -\bar{Y})^2) = E[\sum(Y_i^2 -2Y_i\bar{Y} +\bar{Y})]$
$\displaystyle =\sum[E(Y_i ^2) - [E(\bar(Y))^2] = \sum(VAR(Y)) = nVAR(Y)$
4. ## Re: the expectation of the sum of yi minus ybar squared
This looks right and remember also that expectation is linear so if you have a sigma inside the expectation, you can take it outside the expectation as well.
,
,
,
,
,
,
,
### yhat minus ybar show that is equal to zero
Click on a term to search for related topics.
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# 直圓錐上截橢圓
TOP
## 回復 1# 瓜農自足 的帖子
\begin{align} & {{x}^{2}}+{{y}^{2}}=1 \\ & {{\left( -3x+4y \right)}^{2}}\le \left[ {{\left( -3 \right)}^{2}}+{{4}^{2}} \right]\left( {{x}^{2}}+{{y}^{2}} \right)=25 \\ & -5\le -3x+4y\le 5 \\ \end{align}
Z之最大值為20，最小值為10
\begin{align} & \left\{ \begin{align} & 20=-3x+4y+15 \\ & \frac{-3}{x}=\frac{4}{y} \\ \end{align} \right. \\ & \left\{ \begin{align} & 10=-3x+4y+15 \\ & \frac{-3}{x}=\frac{4}{y} \\ \end{align} \right. \\ \end{align}
TOP
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Bidirectional path tracing and pinhole camera
This topic is 3183 days old which is more than the 365 day threshold we allow for new replies. Please post a new topic.
Recommended Posts
I'm implementing a bidirectional path tracer as described in chapter 10 of the Veach thesis: http://graphics.stanford.edu/papers/veach_thesis/ Now my problem is how to calculate the probability density function (PDF) with respect to projected solid angle for the primary ray (z0->z1) for a pinhole camera. This is important to calculate the relative probabilities with which the other sampling techniques would have created the same path and needed to calculate the weight for the contribution of the path. According to equations (10.9) from the thesis the relative probabilities can be calculated. As an example consider the path generated by a (1,2) sampling technique. Then the ratio p2/p1 = ( P(x0->x1) * G(x0<->x1) ) / ( P(x2->x1) * G(x2<->x1) ), where P(x->x') is the PDF with respect to projected solid angle for extending the path at x into the direction to x'. Now the term P(x2->x1) is the same as P(z0->z1), where z0 is the location of the camera (and origin of all primary rays) and z1 is the vertex hit by the primary ray. So how to compute the value of P(z0->z1) ? [Edited by - nmi on February 27, 2010 2:05:06 AM]
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# Positioning problem
## Recommended Posts
Hi all I'm getting started with phaser and I was working on a candy crush like game, I found this really cool tutorial http://www.joshmorony.com/how-to-create-a-candy-crush-or-bejeweled-game-in-phaser/ and it all works great I even manage to add on some particle effects and I'm starting to play with screen sizes etc, not I have a huge problem, I have been looking at it for over a week and no luck, basically my problem es the placement of the tileGrid (see code below) the problem is that this is generated on the (0,0) point but I would like to move it a little bit down and maybe even be able to center it, is this something that is possible to do? my whole game is base on tutorial except for all the extra stuff that I'm adding and changing, please help.
`me.tileGrid = [`
` ``[``null``, ``null``, ``null``, ``null``],`
` ``[``null``, ``null``, ``null``, ``null``],`
` ``[``null``, ``null``, ``null``, ``null``],`
` ``[``null``, ``null``, ``null``, ``null``],`
` ``[``null``, ``null``, ``null``, ``null``],`
` ``[``null``, ``null``, ``null``, ``null``],`
` ``[``null``, ``null``, ``null``, ``null``],`
` ``[``null``, ``null``, ``null``, ``null``]`
`];`
##### Share on other sites
I have achieved this in the past by calculating the grid size, and subtracting it from your center point. So lets say you want the centre of your grid to sit at 300 x 300. You may use something like this. (a few presumptions have to be made in order to achieve this).
``````centerGrid(row, col, centerX, centerY) {
// row = 4;
// col = 5;
// centerX = 300;
// centerY = 300;
let gridWidth = 64;
let gridHeight = 64;
let gridSpace = 20;
let gridMaxWidth = (gridWidth + gridSpace) * row; //336
let gridMaxHeight = (gridHeight + gridSpace) * col; // 420
let gridCenterWidthX = gridMaxWidth / 2; // 168
let gridCenterHeightY = gridMaxHeight / 2; // 210
let gridCenterX = gridCenterWidthX + centerX; // 468
let gridCenterY = gridCenterHeightY + centerY; // 510
return {x: gridCenterX, y: gridCenterY};
}``````
You can make this much simpler however, by performing the calculations in one line like so:
``````centerGrid(row, col, centerX, centerY, gridWidth, gridHeight, gridSpace) {
let gridCenterX = (((gridWidth + gridSpace) * row) / 2) + centerX);
let gridCenterY = (((gridHeight + gridSpace) * col) / 2) + centerY);
return {x: gridCenterX, y: gridCenterY};
}``````
##### Share on other sites
Omg thanks so so much I will try tonight
##### Share on other sites
• 1 month later...
May I ask how do I apply the calculations to the Grid?
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Subject Re: [firebird-support] Re: Hourly rate Martijn Tonies 2005-10-23T09:35:36Z
> > >TIME is not a cosmic static point!
> >
> > A TIME value represents a static point in a day (according to the
> 24-hour
> > clock convention observed by many mortals and computing devices on
> this
> > planet).
>
> Yup, meaning of static is depends on perspectif.
> It is static becouse the countries static.
> It is not static because, humans are portable with some objects.
> It is static because local users focused on local time.
> It is not static because there is Daylight savings and time tunings
> about world turning around speed lost.
> etc etc..
I fail to see what this has to do with the TIME datatype?
> > In Firebird, TIME is *stored* as a 32-bit unsigned integer (not
> BigInt),
> > representing the number of ten-thousandths of a second that have
> elapsed
> > since midnight. It will never be higher than 863,999,999 because
> it is
> > constrained to this limit. Internally, this number is handled as
> DECIMAL(9,4).
>
> I know this. it is what i said to Martijn.
> TIME is a numeric value >= 0 and <86400
TIME - the datatype - is NOT a numeric value, it's "time of day" as Helen
explained. How it is stored internally does not mean that "dividing" a TIME
will give you a TIME as a result, although you can assign the result to the
internal storage mechanism of TIME.
08:00 divided by 2 does not give you "time of day" 04:00 ... Yes, you can
assign it internally, on the low-level of "numeric value", but it's not
right.
> How can you will calculate it without decimal value?
>
> Some digital systems can keep TIME parts in different registers.
>
> Time.Hours register between 0 and 23
> Time.Minutes Register between 0 and 59
> Time.Seconds Register between 0 and 59
> Time.Milliseconds Register between 0 999
>
> Milliseconds counter register feeding by freq. oscilator.
> when it become 1000, logic gates resets it to zero and
> adds 1 to Seconds,
> when seconds become 60, logic gates resets it to zero and
> adds 1 to minutes
> ...
All this has nothing to do with maths on TIME.
> Or there is only one big integer register counts oscilator signals.
> You can build TIME or TIMESTAMP from this number.
>
> It is not importand how you keep time numbers.
> But, if you want to some calculation with it you must to be obey
> math rules.
There are different rules. The concept of TIME is such that
TIME cannot be 25:00 ... I said this before and this still holds.
> > And the Firebird engine does its TIME and DATE computations in C
> in order
> > to store and retrieve date/time data. But data-users do their
> computations
> > in SQL. SQL uses symbols taken from English, but that does not
> mean that
> > SQL is semantically mapped to English. In English, if we want to
> know the
> > time-of-day, we ask "What is the TIME?" It is that gloss that SQL
> has
> > borrowed (and not "Can you estimate the TIME it will take?")
> >
>
> Helen it is not about English or any other language.
> It is about Logic, Digital technology depends on math and other
> sciences.
> We are lowest level computer users, if we don't intersted about
> technical calculation details, who will does it?
> Gates/Ellison/Dell/Honda..?
>
> Don't say me SQL is not related about
> logic or technology or programming knowladge.
> You can't be good RDBMS user without modern math and programming
> knowlange.(algorythm)
>
> > In Firebird, in SQL, you can add a numeric(9,4) number n
> (seconds) to a
> > TIME to get a result of TIME type that is that time-of-day + n
> seconds (or
> > an overflow), you can subtract n seconds to get a TIME type result
> (or an
> > underflow) and you can subtract one TIME type from another TIME
> type to get
> > a difference interval in seconds (or an underflow).
> >
> > >if you know a quick way about TIME,DATE,TIMESTAMP calculation,
> > >i will happy to learn it from you.
> >
> > To do what? In SQL you can "add" a DATE type to a TIME type to
> give a
> > TIMESTAMP result. Under the hood, this concatenates the DATE
> > representation (also a constrained 32-bit integer with a scale of
> 4) to the
> > TIME representation. However, since this is a concatenation
> operation
> > (appending time-of-day to a date-only value), you
> cannot "subtract" a TIME
> > from a TIMESTAMP to get a DATE type.
> >
>
> It is written as possible in my book.
> You can substract a time value from a timestamp and can assign
> result to a date datatype. They are numeric values and i can play
> with them what i want. I dont need any high level support.
>
> You know, there was no TIME datatype before IBv6.
> we keept times in smallint fields with check constraints.
> and there was no DATE only datatype also.
> we keept dates in smallint fields with...
> and there was no extract also...
> we calculted them ourself.
Oh brilliant. Then you should know that there's no concept of
"time" that holds 25 hours. That's a duration/interval etc...
The "check constraints" should tell you you're using a specific
domain for your internal values. A domain that makes sense
for your "time value".
But just because some operation (eg: dividing) results in a numerical
value that is inside your constraints does NOT mean it's a valid
result on the conceptual level.
> > You can add a number n to or subtract a number n from a DATE or
> TIMESTAMP
> > to advance or reduce the value by n days. The stored value of a
> DATE or
> > TIMESTAMP can be negative: this will happen for anything earlier
> than
> > November 17, 1898. In resolving these numbers, the engine counts
> backwards
> > until it hits the lower limit of numeric(9,4), which is some point
> in time
> > around AD 10, at which point (if it hasn't finished) it will throw
> an
> > underflow exception. That means Firebird can't store dates
> earlier than
> > that limit. (Task for you, Ali: compute Firebird's lowest and
> highest dates!)
>
> heh! you saying this to me Helen???
> Instead of Martijn or Arnoldo? LOL!
Now, I asked you before: if you think I'm wrong, then proof me wrong.
With regards,
Martijn Tonies
Database Workbench - tool for InterBase, Firebird, MySQL, Oracle & MS SQL
Server
Upscene Productions
http://www.upscene.com
Database development questions? Check the forum!
http://www.databasedevelopmentforum.com
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FAQ
# what is the average distance in millions of kilometers from the sun to the asteroid belt
## What Is The Average Distance In Millions Of Kilometers From The Sun To The Asteroid Belt?
Located between Mars and Jupiter, the belt ranges in distance between 2.2 and 3.2 astronomical units (AU) from the Sun – 329 million to 478.7 million km (204.43 million to 297.45 million mi).Dec 2, 2016
## What is the average distance from the Sun to the asteroid belt?
between 2.2 and 3.2 astronomical units
The asteroid belt lies between 2.2 and 3.2 astronomical units (AU) from our sun. One AU is the distance between the Earth and sun. So the width of the asteroid belt is roughly 1 AU, or 92 million miles (150 million km).Nov 3, 2021
## How many million kilometers from the Sun is the asteroid belt?
The asteroid belt is between the planets Mars and Jupiter. It is located about 2.2 to 3.2 Astronomical Units (AU) from the Sun. That is somewhere between 329-478 million km away. The asteroid belt is huge and the space between each of the asteroids is over 600,000 miles.
## What is the average distance in millions of kilometers from the Sun to the asteroid belt quizlet?
A belt of asteroids is located an average distance of 503 million kilometers from the Sun.
## What is the approximate distance from the Sun to the asteroid belt quizlet?
The center of the asteroid belt if approximately 404 million kilometers from the sun. state the name of the planet that is closest to the center of the asteroid belt.
## What is 503 million km from the sun?
A belt of asteroids is located an average distance of 503 million kilometers from the sun.
## How far is the Kuiper Belt from the sun in KM?
4,400,000,000 to 14,900,000,000 km
The Kuiper Belt is a disk-shaped region past the orbit of Neptune, roughly 4,400,000,000 to 14,900,000,000 km (30 to 100 AU) from the Sun, that consists mainly of small bodies which are the remnants from the Solar System’s formation. It also contains at least one dwarf planet – Pluto.Oct 22, 2020
## Is the asteroid belt a failed planet?
A region between Mars and Jupiter became the asteroid belt. Occasionally people wonder whether the belt was made up of the remains of a destroyed planet, or a world that didn’t quite get started. However, according to NASA, the total mass of the belt is less than the moon, far too small to weigh in as a planet.
## What is the distance between the orbits of Mars and Jupiter?
No planet has a perfectly circular orbit around the sun and therefore the distance between each planet can take on two very extreme numbers. The AVERAGE distance between Mars and Jupiter is 3.68 AU. That’s approximately 550,390,000 km or 342,012,346 miles.
## How long would it take to get to the asteroid belt from Earth?
Adjusted for a trip to the Asteroid Belt, so a spacecraft equipped with an EM drive would take an estimated 32.5 days to reach the Asteroid Belt.
## Which planet has the longest day?
Venus
It was already known that Venus has the longest day – the time the planet takes for a single rotation on its axis – of any planet in our solar system, though there were discrepancies among previous estimates.May 3, 2021
## Why won’t Pluto collide with Neptune Why won’t Pluto collide with Neptune?
Why won’t Pluto collide with Neptune? Pluto orbits the Sun exactly 2 times for every 3 Neptune orbits, which ensures they never come close together. … Pluto is always much farther from the Sun than Neptune. Pluto orbits the Sun exactly 2 times for every 3 Neptune orbits, which ensures they never come close together.
## What are all the planets called?
The order of the planets in the solar system, starting nearest the sun and working outward is the following: Mercury, Venus, Earth, Mars, Jupiter, Saturn, Uranus, Neptune and then the possible Planet Nine.
## How much total mass is contained in the asteroid belt?
The total mass of the Asteroid belt is estimated to be 3.0 to 3.6×1021 kilograms, which is 4 percent of the Earth’s Moon. Of that total mass, one third is accounted for by Ceres alone. The high population makes for a very active environment, where collisions between asteroids occur very often (in astronomical terms).
## During which northern hemisphere season is Earth closest to the sun?
winter
However, in the Northern Hemisphere, we are having winter when Earth is closest to the Sun and summer when it is farthest away!
See also what does it mean for an enzyme to be denatured
## Why does no major planet orbit the sun at the location of the asteroid belt?
space debris left over after the formation of the planets. … Why does no major planet orbit the Sun at the location of the asteroid belt? Jupiter’s gravitational pull prevented the planetesimals in this region from coalescing into a single, large object.
## Is Comet a planet?
They range from a few miles to tens of miles wide, but as they orbit closer to the Sun, they heat up and spew gases and dust into a glowing head that can be larger than a planet. … Comets are cosmic snowballs of frozen gases, rock, and dust that orbit the Sun. When frozen, they are the size of a small town.
## Why is Pluto not a planet?
Answer. The International Astronomical Union (IAU) downgraded the status of Pluto to that of a dwarf planet because it did not meet the three criteria the IAU uses to define a full-sized planet. Essentially Pluto meets all the criteria except one—it “has not cleared its neighboring region of other objects.”
## How big is an asteroid?
Asteroids range in size from Vesta – the largest at about 329 miles (530 kilometers) in diameter – to bodies that are less than 33 feet (10 meters) across. The total mass of all the asteroids combined is less than that of Earth’s Moon.
## How is the distance to the Sun measured?
By measuring the time taken for the radar echo to come back, the distance can be calculated, since radio waves travel at the speed of light. Once this Earth-Venus distance is known, the distance between Earth and the Sun can be calculated.
## How do you calculate orbital distance?
Formula: P2=ka3 where: P = period of the orbit, measured in units of time. a = average distance of the object, measured in units of distance.
Formula: F = G M1M2/R2 where:
1. F = force of gravity.
2. M1,M2 = masses of the objects involved.
3. R = distance between their centers of mass (usually just their centers)
4. G = a constant.
## How far apart are objects in the Kuiper Belt?
Detached Kuiper Belt objects have orbits that never come closer to the Sun than about 40 AU. This sets them apart from most other KBOs, which spend at least part of their orbits in the region between 40 and 50 AU from the Sun.
## What is the planet that exploded?
In 1953, Soviet Russian astronomer I. I. Putilin suggested that Phaeton was destroyed due to centrifugal forces, giving it a diameter of approximately 6,880 kilometers and a rotational speed of 2.6 hours. Eventually, the planet became so distorted that parts of it near its equator were spun off into space.
## Is Pluto explode?
What happened to Pluto? Did it blow up, or go hurtling out of its orbit? Pluto is still very much a part of our Solar System, it’s just no longer considered a planet. In 2006, the International Astronomical Union created a new category for classifying bodies in space: the dwarf planet.
## Which planet is really hot and really cold?
On its sunny side, Mercury can reach a scorching 800 degrees Fahrenheit! (But Mercury is not the hottest planet in the solar system. The hottest planet is Venus.) On its dark side, Mercury gets very cold because it has almost no atmosphere to hold in heat and keep the surface warm.
## What is the distance between Jupiter and Saturn in km?
When Saturn and Jupiter are lined up, there is only a distance of 655 million km. That’s fairly close, from a planetary perspective. When Jupiter and Saturn are on opposite sides they are 2.21 billion km apart. During that alignment, all of the other planets are actually closer to Saturn than to Jupiter.
## How far are the planets from the sun in km?
Distance of the planets from the Sun
Planet Distance from the Sun Diameter
Venus 108,200,000 km (0.723 AU) 12,104 km
Earth 149,600,000 km (1.000 AU) 12,756 km
Mars 227,940,000 km (1.524 AU) 6,805 km
Jupiter 778,330,000 km (5.203 AU) 142,984 km
## What are the distances between the planets?
Planet distance table
From To AU
Venus Neptune 29.37
Earth Mars 0.52
Earth Jupiter 4.2
Earth Saturn 8.52
## How long does it take to get to Eris?
It was calculated that a flyby mission to Eris would take 24.66 years using a Jupiter gravity assist, based on launch dates of April 3, 2032, or April 7, 2044. Eris would be 92.03 or 90.19 AU from the Sun when the spacecraft arrives.
## How long would it take to get to Pluto from Earth?
New Horizons launched on January 19, 2006, and it’ll reach Pluto on July 14, 2015. Do a little math and you’ll find that it has taken 9 years, 5 months and 25 days. The Voyager spacecraft did the distance between Earth and Pluto in about 12.5 years, although, neither spacecraft actually flew past Pluto.
## How long is a day on Ceres?
9 hours
Ceres takes 1,682 Earth days, or 4.6 Earth years, to make one trip around the Sun. As Ceres orbits the Sun, it completes one rotation every 9 hours, making its day length one of the shortest in the solar system. Ceres’ axis of rotation is tilted just 4 degrees with respect to the plane of its orbit around the Sun.
## What planet has 100 hours in a day?
Just to be clear, this answer to ‘which planet has the longest day’ is based on this criteria: a planets day is how long it takes it to complete one rotation on its axis. This is also referred to as its rotational period. So, Venus has the longest day of any planet in our solar system.
## Which planet has largest moon?
Jupiter’s
One of Jupiter’s moons, Ganymede, is the largest moon in the Solar System. Ganymede has a diameter of 3270 miles (5,268 km) and is larger than the planet Mercury.
## What is the hottest planet?
Venus
Planetary surface temperatures tend to get colder the farther a planet is from the Sun. Venus is the exception, as its proximity to the Sun and dense atmosphere make it our solar system’s hottest planet.Jan 30, 2018
## Mysterious Objects in Space We Can’t Explain & Other Space Videos (Space Compilation)
Related Searches
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See more articles in category: FAQ
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# Questions on Algebra: Radicals -- complicated equations involving roots answered by real tutors!
Algebra -> Algebra -> Radicals -> Questions on Algebra: Radicals -- complicated equations involving roots answered by real tutors! Log On
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Algebra: Radicals -- complicated equations involving roots Solvers Lessons Answers archive Quiz In Depth
Question 38242: Find the following product and express answer in simplest radical form. All variables represent non negative real numbers. ((SQRT of 17a) + (SQRT of 7y))((SQRT of 17a)-(SQRT of 7y)) Click here to see answer by fractalier(2101)
Question 38471: The base of a triangle is (c+2) ft and the area is (2c*c-8) ft*ft. What is the height? Click here to see answer by Nate(3500)
Question 38515: Simplify the following radical expressions. Show all work! I dont understand the method to use to solve the problem. (squareroot 7+5)squared Click here to see answer by Alwayscheerful(414)
Question 38556: 32. Find all real or imaginary solutions. 9y^2 + 16 = 0 36. Show a complete solution. Two positive number differ by 11, and their square roots differ by 1. Find the numbers. Click here to see answer by lyra(94)
Question 38572: 36. Show a complete solution. Two positive numbers differ by 11, and their square roots differ by 1. Find the numbers. Click here to see answer by fractalier(2101)
Question 38572: 36. Show a complete solution. Two positive numbers differ by 11, and their square roots differ by 1. Find the numbers. Click here to see answer by AnlytcPhil(1277)
Question 38644: I have a homework problem i cant seem to figure out. I need to perform the operation, write the answer without negative exponents. Assume all variables represent positive numbers. (27a^3b)^-1/3(9a^-2b^2)^-1/2 Can anyone please help me with this, and show me in detail to break it down to get the answer :) Thanks Click here to see answer by Nate(3500)
Question 38644: I have a homework problem i cant seem to figure out. I need to perform the operation, write the answer without negative exponents. Assume all variables represent positive numbers. (27a^3b)^-1/3(9a^-2b^2)^-1/2 Can anyone please help me with this, and show me in detail to break it down to get the answer :) Thanks Click here to see answer by hellyeah(21)
Question 38666: A right triangle has its vertical side equal to 7, while the horizontal base is equal to 10 in length. Find the length x of the third side. Round your answer to the nearest tenth. Click here to see answer by checkley71(8403)
Question 38668: Solve the equation (w-3)^2 minus 36 = 0 where w is a real number. Simplify your answer as much as possible. Click here to see answer by checkley71(8403)
Question 38819: This is a question on my Algebra Homework. I can't understand how to figure it out. My original answer was "-13xthe square root of 2. My teacher told me that was wrong. Can you help me FAST with the answer to 3x(the square root of)2-7x(the square root of)9+5x(the square root of)2. Thank You for your help Click here to see answer by Nate(3500)
Question 38944: Solve this problem please. 5 radical 6 minus 1 all over 5 and i was wondering if you could simplify any. Thank You! Click here to see answer by fractalier(2101)
Question 38970: Hi,can you please explain in detail and help solve these? X:9^x=3 x:2^9x-4=32 4 ___ Express as a product: log 3 /x Thank you so much! Jackee Click here to see answer by stanbon(57347)
Question 38983: I was wondering if you could simplify this problem any more. 5 radical 6 minus 5 all over 5. Please e-mail back a simplified answer ASAP. thank you! I was also wondering if there are two consecutive even interegers whose sum is 84. Thank You also! Click here to see answer by Nate(3500)
Question 38983: I was wondering if you could simplify this problem any more. 5 radical 6 minus 5 all over 5. Please e-mail back a simplified answer ASAP. thank you! I was also wondering if there are two consecutive even interegers whose sum is 84. Thank You also! Click here to see answer by checkley71(8403)
Question 38973: Okay, I am so lost. Please help me with these? Write as the sum or difference of a single logarithms of x,y and z: __ log 8{(3x^3/y ) /z^5 } and Solve for x: log10^7=log 10^x-log10^2 Please help me in detail so I can understand these? Thank you. Click here to see answer by Nate(3500)
Question 39093: 4x(squared) - 17x + 4 = 0 Solved by completing the square. Not sure what to do with the coefficient 4 Click here to see answer by fractalier(2101)
Question 39133: i'm glad i found you guys. i need some help solving 3x(x=4)=0. Can I get a step-by-step? Thanks :^) Click here to see answer by fractalier(2101)
Question 39139: Could you help me rewrite the expression 2sqrt3+sqrt2/sqrt3-sqrt2 so there are no radicals in the denominator? I guess we could say rationalize the denominator. Thank you so much. Click here to see answer by fractalier(2101)
Question 38663: Simplify the following expression as much as possible: 3 square root of 48 x square root of 12 Click here to see answer by fractalier(2101)
Question 39148: ```I need help with this. Simplify the expression Simplify the expression. __ __ _ Ö63 - 2Ö28 + 5Ö7``` Click here to see answer by Nate(3500)
Question 39148: ```I need help with this. Simplify the expression Simplify the expression. __ __ _ Ö63 - 2Ö28 + 5Ö7``` Click here to see answer by AnlytcPhil(1277)
Question 39147: Please help with this. Find the time required for an object to fall to the ground from a building that is 1400 ft. high using the formula t = 1/4 sqrt s in which s is the distance fallen. Click here to see answer by Nate(3500)
Question 39166: How do you solve 3 over the square root of 48 Click here to see answer by checkley71(8403)
Question 39193: The access ramp to an expressway is 850 feet long. The measure of the angle of elevation is 4 degrees.About how long is the expressway from the horizontal at the beginning of the ramp? Click here to see answer by Nate(3500)
Question 39239: Please help me with this. Find the perimeter of the triangle if the three sides are: 4, sqrt 5 - sqrt 5, sqrt 5 + sqrt 5 Thanks Click here to see answer by AnlytcPhil(1277)
Question 39240: If one side of a rectangle is sqrt 3 + sqrt 5, and the second side is sqrt 3 + sqrt 5, find the area of the rectangle. Thanks Click here to see answer by fractalier(2101)
Question 39329: I posted this earlier incorrectly! Find the perimeter of the triangle if one side is 4, one side is sqrt 5 + sqrt 3, and the third side is sqrt 5 - sqrt 3. Thanks again! Click here to see answer by Nate(3500)
Question 39357: I found a problem in a text book, but I wanted to know if I was correct in determining that there was no solution to the problem. The problem is the cuber root of X=x-6; the book I found stated that the answer was 8, but I have tried every approach to solve this problem and maybe I am missing a step. I tried to take the difference of two cubes and set the equation equal to zero, but that didn't work. I also expanded the equation and set it equal to zero, but that didn't work either. I don't know how they came up with the answer 8, but I hope you can help. Click here to see answer by stanbon(57347)
Question 39373: Rationalize the denominator of each radical expression. Assume all variables represent nonnegative numbers and that no denominitors are 0. 1 + Square root 3 __________________ 3 Square root of 5 + 2 Square Root 3 Click here to see answer by Nate(3500)
Question 39381: Perform all indicated operations and write each answer with positive intgers exponents. -1 (m + n) ____________ -2 -2 m - n Click here to see answer by longjonsilver(2297)
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Stupid Möbius inversion problem
I feel that this is a very stupid question to be asking, but I can't figure it out. I've been trying to figure out the Möbius inversion formula, with pretty much no experience in this direction at all, and ran into the following problem:
So according to the Möbius inversion formula, $$g(n)=\sum_{d|n}f(d) \Longleftrightarrow f(n)=\sum_{d|n}g\left(\frac nd\right)\mu(d)$$. Yes? Since this is an equivalence, it should work both ways. Let's say $g(n)=1$, in that case, plug it into the right-hand formula and find that $f(n)=0$, or at least Wikipedia tells me that $\sum_{d|n}\mu(d)=0$. Problem: when we look at the formula on the left, we see that $g(n)$ is now equal to the sum of a finite number of zeros, even though we defined it as $1$. What stupid mistake am I making here?
• $\sum_{d\mid n} \mu(d) = 0$ for $n > 1$. But $\sum_{d\mid 1} \mu(d) = 1$. – Daniel Fischer Jul 16 '14 at 21:37
• Yes, but $n$ is not necessarily $1$, I think; it's the definition of $f(n)$, for $n \in \mathbb N$, so why should $n$ be restricted to $1$? – Laertes Jul 16 '14 at 21:39
• But $$f(1) = \sum_{d\mid 1} g\left(\frac{1}{d}\right)\mu(d) = g(1)\mu(1) = g(1) = 1.$$ For $n > 1$, you have $f(n) = 0$. – Daniel Fischer Jul 16 '14 at 21:40
• $f(n)$ is a function, it changes depending on $n$. So when $n=1$ you get $1$, only otherwise do you get $0$. – Adam Hughes Jul 16 '14 at 21:40
• But $g(n)$ is supposed to be $1$ for all $n$. – Laertes Jul 16 '14 at 21:43
The point is that
$$\sum_{d\mid n} \mu(d) = \begin{cases} 1 &, n = 1\\ 0 &, n > 1. \end{cases}$$
Thus, for $g \equiv 1$ you have
$$f(n) = \sum_{d\mid n} g\left(\frac{n}{d}\right)\mu(d) = \sum_{d\mid n} \mu(d) = \begin{cases} 1 &, n = 1\\ 0 &, n > 1 \end{cases}$$
and hence
$$g(n) = \sum_{d\mid n} f(d) = f(1) + \sum_{\substack{d\mid n\\ d > 1}} f(d) = f(1) + \sum_{\substack{d\mid n\\ d > 1}} 0 = f(1) = 1,$$
as it should be.
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# The prisoner was expedited from California to Florida.
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The prisoner was expedited from California to Florida. [#permalink]
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The prisoner was expedited from California to Florida.
(A) The prisoner was expedited from California to Florida.
(B) From California to Florida, the prisoner was expedited.
(C) The prisoner from California was extradited to Florida.
(D) The prisoner was extradited from California to Florida.
(E) From California, the prisoner was expedited to Florida.
Source: Master GMAT
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Joined: 10 Sep 2019
Posts: 2
Re: The prisoner was expedited from California to Florida. [#permalink]
### Show Tags
06 Dec 2019, 08:01
This is a really hard one.
C and D both seem good but only D conveys the meaning of the original statement. C had a different meaning.
Posted from my mobile device
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Posts: 455
Re: The prisoner was expedited from California to Florida. [#permalink]
### Show Tags
06 Dec 2019, 14:41
(A) The prisoner was expedited from California to Florida.
Expedited is the wrong word [make (an action or process) happen sooner or be accomplished more quickly]
(B) From California to Florida, the prisoner was expedited.
As above
(C) The prisoner from California was extradited to Florida.
This makes sense. The prisoner was extradited to Florida with from California as a modifier. It would be nice to know where she was extradited from though. Don't think there's anything grammatically wrong with this sentence.
(D) The prisoner was extradited from California to Florida.
Again this makes sense - sent from Cali to Flo. It also makes more sense than C, but isn't grammatically any better. Poor way to differentiate, but D is better than C
(E) From California, the prisoner was expedited to Florida.
Active voice is preferred over passive voice. D
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Re: The prisoner was expedited from California to Florida. [#permalink]
### Show Tags
07 Dec 2019, 03:24
Guys! is it really a GMAT type question. It seems the word meaning is the key to solve. Does GMAT test Vocabulary in SC questions.
Re: The prisoner was expedited from California to Florida. [#permalink] 07 Dec 2019, 03:24
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# Galois and Sequencing
It is not often anyone will hear the phrase “Galois field” and “DNA” together, but this paper from my colleagues, Tim Massingham and Nick Goldman provide a great link between these topics. Some other authors have used Galois fields in DNA analysis, but this is the first time I have seen a practical application of this level of mathematics in bioinformatics. It’s a tour de force by Tim, and although only in a lowly BMC Bioinformatics journal I think should be celebrated for its sheer chuptaz in cross scientific – indeed academic – domains.
If you have not met Galois field, then a small crash course in some pretty dense pure maths. Fields are the rather glorious, whimsical world of pure maths where one gets to fool around with the fundamental of maths – in this case redefining addition and multiplication; if certain conditions are met (that one can “add” and “multiply” in any order, and that multiplying an addition is the same as adding together the multiplication of each element, and a couple of other requirements), one has a “Ring”, (with the wonderful Tolkein like world of ring theory). With a couple more criteria to meet, in particular that multiplication doesn’t care about the order one has a Field. Numbers are fields – real numbers, complex numbers, rational numbers… but so are all sorts of other things – vectors, and exotic beasts like p-adic numbers which somehow involve primes in a suitably Alice-in-wonderland like way. Importantly, one can also have finite element fields, in which there is a limited number of elements. A gifted, young frenchman, Galois, explored the properties of these finite elment fields and showed that there is a limited number of fields – in fact, for finite elements with 2, 3 or 4 members there is only one field – ie, only one way to define “addition” and “multiplication” and satisify the criteria of a Field. (If you are curious, the 2 element finite field is like an XOR and an AND in logic for “add” and “multiply”). Many wonderful and deep things have been proved using Galois Fields, in both pure and applied maths.
So much for Maths. Now onto sequencing chemistry.
At the EBI we were funded to explore Lifetech’s new “Exact Call Chemistry” (ECC). Normal Lifetech Solid chemistry reads bases in pairs, where two adjacent bases gives one read out. Because there are 16 possibilities for two adjacent bases, but only 4 fluorophore read outs, each read out is ambigous, representing two possible scenarios. The sequence of these ambigous calls is called “Colorspace”, with the set {0,1,2,3} to distinguish it from the underlying bases (“Basespace” in Solid-speak, with {A,C,T,G} ). As the very first, primer, base is known due to the way the Solid chemistry works, this means one can in theory work out the next base (first in the read), and chain down the entire read. But people rarely do this because if you make an error in one position, the error propagates throughout the rest of the read. It is far more appropriate to do all the calculations (such as alignment, snp calling and even assembly) in colorspace, and then “key” the answer into basespace right at the end. A whole host of tools have sprung up around the colorspace world.
Exact Call Chemistry added another ligation-read step where rather than interrogating the sequence in adjacent pairs, interrogated it in a series of read-2, skip 1, read 1, skip 1 – a complex overlay. This pattern was chosen for its error correcting properties, and had the useful side effect that one could easily map the read directly into basespace. Using this though much of the sequencing errors can be corrected, meaning that one could get to something like 10-6 error rate on the chemistry. But this poses lots of questions – the error model is no longer a simple process associated with each base, rather one has to take a rather gestalt view of the error process across the entire read.
But how does one do this? How does one represent the combination of this colorspace plus this 2on-1off-1on-1off read? Here Tim rather beautifully brings in a Galois Field – remember that there is only one 4 element Galois field, and the design of colorspace allows for a one to one mapping of the 4 elements, traditionally called {0,1,a,b} (or sometimes alpha and beta) to both colors {0,1,2,3} and bases {A,T,G,C}. The color that occurs between two bases is just “addition” in the Galois field. This is a very elegant way to consider colorspace, but really comes into its own for ECC space. The additional 7th read of ECC space with this complex structure is a type of matrix multiplication in the Galois field. (at this point by the way my mathematically abilities have been stretched to breaking point by Tim and Nick and I just have trust them). By using this transformation therefore a lot of the things you might want to do with ECC that can be now written down as “straightforward” maths in this transformed space.
So now Tim can explore the impact say of an error model considering all the separate reads independently (what he calls a “trellis” model), or other representations of the data, most obviously a base-space model of the data with independent error – ie, the traditional “fastq” model. Unsurprisingly the trellis model gives you far more information in, say, calling SNPs than the “fastq” model, because the underlying data has complex interdependancies between the errors – for example, a mistaken T for a G in position 1 implies also a particular error in position 3. However, a different representation, corrected colorspace, in which one stores an errorcorrected colour space read in fact maintains the majority of the information, but is far more compact. Furthermore, as it is error corrected, it will compress better into reference based schemes (something we’re pretty obessed with at the EBI: see previous blog posts), and, indeed this compression is best thought of as a compression of the Galois field elements on the reference, which “naturally” compresses well. There are a bunch of other things that Tim can do in this framework, for example understand the number of errors that can be detected (up to 2) and the number that can be definitely corrected (only 1).
It’s not clear how much of a future ECC chemistry has out there. Lifetech recently bought ion torrent (with a completely different, 454 like chemistry, with very different error properties). It’s clear to me that if colorspace and friends (like ECC) was the only way to sequence DNA, we’d all know this backwards, and “Galois field” would become as commonplace as “Dynamic programming” in bioinformatics circles. But whatever the future of the chemistry, it’s great to see a relatively deep piece of maths (and pretty modern – from the 17th Centuary – in terms of pure maths) being used in a totally profound way in bioinformatics. I have no doubt that over time biology and bioinformatics will end up using more and more of the mathematical toolkit developed over the years.
Who knows, we (bioinformatics) might even inspire the development of new maths sometime in the future.
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File Exchange
## deg2utm
version 1.0.0.0 (2.61 KB) by
Function to convert lat/lon vectors into UTM coordinates (WGS84)
Updated 17 Aug 2006
This function is based on UTM.m function by Gabriel Ruiz Martinez, but instead of providing a GUI it works with vectors of coordinates.
[x,y,utmzone] = deg2utm(Lat,Lon)
% Example 1:
% Lat=[40.3154333; 46.283900; 37.577833; 28.645650; 38.855550; 25.061783];
% Lon=[-3.4857166; 7.8012333; -119.95525; -17.759533; -94.7990166; 121.640266];
% [x,y,utmzone] = deg2utm(Lat,Lon);
% fprintf('%7.0f ',x)
% 458731 407653 239027 230253 343898 362850
% fprintf('%7.0f ',y)
% 4462881 5126290 4163083 3171843 4302285 2772478
% utmzone =
% 30 T
% 32 T
% 11 S
% 28 R
% 15 S
% 51 R
%
% Example 2: If you have Lat/Lon coordinates in Degrees, Minutes and Seconds
% LatDMS=[40 18 55.56; 46 17 2.04];
% LonDMS=[-3 29 8.58; 7 48 4.44];
% Lat=dms2deg(mat2dms(LatDMS)); %convert into degrees
% Lon=dms2deg(mat2dms(LonDMS)); %convert into degrees
% [x,y,utmzone] = deg2utm(Lat,Lon)
%
### Cite As
Rafael Palacios (2020). deg2utm (https://www.mathworks.com/matlabcentral/fileexchange/10915-deg2utm), MATLAB Central File Exchange. Retrieved .
RAJIVE KRISHNAN
Unable to perform assignment because the indices on the left side are not
compatible with the size of the right side.
Error in deg2utm (line 131)
utmzone(i,:)=sprintf('%02d %c',Huso,Letra);
Error in Untitled2 (line 1)
[x,y,utmzone] = deg2utm(Lat,Lon)
Sir this is not working for me in matlab R2019b.
Very nice function, and easy to use!
However, I see a little deviation compared to other calculations available online. For instance at https://www.kartverket.no/en/Maps--Nautical-Charts/transform-coordinates/
If I convert the following:
Lat_DMS = 10.09
Lon_DMS = 59.04
deg2utm gives:
E: 562548.596426461
N: 6545016.26221026
While https://www.kartverket.no/en/Maps--Nautical-Charts/transform-coordinates/ gives:
E: 562548.596
N: 6545016.276
As it can be seen there is a slight deviation on the north coordinate, do you know why?
cn1992
Hello, I wish to the equations in details, can someone provide comments or documents related to the equations.
Jonathan Demmer
How can we choose the utmzone, because when i run it the function the wrong utm zone which give me wrong results....
BlueEyes
Solved your mismatch, Lea. The utm2deg function values should be filled with decimals, as shown below where I've replicated the author's example. Cheers
----------
1.. Direct function:
>> format long g
>> Lat=[40.3154333; 46.283900; 37.577833; 28.645650; 38.855550; 25.061783];
>> Lon=[-3.4857166; 7.8012333; -119.95525; -17.759533; -94.7990166; 121.640266];
>> [x,y,utmzone] = deg2utm(Lat,Lon)
x =
458730.69991953
407652.894658174
239026.690338915
230253.368028571
343898.229457034
362850.016911846
y =
4462881.35112379
5126289.79627619
4163082.87564856
3171843.30844493
4302284.74761666
2772478.3578643
utmzone =
30 T
32 T
11 S
28 R
15 S
51 R
2.. Reverse function
format long g
x=[ 458730.69991953; 407652.894658174; 239026.690338915; 230253.368028571; 343898.229457034; 362850.016911846];
y=[4462881.35112379; 5126289.79627619; 4163082.87564856; 3171843.30844493; 4302284.74761666; 2772478.3578643];
utmzone=['30 T'; '32 T'; '11 S'; '28 R'; '15 S'; '51 R'];
[Lat, Lon]=utm2deg(x,y,utmzone)
Output
Lat =
40.315433306332
46.2839000044623
37.5778330030868
28.6456500029124
38.8555500050597
25.0617830039778
Lon =
-3.48571659470842
7.80123331162143
-119.955249956191
-17.7595329620273
-94.799016576718
121.640266014044
----------
BlueEyes
Hi, Lea, in your test you mentioned the function utm2deg -> [Lat Lon] = utm2deg(x,y,zone)
which is unavailable here, where deg2utm is uploaded. Please, could you post it in order to allow someone (like me) to replicate your test?
Thanks
Lea Lange
I need to convert deg to utm and utm back to deg again. This is why I checked, what happens if I convert from one to the other over and over again:
Lat=54+47/60
Lon=8+16/60
for i=1:10
[x y zone] = deg2utm(Lat,Lon);
disp([num2str(i) ' ' num2str(x) ' ' num2str(y)]);
[Lat Lon] = utm2deg(x,y,zone);
end
1 452837.7301 6070927.9726
...
10 452837.7325 6070927.9747
This is not an issue to my work since the error is negligible in the scope of my work, but I am wondering if this is due to Matlab limitations in precision or due to the formulation of the transformation rule (or any other reason)?
Thanks
Desta Ekaso
how can i get it running because I wrote [x,y,utmzone] = deg2utm(355497.97,5787938.89) on the command window and its is giving me an error message
Excellent code
Colleen Black
Excellent and easy to use!
Aideliz Montiel
Como descargo la función?
Same issue as tuan,
more than one utm zone gives
wrong X coordinates.
can someone help.
Fabio Retorta
where is the function mat2dms and dms2deg?
tuan
It works fine for an area with single utmzone. But for area with location has more than one utmzone it produces X (East direction) not correctly.
Could anyone help me to figure out?
Thanks
It was okay.. My mistake! Thanks
Federico
The function works perfectly, but I have doubts on the outputs: which one represents easting (i.e. longitudinal information) and which one the northing (latitudinal information)?
Kyle
Oops, I got it to work - - just my newness to the software.
Forgot to specify the outputs I wanted.
For people like me (getting into matlab), make sure to put your outputs you want in the command window.
IE: [x,y,utmzone] = deg2utm(Lat,Lon), where if you want x, y, and the utmzone you have to tell matlab that. Else it will only give you x (which is what I was getting before hand).
EDU>> [x,y,utmzone] = deg2utm(38.130587, -99.081752)
x = 4.9284e+05
y = 4.2203e+06
utmzone = 14 S
Aleksander
Hi.
i am unsure if this is a issue or not.
if i input the following:
[x,y]=deg2utm([5153.7870,5153.7870],[00425.8690,00425.8685])
i expect that the x output is the samme for both cordinates, and that i have a diferense of only 0.58m in the y, how ever the diference in x is about 1M and in y about 22,5M. any thoughts ?
jack
thank you very much.
r nawaz
Henry:
I have converted column 1 and 2, the deg2utm works perfectly fine. The lat/longs are from UTM zone 16 to zone 14 (from top to bottom). Note that easting suddenly increases as you move from a higher UTM zone number to lower one. There is not much change in Lats so that zone northing remains from letter Q. Column 3 and 4 has wrong transformed values.
Saúl
Henry
I have problems converting the 1st and 2nd columns. The 3rd and 4th columns are the correct transformed values.. Any help?
18.9000000000000 -85.6000000000000 1016636.08476000 2097349.08510000
19.1000000000000 -86.2000000000000 952730.286652000 2117881.16418000
19.3000000000000 -86.7000000000000 899537.751857000 2138832.87892000
19.5000000000000 -87.2000000000000 846493.267934000 2159917.67197000
19.6000000000000 -87.5000000000000 814775.913438000 2170418.08636000
19.8000000000000 -87.9000000000000 772442.462010000 2191882.52618000
20.3000000000000 -88.8000000000000 677544.995001000 2246042.98172000
20.5000000000000 -89.8000000000000 573005.667765000 2267422.61128000
20.5000000000000 -91.0000000000000 447853.577293000 2267346.10988000
20.3000000000000 -91.7000000000000 374681.174733000 2245584.33154000
20.1000000000000 -92.2000000000000 322228.063533000 2223898.57802000
19.8000000000000 -93.2000000000000 217073.073883000 2192046.90591000
19.2000000000000 -94.1000000000000 121283.019710000 2127298.41915000
18.6000000000000 -94.9000000000000 35345.9851048000 2062679.39657000
17.9000000000000 -95.6000000000000 -40894.3144760000 1986930.91879000
17.2000000000000 -96.4000000000000 -128393.538814000 1911652.44510000
16.9000000000000 -97.0000000000000 -193630.694790000 1880339.46191000
thanks!
arg, sorry my fault, its no bug
Gerald Lodron Lodron
arg, sorry my fault, its no bug
Gerald Lodron Lodron
I found a bug, in this example the first point P95 is mapped correctly to google maps, but EP2 differs about 70 Meters from GPS coordinates in carthesian:
P95 = [15.425789 47.030778];
[P95_Carthesian(1) P95_Carthesian(2)] = deg2utm(P95(2), P95(1));
EP2 = [15.42604688057756 47.03090763127457];
[EP2_Carthesian(1) EP2_Carthesian(2)] = deg2utm(EP2(2), EP2(1));
S. A. van der Wulp
r nawaz
Anna
Hello!
I want to convert lat lon points in the Northern Shelf Sea (Europe) to coordinates which have the unit meters. So I thought using UTM coordinates was the best solution.
However, the area is quite big (1000km x 1000km) and contains different zones. So after converting, my points are not in the right order any more. So either it is simply not possible to have points of different zones in one data base or the script makes some mistake.
Can somebody help me with this issue?
Thank you. Anna
E Bobane
I have my Lat/Lon coordinates in Degrees, Minutes and Seconds (as in your Example 2). However I am missing the 'dms2deg' and 'mat2dms' functions. Are these in some toolbox or part of this submission.
Tony Gibb
Andy Gardos
Excellent Script! I wonder if you would provide some comments for the calculation section. I've compared some lat/lon data to actual x,y profiles; however, I've found that I must multiply the x and y valuse derived from deg2utm by approximately 3.2 to match the actual values?! Thanks! Again, great script!!!
Junseok Park
Excellent!! And Thank you. This is applied for Korea.
Anthony Liou
Great stuff~ Thank you very much.
Philippe Blondel
Very good ! Thank you very much !
oh kwoun
Thanks a lot!
##### MATLAB Release Compatibility
Created with R14SP3
Compatible with any release
##### Platform Compatibility
Windows macOS Linux
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MATLAB Answers
# Frequency sweeping in a sinusoidal signal
21 views (last 30 days)
Nneka Onubogu on 28 May 2021
Commented: Nneka Onubogu on 7 Jun 2021
I want to generate a chirp signal by sweeping my frequency from 0 kHz to 30 kHz. Can i achieve it this way?
a1=2.4;
a2=6.9*10^-13;
a3=5.1*10^-13;
b1=3.5*10^-7;
b2=2.6*10^-7;
b3=0.5;
dt=10^-9;
x(1)=0;
y(1)=0;
j =1*10^8;%iterations for accuracy
fs=1/dt; %sampling frequency
t=(0:j)*dt/2.4156e8;
y1 = wgn(j,1,-35,50); %white noise
for k=20.5 % pumping power
for i=1:j
pinitial=(k/10)*1e19; %pump power
p(i)=(2.7324*10^-23)*pinitial*(1);
x(i+1)=(((x(i)*y(i))/b2)-(((a1+y1(i))/b2)*x(i))+((a2/b2)*y(i))+(a3/b2))*dt+x(i);
y(i+1)=(-((x(i)*y(i))/b2)-((b1/b2)*y(i))-1+((p(i)/b2)*(1-b3*(1))))*dt+y(i);
end
[bb,aa]=pwelch(x./9.8317e-17,[],[],[],fs); % aa is frequency, bb is amplitude
figure
semilogy(aa(1:3357),bb(1:3357))
grid minor
end
for fp = 1000+((30000-1000)/(j*dt/3.33e-8))*t(1:end-1)
for u=0 % amplitude (2nd order)
f2amp=u/100;
for r=40 % amplitude (1st order)
f1amp=r/100;
for i=1:j
p(i)=(2.7324*10^-22)*pinitial*(1+(f1amp*sin(2*pi*dt*i.*fp/1))+(f2amp*sin(2*pi*dt*i.*fp/2)));
x(i+1)=(((x(i)*y(i))/b2)-(((a1+y1(i))/b2)*x(i))+((a2/b2)*y(i))+(a3/b2))*dt+x(i);
y(i+1)=(-((x(i)*y(i))/b2)-((b1/b2)*y(i))-1+((p(i)/b2)*(1-b3*(1))))*dt+y(i);
end
end
end
end
downsampling=1;
xx=downsample(x./9.8317e-17,downsampling);
[bbb,aaa]=pwelch(xx,[],[],[],fs/downsampling);
figure()
subplot(2,1,1)
plot(t,xx) % time domain
grid minor
subplot(2,1,2)
semilogy(aaa(1:3357),bbb(1:3357))% frequency domain
grid minor
##### 0 CommentsShowHide -1 older comments
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### Accepted Answer
William Rose on 31 May 2021
@Nneka Onubogu, You can initialize the vectors p, x, and y before the loop as follows:
dt=4e-6; %(s)
fs=1/dt; %(Hz)
j=25000; %make this 250K if you want 1 sec duration
t=(0:j)*dt; %time vector
cp=1; %Define cp before the loop.
p=zeros(1,j); %or p=5*ones(1,j) if you want a vector of all fives, etc.
x=zeros(1,j);
y=zeros(1,j);
pinitial=2.05e19;
a1=1; a2=1; a3=1; b1=1; b2=1; b3=1; %define these constants as desired
for i=1:j
p(i)=(2.7324*10^-22)*pinitial*cp*i;
x(i+1)=(((x(i)*y(i))/b2)-(((a1+y(i))/b2)*x(i))+((a2/b2)*y(i))+(a3/b2))*dt+x(i);
y(i+1)=(-((x(i)*y(i))/b2)-((b1/b2)*y(i))-1+((p(i)/b2)*(1-b3*(1))))*dt+y(i);
end
%downsampling=1;
%xx=downsample(x./9.8317e-17,downsampling);
%Downsampling by 1 does nothing. When you divide a vector by a constant,
%you do not need to "dot-divide". Regular division is fine.
xx=x/9.8317e-17;
[bbb,aaa]=pwelch(xx,[],[],[],fs);
figure
subplot(2,1,1)
plot(t,xx) % time domain
grid minor
subplot(2,1,2)
semilogy(aaa,bbb)% frequency domain
grid minor
Code above runs without error. I changed y1(i) to y(i) in the line x(i+1)=... because y1(i) was probably a mistake.
##### 12 CommentsShowHide 11 older comments
Nneka Onubogu on 7 Jun 2021
@William Rose, ok. Thank you very much. I will accept your answer as you have actually shown me how to generate the chirp frequency.
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### More Answers (3)
William Rose on 28 May 2021
If you highlight your code in your posting, and then click the "code" icon at the top of the pane, it will format your code nicely, and more importantly, it will allow others to run your code. Therefore please do this - and check that your code is in fact runnable, or if it isn't, tell us what error you get.
For generating a chirp and other signals with varying frequency, the built-in vco() function (here) is very convenient.
Good luck.
##### 1 CommentShowHide None
Nneka Onubogu on 28 May 2021
@William Rose thank you so much for your comment and thanks for teaching me on how to format my code as I am new to Matlab. I will do that right now and will post the complete code so that it can be run by others.
When I run the second section of my code, it takes forever and after i get a result, my (t,xx) plot is not a chirp signal but rather a signal for only one frequency.
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William Rose on 29 May 2021
Thak you for reformatting the code part of your original post.
The original post asks how do I do a chirp from 0 to 30 kHz, but there is a lot more going on here than just a chirp.
Matlab has a chirp() command and a vco() command. vco() is more general than chirp, but can be used to make a chirp.
Your code will crash my machine if I try to run it, or it will take forever, because you have a for loop that is set to run 100 million times. Each time, the x and y arrays grow by one element. By the end, x() and y() will have 100 million elements each. That will never work. You shoud initialize arrays before you run the loop, because otherwise, each array gets copied to a new array on each loop pass, which gets very slow when the array is huge. And arrays that huge will probably not work anyway.
You also have a lot of weird constants. For example:
t=(0:j)*dt/2.4156e8;
Why do you divide by 241 million? If you define the t vector this way, then it will not have the spacing "dt" that you have specified previously.
Another example of weird constants:
pinitial=(k/10)*1e19; %pump power (k=20.5 defined previously)
p(i)=(2.7324*10^-23)*pinitial*(1);
In what possible units is the pump power 2x10^19? In the next line, you multiply that gigantic number by a very tiny number, resulting in p(i)=0.0005 (approximately). Why use such extreme and offsetting constants? Also, p(i) defined this way in this loop is the same on every loop iteration, so you should define it once, outside the loop, to save time.
Other things:
for k=20.5, % pumping power ...
end
for u=0, % amplitude (2nd order) ...
end
for r=40, % amplitude (1st order) ...
end
The for loops above have only a single value for the loop variable, so each loop will only execute once. This will run, but it does not make sense to set up for loops that only run once.
I suspect the following code is where you try to generate a chirp. I have added a few comments.
for fp = 1000+((30000-1000)/(j*dt/3.33e-8))*t(1:end-1)
for u=0 % amplitude (2nd order) (This will only run once)
f2amp=u/100; %(f2amp will equal zero, so why bother?)
for r=40 % amplitude (1st order) (this will only run once)
f1amp=r/100; %(f1amp=0.4, so define it outside the loop)
for i=1:j
p(i)=(2.7324*10^-22)*pinitial*(1+(f1amp*sin(2*pi*dt*i.*fp/1))+(f2amp*sin(2*pi*dt*i.*fp/2)));
x(i+1)=(((x(i)*y(i))/b2)-(((a1+y1(i))/b2)*x(i))+((a2/b2)*y(i))+(a3/b2))*dt+x(i);
y(i+1)=(-((x(i)*y(i))/b2)-((b1/b2)*y(i))-1+((p(i)/b2)*(1-b3*(1))))*dt+y(i);
end
end
end
end
It appears that p(i) is the variable that is supposed to chirp, and the chirp frequency is supposed to vary from 1000 to 30000. The code above will not work, for various reasons which would take a while to explain.
Let's go back to the beginning. The min and max chirp frequencies are and . You specify dt=1e-9, but you can use a much longer step, which will allow many fewer samples and therefore faster execution and less risk of memory overflow. The fastest frequency is 30 kHz, which has a perod of 1/(30kHz)=33 microseconds. Therefore, if we choose dt=4 microseconds, we will have about 8 samples per cycle at the highest frequency, which is enough, and more samples than that at the lower frequencies of the chirp. So let's specify dt=4e-6.
We want frequency f to equal fc1 at t=0 and f=fc2 at t=Tend. We can write the frequency function as follows:
How long should the chirp take? Let's try one second, i.e. . This is not a crazy duration, since it only takes 1 msec to get through 1 cycle at the slowest frequency.
To make a chirp, you need a signal such as
where the rate of change of is steadily increasing. ( is the mean value of p and is the amplitude of the sinusoidal part.) For a regular un-chirped sine wave, the rate of change of is constant:
, from which it follows that .
For a chirp, f is not constant. It is given by the equation forabove. Therefore we have
This is a simple and solvable differential equation:
Integrate both sides, and assume when t=0, and you get
Therefore the Matlab code to make a chirp from 1 kHz to 30 kHz, lasting one second, is
fc1=1000; %start frequency (Hz)
fc2=30000; %end frequency (Hz)
dt=4e-6; %time step (s), should be at least 5x smaller than 1/fc2
A0=0; %mean value of signal
A1=1; %amplitude of oscillation
Tend=1; %chirp duration (s)
t=0:dt:Tend; %time vector
theta=2*pi*(fc1*t+(fc2-fc1)*t.^2/(2*Tend)); %theta vector
p=A0+A1*sin(theta); %signal vector
figure;
subplot(3,1,1); plot(t,p,'r'); %plot entire signal
subplot(3,1,2); plot(t(1:1000),p(1:1000),'r'); %plot beginning part
subplot(3,1,3); plot(t(end-100:end),p(end-100:end),'r'); %plot end part
figure;
spectrogram(p,1024,512,1024,1/dt,'yaxis'); %plot spectrogram
The code generates 2 figures. The first figure shows the whole signal, the first 4 milliseconds, and the last 0.4 milliseconds. This figure show that frequency is 1 kHz at the start and 30 kHz at the end, as desired. The second figure is the spectrogram, or time-dependent power spectrum. It shows that the frequency increases from 1 kHz to 30 kHz from t=0 to t=1 s.
Change Tend, A0, or A1 to alter the chirp duration, mean, or amplitude.
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Nneka Onubogu on 31 May 2021
Thank you very much for taking your time to look into my code and making corrections line by line. I am very grateful. Actually, those weird constants were gotten from some equations based on my laser set-up and rate equations for laser. The pump power of 2*10^19 gives a resonance frequency of 10 kHz (when you plot the graph, you can see a peak at 10 kHz), so I change the value of ‘k’ to be able to get different resonance frequencies. I forgot to indicate that the code is in two parts. I actually run the first part of the code that stops at the “end” before the ‘fp’ for loop.
Please how can I initialize arrays before running the loop? Please can you show me an example?
I will remove unnecessary for loops. Thanks for the advice and corrections. Thanks also for explaining in details on how to obtain the chirp and for plotting it out to show me too.
The main aim of my simulation is to obtain the dynamic behavior of my fiber laser system. To do this, I needed to sweep my modulation frequency (0 to 30kHz) so as to get the time domain which should be the chirp signal and then measure the maximum peak amplitude at each frequency from which I will plot a bifurcation diagram. That Is the essence of this part of my code:
for i=1:j
p(i)=(2.7324*10^-22)*pinitial*(1+(f1amp*sin(2*pi*dt*i.*fp/1))+(f2amp*sin(2*pi*dt*i.*fp/2)));
x(i+1)=(((x(i)*y(i))/b2)-(((a1+y1(i))/b2)*x(i))+((a2/b2)*y(i))+(a3/b2))*dt+x(i);
y(i+1)=(-((x(i)*y(i))/b2)-((b1/b2)*y(i))-1+((p(i)/b2)*(1-b3*(1))))*dt+y(i);
end
end
end
end
downsampling=1;
xx=downsample(x./9.8317e-17,downsampling);
[bbb,aaa]=pwelch(xx,[],[],[],fs/downsampling);
figure()
subplot(2,1,1)
plot(t,xx) % time domain
grid minor
subplot(2,1,2)
semilogy(aaa(1:3357),bbb(1:3357))% frequency domain
grid minor
Based on your corrections, would it be correct to change the above section of my code in this way:
Assuming I call the signal vector cp,
for i=1:j
p(i)=(2.7324*10^-22)*pinitial*cp*i;
x(i+1)=(((x(i)*y(i))/b2)-(((a1+y1(i))/b2)*x(i))+((a2/b2)*y(i))+(a3/b2))*dt+x(i);
y(i+1)=(-((x(i)*y(i))/b2)-((b1/b2)*y(i))-1+((p(i)/b2)*(1-b3*(1))))*dt+y(i);
end
downsampling=1;
xx=downsample(x./9.8317e-17,downsampling);
[bbb,aaa]=pwelch(xx,[],[],[],fs/downsampling);
figure()
subplot(2,1,1)
plot(t,xx) % time domain
grid minor
subplot(2,1,2)
semilogy(aaa(1:3357),bbb(1:3357))% frequency domain
grid minor
When I run it, I get an error on the p(i) line. I want to be ale to plot :
plot(t,xx) % time domain and semilogy(aaa(1:3357),bbb(1:3357))% frequency domain.
Thanks a lot.
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# Derive an expression for the magnetic field at a point on the axis of a current carrying circulat loop.
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Solution : Expression for the magnetic field at a point on the axis of a current carrying circular loop : <br> <img src="https://d10lpgp6xz60nq.cloudfront.net/physics_images/VIK_PHY_QB_C08_E03_001_S01.png" width="80%"> <br> (1) Consider 'O' is the centre of a circular coil of one turn and radius 'a'. <br> (2) Let P is a point at a distance r from the centre, along the axis of coil. <br> (3) The plane of the coil is _|_^(r) to the plane of paper. <br> (4) Consider two elements AB and A'B' each of length dl which are diameterically opposite. <br> (5) Then, the magnetic fields at P due to these two elements will be dB and dB in the direction PM and PN respectively. <br> (6) These directions are _|_^(r) to the lines joining the mid-points of the elements with the point P. <br> (7) Resolve these fields into two components parallel (dB sin theta) and perpendicular (dB cos theta) to the axis of the coil. <br> (8) The dB cos theta components cancel one another and dB sin theta components are in the same direction and add up due to the symmetric elements of the circular coil. <br> (9) Therefore, the total magnetic field along the axis = B = int dB sin theta of the circular coil along PC - (I) <br> (10) The magnetic field at 'P' due to current element of length 'dl' is <br> 'dB' = (mu_(0))/(4pi)(I dl sin phi)/((a^(2)+ x^(2))) = (mu_(0))/(4pi) (I dl)/((a^(2) + r^(2))) - (II) [because phi = 90^(@)] <br> (11) From equations (I) and (II), B = int(mu_(0))/(4pi) (I dl)/((a^(2) + r^(2))) sin theta <br> From Delta^(l e) OPE, sin theta = (a)/(sqrt(a^(2)+r^(2))) <br> rArr B = int(mu_(0))/(4pi) (I dl a)/((a^(2)+r^(2))^(3//2)) = (mu_(0) I a)/(4pi(a^(2)+r^(2))^(3//2)) int dl <br> But int dl = circumference of the coil = 2pi a <br> :. B = (mu_(0) I a)/(4pi(a^(2) + r^(2))^(3//2)) xx 2pi a = (mu_(0) I a^(2))/(2(a^(2) + r^(2))^(3//2)) <br> (12) If the coil contains N turns, then B = (mu_(0) NI a^(2))/(2(a^(2)+r^(2))^(3//2)) <br> (13) AT the centre of the coil r = 0, B = (mu_(0) NIa^(2))/(2a^(3)) = (mu_(0) NI_(A))/(2pi r^(3))
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## [Confab] Circle-Square
The following problem has obsessed me since I first heard about it several months ago from a workshop participant in Boston. I believe it originates from The Stanford Mathematics Problem Book, though I’ve seen it elsewhere in other forms.
Given an arbitrary point P on a line segment AB, let AP form the perimeter of a square and PB form the circumference of a circle. Find P such that the area of the square and circle are equal.
Here’s why I’m obsessed. In the first place, the task involves a lot of important mathematics:
1. making sense of precise mathematical language,
2. connecting the verbal representation to a geometric representation,
3. reasoning quantitatively by estimating a guess at the answer,
4. reasoning abstractly by assigning a variable to a changing quantity in the problem,
5. constructing an algebraic model using that variable and the formulas for the area of a square and a circle,
6. performing operations on that model to find a solution,
7. validating that solution, ensuring that it doesn’t conflict with your estimation from #3.
Great math. But here’s the interesting part. Students won’t do any of it if they can’t get past #1. If the language knocks them down (and we know how often it does) we’ll never know if they could perform the other tasks.
What can you do with this? How can you improve the task?
I’m going to update this post periodically over the next few days with the following:
• two resources I’ve created that may be helpful,
• commentary from some very smart math educators on the original problem and those resources.
Help us out. Come check back in.
Previous Confab
The Desmos team asked you what other Function Carnival rides you’d like to see. You suggested a bunch, and the Desmos team came through.
Man did you guys came to play. Loads of commentary. I’ve read it all and tried to summarize, condense, and respond. Here are your big questions as I’ve read them:
• Is learning to translate mathematical language the goal here? Or can we exclude that goal?
• What role can an animation play here? Do we want students to create an animation?
• What kinds of scaffolds can make this task accessible without making it a mindless walk from step to step? On the other end, how can we extend this task meaningfully?
There was an important disagreement on our mission here, also:
Mr. K takes one side:
It took me about 3-4 minutes to solve – the math isn’t the hard part. The hard part is making it accessible to students.
Gerry Rising takes the other …
If we want students to solve challenging exercises, we should not seek out ways to make the exercises easier; rather, we should seek ways to encourage the students to come up with their own means of addressing them in their pristine form.
… along with Garth:
Put it to the kids to make it interesting.
I’ll point out that making a task “accessible” (Mr. K’s word) is different than making it “easier” (Gerry’s). Indeed, some of the proposed revisions make the task harder and more accessible simultaneously.
I’ll ask Gerry and Garth also to consider that their philosophy of task design gives teachers license to throw any task at students, however lousy, and expect them to find some way to enjoy it. This seems to me like it’s letting teachers take the easy way out.
Lots of you jumped straight to creating a Geogebra / Desmos / Sketchpad / Etoys animation. (Looking at Diana Bonney, John Golden, Dan Anderson, Stephen Thomas, Angelo L., Dave, Max Ray here.) I’ve done the same. But very few of these appleteers have articulated how those interactives should be used in the classroom, though. Do you just give it to your students on computers? To what end? Do you have them create the applet?
Stephen Thomas asks two important questions here:
1. How easy is using [Geogebra, Desmos, Etoys, Scratch] for kids to construct their own models?
2. When would you want (and not want) the kids to construct their own models?
My own Geogebra applet required lots of knowledge of Geogebra that may be useful in general but which certainly wasn’t germane to the solution of the original task. It adds “constructions with a straightedge and compass” to the list of prerequisites also, which doesn’t strike me as an obviously good decision.
Lots of people have changed the wording of the problem, replacing the mathematical abstractions of points and line segments to rope (Eddi, Angelo L) and ribbon (Lisa Lunney Borden) and fencing (Howard Phillips).
This makes the context less abstract, yes, but the student’s work remains largely the same: students assign variables to a changing quantity on the line segment, then construct an algebraic model, and then solve it. The same is true for some suggestions (though not all) of giving the students actual rope or ribbon or wire.
So I’m interested now in suggestions that change the students’ work.
Kenneth Tilton proposes a “stack” of scaffolding questions:
1. If the length of AB is 1, what is the length of AP?
2. What is the ratio of AP to PB?
3. Given Ps, the perimeter of a square, what is the area of the square?
4. Given Pc, the perimeter of a circle, what is the area of a circle?
5. How would you express “the two areas are equal” algebraically?
The trouble with scaffolds arises when a) they do important thinking for students, and b) when they morsel the task to such a degree it becomes tasteless. Tilton may have dodged both of those troubles. I don’t know.
David Taub lets students choose a point to start with. Choosing is new work.
Mr K asks students to start by correcting a wrong answer. Correcting is new work.
It seems to me that a simple model of the problem, (picture of a string) with a failed attempt (string cut into two equal parts) should be enough to pique the kids “I can do better” mode. Providing actual string with only one chance to cut raises the stakes above it being a guessing game.
I think more important would be to start with some “random” points and some concrete numbers and see what happens.
Max Ray builds fluency in mathematical language into the end of the problem:
So I would have my students solve the problem as a rope-cutting problem. Then I would invent or find a mathematical pen-pal and have them try to pose the rope problem to them.
If our mathematical language is as efficient and precise as we like to believe, its appeal should be more evident to the students at the end of the task than if we put it on them at the start of the task.
Gerry Rising offers us an extension question, which we could call “Circle-Triangle.” I’d propose “Circle-Circle,” also, and more generally “Circle-Polygon.” What happens to the ratio on the line as the number of sides of the regular polygon increases? (h/t David Taub.)
On their own blogs:
• Justin Lanier offers a redesign that starts with a general case and then becomes more precise. I’m curious about his rationale for that move.
• Jim Doherty runs the task with his Calculus BC students and reports the results.
• Mike Lawler gives us video of his son working through the problem.
2014 Feb 26. Some of my own resources.
Here’s one way this problem could begin:
1. Show this video. Ask students to tell each other what’s happening. What’s controlling how the square and circle change?
2. Then show this video. Ask students to write down and share their best guess where they are equal.
The problem could then proceed with students calculating whether or not they were right, formulating an algebraic model, solving it, checking their answer against their guess, generalizing their solution, and communicating the original problem in formal mathematical language.
Mr. K has already anticipated my redesign and raised some concerns, all fair. My intent here is more to provoke and less to settle anything.
I’m going to link up this video also without commentary.
2014 Feb 27. Other smart people.
I asked some people to weigh in on this redesign. I showed the following people the original task and the videos I created later.
• Jason Dyer, math teacher and author of the great math education blog Number Warrior.
• Keith Devlin, mathematician at Stanford University.
• Two sharp curriculum designers on the ISDDE mailing list, whose comments I’m reproducing with permission.
Here’s video of a conversation I had with Jason where he processes and redesigns the original version of the task in realtime. It’s long, but worth your time.
I immediately drew a simple sketch – divide the interval, fold a square from one segment, wrap a circle from the other, and then dive straight into the algebraic formulas for the areas to yield the quadratic. I was hoping that the quadratic or its solution (by the formula) would give me a clue about some neat geometric solution, but both looked a mess. No reason to assume there is a neat solution. The square has a rational area, the circle irrational, relative to the break point.
So in the end I just computed. I got an answer but no insight. I guess that reveals something of a mathematician’s meta cognitive arsenal. You can compute without insight, so when you don’t have initial insight, do the computation and see if that leads to any insight.
In the case of the obviously similar golden ratio construction, the analogous initial computation does lead to insight, because the equation is so simple, and you see the wonderful relationship between the roots
So in one case, computation just gives you a number, in the other it yields deep understanding.
Off the ISDDE mailing list, Freudenthal Institute curriculum designer Peter Boon had some useful comments on the use of interactives and videos:
I would like to investigate the possibility of giving students tools that enable them to create those videos or something similar themselves. As a designer of technology-rich materials I often betray myself by keeping the nice math (necessary for constructing these interactive animations) for myself and leaving student with only the play button or sliders. I can imagine logo-like tools that enable students to create something like this and by doing so play with the concept variable as tools (and actually create a need for these tools).
Leslie Dietiker (Boston University) describes how you can make an inaccessible task more accessible by giving students more work to do (more interesting work, that is) rather than less:
If the need for the task is not to generate a quadratic but rather challenge students to analyze a situation, quantify with variables, and apply geometric reasoning with given constraints, then I’m pretty certain that my students would appreciate a problem of cutting and reforming wire for the sake of doing exactly that …
I disagree with people who are saying that this problem as written is inherently bad or artificial. As an undergrad math major, a big part of the learning for me was figuring out that statements worded like this problem were very precise formulations of fundamental insights — insights that often had tangible models or visualizations at their core.
I remember lectures about knots, paper folding, determinants, and crazy algebras that the lecturers took the time to connect to interesting physical situations, or even silly but understandable situations about ants taking random walks on a picnic blanket. For a moment I even entertained the idea of graduate work in mathematics, because I realized that math was actually a pretty neat dance between thinking intuitively and thinking precisely.
Terrence Tao writes about that continuum here.
tl;dr version: Translating this problem from precise to intuitive and intuitive to precise, is part of the real work that research mathematicians (and their college students) do, and not something we should always keep from our students. It’s a skill we should help them hone.
2014 Mar 4. As usual Tim Erickson got here first.
Tagged in:
I'm Dan and this is my blog. I'm a former high school math teacher and current head of teaching at Desmos. More here.
1. #### eddi
February 24, 2014 - 8:28 pm -
Dan,
I would think starting off with easier shapes to work with, like rectangles, would help. But then the exact question is too easy (“cut it in half, so you get same size squares”). So modify that a little: You’ve got a rope 12 inches long. Cut it into 2 pieces and make a rectangle out of each piece. Where do you have to cut it so that the bigger rectangle is twice the size of the smaller rectangle.
Once you’ve got that idea down, you can change the shape to square and circle, or equilateral triangle and circle, or what have you.
Great problem, and interesting to think about approaching.
Eddi
2. #### Diana Bonney
February 24, 2014 - 8:36 pm -
It would be cool to use some geometry software like GSP (Geometer’s Sketchpad) to help students play around with the idea and think about it. I made a GSP file… but I’m not sure if it represents the situation accurately. I’ll email it to you so you can take a look.
3. #### Jim Doherty
February 24, 2014 - 8:39 pm -
My first thought is that a visual representation will go a long way toward nudging the students to a place where they can begin to construct a model. I visualize this as soon as I read it. I know many of my students will not.
4. #### Annette Lievaart
February 25, 2014 - 12:00 am -
The size of your problem depends on the goal you want to reach with students doing this exercise. Is it the start of something new? Is it to test if they understood what they’ve learned before? …?
Annette
5. #### David Taub
February 25, 2014 - 12:49 am -
The need to change the language would depend on what a particular class of students is used to – more “everyday” language vs. more “mathematical” language.
Visual aids would be helpful in the beginning, Geogebra after the students worked a little with it first.
I think more important would be to start with some “random” points and some concrete numbers and see what happens.
Something like:
You have a string that is 36 cm long. Choose somewhere to cut the string into two pieces. Make a square out of one piece and a circle out of the other piece. Which has a bigger area? How much bigger is it?
Now switch pieces – make a circle out of the piece you made a square from and a square from the piece you made a circle from. Which has a bigger area? How much bigger is it?
Cut the string at another place and do the same thing again.
Is there somewhere you can cut the string so that the circle and square will have the same area?
6. #### Lisa Lunney Borden
February 25, 2014 - 3:37 am -
Hi Dan,
This is a great problem but there are two things I would do to transform it: Contextualize and Verbify! First I would give it a context (a piece of ribbon) and then I would make it active instead of so static (cutting ribbon to make shapes). So:
I have been given a certain length of ribbon. I want to cut the ribbon into two pieces to make (outline) two shapes, a circle and a square. I want the shapes to have equal areas. Where should I cut the ribbon?
Dan, I envision you creating a little video to show the active nature of this problem – you sitting at your table cutting ribbon and trying to make circles and squares from the pieces.
The problem you present is a classic example of how we can make math problem linguistically challenging by an over reliance on static or nominalized language. When we make it active, it opens it up to more students. I have actually written about the need to “Verbify” mathematics in a “For the Learning of Mathematics” article which you can download here: http://showmeyourmath.ca/sites/default/files/02-Borden.pdf. Feel free to share.
7. #### Howard Phillips
February 25, 2014 - 4:54 am -
A rancher needs to build two new corrals for his cows. Each is to contain 100 cows. He gives the job to his two sons, Joe and Wayne. Joe makes a square corral and Wayne makes a circular one, and between them they use 400 yards of fencing. How much fencing did Joe use?
At least this version is more practical (slightly!).
The difficulty with the mathematical description of the problem is its static nature. The ‘line segment AB’ bit is designed to muddy the problem. Once one sees that a piece of string is what is meant the difficulties vanish.
How about putting a square on the line at one end and a circle of the same area at the other end, rolling them towards each other by one full turn and seeing what you get (a gap which can be thrown away or a change in the area, so that the objects meet, at a sensible point of course).
Also, the problem as given is as artificial as some of the worst ‘Word problems’. Clearly designed for students who are already ‘into’ math.
8. #### Gerry Rising
February 25, 2014 - 5:59 am -
It seems to me that many of the responses miss the point of the exercise. If we want students to solve challenging exercises, we should not seek out ways to make the exercises easier; rather, we should seek ways to encourage the students to come up with their own means of addressing them in their pristine form. We want them, not us, to generate the interesting ways cited by responders of exploring the problem.
9. #### Stephen Thomas
February 25, 2014 - 6:52 am -
So I played around for an hour and created a simple visualization for this problem, that given the proper guidance, kids could easily construct themselves using Etoys. Which in addition to the “important mathematics” listed in 1-6, would add item 7 Computational Thinking (using computers to think and model with).
Here is a post to a video of the project: http://youtu.be/8Nl-XytAtMc
FYI Tried posting this before but didn’t see it, then got a “duplicate comment detected”
Cheers,
Mr. Steve
10. #### Garth
February 25, 2014 - 7:34 am -
My first question was “Why would a kid want to do this in the first place?” As a math geek (most math teacher seem to be math geeks) I think the question looks fun. Most kids are not math geeks. Gerry Rising has the point on a problem like this. Put it to the kids to make it interesting.
11. #### Kenneth Tilton
February 25, 2014 - 8:03 am -
Let’s make it more “real world”: “Two trains leave NYC and Boston. When they meet….”
Seriously, it is a great teaching problem. I am imagining a Socratic “stack” of questions leading them to a solution providing the minimum of help needed by each learner, with the idea that, if they cannot answer a question, they push it onto the unanswered stack and go on to the next slightly easier question, popping the stack of unanswered questions after answering any successfully to see if they can now handle it:
0. If the length of AB is 1, what is the length of AP?
1. What is the ratio of AP to PB?
2. Given Ps, the perimeter of a square, what is the area of the square?
3. Given Pc, the perimeter of a circle, what is the area of a circle.
4. How would you express “the two areas are equal” algebraically?
That’s just off the top of my head. I see a gap already been 0 and 1.
12. #### Dan Allen
February 25, 2014 - 8:10 am -
I like the idea of the Geogebra applet but it’s also important for students to look at general cases in higher level courses. The geogrbea applet could be misunderstood as “the answer” by students who do not know the meaning of “arbitrary” and the possibility of multiple answers.
13. #### Angelo L.
February 25, 2014 - 10:28 am -
Dan,
I also jumped straight to the technology to make a simulation (on GSP). If I were to do this with my class, I think that, as Eddi said, I would use rope.
Give each pair of students a piece of rope or string and a pair of scissors. Give them the problem and challenge them to find the maximum area for a square and a circle. Go from there to the technology (GSP, DESMOS etc.) and the more formal algebra.
14. #### Stephen Thomas
February 25, 2014 - 10:42 am -
@Dan Anderson
Nice work with Desmos. Couple of thoughts:
1) Have a version where the Square and Circle overlap
2) Make it easier to change line length in one step
2) show/hide gridlines option
15. #### Joe Mako
February 25, 2014 - 11:06 am -
My first question was: What is the ratio of AP to PB? That way we could know where P is for any length of AB.
Here was my process:
PB = 2 *pi*r
area = pi*r^2
AP = 4*x
area = x^2
x^2 = pi*r^2
x = sqrt(pi)*r
AP = 4*sqrt(pi)*r
AP/PB = (4*sqrt(pi)*r)/(2 *pi*r)
simplified to:
AP/PB = 2/sqrt(pi)
(where r>0)
so for any positive length of AB,
length of AP = (2*AB)/(2+sqrt(pi))
16. #### EC
February 25, 2014 - 11:15 am -
I have a comment re: use of the word “form” here. Substituting “is equal to” makes the meaning more clear for me- is this reducing the required complexity/ use of mathematical language?
17. #### Howard Phillips
February 25, 2014 - 11:27 am -
Another point – Why does the problem formulation have names for the points. This gets in the way.
Without mentioning the words rope or string it would still be the same problem if it said
Here is a line ______________________
Pick a point on the line.
Make a square from the piece on the left.
Make a circle from the piece on the right.
Will these have the same area?
…..
The question as originally put should now scream at you!!!
18. #### Stephen Thomas
February 25, 2014 - 11:31 am -
@Dan Anderson
Nice! and quick ;)
@John Golden
Your Geogebra is also very good.
Now a few more comments/questions for all solutions:
1) How easy is using <pick your tool, Geogebra, Desmos, Etoys, Scratch, etc) is it for kids to construct their own models
2) When would you want (and NOT want) the kids to construct their own models.
3) Challenge: Can you make a version where they find the answer via a series of successive approximations? Say where they pick the middle of the segment then can decided Left or Right and go half way each time until they get "close enough".
Cheers,
Mr. Steve
19. #### David Taub
February 25, 2014 - 11:41 am -
I got curious about extending the problem to cutting the string to make an arbitrary polygon and a circle with the same area (just for fun). This is what I came up with for where to put the point (given as the fraction of the polygon section). I apologize for the cumbersome “in text” mathematical notation:
1/(1+sqrt((pi/n)cot(pi/n)))
20. #### Jon Tyndall
February 25, 2014 - 11:56 am -
I’d say Howard Phillips has nailed it perfectly when it comes to lowering the barrier to entry and by getting the kids to pose the question you want them to answer. At this stage no tech is necessary, just a board and a marker. I think I’d only use tech in act3 to show that their answer looks right.
21. #### Dave
February 25, 2014 - 12:24 pm -
Read the question this afternoon and did a little GeoGebra doodling as a means of getting students to understand the question without giving the answer.
A great part of learning math is learning how to read, and visuals can only help this process.
22. #### Dave
February 25, 2014 - 12:25 pm -
Read the question this afternoon and did a little GeoGebra doodling as a means of getting students to understand the question without giving the answer.
A great part of learning math is learning how to read, and visuals can only help this process.
Here is my doodle.
http://ggbtu.be/m90339
23. #### Jered
February 25, 2014 - 2:42 pm -
I played with it for about 15 minutes before finding the ratio of AP to PB. I’m sure there would be many varied approaches, but they are related as follows:
2:√(pi)
24. #### Max Ray (@maxmathforum)
February 25, 2014 - 3:21 pm -
I too went quickly to a visual model, using Geometer’s Sketchpad. Because Sketchpad, like Geogebra and Desmos and ETools(?) is a geometric construction tool, I ended up with a square and a circle based on the measurements of AP and PB, both of which kind of hang out at the end of the line. The whole time my brain is going “can’t the line kind of wrap up to make the square and the circle?” but because I couldn’t model it that way with my tool, I couldn’t see the interpretation of using part of the line (or rope, or ribbon, or whatever) to physically form the circle or the square.
Gerry wonders if our goal is to get the kids to do the problem or to get them to translate the precise mathematical language… and if it is to get them to translate the language, what we would do then.
If the goal is to get kids to do the math, I really like the simplicity of having some rope and asking kids to make guesses about where to cut the rope so that a circle formed from one and a square from another have equal area. It lets them get into the problem, make some guesses, test possibilities, and will require them to set up the exact same algebraic expressions to eventually solve the problem. This also lowers the cost of entry by a LOT. Any kid can point to a spot on a rope. Very few kids can start working on expressions for area in terms of PA/AB (or whatever it is y’all did to solve this).
But… if we do all the interpreting for them, then we will have kids who can solve real problems that they define for themselves, and any problem interestingly posed, but who will look at dry old mathematician talk and turn up their noses. And I DON’T think that the reason kids should learn to read problems like this and get to work is because they’ll have to on tests. That would suggest that there’s no other reason that precise language might be important in math. I think there are good reasons for precise language and students should encounter those reasons.
I also think we learn to read by learning to write, often.
So I would have my students solve the problem as a rope-cutting problem. Then I would invent or find a mathematical pen-pal and have them try to pose the rope problem to them. Think of the Writing Instructions for a Peanut Butter & Jelly Sandwich task so beloved by math & science teachers. This pen pal would come back at them with questions and non-answers until they had mathematized the problem as selecting an arbitrary point P on line AB of arbitrary length, forming a square whose perimeter was equal to PA and a circle whose circumference was equal to PB, such that the areas enclosed by the circle and square were equal. Such nitpicking might include wondering what it means to “form a circle from the rope” or why you couldn’t just measure the length of the rope or how a rope could have area, or whether you could cut the rope along its long axis, thus making two long, skinny pieces of rope, etc. The point of modeling with math is to be able to abstract away those pesky real-world details that get in the way of solving a reasonable version of the question… and I’d want my students to experience that kind of writing, so they could begin to make sense of that kind of reading. I’ve never tried this so I don’t know how it would go, but I could imagine some version of it leading to frustration but also insight.
Last, of course, we’d look at the precise math version Dan presents and compare it to our informal version and think about what benefits and costs each has. We’d practice translating some other informal math presentations of questions into fancy ones, and vice versa (e.g. translating “pick two numbers that add up to 25. Now multiply them. How big an answer can you get?” into “x and y are real numbers such that x + y = 25. Find the maximum value of xy, or explain why there is no upper bound on xy.”
25. #### Sue Popelka
February 25, 2014 - 6:35 pm -
This problem is rich in Algebra and Geometry–it’s got perimeters/areas of squares and circles, the quadratic formula, simplifying radicals, factoring, compound fractions, and simplifying rational expressions. I have not seen it before, but I really enjoyed it. I worked through it, got the correct answer and then later checked your list of seven mathematics ideas involved; I had used all of them. I found the ratio of the length of the “square” piece to the length of the “circle” piece to be
(1-sqrt(pi/4))/(sqrt(pi/4)-pi/4)=1.129. I teach AP Calculus BC and find that what students struggle most with in Calc is the Algebra. This problem really stretches those skills. I am going to use this problem in my Calc classes next week. I plan on presenting it just as it was here and then have the students cut various lengths of strips of paper to try to get the areas to be equal before they attempt the algebra.
26. #### Yvette
February 25, 2014 - 8:15 pm -
If I read the post, it seems to me that the question is “How do we teach/ensure that the language is understood? At some point, a child will not have someone who will simplify the language.
They can only model the problem on their own IF they can understand the question.
I’m just a parent, but it seems to me that the first thing would be to ascertain comprehension.
Get a highlighter, underline key words. Encourage them to look up definitions.
The using those words, illustrate or model the problem. Draw a diagram.
If you can’t get past that point, then finding the right formula is going to be a challenge.
My son is younger, but we’ve been trying to use math language in daily language and find ways for him to retain the meaning. For example: Perimeter …”Secure the perimeter!!!” As in the distance around the object. Which as a typical young boy, he relates to.
Personally, I always found working through math questions it helped to read it once all the way through. Then read it a 2nd or even 3rd time. Draw the diagram one phrase at a time. TMI to retain all that in short term memory. I feel, at least with my kid, that they feel they should get it the first time through. Not sure math is like that. It’s okay to re-read, ponder and draw.
27. #### Jered
February 25, 2014 - 9:13 pm -
I should clarify that I started with a circle and square with congruent areas. I used 36. So the perimeter was 24. I solved for the radius which was 6/(sq rt pi). Circumference then is 12*(sq rt pi). Compare the two lengths [24 : 12*(sq rt pi)] and this simplifies to the ratio I listed above: 2 : (sq rt pi). I ran this by other math teachers and at least one of them used quadratics to get there. I thought this was fascinating, but he was fascinated that I started with the shapes. He called that “working backwards” but it was just the way I naturally came into the problem.
28. #### Avery
February 25, 2014 - 10:15 pm -
I’m going to be Downer Debbie here. This problem doesn’t do it for me. I looked at it using technology and came up with an answer, but didn’t feel that this answer helped me understand the “story” behind the problem any more deeply. I then solved the problem algebraically, but was once again left with a feeling of “ok then” versus “whoa.” I wasn’t surprised by the answer. Using technology, I didn’t find a “simple” answer that just screamed “something else must be going on here” or “there must be a simpler solution.” There aren’t any variations that I am yearning to solve. I’m not sure how a student might use this to explore a new concept or idea. Every visual representation I thought up missed the mark in relating the length of the lines and the “perimeters” of the shapes.
So I wasn’t going to post this comment. I didn’t want to come across as rude or mean, the person yucking other people’s yum. But then I realized this was silly. The fact that this problem didn’t inspire me doesn’t mean that Dan’s an idiot simpleton because it DID inspire him. On the contrary, it’s a great reminder that there isn’t a quick answer to what makes for a good problem, and this answer isn’t the same for everyone.
29. #### Mr K
February 26, 2014 - 7:23 am -
I put this off for a bit, due to grading (Argh!).
It took me about 3-4 minutes to solve – the math isn’t the hard part. The hard part is making it accessible to students.
If I make an interactive visual model (my first instinct) I am depriving them of the opportunity to create their own models. If I give them concrete numbers (The line is one meter long) I eliminate a lot of the value of the problem.
It seems to me that a simple model of the problem, (picture of a string) with a failed attempt (string cut into two equal parts) should be enough to pique the kids “I can do better” mode. Providing actual string with only one chance to cut raises the stakes above it being a guessing game.
Following that, it’s a matter of scaffolding their modeling (CCSMP #4!) encouraging (CCSMP #1!) them to climb the ladder of abstraction (CCSMP #2!).
I’ve got a dead day at the end of this week. Maybe I’ll try this then..
30. #### Max Ray (@maxmathforum)
February 26, 2014 - 7:28 am -
Mr. K, the idea of only one piece of string and only one chance to cut is nice because it makes the concept of “an arbitrary length” meaningful. You won’t tell them the string length in advance, they need to be prepared for any length. Cool!
31. #### Mr K
February 26, 2014 - 7:46 am -
On further thought, I really like the idea of the introduction being:
“I cut this string in half – do the two shapes I made out of it have the same area?”
It asks a simpler question, without doing the work for them, and in answering it they’ll be developing some of the components of the model they’ll need to use later.
32. #### Andy Brown
February 26, 2014 - 8:40 am -
Vary interesting comments. As Don Small probably battled his whole career teaching cadets at West Point, mathematics is only about modeling, even the arithmetic we put on the paper lest we forget what makes us happy to do.
33. #### Ignacio Mancera
February 26, 2014 - 12:00 pm -
I’ve been reading some of the comments and I really don’t think that one should add a real/fake context such as a ribbon, a corral or any of the kind. I do think, on the contrary, that purely mathematical questions should be presented to the students once in a while, so that they can appreciate them “for the sake of it”. I find the problem appealing and I resent the idea that “most” kids won’t find it so.
That being said, the difficulty of the language on which the problem is stated is of course an obstacle. My first idea was also to go to Geogebra and make a visualization, but I still thinks would struggle with the idea of “make a circle out of it”. I imagine that most of my students would misunderstand its meaning with “make a circle using that as its radius/diameter”. The same applies with the square and the perimeter: my first thought was that AP was the side of it, not its perimeter.
1. Draw a segment AB and a point P on it.
2. Draw a square using AP and a circle using PB.
3. Is it possible for the the square and the circle to have the same area? If so, where should we put P?
With this formulation, kids would still have different ideas: they might draw the circle using PB as its radius or as its diameter. Perhaps some kid (let us name him Dan) would draw the square using AP not as the side but as the diagonal of the square. We should take that chance to make them believe that THEY have created a new problem: “Ok class, new situation: what about Dan’s approach? How would that change the problem?”
After that, some leading might carry them towards new possibilities: “What if AP is…?” Or maybe: “What other thing might AP represent in a square?”. From that, we might get our problem as a more elaborated version of the simpler and more intuitive first one (but, actually, it will be much easier to solve it after having solved the rest!)
34. #### Howard Phillips
February 26, 2014 - 2:05 pm -
Me again!
Referring to comment no. 32 Max Ray (@maxmathforum), I like the approach, but he doesn’t go quite far enough. So:
———————–
Kick it around.
Find out where to use a bit of math, and what it entails (representation, solving equations etc)
Get an answer and check it for reasonableness.
Finally, maybe as a separate activity, figure out how to present the problem in mathematical fashion, ie abstraction, use of technical terms etc.
———————-
This way the students will have a much better idea about how to deal with problems already formulated in abstract terms.
Oh, and mathematicians (and engineers) do not happen upon precisely formulated abstract problems and then solve them, they do more or less exactly what I have described above. Also they don’t spend hours trying to prove theorems unless they are convinced that the theorems are true.
The problem as presented is really only suitable for math specialists, and as such quite a nice one.
35. #### M Ruppel
February 26, 2014 - 6:35 pm -
Dan et al,
I think this is a great problem that is unfortunately tied up in some confusing language:
Given an arbitrary point P on a line segment AB, let AP form the perimeter of a square and PB form the circumference of a circle. Find P such that the area of the square and circle are equal.
The biggest thing that stands out to me is what does it mean to “find P” – P isn’t a point with coordinates, it isn’t a fixed distance from A, etc. What we are really after is a ratio between 2 lengths, and the problem doesn’t make that clear.
If I were posing the problem to kids, I would make the question itself much simpler and concrete: Here’s a piece of string. You are going to cut it into 2 pieces so that you create a circle and a square that have the same area.
Setting it up, I would want to establish with the students that chopping it in half makes the circle bigger than the square, and ask why. Then I would ask them to take a guess, and then begin exploring. Each group would only get one piece of string, which would require them to compute the area in terms of the perimeter, rather than experimenting with cutting.
I think a lot of commenters want to use Geogebra or something here, but I’m not sure that gives a satisfying answer, because it’s necessarily approximate, and not based on manipulating the formulas. I would ask kids to do this totally by hand. Some might make successive approximations (what if I do a 60:40 split, 65:35, etc) until they have it, but the challenge of having to do it by hand may motivate them to think about it w/ algebra.
Great problem, not great set up.
36. #### Dan Meyer
February 26, 2014 - 7:36 pm -
You folks have brought a lot of depth here. I’ve done two things now:
One, I’ve summarized, condensed, and responded to a lot of your commentary at this link.
Two, I’ve added several resources I’ve created which I hope will serve less to dampen debate and more to toss a few dry logs onto the fire here.
37. #### Max Ray (@maxmathforum)
February 26, 2014 - 8:05 pm -
Dan, you’re a visual wizard. I was making my Sketchpad sketch in the hopes of making a wordless presentation of the video* but I couldn’t figure out a way to suggest how the location of P was building the circle and square. With your video, I would present the task wordlessly first, using videos 1 and 2 (or 1 and 3? Or 3, 1, and 2?) and then introduce the precise language once we had an intuition and played the “stop me when they’re equal…” game.
But… I’m also not convinced that the video adds much that the rope doesn’t… depending on what language game we’re playing. If my students understand the rope as a physical stand-in for a math line, just like the pixels on the screen are a physical stand-in for the math line, then I don’t see any difference. If they take the rope too literally, then we’ve changed the task in subtle (maybe important, maybe not?) ways.
*Having students make their own sketch felt like one of many problem-solving methods students might use, and not one I’d want to enforce or say is “the” way we’re going to make sense of the problem. It’s part of the solving, not part of the sense-making.
38. #### Mr K
February 26, 2014 - 11:57 pm -
I can’t tell – do you think new work a good thing or a bad thing?
To me it feels like it’s a byproduct of trying to make the problem more accessible without actually making it easier. In that sense, I’m okay with it.
39. #### David Taub
February 27, 2014 - 2:06 am -
It wouldn’t be too hard to throw together a javscript app that presents a line on screen and the students can click (drag?) on a point anywhere on the line.
The line would then break and form a circle and square with the two pieces.
A few questions on whether it is worthwhile and if so some of the details.
1. Difference between Dan’s video is the students get to play with where the points go, so the intro is more interactive. Is this helpful or not?
2. Requires students have access to individual/group devices (advantage of javascript is it will work in any browser)
3. How much information should it give? This could in theory be controlled by the teacher in some way.
4. Could just give relative sizes of the shapes. For example, “the area of the square is 5% bigger than the area of the circle”.
5. Or it could give some kind of actual value based on an arbitrary line length.
6. Options to show a grid in the background for estimating relative sizes.
7. Options to add numbers to the line – or, possible to have an optional box where students can type in a fraction instead of clicking for where the dot will go when they feel ready to try and be more precise.
8. Teacher controlled options to show the calculations being done to find the two areas.
9. Options to change to other polynomials (makes the programming a bit more complicated, but maybe not overly so).
10. If allowing for a range of polygons, perhaps a table listing the percents based on number of sides and watching them get closer and closer to 50% as the number of sides increases (I did this for fun in excel, and was personally surprised at how fast it started to converge).
Oh, and Dan, a small personal recommendation for your video – when you have the lines being “sucked up” into the circle and square, you leave behind a doted line to show where it was, but that is no longer there. I would really recommend changing the color of this dotted line so it is no longer the same color as the original line it is replacing. Possibly the same relative shade but much darker so the difference it more clear. It was kind of hard at first to see what was happening. I think a color change would make that stand out a lot more. Just my 2 cents.
40. #### Justin Lanier
February 27, 2014 - 4:14 am -
Hi Dan,
When you say you’re curious about my rationale for my redesign, do you mean you have questions about it, or just that you thought there was something interesting about it?
41. #### Kenneth Tilton
February 27, 2014 - 4:33 am -
Left unsaid (my fault) was that only the first question is offered up front, and that is for the gold medal (or whatever). The next question/hint must be requested, and that drops them to silver.
I remember watching kids using a Carmen San Diego game in school. They just punched clue-clue-clue until it pretty much gave them the answer. The good news? Kids aren’t dumb. :)
btw, I do not understand the objections to the GeoGebra implementation. It is a great intermediate step in a progression, one in which the kids could play with the value of P and see the area trade-off and then perhaps appreciate, “Oh, so this is why we have Algebra (and Geometry): Instead of wiggling P back and forth we can just calculate the answer directly from the givens.” In their own words, of course. :)
42. #### Mike Lawler
February 27, 2014 - 5:07 am -
My quick reaction to the questions you asked:
(1) Is learning to translate mathematical language the goal here? Or can we exclude that goal?
This problem works well for both cases, I think. If your goal with the students is to learn to translate mathematical language, then I think that both the videos you’ve create, Dan, and the other interactive elements created by commentators are great starting points. They help students see what’s going on in the problem and provide a nice starting point.
However, this problem is also a great question for more advanced settings where the students are already comfortable translating into mathematical language. In these settings I would suggest a modification to the videos / visuals and to the problem to allow the students to explore some interesting connections between the geometry and algebra in this problem:
Rather than taking a point on the line, take a point in the plane. Show the same geometry in the video – how one line segment folds into a square and the other curls up into the circle – without giving any hint of the location of the point on the line the problem is asking about. Now there are three follow up questions from the video:
(a) Find the point on the line segment that causes the square and the circle to have the same area.
(b) If we extend the line segment in both directions, are there any other points on the line segment where the square and circle will have the same area? This question asks for some geometric insight and can be answered without calculation, though finding the exact point obviously requires calculation.
(c) Now find all of the points in the plane where the square and the circle have the same area. The shape of this set of points will likely be a surprise to the students.
(2) What role can an animation play here? Do we want students to create an animation?
I do not believe that the animation is necessary, but I don’t see any down side in the animations. Non-animated pictures drawn by the students would also be helpful, and the suggestions of trying out the original problem with string might also be fun for students. There are many different ways to approach the problem, which is part of what makes it so interesting.
(3) What kinds of scaffolds can make this task accessible without making it a mindless walk from step to step? On the other end, how can we extend this task meaningfully?
My suggestion on extending the problem in the answer to question (1) probably covers most of my thoughts on this one. For me personally, when I read the problem on Tuesday morning, walking through that extension with my son was something I was really excited about.
43. #### Mr K
February 27, 2014 - 9:34 am -
Here’s what I’m trying today. It’s working surprisingly well with my Geometry students. The Algebra students may require some formulas, we’ll see.
The green fields are built in in two separate steps. But aside from that, there’s the whole problem.
Also, stock photo. I’ll take that strike.
44. #### Dan Meyer
February 27, 2014 - 2:13 pm -
Mr. K:
I can’t tell – do you think new work a good thing or a bad thing?
New, more interesting work is a good thing. In its current form the problem asks students to translate, formulate, and solve, in that order. I’d rather move translate to the back end of the task and add new, more interesting work to the front end.
Justin Lanier:
When you say you’re curious about my rationale for my redesign, do you mean you have questions about it, or just that you thought there was something interesting about it?
Both. I’m interested in your rationale for moving from the general to the specific. That seems to make the problem less accessible rather than more, though maybe there’s something to be gained I’m not seeing.
Max Ray:
But… I’m also not convinced that the video adds much that the rope doesn’t… depending on what language game we’re playing. If my students understand the rope as a physical stand-in for a math line, just like the pixels on the screen are a physical stand-in for the math line, then I don’t see any difference. If they take the rope too literally, then we’ve changed the task in subtle (maybe important, maybe not?) ways.
The video and the rope share a lot of features and each has features the other lacks.
I like the rope (the physical one, not a picture or the word) because it places a lot of weight on making the right cut the first time. It discourages guess and check in a great way.
The video, meanwhile, allows students to make a cost-free guess and compare it to the rest of the class. You can guess with the rope but if we all have ropes of different lengths, it’s going to make the share-out less meaningful. (This is to say nothing of the fact that the thought of passing out scissors to everyone in the class gives me a certain queasy feeling the video doesn’t.)
The video also allows me to ask the sense-making question, “What do you see happening here? Put it into words.” It’s hard to ask the same question of a rope that’s sitting on your desk.
45. #### Dan Meyer
February 27, 2014 - 2:47 pm -
Another update:
I interviewed some really smart folk – Jason Dyer, Keith Devlin, and the crew from a curriculum design mailing list I frequent – and asked for their opinions on this task and my redesign.
I recorded video of my interview Jason and pasted all of that in the new update. I found it extremely enlightening and I hope you do too.
46. #### Mr K
February 27, 2014 - 4:23 pm -
BTW – I think Jim’s extension of asking whether that solution to the problem is also the minimum possible area of the two shapes is pretty genius. Definitely calculus level, but it totally justifies calculus.
47. #### Keith Devlin
February 28, 2014 - 6:37 am -
Mike Lawler’s extension is neat. Super question to raise with student: “What about the other root?”
48. #### David Taub
February 28, 2014 - 6:47 am -
In case anyone is interested, here is a quick first draft of a javascript version that allows the student (or teacher) to click on a point on the line and then it will show the two pieces forming a square and a circle.
For now, no additional information is given. This is similar to Dan’s video, but allows the students to choose their own points.
http://hem.bredband.net/taub/circlesquareline.html
49. #### Stephen Thomas
February 28, 2014 - 7:47 am -
Peter Boon’s comment caught my attention, in particular “imagine logo-like tools that enable students to create” well Etoys is a logo like tool that students can easily use to model this problem. I created a a blog post showing how with 3 Objects and 9 lines of tile scripting kids could easily model this problem.
Etoys was designed as an educational tool for teaching children powerful ideas in compelling ways and a media-rich authoring environment and visual programming system for kids to create. So not only can it model this problem but many more Fractions, Multiplication etc. Etoys Illinois has some great resources.
Cheers,
Mr. Steve
50. #### Robert Woodley
February 28, 2014 - 8:08 am -
Not sure if anyone is still reading the comments, but my 2 cents:
The first sentence of the problem makes me feel so sad for the students that have to deal with it.
There are 3 problems here:
Problem 1: What is the radius of a circle with an area equal to a square of side X?
Problem 2: What is the ratio of the perimeters of the 2 objects in Problem 1?
Problem 3: If P1 + P2 = 1, what is P1 and P2?
Problem 1 is the most interesting one. It is a general concept that can find many applications and will appeal to the geeks in the class. (A fun side-line is to extend it to N dimensions – for high schoolers of course). It should be solved using algebra.
But the problem starts off using the language of geometry. It hides the mystery and excitement of solving a more general principle. It hides the beauty of the math behind a nuts-and-bolts problem that is poorly articulated and a problem one would never encounter in real life.
As stated it is neither general nor useful!!!! Neither of theoretical nor applicable use!
Crikey.
51. #### Max Ray (@maxmathforum)
February 28, 2014 - 2:02 pm -
I disagree with people who are saying that this problem as written is inherently bad or artificial. As an undergrad math major, a big part of the learning for me was figuring out that statements worded like this problem were very precise formulations of fundamental insights — insights that often had tangible models or visualizations at their core.
I remember lectures about knots, paper folding, determinants, and crazy algebras that the lecturers took the time to connect to interesting physical situations, or even silly but understandable situations about ants taking random walks on a picnic blanket. For a moment I even entertained the idea of graduate work in mathematics, because I realized that math was actually a pretty neat dance between thinking intuitively and thinking precisely.
tl;dr version: Translating this problem from precise to intuitive and intuitive to precise, is part of the real work that research mathematicians (and their college students) do, and not something we should always keep from our students. It’s a skill we should help them hone.
52. #### Dan Meyer
February 28, 2014 - 2:07 pm -
Max Ray:
Translating this problem from precise to intuitive and intuitive to precise, is part of the real work that research mathematicians (and their college students) do, and not something we should always keep from our students. It’s a skill we should help them hone.
Good word.
53. #### Mike Pac
March 1, 2014 - 8:15 am -
To me, this problem doesn’t beg for a need to be precise in finding P, and therefore doesn’t inspire a need to pursue/develop a generalized answer. Once I can get to the point of visualizing a square in relation to a circle (like in the animations), my intuitive sense of “about equal” in area is satisfying enough. I think the important part of this problem is getting to the visualization itself, and not actually finding the precise answer. In other words, this problem doesn’t create the same need as some of your other problems presented do – I’m thinking about the water tank problem, where no one wants to sit for multiple minutes to see how long the tank is going to take to fill, therefore creating a need to develop a model.
54. #### Jered
March 1, 2014 - 8:31 am -
@Mike Pac – OK, but to be fair that’s just you. I mean, if approximate is OK, couldn’t just splitting the rope in half be “good enough?” After all, the area of a square and of a circle can’t be that far apart.
Or can they?
Let’s say the rope is 48 inches long. The square will end up being 12 x 12, or 144 sq in, and the circle will end up having a circumference of 48, so a radius of about 7.64. The area of this circle is then approximately 183.4 square inches. That’s 27% bigger! But by shifting the cut just under an inch and a half (3% of the rope’s length), the areas become equal.
That’s amazing to me. How does 3% = 27%? Why are the areas of a square and a circle with congruent perimeter/circumference measurements so far apart? And why is it only 3% of the rope that makes the difference? And why do so many fencelines intersect at 90º instead of being rounded?
Maybe this problem in and of itself isn’t interesting, but I bet there are other approaches that could be taken to make it interesting – either to you or to your students.
Sometimes giving a problem to solve just feels like same-ol same-ol. Perhaps you could give them an answer and having them find out how to get it.
I just think there is *something* you can do with just about every problem to make it accessible and interesting.
55. #### Mike Pac
March 1, 2014 - 9:08 am -
@Jered….Your enthusiasm for this problem is great, and I’m sure it’s infectious to your students. “Interesting” is a subjective experience. I am giving my initial impressions of the problem, and also how I think many of my students would view it. One can view “by shifting the cut just under an inch and a half (3% of the rope’s length), the areas become equal” as exciting as you do, or “oh wow, my intuition was only an inch and a half off.” I’m not saying that this problem can’t be interesting. To me, it didn’t strike the need to really dive in and explore any further. Mentioning fencing, to me and many students, also makes the problem less interesting.
56. #### Chris Hill
March 3, 2014 - 1:40 pm -
I’m really surprised no one has mentioned this yet (or I missed it in my skimming comments).
What does that second solution mean? Pose that extension question to your Pre-calculus and above students. I have some initial thoughts, but I haven’t checked up on them with any thorough reasoning.
57. #### Tim Erickson
March 4, 2014 - 9:33 am -
Sorry to be so late to this confab, but so much has been great there’s not much for me to add. Two things: Dan has kindly referenced a page on my blog that talks about ways to attack the max/min version of this problem using data.
http://bestcase.wordpress.com/2012/09/13/reflection-on-modeling/
if that is of any use.
But I also want to give a shout-out to Mr K (27 Feb, #54, above), and here is the link:
http://mathpl.us/posters/circlesquare.001.jpg
What I love is the precise level of vagueness of the question. If that makes sense. You cut the shoelace, make the circle and square, and the question is, “What are the two areas?” It may be a commonplace here (sorry if I’ve missed it), but I think this type of question is important, if only because a really great answer is, “it depends.”
Which motivates the payoff question: “Depends on what?” (Followed by, HOW does it depend…”). I made a bunch of physics labs where the vague question was an essential part of the intro — enough so that I eventually got field-test students to recognize the ploy and shout, “It depends” in unison whenever they sensed one of these.
This is no replacement for a really good first act. But if you don’t have one, a vague question may be a suitable substitute.
58. #### Dan Meyer
March 4, 2014 - 10:44 am -
Tim Erickson:
This is no replacement for a really good first act. But if you don’t have one, a vague question may be a suitable substitute.
I’ve been going around a couple different places and claiming that “asking questions about the question” is an essential modeling act. But this requires a certain level of comfort from teachers with questions that aren’t fully specified. Not just questions where the information exists but has been withheld for a minute, but questions where the question requires follow-up questions. Like, “What do you mean by that? What form is the answer going to take? What does that word mean in this context?”
I find teachers evenly split in their perception of those ill-defined questions as advantages or disadvantages in class, as features or bugs.
59. #### Brian Miller
March 11, 2014 - 6:57 pm -
I used the original question and handed it to students after they turned in a test – and just waited to see what they would do. A lot of them initially punted on it. They didn’t know how to do it and there was no help so they flipped it over and started drawing on it.
After everyone finished the test I decided to try and be a teacher and scaffold a bit. I put the point right in the middle of AB and asked them if the perimeters were the same here. They said “yes”. Then I asked them if the areas were the same – and this when they made me proud and told me “no” because if the perimeters of the square and circumference were the same, then the circle would have a larger area because the circle has the largest possible area for a fixed perimeter. Thus point P had to be closer to B then A. I think that realization made them more proud that it would have if I had already provided them the visual evidence.
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HOT OFFERS!
# The Internet Can’t Figure Out How Many Triangles Are in This Image
### How many can you count?
The Internet Can’t Figure Out How Many Triangles Are in This Image
Calling all brainteaser aficionados; this triangle trick is sure to baffle even the most astute of brains for at least a few minutes.
Kumar Ankit, an engineering student from India, designed and posted this geometric puzzle to Quora, challenging users to find all the hidden triangles in this image.
You’re going to have to really exercise that eye for detail (and a thorough concept of geometry) to solve this visual riddle.
At first glance, the image seems to be a simple illustration with multiple shapes. Now here’s the catch—count how many triangles are in the drawing. People are scratching their heads at the different numbers they are coming up with.
If you’re completely stumped, we’ll give you a hint: The trick lies in factoring multiple smaller triangles into becoming a bigger triangle.
Got an answer? If you solved the math and your answer was 24, congratulations, you’re in the majority. Most people on Quora agreed that the answer is 24, with each row containing six triangles.
However, if you really want to get technical, the truly “correct” answer is 25. The 25th triangle is hidden in the ‘A’ in the artist’s signature in the right-hand corner. An even more nit-picking Quora user noted there was another triangle in the image if you count the word ‘triangles.’
Math expert Martin Silvertant posted a very helpful graphic on the site walking through how to locate the pictorial triangles and provided a logical explanation as to how you can stretch the number to 26 of them.
“A triangle is a mathematical idea rather than something real; physical triangles are by definition not geometrically perfect, but approximations of triangles. In other words, both the pictorial triangles and the words referring to triangles are referents to the concept of a triangle,” said Silvertant.
If you’re really getting into the debate and want to hash out (or defend) your answer, check out the entire Quora discussion for different users’ mathematical methods. (Want more optical puzzlers? See if you can spot the hidden image in these five photos).
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### 45 facts that will make you stop using plastic
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+1-415-315-9853
info@mywordsolution.com
## Statistics
1. The following probability distribution has been assessed for the number of accidents which take place in Midwestern city each day:
Accidents Probability 0 0.25 1 0.20 2 0.30 3 0.15 4 0.10
Based on this distribution, the expected number of accidents in a given day is:
a. 0.30.
b. 1.65.
c. 2.00.
d. 2.50
2. The impact on sampling of increasing the sample size is:
a. the potential for extreme sampling error is reduced.
b. the amount of sampling error is always reduced.
c. the sample mean will always be closer to the population mean.
d. there is no specific relationship between sample size and sampling error.
Statistics and Probability, Statistics
• Category:- Statistics and Probability
• Reference No.:- M921764
Have any Question?
## Related Questions in Statistics and Probability
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# Free math games
Listed below are the educational activities and games that have been developed for the HelpingWithMath.com web site. Also listed are a number of fun math games that have been created for the BBC Schools web site. These games will be especially useful for Kindergarten through 3rd grade although they can be used by any student that wants to practice their math in a fun way.
You will also find some of the games below, along with others, listed on the multiplication games page and also on the addition and subtraction games page.
### Latest Additions
December 2019: The 10 matching games below offer practice with shapes, rounding, and with decimals.
## Multiplication games
### More multiplication games
These are basic math fact games developed for HelpingWithMath.com.
The challenge is to aim the sight and fire at any of four targets – ideally at the one that shows the correct answer to the multiplication problem!
### Build the answer
These multiplication games require the “magnets” to be moved to make the correct answer.
## Algebra Games
### Addition and subtraction games
The game below has options for finding a number that is one or two less than, or one or two more than the given number and also to have both timed and untimed versions. It provides practice with number relationships with numbers to 10 and is typically suited to students around the 1st grade level.
This number partitions game is good for building foundation skills for addition and subtraction.
The games below require the numbers and signs to be moved around to make the correct answer.
#### Making 10 Games
And this addition and subtraction game mixes up numbers and signs for a slightly different challenge
### Math games by the BBC
The BBC Bitewise team have recently updated their math(s) games. Click here to learn more about the Guardians: Defenders of Mathematica.
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# How to use a slider in excel
## Create Dynamic Chart Data Labels with Slicers - Excel Campus
6 Feb 2019 Excel scroll and sort table using dynamic array formulas is far simpler than the old approach which required multiple tables, formulas and Solved: Re: Use decimals with slider control - Power Platform I'm working through setting up a slider to record the body temperature of a patient in Farenheit. I set the slider to default to 986 and then set min. Excel Slider Control: How could I limit the sum of all sliders to
## Excel 2016 from Scratch - Conditional Formatting with Slider
Minimizing Distractions in Microsoft Excel - Accountex Report 24 Jul 2017 The first prompt I'll cover is the Quick Analysis icon in Excel 2013 and As with the Zoom Slider, use the 100% command on the View menu to Decimals and Negative Values for Spinners and Scroll Bars Spinners and scroll bars can only be set to positive whole numbers in Excel. Hence to get decimal values and negative numbers you need to use formulas in a How-to Make an Excel Project Status Spectrum Chart - Excel But as a Project Manager, you would then move each individual slider left or right,. Would you use this as a Project Status Indicator by Phase for your project
Honey, I Shrunk the Scroll-Bar Slider in Excel - CFO.com 5 Oct 2011 The slider on the vertical scroll bar that allows Excel users to to use in an upcoming column, you'll win one of his Excel books as a thank-you. How to Hide Scroll Bars and Reset Slider Range in Excel 13 Nov 2019 Scrolling in Excel refers to moving up-and-down or side-to-side through a worksheet using the scroll bars, the arrow keys on the keyboard,
But as a Project Manager, you would then move each individual slider left or right,. Would you use this as a Project Status Indicator by Phase for your project How to Make Excel Spreadsheets Fit on One Page or Screen 25 Apr 2016 You may already use the Zoom Slider control at the bottom right-hand corner of Excel, as shown in Figure 1. You can use this to quickly Actual vs Targets Chart in Excel - Excel Campus
## Does anyone know how to analyze the online SVO slider
24 Jun 2011 How to add interactive elements to an Excel worksheet using spin buttons The middle slider is not controlled by any setting so a user can get Dynamic Scroll Bar in Excel - Step by Step Tutorial 31 Jul 2017 The dynamic scroll bar belongs to Form Controls. With its use we open up many great opportunities in Excel. We can manage any size lists in a Disabling Excel's Zoom Slider – Accounting Advisors, Inc. Automating Excel Charts in Two Keystrokes · Excel Tip: Automatic Backup of Key Excel Workbooks. Jun 14. Disabling Excel's Zoom Slider. Filed under Excel. [i-excel] Support | WordPress.org
### Excel's Auto Outline quickly hides data details to simplify
Spinners and scroll bars can only be set to positive whole numbers in Excel. Hence to get decimal values and negative numbers you need to use formulas in a How-to Make an Excel Project Status Spectrum Chart - Excel But as a Project Manager, you would then move each individual slider left or right,. Would you use this as a Project Status Indicator by Phase for your project How to Make Excel Spreadsheets Fit on One Page or Screen 25 Apr 2016 You may already use the Zoom Slider control at the bottom right-hand corner of Excel, as shown in Figure 1. You can use this to quickly Actual vs Targets Chart in Excel - Excel Campus 4 Nov 2019 Excel Bar Chart Actual vs Target Forecast Budget Using the slider for the Series Overlap, slide the indicator all the way to the right so that it
You can insert slicers in Excel to quickly and easily filter pivot tables. However, using the report filter gives the exact same result. How to Control percentages with a scroll bar in MS Excel
Create Excel Chart Slider | Dedicated Excel Scroll bars, or Sliders on an Excel Chart are a great addition if you are This guide will walk you through the steps in creating an Excel Chart with a scroll bar…. We use cookies to ensure that we give you the best experience on our website. I Made a Dynamic Hurricane Map with Excel! - Towards Data 25 Jul 2019 I challenged myself and made a data map using Excel. effects of the hurricane trajectory, we can use Slider Control to control the time. How to use spin buttons in Excel, interactive charts 24 Jun 2011 How to add interactive elements to an Excel worksheet using spin buttons The middle slider is not controlled by any setting so a user can get Dynamic Scroll Bar in Excel - Step by Step Tutorial
Thank you. The Mailman is on His Way :)
Sorry, don't know what happened. Try later :(
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# Supply the missing dollar amounts for the income statement of Lewis Retailers for each of the following independent cases:
Supply the missing dollar amounts for the income statement of Lewis Retailers for each of the following independent cases:
Cases Sales Revenue Beginning Inventory Purchases Cost of Goods Available for Sale Cost of Goods Sold Cost of Ending Inventory Gross Profit
A \$1,100 \$500 \$1,200 \$780
B \$1,300 \$600 \$1,200 \$580
C \$500 \$600 \$700 \$800
D \$1,280 \$1,000 \$1,050 \$650
E \$1,400 \$450 \$1,300 \$980
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https://mapleprimes.com/questions/36506-Taylor-With-Cos-theta1ttheta2t
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# Question:taylor with cos (theta1(t)-theta2(t))
## Question:taylor with cos (theta1(t)-theta2(t))
Maple
Hello everybody,
I have a big expression with many cos(theta(t)) sin((theta1(t)-theta2(t))) ...
and I would like to obtain this expression linearized by taylor 1st order, so that there will be only differentials order 2.
I tried with mtaylor and taylor, but my arguments are not accepted.
A:=(1/2)*M[2]*(-2*L[1]*(diff(Theta[1](t), t, t))*e[2]*sin(Theta[1](t)-Theta[2](t))-2*L[1]*(diff(Theta[1](t), t))*e[2]*cos(Theta[1](t)-Theta[2](t))*(diff(Theta[1](t), t)-(diff(Theta[2](t), t)))+2*S[2]*(diff((sin(Theta[2]))(t), t, t))*e[2]*sin(Theta[2](t))+2*S[2]*(diff((sin(Theta[2]))(t), t))*e[2]*cos(Theta[2](t))*(diff(Theta[2](t), t))+2*e[2]^2*(diff(Theta[2](t), t, t))+2*S[2]*(diff((cos(Theta[2]))(t), t, t))*e[2]*cos(Theta[2](t))-2*S[2]*(diff((cos(Theta[2]))(t), t))*e[2]*sin(Theta[2](t))*(diff(Theta[2](t), t)));
should I use some freeze or fronted before?
Many thanks,
Ternox
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Number Sequence Games for Preschool Online
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//
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http://math.stackexchange.com/questions/355843/are-there-any-rules-of-algebra
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# Are there any rules of algebra?
In Roger Penrose's book 'The Road to Reality', while talking about complex numbers, he suggests we test that they work for "all of the necessary rules of algebra".
The rules he stated are:
1. $w+z=z+w$
2. $w+(u+z)=(w+u)+z$
3. $wz=zw$
4. $w(uz)=(wu)z$
5. $w(u+z)=wu+wz$
6. $w+0=w$
7. $w1=w$
8. $(-1)w=-w$
9. $(-1)(-w)=w$
That was all he listed. Are there any more "rules of algebra" in a list form such as this?
Could the following count as rules of algebra: (when /$\times x$ comes after a relation, it means times by x both sides and the like)
1. $\dfrac{a}{b}=c$ $/\times b$ => $a=bc$
2. $ab=c$ /divide b => $a=\dfrac{c}{b}$
3. $a=b$ $/-a$ => $b-a=0$
4. $b-a=0$ $/+a$ => $b=a$
But then again, perhaps this second list isn't so much a list of rules, but more a list of legal actions. However, it is these legal actions that allow one to do simple algebraic rearranging and solving. These are the four actions (multiply, divide, add, subtract) that you are allowed to do with algebraic relations.
1. So would my list of 4 expressions count as rules of algebra? Or would they be laws? Or just simply actions?
2. How far can Roger Penrose's list of rules be expanded? How many more are there of this nature?
3. Is there even a fundamental theory, or rule(s) of algebra? Or is the scope of algebra far too great for there to exist a list of rules and laws without it getting inconceivably messy?
-
I also really enjoyed that book. Penrose has a real talent for conveying physical ideas to mathematicians (unlike many physicists I have read/talked to :/). – rschwieb Apr 9 '13 at 12:17
I think you must be reading too deeply into what Penrose said, because I don't have any idea what would count as an answer to your question. All of those things that are listed are algebraic properties that we find useful in algebraic objects. But not all of those properties need to be shared by all interesting objects, nor is there an exhaustive list of "laws of algebra". Rings and algebras carry enough of those rules to do addition, multiplication and subtraction. A field (or division ring) carries division, also.
The last line makes it seem as if you are taking such "laws" to be members of a collection which we are gathering to describe "all of algebra" as if we were physicists seeking laws of physics. But this is not the case: in mathematics, we pick the rules and determine their consequences. Determining what rules lead to what consequences is a matter of research, but it is not expected that such discoveries lead to "the entirety" of algebra.
One more way to try convey my meaning occurred to me. The addition, subtraction and multiplication axioms for an algebra are, by convention, the shortest list of what we want an algebra to obey. They aren't to be considered as members of an incomplete and growing list of a "grand unified theory of algebra".
-
Those rules (called axioms) in a way are the/a minimal set that describes the numbers we are familiar with and their familiar operations (this is the rational numbers, called $\mathbb{Q}$). This set was extended step by step to "complete" it, making some further operations closed (inside the set), giving the reals $\mathbb{R}$ and complex $\mathbb{C}$ numbers, all following the above axioms.
The other rules you mention don't need to be assumed from the start (like the axioms), as they can be proved from the axioms (they are theorems). This is what mixedmath's answer explains.
As rschwieb's answer says, one of the sports mathemathicians indulge in is to invent new groud rules (or look around for cases where different sets of ground rules are apparent) and see what a minimal set of axioms would be, and where those axioms lead them.
-
All four of the items you proposed follow from Penrose's list. For example, you write $a = b \implies b - a = 0$. This comes from starting with $a = b$, adding the additive inverse of $a$ to get $a -a = b- a$. On the left, though, we have $a - a =0$, as that is what we mean by the additive inverse of $a$. Thus $b - a = 0$.
Similarly, your other rules follow from his. His list is complete, but there are multiple 'things' that follow those rules. The set of real numbers, of the set of fractions, or the complex numbers, for instance.
-
An algebra is a the overarching term for a space, where the elements “interact” with each other according to certain operations/dynamics. What these operations may be, and what properties they may have, lead to the “name-calling” — i. e. naming the algebra as a group, or a ring, or field, or a boolean algebra, or a C* or $\ldots$ etc.
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# 5.3: Vertical Relationships
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Table 5-2 lists the atomic radii and the ionization potentials of the elements found in the first column of the periodic table, the group I elements.
Table 5-2: Atomic Radii and Ionization Potentials of Group I Elements
Element
Li
Na
K
Rb
Cs
1.50
1.86
2.27
2.43
2.62
I1 (ev)
5.4
5.1
4.3
4.2
3.9
The average value of the distance between the electron and the nucleus increases as the value of the principal quantum number is increased. The increase in the atomic diameters down a given group in the periodic table is thus understandable. Each of the group I elements represents the beginning of a new quantum shell. There will be a very sharp decrease in the effective nuclear charge on passing from the preceding closed shell element to a member of group I, as the number of the inner shell electrons is increased by eight. This large sudden reduction in the effective nuclear charge and the fact that the electron, because of the Pauli exclusion principle, must enter a new quantum shell, causes the group I elements to be larger in size and much more readily ionized than the preceding noble gas elements. The decrease in the effective nuclear charge and the increase in the principal quantum number down a given family bring about a steady decrease in the observed ionization potentials. Thus the outer 6s electron in cesium is on the average, further from the nucleus than is the outer 2s electron in lithium. It is also more readily removed.
So far we have considered the periodic variations in the energy required to remove an electron from an atom:
In some favourable cases it is possible to determine the energy released when an electron is added to an atom:
The magnitude of the energy released when an atom captures an extra electron is a measure of the atom's electron affinity.
It might at first seem surprising that a neutral atom may attract an extra electron. Indeed many elements do not have a detectable electron affinity. However, consider the outer electronic configuration of the group VII elements, the halogens:
ns2np5
There is a single vacancy in the outer set of orbitals and the effective nuclear charge experienced by the valence electrons in a halogen atom is almost the maximum value possible for any given row. Because of the incomplete screening of the nuclear charge by the outer electrons, the remaining vacancy in the outer shell will, in effect, exert an attractive force on a free electron large enough to bind it to the atom.
The electron affinities for the rare gas atoms will be effectively zero even though the effective nuclear charge is a maximum for this group of elements there are no vacancies in the outer set of orbitals in a rare gas atom and as a result of the Pauli principle, an extra electron would have to enter an orbital in the next quantum shell. The electron in this orbital will experience only a very small effective nuclear charge as all of the electrons originally present in the atom will be in inner shells with respect to it. Elements to the left of the periodic table, the alkali metals for example, do have vacancies in their outer quantum shell but their effective nuclear charges are very small in magnitude. Thus these elements do not exhibit a measurable electron affinity. The electron affinity increases across a given row of the periodic table and reaches a maximum value with the group VII elements. This is a direct reflection of the variation in the effective nuclear charge.
The orbital vacancy in which the extra electron is placed is found at larger distances from the nucleus when the principal quantum number is increased. Thus the electron affinity should decrease down any given family of elements in the periodic table. For example, the electron affinities for the halogens should decrease in the order F > Cl > Br > I. (Click here for note.)
The variation in the ionization potentials across a given row is reflected in the values shown in the atomic orbital energy level diagram for the elements from hydrogen through to neon (Fig. 5-3).
Fig. 5-3. An orbital energy level diagram for the elements H to Ne
(Note that the energy scale used for the 1s orbital differs by a factor of ten from that for the 2s and 1p orbitals.) The orbital energies show a uniform decrease when the nuclear charge is increased, reflecting an increase in the binding of the electrons. The total energy of a many-electron atom is not simply the sum of the orbital energies. Summing the orbital energies does not take proper account of the repulsions between the electrons. The orbital energies do, however, provide approximate estimates of the ionization potentials. The ionization potential is the energy required to remove one electron from an atom, and an orbital energy is a measure of the binding of a single electron in a given orbital. Thus the ionization potential should be approximately equal to minus the orbital energy. For example, the ionization potential of lithium is 5.39 ev and the 2s orbital energy is -5.34 ev. Similarly I1, for neon is 21.56 ev and the 2p orbital energy is -23.14 ev.
Shell structure is also evident in the ionization potentials and orbital energies of atoms. By exposing the atom to light of very short wavelength (in the X-ray region of the spectrum), it is possible to ionize one of the inner shell electrons, rather than a valence electron. That is, the energy of an X-ray photon is comparable to the binding energy of an inner shell electron. The resulting ion is in a very unstable configuration and after a very brief period of time an electron from the outer shell "falls" into the vacancy in the inner shell. In falling from an outer to an inner shell the binding of the electron is greatly increased and a photon is emitted. The energy of this photon should be approximately equal to the difference in energies of the outer shell and inner shell orbitals. For example, the photon emitted when neon loses an inner shell electron has an energy of 849 ev. The difference in energy between the 2p and 1s orbitals of neon is 869 ev. Photons with energies in this range occur in the X-ray region of the spectrum. It is apparent from the variation in the 1s orbital energies shown in Fig. 5-3 that the energies and hence the frequencies of the X-ray photons will increase as the nuclear charge is increased. It was from a study of the X-ray photons emitted by the elements that Moseley was first able to determine the nuclear charges (the atomic numbers) of the elements.
This page titled 5.3: Vertical Relationships is shared under a CC BY-NC-SA 4.0 license and was authored, remixed, and/or curated by Richard F. W. Bader via source content that was edited to the style and standards of the LibreTexts platform; a detailed edit history is available upon request.
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| 793,525,791
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## Class/Course - Class XII
### Subject - Physics
#### Chapter - Electric Charges and Fields
Attention Class XII standard Aspirants! Just Exam provide a platform to all students who want to make practice for various subject online. We prepare this platform on the base of CBSE. If you want to join our unlimited test series, then please write us at info@justexam.in about your class/course
Dear User, Kindly login/register to view answer & explanation of each question. Click here to Login/Sign Up. Q.1 A tiny electric dipole of dipole moment $\vec{P}$ = $P_{0}\hat{j}$ is placed at point (l,p). There exists an electric field $\vec{E}$ = 2ax2$\hat{i}$ +(2by2 + 2cy)$\hat{j}$. Force on dipole is $2P_{0}a\hat{i}$ Force on dipole is $2P_{0}b\hat{i}$ Force on dipole is $2P_{0}c\hat{i}$ Force on dipole is $-2P_{0}c\hat{i}$
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| 2.96875
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CC-MAIN-2020-40
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latest
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en
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https://www.stockpricetrends.com/company/ILMN
| 1,723,179,713,000,000,000
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text/html
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crawl-data/CC-MAIN-2024-33/segments/1722640759711.49/warc/CC-MAIN-20240809044241-20240809074241-00238.warc.gz
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|
# Illumina Inc (ILMN)
Illumina, Inc. is an American company. Incorporated on April 1, 1998, Illumina develops, manufactures, and markets integrated systems for the analysis of genetic variation and biological function. The company provides a line of products and services that serves the sequencing, genotyping and gene expression, and proteomics markets. Its headquarters are located in San Diego, California.
## Stock Price Trends
Stock price trends estimated using linear regression.
## Paying users area
The data is hidden behind and trends are not shown in the charts.
Unhide data and trends.
This is a one-time payment. There is no automatic renewal.
#### Key facts
• The primary trend is decreasing.
• The decline rate of the primary trend is 36.07% per annum.
• ILMN price at the close of August 7, 2024 was \$121.10 and was inside the primary price channel.
• The secondary trend is increasing.
• The growth rate of the secondary trend is 138.07% per annum.
• ILMN price at the close of August 7, 2024 was inside the secondary price channel.
• The direction of the secondary trend is opposite to the direction of the primary trend. This indicates a possible reversal in the direction of the primary trend.
### Linear Regression Model
Model equation:
Yi = α + β × Xi + εi
Top border of price channel:
Exp(Yi) = Exp(a + b × Xi + 2 × s)
Bottom border of price channel:
Exp(Yi) = Exp(a + b × Xi – 2 × s)
where:
i - observation number
Yi - natural logarithm of ILMN price
Xi - time index, 1 day interval
σ - standard deviation of εi
a - estimator of α
b - estimator of β
s - estimator of σ
Exp() - calculates the exponent of e
### Primary Trend
Start date:
End date:
a =
b =
s =
Annual growth rate:
Exp(365 × b) – 1
= Exp(365 × ) – 1
=
Exp(4 × s) – 1
= Exp(4 × ) – 1
=
#### January 27, 2021 calculations
Top border of price channel:
Exp(Y)
= Exp(a + b × X + 2 × s)
= Exp(a + b × + 2 × s)
= Exp( + × + 2 × )
= Exp()
= \$
Bottom border of price channel:
Exp(Y)
= Exp(a + b × X – 2 × s)
= Exp(a + b × – 2 × s)
= Exp( + × – 2 × )
= Exp()
= \$
#### August 6, 2024 calculations
Top border of price channel:
Exp(Y)
= Exp(a + b × X + 2 × s)
= Exp(a + b × + 2 × s)
= Exp( + × + 2 × )
= Exp()
= \$
Bottom border of price channel:
Exp(Y)
= Exp(a + b × X – 2 × s)
= Exp(a + b × – 2 × s)
= Exp( + × – 2 × )
= Exp()
= \$
#### Description
• The primary trend is decreasing.
• The decline rate of the primary trend is 36.07% per annum.
• ILMN price at the close of August 7, 2024 was \$121.10 and was inside the primary price channel.
### Secondary Trend
Start date:
End date:
a =
b =
s =
Annual growth rate:
Exp(365 × b) – 1
= Exp(365 × ) – 1
=
Exp(4 × s) – 1
= Exp(4 × ) – 1
=
#### May 21, 2024 calculations
Top border of price channel:
Exp(Y)
= Exp(a + b × X + 2 × s)
= Exp(a + b × + 2 × s)
= Exp( + × + 2 × )
= Exp()
= \$
Bottom border of price channel:
Exp(Y)
= Exp(a + b × X – 2 × s)
= Exp(a + b × – 2 × s)
= Exp( + × – 2 × )
= Exp()
= \$
#### August 7, 2024 calculations
Top border of price channel:
Exp(Y)
= Exp(a + b × X + 2 × s)
= Exp(a + b × + 2 × s)
= Exp( + × + 2 × )
= Exp()
= \$
Bottom border of price channel:
Exp(Y)
= Exp(a + b × X – 2 × s)
= Exp(a + b × – 2 × s)
= Exp( + × – 2 × )
= Exp()
= \$
#### Description
• The secondary trend is increasing.
• The growth rate of the secondary trend is 138.07% per annum.
• ILMN price at the close of August 7, 2024 was inside the secondary price channel.
| 1,071
| 3,455
|
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| 2.796875
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CC-MAIN-2024-33
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latest
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en
| 0.866737
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https://paperzz.com/doc/9071232/curse.demo
| 1,670,142,520,000,000,000
|
text/html
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crawl-data/CC-MAIN-2022-49/segments/1669446710968.29/warc/CC-MAIN-20221204072040-20221204102040-00581.warc.gz
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| 6,579
|
### Curse.demo
```Curse of dimensionality
Applying ML algorithms to highlydimensional data – an illustration
Barbora Hladká and Martin Holub, Charles University in Prague, NPFL 054, 2016/17
I. Distribution of feature vectors
– each feature is binary, with the same binomial distribution
N
dim
prob
<- 10^6
<- 7
<- 1/10
# number of observations
# number of dimensions
# probability of Wi=1
binom_vector <- character(N)
for(i in 1:N) binom_vector[i] <- paste(rbinom(dim,1,prob), collapse="")
expected_values = 2^dim
emerged_values = length(unique(binom_vector))
print( sort(table(binom_vector), dec=T) )
observations = 1,000,000
dimensions = 7
p(x = 1) = 0.1
number of possible different values = 128
number of emerged different values = 125
0000000
477890
1000100
5981
0100010
5867
1001010
704
0101010
667
1001001
639
0100101
622
0010111
76
0011101
72
0110011
66
1101011
13
1110110
8
1111001
3
0010000
53566
0010001
5967
1000010
5844
1000110
700
1100010
665
0110010
637
0001011
616
0101101
76
1000111
72
1100101
66
1001111
12
1010111
7
0111111
2
0100000
53305
0000011
5961
1100000
5843
0010110
692
1010010
663
0010101
635
1011000
614
1011001
75
1010011
72
0110101
65
1101110
11
1111010
7
1111011
2
0001000
53303
0000101
5923
0101000
5815
0101100
690
0000111
656
1010100
635
0110001
589
1011100
75
1010110
72
0101110
63
0110111
10
1110011
6
1101111
1
0000001
53279
0001001
5901
0100001
5814
0100011
686
0011100
656
0011001
634
1011010
95
1101001
75
1111000
72
0111010
62
0111101
10
1111100
6
1111110
1
0000010
53272
0011000
5899
0100100
5810
1001100
680
0100110
652
1100001
634
1001011
87
0011110
73
0100111
71
1110010
62
0101111
9
1011011
5
1000000
53249
0110000
5898
1010000
5779
0001101
673
0110100
648
1000101
632
0110110
83
0111001
73
0101011
71
0001111
56
1101101
9
1100111
5
0000100
53023
0010100
5896
1000001
5778
1100100
673
0111000
648
1101000
630
1001101
78
1101100
73
1100011
70
0111100
56
0011111
8
0111110
4
0001010
6100
0001100
5893
0010010
5722
1010001
672
0011010
646
1110000
630
1001110
77
1110100
73
1101010
69
1100110
55
0111011
8
1011101
4
1001000
5995
0000110
5886
0010011
714
0001110
669
0101001
640
1000011
626
1010101
77
0011011
72
1110001
68
1011110
14
1110101
8
1110111
3
II. Distribution of distances from 0 of randomly distributed points in a unit cube
dim
n
<- 6
<- 10000
cube <- data.frame(
x1 = runif(n),
x2 = runif(n),
x3 = runif(n),
x4 = runif(n),
x5 = runif(n),
x6 = runif(n)
)
distances <- numeric(n)
for(i in 1:n) distances[i] <- sqrt(sum(cube[i,]^2))
greater_than_1 <- sum(distances > 1)
message("Most of the distances (",
format(greater_than_1/n*100, digits=3), "%) are greater than 1.")
message("Frequency of distances in intervals:")
print(table(cut(distances, breaks=seq(0, 2.5, 0.5))))
---------------------------------------------------------------This program generates 10,000 random 6-dimensional sample points in a unit cube.
Maximum possible distance from 0 is: 2.45
Distances from 0 in the sample of 10000 points:
Min. 1st Qu. Median
Mean 3rd Qu.
Max.
0.324
1.214
1.407
1.390
1.578
2.183
Most of the distances (91.7%) are greater than 1.
Frequency of distances in intervals:
(0,0.5] (0.5,1] (1,1.5] (1.5,2] (2,2.5]
7
824
5591
3529
49
III. Distribution of mutual distances between randomly distributed points in a unit cube
dim
n
d
lim
<<<<-
6
150
choose(n,2)
0.5
message("Maximum possible distance between two points is: ",
format(sqrt(6), digits=3) )
message("Number of different pairs is: ", d)
cube <- data.frame(
x1 = runif(n),
x2 = runif(n),
x3 = runif(n),
x4 = runif(n),
x5 = runif(n),
x6 = runif(n)
)
distances <- numeric(d)
k <- 1
for(i in 1:(n-1) ) for(j in (i+1):n ) {
distances[k] <- sqrt( sum((cube[i,]-cube[j,])^2) ); k <- k+1
}
greater_than_lim <- sum(distances > lim)
message("Most of the distances (",
format(greater_than_lim/d*100, digits=3), "%) are greater than ", lim, ".")
message("Frequency of distances in intervals:")
print(table(cut(distances, breaks=seq(0, 2.5, 0.25))))
---------------------------------------------------------------This program generates 150 random 6-dimensional sample points in a unit cube.
Maximum possible distance between two points is: 2.45
Number of different pairs is: 11,175
Mutual distances in the sample of 150 points:
Min. 1st Qu. Median
Mean 3rd Qu.
Max.
0.1173 0.8098 0.9797 0.9732 1.1420 1.8770
Most of the distances (97%) are greater than 0.5.
Frequency of distances in intervals:
(0,0.25] (0.25,0.5] (0.5,0.75]
9
322
1704
(1.75,2]
4
(2,2.25] (2.25,2.5]
0
0
(0.75,1]
3914
(1,1.25] (1.25,1.5] (1.5,1.75]
3793
1301
128
```
| 1,806
| 4,530
|
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| 2.625
| 3
|
CC-MAIN-2022-49
|
latest
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en
| 0.155176
|
https://www.atilim.edu.tr/en/ects/site-courses/208/9454/detail
| 1,722,991,097,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2024-33/segments/1722640523737.31/warc/CC-MAIN-20240806224232-20240807014232-00612.warc.gz
| 521,919,983
| 21,226
|
# Probability and Statistics I (IE201) Course Detail
Course Name Course Code Season Lecture Hours Application Hours Lab Hours Credit ECTS
Probability and Statistics I IE201 3. Semester 3 0 0 3 7
Pre-requisite Course(s)
N/A
Course Language English Compulsory Departmental Courses Bachelor’s Degree (First Cycle) Face To Face Lecture, Question and Answer, Problem Solving. Prof. Dr. Serkan Eryılmaz The course aims to expose students to basic concepts of probability and statistics and to use basic modeling and statistical decision techniques in Industrial Engineering applications. The students who succeeded in this course; Students will have a working knowledge of probability theory. Students will acquire the basic concepts of probability theory and apply them to engineering problems. Students will understand the relationship between random variables and their distribution functions. Students will improve their problem solving skills and their analytical thinking ability. The role of statistics in engineering, probability, random variables, expected value and variance of a random variable, discrete and continuous probability distributions, joint probability distributions
### Weekly Subjects and Releated Preparation Studies
Week Subjects Preparation
1 The role of probability, statistics and data analysis in engineering
2 Sample space, Probability of an event
3 Axioms of probability, Conditional probability
4 Independence, Bayes’ Rule
5 Random variables and probability distributions
6 Joint probability distributions
7 Mathematical expectation, Variance
8 Midterm
9 Covariance and correlation
10 Some discrete probability distributions
11 Some continuous probability distributions
12 Functions of random variables
13 Sampling distributions
14 Sampling distributions (cont.)
15 Final Examination Period
16 Final Examination Period
### Sources
Course Book 1. Walpole, R.E., Myers, R.H., Myers, S.L. and Ye, K., Probability and Statistics for Engineers and Scientists, Prentice Hall, 2007. 2. Montgomery, D.C., and Runger, G.C., Applied Statistics and Probability for Engineers, 5th Edition, John Wiley and Sons, 2011.
### Evaluation System
Attendance/Participation - -
Laboratory - -
Application - -
Field Work - -
Special Course Internship - -
Quizzes/Studio Critics 2 20
Homework Assignments - -
Presentation - -
Project - -
Report - -
Seminar - -
Midterms Exams/Midterms Jury 1 30
Final Exam/Final Jury 1 50
Toplam 4 100
Percentage of Semester Work 50 50 100
### Course Category
Core Courses X
### The Relation Between Course Learning Competencies and Program Qualifications
# Program Qualifications / Competencies Level of Contribution
1 2 3 4 5
1 Acquires sufficient knowledge in mathematics, natural sciences, and related engineering disciplines; gains the ability to use theoretical and applied knowledge in these fields in solving complex engineering problems. X
2 Gains the ability to identify, define, formulate, and solve complex engineering problems; acquires the skill to select and apply appropriate analysis and modeling methods for this purpose. X
3 Gains the ability to design a complex system, process, device, or product to meet specific requirements under realistic constraints and conditions, and applies modern design methods for this purpose.
4 Develops the skills to develop, select, and use modern techniques and tools necessary for the analysis and solution of complex problems encountered in industrial engineering applications; gains the ability to effectively use information technologies.
5 Gains the ability to design experiments, conduct experiments, collect data, analyze and interpret results for the investigation of complex engineering problems or discipline-specific research topics.
6 Acquires the ability to work effectively in intra-disciplinary and multidisciplinary teams, as well as individual work skills.
7 Acquires effective oral and written communication skills in Turkish; at least one foreign language proficiency; gains the ability to write effective reports, understand written reports, prepare design and production reports, make effective presentations, and give and receive clear instructions.
8 Develops awareness of the necessity of lifelong learning; gains the ability to access information, follow developments in science and technology, and continuously renew oneself.
9 Acquires the consciousness of adhering to ethical principles, and gains professional and ethical responsibility awareness. Gains knowledge about the standards used in industrial engineering applications.
10 Gains knowledge about practices in the business life such as project management, risk management, and change management. Develops awareness about entrepreneurship and innovation. Gains knowledge about sustainable development.
11 Gains knowledge about the universal and social dimensions of the impacts of industrial engineering applications on health, environment, and safety, as well as the problems reflected in the engineering field of the era. Gains awareness of the legal consequences of engineering solutions.
12 Gains skills in the design, development, implementation, and improvement of integrated systems involving human, material, information, equipment, and energy.
13 Gains knowledge about appropriate analytical and experimental methods, as well as computational methods, for ensuring system integration.
Activities Number Duration (Hours) Total Workload
Course Hours (Including Exam Week: 16 x Total Hours) 16 3 48
Laboratory
Application
Special Course Internship
Field Work
Study Hours Out of Class 14 3 42
Presentation/Seminar Prepration
Project
Report
Homework Assignments 7 5 35
Quizzes/Studio Critics
Prepration of Midterm Exams/Midterm Jury 1 25 25
Prepration of Final Exams/Final Jury 1 25 25
| 1,131
| 5,778
|
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| 2.8125
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CC-MAIN-2024-33
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latest
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| 0.854464
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http://letsmakerobots.com/node/16974
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crawl-data/CC-MAIN-2015-18/segments/1429246643283.35/warc/CC-MAIN-20150417045723-00159-ip-10-235-10-82.ec2.internal.warc.gz
| 51,433,536
| 7,862
|
# A multimeter question
Quick question- Why is it that when I set my multimeter to 9v battery it says my solar panel is putting out 0.89v and when I put it to 20v (not battery) it says my solar panel is putting out 2.10v?
## Comment viewing options
im thinking you are not familiar with range when using a multimeter . set the multimeter range to a slightly higher voltage than what you are measuring , for example if your solar panel (source) is rated at 20 V set the range of the multimeter to that . some multimeters have an auto-range function but as it appears yours don't , a picture of the multimeter or its make will help us guide you better :)
I think I understand now... So to measure the real voltage put out by the solar panel, I would put it on the ~V (20) setting instead of the Battery (9v), right?
Thanks!
Does your multimeter have a ±V(20) setting? The little '~' in front of the V normally means you're measuring alternating current, you want DC.
Do you actually have a setting called "9V Battery" on your VOM? You need to set it to DC voltage above the solar panels voltage output. If indeed you have a "9V battery" setting, then you have a meter that stress tests batteries. Which means that it connects the voltage source to a load resistor to test the strength/life of the 9V battery. It is not intended to measure DC voltage.
I've got a \$3 harbor freight special multimeter with a "battery checker" setting. It actually says "1.5V(4.0mA)" and "9V(25mA)" on the dial, so I've always taken it to mean "how much balls the battery actually has left" (like you said, a stress test). Just cause a battery still puts out 9V doesn't mean it has any oomph behind it.
What I don't get why it's labelled "mA" though. A standard AA should be capable of putting out quite a bit more than 4 milliamps I thought.
Fun with cheap ass tools. That's why I also have a Fluke that's near as old as I am for checking real things ;)
Ya I bought a cheap \$20 Meter from Amazon.com that tests the usual Voltage Amps and Ohms plus it tests Capacitors and Transistors and has a larger range in each scope than other more expensive ones that were available at Amazon.
It sure beat my old \$20 Radio Shack one hands down.
I think you have a weird MM....
| 550
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| 2.59375
| 3
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CC-MAIN-2015-18
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latest
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en
| 0.952302
|
bayarearealestatemom.com
| 1,675,837,218,000,000,000
|
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crawl-data/CC-MAIN-2023-06/segments/1674764500719.31/warc/CC-MAIN-20230208060523-20230208090523-00316.warc.gz
| 140,938,662
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# Debt to Income
Click Here for Debt to Income Calculator
So in a previous video you might remember that I talked about improving your credit score in order to qualify for a mortgage. In that video I talked about debt to income ratios. So in this video I want to go a little deeper into that and help you calculate what your debt to income ratio is.
I’ve created a super simple spreadsheet that I’m going to share with you in the description I’ll put a link to it in the description below. Where you can figure out your debt to income what it is today. I also want to caution you because you know the lender’s going to tell you that they need to see a debt to income ratio of a certain amount it might be you know 42 it could be all the way up to like 55 percent but even if they say that you can afford it I want you to think about it on your own and for yourself. You need to decide if you can afford it because you’re the one committing in most cases 30 years of paying that debt back.
So you got to be comfortable with it and a lot of people just take whatever the lender tells them and says “Oh look I can buy this house and this is my debt and they say I can afford it so I’m good”. But a lot of people find that that wasn’t the best way to go and that you might have other plans you might not want to keep you know working as hard as you are right now for the next 30 years and you might want to just buy a little less house than you can qualify for and that’s perfectly okay to do just because they tell you this is the maximum amount you can get does not mean that you have to spend the maximum amount and I want you to not spend the maximum amount if you don’t have to.
So let’s go ahead and take a look at the spreadsheet I have for determining your debt to income ratio and then you’ll be able to download it yourself and try it out for yourself so here’s the spreadsheet for the debt to income ratio.
As you can see up here at the top it says income sources so this is any income that you have that you’re going to use to qualify for that mortgage you want to put that in here. To make it simple on this video let’s just go with an annual salary. Let’s say you have an annual salary of a \$100,000 a year.
There’s another category here because sometimes a lot of people have extra income. Maybe you get an annual bonus and it’s really consistent and you’ve gotten it every year. For the last 10 years maybe you have a side job maybe you have an extra weekend job that you do, it is if you’ve been doing it for a few years and it’s something that you know you’re going to continue doing then let’s go ahead and add that in as well so let’s just say you make another \$10,000 a year doing whatever that extra income is.
Let’s look at debt sources so we’re going to leave this blank right here for the mortgage because we don’t yet know how much money you’re going to spend on a mortgage so we’ll get to that at the end but let’s take a look at other debts that you have and you know car loans maybe you have an RV or a boat or a motorcycle.
Let’s make some assumptions let’s say assume you have a car loan of \$300 a month and we’re gonna put that in the monthly category so it calculates correctly. Let’s say you have I don’t know you bought a boat and it’s \$200 a month and let’s say you have some a little bit of credit card dad that you carry over and you’re not planning to pay it off before you get a mortgage and let’s say that’s another \$200 a month and let’s say that you went to college and you didn’t pay off all your debt yet and you still have several years left before that gets paid off and you have let’s say \$400 a month of student loan debt.
If there’s any other debt that you can think of that’s monthly that’s going to come up and I’m not talking about electricity bills or gas for your car or maintenance for your car or cellphone. Nothing like that those aren’t monthly debts from money that you borrowed those are just living expenses so let’s not calculate that in. So what we have there is \$1,100 a month in debt and an income of \$110,000.
Remember this income is before taxes this is not your after tax income so this is your gross income. That gives you a debt to income ratio of 12% which is amazing it’s 12% very low. However we’re not factoring in what your mortgage is or what your rent might be so we’re not even considering that yet at this point.
So let’s say you have save some money for a down payment and you’re ready to buy a house and in California things are expensive. Let’s say you are going to buy a house for \$700,000 and you need a down payment of 20% now not every loan requires 20% down. Let’s say you have a 20% down payment so you’re looking at a mortgage amount of \$560,000 let’s plug that in. Remember it’s a 30-year loan in most cases unless you have a really big down payment or a really big income and you can afford a 15-year mortgage let’s go with a 30-year mortgage.
At an interest rate let’s say it’s 3.5 because that’s what’s projected by California association realtors to happen by the end of 2022 they anticipate that 3.5 is where mortgage rates are going to be in 2022.
Now let’s add in property tax it’s going to be different depending on what county, what city, what state it’s going to depend on that. I’m going to put \$8,400 for a \$700,000 house and annual homeowners insurance let’s just say a \$1,000 depends on what options you pick how much coverage you have, how low your deductible is. Those are all variables when you choose insurance so let’s just go with a round number of a thousand. And what we have here is a payment a monthly payment of \$2514.65. That is just principal and interest that doesn’t include the taxes and the insurance. So if we want that number which is the PITI that goes to \$3297.98
Let’s take that number and edit it back into the spreadsheet and see what the debt to income ratio is now. Our debt to income ratio now is 47.98%. That’s a little bit high and remember that this is an amount of your based on your income before your income taxes taken from your paycheck.
Could you qualify at this amount? Yes! There are lenders out there that will lend to you at this debt to income ratio. Should you take a loan at this amount? I’d be real hesitant to do it because you’re committing to 30 years and once you factor in your income tax, that’s going to be way more than 50 of your income so really be cautious
What could you do to change this? Let’s say in your area that \$700,000 price point is barely gonna get you into a house that you want. What can you do to fix that? Well one thing that you can do is get rid of some of this debt.
I know that everybody likes a new car but you can’t live in your car. Your car is not going to help you and that car is only going to go down in value so if you could pay off that \$300 a month car loan and still keep driving that car. That would help significantly.
Let’s say you have no car loans take that out how does that change it so it changes it down to 44.71%. That’s a big improvement. What if you just get rid of that whatever you got, your boat, your motorcycle, your RV, whatever that is, do you really need it? Wouldn’t you rather put that \$200 a month towards your house by maybe fixing it up or saving for a rainy day in case the roof leaks or something happens?
Let’s take that out now we’re getting down to close to 42% and this is right in the number where I think most lenders prefer to see. If you could pay off those credit cards, how much better would that be for you? Now you’re at 40% which is much more doable than than close to 50.
I think this is a good number to be at and I just wanted to show you how much it can matter for you by having that car loan, the credit card debt, it makes a significant impact on the house you can buy and you got to decide like what is more important to you and that’s how you use a debt to income calculator.
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# Mathematics 1. (Integration)
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1 Mthemtics 1. (Integrtion) University of Debrecen fll
2 Definition Let I R be n open, non-empty intervl, f : I R be function. F : I R is primitive function of f if F is differentible nd F = f on I.
3 Definition Let I R be n open, non-empty intervl, f : I R be function. F : I R is primitive function of f if F is differentible nd F = f on I. If F, G : I R re primitive functions of f, then F G is constnt.
4 Definition Let I R be n open, non-empty intervl, f : I R be function. F : I R is primitive function of f if F is differentible nd F = f on I. If F, G : I R re primitive functions of f, then F G is constnt. Nottions for the primitive function f, or f (x)dx
5 Fundmentl primimtive functions: 1dx = x + c; x r dx = x r+1 r+1 + c; r 1 1 x = ln x + c; e x dx = e x + c; sin(x)dx = cos(x) + c; cos(x)dx = sin(x) + c; x dx = x ln dx, > 0 1 cos 2 (x) dx = tn(x) + c; 1 1+x 2 dx = rctn(x) + c
6 Let f, g : I R be functions nd, b R be constnts. Assume tht there exist primitives of f nd g, (f + bg) = f + b g.
7 Methods for integrtion: I.) If F is the primitive function of f, then for ll, b R 0, then f (x + b)dx = F (x + b) + c. Exmples () (2x 3) 10 dx = (2x 3) c; (b) 1 2 5x dx = (2 5x) c; 2 ( 5) (c) cos(4x 7)dx = sin(4x 7) 4 + c; (d) 3 2 x dx = 32 x ln(3)( 1) + c; (e) e 2x+1 dx = e 2x c.
8 Methods for integrtion: II.) If f : I R is differentible function, then f (x) dx = ln f (x) + c. f (x) Exmples () x+1 x 2 +2x 1 dx = 1 2 ln x 2 + 2x 1 + c; (b) x 2 x(x 4) dx = 1 2 ln x 2 4x + c; (c) 1 x ln x dx = ln ln(x) + c; (d) tn(x)dx = ln cos(x) + c; (e) e 2x 1+e dx = 1 2x 2 ln(1 + e2x ) + c.
9 Methods for integrtion: III.) If f : I R is differentible function, then for ll k 1 f k (x) f (x)dx = f k+1 (x) k c. If k = 1, then we cn pply Method II. Exmples () 2x(1 + x 2 ) 3 dx = (1+x 2 ) c; (b) x 2 (2x 3 + 4)dx = 1 (2x 3 +4) c; (c) sin 3 (x) cos(x)dx = sin4 (x) 4 + c; (d) 4x 3 + 2x 2 dx = (3+2x 2 ) c; (e) x (1+x 2 ) dx = 1 (1+x 2 ) c.
10 Methods for integrtion: IV.) If f : I R is differentible function, then e f (x) f (x)dx = e f (x) + c. Exmples () (1 + e x 1 )dx = x + e x 1 + c; (b) x e x 2 dx = 1 2 e x 2 + c.
11 (Integrtion by prts) Assume tht there exist primitives of f nd g. If dditionlly f nd g re differentible, then f (x)g(x)dx = f (x)g(x) f (x)g (x)dx. Exmples () e 2x (2x + 3)dx = e2x e2x 2 (2x + 3) 2 ; (b) e x (x 2 + 3)dx = e x (x 2 + 3) 2xe x + 2e x + c; (c) sin(x)(x + 14)dx = cos(x)(x + 14) + sin(x) + c; (d) ln(x)dx = x ln(x) x + c; (e) x 2 ln(x)dx = x 3 3 ln(x) x c.
12 (Chnge of vribles) Let g : I R be strictly monotone, differentible function. If there is primitive of f : g(i ) R, then there exists primitive of (f g)g nd ( ) f (x)dx g(x) = (f g)(x)g (x)dx. Exmples () sin(2x + 1)dx; (b) x cos(x 2 + 1)dx; (c) e 2x e x +1 dx.
13 Integrtion, Newton-Leibniz formul F (x) := x f (t)dt
14 Integrtion, Newton-Leibniz formul F (x) := x f (t)dt If f is Riemnn integrble, then F is continuous.
15 Integrtion, Newton-Leibniz formul F (x) := x f (t)dt If f is Riemnn integrble, then F is continuous. If f is continuous, then F is differentible, nd F (x) = f (x).
16 Integrtion, Newton-Leibniz formul F (x) := x f (t)dt If f is Riemnn integrble, then F is continuous. If f is continuous, then F is differentible, nd F (x) = f (x). (Fundmentl theorem of clculus) If f is continuous nd F is primitive function of f, then b f (x)dx = F (b) F ().
17 Integrtion, Newton-Leibniz formul F (x) := x f (t)dt If f is Riemnn integrble, then F is continuous. If f is continuous, then F is differentible, nd F (x) = f (x). (Fundmentl theorem of clculus) If f is continuous nd F is primitive function of f, then b f (x)dx = F (b) F (). It is enough to ssume tht f is Riemnn integrble, nd it hs differentible primitive function F.
18 Integrtion, Integrtion by prts, Chnge of vribles Let g : [A, B] R be strictly monotone incresing, continuously differentible function nd g(a) =, g(b) = b. If f : [, b] R is continuous, then b f (x)dx = B A f (g(y))g (y)dy.
19 Integrtion, Integrtion by prts, Chnge of vribles Let g : [A, B] R be strictly monotone incresing, continuously differentible function nd g(a) =, g(b) = b. If f : [, b] R is continuous, then b f (x)dx = B A f (g(y))g (y)dy. If g, f : [, b] R re differentible functions with integrble derivtives, then b f (x)g(x)dx = f (b)g(b) f ()g() b f (x)g (x)dx.
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Properties of the Riemnn Stieltjes Integrl Theorem (Linerity Properties) Let < c < d < b nd A,B IR nd f,g,α,β : [,b] IR. () If f,g R(α) on [,b], then Af +Bg R(α) on [,b] nd [ ] b Af +Bg dα A +B (b) If
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### Assignment 6 Solution. Please do not copy and paste my answer. You will get similar questions but with different numbers!
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Czechoslovk Mthemticl Journl, 55 (130) (2005), 933 940 ESTIMATES OF THE REMAINDER IN TAYLOR S THEOREM USING THE HENSTOCK-KURZWEIL INTEGRAL, Abbotsford (Received Jnury 22, 2003) Abstrct. When rel-vlued
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Riemnn Sums nd Riemnn Integrls Jmes K. Peterson Deprtment of Biologicl Sciences nd Deprtment of Mthemticl Sciences Clemson University August 26, 203 Outline Riemnn Sums Riemnn Integrls Properties Abstrct
### Harman Outline 1A1 Integral Calculus CENG 5131
Hrmn Outline 1A1 Integrl Clculus CENG 5131 September 5, 213 III. Review of Integrtion A.Bsic Definitions Hrmn Ch14,P642 Fundmentl Theorem of Clculus The fundmentl theorem of clculus shows the intimte reltionship
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## Integrals involving roots
by: Alejandra D. Rocha
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# Integrals involving roots 2314
Marketplace > University of Texas at El Paso > Math > 2314 > Integrals involving roots
Alejandra D. Rocha
UTEP
GPA 3.1
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we’re going to look at an integration technique that can be useful for some integrals with roots in them. We’ve already seen some integrals with roots in them. Some can be done quickly with a sim...
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Date Created: 06/10/16
Integrals involving roots Let’s look at a couple of examples to see another technique that can be used on occasion to help with these integrals. Example 1 Evaluate the following integral. Solution Sometimes when faced with an integral that contains a root we can use the following substitution to simplify the integral into a form that can be easily worked with. So, instead of letting u be the stuff under the radical as we often did in Calculus I we let u be the whole radical. Now, there will be a little more work here since we will also need to know what x is so we can substitute in for that in the numerator and so we can compute the differential, dx. This is easy enough to get however. Just solve the substitution for x as follows, Using this substitution the integral is now, So, sometimes, when an integral contains the root the substitution, can be used to simplify the integral into a form that we can deal with. Let’s take a look at another example real quick. Example 2 Evaluate the following integral. Solution We’ll do the same thing we did in the previous example. Here’s the substitution and the extra work we’ll need to do to get x in terms of u. With this substitution the integral is, This integral can now be done with partial fractions. Setting numerators equal gives, Picking value of u gives the coefficients. The integral is then, So, we’ve seen a nice method to eliminate roots from the integral and put it into a form that we can deal with. Note however, that this won’t always work and sometimes the new integral will be just as difficult to do.
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# The Farmer's Puzzle Years 4 - 10
A mathematician's work begins with an interesting problem.
The focus of this activity is on learning the steps a mathematician takes when they find a problem that interests them.
### Preparation
G'day young mathematician.
You are just starting this problem, so you really don't know what you need yet.
Some stuff you could have around just in case might be:
• Objects like a collection of plastic screw caps that can be used to represent things in the problem.
• Blank paper and markers for drawing and doodling as you think through the problem.
• A piece or two of this Square Line Paper.
Two (2) things you know you always need when you are learning to work like a mathematician are:
• The Working Mathematically page. Printing it for this activity is probably best.
• Your Maths Journal with the name of the activity and today's date.
Mathematicians never solve a problem straight away. So, they need to keep a diary of their explorations.
That's what happens in your Maths Journal.
These are the steps in solving all problems:
Have fun exploring The Farmer's Puzzle
while you are learning to work like a mathematician.
• In a moment you are going to click a link to see a slide of the farmer's puzzle.
• Read it through once or twice - no more ... then close the file so you can't see the writing.
• Use a marker on blank paper to write or draw or doodle things you remember from the puzzle.
• Then open the link again and check and change what you remembered.
Here is The Farmer's Puzzle ...
click the drawing to find it.
Artist: Rob Mullarvey
Allow full screen when asked by the slide.
Hint
• What did she do?
• She bought some animals. Right?
• Sure, but what was special about how she spent her money?
Copy and complete this sentence in your journal.
I have to find out...
### 2. Plan A Strategy
NO PENS FOR THIS PART.
If you have a pen in your hand you will start doing something before you have planned what to do. Sit on your hands if you have to.
• Read through the strategy toolbox on the Working Mathematically page.
• Which strategies do you think might work? - guess & check or make a model or draw a diagram or make a table or write an equation or ...?
• When you choose one, what will be the first thing you do when you are allowed to pick up a pen?
... (Pause for thinking) ...
Okay. If you know where you are going to start...
### 3. Do It
Pick up a pen or anything else you need and start.
• Starting is all a mathematician can do.
• They know what they are aiming for
- buy 100 animals and spend \$100
- but they don't know how they will get there or how long it will take to get there.
• If they did ... it wouldn't be a problem.
Enjoy the journey. You will know when you have the answer because you will know how many of each animal the farmer bought.
There is one hint at the bottom of the page, but don't look unless you really, really, have to.
### 4. Check & Reflect
Check
So you spent \$100 and you know how many of each animal and there are a total of 100 animals. You should be right, but...
Reflect
• Is there another solution? If not, how can you be sure? If so, how can you be sure you have found them all?
• What happens if the problem is still 100 animals and 100 dollars, but the prices are different?
• Is 100 such a special number? Could a similar problem be constructed with say 80 animals for \$80. How about for other numbers?
One class decided to try explore this last one. These three (3) slides will show you something about what happened. If you have the time, you could try to solve at least one of their puzzles.
### Just Before You Finish
Prepare a report for your colleagues to explain what the problem is and how you solved it.
Perhaps you could make a video with your phone.
You might also find a way to use this Professor Morris Puzzle Poster.
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Cody
# Problem 154. Reverse Boggle
Solution 192313
Submitted on 17 Jan 2013
This solution is locked. To view this solution, you need to provide a solution of the same size or smaller.
### Test Suite
Test Status Code Input and Output
1 Fail
%% x = ['TIPE' 'YECV' 'LSRA' 'WOTU']; y = 'RACIEST'; assert(isequal(boggle_checker(x,y),true))
2 Fail
%% x = ['TIPE' 'YECV' 'LSRA' 'WOTU']; y = 'RACIESTS'; assert(isequal(boggle_checker(x,y),false))
3 Fail
%% x = ['TIPE' 'YECV' 'LSRA' 'WOTU']; y = 'RACIESTW'; assert(isequal(boggle_checker(x,y),false))
4 Fail
%% x = ['TIPE' 'YECV' 'LSRA' 'WOTU']; y = 'AUTOLYTIC'; assert(isequal(boggle_checker(x,y),true))
5 Fail
%% x = ['TIPE' 'YECV' 'LSRA' 'WOTU']; y = 'RESTAR'; assert(isequal(boggle_checker(x,y),false))
6 Fail
%% x = ['OCEW' 'LRIR' 'GYSI' 'KREM']; y = 'SIRI'; assert(isequal(boggle_checker(x,y),true))
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# Example: Modeling Base Rate Fallacy
The base rate fallacy is also known as base rate neglect or base rate bias.
When evaluating the probability of an event―for instance, diagnosing a disease, there are two types of information that may be available.
1. Generic information about how frequently an event occurs naturally. Suppose, according to the statistics, 1% of women have breast cancer. So, this information is general information.
2. Specific information about an event in a given context. For example, 80% of mammograms detect breast cancer when a woman really has breast cancer.
When presented with both types of information at the same time, type 1 information is called "base rate" information.
When we have just the generic information, it is okay to assume the probability of an event based on that generic information. But when we have more specific information, our brain tends to judge the probability of an event based on that specific information and neglect the base rate information. That's why it is called base rate neglect too. (neglecting the base rate). Neglecting the base rate information in this way is called Base Rate Fallacy.
Using Rational Will or the Bayesian Network Software from SpiceLogic, you can incorporate these 2 types of information to judge a probability of an event or a hypothesis.
### Real-Life Example
Let's apply that concept in a real-world example. Suppose, we have generic information, "1% of women have breast cancer". Also, we have specific information - "80% of mammograms detect breast cancer when a woman really has breast cancer". Another specific information we collected was that "9.6% of mammograms detect breast cancer when it's not there (false positive)".
Now suppose a woman gets a positive test result. What are the chances that she has cancer?
#### [Calculation]
Let's define some variables.
C = "Cancer".
R = "Positive Test Result"
As 1% of women have breast cancer. Probability of Cancer in general = Pr(C) = 0.01. This is what we call base rate.
Pr(R|C) = Probability of the positive test result (X) given that the woman has cancer (C). This is the probability of a true positive. According to our information,
Pr(R|C) = 0.8.
Pr(not C) = Probability of not having cancer = 1 - 0.01 = 0.99
Pr(R|not C) = Probability of a positive test result (R) given that the woman does not have cancer. (~C). This is the false positive. = 9.6% = 0.096
Now, we need to find out Pr(C|R) = the probability of having cancer (C) given a positive test result (R).
According to Baye's theorem,
Pr(C|R) = Probability of the woman has cancer given the positive test result
= Pr(R|C) * Pr(C) / (Pr(R|C) * Pr(C) + Pr(R|not C) * Pr(not C))
= 0.8 * 0.01 / ( 0.8 * 0.01 + 0.096 * 0.99)
= 0.0776
= 7.76%.
You can model this problem in the Bayesian Network Software or the Rational Will and get the same result easily without doing the calculation by hand. The software will give you a pleasing way to visually depict the problem and see the result in the graphical interface.
### Modeling base rate fallacy in Bayesian Inference
Start the software and choose the "Bayesian Inference"
Now, you are In the Bayesian Inference area. Add your Hypothesis that the woman has cancer.
Now, click the Lock button to "Lock" your prior beliefs.
Now, in the Experiments and Observations panel, add a new experiment as "Mammogram test". Under that experiment, add observation "positive test result".
Then, under the added experiment, add a new observation "positive test result".
Finally, concentrate on the Causal Discovery panel.
#### Finding the result
Once you set the True positive and False positive probabilities, click the "Update Beliefs" button. The software will calculate the updated belief based on this information using Bayes Theorem and update the chart of 'Updated Beliefs'. We can see that the probability of the woman having cancer is calculated as 7.76%. This is the new calculated belief that incorporated the base rate in the calculation. Remember that, this is the value we got from our hand calculation.
Notice the belief history chart. It shows, how your belief is updated over time, upon evidence. This is an example of Diachronic Interpretation. In the Hypotheses panel, your hypothesis probability is updated as well.
Now, if you observe any new evidence (say, another test result), your prior belief will be this calculated belief, and incorporating this newly calculated belief and your next test result, you can have a new belief. In that way, you can continuously keep updating your beliefs upon pieces of evidence you observe one by one.
If you want to add a new hypothesis or override the hypothesis belief manually, you can click the Lock button to unlock the hypotheses panel, and then change the hypotheses, and then lock again to proceed to causal discovery.
### Modeling base rate fallacy in Bayesian Network
You can model the same problem in a Bayesian Network as well. Start the Bayesian Network tool and drag and drop two random variable nodes as shown below. A random variable that represents the woman has cancer. Another random variable represents the positive test result from the mammogram test.
We have a base rate information that 1% of the woman has cancer. So, set the True state variable for 'Woman has cancer' = 0.01. The False state probability will be calculated automatically as 1 - 0.01 = 0.99
We want to incorporate this base rate information in our judgment. As this base rate information influences the probability of positive test results, draw an arrow connecting the Cancer node to the Positive test result node. Then, select the variable 'Positive test result from mammogram'. You will see the following conditional probability table displayed for this variable. As we know, the mammogram test results positive probability is 0.8 when the woman has cancer. And when the woman does not have cancer, the probability of false-positive is 0.096. So, enter the probabilities accordingly.
Thus, we have modeled the Bayesian network for this problem. Now, we want to find out what is the probability of the woman having cancer if we observe a positive test result. In order to find that out, select the node "Positive test result" and check the checkbox "Instantiate..."
Notice that, as soon as you instantiate the variable, the "Woman has Cancer" node's marginal probability is displayed as 0.0776. This is the number we got from our hand calculation. So, the diagram confirms that our calculation result was correct. That is the number we were looking for. That means, the Bayesian network calculates the probability of Cancer given that a Positive test result was observed.
There is another way to find out the probability without instantiating in the diagram. You can open the Query window by clicking the Query button. Then, in the query window, in the top panel, you can check the "Woman has Cancer" and select "True" in the drop-down for Cancer. Then, in the bottom panel, check "positive test result..." and select "True" in the corresponding drop-down. You will see the calculated probability value will be shown as P(X).
Last updated on Feb 24, 2022
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Cody
# Problem 68. Kaprekar Steps
Solution 2152954
Submitted on 7 Mar 2020 by Søren Holm-Petersen
This solution is locked. To view this solution, you need to provide a solution of the same size or smaller.
### Test Suite
Test Status Code Input and Output
1 Pass
x = 3276; y_correct = 5; assert(isequal(KaprekarSteps(x),y_correct))
str = '5265' str = '3996' str = '6264' str = '4176' str = '6174'
2 Pass
x = 3; y_correct = 6; assert(isequal(KaprekarSteps(x),y_correct))
str = '2997' str = '7173' str = '6354' str = '3087' str = '8352' str = '6174'
3 Pass
x = 691; y_correct = 7; assert(isequal(KaprekarSteps(x),y_correct))
str = '9441' str = '7992' str = '7173' str = '6354' str = '3087' str = '8352' str = '6174'
4 Pass
x = 3333; y_correct = Inf; assert(isequal(KaprekarSteps(x),y_correct))
5 Pass
x = 1; y_correct = 5; assert(isequal(KaprekarSteps(x),y_correct))
str = '0999' str = '8991' str = '8082' str = '8532' str = '6174'
6 Pass
x = 6174; y_correct = 0; assert(isequal(KaprekarSteps(x),y_correct))
7 Pass
x = 1234; y_correct = 3; assert(isequal(KaprekarSteps(x),y_correct))
str = '3087' str = '8352' str = '6174'
8 Pass
x = 3141; y_correct = 5; assert(isequal(KaprekarSteps(x),y_correct))
str = '3177' str = '6354' str = '3087' str = '8352' str = '6174'
9 Pass
x = 8080; y_correct = 6; assert(isequal(KaprekarSteps(x),y_correct))
str = '8712' str = '7443' str = '3996' str = '6264' str = '4176' str = '6174'
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THE GAME PIECES ARE BEARS - JohnSchneider [more]
Chief Delphi First Time PID user
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#1
02-13-2018, 11:12 PM
klhutchi Registered User FRC #5736 Join Date: Feb 2016 Location: New York Posts: 9
First Time PID user
Our team is using a rotating arm to grab cubes off the floor and lift them for switch height. In order to hold position we are trying out a PID controller, We have a 30in arm ~10lbs(+4 with cube) that rotates from 0-120 degrees.
A Mini Cim geared down to 140:1 and chain driven sprockets at 12:28 to slow it down to manageable levels attached to a Analog Pot.
I've read a few forums about tuning a PID loop and was hoping for some experienced advice for this application. We added torsion springs to try and assist the arm, but the speed difference between rising and falling are not quite what we want. Currently the motors run a 75% with P at 2.25, I at .008, and D at 0. It holds position but doesn't find it very nicely (Sudden stops not much oscillation).
I was thinking of trying the AutoTuning function in Labview but not to sure of its effectiveness. Any ideas?
Additionally the Motor controller is set to brake, does that work with or against the PID control?
#2
02-14-2018, 12:00 AM
Classified* Drive it like you stole it AKA: Julia Cecchetti FRC #0291 (CIA: Creativity in Action) Team Role: Leadership Join Date: Nov 2015 Rookie Year: 2015 Location: Erie PA Posts: 222
Re: First Time PID user
We have an arm this year as well. It all comes down to telling your PID controller everything you can about the dynamics of your system. The simple answer is to do the math. Calculate the torque due to gravity at a given angle of the arm (this will be the weight of your arm times the distance from the pivot to the center of mass of your arm). Remember gravity always points down, so you need to factor in the angle of your arm and only take the component that points down. So this means if you define theta (the angle) with zero degrees being your arm straight up and down and 90 degrees being parallel to the floor, the torque due to gravity on your arm is:
Torque = ArmWeight*sin(theta)*DistanceOfCoM
So in your case this is (14 lbs)*sin(theta)*(20 in). I made a guess on where your center of mass would be, as it most likely isn't all the way out on the end of the arm.
Then you calculate the power you need to send to the motors to hold the arm stationary using your gear ratio and the type of motor you have. Then simply add this to your PID output! What this does is it removes the difference in the apparent weight of the system to your PID loop. So in your code, you would calculate the PID output, add the "hold Position" power you calculated, and send that to the motors.
This is exactly what we did and you can see the results in this video! Our math was a bit more complicated because we had a gas spring to help the arm up (think 4 nested trig functions) , but it works super well!
Break mode won't really affect anything, It only applies when you send zero power to the motors.
If you have further questions, just ask! I hope that helps!
__________________
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#3
02-14-2018, 10:20 AM
klhutchi Registered User FRC #5736 Join Date: Feb 2016 Location: New York Posts: 9
Re: First Time PID user
Thank you I figured it was some scaling factor didn't quite see how to apply it. I'll try it out today
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# Properties
Base field $$\Q(\sqrt{-7})$$ Weight 2 Level norm 14641 Level $$\left(121\right)$$ Label 2.0.7.1-14641.3-d Dimension 1 CM no Base-change yes Sign -1 Analytic rank odd
# Related objects
## Base Field: $$\Q(\sqrt{-7})$$
Generator $$a$$, with minimal polynomial $$x^2 - x + 2$$; class number $$1$$.
## Form
Weight 2 Level 14641.3 = $$\left(121\right)$$ Label 2.0.7.1-14641.3-d Dimension: 1 CM: no Base change: yes 5929.2.a.h , 121.2.a.d Newspace: 2.0.7.1-14641.3 (dimension 24) Sign of functional equation: -1 Analytic rank: odd
## Hecke eigenvalues
The Hecke eigenvalue field is $\Q$.
Norm Prime Eigenvalue
$$2$$ 2.1 = ($$a$$) $$2$$
$$2$$ 2.2 = ($$-a + 1$$) $$2$$
$$7$$ 7.1 = ($$-2 a + 1$$) $$2$$
$$9$$ 9.1 = ($$3$$) $$-5$$
$$11$$ 11.1 = ($$-2 a + 3$$) $$0$$
$$11$$ 11.2 = ($$2 a + 1$$) $$0$$
$$23$$ 23.1 = ($$-2 a + 5$$) $$-1$$
$$23$$ 23.2 = ($$2 a + 3$$) $$-1$$
$$25$$ 25.1 = ($$5$$) $$-9$$
$$29$$ 29.1 = ($$-4 a + 1$$) $$0$$
$$29$$ 29.2 = ($$4 a - 3$$) $$0$$
$$37$$ 37.1 = ($$-4 a + 5$$) $$3$$
$$37$$ 37.2 = ($$4 a + 1$$) $$3$$
$$43$$ 43.1 = ($$-2 a + 7$$) $$6$$
$$43$$ 43.2 = ($$2 a + 5$$) $$6$$
$$53$$ 53.1 = ($$-4 a - 3$$) $$-6$$
$$53$$ 53.2 = ($$4 a - 7$$) $$-6$$
$$67$$ 67.1 = ($$-6 a + 1$$) $$-7$$
$$67$$ 67.2 = ($$6 a - 5$$) $$-7$$
$$71$$ 71.1 = ($$-2 a + 9$$) $$-3$$
$$71$$ 71.2 = ($$2 a + 7$$) $$-3$$
$$79$$ 79.1 = ($$-6 a + 7$$) $$10$$
$$79$$ 79.2 = ($$6 a + 1$$) $$10$$
$$107$$ 107.1 = ($$-2 a + 11$$) $$-18$$
$$107$$ 107.2 = ($$2 a + 9$$) $$-18$$
## Atkin-Lehner eigenvalues
Norm Prime Eigenvalue
$$11$$ 11.1 = ($$-2 a + 3$$) $$-1$$
$$11$$ 11.2 = ($$2 a + 1$$) $$-1$$
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Some problems involve the simple counting of integers to reveal the structure of molecules. Other problems explore more complicated methods of statistical analysis, algebra, geometry and, yes, even calculus! All problems are matrixed into the national math and science standards for a quick review.
Created Date: 6/1/2004 11:49:57 AM
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• GRE Math Review 4 . Example 1.1.2: 25 is a multiple of only six integers: 1, 5, 25, and their negatives. Example 1.1.3:
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# 2 By 2 Multiplication Worksheets
2 By 2 Multiplication Worksheets – One of the more tough and difficult stuff you can do with elementary school individuals is have them to enjoy math. Addition worksheets and subtraction worksheets aren’t what most children wish to be performing throughout their day. Math in standard phrases is normally all of that an easy task to instruct, but in relation to training the better complex concepts of mathematics, about thirdly quality, it may be a somewhat more challenging to keep the individuals intrigued. It might be even more complicated to apply means of teaching that can help students know the ideas of multiplication without receiving puzzled.
Multiplication tables are a great way to get rid of into educating multiplication. The 5 times tables and also other modest numbers confirm much easier to teach most elementary school individuals. Pupils will see it simple just to keep adding an additional digit to their last end result to obtain the following number in the time tables. By educating this primary, you can find students from the mood and comfortable with multiplying. The difficulties may arise, however, once you’ve reached a higher times tables than several. The key of just adding another digit on the final outcome, to discover the up coming number from the table will become much more hard.
There is available a level where individuals will need to commence memorizing the times tables in order to bear in mind them, in contrast to having the capacity to use a particular method. Right here you should be able to apply fun ways of instructing to create memorizing the times tables a significantly much less overwhelming task. This is an important part of mathematics. Should your pupils forget to grow to be efficient because of their times tables soon they’ll have a problem in the later marks of school. So what can you truly do so as to improve their ability with this area?
## 2 By 2 Multiplication Worksheets is Important to Improve your Math Skills
You will find a number of items that both mom and dad and professors alike can perform to aid the pupil succeed. With respect to their moms and dads, they should conduct a number of points. Firstly, a research method has to be integrated in your house to make certain that the little one is not neglecting their scientific studies in the home. Overlooking their work at home and trying to do everything at school never ever helps! By sitting with your kids in the evening, transforming off the television and supporting them grow to be acquainted with multiplication you can expect to finally be assisting them perform much better in course. I often suggest flash cards for multiplication at home.
If you feel your son or daughter may benefit, you can find things you can purchase to boost their reports. You are able to clearly get books to help them, but many young children see reading as a chore today. You can choose from a number of pc software program CDs and math web sites that can make multiplication entertaining. These math games enable your little one to get fun actively playing games while understanding multiplication tables simultaneously.
Professors have to clearly set up suitable due diligence, and make certain that it’s not very a lot. Homework is an essential part of the student’s reports, but environment a lot of can lower morale making a youngster really feel overloaded. Setting exciting and reasonable levels of research can seriously assist a young child on his or her way to memorizing and learning the multiplication tables.he most important issue to remember when attemping to help your kids is in order to ensure that it stays enjoyable. Math should be shown to young children in a fun way. They should not actually understand they can be discovering.
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| 2.65625
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CC-MAIN-2024-26
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https://oeis.org/A274716
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The OEIS is supported by the many generous donors to the OEIS Foundation.
Hints (Greetings from The On-Line Encyclopedia of Integer Sequences!)
A274716 a(2*n+1) = a(2*floor(n/2)+1) + n, a(2*n) = a(n), for n>=1 with a(1)=0. 3
0, 0, 1, 0, 3, 1, 4, 0, 7, 3, 8, 1, 10, 4, 11, 0, 15, 7, 16, 3, 18, 8, 19, 1, 22, 10, 23, 4, 25, 11, 26, 0, 31, 15, 32, 7, 34, 16, 35, 3, 38, 18, 39, 8, 41, 19, 42, 1, 46, 22, 47, 10, 49, 23, 50, 4, 53, 25, 54, 11, 56, 26, 57, 0, 63, 31, 64, 15, 66, 32, 67, 7, 70, 34, 71, 16, 73, 35, 74, 3, 78, 38, 79, 18, 81, 39, 82, 8, 85, 41, 86, 19, 88, 42, 89, 1, 94, 46, 95, 22, 97, 47, 98, 10, 101, 49, 102, 23, 104, 50, 105, 4, 109, 53, 110, 25, 112, 54, 113, 11, 116, 56, 117, 26, 119, 57, 120, 0 (list; graph; refs; listen; history; text; internal format)
OFFSET 1,5 COMMENTS The values {2^a(n)} form the denominators of the coefficients in the g.f. of A274717, which satisfies G(x) = G(x^2) + sqrt( G(x^2) ). LINKS Table of n, a(n) for n=1..128. FORMULA a(2^n) = 0 for n>=0. a(2*n-1) = A005187(n-1) for n>=1. PROG (PARI) {a(n) = if(n==1, 0, if(n%2==0, a(n/2), a(2*(n\4)+1) + n\2 ) )} for(n=1, 32, print1(a(n), ", ")) CROSSREFS Cf. A274717, A005187. Sequence in context: A076816 A021765 A267187 * A245434 A305100 A051512 Adjacent sequences: A274713 A274714 A274715 * A274717 A274718 A274719 KEYWORD nonn AUTHOR Paul D. Hanna, Jul 07 2016 STATUS approved
Lookup | Welcome | Wiki | Register | Music | Plot 2 | Demos | Index | Browse | More | WebCam
Contribute new seq. or comment | Format | Style Sheet | Transforms | Superseeker | Recents
The OEIS Community | Maintained by The OEIS Foundation Inc.
Last modified September 16 04:06 EDT 2024. Contains 375959 sequences. (Running on oeis4.)
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https://careercup.com/question?id=12549704
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## Amazon Interview Question for Software Engineer / Developers
Team: SDE
Country: India
Interview Type: In-Person
Comment hidden because of low score. Click to expand.
3
of 3 vote
``````public static void printBrackets(int left, int right, char[] str, int count)
{
if(left == 0 && right == 0)
System.out.println(str);
else
{
if(left > 0)
{
str[count] = '{';
printBrackets(left -1, right, str, count + 1);
}
if(right > left)
{
str[count] = '}';
printBrackets(left, right - 1, str, count + 1);
}
}
}``````
Comment hidden because of low score. Click to expand.
0
can you explain the logic? atleast how to call the function initially with correct params
Comment hidden because of low score. Click to expand.
0
of 0 vote
for 2N braces, it can only be organized in two fashions:
(.....)(...) and (........)
In the first case, first ( and first ) meant to be a pair, second ( and second ) DOES NOT have to be a pair. so ()()() should be viewed as () ()()
In the second case, two ( ) is a pair, so ( ()() ) is valid, but not ()()()
In the first case, the first ')' can occur at i = 2, 4, .... 2N-2. For a fixed i, there are
f(i-2)*f(2n-i) cases
and for case two, f(2n-2) cases.
So f(n) = sum_{i=2, 4, ... 2N-2} ( f(i-2)*f(2n-i) ) + f(2n-2)
Comment hidden because of low score. Click to expand.
0
of 0 vote
More details at stackoverflow.com/questions/3172179/valid-permutation-of-parenthesis
Comment hidden because of low score. Click to expand.
0
of 0 vote
awesome answer.how do you think like that.
Comment hidden because of low score. Click to expand.
0
of 0 vote
``````def paren(n):
if n == 1: return ['()']
else:
pnless1 = paren(n-1)
pn = []
for i in pnless1:
pn.append('()'+i)
pn.append(i+'()')
pn.append('('+i+')')
pn = pn[1:] # to avoid the repeated pattern
return pn``````
Comment hidden because of low score. Click to expand.
0
of 0 vote
The solution provided by Hello World is most optimum one. We go on recursivelya dding left and right brackets to form valid strings.
We can add left brackets as long as we have left brackets remaining with us.
We can add right brackets as long as the number of right brackets remaining is > number of left brackets remaining.
When the number of right brackets and left brackets is exhausted, print the string
Name:
Writing Code? Surround your code with {{{ and }}} to preserve whitespace.
### Books
is a comprehensive book on getting a job at a top tech company, while focuses on dev interviews and does this for PMs.
### Videos
CareerCup's interview videos give you a real-life look at technical interviews. In these unscripted videos, watch how other candidates handle tough questions and how the interviewer thinks about their performance.
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https://www.transtutors.com/questions/problem-11-22-speclal-order-decisions-lo11-4-polaski-company-manufactures-and-sells--5140471.htm
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# Problem 11-22 Speclal Order Decisions [LO11-4] Polaski Company manufactures and sells a single produ 1 answer below »
Problem 11-22 Speclal Order Decisions [LO11-4] Polaski Company manufactures and sells a single product called a Ret. Operating at capacity, the company can 50,000 Rets per year. Costs associated with this level of production and sales are given below: produce and sell Unit Total Direct materials Direct labor 15 750,eee 3ee, eee 150,e0e 350, eee 6 Variable manufacturing overhead Fixed manufacturing overhead Variable selling expense Fixed selling expense 2 100,eee 300.eee Total cost 39 1,950,eee The Rets normally sell for $44 each. Fixed manufacturing overhead is$350,000 per year within the range of 42,000 through 50,000 Rets per year Required: 1. Assume that due to a recession, Polaski Company expects to sell only 42,000 Rets through regular channels next year. A large retail chain has offered to purchase 8,000 Rets if Polaski is willing to accept a 16% discount off the regular price. There would be no sales commissions on this order, thus, variable selling expenses would be slashed by 75 % . However, Polaski Company would have to purchase a special machine to engrave the retail chain's name on the 8,000 units. This machine would cost $16,000. Polaski Company has no assurance that the retail chain will purchase additional units in the future. What is the financial advantage (disadvantage) of accepting the special order? (Round your Intermedlate calculations to 2 declmal places.) 2. Refer to the original data. Assume again that Polaski Company expects to sell only 42,000 Rets through regular channels next year The U.S. Army would like to make a one-time-only purchase of 8,000 Rets. The Army would pay a fixed fee of$2.00 per Ret, and it would reimburse Polaski Company for all costs of production (variable and fixed) associated with the units. Because the army would pick up the Rets with its own trucks, there would be no variable selling expenses associated with this order. What is the financial advantage (disadvantage) of accepting the U.S. Army's special order? 3. Assume the same situation as described in (2) above, except that the company expects to sell 50,000 Rets through reqular channels next year. Thus, accepting the U.S. Army's order would require giving up regular sales of 8,000 Rets. Given this new information, what is the financial advantage (disadvantage) of accepting the U.S. Army's special order? 1 2 3
Akash V
Sales 8,000 * 36.96 2,95,680 Less: relevant Costs Direct material 8,000 *...
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## Recent Questions in Accounting - Others
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CC-MAIN-2020-50
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https://www.khanacademy.org/computing/pixar/virtual-cameras/depth-of-field/v/optics9-final
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If you're seeing this message, it means we're having trouble loading external resources on our website.
If you're behind a web filter, please make sure that the domains *.kastatic.org and *.kasandbox.org are unblocked.
## Pixar in a Box
### Course: Pixar in a Box>Unit 6
Lesson 2: Mathematics of depth of field
# Out of focus objects
In this video we'll explore what happens when an object is out of focus from a geometric perspective.
## Want to join the conversation?
• At , is that the radius of the blur, or the diameter?
• It is actually the radius of the circle of confusion and not the diameter. It is easy to be confused, because I think the sketch is a bit misleading. It depicts only half of the ray bundle that is emitted from point P, namely the bundle between points D and E. One should imagine another point, let's call if F, symmetrical to point D with point E as the center of symmetry. Now, imagine the rest of the rays that originate from point P and pass through the lens with outermost the ray passing through F. These rays obviously also converge on point A by the bending of the lens. Then, if you extrude this outermost ray to meet the image plane, you will have the full diameter of the circle. I understand, it sounds much more complicated in writing than in sketches, but hopefully, it helped.
• He talks about "moving the Image Plane". I forgot - what exactly determines the distance from the pinhole to the image plane? Because in one video I thought he said the Image Plane was positioned at the focal point. What did I miss? Because that is obviously wrong.
A quick explanation is all I need, thanks!
I have been watching these videos in a blur of confusion, I'm seeing circles all around me. :\
In all seriousness, I really enjoyed all these videos! :-)
• What would the out of focus do?
(1 vote)
• I am so lost thaat i do not now what to do
(1 vote)
• At , is that the radius of the blur, or the diameter?
(1 vote)
• I'm confused. I believe I understand what he is saying in the video, but having trouble picturing this in a real camera. I assume the focal point is fixed for a given lens. Does the image plane move? Is that how you can get a lens with a range of say 35mm to 70mm? Which part does the f-stop control, the aperture? And the focus control the image plane?
(1 vote)
• great video! Thanks Mr VanSchieck
(1 vote)
• How is BC the radius of the circle of confusion, and not the diameter?
(1 vote)
• At , is that the radius of the blur, or the diameter?
(1 vote)
• It is the radius of the blur, not the diameter
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http://math.stackexchange.com/questions/238728/finding-summation/238792
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# Finding summation
I have been having problem with calculating the following summation: $$\sum_{i=1}^n {1\over 4i^2-1} = {1\over3} + {1\over15} + {1\over35} + \cdots + {1\over 4n^2-1}$$ I do know the answer, but just can not find the way to get it.
-
There is a common trick here: use the fact that $$\frac{1}{4n^2-1} = \frac{1}{2}\left(\frac{1}{2n-1} - \frac{1}{2n+1}\right).$$
UPDATE: Also, there's often another way to go about it. If you already know the answer, there is a good chance that you can prove it by induction. It does work in this problem.
-
If you use partial fractions, you get a telescoping series: $$\sum_{i=1}^n {1\over 4i^2-1}={1\over 2}\sum_{i=1}^n \Bigl({1\over 2i-1}-{1\over 2i+1}\Bigr)$$ $$={1\over 2}\Bigl(\Bigl(1-{1\over 3}\Bigr)+\Bigl({1\over 3}-{1\over 5}\Bigr)+\cdots +\Bigl({1\over 2n-1}-{1\over 2n+1}\Bigr)\Bigr)$$ $$={1\over 2}\Bigl(1-{1\over 2n+1}\Bigr)={n\over 2n+1}$$
-
You have $$\sum_{i=1}^n {1\over 4i^2-1} = \sum_{i=1}^n\frac{1}{{(2i + 1)(2i - 1)}}$$ $$= \sum_{i=1}^n \left(\frac{1}{{2(2i - 1)}} - \frac{1}{{2(2i + 1)}}\right) = \frac{1}{2}\left(\sum_{1\leq i\leq n} \frac{1}{{(2i - 1)}} - \sum_{1\leq i\leq n}\frac{1}{{(2i + 1)}}\right)$$ $$=\frac{1}{2}\left( \frac{1}{1}+\sum_{2\leq i\leq n} \frac{1}{{(2i - 1)}} - \sum_{1\leq i\leq n-1}\frac{1}{{(2i + 1)}}- \frac{1}{2n+1}\right)$$ By translation of index i=k+1 you have $$=\frac{1}{2}\left( \frac{1}{1}+\sum_{2\leq k+1\leq n} \frac{1}{{(2[k+1] - 1)}} - \sum_{1\leq i\leq n-1}\frac{1}{{(2i + 1)}}- \frac{1}{2n+1}\right).$$ Remember that $2\leq k+1\leq n$ if, onli if, $2-1\leq k\leq n-1$. Then $$=\frac{1}{2}\left( \frac{1}{1}+\sum_{1\leq k\leq n-1} \frac{1}{{(2k+1)}} - \sum_{1\leq i\leq n-1}\frac{1}{{(2i + 1)}}- \frac{1}{2n+1}\right)$$ $$=\frac{1}{2}-\frac{1}{4n+2}.$$
-
| 881
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| 4.65625
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https://help.scilab.org/docs/6.0.0/en_US/mrfit.html
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Scilab Home page | Wiki | Bug tracker | Forge | Mailing list archives | ATOMS | File exchange
Scilab 6.0.0
Change language to: Français - Português - 日本語 - Русский
See the recommended documentation of this function
# mrfit
frequency response fit
### Syntax
```sys=mrfit(w,mag,order)
[num,den]=mrfit(w,mag,order)
sys=mrfit(w,mag,order,weight)
[num,den]=mrfit(w,mag,order,weight)```
### Arguments
w
positive real vector of frequencies (Hz)
mag
real vector of frequency responses magnitude (same size as `w`)
order
integer (required order, degree of `den`)
weight
positive real vector (default value `ones(w)`).
num,den
stable polynomials
### Description
`sys=mrfit(w,mag,order,weight)` returns a bi-stable transfer function `G(s)=sys=num/den`, of of given `order` such that its frequency response magnitude `abs(G(w(i)))` matches `mag(i)` i.e. `abs(freq(num,den,%i*w))` should be close to `mag`. `weight(i)` is the weight given to `w(i)`.
### Examples
```w=0.01:0.01:2;
s=poly(0,'s');
G=syslin('c',2*(s^2+0.1*s+2),(s^2+s+1)*(s^2+0.3*s+1)); // syslin('c',Num,Den);
fresp=repfreq(G,w);
mag=abs(fresp);
Gid=mrfit(w,mag,4);
frespfit=repfreq(Gid,w);
plot2d([w',w'],[mag(:),abs(frespfit(:))])```
• cepstrum — cepstrum calculation
• frfit — frequency response fit
• freq — frequency response
• calfrq — frequency response discretization
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CC-MAIN-2019-04
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https://freestatistics.org/blog/index.php?v=date/2011/Dec/01/t1322753322kbxmjygc6tyb409.htm/#T=0
| 1,685,310,718,000,000,000
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## Free Statistics
of Irreproducible Research!
Author's title
Author*The author of this computation has been verified*
R Software Modulerwasp_decomposeloess.wasp
Title produced by softwareDecomposition by Loess
Date of computationThu, 01 Dec 2011 10:28:09 -0500
Cite this page as followsStatistical Computations at FreeStatistics.org, Office for Research Development and Education, URL https://freestatistics.org/blog/index.php?v=date/2011/Dec/01/t1322753322kbxmjygc6tyb409.htm/, Retrieved Sun, 28 May 2023 21:51:58 +0000
Statistical Computations at FreeStatistics.org, Office for Research Development and Education, URL https://freestatistics.org/blog/index.php?pk=149807, Retrieved Sun, 28 May 2023 21:51:58 +0000
QR Codes:
Original text written by user:
IsPrivate?No (this computation is public)
User-defined keywords
Estimated Impact63
Family? (F = Feedback message, R = changed R code, M = changed R Module, P = changed Parameters, D = changed Data)
- [Classical Decomposition] [HPC Retail Sales] [2008-03-02 16:19:32] [74be16979710d4c4e7c6647856088456]
- RMPD [Decomposition by Loess] [] [2011-11-26 21:27:22] [ee8c3a74bf3b349877806e9a50913c60]
- R PD [Decomposition by Loess] [] [2011-12-01 15:28:09] [7dc03dd48c8acabd98b217fada4a6bc0] [Current]
Feedback Forum
Post a new message
Dataseries X:
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Summary of computational transaction Raw Input view raw input (R code) Raw Output view raw output of R engine Computing time 2 seconds R Server 'Herman Ole Andreas Wold' @ wold.wessa.net
\begin{tabular}{lllllllll}
\hline
Summary of computational transaction \tabularnewline
Raw Input & view raw input (R code) \tabularnewline
Raw Output & view raw output of R engine \tabularnewline
Computing time & 2 seconds \tabularnewline
R Server & 'Herman Ole Andreas Wold' @ wold.wessa.net \tabularnewline
\hline
\end{tabular}
%Source: https://freestatistics.org/blog/index.php?pk=149807&T=0
[TABLE]
[ROW][C]Summary of computational transaction[/C][/ROW]
[ROW][C]Raw Input[/C][C]view raw input (R code) [/C][/ROW]
[ROW][C]Raw Output[/C][C]view raw output of R engine [/C][/ROW]
[ROW][C]Computing time[/C][C]2 seconds[/C][/ROW]
[ROW][C]R Server[/C][C]'Herman Ole Andreas Wold' @ wold.wessa.net[/C][/ROW]
[/TABLE]
Source: https://freestatistics.org/blog/index.php?pk=149807&T=0
Globally Unique Identifier (entire table): ba.freestatistics.org/blog/index.php?pk=149807&T=0
As an alternative you can also use a QR Code:
The GUIDs for individual cells are displayed in the table below:
Summary of computational transaction Raw Input view raw input (R code) Raw Output view raw output of R engine Computing time 2 seconds R Server 'Herman Ole Andreas Wold' @ wold.wessa.net
Seasonal Decomposition by Loess - Parameters Component Window Degree Jump Seasonal 1321 0 133 Trend 19 1 2 Low-pass 13 1 2
\begin{tabular}{lllllllll}
\hline
Seasonal Decomposition by Loess - Parameters \tabularnewline
Component & Window & Degree & Jump \tabularnewline
Seasonal & 1321 & 0 & 133 \tabularnewline
Trend & 19 & 1 & 2 \tabularnewline
Low-pass & 13 & 1 & 2 \tabularnewline
\hline
\end{tabular}
%Source: https://freestatistics.org/blog/index.php?pk=149807&T=1
[TABLE]
[ROW][C]Seasonal Decomposition by Loess - Parameters[/C][/ROW]
[ROW][C]Component[/C][C]Window[/C][C]Degree[/C][C]Jump[/C][/ROW]
[ROW][C]Seasonal[/C][C]1321[/C][C]0[/C][C]133[/C][/ROW]
[ROW][C]Trend[/C][C]19[/C][C]1[/C][C]2[/C][/ROW]
[ROW][C]Low-pass[/C][C]13[/C][C]1[/C][C]2[/C][/ROW]
[/TABLE]
Source: https://freestatistics.org/blog/index.php?pk=149807&T=1
Globally Unique Identifier (entire table): ba.freestatistics.org/blog/index.php?pk=149807&T=1
As an alternative you can also use a QR Code:
The GUIDs for individual cells are displayed in the table below:
Seasonal Decomposition by Loess - Parameters Component Window Degree Jump Seasonal 1321 0 133 Trend 19 1 2 Low-pass 13 1 2
Seasonal Decomposition by Loess - Time Series Components t Observed Fitted Seasonal Trend Remainder 1 274 258.825969297934 26.4236653861948 262.750365315871 -15.174030702066 2 291 291.074065711283 31.2979190176708 259.628015271047 0.0740657112825716 3 280 276.049434712804 27.4449000609741 256.505665226222 -3.9505652871961 4 258 258.212634364351 4.67600235024529 253.111363285404 0.212634364350805 5 252 253.830373753073 0.452564902341024 249.717061344586 1.83037375307325 6 251 259.136401466967 -3.24364100523016 246.107239538264 8.13640146696659 7 224 233.533349901256 -28.030767633197 242.497417731941 9.5333499012556 8 225 236.391724668407 -25.1501097519631 238.758385083556 11.3917246684067 9 234 239.250102593327 -6.26945502849838 235.019352435172 5.25010259332686 10 233 238.808370173524 -3.26430176703221 230.455931593509 5.80837017352368 11 229 232.184825024964 -0.0773357768091877 225.892510751846 3.18482502496363 12 208 218.819406060445 -24.2594446230739 221.440038562629 10.8194060604446 13 224 204.588768240392 26.4236653861948 216.987566373413 -19.4112317596078 14 226 207.046092187521 31.2979190176708 213.655988794808 -18.9539078124792 15 223 208.230688722822 27.4449000609741 210.324411216204 -14.769311277178 16 205 196.699714448582 4.67600235024529 208.624283201172 -8.30028555141757 17 201 194.623279911518 0.452564902341024 206.924155186141 -6.37672008848168 18 202 199.297444563401 -3.24364100523016 207.946196441829 -2.70255543659889 19 183 185.06252993568 -28.030767633197 208.968237697517 2.06252993567958 20 188 189.345534155014 -25.1501097519631 211.804575596949 1.34553415501392 21 200 191.628541532117 -6.26945502849838 214.640913496381 -8.3714584678826 22 206 197.389114317507 -3.26430176703221 217.875187449526 -8.61088568249343 23 211 200.967874374139 -0.0773357768091877 221.10946140267 -10.0321256258611 24 201 201.30198112077 -24.2594446230739 224.957463502304 0.301981120769597 25 299 342.770869011867 26.4236653861948 228.805465601938 43.7708690118669 26 244 223.032878673122 31.2979190176708 233.669202309207 -20.9671213268776 27 251 236.022160922551 27.4449000609741 238.532939016475 -14.9778390774493 28 241 233.353851955543 4.67600235024529 243.970145694212 -7.64614804445725 29 244 238.14008272571 0.452564902341024 249.407352371949 -5.85991727428964 30 252 252.220130901067 -3.24364100523016 255.023510104163 0.220130901067392 31 234 235.39109979682 -28.030767633197 260.639667836377 1.39109979682013 32 246 250.233914402013 -25.1501097519631 266.91619534995 4.23391440201272 33 265 263.076732164974 -6.26945502849838 273.192722863524 -1.92326783502557 34 277 277.185614884768 -3.26430176703221 280.078686882264 0.185614884768427 35 287 287.112684875805 -0.0773357768091877 286.964650901004 0.112684875805485 36 275 280.380009963114 -24.2594446230739 293.87943465996 5.38000996311411 37 320 312.782116194889 26.4236653861948 300.794218418916 -7.21788380511066 38 338 336.6629188195 31.2979190176708 308.039162162829 -1.33708118049992 39 342 341.270994032284 27.4449000609741 315.284105906742 -0.729005967716489 40 322 316.628701218287 4.67600235024529 322.695296431467 -5.37129878171265 41 323 315.440948141467 0.452564902341024 330.106486956192 -7.55905185853334 42 343 351.69690644127 -3.24364100523016 337.54673456396 8.69690644126996 43 315 313.043785461469 -28.030767633197 344.986982171728 -1.95621453853107 44 334 340.251178329765 -25.1501097519631 352.898931422198 6.2511783297648 45 359 363.45857435583 -6.26945502849838 360.810880672669 4.45857435582985 46 362 358.237759334306 -3.26430176703221 369.026542432726 -3.76224066569392 47 378 378.835131584025 -0.0773357768091877 377.242204192784 0.835131584025419 48 345 329.573236907489 -24.2594446230739 384.686207715585 -15.4267630925113 49 422 425.446123375419 26.4236653861948 392.130211238387 3.44612337541861 50 430 430.640054905224 31.2979190176708 398.062026077105 0.640054905223906 51 443 454.561259023202 27.4449000609741 403.993840915824 11.561259023202 52 431 448.605803426908 4.67600235024529 408.718194222847 17.6058034269078 53 425 436.104887567789 0.452564902341024 413.44254752987 11.1048875677891 54 432 449.872466923501 -3.24364100523016 417.371174081729 17.8724669235008 55 387 380.730966999608 -28.030767633197 421.299800633589 -6.26903300039191 56 396 392.480656576576 -25.1501097519631 424.669453175387 -3.51934342342429 57 411 400.230349311313 -6.26945502849838 428.039105717186 -10.7696506886875 58 421 414.31078730989 -3.26430176703221 430.953514457142 -6.68921269010985 59 424 414.209412579711 -0.0773357768091877 433.867923197098 -9.79058742028917 60 410 407.961424857112 -24.2594446230739 436.298019765962 -2.03857514288848 61 464 462.848218278979 26.4236653861948 438.728116334826 -1.15178172102111 62 486 500.221987712594 31.2979190176708 440.480093269735 14.2219877125943 63 490 510.323029734382 27.4449000609741 442.232070204644 20.3230297343824 64 459 470.662505464168 4.67600235024529 442.661492185587 11.6625054641681 65 454 464.456520931129 0.452564902341024 443.09091416653 10.4565209311294 66 446 453.905909070272 -3.24364100523016 441.337731934958 7.90590907027206 67 406 400.44621792981 -28.030767633197 439.584549703387 -5.55378207018958 68 412 413.725304411208 -25.1501097519631 435.424805340755 1.72530441120836 69 428 431.004394050375 -6.26945502849838 431.265060978123 3.00439405037548 70 429 436.04677737402 -3.26430176703221 425.217524393013 7.04677737401954 71 425 430.907347968907 -0.0773357768091877 419.169987807902 5.90734796890678 72 396 404.263131901301 -24.2594446230739 411.996312721773 8.26313190130094 73 429 426.753696978162 26.4236653861948 404.822637635643 -2.24630302183817 74 439 449.548449532047 31.2979190176708 397.153631450282 10.5484495320469 75 424 431.070474674105 27.4449000609741 389.484625264921 7.07047467410462 76 379 371.177422849211 4.67600235024529 382.146574800543 -7.82257715078867 77 370 364.738910761494 0.452564902341024 374.808524336165 -5.26108923850643 78 353 340.925770581175 -3.24364100523016 368.317870424055 -12.0742294188252 79 322 310.203551121252 -28.030767633197 361.827216511945 -11.7964488787481 80 322 312.813110314646 -25.1501097519631 356.336999437317 -9.18688968535366 81 338 331.42267266581 -6.26945502849838 350.846782362688 -6.57732733418999 82 348 353.305919340063 -3.26430176703221 345.958382426969 5.3059193400631 83 350 359.007353285559 -0.0773357768091877 341.06998249125 9.00735328555925 84 312 311.803446859452 -24.2594446230739 336.455997763622 -0.196553140548474 85 358 357.73432157781 26.4236653861948 331.842013035995 -0.26567842218958 86 378 397.584974570901 31.2979190176708 327.117106411428 19.5849745709013 87 352 354.162900152165 27.4449000609741 322.392199786861 2.16290015216487 88 312 301.854527226849 4.67600235024529 317.469470422905 -10.1454727731508 89 310 307.000694038709 0.452564902341024 312.54674105895 -2.99930596129093 90 292 279.718127475333 -3.24364100523016 307.525513529898 -12.2818725246674 91 276 277.526481632352 -28.030767633197 302.504286000845 1.52648163235176 92 269 265.426095192001 -25.1501097519631 297.724014559962 -3.57390480799938 93 286 285.325711909419 -6.26945502849838 292.94374311908 -0.674288090581342 94 292 298.181810459289 -3.26430176703221 289.082491307743 6.18181045928901 95 288 290.856096280402 -0.0773357768091877 285.221239496407 2.85609628040243 96 255 252.203693453886 -24.2594446230739 282.055751169188 -2.79630654611429 97 304 302.686071771836 26.4236653861948 278.89026284197 -1.3139282281644 98 299 290.633599781964 31.2979190176708 276.068481200366 -8.36640021803646 99 293 285.308400380264 27.4449000609741 273.246699558762 -7.69159961973583 100 275 274.176820941592 4.67600235024529 271.147176708163 -0.823179058408414 101 272 274.499781240094 0.452564902341024 269.047653857565 2.49978124009448 102 264 263.059390619822 -3.24364100523016 268.184250385408 -0.940609380178046 103 234 228.709920719945 -28.030767633197 267.320846913252 -5.29007928005484 104 231 218.995141542205 -25.1501097519631 268.154968209758 -12.0048584577947 105 263 263.280365522234 -6.26945502849838 268.989089506264 0.28036552223449 106 264 259.556699251239 -3.26430176703221 271.707602515793 -4.44330074876081 107 264 253.651220251487 -0.0773357768091877 274.426115525322 -10.348779748513 108 245 235.207207950387 -24.2594446230739 279.052236672687 -9.79279204961341 109 297 283.897976793753 26.4236653861948 283.678357820052 -13.1020232062472 110 317 312.443002980782 31.2979190176708 290.259078001547 -4.55699701921753 111 318 311.715301755985 27.4449000609741 296.839798183041 -6.2846982440152 112 315 320.343692711929 4.67600235024529 304.980304937826 5.34369271192884 113 312 310.426623405048 0.452564902341024 313.120811692611 -1.57337659495164 114 310 300.888610357863 -3.24364100523016 322.355030647367 -9.11138964213688 115 306 308.441518031074 -28.030767633197 331.589249602123 2.44151803107354 116 313 310.416618571051 -25.1501097519631 340.733491180912 -2.58338142894894 117 350 356.391722268798 -6.26945502849838 349.877732759701 6.39172226879776 118 354 353.521072899121 -3.26430176703221 357.743228867911 -0.478927100878593 119 371 376.468610800688 -0.0773357768091877 365.608724976121 5.46861080068823 120 357 366.59329110528 -24.2594446230739 371.666153517794 9.59329110527977 121 419 433.852752554338 26.4236653861948 377.723582059467 14.8527525543379 122 425 437.009843201092 31.2979190176708 381.692237781237 12.0098432010923 123 424 434.894206436019 27.4449000609741 385.660893503007 10.8942064360194 124 399 407.126461745117 4.67600235024529 386.197535904637 8.12646174511747 125 393 398.813256791391 0.452564902341024 386.734178306268 5.81325679139115 126 378 372.354470058853 -3.24364100523016 386.889170946377 -5.64552994114678 127 371 382.986604046711 -28.030767633197 387.044163586486 11.986604046711 128 364 366.199063706569 -25.1501097519631 386.951046045394 2.19906370656878 129 384 387.411526524196 -6.26945502849838 386.857928504303 3.41152652419578 130 377 370.756927114394 -3.26430176703221 386.507374652638 -6.24307288560584 131 383 379.920514975836 -0.0773357768091877 386.156820800973 -3.07948502416428 132 352 342.645037288947 -24.2594446230739 385.614407334126 -9.35496271105256
\begin{tabular}{lllllllll}
\hline
Seasonal Decomposition by Loess - Time Series Components \tabularnewline
t & Observed & Fitted & Seasonal & Trend & Remainder \tabularnewline
1 & 274 & 258.825969297934 & 26.4236653861948 & 262.750365315871 & -15.174030702066 \tabularnewline
2 & 291 & 291.074065711283 & 31.2979190176708 & 259.628015271047 & 0.0740657112825716 \tabularnewline
3 & 280 & 276.049434712804 & 27.4449000609741 & 256.505665226222 & -3.9505652871961 \tabularnewline
4 & 258 & 258.212634364351 & 4.67600235024529 & 253.111363285404 & 0.212634364350805 \tabularnewline
5 & 252 & 253.830373753073 & 0.452564902341024 & 249.717061344586 & 1.83037375307325 \tabularnewline
6 & 251 & 259.136401466967 & -3.24364100523016 & 246.107239538264 & 8.13640146696659 \tabularnewline
7 & 224 & 233.533349901256 & -28.030767633197 & 242.497417731941 & 9.5333499012556 \tabularnewline
8 & 225 & 236.391724668407 & -25.1501097519631 & 238.758385083556 & 11.3917246684067 \tabularnewline
9 & 234 & 239.250102593327 & -6.26945502849838 & 235.019352435172 & 5.25010259332686 \tabularnewline
10 & 233 & 238.808370173524 & -3.26430176703221 & 230.455931593509 & 5.80837017352368 \tabularnewline
11 & 229 & 232.184825024964 & -0.0773357768091877 & 225.892510751846 & 3.18482502496363 \tabularnewline
12 & 208 & 218.819406060445 & -24.2594446230739 & 221.440038562629 & 10.8194060604446 \tabularnewline
13 & 224 & 204.588768240392 & 26.4236653861948 & 216.987566373413 & -19.4112317596078 \tabularnewline
14 & 226 & 207.046092187521 & 31.2979190176708 & 213.655988794808 & -18.9539078124792 \tabularnewline
15 & 223 & 208.230688722822 & 27.4449000609741 & 210.324411216204 & -14.769311277178 \tabularnewline
16 & 205 & 196.699714448582 & 4.67600235024529 & 208.624283201172 & -8.30028555141757 \tabularnewline
17 & 201 & 194.623279911518 & 0.452564902341024 & 206.924155186141 & -6.37672008848168 \tabularnewline
18 & 202 & 199.297444563401 & -3.24364100523016 & 207.946196441829 & -2.70255543659889 \tabularnewline
19 & 183 & 185.06252993568 & -28.030767633197 & 208.968237697517 & 2.06252993567958 \tabularnewline
20 & 188 & 189.345534155014 & -25.1501097519631 & 211.804575596949 & 1.34553415501392 \tabularnewline
21 & 200 & 191.628541532117 & -6.26945502849838 & 214.640913496381 & -8.3714584678826 \tabularnewline
22 & 206 & 197.389114317507 & -3.26430176703221 & 217.875187449526 & -8.61088568249343 \tabularnewline
23 & 211 & 200.967874374139 & -0.0773357768091877 & 221.10946140267 & -10.0321256258611 \tabularnewline
24 & 201 & 201.30198112077 & -24.2594446230739 & 224.957463502304 & 0.301981120769597 \tabularnewline
25 & 299 & 342.770869011867 & 26.4236653861948 & 228.805465601938 & 43.7708690118669 \tabularnewline
26 & 244 & 223.032878673122 & 31.2979190176708 & 233.669202309207 & -20.9671213268776 \tabularnewline
27 & 251 & 236.022160922551 & 27.4449000609741 & 238.532939016475 & -14.9778390774493 \tabularnewline
28 & 241 & 233.353851955543 & 4.67600235024529 & 243.970145694212 & -7.64614804445725 \tabularnewline
29 & 244 & 238.14008272571 & 0.452564902341024 & 249.407352371949 & -5.85991727428964 \tabularnewline
30 & 252 & 252.220130901067 & -3.24364100523016 & 255.023510104163 & 0.220130901067392 \tabularnewline
31 & 234 & 235.39109979682 & -28.030767633197 & 260.639667836377 & 1.39109979682013 \tabularnewline
32 & 246 & 250.233914402013 & -25.1501097519631 & 266.91619534995 & 4.23391440201272 \tabularnewline
33 & 265 & 263.076732164974 & -6.26945502849838 & 273.192722863524 & -1.92326783502557 \tabularnewline
34 & 277 & 277.185614884768 & -3.26430176703221 & 280.078686882264 & 0.185614884768427 \tabularnewline
35 & 287 & 287.112684875805 & -0.0773357768091877 & 286.964650901004 & 0.112684875805485 \tabularnewline
36 & 275 & 280.380009963114 & -24.2594446230739 & 293.87943465996 & 5.38000996311411 \tabularnewline
37 & 320 & 312.782116194889 & 26.4236653861948 & 300.794218418916 & -7.21788380511066 \tabularnewline
38 & 338 & 336.6629188195 & 31.2979190176708 & 308.039162162829 & -1.33708118049992 \tabularnewline
39 & 342 & 341.270994032284 & 27.4449000609741 & 315.284105906742 & -0.729005967716489 \tabularnewline
40 & 322 & 316.628701218287 & 4.67600235024529 & 322.695296431467 & -5.37129878171265 \tabularnewline
41 & 323 & 315.440948141467 & 0.452564902341024 & 330.106486956192 & -7.55905185853334 \tabularnewline
42 & 343 & 351.69690644127 & -3.24364100523016 & 337.54673456396 & 8.69690644126996 \tabularnewline
43 & 315 & 313.043785461469 & -28.030767633197 & 344.986982171728 & -1.95621453853107 \tabularnewline
44 & 334 & 340.251178329765 & -25.1501097519631 & 352.898931422198 & 6.2511783297648 \tabularnewline
45 & 359 & 363.45857435583 & -6.26945502849838 & 360.810880672669 & 4.45857435582985 \tabularnewline
46 & 362 & 358.237759334306 & -3.26430176703221 & 369.026542432726 & -3.76224066569392 \tabularnewline
47 & 378 & 378.835131584025 & -0.0773357768091877 & 377.242204192784 & 0.835131584025419 \tabularnewline
48 & 345 & 329.573236907489 & -24.2594446230739 & 384.686207715585 & -15.4267630925113 \tabularnewline
49 & 422 & 425.446123375419 & 26.4236653861948 & 392.130211238387 & 3.44612337541861 \tabularnewline
50 & 430 & 430.640054905224 & 31.2979190176708 & 398.062026077105 & 0.640054905223906 \tabularnewline
51 & 443 & 454.561259023202 & 27.4449000609741 & 403.993840915824 & 11.561259023202 \tabularnewline
52 & 431 & 448.605803426908 & 4.67600235024529 & 408.718194222847 & 17.6058034269078 \tabularnewline
53 & 425 & 436.104887567789 & 0.452564902341024 & 413.44254752987 & 11.1048875677891 \tabularnewline
54 & 432 & 449.872466923501 & -3.24364100523016 & 417.371174081729 & 17.8724669235008 \tabularnewline
55 & 387 & 380.730966999608 & -28.030767633197 & 421.299800633589 & -6.26903300039191 \tabularnewline
56 & 396 & 392.480656576576 & -25.1501097519631 & 424.669453175387 & -3.51934342342429 \tabularnewline
57 & 411 & 400.230349311313 & -6.26945502849838 & 428.039105717186 & -10.7696506886875 \tabularnewline
58 & 421 & 414.31078730989 & -3.26430176703221 & 430.953514457142 & -6.68921269010985 \tabularnewline
59 & 424 & 414.209412579711 & -0.0773357768091877 & 433.867923197098 & -9.79058742028917 \tabularnewline
60 & 410 & 407.961424857112 & -24.2594446230739 & 436.298019765962 & -2.03857514288848 \tabularnewline
61 & 464 & 462.848218278979 & 26.4236653861948 & 438.728116334826 & -1.15178172102111 \tabularnewline
62 & 486 & 500.221987712594 & 31.2979190176708 & 440.480093269735 & 14.2219877125943 \tabularnewline
63 & 490 & 510.323029734382 & 27.4449000609741 & 442.232070204644 & 20.3230297343824 \tabularnewline
64 & 459 & 470.662505464168 & 4.67600235024529 & 442.661492185587 & 11.6625054641681 \tabularnewline
65 & 454 & 464.456520931129 & 0.452564902341024 & 443.09091416653 & 10.4565209311294 \tabularnewline
66 & 446 & 453.905909070272 & -3.24364100523016 & 441.337731934958 & 7.90590907027206 \tabularnewline
67 & 406 & 400.44621792981 & -28.030767633197 & 439.584549703387 & -5.55378207018958 \tabularnewline
68 & 412 & 413.725304411208 & -25.1501097519631 & 435.424805340755 & 1.72530441120836 \tabularnewline
69 & 428 & 431.004394050375 & -6.26945502849838 & 431.265060978123 & 3.00439405037548 \tabularnewline
70 & 429 & 436.04677737402 & -3.26430176703221 & 425.217524393013 & 7.04677737401954 \tabularnewline
71 & 425 & 430.907347968907 & -0.0773357768091877 & 419.169987807902 & 5.90734796890678 \tabularnewline
72 & 396 & 404.263131901301 & -24.2594446230739 & 411.996312721773 & 8.26313190130094 \tabularnewline
73 & 429 & 426.753696978162 & 26.4236653861948 & 404.822637635643 & -2.24630302183817 \tabularnewline
74 & 439 & 449.548449532047 & 31.2979190176708 & 397.153631450282 & 10.5484495320469 \tabularnewline
75 & 424 & 431.070474674105 & 27.4449000609741 & 389.484625264921 & 7.07047467410462 \tabularnewline
76 & 379 & 371.177422849211 & 4.67600235024529 & 382.146574800543 & -7.82257715078867 \tabularnewline
77 & 370 & 364.738910761494 & 0.452564902341024 & 374.808524336165 & -5.26108923850643 \tabularnewline
78 & 353 & 340.925770581175 & -3.24364100523016 & 368.317870424055 & -12.0742294188252 \tabularnewline
79 & 322 & 310.203551121252 & -28.030767633197 & 361.827216511945 & -11.7964488787481 \tabularnewline
80 & 322 & 312.813110314646 & -25.1501097519631 & 356.336999437317 & -9.18688968535366 \tabularnewline
81 & 338 & 331.42267266581 & -6.26945502849838 & 350.846782362688 & -6.57732733418999 \tabularnewline
82 & 348 & 353.305919340063 & -3.26430176703221 & 345.958382426969 & 5.3059193400631 \tabularnewline
83 & 350 & 359.007353285559 & -0.0773357768091877 & 341.06998249125 & 9.00735328555925 \tabularnewline
84 & 312 & 311.803446859452 & -24.2594446230739 & 336.455997763622 & -0.196553140548474 \tabularnewline
85 & 358 & 357.73432157781 & 26.4236653861948 & 331.842013035995 & -0.26567842218958 \tabularnewline
86 & 378 & 397.584974570901 & 31.2979190176708 & 327.117106411428 & 19.5849745709013 \tabularnewline
87 & 352 & 354.162900152165 & 27.4449000609741 & 322.392199786861 & 2.16290015216487 \tabularnewline
88 & 312 & 301.854527226849 & 4.67600235024529 & 317.469470422905 & -10.1454727731508 \tabularnewline
89 & 310 & 307.000694038709 & 0.452564902341024 & 312.54674105895 & -2.99930596129093 \tabularnewline
90 & 292 & 279.718127475333 & -3.24364100523016 & 307.525513529898 & -12.2818725246674 \tabularnewline
91 & 276 & 277.526481632352 & -28.030767633197 & 302.504286000845 & 1.52648163235176 \tabularnewline
92 & 269 & 265.426095192001 & -25.1501097519631 & 297.724014559962 & -3.57390480799938 \tabularnewline
93 & 286 & 285.325711909419 & -6.26945502849838 & 292.94374311908 & -0.674288090581342 \tabularnewline
94 & 292 & 298.181810459289 & -3.26430176703221 & 289.082491307743 & 6.18181045928901 \tabularnewline
95 & 288 & 290.856096280402 & -0.0773357768091877 & 285.221239496407 & 2.85609628040243 \tabularnewline
96 & 255 & 252.203693453886 & -24.2594446230739 & 282.055751169188 & -2.79630654611429 \tabularnewline
97 & 304 & 302.686071771836 & 26.4236653861948 & 278.89026284197 & -1.3139282281644 \tabularnewline
98 & 299 & 290.633599781964 & 31.2979190176708 & 276.068481200366 & -8.36640021803646 \tabularnewline
99 & 293 & 285.308400380264 & 27.4449000609741 & 273.246699558762 & -7.69159961973583 \tabularnewline
100 & 275 & 274.176820941592 & 4.67600235024529 & 271.147176708163 & -0.823179058408414 \tabularnewline
101 & 272 & 274.499781240094 & 0.452564902341024 & 269.047653857565 & 2.49978124009448 \tabularnewline
102 & 264 & 263.059390619822 & -3.24364100523016 & 268.184250385408 & -0.940609380178046 \tabularnewline
103 & 234 & 228.709920719945 & -28.030767633197 & 267.320846913252 & -5.29007928005484 \tabularnewline
104 & 231 & 218.995141542205 & -25.1501097519631 & 268.154968209758 & -12.0048584577947 \tabularnewline
105 & 263 & 263.280365522234 & -6.26945502849838 & 268.989089506264 & 0.28036552223449 \tabularnewline
106 & 264 & 259.556699251239 & -3.26430176703221 & 271.707602515793 & -4.44330074876081 \tabularnewline
107 & 264 & 253.651220251487 & -0.0773357768091877 & 274.426115525322 & -10.348779748513 \tabularnewline
108 & 245 & 235.207207950387 & -24.2594446230739 & 279.052236672687 & -9.79279204961341 \tabularnewline
109 & 297 & 283.897976793753 & 26.4236653861948 & 283.678357820052 & -13.1020232062472 \tabularnewline
110 & 317 & 312.443002980782 & 31.2979190176708 & 290.259078001547 & -4.55699701921753 \tabularnewline
111 & 318 & 311.715301755985 & 27.4449000609741 & 296.839798183041 & -6.2846982440152 \tabularnewline
112 & 315 & 320.343692711929 & 4.67600235024529 & 304.980304937826 & 5.34369271192884 \tabularnewline
113 & 312 & 310.426623405048 & 0.452564902341024 & 313.120811692611 & -1.57337659495164 \tabularnewline
114 & 310 & 300.888610357863 & -3.24364100523016 & 322.355030647367 & -9.11138964213688 \tabularnewline
115 & 306 & 308.441518031074 & -28.030767633197 & 331.589249602123 & 2.44151803107354 \tabularnewline
116 & 313 & 310.416618571051 & -25.1501097519631 & 340.733491180912 & -2.58338142894894 \tabularnewline
117 & 350 & 356.391722268798 & -6.26945502849838 & 349.877732759701 & 6.39172226879776 \tabularnewline
118 & 354 & 353.521072899121 & -3.26430176703221 & 357.743228867911 & -0.478927100878593 \tabularnewline
119 & 371 & 376.468610800688 & -0.0773357768091877 & 365.608724976121 & 5.46861080068823 \tabularnewline
120 & 357 & 366.59329110528 & -24.2594446230739 & 371.666153517794 & 9.59329110527977 \tabularnewline
121 & 419 & 433.852752554338 & 26.4236653861948 & 377.723582059467 & 14.8527525543379 \tabularnewline
122 & 425 & 437.009843201092 & 31.2979190176708 & 381.692237781237 & 12.0098432010923 \tabularnewline
123 & 424 & 434.894206436019 & 27.4449000609741 & 385.660893503007 & 10.8942064360194 \tabularnewline
124 & 399 & 407.126461745117 & 4.67600235024529 & 386.197535904637 & 8.12646174511747 \tabularnewline
125 & 393 & 398.813256791391 & 0.452564902341024 & 386.734178306268 & 5.81325679139115 \tabularnewline
126 & 378 & 372.354470058853 & -3.24364100523016 & 386.889170946377 & -5.64552994114678 \tabularnewline
127 & 371 & 382.986604046711 & -28.030767633197 & 387.044163586486 & 11.986604046711 \tabularnewline
128 & 364 & 366.199063706569 & -25.1501097519631 & 386.951046045394 & 2.19906370656878 \tabularnewline
129 & 384 & 387.411526524196 & -6.26945502849838 & 386.857928504303 & 3.41152652419578 \tabularnewline
130 & 377 & 370.756927114394 & -3.26430176703221 & 386.507374652638 & -6.24307288560584 \tabularnewline
131 & 383 & 379.920514975836 & -0.0773357768091877 & 386.156820800973 & -3.07948502416428 \tabularnewline
132 & 352 & 342.645037288947 & -24.2594446230739 & 385.614407334126 & -9.35496271105256 \tabularnewline
\hline
\end{tabular}
%Source: https://freestatistics.org/blog/index.php?pk=149807&T=2
[TABLE]
[ROW][C]Seasonal Decomposition by Loess - Time Series Components[/C][/ROW]
[ROW][C]t[/C][C]Observed[/C][C]Fitted[/C][C]Seasonal[/C][C]Trend[/C][C]Remainder[/C][/ROW]
[ROW][C]1[/C][C]274[/C][C]258.825969297934[/C][C]26.4236653861948[/C][C]262.750365315871[/C][C]-15.174030702066[/C][/ROW]
[ROW][C]2[/C][C]291[/C][C]291.074065711283[/C][C]31.2979190176708[/C][C]259.628015271047[/C][C]0.0740657112825716[/C][/ROW]
[ROW][C]3[/C][C]280[/C][C]276.049434712804[/C][C]27.4449000609741[/C][C]256.505665226222[/C][C]-3.9505652871961[/C][/ROW]
[ROW][C]4[/C][C]258[/C][C]258.212634364351[/C][C]4.67600235024529[/C][C]253.111363285404[/C][C]0.212634364350805[/C][/ROW]
[ROW][C]5[/C][C]252[/C][C]253.830373753073[/C][C]0.452564902341024[/C][C]249.717061344586[/C][C]1.83037375307325[/C][/ROW]
[ROW][C]6[/C][C]251[/C][C]259.136401466967[/C][C]-3.24364100523016[/C][C]246.107239538264[/C][C]8.13640146696659[/C][/ROW]
[ROW][C]7[/C][C]224[/C][C]233.533349901256[/C][C]-28.030767633197[/C][C]242.497417731941[/C][C]9.5333499012556[/C][/ROW]
[ROW][C]8[/C][C]225[/C][C]236.391724668407[/C][C]-25.1501097519631[/C][C]238.758385083556[/C][C]11.3917246684067[/C][/ROW]
[ROW][C]9[/C][C]234[/C][C]239.250102593327[/C][C]-6.26945502849838[/C][C]235.019352435172[/C][C]5.25010259332686[/C][/ROW]
[ROW][C]10[/C][C]233[/C][C]238.808370173524[/C][C]-3.26430176703221[/C][C]230.455931593509[/C][C]5.80837017352368[/C][/ROW]
[ROW][C]11[/C][C]229[/C][C]232.184825024964[/C][C]-0.0773357768091877[/C][C]225.892510751846[/C][C]3.18482502496363[/C][/ROW]
[ROW][C]12[/C][C]208[/C][C]218.819406060445[/C][C]-24.2594446230739[/C][C]221.440038562629[/C][C]10.8194060604446[/C][/ROW]
[ROW][C]13[/C][C]224[/C][C]204.588768240392[/C][C]26.4236653861948[/C][C]216.987566373413[/C][C]-19.4112317596078[/C][/ROW]
[ROW][C]14[/C][C]226[/C][C]207.046092187521[/C][C]31.2979190176708[/C][C]213.655988794808[/C][C]-18.9539078124792[/C][/ROW]
[ROW][C]15[/C][C]223[/C][C]208.230688722822[/C][C]27.4449000609741[/C][C]210.324411216204[/C][C]-14.769311277178[/C][/ROW]
[ROW][C]16[/C][C]205[/C][C]196.699714448582[/C][C]4.67600235024529[/C][C]208.624283201172[/C][C]-8.30028555141757[/C][/ROW]
[ROW][C]17[/C][C]201[/C][C]194.623279911518[/C][C]0.452564902341024[/C][C]206.924155186141[/C][C]-6.37672008848168[/C][/ROW]
[ROW][C]18[/C][C]202[/C][C]199.297444563401[/C][C]-3.24364100523016[/C][C]207.946196441829[/C][C]-2.70255543659889[/C][/ROW]
[ROW][C]19[/C][C]183[/C][C]185.06252993568[/C][C]-28.030767633197[/C][C]208.968237697517[/C][C]2.06252993567958[/C][/ROW]
[ROW][C]20[/C][C]188[/C][C]189.345534155014[/C][C]-25.1501097519631[/C][C]211.804575596949[/C][C]1.34553415501392[/C][/ROW]
[ROW][C]21[/C][C]200[/C][C]191.628541532117[/C][C]-6.26945502849838[/C][C]214.640913496381[/C][C]-8.3714584678826[/C][/ROW]
[ROW][C]22[/C][C]206[/C][C]197.389114317507[/C][C]-3.26430176703221[/C][C]217.875187449526[/C][C]-8.61088568249343[/C][/ROW]
[ROW][C]23[/C][C]211[/C][C]200.967874374139[/C][C]-0.0773357768091877[/C][C]221.10946140267[/C][C]-10.0321256258611[/C][/ROW]
[ROW][C]24[/C][C]201[/C][C]201.30198112077[/C][C]-24.2594446230739[/C][C]224.957463502304[/C][C]0.301981120769597[/C][/ROW]
[ROW][C]25[/C][C]299[/C][C]342.770869011867[/C][C]26.4236653861948[/C][C]228.805465601938[/C][C]43.7708690118669[/C][/ROW]
[ROW][C]26[/C][C]244[/C][C]223.032878673122[/C][C]31.2979190176708[/C][C]233.669202309207[/C][C]-20.9671213268776[/C][/ROW]
[ROW][C]27[/C][C]251[/C][C]236.022160922551[/C][C]27.4449000609741[/C][C]238.532939016475[/C][C]-14.9778390774493[/C][/ROW]
[ROW][C]28[/C][C]241[/C][C]233.353851955543[/C][C]4.67600235024529[/C][C]243.970145694212[/C][C]-7.64614804445725[/C][/ROW]
[ROW][C]29[/C][C]244[/C][C]238.14008272571[/C][C]0.452564902341024[/C][C]249.407352371949[/C][C]-5.85991727428964[/C][/ROW]
[ROW][C]30[/C][C]252[/C][C]252.220130901067[/C][C]-3.24364100523016[/C][C]255.023510104163[/C][C]0.220130901067392[/C][/ROW]
[ROW][C]31[/C][C]234[/C][C]235.39109979682[/C][C]-28.030767633197[/C][C]260.639667836377[/C][C]1.39109979682013[/C][/ROW]
[ROW][C]32[/C][C]246[/C][C]250.233914402013[/C][C]-25.1501097519631[/C][C]266.91619534995[/C][C]4.23391440201272[/C][/ROW]
[ROW][C]33[/C][C]265[/C][C]263.076732164974[/C][C]-6.26945502849838[/C][C]273.192722863524[/C][C]-1.92326783502557[/C][/ROW]
[ROW][C]34[/C][C]277[/C][C]277.185614884768[/C][C]-3.26430176703221[/C][C]280.078686882264[/C][C]0.185614884768427[/C][/ROW]
[ROW][C]35[/C][C]287[/C][C]287.112684875805[/C][C]-0.0773357768091877[/C][C]286.964650901004[/C][C]0.112684875805485[/C][/ROW]
[ROW][C]36[/C][C]275[/C][C]280.380009963114[/C][C]-24.2594446230739[/C][C]293.87943465996[/C][C]5.38000996311411[/C][/ROW]
[ROW][C]37[/C][C]320[/C][C]312.782116194889[/C][C]26.4236653861948[/C][C]300.794218418916[/C][C]-7.21788380511066[/C][/ROW]
[ROW][C]38[/C][C]338[/C][C]336.6629188195[/C][C]31.2979190176708[/C][C]308.039162162829[/C][C]-1.33708118049992[/C][/ROW]
[ROW][C]39[/C][C]342[/C][C]341.270994032284[/C][C]27.4449000609741[/C][C]315.284105906742[/C][C]-0.729005967716489[/C][/ROW]
[ROW][C]40[/C][C]322[/C][C]316.628701218287[/C][C]4.67600235024529[/C][C]322.695296431467[/C][C]-5.37129878171265[/C][/ROW]
[ROW][C]41[/C][C]323[/C][C]315.440948141467[/C][C]0.452564902341024[/C][C]330.106486956192[/C][C]-7.55905185853334[/C][/ROW]
[ROW][C]42[/C][C]343[/C][C]351.69690644127[/C][C]-3.24364100523016[/C][C]337.54673456396[/C][C]8.69690644126996[/C][/ROW]
[ROW][C]43[/C][C]315[/C][C]313.043785461469[/C][C]-28.030767633197[/C][C]344.986982171728[/C][C]-1.95621453853107[/C][/ROW]
[ROW][C]44[/C][C]334[/C][C]340.251178329765[/C][C]-25.1501097519631[/C][C]352.898931422198[/C][C]6.2511783297648[/C][/ROW]
[ROW][C]45[/C][C]359[/C][C]363.45857435583[/C][C]-6.26945502849838[/C][C]360.810880672669[/C][C]4.45857435582985[/C][/ROW]
[ROW][C]46[/C][C]362[/C][C]358.237759334306[/C][C]-3.26430176703221[/C][C]369.026542432726[/C][C]-3.76224066569392[/C][/ROW]
[ROW][C]47[/C][C]378[/C][C]378.835131584025[/C][C]-0.0773357768091877[/C][C]377.242204192784[/C][C]0.835131584025419[/C][/ROW]
[ROW][C]48[/C][C]345[/C][C]329.573236907489[/C][C]-24.2594446230739[/C][C]384.686207715585[/C][C]-15.4267630925113[/C][/ROW]
[ROW][C]49[/C][C]422[/C][C]425.446123375419[/C][C]26.4236653861948[/C][C]392.130211238387[/C][C]3.44612337541861[/C][/ROW]
[ROW][C]50[/C][C]430[/C][C]430.640054905224[/C][C]31.2979190176708[/C][C]398.062026077105[/C][C]0.640054905223906[/C][/ROW]
[ROW][C]51[/C][C]443[/C][C]454.561259023202[/C][C]27.4449000609741[/C][C]403.993840915824[/C][C]11.561259023202[/C][/ROW]
[ROW][C]52[/C][C]431[/C][C]448.605803426908[/C][C]4.67600235024529[/C][C]408.718194222847[/C][C]17.6058034269078[/C][/ROW]
[ROW][C]53[/C][C]425[/C][C]436.104887567789[/C][C]0.452564902341024[/C][C]413.44254752987[/C][C]11.1048875677891[/C][/ROW]
[ROW][C]54[/C][C]432[/C][C]449.872466923501[/C][C]-3.24364100523016[/C][C]417.371174081729[/C][C]17.8724669235008[/C][/ROW]
[ROW][C]55[/C][C]387[/C][C]380.730966999608[/C][C]-28.030767633197[/C][C]421.299800633589[/C][C]-6.26903300039191[/C][/ROW]
[ROW][C]56[/C][C]396[/C][C]392.480656576576[/C][C]-25.1501097519631[/C][C]424.669453175387[/C][C]-3.51934342342429[/C][/ROW]
[ROW][C]57[/C][C]411[/C][C]400.230349311313[/C][C]-6.26945502849838[/C][C]428.039105717186[/C][C]-10.7696506886875[/C][/ROW]
[ROW][C]58[/C][C]421[/C][C]414.31078730989[/C][C]-3.26430176703221[/C][C]430.953514457142[/C][C]-6.68921269010985[/C][/ROW]
[ROW][C]59[/C][C]424[/C][C]414.209412579711[/C][C]-0.0773357768091877[/C][C]433.867923197098[/C][C]-9.79058742028917[/C][/ROW]
[ROW][C]60[/C][C]410[/C][C]407.961424857112[/C][C]-24.2594446230739[/C][C]436.298019765962[/C][C]-2.03857514288848[/C][/ROW]
[ROW][C]61[/C][C]464[/C][C]462.848218278979[/C][C]26.4236653861948[/C][C]438.728116334826[/C][C]-1.15178172102111[/C][/ROW]
[ROW][C]62[/C][C]486[/C][C]500.221987712594[/C][C]31.2979190176708[/C][C]440.480093269735[/C][C]14.2219877125943[/C][/ROW]
[ROW][C]63[/C][C]490[/C][C]510.323029734382[/C][C]27.4449000609741[/C][C]442.232070204644[/C][C]20.3230297343824[/C][/ROW]
[ROW][C]64[/C][C]459[/C][C]470.662505464168[/C][C]4.67600235024529[/C][C]442.661492185587[/C][C]11.6625054641681[/C][/ROW]
[ROW][C]65[/C][C]454[/C][C]464.456520931129[/C][C]0.452564902341024[/C][C]443.09091416653[/C][C]10.4565209311294[/C][/ROW]
[ROW][C]66[/C][C]446[/C][C]453.905909070272[/C][C]-3.24364100523016[/C][C]441.337731934958[/C][C]7.90590907027206[/C][/ROW]
[ROW][C]67[/C][C]406[/C][C]400.44621792981[/C][C]-28.030767633197[/C][C]439.584549703387[/C][C]-5.55378207018958[/C][/ROW]
[ROW][C]68[/C][C]412[/C][C]413.725304411208[/C][C]-25.1501097519631[/C][C]435.424805340755[/C][C]1.72530441120836[/C][/ROW]
[ROW][C]69[/C][C]428[/C][C]431.004394050375[/C][C]-6.26945502849838[/C][C]431.265060978123[/C][C]3.00439405037548[/C][/ROW]
[ROW][C]70[/C][C]429[/C][C]436.04677737402[/C][C]-3.26430176703221[/C][C]425.217524393013[/C][C]7.04677737401954[/C][/ROW]
[ROW][C]71[/C][C]425[/C][C]430.907347968907[/C][C]-0.0773357768091877[/C][C]419.169987807902[/C][C]5.90734796890678[/C][/ROW]
[ROW][C]72[/C][C]396[/C][C]404.263131901301[/C][C]-24.2594446230739[/C][C]411.996312721773[/C][C]8.26313190130094[/C][/ROW]
[ROW][C]73[/C][C]429[/C][C]426.753696978162[/C][C]26.4236653861948[/C][C]404.822637635643[/C][C]-2.24630302183817[/C][/ROW]
[ROW][C]74[/C][C]439[/C][C]449.548449532047[/C][C]31.2979190176708[/C][C]397.153631450282[/C][C]10.5484495320469[/C][/ROW]
[ROW][C]75[/C][C]424[/C][C]431.070474674105[/C][C]27.4449000609741[/C][C]389.484625264921[/C][C]7.07047467410462[/C][/ROW]
[ROW][C]76[/C][C]379[/C][C]371.177422849211[/C][C]4.67600235024529[/C][C]382.146574800543[/C][C]-7.82257715078867[/C][/ROW]
[ROW][C]77[/C][C]370[/C][C]364.738910761494[/C][C]0.452564902341024[/C][C]374.808524336165[/C][C]-5.26108923850643[/C][/ROW]
[ROW][C]78[/C][C]353[/C][C]340.925770581175[/C][C]-3.24364100523016[/C][C]368.317870424055[/C][C]-12.0742294188252[/C][/ROW]
[ROW][C]79[/C][C]322[/C][C]310.203551121252[/C][C]-28.030767633197[/C][C]361.827216511945[/C][C]-11.7964488787481[/C][/ROW]
[ROW][C]80[/C][C]322[/C][C]312.813110314646[/C][C]-25.1501097519631[/C][C]356.336999437317[/C][C]-9.18688968535366[/C][/ROW]
[ROW][C]81[/C][C]338[/C][C]331.42267266581[/C][C]-6.26945502849838[/C][C]350.846782362688[/C][C]-6.57732733418999[/C][/ROW]
[ROW][C]82[/C][C]348[/C][C]353.305919340063[/C][C]-3.26430176703221[/C][C]345.958382426969[/C][C]5.3059193400631[/C][/ROW]
[ROW][C]83[/C][C]350[/C][C]359.007353285559[/C][C]-0.0773357768091877[/C][C]341.06998249125[/C][C]9.00735328555925[/C][/ROW]
[ROW][C]84[/C][C]312[/C][C]311.803446859452[/C][C]-24.2594446230739[/C][C]336.455997763622[/C][C]-0.196553140548474[/C][/ROW]
[ROW][C]85[/C][C]358[/C][C]357.73432157781[/C][C]26.4236653861948[/C][C]331.842013035995[/C][C]-0.26567842218958[/C][/ROW]
[ROW][C]86[/C][C]378[/C][C]397.584974570901[/C][C]31.2979190176708[/C][C]327.117106411428[/C][C]19.5849745709013[/C][/ROW]
[ROW][C]87[/C][C]352[/C][C]354.162900152165[/C][C]27.4449000609741[/C][C]322.392199786861[/C][C]2.16290015216487[/C][/ROW]
[ROW][C]88[/C][C]312[/C][C]301.854527226849[/C][C]4.67600235024529[/C][C]317.469470422905[/C][C]-10.1454727731508[/C][/ROW]
[ROW][C]89[/C][C]310[/C][C]307.000694038709[/C][C]0.452564902341024[/C][C]312.54674105895[/C][C]-2.99930596129093[/C][/ROW]
[ROW][C]90[/C][C]292[/C][C]279.718127475333[/C][C]-3.24364100523016[/C][C]307.525513529898[/C][C]-12.2818725246674[/C][/ROW]
[ROW][C]91[/C][C]276[/C][C]277.526481632352[/C][C]-28.030767633197[/C][C]302.504286000845[/C][C]1.52648163235176[/C][/ROW]
[ROW][C]92[/C][C]269[/C][C]265.426095192001[/C][C]-25.1501097519631[/C][C]297.724014559962[/C][C]-3.57390480799938[/C][/ROW]
[ROW][C]93[/C][C]286[/C][C]285.325711909419[/C][C]-6.26945502849838[/C][C]292.94374311908[/C][C]-0.674288090581342[/C][/ROW]
[ROW][C]94[/C][C]292[/C][C]298.181810459289[/C][C]-3.26430176703221[/C][C]289.082491307743[/C][C]6.18181045928901[/C][/ROW]
[ROW][C]95[/C][C]288[/C][C]290.856096280402[/C][C]-0.0773357768091877[/C][C]285.221239496407[/C][C]2.85609628040243[/C][/ROW]
[ROW][C]96[/C][C]255[/C][C]252.203693453886[/C][C]-24.2594446230739[/C][C]282.055751169188[/C][C]-2.79630654611429[/C][/ROW]
[ROW][C]97[/C][C]304[/C][C]302.686071771836[/C][C]26.4236653861948[/C][C]278.89026284197[/C][C]-1.3139282281644[/C][/ROW]
[ROW][C]98[/C][C]299[/C][C]290.633599781964[/C][C]31.2979190176708[/C][C]276.068481200366[/C][C]-8.36640021803646[/C][/ROW]
[ROW][C]99[/C][C]293[/C][C]285.308400380264[/C][C]27.4449000609741[/C][C]273.246699558762[/C][C]-7.69159961973583[/C][/ROW]
[ROW][C]100[/C][C]275[/C][C]274.176820941592[/C][C]4.67600235024529[/C][C]271.147176708163[/C][C]-0.823179058408414[/C][/ROW]
[ROW][C]101[/C][C]272[/C][C]274.499781240094[/C][C]0.452564902341024[/C][C]269.047653857565[/C][C]2.49978124009448[/C][/ROW]
[ROW][C]102[/C][C]264[/C][C]263.059390619822[/C][C]-3.24364100523016[/C][C]268.184250385408[/C][C]-0.940609380178046[/C][/ROW]
[ROW][C]103[/C][C]234[/C][C]228.709920719945[/C][C]-28.030767633197[/C][C]267.320846913252[/C][C]-5.29007928005484[/C][/ROW]
[ROW][C]104[/C][C]231[/C][C]218.995141542205[/C][C]-25.1501097519631[/C][C]268.154968209758[/C][C]-12.0048584577947[/C][/ROW]
[ROW][C]105[/C][C]263[/C][C]263.280365522234[/C][C]-6.26945502849838[/C][C]268.989089506264[/C][C]0.28036552223449[/C][/ROW]
[ROW][C]106[/C][C]264[/C][C]259.556699251239[/C][C]-3.26430176703221[/C][C]271.707602515793[/C][C]-4.44330074876081[/C][/ROW]
[ROW][C]107[/C][C]264[/C][C]253.651220251487[/C][C]-0.0773357768091877[/C][C]274.426115525322[/C][C]-10.348779748513[/C][/ROW]
[ROW][C]108[/C][C]245[/C][C]235.207207950387[/C][C]-24.2594446230739[/C][C]279.052236672687[/C][C]-9.79279204961341[/C][/ROW]
[ROW][C]109[/C][C]297[/C][C]283.897976793753[/C][C]26.4236653861948[/C][C]283.678357820052[/C][C]-13.1020232062472[/C][/ROW]
[ROW][C]110[/C][C]317[/C][C]312.443002980782[/C][C]31.2979190176708[/C][C]290.259078001547[/C][C]-4.55699701921753[/C][/ROW]
[ROW][C]111[/C][C]318[/C][C]311.715301755985[/C][C]27.4449000609741[/C][C]296.839798183041[/C][C]-6.2846982440152[/C][/ROW]
[ROW][C]112[/C][C]315[/C][C]320.343692711929[/C][C]4.67600235024529[/C][C]304.980304937826[/C][C]5.34369271192884[/C][/ROW]
[ROW][C]113[/C][C]312[/C][C]310.426623405048[/C][C]0.452564902341024[/C][C]313.120811692611[/C][C]-1.57337659495164[/C][/ROW]
[ROW][C]114[/C][C]310[/C][C]300.888610357863[/C][C]-3.24364100523016[/C][C]322.355030647367[/C][C]-9.11138964213688[/C][/ROW]
[ROW][C]115[/C][C]306[/C][C]308.441518031074[/C][C]-28.030767633197[/C][C]331.589249602123[/C][C]2.44151803107354[/C][/ROW]
[ROW][C]116[/C][C]313[/C][C]310.416618571051[/C][C]-25.1501097519631[/C][C]340.733491180912[/C][C]-2.58338142894894[/C][/ROW]
[ROW][C]117[/C][C]350[/C][C]356.391722268798[/C][C]-6.26945502849838[/C][C]349.877732759701[/C][C]6.39172226879776[/C][/ROW]
[ROW][C]118[/C][C]354[/C][C]353.521072899121[/C][C]-3.26430176703221[/C][C]357.743228867911[/C][C]-0.478927100878593[/C][/ROW]
[ROW][C]119[/C][C]371[/C][C]376.468610800688[/C][C]-0.0773357768091877[/C][C]365.608724976121[/C][C]5.46861080068823[/C][/ROW]
[ROW][C]120[/C][C]357[/C][C]366.59329110528[/C][C]-24.2594446230739[/C][C]371.666153517794[/C][C]9.59329110527977[/C][/ROW]
[ROW][C]121[/C][C]419[/C][C]433.852752554338[/C][C]26.4236653861948[/C][C]377.723582059467[/C][C]14.8527525543379[/C][/ROW]
[ROW][C]122[/C][C]425[/C][C]437.009843201092[/C][C]31.2979190176708[/C][C]381.692237781237[/C][C]12.0098432010923[/C][/ROW]
[ROW][C]123[/C][C]424[/C][C]434.894206436019[/C][C]27.4449000609741[/C][C]385.660893503007[/C][C]10.8942064360194[/C][/ROW]
[ROW][C]124[/C][C]399[/C][C]407.126461745117[/C][C]4.67600235024529[/C][C]386.197535904637[/C][C]8.12646174511747[/C][/ROW]
[ROW][C]125[/C][C]393[/C][C]398.813256791391[/C][C]0.452564902341024[/C][C]386.734178306268[/C][C]5.81325679139115[/C][/ROW]
[ROW][C]126[/C][C]378[/C][C]372.354470058853[/C][C]-3.24364100523016[/C][C]386.889170946377[/C][C]-5.64552994114678[/C][/ROW]
[ROW][C]127[/C][C]371[/C][C]382.986604046711[/C][C]-28.030767633197[/C][C]387.044163586486[/C][C]11.986604046711[/C][/ROW]
[ROW][C]128[/C][C]364[/C][C]366.199063706569[/C][C]-25.1501097519631[/C][C]386.951046045394[/C][C]2.19906370656878[/C][/ROW]
[ROW][C]129[/C][C]384[/C][C]387.411526524196[/C][C]-6.26945502849838[/C][C]386.857928504303[/C][C]3.41152652419578[/C][/ROW]
[ROW][C]130[/C][C]377[/C][C]370.756927114394[/C][C]-3.26430176703221[/C][C]386.507374652638[/C][C]-6.24307288560584[/C][/ROW]
[ROW][C]131[/C][C]383[/C][C]379.920514975836[/C][C]-0.0773357768091877[/C][C]386.156820800973[/C][C]-3.07948502416428[/C][/ROW]
[ROW][C]132[/C][C]352[/C][C]342.645037288947[/C][C]-24.2594446230739[/C][C]385.614407334126[/C][C]-9.35496271105256[/C][/ROW]
[/TABLE]
Source: https://freestatistics.org/blog/index.php?pk=149807&T=2
Globally Unique Identifier (entire table): ba.freestatistics.org/blog/index.php?pk=149807&T=2
As an alternative you can also use a QR Code:
The GUIDs for individual cells are displayed in the table below:
Seasonal Decomposition by Loess - Time Series Components t Observed Fitted Seasonal Trend Remainder 1 274 258.825969297934 26.4236653861948 262.750365315871 -15.174030702066 2 291 291.074065711283 31.2979190176708 259.628015271047 0.0740657112825716 3 280 276.049434712804 27.4449000609741 256.505665226222 -3.9505652871961 4 258 258.212634364351 4.67600235024529 253.111363285404 0.212634364350805 5 252 253.830373753073 0.452564902341024 249.717061344586 1.83037375307325 6 251 259.136401466967 -3.24364100523016 246.107239538264 8.13640146696659 7 224 233.533349901256 -28.030767633197 242.497417731941 9.5333499012556 8 225 236.391724668407 -25.1501097519631 238.758385083556 11.3917246684067 9 234 239.250102593327 -6.26945502849838 235.019352435172 5.25010259332686 10 233 238.808370173524 -3.26430176703221 230.455931593509 5.80837017352368 11 229 232.184825024964 -0.0773357768091877 225.892510751846 3.18482502496363 12 208 218.819406060445 -24.2594446230739 221.440038562629 10.8194060604446 13 224 204.588768240392 26.4236653861948 216.987566373413 -19.4112317596078 14 226 207.046092187521 31.2979190176708 213.655988794808 -18.9539078124792 15 223 208.230688722822 27.4449000609741 210.324411216204 -14.769311277178 16 205 196.699714448582 4.67600235024529 208.624283201172 -8.30028555141757 17 201 194.623279911518 0.452564902341024 206.924155186141 -6.37672008848168 18 202 199.297444563401 -3.24364100523016 207.946196441829 -2.70255543659889 19 183 185.06252993568 -28.030767633197 208.968237697517 2.06252993567958 20 188 189.345534155014 -25.1501097519631 211.804575596949 1.34553415501392 21 200 191.628541532117 -6.26945502849838 214.640913496381 -8.3714584678826 22 206 197.389114317507 -3.26430176703221 217.875187449526 -8.61088568249343 23 211 200.967874374139 -0.0773357768091877 221.10946140267 -10.0321256258611 24 201 201.30198112077 -24.2594446230739 224.957463502304 0.301981120769597 25 299 342.770869011867 26.4236653861948 228.805465601938 43.7708690118669 26 244 223.032878673122 31.2979190176708 233.669202309207 -20.9671213268776 27 251 236.022160922551 27.4449000609741 238.532939016475 -14.9778390774493 28 241 233.353851955543 4.67600235024529 243.970145694212 -7.64614804445725 29 244 238.14008272571 0.452564902341024 249.407352371949 -5.85991727428964 30 252 252.220130901067 -3.24364100523016 255.023510104163 0.220130901067392 31 234 235.39109979682 -28.030767633197 260.639667836377 1.39109979682013 32 246 250.233914402013 -25.1501097519631 266.91619534995 4.23391440201272 33 265 263.076732164974 -6.26945502849838 273.192722863524 -1.92326783502557 34 277 277.185614884768 -3.26430176703221 280.078686882264 0.185614884768427 35 287 287.112684875805 -0.0773357768091877 286.964650901004 0.112684875805485 36 275 280.380009963114 -24.2594446230739 293.87943465996 5.38000996311411 37 320 312.782116194889 26.4236653861948 300.794218418916 -7.21788380511066 38 338 336.6629188195 31.2979190176708 308.039162162829 -1.33708118049992 39 342 341.270994032284 27.4449000609741 315.284105906742 -0.729005967716489 40 322 316.628701218287 4.67600235024529 322.695296431467 -5.37129878171265 41 323 315.440948141467 0.452564902341024 330.106486956192 -7.55905185853334 42 343 351.69690644127 -3.24364100523016 337.54673456396 8.69690644126996 43 315 313.043785461469 -28.030767633197 344.986982171728 -1.95621453853107 44 334 340.251178329765 -25.1501097519631 352.898931422198 6.2511783297648 45 359 363.45857435583 -6.26945502849838 360.810880672669 4.45857435582985 46 362 358.237759334306 -3.26430176703221 369.026542432726 -3.76224066569392 47 378 378.835131584025 -0.0773357768091877 377.242204192784 0.835131584025419 48 345 329.573236907489 -24.2594446230739 384.686207715585 -15.4267630925113 49 422 425.446123375419 26.4236653861948 392.130211238387 3.44612337541861 50 430 430.640054905224 31.2979190176708 398.062026077105 0.640054905223906 51 443 454.561259023202 27.4449000609741 403.993840915824 11.561259023202 52 431 448.605803426908 4.67600235024529 408.718194222847 17.6058034269078 53 425 436.104887567789 0.452564902341024 413.44254752987 11.1048875677891 54 432 449.872466923501 -3.24364100523016 417.371174081729 17.8724669235008 55 387 380.730966999608 -28.030767633197 421.299800633589 -6.26903300039191 56 396 392.480656576576 -25.1501097519631 424.669453175387 -3.51934342342429 57 411 400.230349311313 -6.26945502849838 428.039105717186 -10.7696506886875 58 421 414.31078730989 -3.26430176703221 430.953514457142 -6.68921269010985 59 424 414.209412579711 -0.0773357768091877 433.867923197098 -9.79058742028917 60 410 407.961424857112 -24.2594446230739 436.298019765962 -2.03857514288848 61 464 462.848218278979 26.4236653861948 438.728116334826 -1.15178172102111 62 486 500.221987712594 31.2979190176708 440.480093269735 14.2219877125943 63 490 510.323029734382 27.4449000609741 442.232070204644 20.3230297343824 64 459 470.662505464168 4.67600235024529 442.661492185587 11.6625054641681 65 454 464.456520931129 0.452564902341024 443.09091416653 10.4565209311294 66 446 453.905909070272 -3.24364100523016 441.337731934958 7.90590907027206 67 406 400.44621792981 -28.030767633197 439.584549703387 -5.55378207018958 68 412 413.725304411208 -25.1501097519631 435.424805340755 1.72530441120836 69 428 431.004394050375 -6.26945502849838 431.265060978123 3.00439405037548 70 429 436.04677737402 -3.26430176703221 425.217524393013 7.04677737401954 71 425 430.907347968907 -0.0773357768091877 419.169987807902 5.90734796890678 72 396 404.263131901301 -24.2594446230739 411.996312721773 8.26313190130094 73 429 426.753696978162 26.4236653861948 404.822637635643 -2.24630302183817 74 439 449.548449532047 31.2979190176708 397.153631450282 10.5484495320469 75 424 431.070474674105 27.4449000609741 389.484625264921 7.07047467410462 76 379 371.177422849211 4.67600235024529 382.146574800543 -7.82257715078867 77 370 364.738910761494 0.452564902341024 374.808524336165 -5.26108923850643 78 353 340.925770581175 -3.24364100523016 368.317870424055 -12.0742294188252 79 322 310.203551121252 -28.030767633197 361.827216511945 -11.7964488787481 80 322 312.813110314646 -25.1501097519631 356.336999437317 -9.18688968535366 81 338 331.42267266581 -6.26945502849838 350.846782362688 -6.57732733418999 82 348 353.305919340063 -3.26430176703221 345.958382426969 5.3059193400631 83 350 359.007353285559 -0.0773357768091877 341.06998249125 9.00735328555925 84 312 311.803446859452 -24.2594446230739 336.455997763622 -0.196553140548474 85 358 357.73432157781 26.4236653861948 331.842013035995 -0.26567842218958 86 378 397.584974570901 31.2979190176708 327.117106411428 19.5849745709013 87 352 354.162900152165 27.4449000609741 322.392199786861 2.16290015216487 88 312 301.854527226849 4.67600235024529 317.469470422905 -10.1454727731508 89 310 307.000694038709 0.452564902341024 312.54674105895 -2.99930596129093 90 292 279.718127475333 -3.24364100523016 307.525513529898 -12.2818725246674 91 276 277.526481632352 -28.030767633197 302.504286000845 1.52648163235176 92 269 265.426095192001 -25.1501097519631 297.724014559962 -3.57390480799938 93 286 285.325711909419 -6.26945502849838 292.94374311908 -0.674288090581342 94 292 298.181810459289 -3.26430176703221 289.082491307743 6.18181045928901 95 288 290.856096280402 -0.0773357768091877 285.221239496407 2.85609628040243 96 255 252.203693453886 -24.2594446230739 282.055751169188 -2.79630654611429 97 304 302.686071771836 26.4236653861948 278.89026284197 -1.3139282281644 98 299 290.633599781964 31.2979190176708 276.068481200366 -8.36640021803646 99 293 285.308400380264 27.4449000609741 273.246699558762 -7.69159961973583 100 275 274.176820941592 4.67600235024529 271.147176708163 -0.823179058408414 101 272 274.499781240094 0.452564902341024 269.047653857565 2.49978124009448 102 264 263.059390619822 -3.24364100523016 268.184250385408 -0.940609380178046 103 234 228.709920719945 -28.030767633197 267.320846913252 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Parameters (Session):
par1 = 12 ; par2 = periodic ; par3 = 0 ; par5 = 1 ; par7 = 1 ; par8 = FALSE ;
Parameters (R input):
par1 = 12 ; par2 = periodic ; par3 = 0 ; par4 = ; par5 = 1 ; par6 = ; par7 = 1 ; par8 = FALSE ;
R code (references can be found in the software module):
par1 <- as.numeric(par1) #seasonal period
if (par2 != 'periodic') par2 <- as.numeric(par2) #s.window
par3 <- as.numeric(par3) #s.degree
if (par4 == '') par4 <- NULL else par4 <- as.numeric(par4)#t.window
par5 <- as.numeric(par5)#t.degree
if (par6 != '') par6 <- as.numeric(par6)#l.window
par7 <- as.numeric(par7)#l.degree
if (par8 == 'FALSE') par8 <- FALSE else par9 <- TRUE #robust
nx <- length(x)
x <- ts(x,frequency=par1)
if (par6 != '') {
m <- stl(x,s.window=par2, s.degree=par3, t.window=par4, t.degre=par5, l.window=par6, l.degree=par7, robust=par8)
} else {
m <- stl(x,s.window=par2, s.degree=par3, t.window=par4, t.degre=par5, l.degree=par7, robust=par8)
}
m$time.series m$win
m$deg m$jump
m$inner m$outer
bitmap(file='test1.png')
plot(m,main=main)
dev.off()
mylagmax <- nx/2
bitmap(file='test2.png')
op <- par(mfrow = c(2,2))
acf(as.numeric(x),lag.max = mylagmax,main='Observed')
acf(as.numeric(m$time.series[,'trend']),na.action=na.pass,lag.max = mylagmax,main='Trend') acf(as.numeric(m$time.series[,'seasonal']),na.action=na.pass,lag.max = mylagmax,main='Seasonal')
acf(as.numeric(m$time.series[,'remainder']),na.action=na.pass,lag.max = mylagmax,main='Remainder') par(op) dev.off() bitmap(file='test3.png') op <- par(mfrow = c(2,2)) spectrum(as.numeric(x),main='Observed') spectrum(as.numeric(m$time.series[!is.na(m$time.series[,'trend']),'trend']),main='Trend') spectrum(as.numeric(m$time.series[!is.na(m$time.series[,'seasonal']),'seasonal']),main='Seasonal') spectrum(as.numeric(m$time.series[!is.na(m$time.series[,'remainder']),'remainder']),main='Remainder') par(op) dev.off() bitmap(file='test4.png') op <- par(mfrow = c(2,2)) cpgram(as.numeric(x),main='Observed') cpgram(as.numeric(m$time.series[!is.na(m$time.series[,'trend']),'trend']),main='Trend') cpgram(as.numeric(m$time.series[!is.na(m$time.series[,'seasonal']),'seasonal']),main='Seasonal') cpgram(as.numeric(m$time.series[!is.na(m$time.series[,'remainder']),'remainder']),main='Remainder') par(op) dev.off() load(file='createtable') a<-table.start() a<-table.row.start(a) a<-table.element(a,'Seasonal Decomposition by Loess - Parameters',4,TRUE) a<-table.row.end(a) a<-table.row.start(a) a<-table.element(a,'Component',header=TRUE) a<-table.element(a,'Window',header=TRUE) a<-table.element(a,'Degree',header=TRUE) a<-table.element(a,'Jump',header=TRUE) a<-table.row.end(a) a<-table.row.start(a) a<-table.element(a,'Seasonal',header=TRUE) a<-table.element(a,m$win['s'])
a<-table.element(a,m$deg['s']) a<-table.element(a,m$jump['s'])
a<-table.row.end(a)
a<-table.row.start(a)
a<-table.element(a,m$win['t']) a<-table.element(a,m$deg['t'])
a<-table.element(a,m$jump['t']) a<-table.row.end(a) a<-table.row.start(a) a<-table.element(a,'Low-pass',header=TRUE) a<-table.element(a,m$win['l'])
a<-table.element(a,m$deg['l']) a<-table.element(a,m$jump['l'])
a<-table.row.end(a)
a<-table.end(a)
table.save(a,file='mytable.tab')
a<-table.start()
a<-table.row.start(a)
a<-table.element(a,'Seasonal Decomposition by Loess - Time Series Components',6,TRUE)
a<-table.row.end(a)
a<-table.row.start(a)
a<-table.element(a,x[i]+m$time.series[i,'remainder']) a<-table.element(a,m$time.series[i,'seasonal'])
a<-table.element(a,m$time.series[i,'trend']) a<-table.element(a,m$time.series[i,'remainder'])
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en
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http://www.statemaster.com/encyclopedia/Traveling-salesman
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text/html
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crawl-data/CC-MAIN-2019-43/segments/1570986658566.9/warc/CC-MAIN-20191015104838-20191015132338-00161.warc.gz
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Encyclopedia > Traveling salesman
The traveling salesman problem (TSP), also known as the traveling salesperson problem, is a problem in discrete or combinatorial optimization. It is a prominent illustration of a class of problems in computational complexity theory which are hard to solve.
Given a number of cities and the costs of travelling from any city to any other city, what is the cheapest round-trip route that visits each city once and then returns to the starting city?
An equivalent formulation in terms of graph theory is: Given a complete weighted graph, (where the vertices would represent the cities, the edges would represent the roads, and the weights would be the cost or distance of that road,) find the Hamiltonian cycle with the least weight.
It can be shown that the requirement of returning to the starting city does not change the computational complexity of the problem.
A related problem is the Bottleneck traveling salesman problem (bottleneck TSP): Find the Hamiltonian cycle in a weighted graph with the minimal length of the longest edge.
The problem is of considerable practical importance, apart from evident transportation and logistics areas. A classic example is in printed circuit manufacturing -- scheduling of a route of the drill machine to drill holes in a PCB. In robotic machining or drilling applications, the "cities" are parts to machine or holes (of different sizes) to drill, and the "cost of travel" includes time for retooling the robot (single machine job sequencing problem).
## Computational complexity
The most direct solution would be to try all the combinations and see which one is cheapest (using brute force), but given that the number of combinations is N! (the factorial of the number of cities), this solution rapidly becomes impractical.
## NP-hardness
The problem has been shown to be NP-hard (more precisely, it is complete for the complexity class FPNP; see the function problem article), and the decision problem version ("given the costs and a number x, decide whether there is a roundtrip route cheaper than x") is NP-complete.
The bottleneck traveling salesman problem is also NP-hard.
The problem remains NP-hard even for the case when the cities are in the plane with Euclidean distances, as well as in a number of other restrictive cases.
Removing the condition of visiting each city "only once" doesn't remove the NP-hardness, since it is easily seen that in the planar case an optimal tour visits cities only once (otherwise, by the triangle inequality, a shortcut that skips a repeated visit will decrease the tour length).
## Algorithms
The traditional lines of attack for the NP-hard problems are the following:
• Devising algorithms for finding exact solutions (they will work reasonably fast only for relatively small problem sizes)
• Devising "suboptimal" or heuristic algorithms, i.e., algorithms that deliver either seemingly or provably good solutions, but which could not be proved to be optimal.
• Finding special cases for the problem ("subproblems") for which either exact or better heuristics are possible.
For benchmarking of TSP algorithms, TSPLIB a library of sample instances of the TSP and related problems is maintained, see the TSPLIB external reference. Many of them are lists of actual cities and layouts of actual printed circuits.
### Exact algorithms
• Various branch-and-bound algorithms, which can be used to process TSPs containing 40-60 cities.
• Progressive improvement algorithms which use techniques reminiscent of linear programming. Works well for up to 120-200 cities.
An exact solution for 15,112 German cities from TSPLIB was found in 2001 using the Cutting-plane method proposed by George Dantzig, Ray Fulkerson, and Selmer Johnson in 1954, based on linear programming. The computations were performed on a network of 110 processors located at Rice University and Princeton University, see the Princeton external link. The total computation time was equivalent to 22.6 years on a single 500 MHz Alpha processor. In May 2004, the traveling salesman problem of visiting all 24,978 cities in Sweden was solved: a tour of length approximately 72,500 kilometers was found and it was proven that no shorter tour exists.
### Heuristics
Various approximation algorithms, which "quickly" yield "good" solutions with "high" probability, have been devised. Modern methods can find solutions for extremely large problems (millions of cities) within a reasonable time which are provably 2-3% away from the optimal solution.
Several categories of heuristics are recognized.
#### Constructive heuristics
• The nearest neighbour algorithm, which is normally fairly close to the optimal route, and doesn't take too long to execute. Unfortunately, it is provably reliable only for special cases of the TSP. In the general case, there exists an example for which the nearest neighbour algorithm gives the worst possible route.
#### Iterative improvement
• Pairwise exchange, or Kernighan-Lin heuristics.
• k-opt heuristic:
Take a given tour and delete k mutually disjoint edges.
Reassemble the remaining fragments into a tour, leaving no disjoint subtours (that is, don't connect a fragment's endpoints together).
This in effect simplifies the TSP under consideration into a much simpler problem - for k = 3 (the 3-opt heuristic), each fragment endpoint can be connected to 3 other possibilities (5 total fragment endpoints available, 1 disallowed as it's the other end of the fragment under consideration, and another disallowed because it was previously connected - the endpoint in question is not counted as connecting a vertex to itself is rather pointless).
Such a constrained 6 city TSP can then be brute forced to find the least-cost recombination of the original fragments.
#### Randomized improvement
• Optimised Markov chain algorithms which utilise local searching heuristical sub-algorithms can find a route extremely close to the optimal route for 700-800 cities.
TSP is a touchstone for many general heuristics devised for combinatorial optimization: genetic algorithms, simulated annealing, Tabu search, neural nets, ant system.
### Special cases
#### Restricted locations
• A trivial special case is when all cities are located on the perimeter of a convex polygon.
• A good exercise in combinatorial algorithms is to solve the TSP for a set of cities located along two concentric circles.
#### TSP with triangle inequality
A very natural restriction is the triangle inequality. That is, for any 3 cities A, B and C, the distance between A and C must be at most the distance from A to B plus the distance from B to C. Most natural instances of TSP satisfy this constraint.
In this case, there is an algorithm (due to Christofides, 1975) which always finds a tour of length at most 1.5 times the shortest tour. In the next paragraphs, we explain a weaker (but simpler) algorithm which finds a tour of length at most twice the shortest tour.
The length of the minimum spanning tree of the network is a natural lower bound for the length of the optimal route. In the TSP with triangle inequality case it is possible to prove upper bounds in terms of the minimum spanning tree and design an algorithm that has a provable upper bound on the length of the route. The first published (and the simplest) example follows.
• Step 1: Construct the minimal spanning tree.
• Step 2: Duplicate all its edges. This gives us an Eulerian graph.
• Step 3: Find an Eulerian cycle in it. Clearly, its length is twice the length of the tree.
• Step 4: Convert the Eulerian cycle into the Hamiltonian one in the following way: walk along the Eulerian cycle, and each time you are about to come into an already visited vertex, skip it and try to go to the next one (along the Eulerian cycle).
It is easy to prove that the last step works. Moreover, thanks to the triangle inequality, each skipping at Step 4 is in fact a shortcut, i.e., the length of the cycle does not increase. Hence it gives us a TSP tour no more than twice as long as the optimal one.
Christofides algorithm follows a similar outline but combines the minimum spanning tree with a solution of another problem, minimum-weight perfect matching. This gives a TSP tour which is at most 1.5 times the optimal. It is a long-standing (since 1975) open problem to improve 1.5 to a smaller constant. It is known, however, that there is no polynomial time algorithm that finds a tour of length at most 1+1/219 times the optimal, unless P=NP (Papadimitriou and Vempala, 2000).
#### Euclidean TSP
Euclidean TSP, or planar TSP, is the TSP with the distance being the ordinary Euclidean distance. The problem still remains NP-hard, however many heuristics work better.
Euclidean TSP is a particular case of TSP with triangle inequality, since distances in plane obey triangle inequality. However, it seems to be easier than general TSP with triangle inequality. For any c>0, there is a polynomial time algorithm that finds a tour of length at most (1+c) times the optimal on any graph (Arora, 1997). In practice, the running time of this algorithm is too large and heuristics with weaker guarantees are used but they also perform better on instances of Euclidean TSP than on general instances.
#### Asymmetric TSP
In most cases, the distance between two nodes in the TSP network is the same in both directions - the special case where the distance from A to B is not equal to the distance from B to A is called Asymmetric TSP. An example of a practical application of an asymmetric TSP is route optimization using street-level routing (asymmetric due to one-way streets, slip-roads and motorways).
## References
• G. B. Dantzig, R. Fulkerson, and S. M. Johnson, Solution of a large-scale traveling salesman problem, Operations Research 2 (1954), 393-410.
• S. Arora. Polynomial Time Approximation Schemes for Euclidean Traveling Salesman and other Geometric Problems. Journal of ACM, 45(1998), 753-782.
• N. Christofides, Worst-case analysis of a new heuristic for the travelling salesman problem, Report 388, Graduate School of Industrial Administration, CMU, 1976.
• C. H. Papadimitriou, S. Vempala: On the approximability of the traveling salesman problem (extended abstract). Proceedings of STOC'2000, 126-133.
• D. S. Johnson and L. A. McGeoch, The Traveling Salesman Problem: A Case Study in Local Optimization, Local Search in Combinatorial Optimization, E. H. L. Aarts and J.K. Lenstra (ed), John Wiley and Sons Ltd, 1997, pp 215-310.
## Related articles
Results from FactBites:
Jano van Hemert's Homepage (1612 words) The technique is applied in three important domains of combinatorial optimisation, binary constraint satisfaction, Boolean satisfiability, and the travelling salesman problem. The technique is demonstrated on a number of well known domains of combinatorial optimization, including binary constraint satisfaction and the traveling salesman problem. Although the travelling salesman problem (TSP) is considered NP-hard, not every problem instance is necessarily hard to solve.
The Travelling Salesman Problem (389 words) The Travelling Salesman Problem (TSP) consists in the problem of determining the shortest circuit which can be made visiting a list of cities, in such a way that each city is visited (once and only once). If we choose among them the one that corresponds to the shortest traveled distance, we will obtain in a fast way a solution which, if it is not the best, it "close to" the best. The representation scheme used to display the circuits of the TSP is based in a permutation matrix: there is a line for each city and each column represents the position of the city in the circuit.
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# Mathematics / Year 8 / Measurement and Geometry / Geometric reasoning
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Establish properties of quadrilaterals using congruent triangles and angle properties, and solve related numerical problems using reasoning (ACMMG202)
Elaborations
• establishing the properties of squares, rectangles, parallelograms, rhombuses, trapeziums and kites
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https://coursegraph.com/coursera-specializations-algorithms
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## Algorithms Specialization
所在平台: Coursera专项课程 课程类别: 计算机科学 大学或机构: CourseraNew
#### 课程详情
Algorithms are the heart of computer science, and the subject has countless practical applications as well as intellectual depth. This specialization is an introduction to algorithms for learners with at least a little programming experience. The specialization is rigorous but emphasizes the big picture and conceptual understanding over low-level implementation and mathematical details. After completing this specialization, you will be well-positioned to ace your technical interviews and speak fluently about algorithms with other programmers and computer scientists. About the instructor: Tim Roughgarden has been a professor in the Computer Science Department at Stanford University since 2004. He has taught and published extensively on the subject of algorithms and their applications.
#### 课程大纲
Course: 1
Title:Divide and Conquer, Sorting and Searching, and Randomized Algorithms
Description:The primary topics in this part of the specialization are: asymptotic ("Big-oh") notation, sorting and searching, divide and conquer (master method, integer and matrix multiplication, closest pair), and randomized algorithms (QuickSort, contraction algorithm for min cuts).
Course: 2
Title:Graph Search, Shortest Paths, and Data Structures
Description:The primary topics in this part of the specialization are: data structures (heaps, balanced search trees, hash tables, bloom filters), graph primitives (applications of breadth-first and depth-first search, connectivity, shortest paths), and their applications (ranging from deduplication to social network analysis).
Course: 3
Title:Greedy Algorithms, Minimum Spanning Trees, and Dynamic Programming
Description:The primary topics in this part of the specialization are: greedy algorithms (scheduling, minimum spanning trees, clustering, Huffman codes) and dynamic programming (knapsack, sequence alignment, optimal search trees).
Course: 4
Title:Shortest Paths Revisited, NP-Complete Problems and What To Do About Them
Description:The primary topics in this part of the specialization are: shortest paths (Bellman-Ford, Floyd-Warshall, Johnson), NP-completeness and what it means for the algorithm designer, and strategies for coping with computationally intractable problems (analysis of heuristics, local search).
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https://www.tutorialspoint.com/program-to-find-maximum-profit-by-cutting-the-rod-of-different-length-in-cplusplus
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crawl-data/CC-MAIN-2023-14/segments/1679296948932.75/warc/CC-MAIN-20230329023546-20230329053546-00778.warc.gz
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Program to find maximum profit by cutting the rod of different length in C++
Suppose we have a rod is given of length n. We also have a list, that contains different size and price for each size. We have to find the maximum price by cutting the rod and selling them in the market. To get the best price by making a cut at different positions and comparing the prices after cutting the rod.
So, if the input is like prices = [1, 5, 8, 9, 10, 17, 17, 20], n = 8, then the output will be 22, as by cutting the rod in length 2 and 6. The profit is 5 + 17 = 22.
To solve this, we will follow these steps −
• Define an array profit of size: n+1.
• profit[0] := 0
• for initialize i := 1, when i <= n, update (increase i by 1), do −
• maxProfit := negative infinity
• for initialize j := 0, when j < i, update (increase j by 1), do −
• maxProfit := maximum of maxProfit and price[j] + profit[i − j − 1]
• profit[i] := maxProfit
• return maxProfit
Let us see the following implementation to get better understanding −
Example
Live Demo
#include <bits/stdc++.h>
using namespace std;
int max(int a, int b) {
return (a > b)? a : b;
}
int rodCutting(int price[], int n) {
int profit[n+1];
profit[0] = 0;
int maxProfit;
for (int i = 1; i<=n; i++) {
maxProfit = INT_MIN;
for (int j = 0; j < i; j++)
maxProfit = max(maxProfit, price[j] + profit[i-j-1]);
profit[i] = maxProfit;
}
return maxProfit;
}
int main() {
int priceList[] = {1, 5, 8, 9, 10, 17, 17, 20};
int rodLength = 8;
cout << rodCutting(priceList, rodLength);
}
Input
{1, 5, 8, 9, 10, 17, 17, 20}, 8
Output
22
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http://mathhelpforum.com/pre-calculus/98805-questions.html
| 1,527,311,129,000,000,000
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|
1. ## Questions
Wow, I can't believe I've forgotten how to do these, but I have. Any help would be appreciated.
1) Simplify 7root(x^9)/5root(x^6)
2) Solve the equation both algebraically and graphically (I've done it the graphical way, cannot seem to figure out how to do it algebraically)
3) Rewrite the expression logbase5(x+3) into an equivalent expression using only natural logarithms.
4) Describe the transformations that can be used to transform the graph of log(x) to a graph of f(x) = 4log(x+2)-3.
5) Solve the equation 2sin^2(x)cos(x)=cos(x) algebraically
6) Find all the exact solutions to 2sin^2(x)+3sin(x)-2=0 on the interval [0,2pi)
7) Use a graphing calculator to solve the following for x. e^(2x)=3x^2
I graphed it in my graphing calculator, but I am not sure if you're trying to find the intersection of the two graphs or what? I found the intersection and plugged it into x, but they are not equal, so I am not sure what to do.
8) State the smallest interval (
0 ≤theta ≤k) that gives the complete graph of the polar equation r=4cos(5theta)
By the way, I am not asking you to solve these outright, but just help me. These problems are from PreCalculus, but I just forget how to solve them. Thank you for your time!
2. ## A different approach
Hello zaga04
Welcome to Math Help Forum!
Originally Posted by zaga04
Wow, I can't believe I've forgotten how to do these, but I have. Any help would be appreciated.
1) Simplify 7root(x^9)/5root(x^6)
2) Solve the equation both algebraically and graphically (I've done it the graphical way, cannot seem to figure out how to do it algebraically)
3) Rewrite the expression logbase5(x+3) into an equivalent expression using only natural logarithms.
4) Describe the transformations that can be used to transform the graph of log(x) to a graph of f(x) = 4log(x+2)-3.
5) Solve the equation 2sin^2(x)cos(x)=cos(x) algebraically
6) Find all the exact solutions to 2sin^2(x)+3sin(x)-2=0 on the interval [0,2pi)
7) Use a graphing calculator to solve the following for x. e^(2x)=3x^2
I graphed it in my graphing calculator, but I am not sure if you're trying to find the intersection of the two graphs or what? I found the intersection and plugged it into x, but they are not equal, so I am not sure what to do.
8) State the smallest interval (
0 ≤theta ≤k) that gives the complete graph of the polar equation r=4cos(5theta)
By the way, I am not asking you to solve these outright, but just help me. These problems are from PreCalculus, but I just forget how to solve them. Thank you for your time!
May I suggest that you'll be more likely to get help if you (a) post one question at a time (see Rule 14); (b) show us that you've made some sort of attempt, so that we can help you at the specific point that you're having a problem with.
Having said that, here are some suggestions for the first couple of questions.
(1) I take it that you mean: Simplify $\displaystyle \frac{\sqrt[7]{x^9}}{\sqrt[5]{x^6}}$.
You need to use two rules of indices:
• $\displaystyle \sqrt[a]{x^b} = x^{\frac{b}{a}}$
• $\displaystyle \frac{x^a}{x^b}=x^{a-b}$
So $\displaystyle \sqrt[7]{x^9}=x^{\frac{9}{7}}$, and $\displaystyle \sqrt[5]{x^6}=...$ ?
Then you'll need to subtract the second power of $\displaystyle x$ from the first.
(2) $\displaystyle |4x-3| = 5 \sqrt{x+4}$
$\displaystyle \Rightarrow (4x-3)^2 = 25(x+4)$
$\displaystyle \Rightarrow 16x^2 - 24x +9 = 25x +100$
$\displaystyle \Rightarrow ...$ ?
Solve the quadratic equation. (I don't think it factorises, so you'll have to use the formula.)
3. Sorry about posting so many questions, thought it would be easier than posting so many threads. Thanks for helping me out for the first two problems, make sense. I'll come back with the work I've done for the others.
4. So I think I got #3, log5(x+3) converted to ln would = ln(x+3)/ln5
Would #4 be like the graph of 4log(x+2)-3 is vertically stretched by 4, moved horizontally 2 units to the left, and moved vertically down 3 units? I am not sure if this is what they are asking.
I am not sure how to start #5, #6 or #8 at all. #8 I graph it on my calculator, but I'm not sure what the smallest interval would be that would give the complete graph of the polar equation.
#7 I already explained what I did, I graphed it on my graphing calculator, but I am not sure how you find x from the graphing calculator? Do you find the intersection of the two graphs e^(2x) and 3x^2?
That's the thought process I've put into these problems.
5. Originally Posted by zaga04
So I think I got #3, log5(x+3) converted to ln would = ln(x+3)/ln5
Would #4 be like the graph of 4log(x+2)-3 is vertically stretched by 4, moved horizontally 2 units to the left, and moved vertically down 3 units? I am not sure if this is what they are asking.
I am not sure how to start #5, #6 or #8 at all. #8 I graph it on my calculator, but I'm not sure what the smallest interval would be that would give the complete graph of the polar equation.
#7 I already explained what I did, I graphed it on my graphing calculator, but I am not sure how you find x from the graphing calculator? Do you find the intersection of the two graphs e^(2x) and 3x^2?
That's the thought process I've put into these problems.
Q3: Yes, those two expressions are equal (change of base rule)
Q4: It would be assuming the original graph is log(x)
Q5: Take cos(x) from both sides
$\displaystyle 2sin^2(x)cos(x) - cos(x) = 2sin^2(x)(cos(x)-1) = 0$
and if ab=0 either a=0 or b=0. Normally this kind of question will have limits on solutions because there are an infinite number.
Q6: Let u = sin(x) and solve the quadratic
Q7: Yep that's what you do
$\displaystyle e^{2x} = 3x^2$ cannot be solved algebraically hence the calculator
Q8: It's asking how much space you need to view a complete wave of the equation. Remember cos(ax) is the graph of cos(x) squashed by a factor of a and that cos(x) is periodic over 2pi
6. Thank you so much
For #5, I understand the process, but what kind of answer would you get?
For #6, I got pi/6 and 5pi/6.
For #7, I got the intersection to be x=-.39.
For #8, I got the interval to be (0≤theta≤pi)
Are the ones I did correct?
7. Originally Posted by zaga04
Thank you so much
For #5, I understand the process, but what kind of answer would you get?
For #6, I got pi/6 and 5pi/6.
For #7, I got the intersection to be x=-.39.
For #8, I got the interval to be (0≤theta≤pi)
Are the ones I did correct?
5. $\displaystyle sin^2(x) = 0$ or $\displaystyle cos(x) = 1$
$\displaystyle x = k \pi \: ,\, k\, \in \, \mathbb{Z}$
In this case sin^2(x) is 0 when cos(x) is also 0.
6,7. Put the values you found back into the equation and see if it works out to be true
8. ## Polar equation sketch
Hello zaga04
Originally Posted by zaga04
...
8) State the smallest interval (
0 ≤theta ≤k) that gives the complete graph of the polar equation r=4cos(5theta)
To sketch the polar graph, note that the important values of cosine occur at intervals of $\displaystyle \pi/2$:
$\displaystyle \cos 0 = 1,\,\cos(\pi/2) = 0,\, \cos(\pi) = -1, \,\cos(3\pi/2)=0,\,\cos(2\pi) = 1, ...$
These will occur when $\displaystyle \theta = 0, \pi/10,\, 2\pi/10,\, 3\pi/10,...$
So sketch a diagram with radial lines every $\displaystyle \pi/10$; i.e. $\displaystyle 18^o$ from $\displaystyle 0^o$ to $\displaystyle 180^o$.
Mark the point $\displaystyle (4,0)$. The curve starts here and comes in to the origin with the line $\displaystyle \theta = \pi/10$ as a tangent. Then it continues through negative values of $\displaystyle r$ (i.e. out the other side into the third quadrant) going out to $\displaystyle (-4, \pi/5)$; comes back to the origin tangential to $\displaystyle \theta = 3\pi/10$; etc.
See the attached sketch.
You'll find that you need $\displaystyle 0 \le \theta \le \pi$ to get the complete curve.
Also I was trying to rework #5, would this be right?
$\displaystyle 2sin^2(x)cos(x)=cos(x)$
$\displaystyle 2sin^2(x)=1$
$\displaystyle sin^2(x)=\frac{1}{2}$
$\displaystyle sin(x)=\sqrt\frac{1}{2}$
$\displaystyle x=\arcsin\sqrt\frac{1}{2}$
$\displaystyle x=\frac{\pi}{4}$
10. ## The two great commandments
Hello zaga04
Originally Posted by zaga04
Also I was trying to rework #5, would this be right?
$\displaystyle 2sin^2(x)cos(x)=cos(x)$
$\displaystyle 2sin^2(x)=1$
$\displaystyle sin^2(x)=\frac{1}{2}$
$\displaystyle sin(x)=\sqrt\frac{1}{2}$
$\displaystyle x=\arcsin\sqrt\frac{1}{2}$
$\displaystyle x=\frac{\pi}{4}$
There are two great commandments of mathematics. The first is this:
• Thou shalt not divide by zero.
and the second is:
• When thou takest a square root, forget not thy plus or minus sign.
And I'm afraid you've broken them both!
You've got to be careful you don't divide both sides of an equation by something that might have the value zero. If you do, you might miss possible solutions.
In $\displaystyle 2\sin^2x\cos x = \cos x$, where there's a factor of $\displaystyle \cos x$ on both sides, note that $\displaystyle \cos x = 0$ is a possible solution - it makes both sides have an equal value, $\displaystyle 0$. So before you divide by $\displaystyle \cos x$, say that $\displaystyle \cos x =0$ or ... then do the division. So you get:
$\displaystyle 2\sin^2x\cos x = \cos x$
$\displaystyle \Rightarrow \cos x = 0$ or $\displaystyle 2\sin^2x = 1$
The first possibility gives us $\displaystyle x = \frac{\pi}{2}, \frac{3\pi}{2}, ...$
And the second: $\displaystyle \sin^2x = \frac12$
So don't forget the second commandment:
$\displaystyle \sin x = \pm\frac{1}{\sqrt2}$
So you not only get $\displaystyle x = \frac{\pi}{4},\frac{3\pi}{4}, \frac{9\pi}{4}, ...$ corresponding to $\displaystyle \sin x = +\frac{1}{\sqrt2}$, but also the odd multiples of $\displaystyle \frac{\pi}{4}$ in between:
$\displaystyle x=\frac{5\pi}{4}, \frac{7\pi}{4}, ...$
which come from $\displaystyle \sin x = -\frac{1}{\sqrt2}$
So don't forget to obey the commandments!
11. Thank you again Grandad and as you can tell I am not very good at math. I was just trying to do it by myself, but thank you for correcting me and telling me the right way.
So the dots at the end of each answer would correspond that you have to keep on adding multiples of $\displaystyle \frac{\pi}{4}$? Or as the answer on my sheet can I just leave it as ... at the end?
12. Hello zaga04
Originally Posted by zaga04
Thank you again Grandad and as you can tell I am not very good at math. I was just trying to do it by myself, but thank you for correcting me and telling me the right way.
So the dots at the end of each answer would correspond that you have to keep on adding multiples of $\displaystyle \frac{\pi}{4}$? Or as the answer on my sheet can I just leave it as ... at the end?
Yes, the three dots symbol, ..., is called an ellipsis (not to be confused with an ellipse, which is a squashed circle), and it indicates that things have been left out.
You can use it to indicate a finite sequence, like $\displaystyle 2, 4, 6, ..., 10$, or an infinite one, like $\displaystyle \frac{\pi}{4}, \frac{3\pi}{4}, \frac{5\pi}{4}, \frac{7\pi}{4}, ...$ which continues without ending.
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