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10 days diary

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Manager
Joined: 03 Jan 2008
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11 Apr 2008, 14:39
10days left...

So, its now the very hard time for me. Let's see what I have done in the last 3 weeks:
- Get to know GMAT
- Do SC10 twice, SC 11 twice
- Do 200/1000 SC, dont remember the rate I got right, just remind that it was so baddd
- Take GMAT Prep twice - 700 and 730
- Have never taken any AWA yet
-----

It seems that the work I should do is much much more than I have done. I want to finish 1000SC and RC. Want to practice alot AWA, want to... relax, want to go out of my room at least once a day ..,
Well I forgot that I am a woman these days. Just stay at home and practice, practice.. Dont remember where my mirror is..

----
My weaknesses at this time: SC is the biggest, then CR-sometimes I get mistakes. RC is Ok, because my strategy is to read through and understand the pssg. Quant is Ok.
Besides, my variety of words is soo poor. That will be the big advantanges in AWA. Sometimes I even dont understand the whole meaning of AWA questions.
---

Haizzzz. Today's target: 80 RC in 1000RC, 70 SC in OG10 -third time, and 1 AWA. Then will take 1 Prep ^_^
Now dont care the rate of right and wrong! Just concentrate on why this is wrong or why that is right!

---
P/S: listening to "To be with you" of Mr Big is so relaxxxx... Love it...
AND...
...... it is NOT illegal if you accidently read my diary
Senior Manager
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Schools: UCSD ($) , UCLA, USC ($), Stanford
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11 Apr 2008, 15:41
Good luck!

I'm taking it in 10 days as well and am so behind with everything, no time for a diary or even posting too much any more

......don't think I'll have time for any future re-takes, so it's now or never,,,,which of course puts that much more pressure and freaks me out....

If you can afford the time to re-take, have fun with this one, treat it like a competitive video-game for which you pay $250 to play.....and take full advantage of those 4 hours to, above all, HAVE FUN!! After all, you pay good money and spend a lot of time to be there for just those damn 4 hours. And don't forget to smile ...in the end it's just a game in which you can't lose any of what you already have...You can only WIN. So, feel good about it. Finally, like some say....life is too damn short to worry about things you don't have or can't do - enjoy what you DO have.... _________________ Chinese Democracy is misunderstood...at your nearest BestBuy. Best AWA guide here: http://gmatclub.com/forum/how-to-get-6-0-awa-my-guide-64327.html Intern Joined: 04 Apr 2008 Posts: 12 Followers: 0 Kudos [?]: 1 [0], given: 0 Re: 10 days diary [#permalink] Show Tags 11 Apr 2008, 16:39 Hey...I'm taking mines in 2 days YIKES!! Nice to see females here... TOTALLY in the same boat as you are....I am dying to go out with my friends and have a drink. Been home (besides work of course), hibernated myself, studying, practicing, studying, practicing....etc etc. This will be my first time taking it...I'm so nervous!! ahhh... Good Luck on yours!!! Senior Manager Joined: 07 Feb 2008 Posts: 314 Followers: 2 Kudos [?]: 141 [0], given: 1 Re: 10 days diary [#permalink] Show Tags 11 Apr 2008, 16:53 Good luck to you all!!!! I know you'll be fine. I'll be keepiung my fingers crossed. Don't forget to smile on the day Manager Joined: 03 Jan 2008 Posts: 109 Followers: 1 Kudos [?]: 63 [0], given: 0 Re: 10 days diary [#permalink] Show Tags 11 Apr 2008, 18:18 missbrooklyn wrote: Hey...I'm taking mines in 2 days YIKES!! Nice to see females here... TOTALLY in the same boat as you are....I am dying to go out with my friends and have a drink. Been home (besides work of course), hibernated myself, studying, practicing, studying, practicing....etc etc. This will be my first time taking it...I'm so nervous!! ahhh... Good Luck on yours!!! Tks you guys! And good luck to you guys too ^_^ I do not plan to have it again. Now or Nerver SVP Joined: 04 May 2006 Posts: 1923 Schools: CBS, Kellogg Followers: 23 Kudos [?]: 1066 [0], given: 1 Re: 10 days diary [#permalink] Show Tags 11 Apr 2008, 22:19 Good luck notahug! Where are you from? _________________ CEO Joined: 17 May 2007 Posts: 2986 Followers: 60 Kudos [?]: 591 [0], given: 210 Re: 10 days diary [#permalink] Show Tags 12 Apr 2008, 01:49 Good luck ya'll , you've worked hard, its visible in your posts on the forum. I know you will do well. Keep your chin up and tame that cat. CEO Joined: 17 Nov 2007 Posts: 3588 Concentration: Entrepreneurship, Other Schools: Chicago (Booth) - Class of 2011 GMAT 1: 750 Q50 V40 Followers: 565 Kudos [?]: 3791 [0], given: 360 Re: 10 days diary [#permalink] Show Tags 12 Apr 2008, 02:31 You really have worked hard... GOOD LUCK!!! and be calm on the exam day. notahug wrote: Well I forgot that I am a woman these days.......... I wish you to recollect this as soon as possible. _________________ HOT! GMAT TOOLKIT 2 (iOS) / GMAT TOOLKIT (Android) - The OFFICIAL GMAT CLUB PREP APP, a must-have app especially if you aim at 700+ | PrepGame Director Joined: 05 Jan 2008 Posts: 700 Followers: 5 Kudos [?]: 444 [0], given: 0 Re: 10 days diary [#permalink] Show Tags 12 Apr 2008, 06:17 All the best stay cool Notahug All our wishes with you _________________ Persistence+Patience+Persistence+Patience=G...O...A...L Manager Joined: 03 Jan 2008 Posts: 109 Followers: 1 Kudos [?]: 63 [0], given: 0 Re: 10 days diary [#permalink] Show Tags 12 Apr 2008, 18:04 9 days left.. Today is so bad. I almost study nothing. I can not concentrate. He lies.. YEs he does.. Cant not cry.. Yesterday just finished the Gmatprep 2- 2nd time, 740, 50Q and 41V. Though, Not feeling comfortable. Because I am afraid of the test. It will decide my life afterward..But I did not prepare much.. And now, can not study.. Staying here alone is so horrible. And I will have to stay alone, rite, NoTADY? I will forget, forget, forget all...Are you a lier? Are you really bad? Dont let me lose my trust, pls dont.. ........... Will study, from now on........ CAN NOT BELIEVE IN ANYTHING.. P/S: sorry if this content is irrelevent to the forum. Sooo sory.. Manager Joined: 10 Jul 2006 Posts: 56 Followers: 1 Kudos [?]: 0 [0], given: 0 Re: 10 days diary [#permalink] Show Tags 12 Apr 2008, 19:09 Don't lose focus in the last few days. Don't worry all your hard work will take you through. Senior Manager Joined: 24 Feb 2008 Posts: 348 Schools: UCSD ($) , UCLA, USC (\$), Stanford
Followers: 189

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12 Apr 2008, 20:54
No one ever told me when I was alone
they just thought I'd know better, better.

No one ever told me when I was alone
they just thought I'd know better, better.

The hardest part this troubled heart has never
yet been through now
Is to heal the scars that got their start
inside someone like you now.

Well had I known or I'd been shown back
How long it'd take me.
To break the charms that brought me harm
and all that won't erase me.

I never would, not that I could no
matter what you'd pay me
Replay the part, you stole my heart
I should have known you're crazy.

If all I knew was that with you
I'd want someone to save me.
It'd be enough, but just my luck,
I fell in love and baby all that I wanted was

Now I know you better
You know I know better
Now I know you better

So bittersweet, this tragedy won't ask for absolution
This melody inside of me still searches for solution
A twist of fate, a change of heart
Kills my infatuation,
A broken heart, provides the spark for my determination,

No one ever told me when I was alone
They just thought I'd know better, better
Ooh, no one ever told me when I was alone
They just thought I'd know better, better

All that I wanted was
I know you know you know better
You know I know you know better
Now I know you better

I never wanted you to be so full of anger,
I never wanted you to be somebody else
I never wanted you to be someone afraid to know themselves
I only wanted you to see things for yourself

All that I wanted was
Now I know you better
Now we all know better
All that I wanted was

If I were you
I'd manage to abhor the invitation
Of Promised love that can't keep up
To choose your steps and we won't regret
This kind of aggravation

No one ever told me when I was alone
They just thought I'd know better, better
No one ever told me when I was alone
They just thought I'd know better, better
_________________

Best AWA guide here: http://gmatclub.com/forum/how-to-get-6-0-awa-my-guide-64327.html

Manager
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13 Apr 2008, 13:07
Hihi tks :"> I got him out of my mind right now. Everything is fine..
Senior Manager
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Location: India
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14 Apr 2008, 02:29
Hey All the Best!!!! I'm sure you will hit bull's eye
Re: 10 days diary   [#permalink] 14 Apr 2008, 02:29
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The access times of the main memory and the Cache memory, in a computer system, are 500 n sec and 50 n sec, respectively. It is estimated that 80% of the main memory request are for read the rest for write. The hit ratio for the read access only is 0.9 and a write-through policy (where both main and cache memories are updated simultaneously) is used. Determine the average time of the main memory.
retagged | 2.2k views

Average memory access time =  Time spend for read + Time spend for write

= Read time when cache hit + Read time when cache miss
+Write time when cache hit + Write time when cache miss

= 0.8 ⨯ 0.9 ⨯ 50 + 0.8 ⨯ 0.1 ⨯ (500+50) (assuming hierarchical read from memory and cache as only simultaneous write is mentioned in question)
+ 0.2 ⨯ 0.9 ⨯ 500 + 0.2 ⨯ 0.1 ⨯ 500 (simultaneous write mentioned in question)

= 36 + 44 + 90 + 10 = 180 ns

http://www.howardhuang.us/teaching/cs232/24-Cache-writes-and-examples.pdf
selected
what will be the ans if write-through policy is not mentioned in the question?
Why only 500 ns is consider during write are we not considering cache access during write?
We consider simultaneous access in write operation of write through cache ONLY
Arjun sir what would be the answer in write back??
Simultaneous write will happen for sure in write through,but why are we considering simultaneous memory organization for write?
@Arjun Sir,   "The hit ratio for the read access only is 0.9"    For write operation, the hit ratio is not given, how 0.9 is taken for write too ?

Why are you taking "hit ratio for read access" in cache write? You should've given the reason for doing so. The answer written by sameer2009 looks more clear and understandable.

Practically is it possible to have a cache design that supports hierarchical reads and simultaneous writes?
@ Sushmita In case of write-back policy, we have to consider the time for writing back when a new data word comes and to accommodate it we need to replace one existing data word in the event of miss.
That is,

$T_{avg} = 0.8*(0.9*50+0.1*(500+50+X))\\+0.2*(write_h*50+ write_m*(50+500+X))$

$X=time \ for\ writing\ back\ into\ memory\ since\\ memory\ doesn't\ contain\ updated\ data$
I am totally agree with @Akash Kanase.We should solve the problem according to information given in the question.
They are asking for the average memory access time.

If nothing is mentioned whether it is a simultaneous memory access or hierarchical memory access, we need to consider it as hierarchical memory access only. Here in this question nothing is given about read operation so we need to take it as hierarchical memory access for read operation. But for write it is given as write through policy, which will result in simultaneous memory access. Also for write operations write-through cache always go to the main memory, in main memory its hit rate is 100%. Hence Hit rate for write= 1.

Since there are 80% read operations and 20% write operations,

Average memory access time =  0.8 * Time spent for read + 0.2 * Time spent for write

Time spent for read = Hit-rate-for-read * cache access time + Miss-rate-for-read(cache access time + main memory access time)

= ( 0.9 ⨯ 50 + 0.1 ⨯ (500+50) )

= 45+55 = 100

Time spend for write = 500ns (simultaneous write mentioned in question)

Average memory access time =  0.8 * 100 + 0.2 * 500 = 80+100 = 180ns
answer is 200ns=50+ 0.2x500x0.9(write through that are in cache)+ 0.8x0.1x500(read misses)+0.2x(500+500)x0.1(write through that are not in cache)
why 500+500 on write cache miss?
thanks for the link! But then it should be 50+500+50 as first it checks in cache, then loads(and write also) the desired block from main memory to cache. So in total, twice you are accessing the cache.
then that 50 should also be there for cache miss during read rt?

So, 2 times cache is accessed even during read.
frankly speaking, it should be but I think for sake of brevity(or due to its miniscule nature) it is omitted
Yes. I guess another reason is there is no need for the cache update to be done for the CPU to get the data. From the main memory, data can be simultaneously passed to the CPU and updated in the cache. There is no need to wait for the cache update to be finished.

@Arjun

From the main memory, data can be simultaneously passed to the CPU and updated in the cache.

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## Clifford Algebra Paper

Ideas about how a world with more than three spatial dimensions would work - what laws of physics would be needed, how things would be built, how people would do things and so on.

### Clifford Algebra Paper

Platonic solids generate their four-dimensional analogues
PIERRE-PHILIPPE DECHANT a,b,c*

This is above me. I'm hoping someone can explain it, especially how spinors are the link from 3D to 4D Platonic solids.

https://arxiv.org/abs/1307.6768
PatrickPowers
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### Re: Clifford Algebra Paper

Wendy is probably the right person to ask for this one.

I'm not at all facile with spinors, but in my incomplete biased understanding, the connection here is essentially related to the fact that 3D rotational vectors have a 4-dimensional representation, e.g., via the quaternion representation, or in this case, the spinor representation. Since 3D Platonic solids have rotational symmetries, we can take those symmetries and rewrite them in a 4D representation, then re-interpret the 4D representation as points in 4D Euclidean space. Apparently, the paper is saying that we can derive the 4D regular polytopes from the 3D Platonic solids this way (though I only read the abstract so I'm not clear how exactly this connection is made).

There may be some connection with the Hopf fibration as well (perhaps that's where spinors come in -- since the Hopf fibration can be understood as a particular mapping of a 3D spherical polytope to 2D complex numbers, the latter of which can be interpreted as 4 scalar coordinates of a 4D vector).
quickfur
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### Re: Clifford Algebra Paper

The link lies in complex numbers, and algebraic geometry.

The equation of a line is y=ax+b. In complex euclidean geometry, all the dimensions are doubled, so the 2d space becomes 4d, and the 1d space becomes 2. It still means that you can draw only one 'line' through two given points. If we now consider the lines through 0,0 we get y=ax. Consider now the value w = cis(\omega t), ie a rotation. We can put wy = awx, so this shows that one can have in 4 dimensions, a rotation involving every point tracking in a simple circle around the origin. A clifford rotation.

The value a can take complex values as well, and the slope of a line at right angles has a slope -1/a. We can achieve this result with a sphere, of diameter (0,0,0) to (0,0,1), where the plane is the gradients (r, i, 0) [real, imaginary]. The plane is projected onto this sphere, by the intersection of the line drawn from (r, i, 0) to (0, 0, 1), gives a conformal mapping of the plane, and also preserves the feature where lines at right angles to each other are diametrically opposite.

This creates one swirlybob. Representing 4d space as CE2, has the same effect as 2d space as CE1, in that there is only one real line.

The idea here is that having a sphere, one can project the 62 icosahedral points, or 26 octahedral points, or the 20 tetrahedral points onto this sphere, and they represent clifford-rotations that (of varying orders), leave [3,3,5] or [3,4,3] or [3,3,4] invariant. There is a fourth one not discussed in the paper, formed by [[3,4,3]].

In a clifford rotation, not only do all of the points rotate, but for any given direction (arrow = line with direction), it is immediately passed to every point on the sphere so that we end up with every point has a path and direction. Hopf fibulation. But this solution is not unique, If we suppose the point A,A represents a great arrow (a single arrow passing through a point), then A, X represents all paths that belong to the same swirlybob A. A, -A represents the point diametrically opposite A,A here. Reversing the arrow also changes both signs, so A,X and -A,-X follow the same track in different directions.

The paths of the lines in a clifford-rotation is such that if the lines near A,A "rotate" around this line, in that if one were to divide the 4-space into planes 3d spaces that contain a,a, and radiate outwards like a paddle-steamer wheel, the circle of lines around A,A will cross these paddles in succession, spending half above the 4space and half below. If in the direction of travel, this is a clockwise rotation, then it's a clockwise rotation for every paddle-wheel and circle.

There is a whole corresponding set of clifford parallesl that are anticlockwise. In essence, if A,A is perpendicular to A,-A and -A,A. If the clockwise rotation represents the first symbol, the anticlockwise rotation represents the second. So the space of great arrows is in 6D, a prism-product of two spheres. These are still 4D rotations, but it takes 6 (as in 1,4,6,4,1) dimensions to represent all four rotations.

By varying the intensity around the two axies, one can make all points of six dimensions represent a valid rotation in 4d. Most of these would be Lassajur rotations, ie twice around one axis as once around the other.

Into these bisphere prism we can project the icosahedron, dodecahedron and ID, but if we insert an I * I, D * D and ID * ID we get a series of simple rotations around the {3,3,5} of orders (5, 3, 2 ), For the individual I * pt, D * pt, ID * pt, we have 62 rotations of orders 10, 6, 4.

So we have the 12 great circles of the dodecahedron, representing a set of points A,A, and we can create from this, 5 paths adjacent, 5 paths distant, and one opposite, like the vertices of an icosahedron. We can do it in two different ways.

The regular polytopes {3,3,4}, {3,4,3} and {3,3,5} are formed in this way. The simplex {3,3,3} can not, because opposite a rotation of order 3 is one of order 2. The {4,3,3}, {3,4,3} and {5,3,3} have paths of faces that form Poincare cube, octahedron and dodecahedron, in that if you leave one face you come back on the opposite a half-edge or edge later.

The remaining solid with this property is the octagonny, having 48 faces, each a truncated cube. As with the cube it's an edge, rather than a half-edge, but there is a path of eight faces one covers before returning to the start.
The dream you dream alone is only a dream
the dream we dream together is reality.

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### Re: Clifford Algebra Paper

Consider the rotational symmetries of a cube.

Each of the three axes e₁,e₂,e₃ (passing through the centre of a face) has rotations by 90°,180°,270°. A rotation by 90° around the vector e₃, or in the plane e₁e₂, is represented by the rotor

R = cos 45° + e₁e₂ sin 45° = 1/√2 + e₁e₂/√2.

Remember that this acts on vectors by R⁻¹vR, so its opposite -R produces the same rotation. These 9 different rotations thus correspond to 18 different rotors:

± 1/√2 ± e₁e₂/√2,
± 1/√2 ± e₁e₃/√2,
± 1/√2 ± e₂e₃/√2,
± e₁e₂,
± e₁e₃,
± e₂e₃.

(The last few are for 180° rotations.)

Each of the six axes e₁±e₂, e₁±e₃, e₂±e₃ (passing through the centre of an edge) also has a rotation by 180°. The vector e₁+e₂ should be normalized (divided by its magnitude) and dualized (multiplied by the unit trivector) to get a unit bivector:

B = e₁e₂e₃(e₁ + e₂)/√2 = e₂e₃/√2 - e₁e₃/√2,

and then the rotor is

R = cos 90° + B sin 90° = B.

These 6 rotations correspond to 12 rotors:

± e₁e₂/√2 ± e₁e₃/√2,
± e₁e₂/√2 ± e₂e₃/√2,
± e₁e₃/√2 ± e₂e₃/√2.

Finally, each of the four axes e₁±e₂±e₃ (passing through a vertex) has rotations by 120°,240°. The unit bivector dual to e₁+e₂+e₃ is

B = e₁e₂e₃(e₁+e₂+e₃)/√3 = e₂e₃/√3 - e₁e₃/√3 + e₁e₂/√3,

and the rotor is

R = cos 60° + B sin 60° = 1/2 + B√3/2 = 1/2 + e₂e₃/2 - e₁e₃/2 + e₁e₂/2.

These 8 rotations correspond to 16 rotors:

± 1/2 ± e₁e₂/2 ± e₁e₃/2 ± e₂e₃/2.

And we can't forget the identity! That's just R=±1. In total, we have 18+12+16+2=48 rotors. Using the mapping from 3D even multivectors to 4D vectors

1≅(1,0,0,0),
e₂e₃≅(0,1,0,0),
e₃e₁≅(0,0,1,0),
e₁e₂≅(0,0,0,1),

we get the coordinates of the vertices of a 24-cell and its dual:

(±1/√2, ±1/√2, 0, 0),
(±1/√2, 0, ±1/√2, 0),
(±1/√2, 0, 0, ±1/√2),
(0, ±1/√2, ±1/√2, 0),
(0, ±1/√2, 0, ±1/√2),
(0, 0, ±1/√2, ±1/√2),
(±1, 0, 0, 0),
(0, ±1, 0, 0),
(0 ,0, ±1, 0),
(0, 0, 0, ±1),
(±1/2, ±1/2, ±1/2, ±1/2).

(Or, interpreting these not as vertices but as normal vectors for cells, we get the octagonny.)

Similarly, the tetrahedron's symmetries produce the coordinates of a single 24-cell.
Last edited by mr_e_man on Wed Dec 02, 2020 1:32 am, edited 2 times in total.
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### Re: Clifford Algebra Paper

Gosh, this is so helpful! Now I feel I will understand it in a reasonable amount of time.
PatrickPowers
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### Re: Clifford Algebra Paper

Yes, thanks to you guys I get what is going on. I never would have figured out the original paper. I didn't even know the 24-cell existed.

I was (as usual) wondering how to depict the 24-cell in 3D. The octet truss https://www.google.com/search?q=octet+truss&source=lnms&tbm=isch comes reasonably close. It was invented by Alexander Graham Bell in 1903 for the construction of airplane frames, but didn't catch on. It's mind-boggling to look at. Even closer to 4D is the plate truss, constructed in 2018 via 3D printing. Like a 4D geometric figure, the edges are 2D.
https://ethz.ch/en/news-and-events/eth-news/news/2018/12/stiffest-porous-lightweight-material.html. This has real potential, though I can't see how it is porous. I guess there are non-porous structures that are even better.

Can the 120-cell be constructed in a similar manner? I suppose I could figure it out, and maybe I will, but it's easier to ask.

It seems apparent that this process can be continued indefinitely. The result isn't a Platonic solid, but what of that?

Wendy's response got me interested in ax+b in complex spaces where a and b are complex. Who knows, that might lead somewhere.
PatrickPowers
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### Re: Clifford Algebra Paper

There are only a few types of symmetry a polyhedron can have: 7 infinite families corresponding to the 2D frieze groups, and 7 special groups related to the Platonic solids. Here are representative figures with the frieze groups as symmetries; they're supposed to extend infinitely along one dimension:

Code: Select all
1:⌊ - ⌊ - ⌊ - ⌊ - ⌊ -2:⌊ - ⌊ - ⌊ - ⌊ - ⌊ -⌈ - ⌈ - ⌈ - ⌈ - ⌈ -3:⌊ - ⌊ - ⌊ - ⌊ - ⌊ -- ⌈ - ⌈ - ⌈ - ⌈ - ⌈4:⌊ - ⌊ - ⌊ - ⌊ - ⌊ -⌉ - ⌉ - ⌉ - ⌉ - ⌉ -5:⌊⌋ -- ⌊⌋ -- ⌊⌋ -- ⌊⌋ -- ⌊⌋ --6:⌊⌋ -- ⌊⌋ -- ⌊⌋ -- ⌊⌋ -- ⌊⌋ --⌈⌉ -- ⌈⌉ -- ⌈⌉ -- ⌈⌉ -- ⌈⌉ --7:⌊⌋ -- ⌊⌋ -- ⌊⌋ -- ⌊⌋ -- ⌊⌋ ---- ⌈⌉ -- ⌈⌉ -- ⌈⌉ -- ⌈⌉ -- ⌈⌉

Now imagine wrapping each figure around a cylinder in 3D (like a label on a can), so that ⌊ appears n times in a circle. Then figure 5 has the symmetry of an n-gon pyramid, figure 6 has the symmetry of an n-gon prism, and figure 7 has the symmetry of an n-gon antiprism. Figures 2 and 3 are like the treads on tyres. The symmetry group of figure 1 contains only n rotations. Figures 2 and 3, n rotations and n rotoreflections. Figure 4, n rotations around the main axis and n 180° rotations around perpendicular axes. Figure 5, n rotations and n reflections. Figures 6 and 7, n rotations around the main axis, n 180° rotations around perpendicular axes, n reflections, and n rotoreflections.

(A rotoreflection is a reflection across a plane, followed by a rotation around the axis perpendicular to the plane.)

The 7 special groups are:
Tetrahedral; 8 120° rotations, 3 180° rotations, 1 identity, 6 reflections, 6 90° rotoreflections; total 24
Chiral tetrahedral; 8 120° rotations, 3 180° rotations, 1 identity; total 12
Cubic; 6 90° rotations, 8 120° rotations, 6+3 180° rotations, 1 identity, 6+3 reflections, 6 90° rotoreflections, 8 60° rotoreflections, 1 inversion; total 48
Chiral cubic (the symmetry of the snub cube); 6 90° rotations, 8 120° rotations, 6+3 180° rotations, 1 identity; total 24
Pyritohedral (something between chiral tetrahedral and cubic); 8 120° rotations, 3 180° rotations, 1 identity, 3 reflections, 8 60° rotoreflections, 1 inversion; total 24
Dodecahedral; 12 72° rotations, 12 144° rotations, 20 120° rotations, 15 180° rotations, 1 identity, 15 reflections, 20 60° rotoreflections, 12 36° rotoreflections, 12 108° rotoreflections, 1 inversion; total 120
Chiral dodecahedral; 12 72° rotations, 12 144° rotations, 20 120° rotations, 15 180° rotations, 1 identity; total 60

The inversion can be considered a 180° rotoreflection, but the axis of rotation isn't unique. It just negates the coordinates of everything. Similarly the identity can be considered a 0° rotation, and a reflection can be considered a 0° rotoreflection.

Now quaternions can only represent rotations, not reflections. So, in the above, only the symmetries of figures 1 and 4, and the three chiral special groups, can be made into quaternions. We've already considered the cubic and tetrahedral groups. I believe the dodecahedral rotations generate the cells of a 120-cell, or the vertices of a 600-cell. Figure 1 is rather degenerate; it generates 2n quaternions of the form cos(kπ/n)+isin(kπ/n)+0j+0k, which are the vertices of a polygon, not a polychoron. Figure 4 generates the cells of a 2n,2n-duoprism. This includes the 8-cell (or tesseract, or 4-cube, or hypercube, or 4,4-duoprism) in the case n=2, corresponding to the rotational symmetries of a box with three different edge lengths.

But figure 4 looks nothing like a box!

Well, if you introduce some reflections (and no new rotations) to figure 4, you get figure 6, representing a prism of a regular n-gon, or of an irregular but isogonal 2n-gon. An isogonal 4-gon is just a rectangle. Hence, figure 6 in the case n=2 is a box.
Last edited by mr_e_man on Thu Jan 07, 2021 3:39 am, edited 1 time in total.
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mr_e_man
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### Re: Clifford Algebra Paper

Geometric algebra can handle reflections as well: If the plane has unit normal vector n, then the reflection of a vector v across the plane is -nvn. General rotoreflections are represented by odd multivectors, combinations of e₁, e₂, e₃, e₁e₂e₃. These are orthogonal to the "quaternions" 1, e₁e₂, e₁e₃, e₂e₃; the full space of multivectors is 8-dimensional (even though they're supposed to be objects in 3D). So we could make the other 3D symmetry groups into 8D polytopes, but the results aren't very interesting; they're just products of 4D polytopes.

Generalizing, we can always take an n-dimensional polytope's symmetry group, represent the rotations by even multivectors, and make a polytope in 2n-1 dimensions. (But the polytope may be contained in a lower-dimensional subspace, if the symmetry group is too small.)

I suppose rotation matrices would also work, to make a polytope in n2 dimensions.
ΓΔΘΛΞΠΣΦΨΩ αβγδεζηθϑικλμνξοπρϱσςτυϕφχψωϖ °±∓½⅓⅔¼¾×÷†‡• ⁰¹²³⁴⁵⁶⁷⁸⁹⁺⁻⁼⁽⁾₀₁₂₃₄₅₆₇₈₉₊₋₌₍₎
ℕℤℚℝℂ∂¬∀∃∅∆∇∈∉∋∌∏∑ ∗∘∙√∛∜∝∞∧∨∩∪∫≅≈≟≠≡≤≥⊂⊃⊆⊇ ⊕⊖⊗⊘⊙⌈⌉⌊⌋⌜⌝⌞⌟〈〉⟨⟩
mr_e_man
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### Re: Clifford Algebra Paper

I have enough group theory that this makes sense. Very interesting!
PatrickPowers
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The increase in dry mass seems to be counterbalanced by an increase in volume so that the density stays the same. As was written, almost all of the mass of an atom is located in the nucleus, with a very small contribution from the electron cloud.The mass of the nucleus is associated with the atomic mass number, which is the total number of protons and neutrons in the nucleus of an atom.The mass number is different for each different isotope of a chemical element. what is dead time. For large values of Z, the number of neutrons is greater than the number of protons and the data points fall above the red line. 3.1k SHARES. A helium nucleus is also called an alpha (α) particle. Nuclear density is the density of the nucleus of an atom. However, it is consistent with earlier comments about the nucleus containing nearly all of the mass of the atom in a tiny region of space. Nuclear density is the density of the nucleus of an atom. It is the ratio of mass per unit volume inside the nucleus. Note the presence of circula$$r$$ regions, o$$r$$ nodes, where the probability density is zero. However, it is consistent with earlier comments about the nucleus containing nearly all of the mass of the atom in a tiny region of space. Thus, the density of nuclear material is more than 2.10 14 times greater than that of water. The nucleus of the hydrogen atom is a single proton, which has a radius of about 1.0 x 10-15 m. The single electron in a hydrogen atom normally orbits the nucleus at a … Is the value of the function 0 at r = 0? Nuclear density is the density of the nucleus of an atom. (2.3) In this case, we need to clarify that we are aproximating this calculation near to the nucleus of the atom and therefore we need to evaluate this last expression when R = 0. For finding density we need mass and volume of nucleus. Problem 8P from Chapter 12: What is the ratio of density of the nucleus to that of water? Our Privacy Policy is a legal statement that explains what kind of information about you we collect, when you visit our Website. Nuclear density is the density of the nucleus of an atom, averaging about 2.3×10 17 kg/m 3. The density (mass/volume) of a nucleus is extremely high: about 2.3 × 10 17 kg/m 3 (according to ). A graph of the number N of neutrons versus the number Z of protons for a range of stable nuclei (nuclides) is shown in (Figure). 1) You may use almost everything for non-commercial and educational use. Ans; (4)10 12 kg/cc. This shows that the density of nucleus is very high and is independent with atomic number. The Iron Nucleus Find the radius (r) and approximate density of a Fe-56 nucleus. Rutherford’s gold foil experiment showed us that the atom is mostly empty space with a comparatively tiny, massive, positively charged nucleus in the centre. Density of the nucleus = (\text { mass of nucleus})/(\text { volume of nucleus}) Mass of the nucleus = A amu = A ×1.66 × 10-27 kg. For the unstable nuclei, the number represents the half-life. From the attached images, it is clear that the function is at zero for p and d orbitals, but TENDS to infinity as approaching r = 0 for the s-orbitals. Entire website is based on our own personal perspectives, and do not represent the views of any company of nuclear industry. Isotopes of a given atom share the same chemical properties, since these properties are determined by interactions between the outer electrons of the atom, and not the nucleons. The density found here may seem incredible. ← Prev Question Next Question → +1 vote . The density found here may seem incredible. For example, water that contains deuterium rather than hydrogen (“heavy water”) looks and tastes like normal water. Maths. What is Nuclear Radius - Radius of Nucleus - Definition, What is Discovery of Atomic Nucleus - Definition, What is Volume of an Atom and Nucleus - Definition, What is Ionization Energy of Atoms – Definition, What is Nucleon – Structure of Atomic Nucleus – Definition. The Quantum Tunneling of Particles through Potential Barriers, 58. A nuclide is a stable nucleus. This shows that the density of nucleus is very high and is independent with atomic number. Here's the mathematical proof that the density of a nucleus is constant. Additionally, how does the density of the nucleus compare to the density of the rest of the atom? Physics of Nuclear Kinetics. If an atom were the size of a major league baseball stadium, the nucleus would be roughly the size of the baseball. Partial chart of the nuclides. The rate of sedimentation, as occurs in a centrifuge, is quantified in units of Svedberg which is the origin of … Calculate the energy of a particle that has a wavelength of , small enough to detect details about one-tenth the size of a nucleon. Solution for Part 1. … Nucleus has no definite boundary, but its radius is given by following relation: R = R o A 1/3. Scattering experiments support this general relationship for a wide range of nuclei, and they imply that neutrons have approximately the same radius as protons. The usual definition of nuclear density gives for its density: ρ nucleus = m / V = 238 x 1.66 x 10-27 / (1.73 x 10-42) = 2.3 x 10 17 kg/m 3. DOE Fundamentals Handbook, Volume 1 and 2. Mass number is given as 55.85u (this is mass number of iron nucleus) here "u" is atomic mass unit. (a) What is the approximate density of a nucleus? Thus, the density of nuclear material is more than 2.1014 times greater than that of water. Dentate nucleus neuronal density did not differ between essential tremor cases and controls (P = 0.44). Volume of the nucleus =4/3πR^3 = 4/3π(R_0A^(1/3))^3 A =4/3πR_0^3 A Density of nucleus is independent of mass number. Assume the mass of the Fe-56 nucleus is approximately 56 u. Gm tube, what is dead time. Click hereto get an answer to your question  (b) Show that density of nucleus is independent of its mass number A. (a) Show that the density of a nucleus is independent of A. A density of 300 kg m −3 would require a high mean nucleus porosity, >70%, that could originate by inefficient packing of elongate or irregularly shaped particles and/or macroscopic voids. The values are 6.70 × 10^14"kg/dm"^3 and 2.7"kg/dm"^3. Intutively, as the distance from the nucleus is the least, there should be maximum probability density. The nucleus density was correlated to the optical property, in fact to the haze index of injection-molded plaques in order to qualify the calculated values of N. It was found that N increases more orders of magnitude in nucleated samples and correlates strongly to the optical properties. Define velocity, VELOCITY: The time rate of change of displacement of a... VELOCITY: The time rate of change of displacement of a body is known as velocity. (Here, we are neglecting mass of the orbital electrons) Mass of the nucleus of the atom of the mass number A. Mass of Nucleus – Atomic Mass Unit. A nucleon refers to either a proton or neutron. Conclusions. The nucleus, has atomic mass 106.905095 g/mol with an abundance of ; whereas has atomic mass 108.904754 g/mol with an abundance of . Nuclear density is the density of the nucleus of an atom, averaging about 2.3×10 17 kg/m 3. Nuclear stability occurs when the attractive forces between nucleons compensate for the repulsive, long-range electrostatic forces between all protons in the nucleus. Assume the mass of the Fe-56 nucleus is approximately 56 u. According to scattering experiments, the nucleus is spherical or ellipsoidal in shape, and about 1/100,000th the size of a hydrogen atom. Orbital Magnetic Dipole Moment of the Electron, 60. Because the nuclear force is the strongest of the forces, we might expect that this strong force would cause the protons and neutrons to congregate at the center of the nucleus, giving an increased density in the central region. Show that the density of nucleus over a wide range of nuclei is constant independent of mass number. Find the length of a side of a cube having a mass of 1.0 kg and the density of nuclear matter. As was written, almost all of the mass of an atom is located in the nucleus, with a very small contribution from the electron cloud.The mass of the nucleus is associated with the atomic mass number, which is the total number of protons and neutrons in the nucleus of an atom.The mass number is different for each different isotope of a chemical element. Because of the existence of stable isotopes, we must take special care when quoting the mass of an element. (An atomic mass unit is defined as 1/12th the mass of a nucleus.) A particle has a mass equal to 10 u. EDP Sciences, 2008. Reason : Radius of nucleus is directly proportional to the cube root of mass number. An s orbital is spherically symmetrical around the nucleus. The mass (M) and the radius (r) of a nucleus can be expressed in terms of the mass number, A. Solving for r, we have. The arrows represent attractive forces exerted by a neutron (in red) on its nearest neighbors. J. R. Lamarsh, Introduction to Nuclear Reactor Theory, 2nd ed., Addison-Wesley, Reading, MA (1983). Therefore, the result shown for a mid-sized nucleus is reasonable. The nucleus of an atom has a radius about 10^-5 that of the atom and contains nearly all the mass of the entire atom. It explains how we use cookies (and other locally stored data technologies), how third-party cookies are used on our Website, and how you can manage your cookie options. K. O. Ott, W. A. Bezella, Introductory Nuclear Reactor Statics, American Nuclear Society, Revised edition (1989), 1989, ISBN: 0-894-48033-2. For example, represents the carbon nucleus with six protons and six neutrons (or 12 nucleons). 2) You may not distribute or commercially exploit the content, especially on another website. Since atomic nucleus carries most of atom’s mass and atomic nucleus is very small in comparison to entire atom, the nuclear density is very high. Addison-Wesley Pub. (b) One remnant of a supernova, called a neutron star, can have the density of a nucleus. Where R o is constant for all nuclei and its value is 1.2 x 10-15 m Herein, we report a density-responsive … The constancy of nuclear density implies an A⅓ law for the mean nuclear radius, the proportionality constant being 1.07 x 1O–13 centimeter. The detail that you can observe using a probe is limited by its wavelength. $\begingroup$ The density of nuclei in an atom definitely isn't constant; there's only one nucleus in an atom, and it's located at the center, with no nuclei elsewhere. Volume of the nucleus =4/3πR^3 = 4/3π(R_0A^(1/3))^3 A =4/3πR_0^3 A D. L. Hetrick, Dynamics of Nuclear Reactors, American Nuclear Society, 1993, ISBN: 0-894-48453-2. This orbital motion produces the nucleon with mechanical angular momentum. Why do isotopes of the same atom share the same chemical properties? The atomic nucleus is composed of protons and neutrons. Calculate its volume using the radius found in part (a), and then find its density from, Substituting the values for and A yields. You use the formula for density and the published radii to calculate the densities. Two types of forces exist: (1) the long-range electrostatic (Coulomb) force that makes the positively charged protons repel one another; and (2) the short-range strong nuclear force that makes all nucleons in the nucleus attract one another. The volume V of the nucleus is therefore proportional to the number of nucleons A, expressed by, where r is the radius of a nucleus and k is a constant with units of volume. It is about 2.3 x 10^17 kg/m3. The number of protons in the nucleus is given by the atomic number. Mapping nucleus density is essential to study its roles in fundamental biology and pathological conditions, but ideal tools are still lacking. We use cookies to ensure that we give you the best experience on our website. Since atomic nucleus carries most of atom’s mass and atomic nucleus is very small in comparison to entire atom, the nuclear density is very high. Size of Nucleus: The nuclear radius is the distance from the centre of the nucleus at which the density of nuclear material decreases to one-half of its maximum value at the centre. NCERT DC Pandey Sunil Batra HC Verma Pradeep Errorless. (a) What is the approximate density of a nucleus? nuclei; cbse; class-12; Share It On Facebook Twitter Email. The nucleus of an atom has a radius about 10-5 that of the atom and contains nearly all the mass of the entire atom. Physics. What are isotopes? In any electron shell, there is only one s orbital. All the nuclei have approximately same density. The usual definition of nuclear density gives for its density: ρ nucleus = m / V = 238 x 1.66 x 10-27 / (1.73 x 10-42) = 2.3 x 10 17 kg/m 3. Each cell or tile represents a separate nucleus. The Exclusion Principle and the Periodic Table, 79. These particles are packed together into an extremely small space at the center of an atom. The simplest model of the nucleus is a densely packed sphere of nucleons. The number of protons in the nucleus is given by the atomic number, Z. Compare your answer to that for iron (Fe). The nuclear density for a typical nucleus can be approximately calculated from the size of the nucleus and from its mass. There are 3 different p orbitals, each 90 degrees apart. One cubic meter of nuclear matter has the same mass as a cube of water 61 km on each side. What did RutherFord’s gold foil experiment teach us about the atomic structure? U.S. Department of Energy, Nuclear Physics and Reactor Theory. The formulas to use are "Density" = "mass"/"volume" or ρ = m/V and V = (4/3)πr^3 Hydrogen nuclei A hydrogen nucleus is a proton, and the mass of a proton is 1.67 × 10^(-27)"kg". Class 12 Class 11 Class 10 Class 9 Class 8 Class 7 Class 6. Main purpose of this website is to help the public to learn some interesting and important information about radiation and dosimeters. The number of neutrons is generally greater than the number of protons for. kg / cc kg / cc kg / cc kg / cc Answer : D Solution : The density of neutrons is of the order kg /cc. The number of neutrons in the nucleus is the neutron number, N. The total number of nucleons is the mass number, A. Total dentate nucleus neuronal number correlated with neuronal density (P = 0.007) and did not differ between essential tremor cases and controls (P = 0.95). E. E. Lewis, W. F. Miller, Computational Methods of Neutron Transport, American Nuclear Society, 1993, ISBN: 0-894-48452-4. This density is about a trillion times greater than that of water. The density of the nucleus is approximately 2.3×10 17 kg.m -3. Modern Physics for Scientists and Engineers (4th Edition) Edit edition. Specifies the air density applied for this Maya Nucleus solver. Protons are shown in blue, and neutrons are shown in red. NCERT P Bahadur IIT-JEE Previous Year Narendra Awasthi MS Chauhan. K. O. Ott, R. J. Neuhold, Introductory Nuclear Reactor Dynamics, American Nuclear Society, 1985, ISBN: 0-894-48029-4. The nuclear density for a typical nucleus can be approximately calculated from the size of the nucleus and from its mass. Download PDF's. Density of nucleus is very high, i.e., tonne/cc or 130 trillion tonnes . The density of a nucleus is computed using the ratio of the number of protons and neutrons, A, to the volume of the nucleus (which at my current level, is assumed to be a sphere) The math is trivial. Low Air Density values decrease the amount of air friction acting on an object, which reduces the affect of drag. Nuclei are typically femtometres in diameter - (a femtometre (fm) is 10-15 m). Therefore, the result shown for a mid-sized nucleus is reasonable. 55. I wanted to know if probability density AT the nucleus is maximum, or zero. It is an immense density. W. M. Stacey, Nuclear Reactor Physics, John Wiley & Sons, 2001, ISBN: 0- 471-39127-1. NCERT RD Sharma Cengage KC Sinha. Nuclear Spin or Angular Momentum: Proton and neutron are in continuous motion in discrete quantized orbit. Let radius of nucleus is R. Then,volume of nucleus, =4/3 piR^3 = 4/3 pi (R_0A^(1/3))^3 =4/3 piR_0^3A One cubic meter of nuclear matter has the same mass as a cube of water 61 km on each side. From the attached images, it is clear that the function is at zero for p and d orbitals, but TENDS to infinity as approaching r = 0 for the s-orbitals. The atomic nucleus is composed of protons and neutrons. For heavy nuclei excess neutrons are necessary to keep the electrostatic interactions from breaking the nucleus apart, as shown in (Figure). If you want to get in touch with us, please do not hesitate to contact us via e-mail: Radius and Density of Atomic Nucleus. The arrows represent outward forces on protons (in blue) at the nuclear surface by a proton (also in blue) at the center. Since atomic nucleus carries most of atom’s mass and atomic nucleus is very small in comparison to entire atom, the nuclear density is very high. In this case, as a little sample of a calculation, we can calculate the density of probability of an electron with the following quantum numbers: n = 1, l = 0 and m = 0, around the nucleus. Thus, the density of nuclear material is more than 2.10 14 times greater than that of water. asked Oct 5, 2018 in Physics by Minu (46.0k points) Show that the density of nucleus over a wide range of nuclei is constant independent of mass number. (b) Calculate the density of a gold (Au) nucleus. (Recall that .). Problem:- Given the mass of iron nucleus as 55.85u and A=56,find the nuclear density? Finally, we will need to convert density from units of u/fm 3 to kg/m 3. Where R o is constant for all nuclei and its value is 1.2 x 10-15 m For a given value of Z, multiple values of N (blue points) are possible. The descriptive term nuclear density is also applied to situations where similarly high densities occur, such as within neutron stars. In atomic mass units, the mass of a helium nucleus (A = 4) is approximately 4 u. (a) The electrostatic force is repulsive and has long range. ρ = (m × A)/ (4/3 πR 3) where, m is the mass of one nucleon. The density of the nucleus is approximately 2.3×10 17 kg.m-3. Notably, nuclei in G1, S and G2 phase have different sizes but show a similar RI suggesting a similar density of the nucleus during the cell cycle. Nucleus has no definite boundary, but its radius is given by following relation: R = R o A 1/3. What would be the radius of a neutron star with a mass 10 times that of our Sun (the radius of the Sun is 7 × 10 8)? Biology. NCERT NCERT Exemplar NCERT Fingertips Errorless Vol-1 Errorless Vol-2. The Cookies Statement is part of our Privacy Policy. It is the ratio of mass per unit volume inside the nucleus. The mass (M) of the whole nucleus is the mass of a nucleon (m) multiplied by the number of nucleons (A) Mass M = Am. Is the value of the function 0 at r = 0? Books. Atomic nuclei with the same atomic number. These numbers are related by. Periodic Table, Copyright 2021 Periodic Table | All Rights Reserved |. The density of the nucleus is. Multiply 55.85 and 1.660539 x 10-27 you will get mass of iron nucleus in Kg. Density of nucleus is very high, i.e., tonne/cc or 130 trillion tonnes . Hope it helps:-) Strategy (a) Finding the radius of is a straightforward application of , given (b) To find the approximate density of this nucleus, assume the nucleus is spherical. Chemistry. ISBN: 978-2759800414. What is the ratio of their atomic masses? . The answer to this question requires an understanding of forces inside the nucleus. The nucleus of an atom contains nucleons: protons and neutrons.. Protons repel each other via the Coulomb force but over a very short range (femtometres) there exists a very strong force called the strong nuclear force.This binds the nucleons together into a nucleus. Williams. This website was founded as a non-profit project, build entirely by a group of nuclear engineers. $\endgroup$ – Michael Seifert Mar 25 '18 at 19:46 Atoms that contain nuclei with the same number of protons (Z) and different numbers of neutrons (N) are called isotopes. The mass density of a nucleus varies with mass number A as (a) A2 (b) A (c) constant (d) 1/A Thus, the atomic mass of Cu is The mass of an individual nucleus is often expressed in atomic mass units (u), where . Nuclear Density. For example, Copper (Cu) has two stable isotopes: Given these two “versions” of Cu, what is the mass of this element? For example, hydrogen has three isotopes: normal hydrogen (1 proton, no neutrons), deuterium (one proton and one neutron), and tritium (one proton and two neutrons). It is an immense density. Robert Hofstadter received a Nobel Prize for his research regarding The electron-scattering method and its application to the structure of nuclei and nucleons. Text Solution. The mass number of iron nucleus is 56, the nuclear density is (A) 2.29×1016 kg m-3 (B) 2.29×1017 kg m-3 (C) 2.29×1018 kg m-3 (D) 2.29×1015 kg m-3. The nuclear density for a typical nucleus can be approximately calculated from the size of the nucleus: n = A 4 3πR3 n = A 4 3 π R 3. However, it is consistent with earlier comments about the nucleus containing nearly all of the mass of the atom in a tiny region of space. Nuclear Reactor Engineering: Reactor Systems Engineering, Springer; 4th edition, 1994, ISBN: 978-0412985317, W.S.C. The nuclei are arranged in order of ascending Z (along the horizontal direction) and ascending N (along the vertical direction). If you continue to use this site we will assume that you are happy with it. Nucleus density is an important factor associated with a wide variety of cell behaviors. Air density refers to the mass of the atmosphere in your scene, which affects the relative drag on an object. Co; 1st edition, 1965. This chart is called a chart of the nuclides. The following table shows a list of common isotopes. I answered a similar question a short time ago. All the nuclei have approximately same density. The order of density in nucleus is: (1)10 8 kg/cc (2)10-8 kg/ cc (3)10-9kg/cc (4)10 12 kg/cc. Silver has two stable isotopes. Our Website follows all legal requirements to protect your privacy. Relevant Equations: ρ = m/V I have the solution to the problem but I think I found either a typo or steps that were not included. The diameter of a hydrogen nucleus is 1.0 x10^-3 pm and mass is 1.67x10^-24g.What is the density of the nucleus in grams per cubic centimeter?Assume the nucleus is … ρ = (3m)/ (4πr o3) Putting m = 1.66 × 10 -27 and r o = 1.15 × 10 -15, ρ = 2.7 × 10 17 kgm -3. Get solutions (b) The strong nuclear force acts between neighboring nucleons. You want the density of the carbon nucleus, so the referenced link in the first answer is of no value to you. Express your answer in mega-electron volts (MeV). We hope, this article, Radius and Density of Atomic Nucleus, helps you. By the end of this section, you will be able to: The atomic nucleus is composed of protons and neutrons ((Figure)). Finally, we will need to convert density from units of u/fm 3 to kg/m 3. Frequently Asked Questions – FAQs. The experimentally measured value for is approximately 1.2 femtometer (recall that ). Find the atomic mass of the element silver. It is the ratio of mass per unit volume inside the nucleus. Or something else? Density of the nucleus = (\text { mass of nucleus})/(\text { volume of nucleus}) Mass of the nucleus = A amu = A ×1.66 × 10-27 kg. Typical nuclear radii are of the order 10−14 m. Assuming spherical shape, nuclear radii can be calculated according to following formula: If we use this approximation, we therefore expect the geometrical cross-sections of nuclei to be of the order of πr2 or 4.5×10−30 m² for hydrogen nuclei or 1.74×10−28 m² for 238U nuclei. = A a.m.u = A × 1.660565 × 10 −27 kg. Are you talking about the density of nucleons in a nucleus? (a) The electron probability density in any plane that contains the nucleus is shown. Answer:- Given: - m Fe = 55.85, u = 9.27 × 10 –26 kg For small values of Z, the number of neutrons equals the number of protons and the data fall on the red line. To find the density of nucleus of an atom, we have an atom with mass number lets say A. 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# CAT 2020 Quant Question [Slot 2] with Solution 01

Question

The sum of the perimeters of an equilateral triangle and a rectangle is 90 cm. The area, T, of the triangle and the area, R, of the rectangle, both in sq cm, satisfy the relationship $R = {T^2}$. If the sides of the rectangle are in the ratio 1: 3, then the length, in cm, of the longer side of the rectangle, is

1. 24
2. 27
3. 21
4. 18
Option: 2
Solution:

Let the breadth of the rectangle be b.

Length of the rectangle $= 3b$

Let a be the side of the equilateral triangle.

Given,

$R = {T^2} \Rightarrow 3{b^2} = {\left( {\frac{{\sqrt 3 }}{4} \times {a^2}} \right)^2} \Rightarrow b = {a^2}/4$

Given,

$2(4b) + 3a = 90$

$\Rightarrow 8\left( {{a^2}/4} \right) + 3a - 90 = 0$

$\Rightarrow 2{a^2} + 3a - 90 = 0$

$\Rightarrow (a - 6)(2a + 15) = 0$

$\Rightarrow a = 6$

$\therefore b = 9$

$\Rightarrow 3b = 27$

CAT Online Course @ INR 9999 only

## CAT 2020 Quant questions with Solutions

 Bodhee Prep's YouTube channel CAT Prep Whatsapp Group CAT Prep Telegram Group
X


text:
# Reply To: When the blossom starts to bloom

Welcome to SCGC Players Forum Forums A General Discussion When the blossom starts to bloom Reply To: When the blossom starts to bloom

#1201
Daniel
Participant

The title of this thread reminded me of this song:

My own experience is that my dreadnought was pretty loud when I bought it nearly new in 2002.  But it was brash.  A perfect Rockabilly guitar.  So I banged on it fr about 6 months.  “Wake Up Little Suzy” was indeed one of those songs.  🙂

After 6 months, the tone started to get more complex, but the power did not drop.

For a number of years I couldn’t overdrive the top.  It always just got louder and the topped out without flattening.  These days, like me I guess, it’s more mature, more complex, and a little less powerful.  I can overdrive it if I try.

🙂
D


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# Problem of the Week Problem C and Solution A Square in a Square

## Problem

In the diagram, $$PQRS$$ is a square. Points $$T$$, $$U$$, $$V$$, and $$W$$ are on sides $$PQ$$, $$QR$$, $$RS$$, and $$ST$$, respectively, forming square $$TUVW$$.

If $$PT=QU=RV=SW=4$$ m and $$PQRS$$ has area $$256\text{ m}^2$$, determine the area of $$TUVW$$.

## Solution

The area of square $$PQRS$$ is $$256\text{ m}^2$$. Therefore, square $$PQRS$$ has side length equal to $$16$$ m, since $$16\times 16 = 256$$ and the area of a square is the product of its length and width.

We are given that $$PT=QU=RV=SW=4$$ m. Since $$16-4=12$$, we know that $$TQ = UR = VS = WP = 12$$ m.

We add this information to the diagram.

From this point, we will present two different solutions that calculate the area of square $$TUVW$$.

Solution 1

In $$\triangle WPT$$, $$PT=4$$ and $$WP=12$$. Also, this triangle is right-angled, so we can use one of $$PT$$ and $$WP$$ as the base and the other as the height in the calculation of the area of the triangle, since they are perpendicular to each other. Therefore, the area of $$\triangle WPT$$ is equal to $$\frac{PT\times WP}{2}=\frac{4\times 12}{2}=24\text{ m}^2$$. Since the triangles $$\triangle WPT$$, $$\triangle TQU$$, $$\triangle URV$$, and $$\triangle VSW$$ each have the same base length and height, their areas are equal. Therefore, the total area of the four triangles is $$4\times 24=96\text{ m}^2$$.

The area of square $$TUVW$$ can be determined by subtracting the area of the four triangles from the area of square $$PQRS$$.

Therefore, the area of square $$TUVW$$ is $$256-96=160\text{ m}^2$$.

Solution 2

Some students may be familiar with the Pythagorean Theorem. This theorem states that in a right-angled triangle, the square of the length of the hypotenuse (the longest side) is equal to the sum of the squares of the other two sides. The longest side is located opposite the right angle.

$$\triangle WPT$$ is a right-angled triangle with $$PT=4$$, $$WP=12$$, and $$TW$$ is the hypotenuse. Therefore, \begin{aligned} TW^2&=PT^2+WP^2\\ &=4^2+12^2\\ &=16+144\\ &=160\end{aligned}

Taking the square root, we have $$TW=\sqrt{160}$$ m, since $$TW>0$$.

Now $$TUVW$$ is a square. Therefore, all of its side lengths are equal to $$\sqrt{160}$$. The area of $$TUVW$$ is calculated by multiplying its length by its width.

Therefore, the area of $$TUVW$$ is equal to $$\sqrt{160}\times\sqrt{160}=160\text{ m}^2$$.

Note: Alternatively, we could have found the area of square $$TUVW$$ by noticing that the area of a square is $$s^2$$, where $$s$$ is the side length of the square. For square $$TUVW$$, $$s=TW$$, and therefore the area is $$s^2=TW^2$$. Now, from the Pythagorean Theorem above, we see $$TW^2=160\text{ m}^2$$. Therefore, the area is $$160\text{ m}^2$$.


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43 views
Let $\text{C}1$ and $\text{C}2$ be concentric circles such that the diameter of $\text{C}1$ is $2 \; \text{cm}$ longer than that of $\text{C}2.$ If a chord of $\text{C}1$ has length $6 \; \text{cm}$ and is a tangent to $\text{C}2,$ then the diameter, in $\text{cm},$ of $\text{C}1$ is

Given that, the diameter of $\text{C1}$ is $2 \; \text{cm}$ longer than $\text{C}2.$

So, the radius of $\text{C}1$ should be $1 \; \text{cm}$ longer than $\text{C}2.$

Let the radius of $\text{C}1 = \text{R}\; \text{cm}$

So, the radius of $\text{C}2 = \text{R-1} \; \text{cm}$

The chord of $\text{C}1$ has length $6 \; \text{cm}$ and is a tangent to $\text{C}2.$

Now, we can draw the diagram,

DIAGRAM

In $\triangle \text{OCB},$ using the pythagoras’ theorem,

$\text{(OB)}^{2} = \text{(OC)}^{2} + \text{(CB)}^{2}$

$\Rightarrow \text{R}^{2} = \text{(R-1)}^{2} + 3^{2}$

$\Rightarrow \text{R}^{2} = \text{R}^{2} + 1 – 2 \text{R} + 9$

$\Rightarrow 2 \text{R} = 10$

$\Rightarrow \boxed { \text{R} = 5 \; \text{cm}}$

So, diameter of $\text{C}1 = 2 \text{R} = 2(5) = 10 \; \text{cm}.$

$\therefore$ The diameter of $\text{C}1$ is $10 \; \text{cm}.$

Correct Answer$: 10$
by
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# zbMATH — the first resource for mathematics

Lattice uniformities generated by filters. (English) Zbl 0907.06015
In an earlier paper [Ann. Mat. Pura Appl. 160, 347-370 (1991; Zbl 0790.06006) and ibid. 165, 133-158 (1993; Zbl 0799.06014)], the second author studied uniform lattices as a common generalization of topological Boolean rings and topological Riesz spaces. Lattice uniformities play a similar role in the study of modular functions as FN-topologies.
The main aim of this paper is to give a contribution to the examination of the lattice structure of the space $${\mathcal L}{\mathcal U}_c(L)$$ of all exhaustive lattice uniformities on a lattice $$L$$. In particular, in the case where $$L$$ is a modular sectionally complemented lattice, we get a satisfactory result; in this case $${\mathcal L}{\mathcal U}_c(L)$$ is a complete Boolean algebra (see Corollary 5.11). Such a result can be used as the main tool in obtaining decomposition theorems for modular functions as well as information about the structure of uniform completions of $$L$$.
Important for our investigations is the fact that a lattice uniformity on a sectionally complemented lattice $$L$$ is uniquely determined by its $$0$$-neighbourhood system; moreover the filters which are $$0$$-neighbourhood filters for lattice uniformities are precisely the “distributive” filters, the space of which we indicate by $${\mathcal F}{\mathcal N}{\mathcal D}(L)$$.
We study this space $${\mathcal F}{\mathcal N}{\mathcal D}(L)$$ also for an arbitrary lattice $$L$$. Every filter of $${\mathcal F}{\mathcal N}{\mathcal D}(L)$$ induces on $$L$$ lattice uniformity in a similar way that a distributive ideal in $$L$$ induces a congruence relation. In fact, the concept of “distributive” filters introduced here and some statements about them were suggested by the concept and some properties of distributive ideals.
The main result essentialy says that, for a Hausdorff exhaustive lattice uniformity $$w$$ on $$L$$ generated by some $${\mathcal F}_0\in{\mathcal F}{\mathcal N}{\mathcal D}(L)$$, the space of all lattice uniformities generated by filters of $${\mathcal F}{\mathcal N}{\mathcal D}(L)$$ coarser than $${\mathcal F}_0$$ is isomorphic to the space of all distributive elements of the completion of $$(L, w)$$. As a consequence, we get the result 5.11 mentioned above and the fact proved earlier [H. Weber, Order 12, 295-305 (1995; Zbl 0834.06013)] that the space of all exhaustive lattice uniformities on an orthomodular lattice is a complete Boolean algebra.

##### MSC:
 06F30 Ordered topological structures (aspects of ordered structures) 54E15 Uniform structures and generalizations 54H12 Topological lattices, etc. (topological aspects)
Full Text:
##### References:
 [1] Basile, A.; Traynor, T., Monotonely Cauchy locally solid topologies, Order, 7, 407-416, (1991) · Zbl 0778.46007 [2] Beran, L., Orthomodular lattices, Mathematics and its applications, (1985), Reidel Dordrecht · Zbl 0558.06008 [3] Birkhoff, G., Lattice theory, (1967), Am. Math. Soc Providence · Zbl 0126.03801 [4] Drewnowski, L., Topological rings of sets, continuous set functions, integration I, II, and III, Bull. acad. polon. sci. ser. sci. math. astr. phys., 20, 269-276, (1972) · Zbl 0249.28004 [5] Grätzer, G., General lattice theory, Pure and applied mathematics series, (1978), Academic Press San Diego [6] Maeda, F.; Maeda, S., Theory of symmetric lattices, (1970), Springer-Verlag Berlin/Heidelberg/New York · Zbl 0219.06002 [7] Waelbroecks, Topological vector spaces and algebras, Lecture notes in mathematics, 230, (1971), Springer-Verlag Berlin/New York [8] Weber, H., Group- and vector-valueds, Measure theory, Lecture notes in mathematics, 1089, (1984), Springer-Verlag Berlin/New York, p. 181-198 [9] Weber, H., Uniform lattice: A generalization of topological Riesz spaces and topological Boolean rings; uniform lattice II: order continuity and exhaustivity, Ann. mat. pura appl., 160, 347-370, (1991) · Zbl 0790.06006 [10] Weber, H., Lattice uniformities and modular functions on orthomodular lattices, Order, 12, 295-305, (1995) · Zbl 0834.06013 [11] H. Weber, On modular functions, Funct. Approx. · Zbl 0887.06011 [12] H. Weber
This reference list is based on information provided by the publisher or from digital mathematics libraries. Its items are heuristically matched to zbMATH identifiers and may contain data conversion errors. It attempts to reflect the references listed in the original paper as accurately as possible without claiming the completeness or perfect precision of the matching.


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# Star Wars Solution¶

## Solution 1¶

import re

pattern = re.compile("[^\d]*")
newquotes = [pattern.search(string=x).group() for x in quotes]

print(newquotes)
# [
#     'A ',
#     'I might as well sleep for ',
#     'Hey Patrick, I thought of something funnier than ',
#     'I will have you know that I stubbed my toe last week and only cried for ',
#     'Sandy: Don’t you have to be stupid somewhere else? Patrick: Not until '
# ]

### Explanation¶

Consider the string I might as well sleep for 100 years or so..

1. re.search(pattern="[^\d]", string=...) matches the first non digit character (from the beginning of the string).

I might as well sleep for 100 years or so.

2. re.search(pattern="[^\d]*", string=...) matches zero or more non digit characters (from the beginning of the string).

I might as well sleep for 100 years or so.

3. re.search() returns a match object (assuming a match is found).

re.search(pattern="[^\d]*", string=quote)
# <re.Match object; span=(0, 26), match='I might as well sleep for '>

Match objects have a .group() method which can be used to return the matching substring (or the ith matching capture group).

match = re.search(pattern="[^\d]*", string=quote)
match.group()
# 'I might as well sleep for '

4. This technique can be extended with list comprehension to fetch the matching substring from each string in quotes.

[re.search(pattern="[^\d]*", string=x).group() for x in quotes]

# [
#     'A ',
#     'I might as well sleep for ',
#     'Hey Patrick, I thought of something funnier than ',
#     'I will have you know that I stubbed my toe last week and only cried for ',
#     'Sandy: Don’t you have to be stupid somewhere else? Patrick: Not until '
# ]

5. Since we reuse the same regex multiple times, it'll be more efficient if we compile it!

pattern = re.compile("[^\d]*")
newquotes = [pattern.search(string=x).group() for x in quotes]

print(newquotes)
# [
#     'A ',
#     'I might as well sleep for ',
#     'Hey Patrick, I thought of something funnier than ',
#     'I will have you know that I stubbed my toe last week and only cried for ',
#     'Sandy: Don’t you have to be stupid somewhere else? Patrick: Not until '
# ]

## Solution 2¶

import re

pattern = re.compile("\d.*")
newquotes = [pattern.sub(string=x, repl="") for x in quotes]

print(newquotes)
# [
#     'A ',
#     'I might as well sleep for ',
#     'Hey Patrick, I thought of something funnier than ',
#     'I will have you know that I stubbed my toe last week and only cried for ',
#     'Sandy: Don’t you have to be stupid somewhere else? Patrick: Not until '
# ]

### Explanation¶

The strategy here is to identify the first digit in each string and everything after it, and replace that matching substring with the empty string "".

1. \d matches a digit.

A 5 letter word for happiness... MONEY.
I might as well sleep for 100 years or so.
Hey Patrick, I thought of something funnier than 24... 25!
I will have you know that I stubbed my toe last week and only cried for 20 minutes.
Sandy: Don’t you have to be stupid somewhere else? Patrick: Not until 4

2. \d. matches a digit followed by any character.

A 5 letter word for happiness... MONEY.
I might as well sleep for 100 years or so.
Hey Patrick, I thought of something funnier than 24... 25!
I will have you know that I stubbed my toe last week and only cried for 20 minutes.
Sandy: Don’t you have to be stupid somewhere else? Patrick: Not until 4

Important

Notice that the last quote no longer has a match!

3. \d.* matches a digit followed by zero or more repetitions of any character.

A 5 letter word for happiness... MONEY.
I might as well sleep for 100 years or so.
Hey Patrick, I thought of something funnier than 24... 25!
I will have you know that I stubbed my toe last week and only cried for 20 minutes.
Sandy: Don’t you have to be stupid somewhere else? Patrick: Not until 4

4. Replace the matches in step 3 with empty strings.

import re

pattern = re.compile("\d.*") # (1)!
newquotes = [pattern.sub(string=x, repl="") for x in quotes] # (2)!

print(newquotes)
# [
#     'A ',
#     'I might as well sleep for ',
#     'Hey Patrick, I thought of something funnier than ',
#     'I will have you know that I stubbed my toe last week and only cried for ',
#     'Sandy: Don’t you have to be stupid somewhere else? Patrick: Not until '
# ]

1. Compile the pattern in preparation for repeated use.
2. Use list comprehension combined with Pattern.sub() to replace matching groups with the empty string.


text:
# Finding value for time from equation (DIFFICULT)

Find the time takes skydiver to hit the ground using th following:

y = (t/SQRT(10) - 2ln(1+e^(t/SQRT(10)) - C) x 20SQRT10

## The Attempt at a Solution

Have tried many times to find value for t with little success help needed!

HallsofIvy
Homework Helper
Find the time takes skydiver to hit the ground using th following:

y = (t/SQRT(10) - 2ln(1+e^(t/SQRT(10)) - C) x 20SQRT10

## The Attempt at a Solution

Have tried many times to find value for t with little success help needed!
So you want to solve
$$20\sqrt{10}(\frac{t}{\sqrt{10}}- 2ln(1+ e^{\frac{t}{\sqrt{10}}})- C)= 0$$

I would start by doing two things: divide both sides by $20\sqrt{10}$ and replace $t/\sqrt{10}$ by "u". Now the equation is
$$u- 2ln(1+ e^u)- C= 0$$

Of course, you can't get a specific numerical result without a specic value for C. Even then I suspect that there will be no solution in terms of elementary functions.

Okay well I derived the solution stated from the following:

y' = 20SQRT10 x (1-e^(t/SQRT10))/(1+e(t/SQRT10))

Where C is a constant of integration, is this correct? Becasue the question looks as if a specific numerical value must be found.

Mark44
Mentor
Okay well I derived the solution stated from the following:

y' = 20SQRT10 x (1-e^(t/SQRT10))/(1+e(t/SQRT10))

Where C is a constant of integration, is this correct? Becasue the question looks as if a specific numerical value must be found.
Why would you take the derivative?

The skydiver will be on the ground at the time when y = 0 in the equation in your first post. That's what HallsOfIvy was telling you in his reply. Did you read it?

Since no integration has apparently been done, C is not a constant of integration. Is there any other information in your problem that you haven't posted?

Well in order to obtain the equation :

y = (t/SQRT(10) - 2ln(1+e^(t/SQRT(10)) - C) x 20SQRT10

I had to use seperation of variables and integration on the following:

y' = 20SQRT10 x (1-e^(t/SQRT10))/(1+e(t/SQRT10))

Which is the reason for the constant of integration C, try it yourself and see if you come up with the same solution as me. Only other information is that at y(0) = 4000.

Thanks

Mark44
Mentor
Well in order to obtain the equation :

y = (t/SQRT(10) - 2ln(1+e^(t/SQRT(10)) - C) x 20SQRT10

I had to use seperation of variables and integration on the following:

y' = 20SQRT10 x (1-e^(t/SQRT10))/(1+e(t/SQRT10))

Which is the reason for the constant of integration C, try it yourself and see if you come up with the same solution as me. Only other information is that at y(0) = 4000.

Thanks
There's a reason for the three parts in the problem template, of which the first is the complete problem statement and all pertinent data. As you presented the problem, there was no indication that you were solving a differential equation. You should also have included the initial condition, y(0) = 4000, since that has to be used to find C.

Okay that was my fault but have you had a go at finding a sloution from the equation stated? Im just curious as I spent alot of time on this question and want to know if correct or not?

Mark44
Mentor
I don't have time right now to check your work, but you can do that. Take your solution y(t) and differentiate it. If you don't get 20sqrt(10)(1 -e^(t/sqrt(10))/(1 + e^(t/sqrt(10)), your solution is incorrect.


text:
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finMath lib documentation
net.finmath.marketdata.model.curves

## Class DiscountCurveNelsonSiegelSvensson

• All Implemented Interfaces:
Serializable, Cloneable, ParameterObjectInterface, CurveInterface, DiscountCurveInterface

public class DiscountCurveNelsonSiegelSvensson
extends AbstractCurve
implements Serializable, DiscountCurveInterface
Implementation of a discount factor curve given by a Nelson-Siegel-Svensson (NSS) parameterization. In the NSS parameterization the zero rate $$r(T)$$ is given by $r(T) = \beta_0 + \beta_1 \frac{1-x_0}{T/\tau_0} + \beta_2 ( \frac{1-x_0}{T/\tau_0} - x_0) + \beta_3 ( \frac{1-x_1}{T/\tau_1} - x_1)$ where $$x_0 = \exp(-T/\tau_0)$$ and $$x_1 = \exp(-T/\tau_1)$$. The sub-family of curves with $$\beta_3 = 0$$ is called Nelson-Siegel parameterization. Note: This is a time-parameterized model. The finmath lib library uses an internal mapping from date to times $$t$$. This mapping does not necessarily need to correspond with the curves understanding for the parameter $$T$$. For that reason this class allows to re-scale the time parameter. Currently only a simple re-scaling factor is supported. The parameter T used in the parameterization is given by T = timeScaling * t, where t is the maturity as an ACT/365 year fraction from the given reference date.
Author:
Christian Fries
See Also:
Serialized Form
• ### Constructor Summary

Constructors
Constructor and Description
DiscountCurveNelsonSiegelSvensson(String name, LocalDate referenceDate, double[] parameter, double timeScaling)
Create a discount curve using a Nelson-Siegel-Svensson parametrization.
• ### Method Summary

All Methods
Modifier and Type Method and Description
DiscountCurveNelsonSiegelSvensson clone()
Create a deep copied clone.
CurveBuilderInterface getCloneBuilder()
Returns a curve builder bases on a clone of this curve.
DiscountCurveNelsonSiegelSvensson getCloneForParameter(double[] value)
Create a clone with a modified parameter.
double getDiscountFactor(AnalyticModelInterface model, double maturity)
Return the discount factor within a given model context for a given maturity.
double getDiscountFactor(double maturity)
Returns the discount factor for the corresponding maturity.
double[] getParameter()
Get the current parameter associated with the state of the objects.
double getTimeScaling()
double getValue(AnalyticModelInterface model, double time)
Returns the value for the time using the interpolation method associated with this curve within a given context, i.e., a model.
double getZeroRate(double maturity)
Returns the zero rate for a given maturity, i.e., -ln(df(T)) / T where T is the given maturity and df(T) is the discount factor at time $T$.
double[] getZeroRates(double[] maturities)
Returns the zero rates for a given vector maturities.
void setParameter(double[] parameter)
Deprecated.
String toString()
• ### Methods inherited from class net.finmath.marketdata.model.curves.AbstractCurve

getName, getReferenceDate, getValue, getValues
• ### Methods inherited from class java.lang.Object

equals, finalize, getClass, hashCode, notify, notifyAll, wait, wait, wait
• ### Methods inherited from interface net.finmath.marketdata.model.curves.CurveInterface

getName, getReferenceDate, getValue
• ### Constructor Detail

• #### DiscountCurveNelsonSiegelSvensson

public DiscountCurveNelsonSiegelSvensson(String name,
LocalDate referenceDate,
double[] parameter,
double timeScaling)
Create a discount curve using a Nelson-Siegel-Svensson parametrization.
Parameters:
name - The name of the curve (the curve can be referenced under this name, if added to an AnalyticModel.
referenceDate - The reference date of this curve, i.e. the date associated with t=0.
parameter - The Nelson-Siegel-Svensson parameters in the order $$( \beta_0, \beta_1, \beta_2, \beta_3, \tau_0, \tau_1 )$$.
timeScaling - The time parameter argument rescaling. See getDiscountFactor(AnalyticModelInterface, double).
• ### Method Detail

• #### getDiscountFactor

public double getDiscountFactor(double maturity)
Description copied from interface: DiscountCurveInterface
Returns the discount factor for the corresponding maturity. This getter is not optimized for performance.
Specified by:
getDiscountFactor in interface DiscountCurveInterface
Parameters:
maturity - The maturity for which the discount factor is requested.
Returns:
The discount factor (i.e., price of the zero coupon bond with given maturity and notional 1.
• #### getDiscountFactor

public double getDiscountFactor(AnalyticModelInterface model,
double maturity)
Return the discount factor within a given model context for a given maturity.
Specified by:
getDiscountFactor in interface DiscountCurveInterface
Parameters:
model - The model used as a context (not required for this class).
maturity - The maturity in terms of ACT/365 daycount form this curve reference date. Note that this parameter might get rescaled to a different time parameter.
Returns:
The discount factor (i.e., price of the zero coupon bond with given maturity and notional 1.
See Also:
DiscountCurveInterface.getDiscountFactor(net.finmath.marketdata.model.AnalyticModelInterface, double)
• #### getValue

public double getValue(AnalyticModelInterface model,
double time)
Description copied from interface: CurveInterface
Returns the value for the time using the interpolation method associated with this curve within a given context, i.e., a model. The model (context) is needed only if the curve relies on another curve. Examples are a forward curve which relies on a discount curve or a discount curve which is defined via a spread over another curve.
Specified by:
getValue in interface CurveInterface
Parameters:
model - An analytic model providing a context.
time - Time for which the value should be returned.
Returns:
The value at the give time.
• #### getZeroRate

public double getZeroRate(double maturity)
Returns the zero rate for a given maturity, i.e., -ln(df(T)) / T where T is the given maturity and df(T) is the discount factor at time $T$.
Parameters:
maturity - The given maturity.
Returns:
The zero rate.
• #### getZeroRates

public double[] getZeroRates(double[] maturities)
Returns the zero rates for a given vector maturities.
Parameters:
maturities - The given maturities.
Returns:
The zero rates.
• #### getCloneBuilder

public CurveBuilderInterface getCloneBuilder()
throws CloneNotSupportedException
Description copied from interface: CurveInterface
Returns a curve builder bases on a clone of this curve. Using that curve builder you may create a new curve from this curve by adding points or changing properties. Note: The clone has the same name than this one.
Specified by:
getCloneBuilder in interface CurveInterface
Returns:
An object implementing the CurveBuilderInterface where the underlying curve is a clone of this curve.
Throws:
CloneNotSupportedException - Thrown, when this curve could not be cloned.
• #### getParameter

public double[] getParameter()
Description copied from interface: ParameterObjectInterface
Get the current parameter associated with the state of the objects.
Specified by:
getParameter in interface ParameterObjectInterface
Returns:
The parameter.
• #### setParameter

@Deprecated
public void setParameter(double[] parameter)
Deprecated.
Description copied from interface: ParameterObjectInterface
Set the current parameter and change the state of the objects.
Specified by:
setParameter in interface ParameterObjectInterface
Parameters:
parameter - The parameter associated with the new state of the objects.
• #### getTimeScaling

public double getTimeScaling()
• #### clone

public DiscountCurveNelsonSiegelSvensson clone()
throws CloneNotSupportedException
Description copied from interface: CurveInterface
Create a deep copied clone.
Specified by:
clone in interface CurveInterface
Overrides:
clone in class AbstractCurve
Returns:
A clone (deep copied).
Throws:
CloneNotSupportedException - Thrown, when the curve could not be cloned.
• #### getCloneForParameter

public DiscountCurveNelsonSiegelSvensson getCloneForParameter(double[] value)
throws CloneNotSupportedException
Description copied from interface: ParameterObjectInterface
Create a clone with a modified parameter.
Specified by:
getCloneForParameter in interface ParameterObjectInterface
Specified by:
getCloneForParameter in interface CurveInterface
Overrides:
getCloneForParameter in class AbstractCurve
Parameters:
value - The new parameter.
Returns:
A clone with an otherwise modified parameter.
Throws:
CloneNotSupportedException - Thrown, when the curve could not be cloned.
• #### toString

public String toString()
Overrides:
toString in class AbstractCurve
Skip navigation links
Copyright © 2017 Christian P. Fries.

Copyright © 2017. All rights reserved.


text:
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# Circles: Method for Solving Questions involving Arcs and Angles

If $$B$$ is the center of the circle in the figure above and the area of the shaded region is $$16 \pi$$, what is the length of arc $$ADC$$?

Next use this to find the circle's radius: > $$\pi r^2=64 \pi$$ > $$r^2=64$$ > $$r=8$$

Now, find the circle's circumference:

> $$\text{Circumference} = 2\pi r = 2\pi(8) =16 \pi$$ Since Arc $$ADC$$ is one-fourth of the circumference, >$$\text{Arc } ADC = \frac{1}{4} \times 16 \pi = 4 \pi$$.

Incorrect.

[[snippet]]

Incorrect.

[[snippet]]

This is the value of the circle's circumference:

$$2 \pi r = 2\times8\times \pi = 16 \pi$$

Now, you have one more step . . .

Incorrect.

[[snippet]]

Incorrect.

[[snippet]]
Correct. [[snippet]] When you plug in the given information, you get > $$\displaystyle{\frac{16 \pi}{\text{Circle Area}}=\frac{90^\circ}{360^\circ}}$$ > $$\displaystyle{\frac{16 \pi}{\text{Circle Area}}=\frac{1}{4}}$$ Therefore, > $$\text{Circle Area} = 4\times16 \pi = 64 \pi$$
$$4$$
$$4 \pi$$
$$16$$
$$8 \pi$$
$$16 \pi$$
Continue


text:
Lemma 14.3.2. Let $\mathcal{C}$ be a category.

1. Given a simplicial object $U$ in $\mathcal{C}$ we obtain a sequence of objects $U_ n = U([n])$ endowed with the morphisms $d^ n_ j = U(\delta ^ n_ j) : U_ n \to U_{n-1}$ and $s^ n_ j = U(\sigma ^ n_ j) : U_ n \to U_{n + 1}$. These morphisms satisfy the opposites of the relations displayed in Lemma 14.2.3, namely

1. If $0 \leq i < j \leq n + 1$, then $d^ n_ i \circ d^{n + 1}_ j = d^ n_{j - 1} \circ d^{n + 1}_ i$.

2. If $0 \leq i < j \leq n - 1$, then $d^ n_ i \circ s^{n - 1}_ j = s^{n - 2}_{j - 1} \circ d^{n - 1}_ i$.

3. If $0 \leq j \leq n - 1$, then $\text{id} = d^ n_ j \circ s^{n - 1}_ j = d^ n_{j + 1} \circ s^{n - 1}_ j$.

4. If $0 < j + 1 < i \leq n$, then $d^ n_ i \circ s^{n - 1}_ j = s^{n - 2}_ j \circ d^{n - 1}_{i - 1}$.

5. If $0 \leq i \leq j \leq n - 1$, then $s^ n_ i \circ s^{n - 1}_ j = s^ n_{j + 1} \circ s^{n - 1}_ i$.

2. Conversely, given a sequence of objects $U_ n$ and morphisms $d^ n_ j$, $s^ n_ j$ satisfying (1)(a) – (e) there exists a unique simplicial object $U$ in $\mathcal{C}$ such that $U_ n = U([n])$, $d^ n_ j = U(\delta ^ n_ j)$, and $s^ n_ j = U(\sigma ^ n_ j)$.

3. A morphism between simplicial objects $U$ and $U'$ is given by a family of morphisms $U_ n \to U'_ n$ commuting with the morphisms $d^ n_ j$ and $s^ n_ j$.

Proof. This follows from Lemma 14.2.4. $\square$

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).


text:
# Iteration Method for Solving Recurrences

In this method, we first convert the recurrence into a summation. We do so by iterating the recurrence until the initial condition is reached. In the example given in the previous chapter, $T(1)$ was the time taken in the initial condition. For converting the recurrence of the previous example into summation, we would first break $T(n)$ into $T(\frac{n}{2})$ and then into $T(\frac{n}{4})$ and so on. Let's take this example and see how to apply this method.

In the previous example, we had our equation in the form:

$$T(n) = \begin{cases}\tag{1} T(1), &\text{n=1} \\[2ex] c + T\left(\frac{n}{2}\right), &\text{if n>1} \end{cases}$$ where, $c$ is a constant

We have replaced $\Theta(1)$ from the previous equation with $c$ because $\Theta(1)$ means constant time.

Let's replace $n$ with $n/2$ in the previous equation.

$$T\left( \frac{n}{2} \right) = c + T\left( \frac{n}{4} \right) \tag{2}$$ where, $c$ is a constant

Now, put the value of $T\left( \frac{n}{2} \right)$ from $eq(2)$ in the $eq(1)$, we get:

$$T(n) = c + T\left(\frac{n}{2}\right)$$ $$=> T(n) = c + \underbrace{\left( c + T\left( \frac{n}{4} \right) \right)}_\text{value of T\left( \frac{n}{2} \right)} \tag{3}$$

Again, let's use $T\left( \frac{n}{4} \right)$ in place of $n$ in the $eq(1)$.

$$T\left( \frac{n}{4} \right) = c + T\left( \frac{n}{8} \right)$$

Putting this value in $eq(3)$, we get,

$$T(n) = c+ c + c + T\left( \frac{n}{8} \right) \tag{4}$$

We are breaking our problem into two subproblems of half size as mentioned in the picture given below.

So, we can assume that $n$ is in the form of $2^k$.

It might be possible that $n$ is not exactly in the form of $2^k$ but $2^k \pm k_o$, where $k_o$ is some constant.

But we are concerned about the rate of the growth only and this approximation is going to give us that. So, we will proceed by assuming that $n$ is in the form of $2^k$.

Recalling the $eq(4)$,

$$T(n) = c+ c + c + T\left( \frac{n}{8} \right)$$

We can also write the above equation as

$$T(n) = 3c + T\left( \frac{n}{2^3} \right)$$

Similarly, we can write

$$T(n) = \underbrace{c+c+...+c}_{k\text{ times}}+ T\left( \frac{n}{2^k} \right)$$ $$\text{Since }n = 2^k\text{,}$$ $$T(n) = \underbrace{c+c+...+c}_{k\text{ times}}+ T\left( \frac{2^k}{2^k} \right)$$ $$=kc+T(1) \tag{5}$$

As mentioned above, $T(1)$ is the base case. So, we have reached the base case.

Also,

$$n = 2^k$$ $$=> \log_2(n)=log_2(2^k)$$ $$=>\lg(n) = k$$

Replacing the value of $k$ in the $eq(5)$, we get

$$T(n) = c\lg(n)+T(1)$$ $$=\Theta(\lg(n))$$

Let's take a brief recap of what we did here. We first formed the recurrence equation from the code or the algorithm and then we solved that equation using the iteration method. In this method, we broke down the equation $\left( n\,→ \frac{n}{2}\, → \frac{n}{4}\, → \frac{n}{8}\, → ...\right)$. And after reaching the base case, we back-substituted the equation (value of $k$) to express the equation in the form of $n$ and initial boundary condition

Let's look at one more example of this method.

$$T(n) = n+2T\left( \frac{n}{2} \right) \tag{6}$$

In this example also, $T(n)$ is broken down into $T\left( \frac{n}{2} \right)$, so let's replace $n$ with $\frac{n}{2}$ in the $eq(6)$.

$$T(\frac{n}{2}) = \frac{n}{2}+2T\left( \frac{n}{4} \right)$$

Putting this value in $eq(6)$,

$$T(n) = n + 2\left( \frac{n}{2} + 2T\left( \frac{n}{4} \right) \right)$$ $$=n+n+4T\left( \frac{n}{4} \right)$$

Again, putting $\frac{n}{4}$ instead of $n$ in the $eq(6)$ and getting the value of $T\left( \frac{n}{4} \right)$.

$$=n+n+4 \underbrace{ \left(\frac{n}{4} + 2\left(\frac{n}{8}\right) \right)}_\text{value of T\left( \frac{n}{4} \right)}$$ $$=n+n+n+8T\left( \frac{n}{8} \right) \tag{7}$$

Let's assume $n$ is of the form of $2^k$ i.e., $n = 2^k$ as we did in the previous example.

For 3 terms, $eq(7)$ can be written as

$$=3n + 2^3T\left( \frac{n}{2^3} \right)$$

Similarly, for $k$ terms, we can write,

$$=kn + 2^kT\left( \frac{n}{2^k} \right)$$

Replacing $n$ with $2^k$,

$$=kn + 2^kT\left( \frac{2^k}{2^k} \right)$$ $$=kn + 2^kT(1)$$ $$=kn + nT(1)$$ Replacing $k$ with $\lg(n)$ ($n=2^k => \lg(n)=k$) $$=n\lg(n) + nT(1)$$ Since, $n\lg(n)$ is faster growing than $n$ $$T(n)=\Theta(nlg(n))$$

So, we have obtained the growth order of both the algorithms discussed in the previous chapters. As stated earlier, we are going to learn two more methods to solve recurrences in the next chapters.

Furious activity is no substitute for analytical thought.
- Alastair Pilkington