url stringlengths 6 1.61k | fetch_time int64 1,368,856,904B 1,726,893,854B | content_mime_type stringclasses 3 values | warc_filename stringlengths 108 138 | warc_record_offset int32 9.6k 1.74B | warc_record_length int32 664 793k | text stringlengths 45 1.04M | token_count int32 22 711k | char_count int32 45 1.04M | metadata stringlengths 439 443 | score float64 2.52 5.09 | int_score int64 3 5 | crawl stringclasses 93 values | snapshot_type stringclasses 2 values | language stringclasses 1 value | language_score float64 0.06 1 |
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https://www.codewars.com/users/Davionk/comments | 1,603,810,327,000,000,000 | text/html | crawl-data/CC-MAIN-2020-45/segments/1603107894203.73/warc/CC-MAIN-20201027140911-20201027170911-00123.warc.gz | 657,708,198 | 11,950 | • ###### Davionkcommented on "Find the unique number" java solution
It is not actually. It is more my stupid mistake for saying that in the end of my comment ;P. As you can see the first reply to my comment explains that I am wrong in a funny way :D.
Again sorry for the last part of my comment and thanks for the reply.
• ###### mateenkasimcommented on "Find the unique number" java solution
How is this O(n/2) if you're still doing ~n comparisons? The loop runs n/2 times but you do 2 comparisons per loop. O(n) is still great though, better than sorting
• ###### FArekkusucommented on "Find the unique number" java solution
tecnically speaking it was O(n / 2)
You could improve it drastically by splitting the array into n 1-element-long segments, and have O(n / n) = O(1) time complexity
• ###### Davionkcommented on "Find the unique number" java solution
This is an ok solution. But that's all.
It is the same as all standard solutions, just more nice to the eye.
The kata specifies to give emphasis to efficiency!
Go check my solution, I solved it in O(N). Even though tecnically speaking it was O(N/2), so half the time!!! | 289 | 1,140 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.140625 | 3 | CC-MAIN-2020-45 | latest | en | 0.919137 |
https://euclideanspace.com/software/computation/theory/types/index.htm | 1,696,126,418,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233510734.55/warc/CC-MAIN-20231001005750-20231001035750-00589.warc.gz | 269,307,209 | 4,374 | # Maths - Computation - Data Types
This page looks at data types as mathematical expressions which we can manipulate in equations.
• See page here for more of an introduction to the subject.
• See this page for a more mathematical approach to the subject.
### Algebraic Data Types
Unit - has one value 1 Final ()
Void 0 Initial not in haskell
Boolean 2
Sum - disjoint union - either + Sum
Product - both Product
Exponent - function space ^ Exponent function type
For example: product creates a tuple.
### Inductivly Defined Types
List L(x)=1+x•L(x)
= 1+x+x²+x³...
List x = Nil | Cons x (List x)
Binary Tree T(x)=1+x•T²(x)
=(1 - √(1-4x))/2x
= 1+x+2x²+5x³+14x4...
Tree x = Tip | Node (Tree x) x (Tree x) left branch value right branch
So a list allows us to create a tuple of variable length by using recursion.
### Fixpoint
The fixpoint are the values of x where T(x)=x
see
### Zipper
value + one-hole-context
### One-hole-context
The derirative of a regular type is its type of one-hole contexts (Conor McBride)
1 => 0 x => 1 x² => 2x x³ => 3x²
Using calculus notation
∂1 = 0 ∂x = 1 ∂x² = 2x ∂x³ = 3x² ∂(f+g) = ∂f + ∂g ∂(f•g) = ∂f•g + f•∂g ∂(f(g)) = ∂f(g) •∂g
### Non-Regular Types
Bags (No order)
ULists (No duplicates)
Sets (No order,No duplicates)
setn(x)= set of size n containing type x
set0(x)= 1
set1(x)= x
set2(x)= x(x-1)/2
in general:
setn(x)= x(x-1)...x-(n+1)/n!
=Σx_n_/n!
=1+x+x3/!2+x5/!3
all possible holes =
Δ f(x)=f(x+1)-f(x)
Δ set(x)=set(x)
set(x) = 2x=all maps from x to bool
Cyclic Lists
Deques | 538 | 1,543 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.953125 | 4 | CC-MAIN-2023-40 | latest | en | 0.699956 |
https://math.stackexchange.com/questions/3200064/the-commutator-product-as-a-replacement-for-the-cross-and-wedge-product-in-geome | 1,718,670,996,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861741.14/warc/CC-MAIN-20240617215859-20240618005859-00280.warc.gz | 349,916,944 | 38,771 | # The commutator product as a replacement for the cross and wedge product in geometric algebra?
From an axiomatic approach to geometric algebra, the wedge product of two vectors $$a$$ and $$b$$ is typcially defined as the antisymmetric $$a \wedge b = \frac{1}{2}(ab - ba)$$, where $$ab$$ is the geometric product of $$a$$ and $$b$$, and is used as the "replacement" for the cross product in 3 dimensions. However, the commutator product for two arbitrary multivectors $$A$$ and $$B$$ is also defined to be $$A \times B = \frac{1}{2}(AB - BA)$$, where $$AB$$ is the geometric product of $$A$$ and $$B$$. For two vectors this is the same as the wedge product - a bivector, a pseudovector in 3 dimensions. For vector $$a$$ and bivector $$B$$, the commutator product $$a \times B$$ yields another vector $$d$$, and substituting $$B = b \times c$$ for two vectors $$b$$ and $$c$$, we get $$d = a \times (b \times c)$$, which is the triple vector product, but using the commutator product instead of the cross product.
The dot product is likewise defined from the axiomatic approach as the symmetric $$a \cdot b = \frac{1}{2}(ab + ba)$$ which could also potentially be extended in the same manner as the commutator to general multivectors $$A$$ and $$B$$. In this manner the symmetric dot product of vector $$a$$ and a bivector $$B$$ is a trivector $$\Gamma$$, which in three dimensions is simply a pseudoscalar $$\gamma I$$. Substituting $$B = b \times c$$ into this equation, we get another equation $$\gamma I = a \cdot (b \times c)$$, which is very similar to the triple scalar product using the traditional vector dot and cross products, except yielding a pseudoscalar instead of a scalar.
So with all these similarities between the commutator and the vector cross product, is it possible to use the commutator product used as the replacement for the cross product in geometric algebra in place of the wedge product? And likewise for the symmetric product extension of the dot product $$A \cdot B = \frac{1}{2}(AB + BA)$$ in place of the typical contraction extension? What are the potential disadvantages of using the commutator and the symmetric dot product in place of the wedge product and contraction dot product in geometric algebra, especially in 3 dimensions?
• Your question is almost impossible to read. Can you insert some newlines for paragraph breaks in the appropriate places. Commented Apr 24, 2019 at 13:56
• I've inserted some newlines in the question. Hopefully this helps @PeeterJoot Commented Apr 24, 2019 at 14:22
• @1103_base_6 it seems you are confusing several different operations with your (undefined) product notations $ab$ and $AB$. As far as I can tell, one seems to be a tensor product and the other matrix multiplication Commented Apr 24, 2019 at 14:27
• @Glougloubarbaki $ab$ and $AB$ are the geometric product of two vectors and multivectors respectively in geometric algebra/clifford algebra. Perhaps I should make it clearer in the question. Commented Apr 24, 2019 at 14:28
• I've also edited the title as well. Commented Apr 24, 2019 at 14:31
Welcome to MSE!
So with all these similarities between the commutator and the vector cross product, is it possible to use the commutator product used as the replacement for the cross product in geometric algebra in place of the wedge product?
First, the vector cross product and the geometric algebra commutator use the same symbol $$\times$$ for their operation. This similarity, not mentioned in your post, could be potentially very confusing when most of the mathematicians and physicists learn vector algebra before they learn geometric algebra, as the commutator of two vectors (as you said) would yield a bivector, while the cross product would only yield a vector.
Another subtle difference between the vector cross product and the commutator that could cause confusion is that the result from the triple vector cross product is the negative of the corresponding triple product for the commutator in $$G^{3,0}$$. For vectors $$a$$, $$b$$, $$c$$, $$d$$, and $$e$$, let $$d = a \times (b \times c)$$ and $$e = b \times c$$, where $$\times$$ represents the vector cross product. Now if $$\times$$ were to be the geometric algebra commutator, $$b \times c$$ would instead equal $$Ie$$, and $$a \times (b \times c) = a \times Ie = I(a \times e) = I^2 d = -d$$, due to the properties of the pseudoscaler in $$G^{3,0}$$, causing another potential source of confusion and error.
Due to the possible confusion between the two operators, in geometric algebra the commutator is typically only used for bivectors, while the outer product/wedge product is used for vectors. Perhaps in a future when vector algebra has been superseded by geometric algebra, the commutator can be more widely used (but see below).
And likewise for the symmetric product extension of the dot product $$A \cdot B = \frac{1}{2}(AB + BA)$$ in place of the typical contraction extension?
In a geometric algebra of dimension less than three, this definition would yield a pure-grade multivector, but once you reach four dimensions, this product of two bivectors would yield a mixed-grade multivector, one of grade 0 and another of grade 4. Typically, we only wish to have one of the two multivector terms - the inner product/contraction would yield the grade 0 term, while the outer product would yield the grade 4 term. Thus, for GA applications such as 3D Conformal Geometry ($$G^{4,1}$$) or Spacetime Algebra ($$G^{3, 1}$$), the contraction and outer product would be more useful in that sense.
What are the potential disadvantages of using the commutator and the symmetric dot product in place of the wedge product and contraction dot product in geometric algebra
There are actually two contractions, the left contraction and the right contraction, which do have the less-commonly-used symbols $$\lfloor$$ and $$\rfloor$$ as well, but due to historical reasons and the relationships with the vector dot product and the inner product in linear algebra, the $$\cdot$$ symbol is typically used for the left contraction. So, technically, you're after defining two new binary operations in geometric algebra, which could be used alongside the contractions and the outer product.
But as a replacement for the contractions and the outer product? One of the big advantages of using geometric algebra is that there is a clear intuition between the algebraic objects and operators in geometric algebra and the corresponding geometric objects in the defined space of any dimension. The contractions and outer product have a distinctive geometric interpretation (as do blades) - the contraction of two blades $$A$$ and $$B$$ being the part of $$A$$ most unlike $$B$$, according to Leo Dorst in his textbook Geometric Algebra for Computer Science, while the outer product of $$A$$ and $$B$$ is the blade $$C$$ representing the subspace whose basis is the union of the the bases of the subspaces representing $$A$$ and $$B$$. But it is unclear what the geometric interpretation of either the commutator or your defined symmetric dot product should result in. Thus using your two products in place of the ones currently used in geometric algebra would lose some of the intuitive geometric interpretation of geometric algebra that makes it such a powerful tool in geometry, physics, and engineering.
The commutator product is important for bivectors in geometric algebra primarily not for its associations with geometry, but rather because the bivectors equipped with the commutator product are associated with particular Lie Algebras from abstract algebra, which are used greatly in more advanced physics.
So, while it is possible to use the commutator in place of the wedge product, the wedge product offers many more advantages to geometric algebra compared to the commutator.
especially in 3 dimensions?
I make a separate case for three dimensions, as it seems that, if a different symbol were to be used for the commutator, your algebra does seem to be a midway point between traditional vector algebra and full-blown $$G^{3,0}$$. The main issue with the output of the cross product is that it fails to distinguish between polar and axial vectors, or vectors and pseudovectors, but by introducing an commutative pseudoscalar (complex imaginary) unit whose square is negative one, and defining pseudovectors as the product of a vector and the pseudoscalar, that issue with the output of the cross product is resolved, and furthermore it highlights a link between quaternions and vector algebra. However, this is only the case in three dimensions, and does not generalise to other dimensions, as geometric algebra is intended to, and so suffers from the same limitations as traditional vector algebra.
Edit: That was quite the long response. I wasn't expecting that. | 2,029 | 8,823 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 61, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.96875 | 4 | CC-MAIN-2024-26 | latest | en | 0.908596 |
http://www.longrangehunting.com/threads/effects-of-wind.12824/ | 1,503,512,313,000,000,000 | text/html | crawl-data/CC-MAIN-2017-34/segments/1502886123312.44/warc/CC-MAIN-20170823171414-20170823191414-00125.warc.gz | 607,623,890 | 19,102 | # Effects of wind?
Discussion in 'Rifles, Bullets, Barrels & Ballistics' started by Coues100, Aug 27, 2005.
1. ### Coues100Member
Messages:
14
Joined:
May 10, 2005
I was wondering how wind, straight on or from straight behind effects trajectory?? How do you calculate for long range shooting? Thanks, Mark
2. ### Charles AWell-Known Member
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480
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Nov 7, 2001
From the from the front the impact will be higher, from behind the impact will be lower. Depending on caliber, bc, velocity, wind speed and especially range it could be a 1/2minute difference. The problem isnt when the wind is truely from 12 or 6 o'clock, but when its at 1,5,7, or 11 o'clock.
3. ### 700Well-Known Member
Messages:
138
Joined:
Jun 30, 2002
LR7WSM
This is called "Vertical Wind Component" becĂ use the wind causes your bullet to plot higher or lower, rather than just left or right.
Give me an email at brXXXian_sinXXXnott@hotXXXmail.com
remove the XXXs.
I can then get details of your shooting system and make up some charts showing the Vertical Wind Component at different ranges.
Rgds
700
4. ### Richard338Well-Known Member
Messages:
126
Joined:
Aug 22, 2003
A head wind will LOWER the point of impact. A tail wind will RAISE the point of impact.
Just picture the limiting case of a tail wind equal to muzzle velocity, then you'd have a vacuum trajectory.
The effect is much less than a cross wind, but I recently shot in a switching head/tail wind that was good for 2MOA up/down.
5. ### jb1000brWell-Known Member
Messages:
1,307
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Jul 8, 2003
Well, when this question comes up, you will always get both answers...shown here.
I have seen more evidence of a HEADWIND RAISING POI AT LONG RANGE.
But it makes more sense that it would do the reverse at short range...
WHY?
On its way to 1000yds the bullet spend more time CLIMBING then FALLING, so why wouldnt a headwind give the bullet LOFT?
At close range, the trajectory is probably flat enough that it is as simple as the wind adding and subtracting from velocity, meaning a HEADWIND WOULD LOWER POI
Now how bout 2000yds+ The bullet is much slower further out and angling toward earth, so here, the loft may even out the added drop, or the headwind may end up causing a total effect lowering POI -- i cant say much here because i have not fired in enough varied conditions past 1mile to see a pattern.
Personally i think the answer depends on the distance you are shooting, and the flatness of trajectory...and surely other things as well.
The above is based only on a flat fire situation shooting over FLAT ground.
YMMV,
JB
6. ### Richard338Well-Known Member
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126
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Aug 22, 2003
JB, why do you say it spends more time climbing?
I ran a trajectory calc on JBM's page using 3000fps and a BC of .7.
The apogee of a 1000 yard zeroed trajectory is at 540 yards, but that occurs at less than half the time for the 1000 yards trajectory because the bullet is going faster at first.
I added 20 mph head/tail winds and got predicted lower POI for headwind and raised POI for tailwind.
I had a clear example of a switching head/tail wind at 1000 yards a couple of weeks ago and definately found this to be the case.
Target rifle shooters have a memory trick "heads-up, and tails down" (as in aim up in a head wind, and aim lower for a tail wind).
The effect is small in any case but can get you.
7. ### jb1000brWell-Known Member
Messages:
1,307
Joined:
Jul 8, 2003
[ QUOTE ]
JB, why do you say it spends more time climbing?
I ran a trajectory calc on JBM's page using 3000fps and a BC of .7.
The apogee of a 1000 yard zeroed trajectory is at 540 yards, but that occurs at less than half the time for the 1000 yards trajectory because the bullet is going faster at first.
I added 20 mph head/tail winds and got predicted lower POI for headwind and raised POI for tailwind.
I had a clear example of a switching head/tail wind at 1000 yards a couple of weeks ago and definately found this to be the case.
Target rifle shooters have a memory trick "heads-up, and tails down" (as in aim up in a head wind, and aim lower for a tail wind).
The effect is small in any case but can get you.
[/ QUOTE ]
Richard -- first, forget computer models for this discussion.
I guess i should have been more specific, but left out something
seems TO ME the less flat the traj, the more likely to rise at 1K with a headwind.
It has been observed by those i trust with a 308 and 260, and myself with a 6br.
my 6.5-284 has not been noticably affected my any head/tailwinds encountered at 1K thus far.
IMO 3000fps with a .7BC is FLAT-A\$\$ traj and many rounds we use cant touch that.
in my mind, i see a flatter shooting round more likely to drop in a headwind, and a round with more drop getting loft.
Like i said -- it depends on too many things to make a rule like "heads up whatever down..."
Also, by "more time" i actually meant distance, and you just showed that was true, even with the flat shooter. 540 compared to 460 -- thats an 80yd diff
IMHO, and YMMV,
8. ### Richard338Well-Known Member
Messages:
126
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Aug 22, 2003
JB,
I noticed it with my 260Rem. I just ran the .7 BC 3000 fps, because I'm waiting for my new 7-270WSM to show up!
Even with 20mph head and tail wind, the computed difference is only .5MOA.
Usually you would never get that much difference in one day.
Where I shoot it is also more complicted because it is open behind us, a dip in the range between 200-400 yards, and there is a hill to the left of us.
We get all kinds of up/down drafts as well.
Nevertheless any decent long range cartridge shoots very flat from a geometry point of view.
I still say that the calcs are correct that a strong head wind will lower the POI.
The target rifle guys who told me this all shoot .308 with 155 grain bullets.
Anytime we are seeing enough to be important, there are probably up/down drafts involved as well.
9. ### jb1000brWell-Known Member
Messages:
1,307
Joined:
Jul 8, 2003
[ QUOTE ]
JB,
I noticed it with my 260Rem. I just ran the .7 BC 3000 fps, because I'm waiting for my new 7-270WSM to show up!
Even with 20mph head and tail wind, the computed difference is only .5MOA.
Usually you would never get that much difference in one day.
Where I shoot it is also more complicted because it is open behind us, a dip in the range between 200-400 yards, and there is a hill to the left of us.
We get all kinds of up/down drafts as well.
Nevertheless any decent long range cartridge shoots very flat from a geometry point of view.
I still say that the calcs are correct that a strong head wind will lower the POI.
The target rifle guys who told me this all shoot .308 with 155 grain bullets.
Anytime we are seeing enough to be important, there are probably up/down drafts involved as well.
[/ QUOTE ]
You definately have some updrafts there then /ubbthreads/images/graemlins/shocked.gif
the 260 and 308 info above came from a FLAT range, and the 6br was fired at williamsport 1KBR
the br rose almost 12" with a 15-20MPH head gust.
the 308 (168SMK) and 260(139scenar) rose 1.5-2.0moa with a seemingly constant 20MPH headwind.
I'll stick to my thinking, where you cant make up a hard and fast rule for every round and every range /ubbthreads/images/graemlins/smile.gif
Do update me if you see anything different, and i'll do the same /ubbthreads/images/graemlins/wink.gif
Cheers,
JB
10. ### Richard338Well-Known Member
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126
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Aug 22, 2003
Sounds good JB. Maybe I'll check McCoy's book tonight.
11. ### Richard338Well-Known Member
Messages:
126
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Aug 22, 2003
JB, I flipped through McCoy's book and found explanations that might fit both our observations.
He says that a pure head or tail wind (with no crosswind) lowers or raises (respectively) the trajectory by a small amount.
This agrees with the calcs I ran.
What could explain what you saw is this business of the vertical component from a cross wind.
If a righthand twist barrel is zeroed at in calm conditons, then wind from the left will push the bullets to the right and down.
Wind from the right will push the bullets left and up. If the wind slowly changed, the string of bullets would form a slanted line (about 20 deg for many bullets, twists etc), going from top left to bottom right.
So if you had wind from the left when you got zeroed, then the wind switched to a headwind, the new headwind would want to lower the trajectory (only a small amount). But the loss of the vertical component from the crosswind (which switched away) would be a bigger effect, and the net result would be a raised trajectory.
12. ### jb1000brWell-Known Member
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Jul 8, 2003
Rich -- at williamsport, anything is possible, and that is the prevailing direction, but i honestly dont remember. All i remember was my bud was centered up and down, then a head gust just as he shot and the bullet hit around 10" high
the crosswind though would have had to have well over 10" in the vert component to use that explaination, and i dont see that being possible.
JB
13. ### Richard338Well-Known Member
Messages:
126
Joined:
Aug 22, 2003
At 1000 yards a crosswind worth 3-4MOA (horizontal), would have about 10 inches of vertical. The angle of the line of shots on the target for different winds should be 15-20 degrees...
14. ### Charles AWell-Known Member
Messages:
480
Joined:
Nov 7, 2001
Alright I guess I should have thought about it and explained myself more. A headwind with the target perfectly(or very near) level with the shooter and the ground level from shooter to target will impact lower. A tailwind, higher. Now if you on high ground and shooting over a depression, valley, ravine, etc., a headwind will raise POI due to updrafts. Tailwind will lower.
In my expierence shooting 308's and 300win at LR, almost always a headwind will raise POI, a tailwind will lower POI. But at the "targets" I normaly shoot at I dont really worry abou it, as I haven't seen much more then 1/2MOA deflection either way. | 2,606 | 10,012 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.625 | 3 | CC-MAIN-2017-34 | latest | en | 0.926001 |
https://mikesdotnetting.com/article/208/practical-recursion-in-asp-net-web-pages-with-webmatrix | 1,713,219,362,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296817033.56/warc/CC-MAIN-20240415205332-20240415235332-00781.warc.gz | 368,198,657 | 16,028 | # Practical Recursion in ASP.NET Web Pages with WebMatrix
Recursive methods are those that call themselves, and they can be applied to solve a number of common practical problems, particularly ones that involve the management and display of hierarchical data. This article explores some of those scenarios and shows how to design recursive helpers and functions that can be used in Web Pages sites.
How do methods "call themselves"? Here's an example method called Add():
```@functions{
if(number > 0){
}
return 0;
}
}```
This method takes an int and then repeatedly calls itself reduces the number by 1 until it reaches 0, and returns the sum of all the reductions. So if you passed 4 into the method, it would return 10 (4 + 3 + 2 + 1). You can see this method at work in the Basic example in the article's accompanying code:
```@{
Page.Title = "Basic Example";
var input = Request["number"].AsInt();
}
<div>
Result:
@for(var i = input; i > 0; i--){
@i @(i == 1 ? " = " : " + ")
}
@result
</div>
<form method="post">
@Html.Label("Enter number: ", "number")
@Html.TextBox("number", input, new { size = "5" })
<input type="submit" />
</form>
@functions{
if(number > 0){
}
return 0;
}
}```
The recursive method features two mechanisms for preventing infinite loops resulting in a stack overflow: the first is a condition under which the method continues to recurse. In this case, the number must be greater than 0. This is usally termed the "base case". Second is a means to ensure that the base case is reached. In the Add method, that is achieved by reducing the method input value by one on each recursive call.
Hierarchical data is data that is structured in a tree-like manner. There is a single root. The root will have branches, which can have branches, which can have branches.... and so on infinitely.
Individual nodes in this type of structure can be represented in C# in the following manner:
```public class Node
{
public int Id { get; set; }
public int? ParentId { get; set; }
public string Name { get; set; }
}```
Each item has an Id and a name. It also optionally has a parent (represented by a nullable int). Roots do not have parents so their ParentId value will be null. All other items in the structure should have a ParentId value that matches the Id of an existing node. This pattern is commonly used to express menu systems, tree views, threaded discussion boards and so on. Here is how you might use the Node class to build a menu:
```var menu = new List<Node> {
new Node {Id = 1, ParentId = null, Name = "File"},
new Node {Id = 2, ParentId = 1, Name = "Open"},
new Node {Id = 3, ParentId = 1, Name = "Save"},
new Node {Id = 4, ParentId = 1, Name = "Export"},
new Node {Id = 5, ParentId = 4, Name = "To PDF"},
new Node {Id = 6, ParentId = 4, Name = "To Word"},
new Node {Id = 7, ParentId = 6, Name = "Doc"},
new Node {Id = 8, ParentId = 6, Name = "DocX"},
new Node {Id = 9, ParentId = 1, Name = "Exit"}
};```
"File" is the root node. It has no parent. "Open", "Save", "Export" and "Exit" are all descendants of the root node. "To PDF" and "To Word" are both descendents of "Export", and "Doc" and "Docx" are children of "To Word". What is needed now is a way to generate the HTML to display this menu. A helper is the best way to cater for that:
```@helper BuildMenu(List<Node> menu, int? parentid = null, int level = 0) {
var items = menu.Where(m => m.ParentId == parentid);
if (items.Any()) {
if(items.First().ParentId.HasValue){
level++;
}
foreach (var item in items) {
<div>
@for (var i = 0; i < level; i++) {
@:
}
@item.Name
</div>
}
}
}```
The method is called within the page as follows:
`@BuildMenu(menu)`
All of the Nodes are passed in to the method, which then extracts just those nodes with a ParentId value equal to the one passed in. On the first call, this value is the default value for the paramter - null. So the method extracts the root node if there is one. The the "base case" is established:
`if (items.Any()) {`
If any nodes meet the criteria in the LINQ Where clause, the first item in the resulting sequence is examined to see if it has a ParentId value. If it does, the value of the level variable is increased by one since this is not the root node. Then each node in the sequence is rendered in a div, and then passed into the BuildMenu method itself - resulting in the hierarchical tree being walked until there are no more nodes. As each level is traversed, the number of non-breaking spaces is added to the output to indent the display:
The Menu example in the accompanying sample site includes a second root node to illustrate how to display multiple multiple menus horizontally.
### Working with hierarchical data from a database
You are most likely to store your hierarchical data in a database, so this section looks at how to manage that. The data will represent a simple Tree View. It is stored in a table called Nodes with the following schema:
```Id int NOT NULL IDENTITY Primary Key
ParentId int
Text nvarchar(50)
DisplayOrder int DEFAULT 0```
The schema bears an obvious resemblence to the Node class definition with the addition of a new property - DisplayOrder. This will be used to determine the order in which items are listed under their parent when CRUD is added later. The helper that is used to render the Tree View to the browser is very similar to the previous one:
```@helper BuildTreeView(IEnumerable<dynamic> data, int? parentid = null, int level = 0) {
var nodes = data.Where(n => n.ParentId == parentid).OrderBy(n => n.DisplayOrder);
if (nodes.Any()) {
if(nodes.First().ParentId != null){
level++;
}
foreach (var node in nodes) {
@node.Text
</div>
@BuildTreeView(data, node.Id, level);
}
}
}```
This helper expects the result of a Database.Query call to be passed into it. The two other differences from the Menu helper are that the LINQ query has an OrderBy method chained to the Where method, and the indentation for each level is managed by applying a padding to the div rather than preceding each item with non-breaking spaces.
My first introduction to recursion came by way of a classic ASP threaded discussion board. It worked, but it was terribly slow in terms of performance. This was because the recursive method included a call to the database to get records for a specific level each time the method executed. Calls to databases are relatively expensive, so they should be minimised. Ideally, if you want to iterate over data obtained from a database multiple times in your code, you should only ever obtain the data once. That is exactly how this helper has been designed; the code to obtain the data is only ever executed once, and then all the data is passed into the method:
```@{
var db = Database.Open("TreeView");
var treeView = db.Query("SELECT * FROM Nodes");
}
@BuildTreeView(treeView)```
Here is a simple form for adding new nodes:
```<h1>Add Node</h1>
<div>
@message
</div>
<form method="post">
<div>
@Html.Label("Text: ", "Text")
@Html.TextBox("Text", Request["Text"])
@Html.ValidationMessage("Text")
</div>
@if(nodes.Any()){
<div>
@Html.Label("Parent: ", "ParentId")
@Html.DropDownList("ParentId", nodes)
</div>
}
<div>
@Html.Label("Display Order: ", "DisplayOrder")
@Html.TextBox("DisplayOrder: ", displayOrder, new { size = 2 })
@Html.ValidationMessage("DisplayOrder")
</div>
<div>
<input type="submit" />
</div>
</form>```
There is a text box for the name of the node and a dropdown list includes all existing nodes so that a parent can be selected. Finally, the display order can be provided in another text box. It is prepopulated with the default value of 0. Usually, items are retrieved from the database in the order that they went in if there is no ORDER BY clause specified, but that cannot be relied upon. In any event, if you delete nodes and subsequently add new ones, you need some way of controlling the order in which they are displayed - hence the DisplayOrder column and the OrderBy method in the LINQ query.
Here is the code block at the top of the Add page:
```@{
var db = Database.Open("TreeView");
var sql = string.Empty;
var message = string.Empty;
var displayOrder = Request["DisplayOrder"].IsInt() ? Request["DisplayOrder"].AsInt() : 0;
if(IsPost && Validation.IsValid()){
sql = "INSERT INTO Nodes (ParentId, Text, DisplayOrder) VALUES (@0, @1, @2)";
var text = Request["Text"];
var parentId = Request["ParentId"].IsEmpty() ? DBNull.Value : (object)Request["ParentId"];
db.Execute(sql, parentId, text, displayOrder);
message = text + " added";
}
sql = "SELECT * FROM Nodes";
var treeView = db.Query(sql);
var nodes = treeView.Select(item => new SelectListItem {
Value = item.Id.ToString(),
Text = item.Text,
Selected = item.Id.ToString() == Request["ParentId"]
});
}```
Validators are used to ensure that a value of the Text property is provided and that it doesn't exceed the size of the database column. Also, the DisplayOrder value must be an integer. If validation passes, the data is inserted into the database. If the ParentId value is not provided, DBNull.Value is inserted into the database to ensue a null value. A message is dsiplayed ot indicate which node was successfully added to the database and then the fresh node data is obtained from the database and used to populate the dropdown list ready for the next node to be added.
### Editing and Deleting Nodes
If you are familiar with basic data access using the Database helper, there is nothing particularly interesting about adding nodes. Editing nodes is straightforward too, so long as you are happy for any child nodes to move across with a parent node if you want to change its parent. Deleting nodes is a slightly more complex operation unless you are happy for orphans to be left in the database. Usually this is not a good idea, so if you provide the user with the ability to delete a node, you need to "walk its tree" to identify all child nodes and delete them too.
First, you need to select a node to edit. An easy way to enable users to do this is to display the tree with each node acting as a link. So the previous helper just needs a little modification to the HTML that's generated:
```@helper BuildTreeView(IEnumerable<dynamic> data, int? parentid = null, int level = 0) {
var nodes = data.Where(n => n.ParentId == parentid);
if (nodes.Any()) {
if(nodes.First().ParentId != null){
level++;
}
foreach (var node in nodes.OrderBy(n => n.DisplayOrder)) {
</div>
@BuildTreeView(data, node.Id, level);
}
}
}```
Clicking on a node takes the user to an Edit form for the selected node, which is identified from the query string value:
```@if((Request["id"].IsInt() && !IsPost) || (IsPost && !Validation.IsValid())){
<h1>Edit Node</h1>
<form method="post">
<div>
@Html.Label("Text: ", "Text")
@Html.TextBox("Text", nodeText)
@Html.ValidationMessage("Text")
</div>
@if(parentId.HasValue){
<div>
@Html.Label("Parent: ", "ParentId")
@Html.DropDownList("ParentId", nodes)
</div>
}
<div>
@Html.Label("Display Order: ", "DisplayOrder")
@Html.TextBox("DisplayOrder", displayOrder, new {size = 2})
@Html.ValidationMessage("DisplayOrder")
</div>
<div>
<input type="submit" name="Action" value="Delete" />
<input type="submit" name="Action" value="Edit" />
</div>
</form>
}```
The form is displayed only there is a valid ID in the query string and the form has not been posted, or if the form has been posted but failed validation. Details of the selected node prepopulate the form for editing. The user has two options - edit or delete. Here's the entire code block at the top of the page:
```@{
var db = Database.Open("TreeView");
var nodeSql = "SELECT * FROM Nodes";
IEnumerable<SelectListItem> nodes = null;
var nodeText = string.Empty;
var displayOrder = 0;
int? parentId = null;
var data = Enumerable.Empty<dynamic>();
var refreshData = true;
if(Request["id"].IsInt() && !IsPost){
data = db.Query(nodeSql);
var id = Request["id"].AsInt();
nodeText = data.First(d => d.Id == id).Text;
parentId = data.First(d => d.Id == id).ParentId;
displayOrder = data.First(d => d.Id == id).DisplayOrder;
nodes = data.Select(item => new SelectListItem {
Value = item.Id.ToString(),
Text = item.Text,
Selected = item.Id == parentId
});
refreshData = false;
}
if(IsPost && Validation.IsValid()){
if(Request["Action"] == "Edit"){
var sql = "UPDATE Nodes Set ParentId = @0, Text = @1, DisplayOrder = @2 WHERE Id = @3";
nodeText = Request["Text"];
displayOrder = Request["DisplayOrder"].AsInt();
var pId = Request["ParentId"].IsEmpty() ? DBNull.Value : (object)Request["ParentId"];
db.Execute(sql, pId, nodeText, displayOrder, Request["id"]);
}
if(Request["Action"] == "Delete"){
data = db.Query(nodeSql);
var idList = new List<int>();
var ids = GetIds(data, ref idList, Request["id"].AsInt());
var parms = ids.Select((s, i) => "@" + i.ToString()).ToArray();
var inclause = string.Join(",", parms);
var sql = "DELETE FROM Nodes WHERE Id IN ({0})";
var idsArray = ids.Select(i => i.ToString()).ToArray();
db.Execute(string.Format(sql, inclause), idsArray);
}
}
if (refreshData) {
data = db.Query(nodeSql);
}
}```
There are three main sections in the code, each responsible for managing a different stage in the process. The first block sets the initial page up with the declaration of a number of variables and validation rules. Right at the bottom of the code block, you can see the most recent node data being retrieved from the database if the refreshData boolean flag is true. When the page is first requested (and after the edit/delete form has been posted) a tree view is displayed:
```@if(Request["id"].IsEmpty() || IsPost){
<h1>Select Node</h1>
@BuildTreeView(data)
}
```
The next block of code executes when a link in the tree view is clicked - the ID of the node is present in the query string, but no form has been posted. All of hte nodes are retireved form the database and then the one that is subject to being edited is identified and used to populate the dropdown list to allow for selection of a parent (if that is to change) and the rest of the form. There is no need to retrieve this data again for the page, so a boolean flag is set to false to prevent the data selection at the bottom of the block from executing.
The last block of code deals with successful form submission. If the Edit button was clicked, a simnple UPDATE statement is executed to apply any submitted changes ot the databse. The code that manages a DELETE is a bit more complex. In this case, the business rules dictate that no orphaned nodes should be left in the database, so the selected node and all its children must be deleted. In order to achieve that, all the IDs of the affected nodes need to be collected and passed in to an IN clause as part of the SQL DELETE statement. I have already written about how to manage IN clauses with the Database helper, so if you are unclear on that part, please refer to the previous article.
The code to obtain the IDs consist of another recursive method wrapped as a Razor function:
```@functions {
public static List<int> GetIds(IEnumerable<dynamic> data, ref List<int> ids, int? id = null, int? parentId = null ){
var items = id.HasValue ? data.Where(i => i.Id == id) : data.Where(i => i.ParentId == parentId);
if(items.Any()){
foreach(var item in items){
GetIds(data, ref ids, null, item.Id);
}
}
return ids;
}
} ```
This method has a return type of List<int> which will contain all the IDs. It also features a parameter marked with the ref keyword. The parameter is a List<int>. The ref keyword ensure that a reference to an initialised List<int> is passed in to the method, which means that it will maintain state externally to the method each time the method is called. If the ref keyword was not used, the List<int> would go out of scope each time the method is called and all values except one will be lost. Beyond that, the structure of the method is very similar to the helpers that you have seen already. When the method is first called, it is presented with a List<int> by ref, and the ID of the node to be deleted. On each subsequent method call, all items with a parent id that matches that of the current node are obtained - i.e. its children. Finally, the code converts the List<int> to a comma seaprated string for the database helper.
### Performance
The sample site that accompanies this article includes the scenarios covered in this article along with a basic illustration of how a threaded discussion board might work. Recursion is a very useful tool for some scenarios, but it comes at a price. As I mentioned earlier, you need to be careful what you actually do within a recursive function as calls to external processes such as databases can be expensive and might slow your application right down. Also, recursion is a bit like Regex. When people first encounter it, they are impressed with its power and then try to apply it to every problem they come up against. Just as straightforward string manipulation can often be a better solution than Regex, so can a loop be a better bet than recursion. For example, the first recursive method that you were shown can just as easily be replaced with a simple while loop:
```@functions {
public static int AddWithoutRecursion(int number) {
var temp = 0;
while (number > 0) {
temp += number;
number --;
}
return temp;
}
}```
Recursion uses a stack. Each call to the recursive method is added to the stack, and when recursion is complete, all of the calls are executed. Stacks are limited in size, although you may be able to increase them if you have enough RAM allocated to your shared hosting App pool (or if you have control over the server). If a stack exceeds the size allocated to it, the result is a StackOverflowException:
This was achieved by entering 9000 into the Basic example using the recursive Add method while debugging against IIS Express in Visual Studio. If you try this from WebMatrix, the site will just stop. If the same number is processed using the AddWithoutRecursion method, there is no such problem. If you are not certain many times your method may need to recurse, you should look at alternative approaches using loops and Stacks. However, if you know that your menu or tree view will never approach the kind of size that may cause performance issues, or that the number of posts in a thread will remain within manageable limits, recursion is a neat solution for managing hierarchical data.
The sample site featuring all the examples illustrated here can be found at GitHub. | 4,331 | 18,465 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.78125 | 3 | CC-MAIN-2024-18 | latest | en | 0.862334 |
https://file.scirp.org/Html/2-7800484_83166.htm | 1,716,578,762,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058736.10/warc/CC-MAIN-20240524183358-20240524213358-00002.warc.gz | 202,783,359 | 20,193 | Secure Passwords Using Combinatorial Group Theory
Journal of Information Security
Vol.09 No.02(2018), Article ID:83166,14 pages
10.4236/jis.2018.92011
Secure Passwords Using Combinatorial Group Theory
Gilbert Baumslag1, Benjamin Fine2, Anja Moldenhauer3, Gerhard Rosenberger3
1Department of Mathematics, City University, New York, USA
2Department of Mathematics, Fairfield University, Fairfield, CT, USA
3Department of Mathematics, University of Hamburg, Hamburg, Germany
Received: February 7, 2018; Accepted: March 18, 2018; Published: March 21, 2018
ABSTRACT
Password security is a crucial component of modern internet security. In this paper, we present a provably secure method for password verification using combinatorial group theory. This method relies on the group randomizer system, a subset of the MAGNUS computer algebra system and corrects most of the present problems with challenge response systems, the most common types of password verification. Theoretical security of the considered method depends on several results in asymptotic group theory. We mention further that this method has applications for many other password situations including container security.
Keywords:
Password Security, Combinatorial Group Theory, Free Group Cryptography, Group Randomizer System
1. Introduction
This material essentially appeared in [1] , but we feel it is an important application that should be more widely publicized and is a perfect entry for the present special volume on Cryptography and Internet security.
Secure password identification is a crucial component of modern internet security. Password verification is essential and this requires a backup system. Backup password security is handled most often by a challenge response system (see [2] ) accompanying the password. In the simplest systems, this takes the form of secondary password questions such as the prover’s mother’s maiden name or place of birth. There are many inherent difficulties with these types of challenge response systems such as the trivial problem of the provers remembering their responses. More critical is the problem that this type of information for many people is readily available and easily found or guessed by would-be attackers or eavesdroppers. Challenge response systems are also subject to middleman attacks and replay attacks. There have been many attempts to alleviate these problems, including zero-knowledge password proofs and challenged responses somewhat based on RSA as well as timed out responses (see CRAM-MD5, Password Authenticated Key Agreement, [2] [3] ).
This article presents an alternative method for challenge response password verification using combinatorial group theory. Further, this method is provably secure. It depends upon the theoretical and practical difficulty of solving the search membership problem within a given finitely presented group without knowing the presentation and the difficulty of solving systems of equations within free groups. This latter problem has been proved to be NP-hard. This alternative method uses the group randomizer system; a computer program that is a subset of MAGNUS, a much larger computer algebra system, designed to handle algorithmic problems in combinatorial group theory. MAGNUS was developed at CAISS, the Center for Algorithms and Interactive Scientific Software, a research laboratory housed at City College of the City University of New York and under the direction of the first author. The group randomizer system can be placed on a simple hand held computer device presently under development at CAISS. The system can also be used from computer to computer.
The group theoretic techniques have several major advantages over other challenge response systems. Using standard authentication terminology, the password presenter will be denoted the prover while the presentee is the verifier. The methods we present can be used for two-way authentication, that is the same method authenticates both the prover to the verifier and the verifier to the prover.
From the standpoint of cryptology, the method is a symmetric key authentication protocol. In its application, each prover has a standard password that is a common shared secret with the verifier. In addition, each prover is assigned a finitely presented group G. This group is called the challenge group. The total common shared secret between prover and verifier consists of $\left(P,G\right)$ where P is a standard password and G is the challenge group. The challenge group will provide an unlimited set of back-up challenges to the password. These challenges are in the form of group theoretical questions concerning G. The assignment of the challenge group to a given prover will be done randomly by the group randomizer system which we will explain. Cryptographically, we assume the adversary can steal the encrypted form of the group theoretic responses. From a security viewpoint, this does not present a problem. Each set of back-up challenges forms a virtual one time code as we will explain in the paper. Therefore, the adversary must steal three things―the original password, the challenge group and the group randomizer. Hence there is almost total security in this challenge response system. Further, there is an infinite supply of finitely presented groups to use as challenge groups and an infinite supply of challenge response questions that never have to be duplicated. These will be explained in the paper. Finally in distinction to other backup password protocols the group theoretic method is a two-way protocol between the prover and the verifier; while the verifier authenticates the prover’s password, simultaneously the prover authenticates that he or she is dealing with the true verifier.
A major advantage of this technique it that it is provably secure. The proof of its security depends upon asymptotic group theory which we explain in Section 6. A result of Lysenok [4] implies that stealing the challenge group is NP-hard while a result of Jitsukawa [5] says that the asymptotic density of using homomorphisms (see Section 6) to attack the group randomizer protocol is zero.
In the next section, we provide a brief primer on combinatorial group theory and then a description of the group randomizer system. We then present several variations on how the group randomizer system can be used for secure password verification protocols. After this with we give the security model showing that it is provably secure.
Finally we describe how the group randomizer system and password methods can be used as a secure lock for container security.
This group randomizer system password security approach is part of a large program to use computational combinatorial group theory as a tool in secure data storage and data identification.
Sadly the first author, who inspired much of this work, passed away during the preparation of the paper. We thank him posthumously for his many ideas.
2. Finitely Presented Groups and Combinatorial Group Theory
Combinatorial group theory attempts to study groups via group presentations. A group presentation can be thought of as an encoded method to describe a given group.
A group presentation for a group G consists of a set of generators X for G and a set R of defining relators on the generators X. In this case we write $G=〈X;R〉$ . For purposes of this paper and the challenge response password authentication protocols we propose, we may assume that G is a finitely presented group. By this we mean that both the set of generators X and the set of defining relators are finite. The books by Baumslag [6] , Lyndon and Schupp [7] , Magnus, Karrass and Solitar [8] Camps, GrRebel and Rosenberger [9] are standard references for this material. Another reference for the use of combinatoiral group theory in cryptography is the book by Baumslag, Fine, Kreuzer and Rosenberger [10] .
Consider a finite alphabet, $X=\left\{{x}_{1},\cdots ,{x}_{n}\right\}$ and the formal inverses $\left\{{x}_{1}^{-1},\cdots ,{x}_{n}^{-1}\right\}$ . Then the set $X=\left\{{x}_{1},\cdots ,{x}_{n}\right\}$ is a set of generators for a group G if every element $g\in G$ has an expression as a word in the generators and their inverses. The identity is considered the empty word. We do not assume that this expression is unique. A relator on X is a word $W\left({x}_{1},\cdots ,{x}_{n}\right)$ which represents the identity in G. A relator of the form ${x}_{i}{x}_{i}^{-1}$ or ${x}_{i}^{-1}{x}_{i}$ is called a trivial relator. A set $R=\left\{{R}_{1},\cdots ,{R}_{k}\right\}$ of words on X is a complete set of defining relators if every relator W can be transformed into the empty word by finitely many insertions and deletions of elements of R and trivial relators. Two words ${W}_{1},{W}_{2}$ on the generators represent the same group element G if and only if ${W}_{1}$ can be transformed into ${W}_{2}$ by insertions and deletions of relators in R and trivial relators. If X is a set of generators for G and R is a set of defining relators we say G has the presentation $G=〈X;R〉$ . A finite presentation will always define a group (see [8] ) for which this is the presentation. A set of words ${W}_{1},\cdots ,{W}_{m},\cdots$ in one-to-one correspondence with the elements of G is said to determine a set of normal forms for G.
Presentations are in no way unique, however in principle all group theoretic questions about G should be answerable given a presentation. For a group presentation, the word problem asks whether given any word on X, is there an algorithm to determine in finitely many steps whether this word represents the identity in G. In general the word problem is undecidable. That is there exist group presentations for which it can be proved that no such algorithm exists (see [6] ). A particularly nice class of groups, which have both normal forms and solvable word problems, are the automatic groups (see [11] ).
For a group G and a subgroup H of G the membership problem (also called the generalized word problem) is the problem of determining algorithmically whether a given word W, written in terms of the generators of G, lies in the subgroup H. As with the word problem, the membership problem is in general undecidable. Clearly a solution to the membership problem, in a given group, implies a solution to the word problem.
Fundamental to combinatorial group theory is the concept of a free group. Let A be a set. Then, a group F is free on A if every mapping $f:A\to G$ , where G is a group, can be extended to a unique homomorphism of F to G. We denote this by $F\left[A\right]$ . A group is free if it is free on some set A. It can be proved that given a set A, there exists a free group on A and further if two sets ${A}_{1}$ and ${A}_{2}$ have the same size then the corresponding free groups $F\left[{A}_{1}\right]$ and $F\left[{A}_{2}\right]$ are isomorphic. If $A=\left\{{x}_{1},\cdots ,{x}_{n}\right\}$ is a finite set, we say that $F\left[A\right]$ is a free group of rank n and sometimes denote this by ${F}_{n}$ .
From the viewpoint of group presentations free groups are groups with a presentation with an empty set of defining relators. If F is free on $X=\left\{{x}_{1},\cdots ,{x}_{n}\right\}$ then there are no nontrivial relators on X and a presentation for F is $F=〈{x}_{1},\cdots ,{x}_{n}〉$ . In this case the elements of F can be considered as reduced words on the alphabet ${\left\{{x}_{1},\cdots ,{x}_{n}\right\}}^{±1}$ . The identity element is considered as the empty word. Reduced means that we can cancel any occurrences of ${x}_{i}{x}_{i}^{-1}$ or ${x}_{i}^{-1}{x}_{i}$ . It is clear that each word has a unique reduced form and hence the word problem for F is solvable.
A well-known theorem due to Nielsen and Schreier (see [6] ) says that a proper subgroup of a free group is again a free group, of course on a different set of generators. For our purposes what is important is that a finitely generated subgroup of a free group is completely determined by a finite set of words which have only trivial cancellation between them.
Essential to the group randomizer system protocols is that algorithmically both the word problem and the membership problems are solvable in free groups. That is given a free group $F=F\left({x}_{1},\cdots ,{x}_{n}\right)$ of finite rank n we can decide algorithmically whether or not a word $W=W\left({x}_{1},\cdots ,{x}_{n}\right)$ represents the identity; given two words ${W}_{1}={W}_{1}\left({x}_{1},\cdots ,{x}_{n}\right)$ , ${W}_{2}={W}_{2}\left({x}_{1},\cdots ,{x}_{n}\right)$ we can decide algorithmically whether ${W}_{1}={W}_{2}$ in F; given a finitely generated subgroup H of F and a word W, we can decide algorithmically whether the element defined by W lies in H. Details on these algorithmic procedures can be found in [6] [8] [9] .
3. The Group Randomizer System
The group randomizer system is a computer program that can handle several elementary tasks involving finitely presented groups. It is a subset of MAGNUS, a large computer algebra system, developed at CAISS―the Center for Algorithms and Interactive Scientific Software. The program MAGNUS is specifically designed to handle computations and algorithmic problems in combinatorial group theory. At present there are various versions of the group randomizer, including a portable hand held version now under development.
The scope of a particular group randomizer system will depend on the type of login protocol or cryptographic protocol desired. At the most basic level the group randomizer system has the ability to do do the following things:
1) Recognize a finite presentation of a finitely presented group with a solvable word problem and manipulate arbitrary words in the alphabet of generators according to the rewriting rules of the presentation. In particular if the group is automatic the group randomizer can rewrite an arbitrary word in the generators in terms of its group normal form.
2) Given a finite presentation of a group G, with a solvable word problem, recognize whether two free group words have the same value in the given group when considered in terms of the given generators of the group.
3) Randomly generate free group words on an alphabet of any finite size.
4) Recognize and store sets of free group words ${W}_{1},\cdots ,{W}_{k}$ on an alphabet ${x}_{1},\cdots ,{x}_{n}$ and rewrite words $W\left({W}_{1},\cdots ,{W}_{k}\right)$ as the corresponding word in ${x}_{1},\cdots ,{x}_{n}$ .
5) Given a free group of finite rank on ${x}_{1},\cdots ,{x}_{n}$ and a set of words ${W}_{1},\cdots ,{W}_{k}$ on ${x}_{1},\cdots ,{x}_{n}$ solve the membership problem in F relative to $H=〈{W}_{1},\cdots ,{W}_{k}〉$ , the subgroup of F generated by ${W}_{1},\cdots ,{W}_{k}$ .
6) Given a stored finitely presented group or a stored set of free group words, the randomizer can accept a random free group word and rewrite it as a normal form in the finitely presented group in the former case or as a word in the ambient free group in the latter case.
7) Evaluate a group word on a set of $n×n$ rational matrices.
In the next section we show how this can be used for secure password verification in both directions―verifier to prover and prover to verifier.
We now present several variations on secure password verification using the group randomizer. First we give an overall outline of the protocol.
1) General Outline of the Authentication Protocol
At a theoretical level, this protocol is a symmetric key cryptographic authentication protocol. Both the prover and verifier use a single private key to both encrypt and decrypt within the authentication process. At first, the prover and verifier must communicate directly, either face-to-face or by a public key method, to set the private shared secret. This is the model now used for most password/password back-up schemes. We assume that both the user and verifier have a group randomizer system. For security analysis, we assume that an adversary or eavesdropper has access to the encrypted form of the transmission but is passive in that the adversary will not change any transmissions.
Step (1): The prover and verifier communicate directly to set up a common shared secret $\left(P,G\right)$ where P is a standard password and G is a challenge group. Each prover’s challenge group is unique to that prover. The challenge group is a finitely presented group with a solvable word problem and satisfying the strong generic free group property (see Section 5). The password is chosen by the prover while the challenge group is randomly chosen by the group randomizer system.
Step (2): The prover presents the password to the verifier. The group randomizer of the verifier presents a group theoretic “question” (see parts (2) and (3)) concerning the challenge group G to the prover. The assumption is that this “question” is difficult in the sense that it is infeasible to answer it if the group G is unknown. The question is then answered by the group randomizer. This is repeated a finite number of times. If the answers are correct the prover (and the password) is verified.
Step (3): The protocol is then repeated from the viewpoint of the prover, authenticating the verifier to the prover.
2) Free Subgroup Method
The first method we present uses a free group as the basic group theoretic object.
We assume that both the prover and the verifier has a group randomizer. Each prover has a standard password. Suppose that F is a free group on $\left\{{x}_{1},\cdots ,{x}_{n}\right\}$ . The prover’s password is linked to a finitely generated subgroup of a free group given as words in the generators―that is the prover’s password is linked to ${W}_{1},\cdots ,{W}_{k}$ where each ${W}_{i}$ is a word in ${x}_{1},\cdots ,{x}_{n}$ . The group $G=〈{W}_{1},\cdots ,{W}_{k}〉$ is called the challenge group. In general, $k\ne n$ . The prover doesn’t need to know the generators. The randomizer can randomly choose words from this subgroup and then freely reduce them. The verifier has the challenge group or subgroup also stored in its randomizer.
The prover submits his or her standard password to the verifier. This activates the verifier’s randomizer to the prover’s set of words. The verifier now submits a random free group word on ${y}_{1},\cdots ,{y}_{k}$ to the prover’s randomizer say $W\left({y}_{1},\cdots ,{y}_{k}\right)$ . The prover’s randomizer treats this as $W\left({W}_{1},\cdots ,{W}_{k}\right)$ and then reduces it in terms of the free group generators ${x}_{1},\cdots ,{x}_{n}$ and rewrites it as ${W}^{\star }\left({x}_{1},\cdots ,{x}_{n}\right)$ . The verifier checks that this is correct―that is $W\left({W}_{1},\cdots ,{W}_{k}\right)={W}^{\star }\left({x}_{1},\cdots ,{x}_{n}\right)$ on the free group on ${x}_{1},\cdots ,{x}_{n}$ . If it is the verifier continues and does this three times (or some other finite number of times). There is one proviso. A challenge word or submitted word can never be reused. The prover’s randomizer will recognize if a presented challenge word has been submitted previously and reject it. This is a further authentication to the prover of the verifier and directly hinders middle man attacks.
To verify that the verifier is legitimate the process is repeated from the prover’s randomizer to the verifier.
An attacker only has access to the transmitted words. Given a series of free group words, reconstructing the subgroup involves solving systems of equations in free groups. Solving such systems has been shown to be NP-hard (see Section 6). To prevent an attacker using an already used word to gain access the group randomizer system allows a free group word, submitted as a challenge word, to be used only once. If an attacker gets access to the verifier and submits an already submitted word or vice versa from the prover this will red flag the attempt. We also suggest that if there is a previously used word, indicating perhaps an attack, the group randomizer should change the prover’s group. The beauty of this system is that this can be done extremely easily―change several of the words for example. Essentially this presents an essential one-time keypad each time the prover presents the password. Hence there is very strong security in this back-up system. The map ${y}_{i}\to {W}_{i}$ is a homomorphism and an attacker can manipulate various equations in an attempt to solve. Presumably if there are enough equations the words ${W}_{1},\cdots ,{W}_{k}$ can be discovered. However in section 6 we will present a security proof based on several results in asymptotic group theory showing that this can not happen with asymptotic density one.
We suggest a noise/diffusion enhancement. The prover’s challenge group generator words ${W}_{1},\cdots ,{W}_{k}$ are indexed. With each use the randomizer applies a random permutation $\varphi$ on $\left\{1,\cdots ,k\right\}$ to scramble the indices. These permutations are coded and stored both in the prover’s randomizer and the verifier’s. These coded permutations are set at the time of initialization of the protocol and become part of the common shared secret. This prevents a length based attack by an eavesdropper since discovering for example what ${W}_{37}$ is, is of no use since it will be indexed differently for the next use. The coded permutation is sent as part of the challenge.
3) General Finitely Presented Group Method
Rather than working with an ambient free group we can work with a given finitely presented group with a solvable membership problem. Let $G=〈X;R〉$ be the group. As before we assume that the group G has a solvable word problem and satisfies the strong generic free group property. Further, as before, we assume that both the prover and the verifier has a group randomizer. Each prover has a standard password. Suppose that $X=\left\{{x}_{1},\cdots ,{x}_{n}\right\}$ and F is a free group on $\left\{{x}_{1},\cdots ,{x}_{n}\right\}$ . The prover’s password is linked to a finitely generated subgroup of G, again given as words in the generators X. That is the prover’s password is linked to ${W}_{1},\cdots ,{W}_{k}$ where each ${W}_{i}$ is a word in ${x}_{1},\cdots ,{x}_{n}$ . As before, $k\ne n$ . The randomizer can randomly choose words from this subgroup and then reduce them via the finite presentation. The verifier has the group or subgroup also stored in its randomizer.
The remainder of the procedure is exactly the same as in the free group case. The prover submits his or her standard password to the verifier. This activates the verifier’s randomizer to the prover’s set of words. The verifier now submits a random free group word on ${y}_{1},\cdots ,{y}_{k}$ to the prover’s randomizer say $W\left({y}_{1},\cdots ,{y}_{k}\right)$ . The prover’s randomizer treats this as $W\left({W}_{1},\cdots ,{W}_{k}\right)$ and rewrites it as ${W}^{\star }\left({x}_{1},\cdots ,{x}_{n}\right)$ . The verifier checks that this is correct, that is $W\left({W}_{1},\cdots ,{W}_{k}\right)={W}^{\star }\left({x}_{1},\cdots ,{x}_{n}\right)$ with equality this time in the group G. If it is true then the verifier continues and does this three (or some other finite number) of times. There is one proviso. The verifier submits a word to the prover only once so that a submitted word can never be reused. The prover’s randomizer will recognize if it has ( this is a verification to the prover of the verifier).
To authenticate that the verifier is legitimate the process is repeated from the prover’s randomizer to the verifier.
As in the free group method, an attacker has access only to the transmitted words. Given a series of group words it is infeasible to reconstruct the group. Further, as in the free group method, a given challenge response word is to be used only once. Since we assume that the group has the strong generic free group property, it follows that previous challenge words cannot be used to discover the challenge group or subgroup of the challenge group.
5. The Strong Generic Free Group Property
Part of the theoretical security of the group randomizer protocols depends upon the strong generic free group property and asymptotic density. Asymptotic density is a general method to compute densities and/or probabilities on infinite discrete sets where each individual outcome is tacitly assumed to be equally likely. The origin of asymptotic density lie in the attempt to compute probabilities on the whole set of integers where each integer is considered equally likely. The method can also be used where some probability distribution is assumed on the elements. It has been effectively applied to determining densities within infinite discrete finitely generated groups where random elements are considered as being generated from random walks on the Cayley graph of the group. The paper by Borovik, Myasnikov and Shpilrain [12] provides a good general description of this method in group theory. Let $\mathcal{P}$ be a group property and let G be a finitely generated group. We want to determine the measure of the set of elements which satisfy $\mathcal{P}$ . For each positive integer n let ${B}_{n}$ denote the n-ball in G. Let $|{B}_{n}|$ denote the actual size of ${B}_{n}$ (which is an integer since G is finitely generated) or the measure of $|{B}_{n}|$ if a distribution has been placed on the elements of G. Let S be the set of elements in G satisfying $\mathcal{P}$ . The asymptotic density of S is then
$\underset{n\to \infty }{lim}\frac{|S\cap {B}_{n}|}{|{B}_{n}|}$
provided this limit exists. We say that the property $\mathcal{P}$ is generic if the asymptotic density of the set S of elements satisfying $\mathcal{P}$ is one.
This concept can be easily extended to properties of finitely generated subgroups, We consider the asymptotic density of finite sets of elements that generate subgroups that have a considered property. For example to say that a group has the generic free group property we mean that
$\underset{m,n\to \infty }{lim}\frac{|{S}_{m}\cap {B}_{m,n}|}{|{B}_{m,n}|}=1$
where ${S}_{m}$ is the collection of finite sets of elements of size m that generate a free subgroup and ${B}_{m,n}$ is the collection of m-element subsets within the n-ball. We refer to the papers [12] and [13] for terminology and further definitions.
We say that a group G has the generic free group property if a finitely generated subgroup is generically a free group. For example a result of Epstein [14] says that the group $GL\left(n,R\right)$ satisfies the generic free group property. A group G has the strong generic free group property if given randomly chosen elements ${g}_{1},\cdots ,{g}_{n}$ in G then generically they are a free basis for the free subgroup they generate. Jitsukawa [5] proved that free groups have the strong generic free group property. That is given k random elements ${W}_{1},\cdots ,{W}_{k}$ in the free group on ${y}_{1},\cdots ,{y}_{n}$ then with asymptotic density one ${W}_{1},\cdots ,{W}_{k}$ are a free basis for the subgroup they generate. We compare this with the Nielsen-Shreier theorem that says that ${W}_{1},\cdots ,{W}_{k}$ generate a free group. In the context of the group randomizer protocols the strong generic free group property implies that if ${V}_{1}\left({y}_{1},\cdots ,{y}_{m}\right),\cdots ,{V}_{k}\left({y}_{1},\cdots ,{y}_{m}\right)$ have already been presented as challenge words then the density is zero that a new challenge word $V\left({y}_{1},\cdots ,{y}_{m}\right)$ lies in the subgroup generated by ${V}_{1},\cdots ,{V}_{k}$ and hence a homomorphism attack is nullified.
The strong generic free group property has been extended to arbitrary free products of infinite groups and many other amalgams including surface groupsaid groups and br by Fine. Myasnikov and Rosenberger [13] and Carstensen, Fine and Rosenberger [15] .
6. Security Analysis of the Group Randomizer Protocols
For the security analysis of the group randomizer password protocols, we make the security assumption that an adversary has access to the coded group theoretic responses. The strength of the proposed protocol is that an attacker must steal three things; the original password, the group randomizer and the challenge group. There is no access without all three. This immediately nullifies middelman attacks. If the adversary pretends to be the verifier to obtain the group words the attack is thwarted by the facts that the prover can verify the verifier and further if the attacker just transmits from the middle, nothing can be stolen since each time through a new challenge word must be used. Further the group randomizer has an infinite supply of both subgroups and challenge responses that are done randomly. In addition since a challenge word can be used only once the protocol nullifies replay attacks. Since challenge responses are machine to machine there is an essential probability of zero of an incorrect response. The protocol shuts down with an incorrect response and hence repeat attacks are harmless.
These are in distinction to answer-driven challenge-response systems where a prover often forgets or misspells a response. In these systems a prover is usually permitted several opportunities to answer. This makes these systems susceptible to both middleman and repeat attacks.
There are two theoretical attacks that must be dealt with. Relative to these attacks, the security of the system, and hence a security proof for the protocol, is provided by several results in asymptotic group theory.
The most straightforward attack is for the adversary to collect enough challenge words and responses. This provides a system of equations in a free group (or a finitely presented group)
${y}_{i}={W}_{i}\left({x}_{1},\cdots ,{x}_{n}\right),\text{\hspace{0.17em}}i=1,\cdots ,m.$
An adversary can then break the protocol by solving the system
${y}_{i}={W}_{i}\left({x}_{1},\cdots ,{x}_{m}\right)$
to obtain the challenge group.
A result of Lysenok [4] shows that solving such systems of equations in free groups (and in most finitely presented groups) is NP-hard. Hence this method of attack is impractical in most cases.
A second method of attack is based on the following. The mapping ${y}_{i}\to {W}_{i}$ is a homomorphism. If a challenge word appears in the subgroup generated by previous challenge words then an attacker can use this to answer a challenge without ever solving for the challenge group. However this approach fails due to the strong generic free property. Each set of challenge words is a free basis for the subgroup they generate with asymptotic density 1. Hence as explained in the previous section the probability converges to zero that a new challenge word is in the subgroup generated by previous challenge words.
There are several enhancements that can perhaps improve security:
1) Permutation Diffusion and Noise: Both the prover’s randomizer and the verifier’s randomizer have a fixed coded set of permutations on $\left\{1,\cdots ,k\right\}$ where k is the rank of the challenge group. Each presentation of a challenge word is accompanied by a random one of these permutations. If $\varphi$ is the presented permutation then the challenge word $W\left({y}_{1},\cdots ,{y}_{k}\right)$ is evaluated on $W\left({y}_{\varphi \left(1\right)},\cdots ,{y}_{\varphi \left(k\right)}\right)$ . As mentioned earlier this prevents hinders attacks by an eavesdropper since discovering for example what ${W}_{37}$ is, is of no use since it will be indexed differently for the next use. The set of coded permutations is agreed upon at initialization and becomes part of the common shared secret.
2) Short Challenge Words: In each challenge word $W\left({y}_{1},\cdots ,{y}_{k}\right)$ , we assume that not all k variables are used. In actual implementation we specify that the number of variables t that appear in any challenge word is small relative to k the rank of the challenge group. For example, we may have $5\le t\le 8$ with $k=256$ . Hence each equation that can be stolen by an attacker has only a relatively small number of variables. This increases the number of equations necessary to impact on a homomorphism solution which in turn is NP-hard to solve.
3) Frequent Reset: We recommend that the challenge group be reset relatively often. Since this protocol is symmetric the rest must be done via some sort of direct communication as in the original initialization of the secret key.
7. Actual Implementation of a Group Randomizer System Protocol
The actual implementation of a workable group randomizer system protocol involves several choices of parameters and subprograms. These include
1) The choice of the rank of the ambient free group in the free group method.
2) An enhancement program which takes a random choice of words ${W}_{1},\cdots ,{W}_{k}$ in a free group F and finds a new set of words ${V}_{1},\cdots ,{V}_{k}$ generating the same subgroup for which the words formed in ${V}_{1},\cdots ,{V}_{k}$ have a great deal of free cancellation. This involves what is called Nielsen transformations (see [7] [8] [9] ).
3) The choice of parameter sizes for the lengths of the randomly chosen words. In an actual implementation all words in the generators will have lengths between a and b where a and b are to be determined. All words used as test logins will have lengths between c and d with c and d to be determined, The optimal values for these parameters must be determined.
4) The implementation of a coded permutation system on $\left\{1,\cdots ,k\right\}$ where k is the rank of the challenge group and so that a coded permutation can be sent with each challenge word.
5) The development of an automatic reset protocol for the challenge group. In an ideal situation this can be done without actually communicating the changes between verifier and prover―that is each randomizer system does the same protocol automatically when reset is called for.
8. Alternative Methods Using Rational Matrices
Free groups have faithful representations in terms of rational and integral matrices (see [16] [17] ), or integral $2×2$ matrices there is an algorithm to go back and forth between a free group and the corresponding matrices in its representation. This is explained in [16] . This can be used in several ways in conjunction with the group randomizer.
First the basic free group method can be enhanced with the matrix representation in the following manner. We assume that the group randomizer has been extended to include the algorithm to go back and forth between a free group and $SL\left(2,Z\right)$ mentioned above. Then what can be sent from verifier to use is an integral matirx rather than a free group word. This is then deciphered by the algorithm into a word in the free group. The prover’s group randomizer then proceeds as in the standard free group method rewriting thw word in terms of the stored password subgroup. Finally, this is rewritten in terms of the matrix representation and an integral matrix sent back to the verifier. This presents a further time obstacle to an attacker.
There is a simpler variation of the whole system solely using matrices. Each prover is assigned a set of $m×m$ invertible rational matrices ${M}_{1},{M}_{2},\cdots ,{M}_{k}$ . These are linked to the prover’s standard password as before. Each matrix is assigned a free group variable ${x}_{1}={M}_{1},\cdots ,{x}_{k}={M}_{k}$ . As in the standard free group method when the prover presents a password the verifier sends a a free group word $W\left({x}_{1},\cdots ,{x}_{k}\right)$ . The prover’s group randomizer evaluates this word on ${M}_{1},\cdots ,{M}_{k}$ to obtain a rational matrix $M=W\left({M}_{1},\cdots ,{M}_{k}\right)$ . This matrix M is then sent back to the verifier. The verifier checks to see if the evaluation of the sent word is correct or not. As before to verify that the verifier is legitimate the process is repeated from the prover’s randomizer to the verifier.
An attacker here sees a matrix polynomial equation $W\left({x}_{1},\cdots ,{x}_{k}\right)=M$ . This must be solved for matrices ${M}_{1},\cdots ,{M}_{k}$ in order to obtain access. For $n\ge 3$ there is no factoring algorithm or solution algorithm for such equations and hence if k is large (or even moderately large) the equation is feasibly insolvable. This again present a one-time keypad type of approach. As mentioned earlier, if the matrices are over the reals R, the group $GL\left(n,R\right)$ has the generic free group property.
9. The Group Randomizer and Container Security
Another very common security problem is container safety or container security. Here a container means a large shipping unit and the fear is that some contraband material or dangerous people will be stored or shipped via a container. Our contention is that the group randomizer can be used here as a secure lock.
We make the assumption that the shipper is legitimate and that we are only interested in the main lock―that is we don’t consider the situation where a terrorist group saws through the center of the container. We only want to check that the main lock has not been tampered with.
We assume that the main lock has been outfitted with a group randomizer. When it is sealed, the group randomizer is given a finitely presented group as in the password case. Since the protocol is symmetric key, the group is be transmitted via some secure key exchange to the receiver. The main lock is set so that when it is opened or tampered with the group is lost. When the container gets to its destination, its group is checked by a group randomizer at the far end. This of course can be done electronically. Without stealing the group, as in the password case, the lock cannot be tampered with.
Acknowledgements
The authors would like to thank the referee for many helpful suggestions concerning the exposition of this paper. This is especially true in Section 4 on secure password verification.
Cite this paper
Baumslag, G., Fine, B., Moldenhauer, A. and Rosenberger, G. (2018) Secure Passwords Using Combinatorial Group Theory. Journal of Information Security, 9, 154-167. https://doi.org/10.4236/jis.2018.92011
References
1. 1. Baumslag, G., Brjukhov, Y., Fine, B. and Troeger, D. (2010) Secure Password Verification Using Combinatorial Group Theory. Groups-Complexity-Cryptology, 2, 67-82. https://doi.org/10.1515/gcc.2010.005
2. 2. Wikipedia. Challenge-Response Authentication. The Free Encyclopedia. https://en.wikipedia.org/wiki/Challenge%E2%80%93response_authentication
3. 3. Challenge-Response System Based on RSA. http://www.cag.lcs.mit.edu/rugina/ssh-procedures
4. 4. Lysenok, I. (2006) Equations over Free Groups. Private Communication.
5. 5. Jitsukawa, T. (2002) Malnormal Subgroups of Free Groups. In: Gilman, R., Shpilrain, V. and Myasnikov, A.G., Eds., Computational and Statistical Group Theory, Contemporary Mathematics, Vol. 298, 83-96. https://doi.org/10.1090/conm/298/05115
6. 6. Baumslag, G. (1993) Topics in Combinatorial Group Theory. Birkhauser.
7. 7. Lyndon, R. and Schupp, P. (1978) Combinatorial Group Theory. Springer.
8. 8. Magnus, W., Karass, A. and Solitar, D. (1968) Combinatorial Group Theory. Wiley Interscience, New York.
9. 9. Camps, T., Rebel, V. and Rosenberger, G. (2008) Einführung in die kombinatorische und die geometrische Gruppentheorie. Berliner Studienreihe zur Mathematik Band 19, Heldermann Verlag.
10. 10. Baumslag, G., Fine, B., Kreuzer, M. and Rosenberger, G. (2015) A Course in Mathematical Cryptography. De Gruyter, Berlin.
11. 11. Fine, B. and Rosenberger, G. (1999) Algebraic Generalizations of Discrete Groups. Marcel-Dekker, New York.
12. 12. Borovik, A., Myasnikov, A.G. and Shpilrain, V. (2002) Measuring Sets in Infinite Groups. In: Gilman, R., Shpilrain, V. and Myasnikov, A.G., Eds., Computational and Statistical Group Theory, Contemporary Mathematics, Vol. 298, 21-42. https://doi.org/10.1090/conm/298/05112
13. 13. Fine, B., Miasnikov, A. and Rosenberger, G. (2009) Generic Properties of Amalgams. Groups-Complexity-Cryptography, 1, 51-61.
14. 14. Epstein, D.B.A. (1971) Almost All Subgroups of Lie Group Are Free. Journal of Algebra, 19, 261-262. https://doi.org/10.1016/0021-8693(71)90107-4
15. 15. Carstensen, C., Fine, B. and Rosenberger, G. (2010) On Asymptotic Densities and Generic Properties in Finitely Generated Groups. Groups-Complexity-Cryptology, 2, 113-121. https://doi.org/10.1515/gcc.2010.008
16. 16. Baumslag, G., Fine, B. and Xu, X. (2006) Cryptosystems Using Linear Groups. Applicable Algebra in Engineering, Communication and Computing, 17, 205-217. https://doi.org/10.1007/s00200-006-0003-z
17. 17. Baumslag, G., Fine, B. and Xu, X. (2006) A Proposed Public Key Cryptosystem Using the Modular Group. In: Fine, B., Gaglione, A.M. and Spellman, D., Eds., Combinatorial Group Theory, Discrete Groups, and Number Theory, Contemporary Mathematics, Vol. 421, 35-44. https://doi.org/10.1090/conm/421/08025 | 9,809 | 40,955 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 125, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.671875 | 3 | CC-MAIN-2024-22 | latest | en | 0.907726 |
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13 more than the product of m and 5
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3. 👁 70
1. 5m + 13
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2. Thank you Ms.Sue
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http://math.stackexchange.com/questions/238886/what-are-some-function-spaces-that-are-ufds | 1,469,779,217,000,000,000 | text/html | crawl-data/CC-MAIN-2016-30/segments/1469257829972.49/warc/CC-MAIN-20160723071029-00146-ip-10-185-27-174.ec2.internal.warc.gz | 158,828,045 | 19,205 | # What are some function spaces that are UFDs?
What are some "useful" sets of functions (rings under pointwise multiplication) $\mathbb{R} \rightarrow \mathbb{R}$ that are unique factorization domains, other than the polynomial ring?
I will quantify usefulness in my context: I'm only interested in functions that form an algebra over $\mathbb{R}$ (or $\mathbb{C}$) in the space of continuous functions, and preferably also forms a dense algebra in $C([a,b])$ with respect to the usual supremum norm.
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What do you mean by "useful"? There are a LOT of useful sets of functions, the list would go on and on. Are you just curious or do you have something in mind? – Patrick Da Silva Nov 16 '12 at 20:33
I've added some clarification. I'm mostly curious, I suppose; however, I cannot really think examples other than polynomials, or perhaps just taking the polynomials and tacking something onto it. – Christopher A. Wong Nov 16 '12 at 20:40
This is a very nice question.
I'll boldly assert that outside of algebraic geometry there are probably no examples of spaces of functions that are unique factorization domains!
The reason is that most spaces of functions, be they continuous, differentiable or smooth (=$\mathcal C^\infty$) do not even constitute a domain.
Holomorphic functions on a connected open subset $U\subset \mathbb C$ (or on a connected holomorphic manifold of arbitrary dimension) do constitute a domain $\mathcal O(U)$ , but (unless I'm mistaken) are never unique factorization domains.
This is easy to see for the entire functions $\mathcal O(\mathbb C)$ : irreducible entire functions are of the form $az+b$ and $\sin(z)$ is certainly not a finite product of such affine functions.
In algebraic geometry however there do exist affine varieties $X$ for which the regular functions $\mathcal O(X)$ form a unique factorization domain : for example $\mathbb C$ or $\mathbb C^*$.
And affine varieties for which this is not the case, like the quadric in $\mathbb C^3$ given by the equation $z^2-xy=0$.
Deciding which is which for an affine variety is a very delicate problem.
And to finish on a slightly humorous note, let me remark that the domain of trigonometric polynomials $\mathbb R[\cos(\theta), \sin(\theta)]\subset \mathcal C(\mathbb R)$ is not a unique factorization domain because of the reasonably well known formula $$\cos\theta \cdot \cos\theta=(1-\sin \theta)\cdot (1+\sin \theta)$$
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thanks for your comments. I'd hoped that maybe there some other nice examples, but I guess it makes sense why there wouldn't be any common examples outside the realm of algebra. I'm not really an algebraist so I'm actually not familiar with any of the examples from algebraic geometry, either. – Christopher A. Wong Nov 17 '12 at 9:27
Why irreducible entire functions are of the form $az+b$? This may be easy but I didn't get clarification. (I was trying to post a question related to this; but the answer to my question is hidden in your statement, whose clarification I want.) – Groups Dec 11 '14 at 6:36
@groups: entire functions without zeros are invertible and thus not irreducible. On the other hand an entire function $f$ which does have a zero $c\in \mathbb C$ is divisible by $z-c$ and thus can be written $f=(z-c)g$. Hence $f$ is irreducible iff $g$ is a constant $a\in \mathbb C^*$ i.e. iff $f=(z-c)a=az+b$ . – Georges Elencwajg Dec 11 '14 at 7:35
@Georges: Thanks for clarification. – Groups Dec 11 '14 at 7:37
You are welcome, dear @Groups. – Georges Elencwajg Dec 11 '14 at 7:39
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This is an interesting ring but it is not a ring of functions. – Georges Elencwajg Nov 16 '12 at 21:31
@GeorgesElencwajg I thought it was close :) – rschwieb Nov 17 '12 at 0:44
Dear rschwieb: yes, I agree, it is close. – Georges Elencwajg Nov 17 '12 at 5:46 | 1,007 | 3,771 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.984375 | 3 | CC-MAIN-2016-30 | latest | en | 0.919909 |
https://docs.rs/near-sdk/latest/near_sdk/collections/struct.TreeMap.html | 1,679,951,218,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296948684.19/warc/CC-MAIN-20230327185741-20230327215741-00391.warc.gz | 247,626,256 | 15,934 | # Struct near_sdk::collections::TreeMap
``pub struct TreeMap<K, V> { /* private fields */ }``
Expand description
TreeMap based on AVL-tree
Runtime complexity (worst case):
• `get`/`contains_key`: O(1) - UnorderedMap lookup
• `insert`/`remove`: O(log(N))
• `min`/`max`: O(log(N))
• `above`/`below`: O(log(N))
• `range` of K elements: O(Klog(N))
## Implementations
Makes a new, empty TreeMap
##### Examples
``````use near_sdk::collections::TreeMap;
let mut tree: TreeMap<u32, u32> = TreeMap::new(b"t");``````
Returns the number of elements in the tree, also referred to as its size.
##### Examples
``````use near_sdk::collections::TreeMap;
let mut tree: TreeMap<u32, u32> = TreeMap::new(b"t");
tree.insert(&1, &10);
tree.insert(&2, &20);
assert_eq!(tree.len(), 2);``````
Clears the tree, removing all elements.
##### Examples
``````use near_sdk::collections::TreeMap;
let mut tree: TreeMap<u32, u32> = TreeMap::new(b"t");
tree.insert(&1, &10);
tree.insert(&2, &20);
tree.clear();
assert_eq!(tree.len(), 0);``````
Returns true if the map contains a given key.
##### Examples
``````use near_sdk::collections::TreeMap;
let mut tree: TreeMap<u32, u32> = TreeMap::new(b"t");
assert_eq!(tree.contains_key(&1), false);
tree.insert(&1, &10);
assert_eq!(tree.contains_key(&1), true);``````
Returns the value corresponding to the key.
##### Examples
``````use near_sdk::collections::TreeMap;
let mut tree: TreeMap<u32, u32> = TreeMap::new(b"t");
assert_eq!(tree.get(&1), None);
tree.insert(&1, &10);
assert_eq!(tree.get(&1), Some(10));``````
Inserts a key-value pair into the tree. If the tree did not have this key present, `None` is returned. Otherwise returns a value. Note, the keys that have the same hash value are undistinguished by the implementation.
##### Examples
``````use near_sdk::collections::TreeMap;
let mut tree: TreeMap<u32, u32> = TreeMap::new(b"t");
assert_eq!(tree.insert(&1, &10), None);
assert_eq!(tree.insert(&1, &20), Some(10));
assert_eq!(tree.contains_key(&1), true);``````
Removes a key from the tree, returning the value at the key if the key was previously in the tree.
##### Examples
``````use near_sdk::collections::TreeMap;
let mut tree: TreeMap<u32, u32> = TreeMap::new(b"t");
assert_eq!(tree.remove(&1), None);
tree.insert(&1, &10);
assert_eq!(tree.remove(&1), Some(10));
assert_eq!(tree.contains_key(&1), false);``````
Returns the smallest stored key from the tree
Returns the largest stored key from the tree
Returns the smallest key that is strictly greater than key given as the parameter
Returns the largest key that is strictly less than key given as the parameter
Returns the smallest key that is greater or equal to key given as the parameter
##### Examples
``````use near_sdk::collections::TreeMap;
let mut map: TreeMap<u32, u32> = TreeMap::new(b"t");
let vec: Vec<u32> = vec![10, 20, 30, 40, 50];
for x in vec.iter() {
map.insert(x, &1);
}
assert_eq!(map.ceil_key(&5), Some(10));
assert_eq!(map.ceil_key(&10), Some(10));
assert_eq!(map.ceil_key(&11), Some(20));
assert_eq!(map.ceil_key(&20), Some(20));
assert_eq!(map.ceil_key(&49), Some(50));
assert_eq!(map.ceil_key(&50), Some(50));
assert_eq!(map.ceil_key(&51), None);``````
Returns the largest key that is less or equal to key given as the parameter
##### Examples
``````use near_sdk::collections::TreeMap;
let mut map: TreeMap<u32, u32> = TreeMap::new(b"t");
let vec: Vec<u32> = vec![10, 20, 30, 40, 50];
for x in vec.iter() {
map.insert(x, &1);
}
assert_eq!(map.floor_key(&5), None);
assert_eq!(map.floor_key(&10), Some(10));
assert_eq!(map.floor_key(&11), Some(10));
assert_eq!(map.floor_key(&20), Some(20));
assert_eq!(map.floor_key(&49), Some(40));
assert_eq!(map.floor_key(&50), Some(50));
assert_eq!(map.floor_key(&51), Some(50));``````
Iterate all entries in ascending order: min to max, both inclusive
Iterate entries in ascending order: given key (exclusive) to max (inclusive)
##### Examples
``````use near_sdk::collections::TreeMap;
let mut map: TreeMap<u32, u32> = TreeMap::new(b"t");
let one: Vec<u32> = vec![10, 20, 30, 40, 50,45, 35, 25, 15, 5];
for x in &one {
map.insert(x, &42);
}
assert_eq!(
map.iter_from(29).collect::<Vec<(u32, u32)>>(),
vec![(30, 42), (35, 42), (40, 42), (45, 42), (50, 42)]
)``````
Iterate all entries in descending order: max to min, both inclusive
Iterate entries in descending order: given key (exclusive) to min (inclusive)
##### Examples
``````use near_sdk::collections::TreeMap;
let mut map: TreeMap<u32, u32> = TreeMap::new(b"t");
let one: Vec<u32> = vec![10, 20, 30, 40, 50,45, 35, 25, 15, 5];
for x in &one {
map.insert(x, &42);
}
assert_eq!(
map.iter_rev_from(45).collect::<Vec<(u32, u32)>>(),
vec![(40, 42), (35, 42), (30, 42), (25, 42), (20, 42), (15, 42), (10, 42), (5, 42)]
);``````
Iterate entries in ascending order according to specified bounds.
##### Panics
Panics if range start > end. Panics if range start == end and both bounds are Excluded.
##### Examples
``````use near_sdk::collections::TreeMap;
use std::ops::Bound;
let mut map: TreeMap<u32, u32> = TreeMap::new(b"t");
let one: Vec<u32> = vec![10, 20, 30, 40, 50];
let two: Vec<u32> = vec![45, 35, 25, 15];
for x in &one {
map.insert(x, &0);
}
for x in &two {
map.insert(x, &0);
}
assert_eq!(
map.range((Bound::Included(20), Bound::Excluded(30))).collect::<Vec<(u32, u32)>>(),
vec![(20, 0), (25, 0)]
);``````
Helper function which creates a [`Vec<(K, V)>`] of all items in the `TreeMap`. This function collects elements from `TreeMap::iter`.
## Trait Implementations
Deserializes this instance from a given slice of bytes. Updates the buffer to point at the remaining bytes. Read more
Deserialize this instance from a slice of bytes.
Serialize this instance into a vector of bytes.
Formats the value using the given formatter. Read more
The type of the elements being iterated over.
Which kind of iterator are we turning this into?
Creates an iterator from a value. Read more
## Blanket Implementations
Gets the `TypeId` of `self`. Read more
Immutably borrows from an owned value. Read more
Mutably borrows from an owned value. Read more
Returns the argument unchanged.
Calls `U::from(self)`.
That is, this conversion is whatever the implementation of `From<T> for U` chooses to do.
Should always be `Self`
The type returned in the event of a conversion error.
Performs the conversion.
The type returned in the event of a conversion error.
Performs the conversion. | 1,913 | 6,429 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.515625 | 3 | CC-MAIN-2023-14 | longest | en | 0.505168 |
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TRUTH TABLE allows you to see all possible scenarios in an expression containing compound conditions.
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TRUTH TABLE allows you to see all possible scenarios in an expression containing compound conditions.
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Subtract: 17 - 40
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Confirmed by yumdrea [4/6/2014 11:26:03 AM]
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Weegy: A linear equation is an equation for a straight line. A function is a relation that uniquely associates members of one set with members of another set. (More)
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The floor of a hall is covered with identical tiles which are fis triangles. One such triangle has the vertices at $(-3,2),(-1,-1)$ and $(1,2)$. If the floor of the hall is completely covered by 110 tiles, find the area of the floor. | 127 | 479 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.875 | 3 | CC-MAIN-2021-49 | latest | en | 0.953142 |
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Basic Rectifier Circuits
### Overview
As we have noted when looking at the Elements of a Power Supply, the purpose of the rectifier section is to convert the incoming ac from a transformer or other ac power source to some form of pulsating dc. That is, it takes current that flows alternately in both directions as shown in the first figure to the right, and modifies it so that the output current flows only in one direction, as shown in the second and third figures below.
The circuit required to do this may be nothing more than a single diode, or it may be considerably more complex. However, all rectifier circuits may be classified into one of two categories, as follows:
• Half-Wave Rectifiers. An easy way to convert ac to pulsating dc is to simply allow half of the ac cycle to pass, while blocking current to prevent it from flowing during the other half cycle. The figure to the right shows the resulting output. Such circuits are known as half-wave rectifiers because they only work on half of the incoming ac wave.
• Full-Wave Rectifiers. The more common approach is to manipulate the incoming ac wave so that both halves are used to cause output current to flow in the same direction. The resulting waveform is shown to the right. Because these circuits operate on the entire incoming ac wave, they are known as full-wave rectifiers.
Rectifier circuits may also be further clasified according to their configuration, as we will see below.
### The Half-Wave Rectifier
The simplest rectifier circuit is nothing more than a diode connected in series with the ac input, as shown to the right. Since a diode passes current in only one direction, only half of the incoming ac wave will reach the rectifier output. Thus, this is a basic half-wave rectifier.
The orientation of the diode matters; as shown, it passes only the positive half-cycle of the ac input, so the output voltage contains a positive dc component. If the diode were to be reversed, the negative half-cycle would be passed instead, and the dc component of the output would have a negative polarity. In either case, the DC component of the output waveform is vp/ = 0.3183vp, where vp is the peak voltage output from the transformer secondary winding.
It is also quite possible to use two half-wave rectifiers together, as shown in the second figure to the right. This arrangement provides both positive and negative output voltages, with each output utilizing half of the incoming ac cycle.
Note that in all cases, the lower transformer connection also serves as the common reference point for the output. It is typically connected to the common ground of the overall circuit. This can be very important in some applications. The transformer windings are of course electrically insulated from the iron core, and that core is normally grounded by the fact that it is bolted physically to the metal chassis (box) that supports the entire circuit. By also grounding one end of the secondary winding, we help ensure that this winding will never experience even momentary voltages that might overload the insulation and damage the transformer.
### The Full-Wave Rectifier
While the half-wave rectifier is very simple and does work, it isn't very efficient. It only uses half of the incoming ac cycle, and wastes all of the energy available in the other half. For greater efficiency, we would like to be able to utilize both halves of the incoming ac. One way to accomplish this is to double the size of the secondary winding and provide a connection to its center. Then we can use two separate half-wave rectifiers on alternate half-cycles, to provide full-wave rectification. The circuit is shown to the right.
Because both half-cycles are being used, the DC component of the output waveform is now 2vp/ = 0.6366vp, where vp is the peak voltage output from half the transformer secondary winding, because only half is being used at a time.
This rectifier configuration, like the half-wave rectifier, calls for one of the transformer's secondary leads to be grounded. In this case, however, it is the center connection, generally known as the center tap on the secondary winding.
The full-wave rectifier can still be configured for a negative output voltage, rather than positive. In addition, as shown to the right, it is quite possible to use two full-wave rectifiers to get outputs of both polarities at the same time.
The full-wave rectifier passes both halves of the ac cycle to either a positive or negative output. This makes more energy available to the output, without large intervals when no energy is provided at all. Therefore, the full-wave rectifier is more efficient than the half-wave rectifier. At the same time, however, a full-wave rectifier providing only a single output polarity does require a secondary winding that is twice as big as the half-wave rectifier's secondary, because only half of the secondary winding is providing power on any one half-cycle of the incoming ac.
Actually, it isn't all that bad, because the use of both half-cycles means that the current drain on the transformer winding need not be as heavy. With power being provided on both half-cycles, one half-cycle doesn't have to provide enough power to carry the load past an unused half-cycle. Nevertheless, there are some occasions when we would like to be able to use the entire transformer winding at all times, and still get full-wave rectification with a single output polarity.
### The Full-Wave Bridge Rectifier
The four-diode rectifier circuit shown to the right serves very nicely to provide full-wave rectification of the ac output of a single transformer winding. The diamond configuration of the four diodes is the same as the resistor configuration in a Wheatstone Bridge. In fact, any set of components in this configuration is identified as some sort of bridge, and this rectifier circuit is similarly known as a bridge rectifier.
If you compare this circuit with the dual-polarity full-wave rectifier above, you'll find that the connections to the diodes are the same. The only change is that we have removed the center tap on the secondary winding, and used the negative output as our ground reference instead. This means that the transformer secondary is never directly grounded, but one end or the other will always be close to ground, through a forward-biased diode. This is not usually a problem in modern circuits.
To understand how the bridge rectifier can pass current to a load in only one direction, consider the figure to the right. Here we have placed a simple resistor as the load, and we have numbered the four diodes so we can identify them individually.
During the positive half-cycle, shown in red, the top end of the transformer winding is positive with respect to the bottom half. Therefore, the transformer pushes electrons from its bottom end, through D3 which is forward biased, and through the load resistor in the direction shown by the red arrows. Electrons then continue through the forward-biased D2, and from there to the top of the transformer winding. This forms a complete circuit, so current can indeed flow. At the same time, D1 and D4 are reverse biased, so they do not conduct any current.
During the negative half-cycle, the top end of the transformer winding is negative. Now, D1 and D4 are forward biased, and D2 and D3 are reverse biased. Therefore, electrons move through D1, the resistor, and D4 in the direction shown by the blue arrows. As with the positive half-cycle, electrons move through the resistor from left to right.
In this manner, the diodes keep switching the transformer connections to the resistor so that current always flows in only one direction through the resistor. We can replace the resistor with any other circuit, including more power supply circuitry (such as the filter), and still see the same behavior from the bridge rectifier. | 1,849 | 8,573 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.203125 | 3 | CC-MAIN-2018-05 | latest | en | 0.768226 |
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Curious Ruler is a neat iPad app for measuring objects. The purpose of the app is to provide comparisons between common objects like coins (US, Canadian, and Euros) and the object that a student measures. To use Curious Ruler students select an object from the menu of comparison objects. The list of comparison objects includes […]
## HeyMath! – Math Lessons and Practice Tests
HeyMath! from Singapore is a free iPad app that offers animated math lessons and practice assessments for students. The animations provide instruction for solving word problems. The word problems contain visual aids to help students understand the explanations that they hear and read on the app. At the end of each unit in HeyMath! students […]
## Crackers and Goo – A Fun Math Practice App
Crackers and Goo is a free iPad app that provides fun addition and multiplication practice games. The games resemble of a combination of elements from Frogger and Tetris. Each game uses the same format but each level increases the difficulty of the mental mathematics calculations required of the player. On each level of Crackers and […]
## Thinking Blocks – Free iPad Apps for Modeling Math Problems
Thinking Blocks is a series of free iPad apps based on the web tool of the same name. The apps are designed to help students develop models of word problems. In each of the apps students create block models to work through a series of progressively more difficult word problems. As students work through the […] | 613 | 3,081 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.75 | 3 | CC-MAIN-2017-04 | longest | en | 0.931053 |
https://math.stackexchange.com/questions/688199/radius-of-convergence-proof | 1,709,555,567,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947476442.30/warc/CC-MAIN-20240304101406-20240304131406-00838.warc.gz | 358,102,546 | 34,079 | Let $a,c_0,c_1,...\in \mathbb{R}$ with at least one of $c_0,c_1,c_2,...$ nonzero, and let the power series $\sum_{n=0}^{\infty}c_n(x-a)^n$ have positive radius of convergence $r$. Show that there exists a positive number $\delta<r$ such that the sum of the series is nonzero for every real number $x$ such that $0<|x-a|<\delta.$
Since it $r$ is its radius of convergence then we have that for $r>0$ the series converge for all $x$ such that $|x-a|<r$. But I am not sure how I can prove this.
How can I prove this?
Let $k:=\min \{n:c_n\ne 0\}$ and consider the expression
$$u(t)=t^k\left(|c_k|-\sum_{n=k+1}^\infty |c_n|t^{n+k}\right)$$
and its relation to values of the series for $|x-a|=t$.
Set $s(x)=\sum_{n=0}^{\infty}c_n(x-a)^n$, then by the triangle inequality
$$|s(x)-c_k(x-a)^k|\le \sum_{n=k+1}^{\infty}|c_n|\,|x-a|^n=|c_k|\,|x-a|^k-u(|x-a|).$$
Again by the triangle inequality
$$|s(x)|\ge |c_k(x-a)^k|-|s(x)-c_k(x-a)^k|\ge u(|x-a|).$$
Now apply standard estimates to $u(t)$, for instance $$u(\tfrac r2s)\ge s^k \left(|c_k|\left(\tfrac r2\right)^k-\sup_{n>k}|c_n|\left(\tfrac r2\right)^{n}\frac{s}{1-s} \right),$$ to find that it is positive for small positive arguments $t=\tfrac r2s$, $0\le s<1$, well inside the radius of convergence of the series, to find that $u(t)>0$ for small positive arguments $t$.
• Can you elaborate please? Feb 24, 2014 at 9:22 | 511 | 1,373 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4 | 4 | CC-MAIN-2024-10 | latest | en | 0.705894 |
http://slideplayer.com/slide/6636998/ | 1,571,469,991,000,000,000 | text/html | crawl-data/CC-MAIN-2019-43/segments/1570986692126.27/warc/CC-MAIN-20191019063516-20191019091016-00275.warc.gz | 184,628,502 | 26,062 | # Properties and the Inverse of
## Presentation on theme: "Properties and the Inverse of"— Presentation transcript:
Properties and the Inverse of
Chapter 6: Properties and the Inverse of z-Transform
Properties of z-Transform
Some of the properties of the z-transform are: Linearity Right shift Left shift Final value theorem Z-differentiation (Multiplication by n)
Example: Linearity The z-transform of a unit ramp function r(nT) is Find the z-transform of the function 5r(nT). Solution Using the linearity property of z-transform,
Right-shifting property
Suppose that the z-transform of f(nT) is F(z). Let y(nT) = f(nT-mT), then assuming that f nT =0 for n<0.
Left-shifting property
Suppose that z-transform of f(nT) is F(z). Let y(nT) = f(nT+mT). Then However, if f nT =0 for n=0 to m-1, this simplifies to
Final value theorem Suppose that the z-transform of f (nT ) is F(z). Then the final value of the time response is given by Note that this theorem is valid if the poles of (1−z−1)F(z) are inside the unit circle or at z = 1.
Example 6.5 Given the function find the final value of g(nT).
Solution: Using the final value theorem,
Multiplication by n Let Then Example Given And Then, Z-differentiation
Inverse z-Transforms Given the z-transform, Y(z), of a function, y(n), it is required to find the time-domain function y(n). Here, we will study the following methods: power series (long division). expanding Y (z) into partial fractions and using z-transform tables to find the inverse.
Method 1: Power series (long division)
This method involves dividing the denominator of Y (z) into the numerator such that a power series of the form is obtained. Notice that the values of y(n) are the coefficients in the power series.
Example 6.6 Find the inverse z-transform for the polynomial Solution: Dividing the denominator into the numerator gives
and the coefficients of the power series are y(0) = 1, y(T) = 4, y(2T) = 8, y(3T) = 8, : The required sequence is y(t) = δ(t) + 4δ(t-T) + 8δ(t-2T) + 8δ(t-3T) + … The first few samples of the time sequence y(nT) is shown below
Example 6.7 Find the inverse z-transform for Y(z) given by Solution:
Dividing the denominator into the numerator gives
and the coefficients of the power series are y(0) = 0, y(T) = 1, y(2T) = 3, y(3T) = 7, y(4T) = 15, : The required sequence is y(t) = δ(t-1) + 3δ(t-T) + 7 δ(t-2T) + 15 δ(t-3T) + … The first few samples of the time sequence y(nT) is shown below
The disadvantage of the power series method is that it does not give a closed form of the resulting sequence. We often need a closed-form result, and other methods should be used when this is the case.
Method 2: Partial fractions
A partial fraction expansion of Y (z) is found, and then tables of z-transform can be used to determine the inverse z-transform. Looking at the z-transform tables, we see that there is usually a z term in the numerator. It is therefore more convenient to find the partial fractions of Y (z)/z and then multiply the partial fractions by z to obtain a z term in the numerator.
Example 6.8 Find the inverse z-transform of the function Solution
The above expression can be written as The values of A and B can be found by equating like powers in the numerator, i.e. We find A = − 1, B = 1, giving
From the z-transform tables we find that
and the coefficients of the power series are y(0) = 0, y(T) = 1, y(2T) = 3, y(3T) = 7, y(4T) = 15, : so that the required sequence is y(t) = δ(t-1) + 3δ(t-T) + 7 δ(t-2T) + 15 δ(t-3T) + … Which is the same answer as Example 6.7.
Example 6.9 Find the inverse z-transform of the function Answer:
The above expression can be written as
The values of A, B, C can be found as follows
Using the inverse z-transform, we find
The coefficients of the power series are and the required sequence is
Example 6.11 Using the partial expansion method described above, find the inverse z-transform of Solution Rewriting the function as
we find that Thus
The inverse transform is found from the tables as and the coefficients of the power series are y(0) = 0, y(T) = 1, y(2T) = 3.3, y(3T) = 5.89, : so that the required sequence is y(t) = δ(t-T) + 3.3δ(t-2T) δ(t-3T) + …
Example 6.12 Find the inverse z-transform of Solution
Rewriting the function as
We obtain
We can now write Y(z) as The inverse transform is found from the tables as Note that for the last term, we used the multiplication by n property which is equivalent to a z-differentiation.
Pulse Transfer Function and Manipulation of Block Diagrams
Given the following system: Suppose we are interested in finding the relationship between the sampled output y*(s) and the sampled input u*(s). The continuous output y(s) is given by: Then, sampling the output signal gives:
Note that if at least one of the continuous functions has been sampled, then the z-transform of the product is equal to the product of the z-transforms of each function (note that [e*(s)]* = [e*(s)], since sampling an already sampled signal has no further effect). G(z) is the ratio of the z-transform of the sampled output and the sampled input at the sampling instants, and is called the pulse transfer function. Notice from (6.35) that we have no information about the output y(z) between the sampling instants.
Example 6.17 Derive an expression for the z-transform of the output of the system. Solution The following expressions can be written for the system: Which gives or
For example, if Then and the output function is given by
Example 6.16 The figure shows an open-loop sampled data system. Derive an expression for the z-transform of the output of the system.
The following expression can be written for the system:
Solution The following expression can be written for the system: or And where
For example, if Then, from the z-transform tables, and the output is given by | 1,546 | 5,826 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.71875 | 5 | CC-MAIN-2019-43 | latest | en | 0.827105 |
https://gmatclub.com/forum/mary-a-veterinary-student-has-been-assigned-an-experiment-39715.html | 1,487,764,538,000,000,000 | text/html | crawl-data/CC-MAIN-2017-09/segments/1487501170940.45/warc/CC-MAIN-20170219104610-00104-ip-10-171-10-108.ec2.internal.warc.gz | 713,048,262 | 57,225 | Mary, a veterinary student, has been assigned an experiment : GMAT Critical Reasoning (CR)
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# Mary, a veterinary student, has been assigned an experiment
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07 Dec 2006, 02:28
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Mary, a veterinary student, has been assigned an experiment in mammalian physiology that would require her to take a healthy, anesthetized dog and subject it to a drastic blood loss in order to observe the physiological consequences of shock. The dog would neither regain consciousness nor survive the experiment. Mary decides not to do this assignment.
Mary’s decision most closely accords with which one of the following principles?
(A) All other things being equal, gratuitously causing any animal to suffer pain is unjustified.
(B) Taking the life of an animal is not justifiable unless doing so would immediately assist in saving several animal lives or in protecting the health of a person.
(C) The only sufficient justification for experimenting on animals is that future animal suffering is thereby prevented.
(D) Practicing veterinarians have a professional obligation to strive to prevent the unnecessary death of an animal except in cases of severely ill or injured animals whose prospects for recovery are dim.
(E) No one is ever justified in acting with the sole intention of causing the death of a living thing, be it animal or human.
If you have any questions
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07 Dec 2006, 03:36
I ll take B
A, I don't understand the word" gratuitously ". Hence I avoided
C is dead wrong. SHe should do it if she follow such principle.
D Unrelated
E is too strong.
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livin in a prison island...
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07 Dec 2006, 04:29
I will go with D.
vineetgupta wrote:
Mary, a veterinary student, has been assigned an experiment in mammalian physiology that would require her to take a healthy, anesthetized dog and subject it to a drastic blood loss in order to observe the physiological consequences of shock. The dog would neither regain consciousness nor survive the experiment. Mary decides not to do this assignment.
Mary’s decision most closely accords with which one of the following principles?
(A) All other things being equal, gratuitously causing any animal to suffer pain is unjustified.
Not just pain here, also death.
(B) Taking the life of an animal is not justifiable unless doing so would immediately assist in saving several animal lives or in protecting the health of a person.
We don't know if Mary thinks it is wrong to take the life of an animal, period.(C) The only sufficient justification for experimenting on animals is that future animal suffering is thereby prevented.
If she thought this, she would have conducted the experiment.
(D) Practicing veterinarians have a professional obligation to strive to prevent the unnecessary death of an animal except in cases of severely ill or injured animals whose prospects for recovery are dim.
Although Mary is not a practicing vet, the question only asks what her decision (making) closely resembles.
(E) No one is ever justified in acting with the sole intention of causing the death of a living thing, be it animal or human.
Death is not the sole intention in this case.
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07 Dec 2006, 09:08
hsampath wrote:
I will go with D.
vineetgupta wrote:
Mary, a veterinary student, has been assigned an experiment in mammalian physiology that would require her to take a healthy, anesthetized dog and subject it to a drastic blood loss in order to observe the physiological consequences of shock. The dog would neither regain consciousness nor survive the experiment. Mary decides not to do this assignment.
Mary’s decision most closely accords with which one of the following principles?
(A) All other things being equal, gratuitously causing any animal to suffer pain is unjustified.
Not just pain here, also death.
(B) Taking the life of an animal is not justifiable unless doing so would immediately assist in saving several animal lives or in protecting the health of a person.
We don't know if Mary thinks it is wrong to take the life of an animal, period.(C) The only sufficient justification for experimenting on animals is that future animal suffering is thereby prevented.
If she thought this, she would have conducted the experiment.
(D) Practicing veterinarians have a professional obligation to strive to prevent the unnecessary death of an animal except in cases of severely ill or injured animals whose prospects for recovery are dim.
Although Mary is not a practicing vet, the question only asks what her decision (making) closely resembles.
(E) No one is ever justified in acting with the sole intention of causing the death of a living thing, be it animal or human.
Death is not the sole intention in this case.
It's a tough one. Author decides the answer.
IMHO, B should be the most relevant one.
Veter. Student asked to take a life of an animal or pet dog... so his decision matches with B.
D talks about veterinarians practice , policies and more. And it talks about
already injured or severely suffering animal .... Not talking life of an healthy pet animal.
So B is my answer on the real test
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07 Dec 2006, 15:30
D sounded right at 1st but then realized Mary is a student, not a
practicing veterinarian.
B.
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08 Dec 2006, 06:16
Thanks the OA is B....
08 Dec 2006, 06:16
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Display posts from previous: Sort by | 1,728 | 7,292 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.015625 | 3 | CC-MAIN-2017-09 | longest | en | 0.91904 |
https://www.speedsolving.com/threads/how-fast-are-you-at-3x3.74997/ | 1,585,704,731,000,000,000 | text/html | crawl-data/CC-MAIN-2020-16/segments/1585370505359.23/warc/CC-MAIN-20200401003422-20200401033422-00058.warc.gz | 1,180,303,334 | 20,755 | # How fast are you at 3x3?
## How fast are you at 3x3?
• ### 1 minute or more
• Total voters
63
#### Micah Morrison
##### Member
I'm just taking a poll to see what most people averge on 3x3
#### Filipe Teixeira
##### Member
i'm only sub20
aw well
take that minh thai
#### KingCanyon
##### Member
I'll say I'm sub 15 because that's basically what I average when I'm warmed up and have a sub 15 official average.
#### Iwannaganx
##### Member
Only sub 1min for me......
I'm learning F2L and improving everyday!!
#### PugCuber
##### Member
I am basically Sub-12. It just varies on some days. I know full 2LLL, Full COLL, almost Full ELL (All but about 6 cases), some ZBLL, some OLLCP, some WV and VLS, etc. The 3 year anniversary of when I finally finished learning how to solve the cube was about a week ago (week ago [lol I know the meme is dead]). The 4 year anniversary of when I got my first cube was this past Wednesday.
#### Nmile7300
##### Member
High 9-ish. I only practice on stackmat. If I did keyboard it would probably be low 9.
I know full CFOP but no COLL or anything. I might learn some soon!
Also I can do Roux in about 15 seconds. (2 look cmll )
#### GAN 356 X
##### Member
Does anyone know if there is a difference between sub-x and having an average? Say if I said I was sub 10 seconds, which im not, would that mean my average is below 10 seconds? or above and around 10 seconds?
Also, why has their a difference between stackmat times and computer times?
#### Micah Morrison
##### Member
Does anyone know if there is a difference between sub-x and having an average? Say if I said I was sub 10 seconds, which im not, would that mean my average is below 10 seconds? or above and around 10 seconds?
sub is Latin for below so sub 10 would be under 10 seconds
#### GAN 356 X
##### Member
sub is Latin for below so sub 10 would be under 10 seconds
Ok thanks. I feel like an idiot now....
#### Nmile7300
##### Member
why has their a difference between stackmat times and computer times?
It's a minimal difference that probably won't affect someone above 15 seconds or so, but the simple answer is that it takes a bit longer to reach the cube from a stackmat than when you do keyboard. With keyboard, you can have one hand on the spacebar and one hand on the cube when you start. I just use stackmat because it simulates competition, but either way is technically fine
#### efattah
##### Member
It's a minimal difference that probably won't affect someone above 15 seconds or so, but the simple answer is that it takes a bit longer to reach the cube from a stackmat than when you do keyboard. With keyboard, you can have one hand on the spacebar and one hand on the cube when you start. I just use stackmat because it simulates competition, but either way is technically fine
My stackmat times are always significantly faster than keyboard, and this is because in the heat of max TPS I can never hit the space bar correctly to stop the solve. Every 2nd solve I miss the space bar and have to hit it again, losing lots of time.
#### GAN 356 X
##### Member
My stackmat times are always significantly faster than keyboard, and this is because in the heat of max TPS I can never hit the space bar correctly to stop the solve. Every 2nd solve I miss the space bar and have to hit it again, losing lots of time.
If u use CStimer, you can stop the solve with any key
Also, does anyone know if you need to supply your own statckmat when you go to a comp?
#### Iwannaganx
##### Member
Also, does anyone know if you need to supply your own statckmat when you go to a comp?
Not from what I've seen but I wouldn't know I haven't been. Maybe check the WCA website?
#### pjk
Staff member
Note: if you haven't done it yet, you can add your PB Records to your SpeedSolving profile by going to your profile, and clicking on the Accomplishments tab, then clicking on the Add A Personal Record button:
Here is the link to mine:
https://www.speedsolving.com/members/pjk.1/#record
It makes it easy to share your PB records and for others to see your records. We also have a Rank Board which will rank you on the site if you've entered in your records (going through an update now which will be released soon too).
Also, within the next week or so, if you fill in our WCA ID on your profile:
https://www.speedsolving.com/account/account-details
#### GAN 356 X
##### Member
Note: if you haven't done it yet, you can add your PB Records to your SpeedSolving profile by going to your profile, and clicking on the Accomplishments tab, then clicking on the Add A Personal Record button:
View attachment 10690
Here is the link to mine:
https://www.speedsolving.com/members/pjk.1/#record
It makes it easy to share your PB records and for others to see your records. We also have a Rank Board which will rank you on the site if you've entered in your records (going through an update now which will be released soon too).
Also, within the next week or so, if you fill in our WCA ID on your profile:
https://www.speedsolving.com/account/account-details
Thing is, I've never been to a comp so I don't have an ID. I'll probably add my PB records sometime soonish
pjk
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posted by .
*I am lost on a bunch of different problems, and this is one of them. Please note that I am NOT asking for final answers, just some guidance so that I can figure out how to do the problem!!! I know that a lot of times on this site it is commented that we are fishing for answers, but I just need some guidance on some stuff I honestly cannot figure out how to do, no matter how easy it might seem to others.
A 1.50 kg box moves back and forth on a horizontal frictionless surface between two different springs, as shown in the accompanying figure. The box is initially pressed against the stronger spring, compressing it 4.00 cm, and then is released from rest.
By how much will the box compress the weaker spring?
5.66 cm....I actually got this one!
What is the maximum speed the box will reach?
Figure the maximum stored energy in a spring.... 1/2 kx^2, set that equal to 1/2 mv^2, solve for v.
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These problems are pretty easy, but confusing (as stated in the subject line). |2x+5|-x=10 |3x+4|-2x=11 I know there are supposed to be two answers for each problem, for the first problem I got 5 and for the second problem I got 7. …
7. ### Math
I have a matrix with brackets around [7 6 ][13] [14 -6] [8] and i know there are two separate brackets but on this thing you cant just do one big one.. so if you don't understand the equation that was suppose to go into matricies was …
8. ### Math
I'm doing some word problems on finding the areas of rectangles, its just like all complicated. I don't understand how to do them. For example: A square field had 3 m added to its length and 2m added to its width. The area is 90 m2(squared). …
9. ### Physics
If an object is hung from a cable with a weight of 40 N, what is the tension in the two cables?
10. ### Algebra 1A
Thanks Reiny,that is my only problem trying to figure out the equation. I just cannot figure out for some reason when getting the intersection, I tried several different ways yet my progress is still the same.
More Similar Questions | 811 | 3,335 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.8125 | 3 | CC-MAIN-2018-05 | latest | en | 0.949918 |
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## Question 566
Edit
### Question
MCQ▾
A man completes of a job in 15 days. At this rate how many more days will it take him to finish the job?
### Choices
Choice (4)Response
a.22.5
b.22
c.27
d.25
## Question 567
Edit
### Question
MCQ▾
can do a piece of work in 10 days and can do alone it in 12 days. B works at it for 5 days and then leaves. alone can finish the remaining work in________.
### Choices
Choice (4)Response
a.5.8 days
b.6.5 days
c.6 days
d.7.5 days
## Question 568
Edit
### Question
MCQ▾
If, then the value of a is
### Choices
Choice (4)Response
a.3
b.4
c.1
d.2
## Question 569
Edit
### Question
MCQ▾
1.20 of a number equals 0.36 of another number. The ratio of the number is________.
### Choices
Choice (4)Response
a.3: 2
b.2: 3
c.3: 10
d.10: 3
## Question 570
Edit
### Question
MCQ▾
If 4 men take an hour to dig a ditch, how long should it take 10 men to dig a ditch of the same type?
### Choices
Choice (4)Response
a.23 min
b.20 min
c.30 min
d.24 min
## Question 571
Edit
### Question
MCQ▾
Number of prime factors in is
### Choices
Choice (4)Response
a.20
b.22
c.23
d.28
## Question 572
Edit
### Question
MCQ▾
If A: B = 3: 4 and B: C = 5: 6, then C: A is equal to, ________.
### Choices
Choice (4)Response
a.24: 15
b.12: 23
c.15: 14
d.25: 14
## Question 573
Edit
### Question
MCQ▾
The value of when is________.
### Choices
Choice (4)Response
a.85
b.84
c.80
d.81 | 627 | 1,835 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.84375 | 4 | CC-MAIN-2020-50 | latest | en | 0.682498 |
https://polymathprogrammer.com/tag/coordinate/ | 1,660,215,672,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882571284.54/warc/CC-MAIN-20220811103305-20220811133305-00064.warc.gz | 435,023,073 | 13,778 | A square described in any rotational orientation with 1 equation
What started out as an innocent question meant to poke me became an interesting math problem to ponder. You might want to read the original question and the answers presented. The final question was, can a square be described in any rotational orientation (based on the 2D Cartesian plane) with just 1 equation?
The answer is a resounding YES.
Meet Roie, who proposed this solution:
maximum {|r cos(theta-theta_0)|,|r sin(theta-theta_0)|} = c
This is basically modified from the max-abs solution by Mike Anderson:
max(abs(x),abs(y)) = c
You move from Cartesian coordinates to polar coordinates. Once you’re in polar plane, you can rotate by changing one variable (the angle). I actually wrote something about converting between Cartesian and polar coordinates for the use of image rotations. I can’t believe I forgot about that…
Roie also gave sample (GNU Octave or MATLAB) code:
```theta0 = 0.5;
x = ones(201,1) * (-10:0.1:10);
y = x’;
z = max(abs(sqrt(x.^2+y.^2) .* cos(atan(y./x)-theta0)),abs(sqrt(x.^2+y.^2) .* sin(atan(y./x)-theta0)));
contour(z)
axis square
```
Unfortunately, I am unable to verify that. Fortunately, Cees Meijer confirmed it for me. He introduced me to FreeMat, a free open source software that works like MATLAB. Unfortunately, I cannot install the Windows version because I need an x64 executable (I’m on Windows 7 64-bit), and FreeMat currently only has x86 version (32-bit).
Oh well, if it works, then I can finally do matrix multiplications in one line of code. So there, Will.
Cartesian coordinates and transformation matrices
If you’re doing any work in 3D, you will need to know about the Cartesian coordinate system and transformation matrices. Cartesian coordinates are typically used to represent the world in 3D programming. Transformation matrices are matrices representing operations on 3D points and objects. The typical operations are translation, rotation, scaling.
2 dimensional Cartesian coordinates
You should have seen something like this in your math class:
The Roman letters I, II, III, and IV represent the quadrants of the Cartesian plane. For example, III represents the third quadrant. Not a lot to say here, so moving on…
3 dimensional Cartesian coordinates
And for 3 dimensions, we have this:
I don’t quite like the way the z-axis points upward. The idea probably stems from having a piece of paper representing the 2D plane formed by the x and y axes. The paper is placed on a flat horizontal table, and the z-axis sticks right up.
Mathematically speaking, there’s no difference.
However, I find it easier to look at it this way:
The XY Cartesian plane is upright, representing the screen. The z-axis simply protrudes out of the screen. The viewport can cover all four quadrants of the XY plane. The illustration only covered the first quadrant so I don’t poke your eye out with the z-axis *smile*
There is also something called the right-hand rule, and correspondingly the left-hand rule. The right-hand rule has the z-axis pointing out of the screen, as illustrated above. The left-hand rule has the z-axis pointing into the screen. Observe the right-hand rule:
The thumb represents the x-axis, the index finger represents the y-axis and the middle finger represents the z-axis. As for the left-hand rule, we have:
We’re looking at the other side of the XY plane, but it’s the same thing. The z-axis points in the other direction. And yes, I have long fingers. My hand span can cover an entire octave on a piano.
What’s the big deal? Because your 3D graphics engine might use a certain rule by default, and you must follow. Otherwise, you could be hunting down errors like why an object doesn’t appear on the screen. Because the object was behind the camera when you thought it’s in front. Your selected graphics engine should also allow you to use the other rule if you so choose.
In case you’re wondering, here’s the right-hand rule illustration with the z-axis pointing upwards:
I still don’t like a skyward-pointing z-axis. It irks me for some reason…
Scaling (or making something larger or smaller)
So how do you enlarge or shrink something in 3D? You apply the scaling matrix. Let’s look at the 2D version:
If your scaling factor is greater than 1, you’re enlarging an object. If your scaling factor is less than 1, you’re shrinking an object. What do you think happens when your scaling factor is 1? Or when your scaling factor is negative?
So how does the scaling factor look like in a scaling matrix?
If you don’t know what that means, or don’t know what the result should be like, review the lesson on matrices and the corresponding program code.
You will notice there are separate scaling factors for x- and y- axes. This means you can scale them independently. For example, we have this:
And we only enlarge along the x-axis:
We can also only enlarge along the y-axis:
Yeah, I got tired of drawing 2D pictures, so I decided to render some 3D ones. Speaking of which, you should now be able to come up with the 3D version of the scaling matrix. Hint: just add a scaling factor for the z-axis.
Rotating (or spinning till you puke)
This is what a rotation matrix for 2 dimensions looks like:
That symbol that looks like an O with a slit in the middle? That’s theta (pronounced th-ay-tuh), a Greek alphabet. It’s commonly used to represent unknown angles.
I’ll spare you the mathematical derivation of the formula. Just use it.
You can convince yourself with a simple experiment. Use the vector (1,0), or unit vector lying on the x-axis. Plug in 90 degrees for theta and you should get (0,1), the unit vector lying on the y-axis.
That’s anti-clockwise rotation. To rotate clockwise, just use the value with a change of sign. So you’ll have -90 degrees.
Depending on the underlying math libraries you use, you might need to use radians instead of degrees (which is typical in most math libraries). I’m sure you’re clever enough to find out the conversion formula for degree-to-radian yourself…
Now for the hard part. The 3D version of rotation is … a little difficult. You see, what you’ve read above actually rotates about the implied z-axis. Wait, that means you can rotate about the x-axis! And the y-axis! Sacrebleu! You can rotate about any arbitrary axis!
I’ll write another article on this. If you’re into this, then you might want to take a look at this article on 3D rotation. I’ll also touch on a concept where you rotate about one axis and then rotate about another axis. Be prepared for lots of sine’s and cosine’s in the formula. Stop weeping; it’s unseemly of you.
Translating (nothing linguistic here)
What it means is you’re moving points and objects from one position to another. Let’s look at a 1 dimensional example:
The squiggly unstable looking d-wannabe? It’s the Greek alphabet delta. Delta-x is a standard notation for “change in x”. In this case “x” refers to distance along the x-axis. We’ll use an easier-to-type version called “dx” for our remaining discussion.
In 2 dimensions, we have the corresponding dy, for “change in y”. Note that there’s no stopping you from using negative values for dx or dy. In the illustration above, dx and dy are negative.
You’ll have to imagine the case for 3D because the diagram is likely to be messy. But it’s easy to visualise. You just do the same for z-axis.
So what’s the transformation matrix for translation? First, you need to extend your matrix size and vector size by one dimension. The exact reasons are due to affine transformations and homogeneous coordinates (I’ve mentioned them briefly earlier).
Consider this: You have a point (x,y,z) and you want it to be at the new position (x + dx, y + dy, z + dz). The matrix will then look like this:
Notice that for scaling, the important entries are the diagonal entries. For rotation, there are sine’s and cosine’s and they’re all over the place. But for translation, the “main body” of the matrix is actually an identity matrix. The fun stuff happens in the alleyway column on the extreme right of the matrix.
That reminds me. Because you’ll be using all the transformation matrices together, all matrices must be of the same size. So scaling and rotation matrices need to be 4 by 4 too. Just extend them with zero entries except the bottom right entry, which is 1.
Conclusion
We talked about 2D and 3D Cartesian coordinates. I’ve also shown you the right-hand and left-hand rules. This forms the basis for learning basic transformations such as scaling, rotation and translation.
There are two other interesting transformation matrices: shearing and reflection. I thought what we have is good enough for now. You are free to do your own research. When the situation arise, I’ll talk about those two transformations.
If you enjoyed this article and found it useful, please share it with your friends. You should also subscribe to get the latest articles (it’s free).
You might find this article on converting between raster, Cartesian and polar coordinates useful.
You might find these books on “coordinate transformation” useful. | 2,054 | 9,141 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.640625 | 4 | CC-MAIN-2022-33 | latest | en | 0.895847 |
http://www.123eng.com/placementpapers/isro_placementpapers5.html | 1,542,884,457,000,000,000 | text/html | crawl-data/CC-MAIN-2018-47/segments/1542039746205.96/warc/CC-MAIN-20181122101520-20181122123520-00499.warc.gz | 391,819,386 | 4,822 | 123 Eng
### Engineering the engineers™
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Placement Papers>>ISRO
# ISRO tech Placement Paper
Paper Type : Candidates Experiences
ISRO
Discipline : Computer Science
Total no of questions: 80
Duration: 90 minutes
Marking: +3 for correct answer -1 for wrong
Total sections : 1
Mode : Offcampus
Discipline Wise Questions
--------------------------
Questions on
PASCAL code
Networking
Binary Tree
Stack & Queue concept
GateS (AND OR NAND NOR XOR...)
Data structure basics
C funda
Page fault and other computer architech. fundamentals
Internet basics
Sin Cosine waves based questions
I attended the ISRO test on 8th aug
it was of GATE pattern mostly of Handa book.
80 ques---
Some of the ques i remember....hope it would be useful...
1. The root directory of disk should be placed at_____
--->at a fixed location on the system disk.
2. DNS is basically used to ____
----->i wrote --to have a hierarchical structure.
3.one more ques on DNS
4. abt TCP and UDP
5. TCP's which layer is there for the OSI's first three layers.
6. Multiprocessing models have---
a)symmetric b)unsymmetric models c)none of the above d) both
7. BCDhave binary nos in
a)bis b) bytes c)nibble
8. Top-down parser ....
a) leftmost derivation
b)rightmost derivation
c)leftmost derivation in reverse
d)rightmost derivation in reverse
9.Loop which avoids check at every iteration
a)loop controlling b) loop jamming
10.one tree was given and some expressions were there and we have to
find the correct postfix exprn
11.abt logic gates
12.worst case of Shell sort
13.sine wave can be split into
a)set of sine waves
b)set of sine waves with phase zero
14.full binary tree with n vertices has how many leaf nodes
somethng like this...
15.complete graph with n vertices and somethng
16.in a cyclic graph with n vertices how many minimum edges
a)n b)n/2...
i'll remember the rest give it later
1.The minimum number of edges in a connected cyclic graph on n vertices is
a) n-1 b) n c) n+1 d)none of these
2.A full binary tree with n non leaf nodes contains
a) n nodes b) log n nodes c)2n-1 nodes d)2n nodes
3.The time complexity of shell sort
a) O(n) b) O(log n) c) O(n 1.2 ) d)O(n2)
4.The time taken to insert an element after an element pointed by some pointer
a) O(1) b) O(log n) c) O(n) d) O(nlogn)
5. what is the name given to the first generation computer?
a) Binary language b)Machine language c)Assembly language
6.The root directory of a disk should be placed
a) at a fixed address in main memory
b) at a fixed location on disk
c) anywhere on disk.
7.A top down parser generates
a) right most derivation
b) left most derivation
c) right most derivation in reverse
d) left most derivation in reverse
8.what is the name of the OS that reads and reacts in terms of actual time?
a)batch system
b)time sharing
c)real time
9.FDDI is a
a)ring network
b)star network
c)mesh network
10.Computer memory consists of
a)ROM
b)PROM
c)RAM
d)all the above
*************************************************************************************
I have given info about my ISRO interview in my previous mail i am pasting it again here for reference. Some more questions all of which which was not asked to me but very important for interview because they were repeated for many candidates even on different dates. So i am listing them here
V. IMPORTANT: can this equation be solved tell verbally? x+y=5 ; 4x+4y=12; x+2y+3Z=20 .Answer: no solution since first two are contradicting so not possible
V. IMPORTANT: SQRT(170)=13 is it possible? Answer: it is possible the base may be other then 10 . We can find out the base let it be x. Then convert both side to decimal we get SQRT(x2 +7x+0)=x+3 => x2+7x=(x+3)2 => x=9 so for base 9 it is possible here "x2 etc means x square"
RISC and CISC
difference between array and structure
My Interview
it went for 20 minutes.
There were 8 person in panel and only one panel was there. there was a long round table between candidate and interviewers so we have to explain everything on board.
they asked about GATE score card ,percentile, why did not went for M.Tech. If you are doing M.Tech. or Job they may ask questions in your specialization
then some questions regarding project
favourite subjects ( I said DSA, C and DBMS . I didn't said OS because for OS they were asking w.r.t. Linux. In CN they may also ask general about Network Security ,firewall, encryption etc. CA include Digital, Microprocessor etc)
they are giving everybody a question from mathematics to solve on board, mostly to find the rank of matrix, solving linear equation (some tricky which can be answered without solving it properly) or probability.
They may give a K - Map to solve on board
In C(i told C as My favourite subject) they ask me about extern, static, linking, global variables, run time library , exit etc.
In DSA they ask me about expression evaluation infix ,postfix, how expression is evaluated etc
solve the equation x+y=5 ; 4x+4y=12; x+2y+3Z=20 .Answer: no solution since first two are contradicting so not possible | 1,355 | 5,324 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.71875 | 3 | CC-MAIN-2018-47 | latest | en | 0.815838 |
https://snippets.cacher.io/snippet/8d217ba91bd1888d6875 | 1,670,625,953,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446711552.8/warc/CC-MAIN-20221209213503-20221210003503-00781.warc.gz | 568,860,043 | 4,725 | 04pallav
9/13/2017 - 10:02 PM
## Survival Models
Survival Models
``````https://www.youtube.com/watch?v=Cg0Gl-_6TwM
https://www.analyticsvidhya.com/blog/2014/04/survival-analysis-model-you/
https://www.analyticsvidhya.com/blog/2014/04/solving-survival-model/
https://rpubs.com/mjeagle/Surv
Survival Models
##########################
They model time to an event. after having heart surgery when is the person dying ? T
We can Linear regrression because
1. The reponse variable T is not normal
2. We have partial information about the features
3. Can model even if we don't have an explantory variable
Properties
1. Individuals do not need to enter the study at same time (STaggered entry)
2. Can drop out
3. individual may be still alive at the end of the event. (censored information)
Functions
Density fn f(t) -- probability of event happening
survival fn S(t) -- P(T>=t) Probability that death happended after time t
Hazard fn -- Instantaneous event rate, the rate of death. Given the patient has survived what is the probability of death at time t (conditional prob)
Kaplan Meier esitmate of survival function
For example, age for marriage, time for the customer to buy his first product after visiting the website for the first time, time to attrition of an employee etc. all can be modeled as survival analysis.
Censored Data
Suppose patients are followed in a study for 20 weeks.
A patient who does not experience the event of interest for the duration of the study is said to be right
censored. Censoring that is random and non informative is usually required in order to avoid bias in
a survival analysis
There are three main types of censoring: right, left, and interval. Censoring could occur, for example, when
administering a survey to mothers every other month asking if they are still breast feeding. Right censoring
occurs when mothers are still breast feeding after the last survey, since we do not know exactly how long they
will continue. Left censoring occurs when mothers enter the study after they have stopped breast feeding. We
do not know exactly when they stopped breast feeding, although we know that it happened before their entry to
the study. Interval censoring occurs if the breast feeding ended between two successive surveys since one can
only say that breast feeding ended somewhere within the past two months.
Cox Proportional Hazards Models (Cox PH)
#############################################
Assumption : The hazard ratio doesnt vary with time (Imp)
To watch properly | 579 | 2,519 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.4375 | 3 | CC-MAIN-2022-49 | latest | en | 0.895077 |
https://spark.adobe.com/page/hqKQW2UEfUywF/ | 1,550,758,439,000,000,000 | text/html | crawl-data/CC-MAIN-2019-09/segments/1550247504790.66/warc/CC-MAIN-20190221132217-20190221154217-00514.warc.gz | 713,057,227 | 5,585 | ## KeiraChapter 11 vocab
Non-contact force
A force that acts at a,distance
If you have put a magnet on a sting and let it try to touch another magnet that is non-contact force.
Force
A push or a pull that acts on an object
If you have ever pushes something then you are using force.
Gravity
The force of attraction between any two objects
Sometimes we hate gravity because it pulls stuff down.
Contact force
A force that requires two pieces of matter to touch
When I watch racing the crew has to use Contact force.
Acceleration
A rate at which the speed or direction of motion of an object changes over time
I saw my dad skee and that is what I saw nothing but a lot of him moving fast.
Friction
The force that results when two materials rub against each other or when their contact prevents sliding
Without fristion you could not walk on a base ball field.
Wedge
Simple machine made of one or two inclined planes
One of these simple machines is a stirring weak that makes driving a lot easier.
Pulley
Simple machine consisting of a rope or cable that runs around a groove wheel
In roller coasters that is what they used to use.
Intertia
The tendency of an object to resist any change in motion
Screw
Simple machine consisting of a smooth cylinder with a tiny inclined plane whrapped around it
We need a screw for our new horse painting.
Well and axle
Simple machine made up of a circular object attached to a bar
Praise the Lord we have weals and axles for our cars.
Simple machine
A machine made up of two or more parts
Me and my dad like to use simple machines like a nut cracker
Incline plane
Simple Michener consisting. Of a flat surface with one end higher that the other
You can see inclined planes on a moving truck or on a car that you donate big stuff in
Lever
A simple machine in witch a bar moves around a fixed point called fulcrum
The lever opened up the door when we pulled it.
# Report Abuse
If you feel that this video content violates the Adobe Terms of Use, you may report this content by filling out this quick form. | 459 | 2,079 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.6875 | 3 | CC-MAIN-2019-09 | latest | en | 0.923276 |
https://www.jiskha.com/display.cgi?id=1213923291 | 1,511,482,806,000,000,000 | text/html | crawl-data/CC-MAIN-2017-47/segments/1510934807044.45/warc/CC-MAIN-20171123233821-20171124013821-00394.warc.gz | 837,952,124 | 5,265 | # algebra
posted by .
convert 5.346 to a percent
Convert 38.6% to a decimal.
• algebra -
multiply by 100
divide by 100
• algebra -
Solve the following algebraically. Trial and error is not an appropriate method of solution. You must show all of your work and write answers in rational form.
3 x + 7 = 9
Show your work here: 3 x + 7 = 9
-7 - 7
3/3 x = 2/3
X = 2/3
b)
Show your work here: -3(x +5) + 3 = 12
-3x+15 +3 =12
- 3 x + 18 = 12
- 18 - 18
3/3x = -6/3
X = -2
C)
Show your work here: 3(2/3 x+ 1/6x)=2.3 2x + ½ x = 6 x =6 x =5 x= 12/5
d)
Show your work here: -2 x – 4 < 10 + 4 +4 -2/-2 x < 14/-2 x = - 7
2)
a) Solve for y
Answer: y = 2 – 3 x
Show your work here: 3 x + 4 y = 8
-3 x 3 x
4/4 y = 8 -3 x
Y = 2 – 3 x
b) When graphed this equation would be a line. By examining your answer to part a, what is the slope and y-intercept of this line?
Slope = -3
Y-intercept = 2
c) Using your answer from part a, find the corresponding value of y when x =8.
Show your work here: x =8 y = 2 -3x y = 2 – 3 (8) y = 2 -24 y =22
3) The following graph shows Mary’s salary from the year 2004 to the year 2007. She was hired in the year 2004; therefore x = 0 represents the year 2004.
a) List the coordinates of any two points on the graph in (x, y) form.
(x3, y¹ 5), (x²1-, 1y²)
b) Find the slope of this line:
X² - x¹
Show your work here: m= - 1 + ¯5 = ¯6
1 + ¯3 = ¯2 slope = 3
c) Find the equation of this line in slope-intercept form.
Answer: y =3 x + 1
Show or explain your work here: y = m x + b
Y = 3 x + 1
d) If Mary’s salary trend continued, what would her salary be in the year 2014? Show how you obtained your answer using part c).
Answer: Mary salary will be 43,500 in 2004.
Show or explain your work here: m x b = y
1500(7) + 33,000 = y 10,500 +33,000= y= 43,500
4) Suppose that the width of a rectangle is 2 inches shorter than the length and that the perimeter of the rectangle is 80. P= 80 x = width x +2 = length
P = 2 (1) + 2 (w)
a) Set up an equation for the perimeter involving only L, the length of the rectangle. Answer: x + 2 = length
b) Solve this equation algebraically to find the length of the rectangle. Find the width as well.
Answer: Length _x = 2 = 19 + 2= 21, Width _x = 19
a) Show your work here: 2 (x = 2) + 2x = 80 2x + 4 + 2 x = 80 4x + 4 = 80 -4 -4 4x/4 = 76/4 x = 19 width = x= 19 length = x = 2 =19 + 2 = 21
5) A tennis club offers two payment options:
Option1: \$32 monthly fee plus \$3/hour for court rental
Option 2: No monthly fee but \$6.50/hour for court rental.
Let x = hours per month of court rental time.
a) Write a mathematical model representing the total monthly cost, C, in terms of x for the following:
Option 1: C=32 + 3 (x)
Option 2: C=_\$6.50 (x)
b) How many hours would you have to rent the court so that the monthly cost of option 1, is less than option 2. Set up an inequality and show your work algebraically using the information in part a. Round to 3 decimal places if necessary.
6) Plot the following points on the given rectangular coordinate system by clicking on the given dots and dragging them.
If you were to connect these points with a line, where would the y-intercept be located? Give answer in (x, y) form.
(___, ___)
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More Similar Questions | 1,639 | 5,092 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.03125 | 4 | CC-MAIN-2017-47 | latest | en | 0.861515 |
https://number.academy/1007094 | 1,719,203,744,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198864986.57/warc/CC-MAIN-20240624021134-20240624051134-00597.warc.gz | 373,259,944 | 11,109 | # Number 1007094 facts
The even number 1,007,094 is spelled 🔊, and written in words: one million, seven thousand and ninety-four, approximately 1.0 million. The ordinal number 1007094th is said 🔊 and written as: one million, seven thousand and ninety-fourth. The meaning of the number 1007094 in Maths: Is it Prime? Factorization and prime factors tree. The square root and cube root of 1007094. What is 1007094 in computer science, numerology, codes and images, writing and naming in other languages
## What is 1,007,094 in other units
The decimal (Arabic) number 1007094 converted to a Roman number is (M)(V)MMXCIV. Roman and decimal number conversions.
#### Time conversion
(hours, minutes, seconds, days, weeks)
1007094 seconds equals to 1 week, 4 days, 15 hours, 44 minutes, 54 seconds
1007094 minutes equals to 2 years, 3 weeks, 6 days, 8 hours, 54 minutes
### Codes and images of the number 1007094
Number 1007094 morse code: .---- ----- ----- --... ----- ----. ....-
Sign language for number 1007094:
Number 1007094 in braille:
QR code Bar code, type 39
Images of the number Image (1) of the number Image (2) of the number More images, other sizes, codes and colors ...
## Share in social networks
#### Is Prime?
The number 1007094 is not a prime number. The closest prime numbers are 1007089, 1007099.
#### Factorization and factors (dividers)
The prime factors of 1007094 are 2 * 3 * 11 * 15259
The factors of 1007094 are
1, 2, 3, 6, 11, 22, 33, 66, 15259, 30518, 45777, 91554, 167849, 335698, 503547, 1007094.
Total factors 16.
Sum of factors 2197440 (1190346).
#### Powers
The second power of 10070942 is 1.014.238.324.836.
The third power of 10070943 is 1.021.433.331.512.386.560.
#### Roots
The square root √1007094 is 1003,540732.
The cube root of 31007094 is 100,23591.
#### Logarithms
The natural logarithm of No. ln 1007094 = loge 1007094 = 13,82258.
The logarithm to base 10 of No. log10 1007094 = 6,00307.
The Napierian logarithm of No. log1/e 1007094 = -13,82258.
### Trigonometric functions
The cosine of 1007094 is 0,99728.
The sine of 1007094 is -0,073709.
The tangent of 1007094 is -0,07391.
## Number 1007094 in Computer Science
Code typeCode value
1007094 Number of bytes983.5KB
Unix timeUnix time 1007094 is equal to Monday Jan. 12, 1970, 3:44:54 p.m. GMT
IPv4, IPv6Number 1007094 internet address in dotted format v4 0.15.93.246, v6 ::f:5df6
1007094 Decimal = 11110101110111110110 Binary
1007094 Decimal = 1220011110210 Ternary
1007094 Decimal = 3656766 Octal
1007094 Decimal = F5DF6 Hexadecimal (0xf5df6 hex)
1007094 BASE64MTAwNzA5NA==
1007094 MD5689a5894bae086feb3e0f4d193c997c6
1007094 SHA1bd9a8eb50f3b9bc90990c8270789e20b52025c01
1007094 SHA2561a7acb86453664ec951cc539c66162d063d4bb914a6b52bd1605954d6080f510
More SHA codes related to the number 1007094 ...
If you know something interesting about the 1007094 number that you did not find on this page, do not hesitate to write us here.
## Numerology 1007094
### Character frequency in the number 1007094
Character (importance) frequency for numerology.
Character: Frequency: 1 1 0 3 7 1 9 1 4 1
### Classical numerology
According to classical numerology, to know what each number means, you have to reduce it to a single figure, with the number 1007094, the numbers 1+0+0+7+0+9+4 = 2+1 = 3 are added and the meaning of the number 3 is sought.
## № 1,007,094 in other languages
How to say or write the number one million, seven thousand and ninety-four in Spanish, German, French and other languages. The character used as the thousands separator.
Spanish: 🔊 (número 1.007.094) un millón siete mil noventa y cuatro German: 🔊 (Nummer 1.007.094) eine Million siebentausendvierundneunzig French: 🔊 (nombre 1 007 094) un million sept mille quatre-vingt-quatorze Portuguese: 🔊 (número 1 007 094) um milhão e sete mil e noventa e quatro Hindi: 🔊 (संख्या 1 007 094) दस लाख, सात हज़ार, चौरानवे Chinese: 🔊 (数 1 007 094) 一百万七千零九十四 Arabian: 🔊 (عدد 1,007,094) مليون و سبعة آلاف و أربعة و تسعون Czech: 🔊 (číslo 1 007 094) milion sedm tisíc devadesát čtyři Korean: 🔊 (번호 1,007,094) 백만 칠천구십사 Danish: 🔊 (nummer 1 007 094) en millioner syvtusinde og fireoghalvfems Hebrew: (מספר 1,007,094) מיליון ושבעת אלפים תשעים וארבע Dutch: 🔊 (nummer 1 007 094) een miljoen zevenduizendvierennegentig Japanese: 🔊 (数 1,007,094) 百万七千九十四 Indonesian: 🔊 (jumlah 1.007.094) satu juta tujuh ribu sembilan puluh empat Italian: 🔊 (numero 1 007 094) un milione e settemilanovantaquattro Norwegian: 🔊 (nummer 1 007 094) en million syv tusen og nittifire Polish: 🔊 (liczba 1 007 094) milion siedem tysięcy dziewięćdziesiąt cztery Russian: 🔊 (номер 1 007 094) один миллион семь тысяч девяносто четыре Turkish: 🔊 (numara 1,007,094) birmilyonyedibindoksandört Thai: 🔊 (จำนวน 1 007 094) หนึ่งล้านเจ็ดพันเก้าสิบสี่ Ukrainian: 🔊 (номер 1 007 094) один мільйон сім тисяч дев'яносто чотири Vietnamese: 🔊 (con số 1.007.094) một triệu bảy nghìn lẻ chín mươi bốn Other languages ...
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The content of the comments is the opinion of the users and not of number.academy. It is not allowed to pour comments contrary to the laws, insulting, illegal or harmful to third parties. Number.academy reserves the right to remove or not publish any inappropriate comment. It also reserves the right to publish a comment on another topic. Privacy Policy. | 1,868 | 5,556 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.375 | 3 | CC-MAIN-2024-26 | latest | en | 0.763309 |
https://www.physicsforums.com/threads/splitting-the-derivative.609175/ | 1,519,176,323,000,000,000 | text/html | crawl-data/CC-MAIN-2018-09/segments/1518891813187.88/warc/CC-MAIN-20180221004620-20180221024620-00489.warc.gz | 941,031,244 | 16,061 | # Splitting the derivative?
1. May 26, 2012
### KingBigness
1. The problem statement, all variables and given/known data
From
$\frac{de_{s}}{dT} = \frac{L_{v}e_{s}}{R_{v}T^{2}}$
derive
$e_{s}(T) = 6.11 e^{\frac{L}{RV}(\frac{1}{T}-\frac{1}{273})}$
2. Relevant equations
3. The attempt at a solution
The way my lecturer derived it was he 'split' the derivative and took them to their respective sides and integrated.
So he got
$\frac{de_{s}}{e_{s}} = \frac{LdT}{R_{v}T^{2}}$
However I was under the impression that you can't 'split' a derivative like that. Is this just a shortcut some physicists take to make the maths more simple? If it is what is the correct way of deriving this?
2. May 26, 2012
### Curious3141
3. May 26, 2012
### KingBigness
How accepted is it? I am doing a double degree in pure mathematics and physics. If I was to do this in my pure maths classes would I be stoned?
4. May 26, 2012
### iRaid
I've taken calculus 1 and I've seen it before? :s
5. May 26, 2012
### KingBigness
Sorry I wasn't very clear. I have seen it before and done it countless times I just always have a memory of a teacher saying don't tell a pure mathematician about it.
6. May 26, 2012
### SammyS
Staff Emeritus
In my experience, physicists & engineers are notorious for treating Leibniz's notation for the derivative as if it were a fraction.
A more rigorous handling of this derivative equation might be as follows:
Treating es as a function of T we have
$\displaystyle \frac{d}{dT}(e_{s}) = \frac{L_{v}e_{s}}{R_{v}T^{2}}$
Rewriting this equation gives us
$\displaystyle \frac{1}{e_{s}}\frac{d}{dT}(e_{s}) = \frac{L}{R_{v}T^{2}}$
Integrating w.r.t. T gives
$\displaystyle \int {\frac{1}{e_{s}}\frac{d}{dT}(e_{s})}dT = \int{\frac{L}{R_{v}T^{2}}}dT$
We can rewrite the integral on the left hand side.
$\displaystyle \int {\frac{1}{e_{s}}\frac{d}{dT}(e_{s})}dT = \int {\frac{1}{e_{s}}}\,de_{s}$
Alternatively, if $\displaystyle \frac{d}{dT}(e_{s}) = \frac{L_{v}e_{s}}{R_{v}T^{2}}\,,$
then the differential of es is given by $\displaystyle d\,e_{s} = \frac{L_{v}e_{s}}{R_{v}T^{2}}\,dT$
7. May 27, 2012
### KingBigness
exactly what I wanted. thank you | 744 | 2,182 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.671875 | 4 | CC-MAIN-2018-09 | longest | en | 0.895235 |
https://en.wikipedia.org/wiki/Talk:Exponential_function | 1,519,582,038,000,000,000 | text/html | crawl-data/CC-MAIN-2018-09/segments/1518891816841.86/warc/CC-MAIN-20180225170106-20180225190106-00088.warc.gz | 683,152,926 | 48,326 | # Talk:Exponential function
WikiProject Mathematics (Rated B-class, Top-importance)
This article is within the scope of WikiProject Mathematics, a collaborative effort to improve the coverage of Mathematics on Wikipedia. If you would like to participate, please visit the project page, where you can join the discussion and see a list of open tasks.
Mathematics rating:
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Field: Basics
One of the 500 most frequently viewed mathematics articles.
## Notation
Aren't there multiple forms of the exponential function? Specifically, isn't the most general form kabx ? Admittedly, the differences can be incorporated in to b thus: kebxln(a), but wouldn't the first form be more clear to the laymen?
You don't really need the b; the most general form is kax. I mention these in the "science" paragraph; they are indeed the most useful to the "laymen". However they cannot be defined without exp(x) and in mathematics, exp(x) is tremendously more important than ax, so I started the article with exp(x) and then came to ax as soon as possible. --AxelBoldt
You can define ax without defining exp(x). For positive a, you define a0 = 1, an+1 = a×an for natural n, and an=1/an. Then you define a1/n to be the unique positive real x such as xn = a, and am/n=(a1/n)m. Since any real x can be approximated arbitrarily well by a fraction m/n, and am/n is monotonous, you just need to require that ax be continuous. --Army1987 12:00, 19 September 2006 (UTC)
I've changed the opening paragraphs to introduce the form kax even sooner. - dcljr 12:50, 6 Aug 2004 (UTC)
It shouldn't read: The graph of ex does not ever touch the x axis, although it comes very close.
but rather: The graph of ex does not ever touch the x axis, although it comes arbitrarily close (in a limit sense).
## Coherent Graphs
Many of the math articles have graphs in smooth brown, using the same design all over. (Other colors also exist, like in the article Taylor expansion.) How are these graphs created? Interesting to anyone starting a new maths article. -- Sverdrup 22:40, 9 Dec 2003 (UTC)
The graphs were written in Java, and copied from the screen. The code is at User:Cyp/Java. Κσυπ Cyp 01:27, 10 Dec 2003 (UTC)
## My edits of Aug 6, 2004
• Rearranged things to put off non-real variables as long as possible. I think this will benefit readers who are less mathematically experienced.
• Right-aligned the graph again (see page history). What's wrong with having it right-alinged?
• Consistently replaced <i>italics</i> with ''wikitalics''.
• Included multiple interpretations of ${\displaystyle {d \over dx}e^{x}=e^{x}}$. I know they're redundant, but that's the point: to say it in ways that might be more familiar to the reader. A picture would be particularly helpful here, I think.
• Removed mention of "linear ordinary differential equations" in the similarly named section; if you think it's crucial, add it back (see page history).
• Removed parenthetical remark in "Banach algebras" section:
if ${\displaystyle xy=yx}$ (we should add the general formula involving commutators here.)
After much struggle, I decided not to change anything else of substance in that section or the next one on Lie algebras. I don't trust my understanding of these things (or lack thereof). Speaking of the "Lie groups" section, someone should probably try to make it more clear.
Oh, and BTW: Do we really need the same properties listed 3 times? I know we're talking about different mathematical objects at different places in the article, but still, I find it kinda redundant. Couldn't we name or number the properties and refer to them that way?
- dcljr 12:43, 6 Aug 2004 (UTC)
## information should be reordered
IMHO,
• the very first paragraph should be as short as possible for several reasons
• editing it makes it necessary to edit the whole page, which may become impossible at some point. So it should only contain things that are absolutely necessary and which will (almost) certainly never need any modification.
• it lessens the usefulness of the "Contents" table, which should come before any detailed information, except for a minimalistic explanation of "what is this page about" and "what is found elsewhere".
• there are too many details about the graph of the exp function in the 1st paragraph (postitive with explanation, increasing with explanation), and still its not complete (convexity, asymptotics, ...). The picture itself is enough on top of the article, as it contains all that information.
• It is strange to have the (body of the) article start with "Properties" instead of an (even informal) definition.
• It is even more strange to have the "Properties" to start with a generalization of the exp function.
MFH 15:07, 8 Apr 2005 (UTC)
## very strange
It is taught in China that y = a to the x-th power is the standard definition of exponential function, while y = e to the x-th power is just a particular case of exponential functions. Got confused when try to translate this article to zh.wikipedia.org.
See exponentiation Bo Jacoby 16:36, 23 October 2005 (UTC)
But this is the article about THE exponential fucntion. AN exponential fuction is y=a^x (which can be derived from THE exponential function by imputting x=ln(a)x). I would therefore agree with China's definition of AN exponential function.--Hypergeometric2F1[a,b,c,x] 11:03, 20 December 2005 (UTC)
This article reads like a mathematical textbook for degree students, not an encyclopedia article. What is its intended audience? People who are studying mathmetics will surely have the literature to tell them what Exponential is, and won't be consulting wikipedia. —The preceding unsigned comment was added by 82.46.29.168 (talkcontribs) 23:10, 18 September 2006 (UTC2)
## Comment moved from article
Moved from top of article [The definition below is incomplete and not rigorous (Paulo Eneas, SP, Brazil)] (Enchanter 22:09, 26 November 2005 (UTC))
## Title change
This article should be named "The Exponential Function" or at least there should be a disambiguation page differentiating this from "AN Exponential Function" f(x)=a^x, to avoid possible confusion.--Hypergeometric2F1[a,b,c,x] 11:06, 20 December 2005 (UTC)
I disagree with the move to "The exponential function". Maybe something more can be said about the general a^x function in addition to what already is in here. Oleg Alexandrov (talk) 21:34, 21 December 2005 (UTC)
If the article were to be moved it should be named the natural exponential function in accord with the term "natural logarithm" (a logarithm can be based on any exponential function in the form a^x, e^x is special because of his "natural" properties). --Friðrik Bragi Dýrfjörð 16:30, 27 April 2006 (UTC)
## sin−1x is ambiguous
There is no need for a shorthand of this kind for reciprocal trigonometric functions since they each have their own name and abbreviation already: (sin x)-1 is normally just written as csc x.
Actually, also the arcsine has its own abbreviaton, i.e. arcsin x. And sin−1x is sometimes used to mean 1/sin x. Then we have cscx and arcsinx, which are both unambiguous, whereas sin−1x is ambiguous. See Trigonometric functions#Inverse function. I personally never use the notation sin−1x to avoid misunderstandings. But the current wording implies that it always refers to the arcsine. (And by the way, I've seen the notation fnx for any functions denotated by a string of lowercase latin characters, e.g. log2x to mean (logx)2 rather than log logx. --Army1987 20:41, 10 March 2006 (UTC)
## Only with base of e?
I was under the impression that exponential functions included all functions in the form of f(x) = abx, not necessarily with a base of Euler's number. For example, y = 5(6x). DroEsperanto 18:00, 15 November 2006 (UTC)
- I think "THE exponential function" is definitely 'e^x', whilst a^x is referred to as AN exponential function Fmadd (talk) 18:12, 22 April 2016 (UTC)
## How to compute exp(x) in computers
Something like:
exp(x) is typically computed using a floating point x and giving a floating point result y. A floating point value y is a structure or tuple of 3 values - sign, exponent and mantissa. exp(x) always gives a positive result (exp(x) >= 0 for all x) so sign is always 0 or positive. Thus, we only have to find the mantissa and the exponent.
y = mantissa * 2^exponent.
log(y) = log(mantissa) + exponent * log(2)
However, log(y) is also log(exp(x)) = x so this means x = log(mantissa) + exponent*log(2).
Thus, dividing x with log(2) - a fixed constant - and getting an integer quotient exponent and a fractional part U where 0 <= U < log(2) can therefore be done. We then next compute exp(U) to find the mantissa.
Since U is small (0 <= U < log(2)) this mantissa = exp(U) can be found by using a series computation as described on the page already. Thus, we find mantissa and can compute the result y by simply putting these parts together in a floating point number.
-- I would argue this is better in a separate article. Enough detail to warrant an article, or too much detail in one area for a concept as broad and universal as 'the exponential function'. It can be linked from floating point, and here. Perhaps an article, 'floating point tricks' or whatever?Fmadd (talk) 18:16, 22 April 2016 (UTC)
## Computing exp(x) in computers
When computing y = exp(x) we can first consider how floating point values are represented in computers. A floating point value y is represented as ${\displaystyle y=smb^{n}}$ Here s is +1 for positive values and -1 for negative values. For y = exp(x) we always have y >= 0 and so s is always +1. b is typically 2 for most computers.
If y = exp(x) we also have log(y) = x and so we get:
${\displaystyle x=\log(y)=\log(sm2^{n})=\log(s)+\log(m)+n\log(2)}$
log(s) = 0 since s = +1.
Thus n is found as the integer such that
${\displaystyle n<={\frac {x}{\log(2)}}
We then find log(m) as the value ${\displaystyle \log(m)=x-n\log(2)}$.
Since log(m) is guaranteed to be in the range ${\displaystyle 0<=log(m)<\log(2)}$ we know that log(m) is small enough so that we can use the previously indicated series: ${\displaystyle m=\exp(\log(m))=1+\log(m)(1+\log(m)({\frac {1}{2}}+\log(m)(...)))}$
Thus, the result ${\displaystyle y=exp(x)=m2^{n}}$ where m is found by the series and n is found earlier by the division of x/log(2).
For complex variables ${\displaystyle z=x+yi}$ it is simply an excersize in the identity:
${\displaystyle e^{x+yi}=e^{x}e^{yi}=e^{x}(\cos(y)+i\sin(y))=e^{x}\cos(y)+ie^{x}\sin(y)}$
A general a^x function is defined as exp(x \log(a)) and a^b where both a and b being complex is thus deifned in terms of exp(b \log(a)) where both a and b is complex and log(a) for complex a yield a complex value and the complex multiplication yield a complex result which is then in turn as deifned above.
To outline it:
a = x + yi b = u + vi
Need to compute log(a) first, convert a to polar co-ordinates:
${\displaystyle r={\sqrt {x^{2}+y^{2}}}}$ ${\displaystyle \theta =\arctan 2(y,x)}$
Note that both the square root and the arctan2 here have real values as arguments.
The above gives us ${\displaystyle a=re^{{\theta }i}}$ and so ${\displaystyle \log(a)=\log(r)+i\theta }$ where we only use the principal value of the multi-valued function.
Thus, ${\displaystyle a^{b}=exp(b\log a)=\exp((u+vi)(\log(r)+{\theta }i))=exp(u\log(r)-v\theta +(u\theta +v\log(r))i)}$
If we define ${\displaystyle p=u\log(r)-v\theta }$ and ${\displaystyle q=u\theta +v\log(r)}$ (both p and q are real values) we then get:
${\displaystyle a^{b}=\exp(p+qi)=\exp(p)\exp(qi)=\exp(p)(\cos(q)+i\sin(q))}$
Thus, a^b for complex a and complex b can be defined.
I plan to add the above to the regular page in 3 days - comments are appreciated.
salte 13:06, 4 December 2006 (UTC)
Added several sections. I didn't copy it exactly as I wrote it above since I split the description into several sections. I also added a description of the algorithm for ${\displaystyle a^{n}}$ for positive integers n as well as a short comment on how to expand it for all integers n. I also placed a description of exp(z) for complex z below the description of complex exponential and a description of how to compute ${\displaystyle a^{b}}$ for complex a and complex b. Hope people find this useful.
salte 10:04, 8 December 2006 (UTC)
## incorrect formula
Factorial signs (!) are missing in the denominators in the formula in 'Exponential_function#Computing exp(x) for real x'. The correct formulas is in the subsection just above on 'numerical value'. Bo Jacoby 14:57, 8 February 2007 (UTC).
## Link to proof for article that ${\displaystyle e^{x}}$ is the only non-zero function that is its own derivative?
The article correctly states that "${\displaystyle e^{x}}$ is its own derivative. It is the only function with that property (other than the constant function f(x)=0)." However I noticed it doesn't provide a proof or a link to a proof of that uniqueness. It would be nice if someone could provide a footnote reference at that sentence that links to a published proof that ${\displaystyle e^{x}}$ is the only non-zero function that is its own derivative. Dugwiki 22:54, 4 April 2007 (UTC)
How about just pointing to the Picard-Lindelöf theorem? - Fredrik Johansson 23:06, 4 April 2007 (UTC)
No, I don't think that quite works because the function ${\displaystyle e^{x}}$ isn't Lipschitz continuous as required by the theorem. In order for ${\displaystyle e^{x}}$ to be Lipschitz continuous, there would have to exist a constant ${\displaystyle K\geq 0}$ such that for all ${\displaystyle x_{1},x_{2}}$ in the reals ${\displaystyle |e^{x_{1}}-e^{x_{2}}|\leq K|x_{1}-x_{2}|}$. That isn't true, though. Note that f(t)=0 is the only one of the two functions that is Lipschitz continuous, and so is the unique Lipschitz continuous solution implied by the PLT if f(x)=f'(x) and f(0)=0. Dugwiki 15:48, 5 April 2007 (UTC)
In fact, I just realized that all functions of the form ${\displaystyle Ke^{x}}$ for constant K have this trait. So any constant multiple works. Dugwiki 15:53, 5 April 2007 (UTC)
The exponential function in this case is the "y" for which the equation is being solved. The function f in the case of the differential equation ${\displaystyle y'=y}$ is simply the identity function with respect to the second argument, ${\displaystyle f(t,y)=y}$, which is Lipschitz. As you say, there are infinitely many solutions; to get uniqueness per the Picard theorem, you need to specify the initial value ${\displaystyle y(0)=1}$. Fredrik Johansson 16:19, 5 April 2007 (UTC)
Ah, thanks, that clears it up for me. I was misreading the theorem, basically thinking that "y" was the Lipschitz continuous function. So you're correct, and in fact PLT says that the function ${\displaystyle y=Ke^{t}}$ is the unique function that solves the differential equation ${\displaystyle y'(t)=y(t),y(0)=K}$ (with the identity function f(t,y(t)) = y(t)). I'll include that reference in the article as well. Dugwiki 17:31, 5 April 2007 (UTC)
## vague lede
The lede of this article doesn't tell us what is an exponential function. The lede statement appears at the top of the first section, while the lede includes several statements that tell us facts about exponential functions, but not what is an exponential function.
The rambling lede:
1. It is one of the most important functions in mathematics. (about the function, not a general description of the function)
2. ... is written as... (about representing the function, not a general description of the function)
3. (or) this can be written in the form... (about representing the function, not descriptive of the entire function)
4. ...function is nearly flat ... (about aspects of the function, not a general description of the function)
5. ... the graph of y=ex is always positive... (about aspects of the function, not a general description of the function)
6. Sometimes, the term exponential function is more generally used for functions of the form kax, where a, called the base, is any positive real number not equal to one. (how the function is alternately used)
7. ...This article will focus initially on the exponential function with base e, Euler's number. ( about the article, not about the subject)
8. In general, the variable x can be any real or complex number, or even an entirely different kind of mathematical object; see the formal definition below. (about aspects of the function, not a general description of the function)
Only in the first section do we get the simple answer, couched in a rhetorical apology for not having told us sooner.
• Most simply, exponential functions multiply at a constant rate.
This sentence belongs at the top, but I don't plan to fix it by entering a correct lede sentence: The exponential functions is an (arithmetic? algebraic? mathematic?) function that multiplies values at a constant rate. Last week, or last month, the style would have been "In mathematics, the exponential function is" but this project is so disorganized, there is no way to know if that is the preferred style today. Doubtless there are two groups somewhere in this project that both claim to have reached a consensus, each around contradicting styles. People get so harassed for adding content to Wikipedia articles someone else owns, I would rather expose the weakness and let it be. If you don't want to be associated with weak, vague text, you fix it. Arithmawhiz 18:48, 18 July 2007 (UTC)
Well put. I have not worked on exponential function, but on exponentiation, where you will find a definition , see Exponentiation#Powers_of_e. Bo Jacoby 16:55, 5 August 2007 (UTC).
## Continued fraction for ${\displaystyle e^{x}}$
A generalized continued fraction for ${\displaystyle e^{x}}$ can be constructed by examining the simple continued fractions for ${\displaystyle e^{1/n}\,\!.}$ and ${\displaystyle e^{2/n}\,\!.}$ formulas found in http://en.wikipedia.org/wiki/Continued_fraction#Regular_patterns_in_continued_fractions.
${\displaystyle e^{1/n}=[1;n-1,1,1,3n-1,1,1,5n-1,1,1,7n-1,1,1,...,(2k+1)n-1,1,1,...]\,\!.}$
Taking the 1st, 4th, 7th, and 10th convergents for each value of n, a pattern develops:
${\displaystyle e^{1/1}:{1 \over 1}{3 \over 1}{19 \over 7}{193 \over 71}...=1+{2 \over 1+}{1 \over 6+}{1 \over 10+}{1 \over 14+}...{1 \over 4k+2+}...}$
${\displaystyle e^{1/2}:{1 \over 1}{5 \over 3}{61 \over 37}{1225 \over 743}...=1+{2 \over 3+}{1 \over 12+}{1 \over 20+}{1 \over 28+}...{1 \over 8k+4+}...}$
${\displaystyle e^{1/3}:{1 \over 1}{7 \over 5}{127 \over 91}{3817 \over 2735}...=1+{2 \over 5+}{1 \over 18+}{1 \over 30+}{1 \over 42+}...{1 \over 12k+6+}...}$
Leading to ${\displaystyle e^{1/n}=1+{2 \over 2n-1+}{1 \over 6n+}{1 \over 10n+}{1 \over 14n+}...{1 \over 2n(2k+1)+}...}$
and, for m = 2n, ${\displaystyle e^{2/m}=1+{2 \over m-1+}{1 \over 3m+}{1 \over 5m+}{1 \over 7m+}...{1 \over m(2k+1)+}...}$
${\displaystyle e^{2/m}=[1;(m-1)/2,6m,(5m-1)/2,1,1,\dots ,3k+(m-1)/2,(12k+6)m,3k+(5m-1)/2,1,1,...]\,\!}$
Here, take the 1st, 3rd, 6th, 8th and 11th convergents for each odd value of m, and
multiply the numerator and denominator of the 2nd and 7th convergents for another pattern:
${\displaystyle e^{2/1}:{1 \over 1}{2 \over 0}{7 \over 1}{37 \over 5}{266 \over 36}{2431 \over 329}{27007 \over 3655}...=1+{2 \over 0+}{1 \over 3+}{1 \over 5+}{1 \over 7+}...{1 \over 2k+1+}...}$
${\displaystyle e^{2/3}:{1 \over 1}{4 \over 2}{37 \over 19}{559 \over 287}{11776 \over 6046}...=1+{2 \over 2+}{1 \over 9+}{1 \over 15+}{1 \over 21+}...{1 \over 6k+3+}...}$
${\displaystyle e^{2/5}:{1 \over 1}{6 \over 4}{91 \over 61}{2281 \over 1529}{79926 \over 53576}...=1+{2 \over 4+}{1 \over 15+}{1 \over 25+}{1 \over 35+}...{1 \over 10k+5+}...}$
Leading to ${\displaystyle e^{2/m}=1+{2 \over m-1+}{1 \over 3m+}{1 \over 5m+}{1 \over 7m+}...{1 \over m(2k+1)+}...}$ - what we had for m even!
Now, please notice a pattern for the continued fractions of ${\displaystyle e^{x/3}\,\!.}$, based on ${\displaystyle e^{2/m}\,\!.}$:
${\displaystyle e^{1/3}=e^{2/6}=1+{2 \over 5+}{1 \over 18+}{1 \over 30+}{1 \over 42+}{1 \over 54+}...=1+{2 \over 5+}{1 \over 18+}{1 \over 30+}{1 \over 42+}{1 \over 54+}...}$
${\displaystyle e^{2/3}=e^{2/3}=1+{2 \over 2+}{1 \over 9+}{1 \over 15+}{1 \over 21+}{1 \over 27+}...=1+{4 \over 4+}{4 \over 18+}{4 \over 30+}{4 \over 42+}{4 \over 54+}...}$
${\displaystyle e^{3/3}=e^{2/2}=1+{2 \over 1+}{1 \over 6+}{1 \over 10+}{1 \over 14+}{1 \over 18+}...=1+{6 \over 3+}{9 \over 18+}{9 \over 30+}{9 \over 42+}{9 \over 54+}...}$
From the above comes a similar pattern for ${\displaystyle e^{x}\,\!.}$:
${\displaystyle e^{1}=e^{2/2}=1+{2 \over 1+}{1 \over 6+}{1 \over 10+}{1 \over 14+}...=1+{2 \over 1+}{1 \over 6+}{1 \over 10+}{1 \over 14+}...}$
${\displaystyle e^{2}=e^{2/1}=1+{2 \over 0+}{1 \over 3+}{1 \over 5+}{1 \over 7+}...=1+{4 \over 0+}{4 \over 6+}{4 \over 10+}{4 \over 14+}...}$
${\displaystyle e^{3}=e^{2/(2/3)}=1+{2 \over -1/3+}{1 \over 2+}{1 \over 10/3+}{1 \over 14/3+}...=1+{6 \over -1+}{9 \over 6+}{9 \over 10+}{9 \over 14+}...}$
Leading to ${\displaystyle e^{x}=1+{2x \over 2-x+}{x^{2} \over 6+}{x^{2} \over 10+}{x^{2} \over 14+}...{x^{2} \over 4k+2+}...}$
Special cases for ${\displaystyle e^{x}}$:
For x=1, e^1 = e^(1/1) changes from [1;0,1,1,2,1,1,4,1,1,...,2k-2,1,1,...] to [2;1,2,1,1,4,1,1,...,2k,1,1,...].
This changes ${\displaystyle e^{1}=1+{2 \over 1+}{1 \over 6+}{1 \over 10+}{1 \over 14+}{1 \over 18+}...}$ to ${\displaystyle 2+{1 \over 1+}{2 \over 5+}{1 \over 10+}{1 \over 14+}{1 \over 18+}...}$
For x=2, e^2 = e^(2/1) changes from [1; 0,6,2,1,1, 3,18,5,1,1, ...,3k,12k+6,3k+2,1,1,...] to [7; 2,1,1,3,18, 5,1,1,6,30, ...,3k-1,1,1,3k,12k+6,...].
This changes the cumbersome ${\displaystyle e^{2}=1+{2 \over 0+}{1 \over 3+}{1 \over 5+}{1 \over 7+}{1 \over 9+}{1 \over 11+}...}$ to ${\displaystyle 7+{2 \over 5+}{1 \over 7+}{1 \over 9+}{1 \over 11+}...}$
Glenn L (talk) 03:53, 28 February 2008 (UTC)
Update: Glenn L (talk) 04:16, 11 May 2009 (UTC)
Update #2: These results for e^x are now reflected on the main page. -- Glenn L (talk) 05:12, 11 February 2013 (UTC)
## Division properties
The division properties should be listed along with the other properties; I'm just not sure how to write them the way the others are written.
(a/b)^n = (a^n)/(b^n)
(a^n)/(a^m) = a^(n-m)
Fuzzform (talk) 04:17, 18 December 2007 (UTC)
## Correctness of complex a^b
In section Computation of ${\displaystyle \,a^{b}}$ where both a and b are complex there is a formula
${\displaystyle \,a^{b}=(e^{\log(r)+{\theta }i})^{u+vi}=e^{(\log(r)+{\theta }i)(u+vi)}}$
That generally seems to be wrong. It may be correct in this context (though I think it's not) but in general for complex a, b c it's not true that: ${\displaystyle (a^{b})^{c}=a^{bc}}$ (see http://mathworld.wolfram.com/ExponentLaws.html) —Preceding unsigned comment added by Findepi (talkcontribs) 19:43, 1 February 2008 (UTC)
The first equals sign here just means "write a and b like this". The second equals sign is essentially a definition. The text could (and should) make this a lot clearer than it is, though.
In general much of the section seems to be focused on nitty-gritty numerical issues that I find to be of limited relevance of this article. For example "Watch out for potential overflow though and possibly scale down the x and y prior to computing x²+y² by a suitable power of 2". Even if this is relevant information about the exponential function in general (which I disupte), it is written like an instruction manual, which Wikipedia is not.
I'm inclined to think the section should either be removed (as irrelevant) or rewritten from scratch to be much shorter and focus on the mathematics of complex powers, rather than the minutiae of implementing them on a computer. –Henning Makholm 02:19, 27 April 2008 (UTC)
I would say instead that complex exponentiation is multiple valued, and that you might end up on a different branch. If you take the branch cuts into account, then you should find that the rule still works. If you expect the principle value, then you might find that it doesn't. More specifically, instead of ${\displaystyle \,a^{b}=(e^{\log(r)+{\theta }i})^{u+vi}=e^{(\log(r)+{\theta }i)(u+vi)}}$ you might use ${\displaystyle \,a^{b}=(e^{\log(r)+{\theta }i+2m\pi i})^{u+vi}=e^{(\log(r)+{\theta }i)(u+vi)+2n\pi i}}$ Gah4 (talk) 05:36, 21 July 2008 (UTC)
The power ab is well defined when a is a positive real, or when b is an integer, but in the general case it is multivalued and thus ill-defined. The expression ab is not used in practice for nonpositive a and noninteger b, so the section is not really useful, and I support Henning Makholm's point of view. The case is treated in the article on exponentiation where I think it belongs, but the treatment there is not satisfactory either. The formula of Gah4 above should probably read ${\displaystyle \,a^{b}=(e^{\log(r)+{\theta }i+2m\pi i})^{u+vi}=e^{(\log(r)+{\theta }i+2m\pi i)(u+vi)}.}$ Bo Jacoby (talk) 08:07, 21 July 2008 (UTC).
## incorrect formula in the complex a^b
Somebody please check if "The result a^b is thus p + qi" should be changed to "The result a^b is thus e^(p + qi)". —Preceding unsigned comment added by 194.67.106.32 (talk) 17:25, 3 March 2008 (UTC)
## Image of e^z
The image in the "On the complex plane" shows some unsightly (and misleading) zig-zag artifacts between the real-part values of -2.5 and 1, roughly. Or is it just my monitor that is out of calibration? –Henning Makholm 02:43, 27 April 2008 (UTC)
I didn't see any artifacts, but what I did find is that if you stare near the right side of the image you can see a black to gray gradient, or a vertical white-on-gray square grating, with a rainbow glowing off the edge of the image, and the original image burned into your eyes. I think it's just that the part where the curves transition from thin to thick is close to flat. But I can't tell if it's actually curved at all, because it's a color gradient. I hate complex color plots sometimes. ᛭ LokiClock (talk) 04:01, 9 March 2012 (UTC)
## Delete computational part
I propose to delete the bits about computing the value of ex, i.e. Exponential_function#Numerical value and Exponential_function#Computing exp(x) for real x. Neither of them is particularly noteworthy and exp is normally evaluated using much better approximations using one polynomial dividing another. The only other type of evaluation which might be interesting is one doing a bit at a time. Dmcq (talk) 21:07, 8 November 2008 (UTC)
## Slight muddle?
The article says:
The exponential function is written as an exponentiation because it obeys the basic exponentiation identities, that is:
${\displaystyle \,e^{1}=e,\ e^{x+y}=e^{x}\cdot e^{y}\ .}$
It is the unique continuous function satisfying these identities for real number exponents.
The function b^x, for any base b, satisfies these identities (well, it doesn't literally satisfy b^1 = e, but that's hardly the point), so "the" exponential function, as defined in the introduction, with base-e, is not unique in this respect as is claimed. I think this part has got mixed up in a confusion about whether the term "exponential function" is always base-e or can apply to any base. —Preceding unsigned comment added by 81.152.169.35 (talk) 21:09, 5 January 2009 (UTC)
• There's another problem here actually. The form of the identities to be satisfied presupposes that we already know the answer. It's like saying "2 is the only number that satisfies 2 + 1 = 3", rather than "x = 2 is the only solution to x + 1 = 3". The identities need to be written as, for example, f(x + y) = f(x)f(y), where f is an arbitrary function, and then we can say the exponential function is the only function that works.
• I think the "unique continuous function" line should just be eliminated. To get the unique base here it's probably easiest to require that f'(x) = f(x) for all x in R, but then you don't need the exponentiation identities, you just need to say it's nonzero. Dcoetzee 00:01, 6 January 2009 (UTC)
I agree. I'm tempted to rewrite this section to change the emphasis. I propose we structure it something like this:
1. The exponential function (we don't yet know its form) satisfies f'(x) = f(x). It can be defined and calculated without recourse to any notion of raising to non-integer powers -- e.g. using the series expansion explained later.
2. It turns out that the function so defined can be written as some number, e, raised to the power of its argument, in the sense that it satisfies the identities familiar from the way integer powers work.
3. Given this property, the exponential function and its inverse can then be conveniently used to define what it means to raise any (positive) real number to any real power using a^b = exp(b*log(a)) -- possibly we would mention this is an alternative to raising to integer powers and taking roots (for rational b) or some sort of limit argument (for irrational b). —Preceding unsigned comment added by 81.152.169.35 (talk) 00:20, 6 January 2009 (UTC)
I disagree with most of the above and have put in a bit of explanation. The e1 is the exponential function and e on its own is e. The continuous on the real numbers part is important, it is possible to satisfy the two equations and still get something which isn't the exponential function. This bit is in an overview, you can go and stick in pedantry later on in the article but the intro and overview should deal with the main description without going and trying to prove everything and make it all boilerplated. This bit is necessary otherwise one is just writing ex without explaining why it is written that way. Dmcq (talk) 00:51, 6 January 2009 (UTC)
I pretty much agree with you - I only wanted to remove the last bit saying "it is the unique continuous function..." You don't need a unique characterization of the function to understand why it's written with exponentiation notation. Dcoetzee 01:57, 6 January 2009 (UTC)
There are still a number of problems with this passage. It still says "It is the unique continuous function satisfying these identities for real number exponents" which is either wrong (since exponentiation to other bases satisfies the same identities), or trivially and uninterestingly true if it's supposed to mean the only function that literally satisfies e^1 = e. The other presentational problem that I mentioned is still present. If we say "the only function that satsifies..." then the equation(s) presented must be in terms of a general function f(x), otherwise we are presupposing that the answer is already known. The way it is now simply does not make sense. Additionally, it has not previously been defined what it means to raise a real number to a real power. Later a definition is given in terms of exp and log, but this cannot be applied to the definition of e^x itself -- it's circular reasoning. Either we can say exp(x) defined without reference to raising to the power of a real number satisfies the expected power laws, therefore we use the function to extend the basic concent of integer powers to all real numbers, or we can have already defined what a^b means for real a and b, so we already know what e^x means. Either is fine, but the article is currently muddled about which approach it is taking. 81.152.169.35 (talk) 03:25, 6 January 2009 (UTC)
There are discontinuous functions satisfying those identities. For instance one can define exp2(q)=eq for all the rational numbers q but keep extending exp2 to the other reals by the axiom of choice for any numbers not yet defined by rational combinations of the ones already defined, for instance one could have exp2(π)=23 and exp(√2)=13. It is unnecessary to say it is the unique continuous function satisfying the identities as this bit is simply explaining why the exponential notation is used but this objection would almost certainly be raised against what is written there if the line was removed.
I don't see that it would matter too much if this section described it in terms of exponentiation to a real power since it is just an overview, but defining it that way would lead to circular reasoning and people would confuse describing and defining. The text however doesn't describe it in terms of exponentiation to a real power, it simply says it is written as an exponentiation because it satisfies the exponentiation identity and that doesn't involve any circular reasoning.
I'll add a bit saying that exponentiation to a real number can be defined using the exponential function, perhaps that will make it more obvious. Dmcq (talk) 09:29, 6 January 2009 (UTC)
## Overview and motivation
I've reinstated the overview and motivation section and deleted the bit moved from the original section to the formal definition section.
I think it is silly to launch straight into the formal definition without even saying why it is called the exponential function and overviewing its main features. The section was deleted before with a comment 'first section hopelessly circular'. This sounds like an argument against it because it was not formally derived but instead described what was to follow. That is part of what overview is about.
I think a little could be removed from the leader as it is long for a leader and there is duplication with the overview section. However wikipedia is supposed to give the information first and jumping straight into formal definition is to write a textbook rather than an encyclopaedia. And actually good textbooks try to be readable, making them unreadable is some sort of conceit I feel. Dmcq (talk) 18:23, 28 July 2009 (UTC)
The Overview section has been deleted again, this time with the comment 'This section is a tradgedy and cannot be allowed to be the first section of the page. It requires major editing'. I have posted a note on the edditor's page to be a bit more specific but perhaps someone else can comment on whether the overview section is a useful thing or whether you think the article should start straight with the mathematical definition? Or if that editor doesn't come back what they meant? Dmcq (talk) 12:44, 5 August 2009 (UTC)
To be fair, User:ObsessiveMathsFreak didn't actually delete the Overview and motivation section - he renamed it to History and Properties and moved it to later in the article, after Derivatives and differential equations. But I agree that he hasn't explained what he doesn't like about this section, and it is not clear to me why he wants to move it. Gandalf61 (talk) 12:56, 5 August 2009 (UTC)
You have stated that it is silly to launch straight into the formal definition without giving an overview of the function first. While this may seem like a good idea, in mathematics it is generally a very, very bad one.
The overview in question begins by listing underived properties of a poorly defined function, and then discusses, without proof, its application to first order odes without even mentioning the derivative property of the function. It then finishes by convolving the discussion with general exponentiation and mentions applications without even stating why it is applied. It's all over the place! For anyone who knows nothing about the exponential function, the entire section is less than useless as it will only serve to confuse. There's no information coming in.
The Bernoulli definition is technically correct but it is exceedingly difficult to prove that it even makes sense, let alone present such a proof in an introductory article. The Taylor series proof is precise, robust and can be used to prove the differentiation formula with the minimum of fuss.
It is better to begin with an appropriate definition and proceed from there. Most articles take that view. Whatever useful information exists in the section is best placed in the introduction at the top of the page. When people scroll past the contents, they expect to get to the meat as quickly as possible, and as mathematical definitions go the Taylor series definition of the exponential function is about as pithy an introduction as you will ever come across.
Unless there are any objections here, I will try to make a major edit on this section in 24 hours. The edit will probably involve taking the disparate pieces of the section and putting them in more appropriate place in the article.ObsessiveMathsFreak (talk) 13:09, 5 August 2009 (UTC)
I most certainly do object. Wikipedia is not a textbook. Please see the Mathematics manual of style in particular the section at the start about the structure of articles. Putting in an informal introductory section is actively encouraged. Starting with a mathematical definition and developing logically until you finally get to something that might be readable if you don't know half the answer already is not.
As to the original definition derived from continuous compound interest it is used in the Exponentiation article under 'Powers of e' and has a very short proof there that it satisfies the exponential identity. It may not be ultra rigorous but it was good enough to kick start the subject. You don't get that so easily from the series or differential equation. By the way the one of the points of an 'Overview' is to be all over the place and not be specific. Dmcq (talk) 13:32, 5 August 2009 (UTC)
I wipe my arse with the Mathematics manual of style!! Are you seriously suggesting that a rambling tirade about the advanced properties of a function that has not even been defined is of use to anyone? Read the section! There is already sufficient introductory material at the top of the page without this section utterly confusing matters. How can the identity exp(x+y)=exp(x)*exp(y) be honestly introduced when the reader has no idea what the function exp(x) even is? The discussion of first order ODEs is even worse as the reader has no idea whatsoever how the function behaves under differentiation, much less of its application to the equations shown. This is no way to present mathematics.
The limit definition is entirely inappropriate for any introductory article whatsoever. The short "proof" given in the Exponentiation article is essentially meaningless, with powers being passed through limits to infinity without the slightest regard for whether the series even converges or not! This very question was the reason behind Bernoulli's original investigation into the number! Given its esoteric nature, I doubt a sufficiently rigorous proof even exists on Wikipedia at all.
The "origins and motivations" section begins with Bernoulli's outdated definition and proceeds to list a series of largely unrelated items which will be of no use to anyone. You cannot have any serious discussion about a mathematical quantity before you have defined it, at least in some fashion. The origins section does not even do that.
The single best way of introducing the exponential function is by its Taylor series. The series is absolutely convergent and moreover can be directly used in later definitions such as the exponentiation of complex numbers or square matrices. It's short, sweet, to the point and the derivative formula can be seen immediately using it. The post contents material should open with it and a better place can be found for most of the material in the current train wreck of a fist section.ObsessiveMathsFreak (talk) 15:55, 5 August 2009 (UTC)
I have raised the matter of your attitude to WP:MSM at Wikipedia talk:WikiProject Mathematics#Manual of style disagreement. Do you have the same problem with WP:NOTTEXTBOOK? Dmcq (talk) 16:27, 5 August 2009 (UTC)
Hang them all is my opinion. You can raise that too if you like.ObsessiveMathsFreak (talk) 17:03, 5 August 2009 (UTC)
This is supposed to be an encyclopaedia and the exponential function is one people who aren't mathematicians can come across quite easily. Compound interest and the continuous version is something people are altogether too much familiar with as opposed to series produced from a hat. If we cannot deal with this little bit of maths which is one of the most familiar to uneducated peasants working their fields in Bangladesh then we've an awful lot of problems. Dmcq (talk) 16:41, 5 August 2009 (UTC)
Fine. If you can present a short paragraph or two linking compound interest to the Bernoulli limit I'll gladly drop my objections to the entire section. I'd try to write one myself, but I was never any good at compound interest myself.ObsessiveMathsFreak (talk) 17:03, 5 August 2009 (UTC)
I'll be away till tomorrow so will try if no-one beats me to it. Continuously compounded interest refers to the section in compound interest about it. Dmcq (talk) 17:30, 5 August 2009 (UTC)
Hi Dmcq, I beat you to it! --WardenWalk (talk) 08:02, 6 August 2009 (UTC)
I am posting here after seeing the message at WT:WPM. There is no reason why we need to start with the series definition, or any other definition. A well-written introductory section can be appropriate for an article where we want to maximize accessibility (another example is First-order logic). On the other hand, we are not writing an axiomatic treatment nor a textbook, so it is not a significant issue if we talk about something before defining it. — Carl (CBM · talk) 00:07, 6 August 2009 (UTC)
Exactly. Paul August 14:03, 6 August 2009 (UTC)
I agree as well. Although I don't think that the introduction as it stands is well written. Another problem is what definition is most appropriate. As a physicist the definition as the inverse of the ln or equivalently the solution to the differential equation df/dt = f is sufficient for most of what we do. Sometimes the (Bernoilli?) form of the compound interest pops up but the vast majority of the time exp(x) comes from a differential equation of some sort. The taylor series definition is used in physics to generalization of exp(x) to include complex arguments and arguments with operators. (It can be used to prove Eulers theorem-at least it is a good enough proof for physicists.) I have no clue which definition is more technically correct, though; from a physicist perspective that is usually unimportant. TStein (talk) 19:56, 7 August 2009 (UTC)
All the definitions are correct, and we should present them all. Because we are writing an overview, not a textbook, there is no reason why we need to choose one specific definition here as "the" definition. That would only be needed if we wanted to prove things, which I believe would be out of scope for this particular article. — Carl (CBM · talk) 12:13, 8 August 2009 (UTC)
There most certainly is a reason we should choose one definition and stick with it. Presenting people with half a dozen definitions, all totally different, will serve only to completely confuse anyone who reads the article. They'll have to choose, probably randomly, between them all. More likely, they will simply leave disgusted, dismissing the exponential function as unfathomable. ObsessiveMathsFreak (talk) 04:32, 9 August 2009 (UTC)
## The lede
A recent editor replaced the introductory sentences with his/her own giving no explanation except for "use sensible definition in lede", even though the previous version had a citation of a text on the subject authored by well-respected experts, and the replacement had none. So that there is no misunderstanding, I wanted to explain why the approach in that text is sensible before doing a partial revert of the edit to reinstate the citation. The real issue is how to explain what e is. If one says that it is approximately 2.718281828..., that does not specify it uniquely. If one says that it is the base of the natural logarithm, that does specify it, but it is not ideal in that it is not self-contained: it presumes that the natural logarithm has already been defined, and that is a topic of the same level of difficulty as the exponential function itself. Saying that e is the number such that e^x is its own derivative specifies e precisely (and as a bonus, highlights the reason the exponential function is important in the first place). --WardenWalk (talk) 07:24, 6 August 2009 (UTC)
## Diagram of slope of function
There is a diagram under Exponential function#Derivatives and differential equations showing a base of 1 and the tangent to the function. Unfortunately this could be misleading for people as they might think the vertical line was at 0 and get totally the wrong idea. Anyone like to try their hand at a better diagram? Dmcq (talk) 08:47, 9 August 2009 (UTC)
I totally agree. In fact, I independently came to the same conclusion, and made an edit to its caption before noticing your comment here. Anyway, I too hope that someone will produce a better diagram. It could also be improved by removing the θ and Tan θ, which are irrelevant to what the diagram is intended to show. --WardenWalk (talk) 14:33, 15 August 2009 (UTC)
## Numerical computation missing
Information about the numerical computation of exp(x) is missing. There once was a section about the topic, but is has apparently been removed due to lack of quality.--84.189.68.236 (talk) 18:54, 21 August 2009 (UTC)
Yes the section was rubbish. There's lots of ways of doing it accurately, I'm not sure it is notable though. If you want something like that probably your best bet is just to refer to gnu libc for a good version and perhaps the IBM version for exactly rounded values. Dmcq (talk)
I don't agree that because a section is poorly written that therefore the subject has no place it the article. That seems EXACTLY opposite the Wikipedia philosophy of improving articles rather than tearing them apart. I find it curious Dmcq, that you refer the above user to a different source of information. Are you suggesting that efficient computation is not a notable part of a function?? Risible. You. Are. Wrong. I am not qualified to determine whether or not the fp algorithm outlined in the two (very poorly written, I agree) sections 11 & 12 here on this talk page are accurate, otherwise I'd put a small portion of them back in, a few lines at most. Many functions, numbers, etc., if not most, have a computation section in Wikipedia. Dmcq you made a bad error in judgment. IMHO.72.172.1.28 (talk) 14:31, 14 September 2013 (UTC)
## e^-x
e^-x —Preceding unsigned comment added by 202.44.111.74 (talk) 16:10, 8 November 2009 (UTC)
(?_?) :-\ Dmcq (talk) 17:19, 8 November 2009 (UTC)
$<br\>$l | $<br\>$ \| $<br\>$ | - $<br\> $` <br\> There you go! ᛭ LokiClock (talk) 23:00, 25 November 2013 (UTC)
## Too technical?
Someone stuck on a too technical tag at the top last month. Personally I feel it addresses that issue fairly well and much better than most other articles at its difficulty level and have removed the tag. No specific issue was mentioned. Any specific issues people see with it? Dmcq (talk) 11:46, 8 March 2011 (UTC)
## formal definition all wrong
we should define a function exp(x)
infinite series or differential equation
• then* show it is e^x for some e
otherwise it is fallacious
double definition —Preceding unsigned comment added by 70.189.170.229 (talk) 13:42, 13 May 2011 (UTC)
Have you looked at Characterizations of the exponential function? Dmcq (talk) 15:07, 13 May 2011 (UTC)
## Exponential function and human understanding
I am wondering if there is also space in this article to discuss the suggested "inability of human beings to understand the exponential function" (Albert Bartlett). There is also a lecture on this topic from back in 2004 titled Arithmetic, Population and Energy. --spitzl (talk) 21:53, 13 June 2011 (UTC)
It's just a bit of rhetoric. There's an article Exponential growth where it might go as some sort of remark somewhere I guess. Dmcq (talk) 21:58, 13 June 2011 (UTC)
## general exponential
I've removed some stuff about a to the power of x in the derivatives section as the article is about the exponential function and there's other articles exponential growth and exponentiation about such things. However I think there should probably be a duplication here of the definition of ax using the exponential function. There is a short section of what I'm thinking of for the complex version of but that's too late. The alternatives are to have another section like it for reals or othewise move that up and talk about complex numbers before getting to dealing with them in general here. Dmcq (talk) 23:00, 6 October 2011 (UTC)
## Matrix
So this article doesn't have almost anything about exponentiation to a matrix. Maybe we should expand it a little bit and possibly include an example. It is right now merged with the section about Banach spaces, which I don't really understand. Banach algebra seems to be a much more abstract concept. Matrix exponentiation has a lot of application in systems engineering, circuit design, biology, climate science and economy so I think it deserves an example. Anybody agree? In my opinion, the optimal thing would be to have an example of both e to a 2x2 Matrix and e to a 3x3 Matrix. Just to illustrate how much more complicated it gets with a larger matrix.
You're looking for Matrix exponential. ᛭ LokiClock (talk) 01:23, 19 April 2013 (UTC)
## Why was computation of e^x removed?
I agree that the section as written here ( see Talk page sections 11 How to compute exp(x) in computers AND 12 Computing exp(x) in computers ) is opaque, poorly written, verbose, etc. (ie. terrible). However, why isn't there a short section in the article detailing efficient algorithms for computing e^x? There should be.72.172.1.28 (talk) 14:24, 14 September 2013 (UTC)
## Non-standard analysis
I think the text about doing a non-standard analysis version of differentiation is an undue digression in this article. Wikipedia isn't about loads of proofs. I think what perhaps could be done in that line is to give the non-standard analysis definition of ${\displaystyle e^{x}}$ but any example should be move to the non-standard analysis article which has no examples in it. Dmcq (talk) 09:17, 30 April 2014 (UTC)
In this edit, Michael Hardy introduced a more general function at the start of the lead. This conflicts with the hatnote, as well as not reflecting the body of the article (aside from a couple of minor digressions, which seem to confusingly conflate the concepts). An article in WP is intended to be about a topic (in this case evidently the natural exponential function), and as per WP:DICT should not try to define diverse uses of a term. Should we remove the first paragraph (and optionally rename/move the article to Natural exponential function), or rewrite the article to be about the function family defined in the first paragraph? I do not see a middle path (covering both topics) as being sensible. —Quondum 18:48, 5 March 2016 (UTC)
Yes I agree with you. The topic should be made clear in the very first paragraph. There is no need to change the title, in most sources exponential function refers to what this article is about and not to general exponentiation. Dmcq (talk) 20:19, 5 March 2016 (UTC)
I think you're both wrong and I've changed the hatnote accordingly. This article is not about general exponentiation, but that does not mean it includes only the natural exponential function and not other bases.
{\displaystyle {\begin{aligned}(b,c)\mapsto b^{c}&&&{\text{This is exponentiation in general.}}\\[10pt]x\mapsto b^{x}&&&{\text{This is an exponential function}}\\&&&{\text{(the base is constant; the input variable is the exponent).}}\end{aligned}}}
This article does not and should not concern exponentiation in general; it does and should include exponential functions with arbitrary positive bases. All of these have essential things in common, including being scalar multiples of their own derivatives. Michael Hardy (talk) 01:25, 6 March 2016 (UTC)
I have started this discussion of this matter. Michael Hardy (talk) 01:32, 6 March 2016 (UTC)
I think this is the appropriate location for the discussion, with the WikiProject section that you linked serving to point others here. I feel that you have mischaracterized my position: I pointed out inconsistencies, and gave two alternatives, your preference being one of them. If this is the chosen alternative, the article needs rewriting to focus in the exponential function as you define it; even with your latest edits, it overwhelmingly focuses on only the natural exponential function. —Quondum 04:16, 6 March 2016 (UTC)
I think the most sensible course of action is to move the present article to natural exponential function, removing the first paragraph. There is also scope for a separate article exponential function, to refer to exponential functions in arbitrary bases. 13:25, 6 March 2016 (UTC)
I agree that there is scope for both articles. —Quondum 15:12, 6 March 2016 (UTC)
How about an RfC on this? It is a topic of general interest. Dmcq (talk) 10:23, 16 March 2016 (UTC)
There are arguments on both sides and it is notable enough, so I think it makes sense to seek a consensus, which does not seem to be emerging on this talk page. —Quondum 15:41, 16 March 2016 (UTC)
### RfC: Should exponential function be about exponentiation to any base?
There is consensus that this article should be about ex with additional sections detailing how ecx is equivalent to bx. The lead should be rewritten accordingly. (non-admin closure) ~ RobTalk 17:58, 22 April 2016 (UTC)
The following discussion is closed. Please do not modify it. Subsequent comments should be made on the appropriate discussion page. No further edits should be made to this discussion.
The article exponential function has for a long time been about ex with generalizations of x. Should it instead be about bx where b is any constant rather than Euler's number e? If so should the current contents be moved to natural exponential function or kept here? There is discussion is immediately above. Dmcq (talk) 10:34, 17 March 2016 (UTC)
• Oppose There are articles exponential growth and exponential decay which deal with bx where both b and x are real numbers and in practically all other cases where a constant b is not e the x is an integer - i.e. it is a power. exponentiation covers the general case of bx.
On the second point if this article does become about bx I would agree the contents here be moved to 'natural exponential function' with a hatnote here and I think the contents of exponential growth and exponential decay should be moved to this article and those articles turned into redirects. Dmcq (talk) 10:51, 17 March 2016 (UTC)
• Conditional oppose, i.e. oppose if there is not an independent article on ex with generalizations of x. If there is such an independent article, it makes sense to merge three articles as suggested by Dmcq above. The function ex cannot be treated as a special case of RR : xbx, since ex in its full generality covers other domains. —Quondum 04:10, 18 March 2016 (UTC)
• Oppose. The article should be about the exponential function, which is ex. It should mention, in the lead, that bx is equivalent to ex.ln(b). Maproom (talk) 06:33, 22 March 2016 (UTC)
• Oppose. It seems silly to have both "natural exponential function" as well as "exponential function" when they would have so much overlap. As one of my first grad school professors said "We call it the exponential function because there's only one!" In that light, I'd say the article should be written about exp(cx). The way I'd do it is set c=1 for the first chunk of the discussion, then write c=ln(b) and note that the presentation applies to any b^x as well. SemanticMantis (talk) 19:43, 30 March 2016 (UTC)
If I read you right that's a support for this article being about exponentiation to any base rather than an oppose. Dmcq (talk) 20:58, 30 March 2016 (UTC)
I know, it's confusing. Sorry. I don't think the article should be about b^x. I think it should be about e^x. I think it should clearly make the point that b^x is just another way of writing e^cx. Maybe I'm too off track in my thinking about aboutness :) Basically, I think there's only one worth having an article on, and the fact that it subsumes all the others is just a neat philosophical perspective that doesn't have any large effect on the math. (Incidentally I think the quote I gave above was from this guy [1]). SemanticMantis (talk) 16:18, 31 March 2016 (UTC)
• Oppose. I think the first oppose opinion has it right. "The exponential function" always means exp, and everything else is adequately covered at exponentiation (for the function) or exponential growth (for most situations involving a fixed base that is not e). Exponentiating to base 2 can also be important in computer science, but mainly with integer powers, and is adequately covered at power of two. —David Eppstein (talk) 05:11, 6 April 2016 (UTC)
• Oppose. Of course! There is only one exponential function. From an encyclopedic and mathematical approach there is no reason to have two (or even three) articles. They should be all merged. Many people are unaware that bx = ex ln(b). Well, this fact should be immediately clarified in the lead. Silvio1973 (talk) 07:01, 11 April 2016 (UTC)
That sounds the same as what was talking about above. It is true one can calculate a power that way and it should be mentioned somewhere - but it is not why the exponential function is important. Any old base could be used besides e for the purpose you talk about and historically base 10 was most often used. Dmcq (talk) 14:13, 11 April 2016 (UTC)
Right, any one could be used, and 10 has some historical weight, but e is special and in fact unique as a base for other reasons, so we call exp(x) the exponential function, which normally causes no confusion or ambiguity. SemanticMantis (talk) 14:38, 11 April 2016 (UTC)
Well, this is English Wikipedia and I do not want to insist on an approach that is perhaps regional. In Italy and in France the exponential function is intended exclusively with base e. Only this function satisfies the equivalence df(x)/dx = f(x). --Silvio1973 (talk) 12:42, 16 April 2016 (UTC)
The discussion above is closed. Please do not modify it. Subsequent comments should be made on the appropriate discussion page. No further edits should be made to this discussion.
## Definition of the family of exponential functions
Assuming that this article will remain about the family of exponential functions and that a separate article on the natural exponential function will be branched from it, it seems appropriate to characterize these functions better. Treating the topic generally seems to be quite tricky (see List of exponential topics). The lead seems to be correct in the context of real analysis, but does not state such a restriction, nor does it seem appropriate, since complex analysis (and no doubt other fields) is relevant.
If we characterize a complex function as an exponential function if it obeys f′(x) = kf(x) for some constant k, or of it obeys any (or all) of several exponential identities (e.g. f(x + y) = kf(x)f(y) for some constant k), we obtain a more general solution than abx or bx+c (in both, b is restricted to positive real numbers for technical reasons): this more general solution is exp(ax + c), where a and c are arbitrary complex numbers. (The case f(x) = 0, if included, must be handled specially, and for the real case we probably need to change this to ±exp(ax + c), but the expression bx+c already has this issue.)
So the question is: Is the phrase "exponential function" restricted to real analysis when referring to exponential functions is this sense? If not, what is the phrase taken to refer to in complex analysis? I think that the article should make this clearer. —Quondum 01:41, 16 March 2016 (UTC)
I'm not assuming what you are assuming. Look up exponential function anywhere and you'll see that by far the majority of cases refers to e^x. Wikipedia's policies are what should be followed rather than talking about exponential functions is in general. Lets get the debate above closed one way or the other first. Dmcq (talk) 10:20, 16 March 2016 (UTC)
For clarity, I was not assuming it. I was trying to make it the premise of an "if-then", which premise has not been established, and trying to get comment on what we are talking about if that premise were valid. —Quondum 15:30, 16 March 2016 (UTC)
I don't think exponential is applied to anything except real numbers in the way you say, we don't even normally come across complex numbers used in that sense. And we don't get b to the power of a matrix, only e. We have matrices to a power but the powers are integers. Except for exponential growth and decay, exponentiation and powers and things like geometric series the number of times there is ever an exponential function not involving e is pretty tiny. Dmcq (talk) 15:57, 16 March 2016 (UTC)
That's fair. It makes sense to resolve the debate above first, as you suggested (though perhaps I'd hoped that lack of a clear notable definition here might nudge the debate). —Quondum 01:13, 17 March 2016 (UTC)
I think it's best for encyclopedic purposes to mostly restrict the article to functions f:R->R. Of course extensions and generalization can be mentioned at the end. We already have a huge image problem. So many people on and of WP tell me that most/all of our math articles are not helpful because they seem to be written by and for math PhDs. So let's keep this article (which will be commonly read by middle school and high school students in the USA) at a fairly basic level. SemanticMantis (talk) 16:29, 31 March 2016 (UTC)
The real version of the function is interesting but not especially so. If one didn't know about the complex case it would just be a version of exponential growth where one could get rid of a multiplying constant in the expression when differentiating, more of a trick than anything else. When one extends it to complex numbers though it shows itself in its true light as a central function of mathematics. Dmcq (talk) 11:03, 6 April 2016 (UTC)
## Assessment comment
The comment(s) below were originally left at Talk:Exponential function/Comments, and are posted here for posterity. Following several discussions in past years, these subpages are now deprecated. The comments may be irrelevant or outdated; if so, please feel free to remove this section.
The fact that the slope of an exponential curve equals its local value can be illustrated by showing the tangent line at some arbitrary point. That line will always intersect the y=0 axis at the same (unit) distance from the tangent point. If the exponent includes a scale parameter, the intersect distance equals that parameter. For example, in exponential decay with time constant t0, y = y0exp(-t/t0), the tangent line crosses the time axis at a time interval t0 from the tangent point. "If the initial rate of a capacitor's RC discharge were maintained constant, the discharge would be complete after one RC time constant." Elexmax 15:21, 5 June 2007 (UTC) Clearly Top-priority. Geometry guy 19:34, 10 June 2007 (UTC)
Last edited at 19:34, 10 June 2007 (UTC). Substituted at 02:05, 5 May 2016 (UTC)
The External Link for "Complex exponential interactive graphic", linking to: http://www-math.mit.edu/daimp/ComplexExponential.html just redirects to http://mathlets.org/. Should it be swapped for a new one, or maybe an archived version? Ethanicus (talk) 21:52, 21 July 2017 (UTC)
I have disabled it for now. Thank you for bringing it to our attention.—Anita5192 (talk) 00:19, 22 July 2017 (UTC) | 16,500 | 63,489 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 76, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.59375 | 4 | CC-MAIN-2018-09 | latest | en | 0.929031 |
http://www.convertit.com/Go/Entisoft/Measurement/Converter.ASP?From=link&To=length | 1,611,708,659,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610704804187.81/warc/CC-MAIN-20210126233034-20210127023034-00199.warc.gz | 123,477,101 | 3,645 | New Online Book! Handbook of Mathematical Functions (AMS55)
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Please read our Help Page and FAQ Page then post a message or send e-mail. Thanks! | 449 | 1,599 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.859375 | 3 | CC-MAIN-2021-04 | latest | en | 0.681473 |
https://ell.stackexchange.com/questions/147047/meaning-of-plural-of-plural-v-s-singular-of-plural | 1,582,174,317,000,000,000 | text/html | crawl-data/CC-MAIN-2020-10/segments/1581875144637.88/warc/CC-MAIN-20200220035657-20200220065657-00511.warc.gz | 380,193,585 | 30,632 | # meaning of plural of plural v.s. singular of plural
I am confused of the followings:
The locations of nodes are given.
The location of nodes is given.
Node locations are given.
Nodes' location is given.
Nodes' locations are given.
Is there any difference among these?
SovereignSun's explanations are great except this one:
The "Nodes' locations" is wrong. The locations don't belong to the nodes. It should be "Nodes locations".).
The locations do belong to the nodes. The locations are where they are located. The location(s) of the nodes, making them possessive. Both of these are correct: * Nodes' location is given. * Nodes' locations are given.
• When I say a set X={(x1,y1), (x2,y2), ..., (xn,yn)}, where (xi,yi) is the location of the node i, do I have to write like "a set X of the locations of all the nodes"? – Danny_Kim Dec 20 '17 at 2:55
You need to understand that you can have one or many nodes and one or many locations.
• The location of a node. (one node has one location)
• The location of nodes. (one location for many nodes that are probably in one place)
• The locations of nodes. (many nodes and many locations, but this may note imply that each location can have only one node)
• The locations of a node. (this is doubtly possible in any context except if we are tracing a node)
The "Nodes' locations" is wrong. The locations don't belong to the nodes. It should be "Nodes locations".).
• Then, both node locations and nodes location are correct?? – Danny_Kim Nov 13 '17 at 9:11 | 375 | 1,518 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.21875 | 3 | CC-MAIN-2020-10 | latest | en | 0.942836 |
https://physics.stackexchange.com/questions/270728/can-this-friction-be-ignored | 1,718,797,253,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861817.10/warc/CC-MAIN-20240619091803-20240619121803-00884.warc.gz | 406,030,700 | 39,828 | # Can this friction be ignored?
I ran across this problem along with an answer. The answer says that a certain force due to friction can be ignored, but I question whether that is correct.
The question is: find the acceleration of block 2. There is friction between block 2 and the floor with coefficient of friction $\mu_{2f}$, and there is friction between block 1 and block 2 with coefficient $\mu_{12}$.
The given answer starts $$(m_1 + m_2)a = F_a - F_\mathrm{friction\; on\; 2\; due\; to\; floor} - F_\mathrm{friction\;on\; 1\; due\;to\; 2}$$ $$(m_1+ m_2)a = F_a - \mu_{2f}(m_1 + m_2)g - \mu_{12}m_1g$$ along with an explanation "There is also a friction force on 2 due to 1, but that can be ignored."
I don't see how we can ignore the friction force on 2 due to 1. In order to make things as clear as possible, I might treat each mass separately in order to clarify the forces on each mass.
$$m_1a_1 = -T +\mu_{12}m_1g$$ $$m_2 a_2 = F_a -T -\mu_{2f}(m_1+m_2) - \mu_{12}m_1g$$ $$a_2 = -a_1 = a$$ The last term in the second equation explicitly accounts for the force on 2 due to 1. This is the force that the given answer says can be ignored. I'm wondering if I'm double counting the friction. I saw this problem and answer in print, so I'm inclined to doubt my analysis. Is the friction force on 2 due to 1 ignorable?
• i dont think its ignorable. Commented Jul 30, 2016 at 14:56
• you have to substitute for T as well in the final expression. So the friction force on 2 due to 1 must be present, as you can see from the last 3 equations. Commented Jul 30, 2016 at 15:00
• "There is also a friction force on 2 due to 1, but that can be ignored." I don't see how we can ignore the friction force on 1. I am confused now; is it a friction force on box 1 or on box 2 that should be ignored? Commented Jul 30, 2016 at 15:30
• It might mean that in this particular problem, that you ignore that component of the friction.
– jim
Commented Jul 30, 2016 at 16:10
• The one that they are suggesting can be ignored is the friction force on box 2, the lower box, due to box 1, the upper box. Commented Jul 30, 2016 at 17:45
Your equation is written incorrectly confusingly for me as:
$$m_1a_1 = -T +\mu_{12}m_1g$$
I would write it as:
$$T = m_1a_1 + \mu_{12}m_1g$$
Then:
$$F_a - T-\mu_{2f}(m_1+m_2)g - \mu_{12}m_1g = m_2 a_2$$
Then replace T as follows:
$$F_a - m_1a_1 - \mu_{12}m_1g -\mu_{2f}(m_1+m_2)g - \mu_{12}m_1g = m_2 a_2$$
This yields:
$$F_a -\mu_{2f}(m_1+m_2)g - 2\mu_{12}m_1g = (m_1 + m_2)a$$
Where:
$$a = a_1 = a_2$$
You can't ignore the friction between 2 and 1 because that friction is working on each of the masses in such a way as to resist the force, $F_a$. Instead, that frictional force should actually be accounted for TWICE.
• My first equation is not incorrect. I just prefer to have the positive axis point to the right in all cases. Then I don't have to remember which coordinate frame points left and which points right. And minus signs always mean "to the left". Commented Jul 30, 2016 at 17:44
• @garyp I understand your point Gary. I can't do it your way because then I end up getting confused. I edited my wording above. Commented Jul 30, 2016 at 17:48
I think your analysis is correct. The 'given answer' provides no reason for ignoring the friction force on block 2 due to block 1. Such 'hand-waving' ought to ring alarm bells.
Suppose blocks 1 & 2 are vertically displaced so that, instead of block 1 resting on block 2, it rests on a fixed ledge. Further suppose that another block of mass $m_1$ rests on block 2, and this block is pushed to the left so that it remains vertically below block 1 as that block moves. The same coefficient of friction $\mu_{12}$ exists between each surface.
The forces on each block remain exactly the same as before, but it is now clear that the friction force $\mu_{12}mg$ acts on each block and must be counted twice, once for each.
However, this way of looking at the problem is no better than (and no different from) drawing separate Free Body Diagrams for each block.
I think it refers to the friction on the wheel axis, which actually will be absent if the m1 is absent, and can be ignored because it should be very small if compared with the other frictions. I admit it is a weird interpretation, but I don't see any other friction
• The complete wording of the problem makes it clear that they are ignoring the friction force on the lower box due to the upper box. Commented Jul 30, 2016 at 17:47
if we specify sign to be positive when directing to the right,
$$m_1(-a) = -T +\mu_{12}m_1g$$
$$m_2 a = F_a -T -\mu_{2f}(m_1+m_2)g - \mu_{12}m_1g$$
subtract the first equation from the second equation,
$$(m_1+m_2) a = F_a -\mu_{2f}(m_1+m_2)g - 2\mu_{12}m_1g$$ | 1,461 | 4,755 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.9375 | 4 | CC-MAIN-2024-26 | latest | en | 0.93562 |
https://www.slideserve.com/hasad-beasley/photon-angular-momentum-and-geometric-gauge | 1,513,429,896,000,000,000 | text/html | crawl-data/CC-MAIN-2017-51/segments/1512948588072.75/warc/CC-MAIN-20171216123525-20171216145525-00019.warc.gz | 805,814,258 | 13,734 | 1 / 23
# Photon angular momentum and geometric gauge - PowerPoint PPT Presentation
Photon angular momentum and geometric gauge. Margaret Hawton, Lakehead University Thunder Bay, Ontario, Canada William Baylis, U. of Windsor, Canada. Outline. photon r operators and their localized eigenvectors leads to transverse bases and geometric gauge transformations,
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Presentation Transcript
### Photon angular momentum and geometric gauge
Margaret Hawton, Lakehead University
Thunder Bay, Ontario, Canada
William Baylis, U. of Windsor, Canada
• photon roperators and their localized eigenvectors
• leads to transverse bases and geometric gauge transformations,
• then to orbital angular momentum of the bases, connection with optical beams
• conclude
pz or z
q
py
f
px
Notation: momentum space
### Is the position of the photon an observable?
In quantum mechanics, any observable requires a Hermitian operator
a =1/2 for F=E+icB ~ p1/2 as in QED to normalize
last term maintains transversality of rP(F)
but the components of rP don’t commute!
thus “the photon is not localizable”?
### A photon is asymptotically localizable
Is there a photon position operator with commuting components and exactly localized eigenvectors?
It has been claimed that since the early day of quantum mechanics that there is not.
Surprisingly, we found a family of r operators,
Hawton, Phys. Rev. A 59, 954 (1999).
Hawton and Baylis, Phys. Rev. A 64, 012101 (2001).
and, not surprisingly, some are sceptical!
p components and z
q
c
py
f
px
### New position operator becomes: components and
its components commute
eigenvectors are exactly localized states
it depends on “geometric gauge”, c,that is on choice of transverse basis
Topology: You can’t comb the hair on a fuzz ball without creating a screw dislocation.
Phase discontinuity at origin gives d-function string when differentiated.
Geometric gauge transformation creating a
no +z singularity
q=p creating a
f
q=0
Is the physics creating a c-dependent?
Localized basis states depend on choice of c, e.g. el(0) or el(-f) localized eigenvectors look physically different in terms of their vortices.
This has been given as a reason that our position operator may be invalid.
The resolution lies in understanding the role ofangular momentum (AM). Note: orbital AM rxp involves photon position.
For an exactly localized state creating a
### “Wave function”, e.g. F=E+icB
Any field can be expanded in plane wave using the transverse basis determined by c:
f(p) will be called the (expansion) coefficient. For F describing a specific physical state, change of el(c) must be compensated by change in f.
Interpretation for helicity creating a l=1, single valued, dislocation on -ve z-axis
sz=1, lz= 0
sz= -1, lz= 2
sz=0, lz= 1
Basis has uncertain spin and orbital AM, definitejz=1.
Position space creating a
Beams creating a
Any Fourier expansion of the fields must make use of sometransverse basis to write
and the theory of geometric gaugetransformations presented so far in the context of exactly localized states applies - in particular it applies to optical beams.
Some examples involving beams follow:
Elimination of creating a e2if term requires linear combination of RH and LH helicity basis states.
Partition of creating a J between basis and coefficient
Dc to rotate axis is also possible, but inconvenient.
Commutation relations creating a
L(c) is a true angular momentum.
Confirms that localized photon has a definite z-component of total angular momentum.
Summary creating a
• Localized photon states have orbital AM and integral total AM, jz, in any chosen direction.
• These photons are not just fuzzy balls, they contain a screw phase dislocation.
• A geometric gauge transformation redistributes orbital AM between basis and coefficient, but leave jz invariant.
• These considerations apply quite generally, e.g. to optical beam AM. Position and orbital AM related through L=rxp. | 1,083 | 4,707 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.5625 | 3 | CC-MAIN-2017-51 | latest | en | 0.846335 |
https://www.lmfdb.org/ModularForm/GL2/Q/holomorphic/2888/2/a/e/1/1/ | 1,719,267,184,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198865490.6/warc/CC-MAIN-20240624214047-20240625004047-00313.warc.gz | 746,779,416 | 77,510 | # Properties
Label 2888.2.a.e.1.1 Level $2888$ Weight $2$ Character 2888.1 Self dual yes Analytic conductor $23.061$ Analytic rank $1$ Dimension $1$ CM no Inner twists $1$
# Related objects
Show commands: Magma / PariGP / SageMath
## Newspace parameters
comment: Compute space of new eigenforms
[N,k,chi] = [2888,2,Mod(1,2888)]
mf = mfinit([N,k,chi],0)
lf = mfeigenbasis(mf)
from sage.modular.dirichlet import DirichletCharacter
H = DirichletGroup(2888, base_ring=CyclotomicField(2))
chi = DirichletCharacter(H, H._module([0, 0, 0]))
N = Newforms(chi, 2, names="a")
//Please install CHIMP (https://github.com/edgarcosta/CHIMP) if you want to run this code
chi := DirichletCharacter("2888.1");
S:= CuspForms(chi, 2);
N := Newforms(S);
Level: $$N$$ $$=$$ $$2888 = 2^{3} \cdot 19^{2}$$ Weight: $$k$$ $$=$$ $$2$$ Character orbit: $$[\chi]$$ $$=$$ 2888.a (trivial)
## Newform invariants
comment: select newform
sage: f = N[0] # Warning: the index may be different
gp: f = lf[1] \\ Warning: the index may be different
Self dual: yes Analytic conductor: $$23.0607961037$$ Analytic rank: $$1$$ Dimension: $$1$$ Coefficient field: $$\mathbb{Q}$$ Coefficient ring: $$\mathbb{Z}$$ Coefficient ring index: $$1$$ Twist minimal: no (minimal twist has level 152) Fricke sign: $$+1$$ Sato-Tate group: $\mathrm{SU}(2)$
## Embedding invariants
Embedding label 1.1 Character $$\chi$$ $$=$$ 2888.1
## $q$-expansion
comment: q-expansion
sage: f.q_expansion() # note that sage often uses an isomorphic number field
gp: mfcoefs(f, 20)
$$f(q)$$ $$=$$ $$q+1.00000 q^{3} +3.00000 q^{5} -2.00000 q^{9} +O(q^{10})$$ $$q+1.00000 q^{3} +3.00000 q^{5} -2.00000 q^{9} -4.00000 q^{11} -5.00000 q^{13} +3.00000 q^{15} -5.00000 q^{17} -1.00000 q^{23} +4.00000 q^{25} -5.00000 q^{27} +3.00000 q^{29} +4.00000 q^{31} -4.00000 q^{33} +2.00000 q^{37} -5.00000 q^{39} -5.00000 q^{41} -11.0000 q^{43} -6.00000 q^{45} -5.00000 q^{47} -7.00000 q^{49} -5.00000 q^{51} -9.00000 q^{53} -12.0000 q^{55} +13.0000 q^{59} -1.00000 q^{61} -15.0000 q^{65} -5.00000 q^{67} -1.00000 q^{69} +1.00000 q^{71} -9.00000 q^{73} +4.00000 q^{75} +17.0000 q^{79} +1.00000 q^{81} +16.0000 q^{83} -15.0000 q^{85} +3.00000 q^{87} +3.00000 q^{89} +4.00000 q^{93} -13.0000 q^{97} +8.00000 q^{99} +O(q^{100})$$
## Coefficient data
For each $$n$$ we display the coefficients of the $$q$$-expansion $$a_n$$, the Satake parameters $$\alpha_p$$, and the Satake angles $$\theta_p = \textrm{Arg}(\alpha_p)$$.
Display $$a_p$$ with $$p$$ up to: 50 250 1000 Display $$a_n$$ with $$n$$ up to: 50 250 1000
$$n$$ $$a_n$$ $$a_n / n^{(k-1)/2}$$ $$\alpha_n$$ $$\theta_n$$
$$p$$ $$a_p$$ $$a_p / p^{(k-1)/2}$$ $$\alpha_p$$ $$\theta_p$$
$$2$$ 0 0
$$3$$ 1.00000 0.577350 0.288675 0.957427i $$-0.406785\pi$$
0.288675 + 0.957427i $$0.406785\pi$$
$$4$$ 0 0
$$5$$ 3.00000 1.34164 0.670820 0.741620i $$-0.265942\pi$$
0.670820 + 0.741620i $$0.265942\pi$$
$$6$$ 0 0
$$7$$ 0 0 1.00000i $$-0.5\pi$$
1.00000i $$0.5\pi$$
$$8$$ 0 0
$$9$$ −2.00000 −0.666667
$$10$$ 0 0
$$11$$ −4.00000 −1.20605 −0.603023 0.797724i $$-0.706037\pi$$
−0.603023 + 0.797724i $$0.706037\pi$$
$$12$$ 0 0
$$13$$ −5.00000 −1.38675 −0.693375 0.720577i $$-0.743877\pi$$
−0.693375 + 0.720577i $$0.743877\pi$$
$$14$$ 0 0
$$15$$ 3.00000 0.774597
$$16$$ 0 0
$$17$$ −5.00000 −1.21268 −0.606339 0.795206i $$-0.707363\pi$$
−0.606339 + 0.795206i $$0.707363\pi$$
$$18$$ 0 0
$$19$$ 0 0
$$20$$ 0 0
$$21$$ 0 0
$$22$$ 0 0
$$23$$ −1.00000 −0.208514 −0.104257 0.994550i $$-0.533247\pi$$
−0.104257 + 0.994550i $$0.533247\pi$$
$$24$$ 0 0
$$25$$ 4.00000 0.800000
$$26$$ 0 0
$$27$$ −5.00000 −0.962250
$$28$$ 0 0
$$29$$ 3.00000 0.557086 0.278543 0.960424i $$-0.410149\pi$$
0.278543 + 0.960424i $$0.410149\pi$$
$$30$$ 0 0
$$31$$ 4.00000 0.718421 0.359211 0.933257i $$-0.383046\pi$$
0.359211 + 0.933257i $$0.383046\pi$$
$$32$$ 0 0
$$33$$ −4.00000 −0.696311
$$34$$ 0 0
$$35$$ 0 0
$$36$$ 0 0
$$37$$ 2.00000 0.328798 0.164399 0.986394i $$-0.447432\pi$$
0.164399 + 0.986394i $$0.447432\pi$$
$$38$$ 0 0
$$39$$ −5.00000 −0.800641
$$40$$ 0 0
$$41$$ −5.00000 −0.780869 −0.390434 0.920631i $$-0.627675\pi$$
−0.390434 + 0.920631i $$0.627675\pi$$
$$42$$ 0 0
$$43$$ −11.0000 −1.67748 −0.838742 0.544529i $$-0.816708\pi$$
−0.838742 + 0.544529i $$0.816708\pi$$
$$44$$ 0 0
$$45$$ −6.00000 −0.894427
$$46$$ 0 0
$$47$$ −5.00000 −0.729325 −0.364662 0.931140i $$-0.618816\pi$$
−0.364662 + 0.931140i $$0.618816\pi$$
$$48$$ 0 0
$$49$$ −7.00000 −1.00000
$$50$$ 0 0
$$51$$ −5.00000 −0.700140
$$52$$ 0 0
$$53$$ −9.00000 −1.23625 −0.618123 0.786082i $$-0.712106\pi$$
−0.618123 + 0.786082i $$0.712106\pi$$
$$54$$ 0 0
$$55$$ −12.0000 −1.61808
$$56$$ 0 0
$$57$$ 0 0
$$58$$ 0 0
$$59$$ 13.0000 1.69246 0.846228 0.532821i $$-0.178868\pi$$
0.846228 + 0.532821i $$0.178868\pi$$
$$60$$ 0 0
$$61$$ −1.00000 −0.128037 −0.0640184 0.997949i $$-0.520392\pi$$
−0.0640184 + 0.997949i $$0.520392\pi$$
$$62$$ 0 0
$$63$$ 0 0
$$64$$ 0 0
$$65$$ −15.0000 −1.86052
$$66$$ 0 0
$$67$$ −5.00000 −0.610847 −0.305424 0.952217i $$-0.598798\pi$$
−0.305424 + 0.952217i $$0.598798\pi$$
$$68$$ 0 0
$$69$$ −1.00000 −0.120386
$$70$$ 0 0
$$71$$ 1.00000 0.118678 0.0593391 0.998238i $$-0.481101\pi$$
0.0593391 + 0.998238i $$0.481101\pi$$
$$72$$ 0 0
$$73$$ −9.00000 −1.05337 −0.526685 0.850060i $$-0.676565\pi$$
−0.526685 + 0.850060i $$0.676565\pi$$
$$74$$ 0 0
$$75$$ 4.00000 0.461880
$$76$$ 0 0
$$77$$ 0 0
$$78$$ 0 0
$$79$$ 17.0000 1.91265 0.956325 0.292306i $$-0.0944227\pi$$
0.956325 + 0.292306i $$0.0944227\pi$$
$$80$$ 0 0
$$81$$ 1.00000 0.111111
$$82$$ 0 0
$$83$$ 16.0000 1.75623 0.878114 0.478451i $$-0.158802\pi$$
0.878114 + 0.478451i $$0.158802\pi$$
$$84$$ 0 0
$$85$$ −15.0000 −1.62698
$$86$$ 0 0
$$87$$ 3.00000 0.321634
$$88$$ 0 0
$$89$$ 3.00000 0.317999 0.159000 0.987279i $$-0.449173\pi$$
0.159000 + 0.987279i $$0.449173\pi$$
$$90$$ 0 0
$$91$$ 0 0
$$92$$ 0 0
$$93$$ 4.00000 0.414781
$$94$$ 0 0
$$95$$ 0 0
$$96$$ 0 0
$$97$$ −13.0000 −1.31995 −0.659975 0.751288i $$-0.729433\pi$$
−0.659975 + 0.751288i $$0.729433\pi$$
$$98$$ 0 0
$$99$$ 8.00000 0.804030
$$100$$ 0 0
$$101$$ 19.0000 1.89057 0.945285 0.326245i $$-0.105783\pi$$
0.945285 + 0.326245i $$0.105783\pi$$
$$102$$ 0 0
$$103$$ 0 0 1.00000i $$-0.5\pi$$
1.00000i $$0.5\pi$$
$$104$$ 0 0
$$105$$ 0 0
$$106$$ 0 0
$$107$$ −12.0000 −1.16008 −0.580042 0.814587i $$-0.696964\pi$$
−0.580042 + 0.814587i $$0.696964\pi$$
$$108$$ 0 0
$$109$$ 7.00000 0.670478 0.335239 0.942133i $$-0.391183\pi$$
0.335239 + 0.942133i $$0.391183\pi$$
$$110$$ 0 0
$$111$$ 2.00000 0.189832
$$112$$ 0 0
$$113$$ 6.00000 0.564433 0.282216 0.959351i $$-0.408930\pi$$
0.282216 + 0.959351i $$0.408930\pi$$
$$114$$ 0 0
$$115$$ −3.00000 −0.279751
$$116$$ 0 0
$$117$$ 10.0000 0.924500
$$118$$ 0 0
$$119$$ 0 0
$$120$$ 0 0
$$121$$ 5.00000 0.454545
$$122$$ 0 0
$$123$$ −5.00000 −0.450835
$$124$$ 0 0
$$125$$ −3.00000 −0.268328
$$126$$ 0 0
$$127$$ 15.0000 1.33103 0.665517 0.746382i $$-0.268211\pi$$
0.665517 + 0.746382i $$0.268211\pi$$
$$128$$ 0 0
$$129$$ −11.0000 −0.968496
$$130$$ 0 0
$$131$$ −15.0000 −1.31056 −0.655278 0.755388i $$-0.727449\pi$$
−0.655278 + 0.755388i $$0.727449\pi$$
$$132$$ 0 0
$$133$$ 0 0
$$134$$ 0 0
$$135$$ −15.0000 −1.29099
$$136$$ 0 0
$$137$$ −5.00000 −0.427179 −0.213589 0.976924i $$-0.568515\pi$$
−0.213589 + 0.976924i $$0.568515\pi$$
$$138$$ 0 0
$$139$$ 15.0000 1.27228 0.636142 0.771572i $$-0.280529\pi$$
0.636142 + 0.771572i $$0.280529\pi$$
$$140$$ 0 0
$$141$$ −5.00000 −0.421076
$$142$$ 0 0
$$143$$ 20.0000 1.67248
$$144$$ 0 0
$$145$$ 9.00000 0.747409
$$146$$ 0 0
$$147$$ −7.00000 −0.577350
$$148$$ 0 0
$$149$$ −17.0000 −1.39269 −0.696347 0.717705i $$-0.745193\pi$$
−0.696347 + 0.717705i $$0.745193\pi$$
$$150$$ 0 0
$$151$$ 16.0000 1.30206 0.651031 0.759051i $$-0.274337\pi$$
0.651031 + 0.759051i $$0.274337\pi$$
$$152$$ 0 0
$$153$$ 10.0000 0.808452
$$154$$ 0 0
$$155$$ 12.0000 0.963863
$$156$$ 0 0
$$157$$ −13.0000 −1.03751 −0.518756 0.854922i $$-0.673605\pi$$
−0.518756 + 0.854922i $$0.673605\pi$$
$$158$$ 0 0
$$159$$ −9.00000 −0.713746
$$160$$ 0 0
$$161$$ 0 0
$$162$$ 0 0
$$163$$ −4.00000 −0.313304 −0.156652 0.987654i $$-0.550070\pi$$
−0.156652 + 0.987654i $$0.550070\pi$$
$$164$$ 0 0
$$165$$ −12.0000 −0.934199
$$166$$ 0 0
$$167$$ −5.00000 −0.386912 −0.193456 0.981109i $$-0.561970\pi$$
−0.193456 + 0.981109i $$0.561970\pi$$
$$168$$ 0 0
$$169$$ 12.0000 0.923077
$$170$$ 0 0
$$171$$ 0 0
$$172$$ 0 0
$$173$$ −5.00000 −0.380143 −0.190071 0.981770i $$-0.560872\pi$$
−0.190071 + 0.981770i $$0.560872\pi$$
$$174$$ 0 0
$$175$$ 0 0
$$176$$ 0 0
$$177$$ 13.0000 0.977140
$$178$$ 0 0
$$179$$ −12.0000 −0.896922 −0.448461 0.893802i $$-0.648028\pi$$
−0.448461 + 0.893802i $$0.648028\pi$$
$$180$$ 0 0
$$181$$ 7.00000 0.520306 0.260153 0.965567i $$-0.416227\pi$$
0.260153 + 0.965567i $$0.416227\pi$$
$$182$$ 0 0
$$183$$ −1.00000 −0.0739221
$$184$$ 0 0
$$185$$ 6.00000 0.441129
$$186$$ 0 0
$$187$$ 20.0000 1.46254
$$188$$ 0 0
$$189$$ 0 0
$$190$$ 0 0
$$191$$ −16.0000 −1.15772 −0.578860 0.815427i $$-0.696502\pi$$
−0.578860 + 0.815427i $$0.696502\pi$$
$$192$$ 0 0
$$193$$ 15.0000 1.07972 0.539862 0.841754i $$-0.318476\pi$$
0.539862 + 0.841754i $$0.318476\pi$$
$$194$$ 0 0
$$195$$ −15.0000 −1.07417
$$196$$ 0 0
$$197$$ 10.0000 0.712470 0.356235 0.934396i $$-0.384060\pi$$
0.356235 + 0.934396i $$0.384060\pi$$
$$198$$ 0 0
$$199$$ 3.00000 0.212664 0.106332 0.994331i $$-0.466089\pi$$
0.106332 + 0.994331i $$0.466089\pi$$
$$200$$ 0 0
$$201$$ −5.00000 −0.352673
$$202$$ 0 0
$$203$$ 0 0
$$204$$ 0 0
$$205$$ −15.0000 −1.04765
$$206$$ 0 0
$$207$$ 2.00000 0.139010
$$208$$ 0 0
$$209$$ 0 0
$$210$$ 0 0
$$211$$ 9.00000 0.619586 0.309793 0.950804i $$-0.399740\pi$$
0.309793 + 0.950804i $$0.399740\pi$$
$$212$$ 0 0
$$213$$ 1.00000 0.0685189
$$214$$ 0 0
$$215$$ −33.0000 −2.25058
$$216$$ 0 0
$$217$$ 0 0
$$218$$ 0 0
$$219$$ −9.00000 −0.608164
$$220$$ 0 0
$$221$$ 25.0000 1.68168
$$222$$ 0 0
$$223$$ −11.0000 −0.736614 −0.368307 0.929704i $$-0.620063\pi$$
−0.368307 + 0.929704i $$0.620063\pi$$
$$224$$ 0 0
$$225$$ −8.00000 −0.533333
$$226$$ 0 0
$$227$$ 12.0000 0.796468 0.398234 0.917284i $$-0.369623\pi$$
0.398234 + 0.917284i $$0.369623\pi$$
$$228$$ 0 0
$$229$$ −22.0000 −1.45380 −0.726900 0.686743i $$-0.759040\pi$$
−0.726900 + 0.686743i $$0.759040\pi$$
$$230$$ 0 0
$$231$$ 0 0
$$232$$ 0 0
$$233$$ 11.0000 0.720634 0.360317 0.932830i $$-0.382669\pi$$
0.360317 + 0.932830i $$0.382669\pi$$
$$234$$ 0 0
$$235$$ −15.0000 −0.978492
$$236$$ 0 0
$$237$$ 17.0000 1.10427
$$238$$ 0 0
$$239$$ 12.0000 0.776215 0.388108 0.921614i $$-0.373129\pi$$
0.388108 + 0.921614i $$0.373129\pi$$
$$240$$ 0 0
$$241$$ −21.0000 −1.35273 −0.676364 0.736567i $$-0.736446\pi$$
−0.676364 + 0.736567i $$0.736446\pi$$
$$242$$ 0 0
$$243$$ 16.0000 1.02640
$$244$$ 0 0
$$245$$ −21.0000 −1.34164
$$246$$ 0 0
$$247$$ 0 0
$$248$$ 0 0
$$249$$ 16.0000 1.01396
$$250$$ 0 0
$$251$$ −9.00000 −0.568075 −0.284037 0.958813i $$-0.591674\pi$$
−0.284037 + 0.958813i $$0.591674\pi$$
$$252$$ 0 0
$$253$$ 4.00000 0.251478
$$254$$ 0 0
$$255$$ −15.0000 −0.939336
$$256$$ 0 0
$$257$$ −25.0000 −1.55946 −0.779729 0.626118i $$-0.784643\pi$$
−0.779729 + 0.626118i $$0.784643\pi$$
$$258$$ 0 0
$$259$$ 0 0
$$260$$ 0 0
$$261$$ −6.00000 −0.371391
$$262$$ 0 0
$$263$$ −19.0000 −1.17159 −0.585795 0.810459i $$-0.699218\pi$$
−0.585795 + 0.810459i $$0.699218\pi$$
$$264$$ 0 0
$$265$$ −27.0000 −1.65860
$$266$$ 0 0
$$267$$ 3.00000 0.183597
$$268$$ 0 0
$$269$$ −17.0000 −1.03651 −0.518254 0.855227i $$-0.673418\pi$$
−0.518254 + 0.855227i $$0.673418\pi$$
$$270$$ 0 0
$$271$$ 29.0000 1.76162 0.880812 0.473466i $$-0.156997\pi$$
0.880812 + 0.473466i $$0.156997\pi$$
$$272$$ 0 0
$$273$$ 0 0
$$274$$ 0 0
$$275$$ −16.0000 −0.964836
$$276$$ 0 0
$$277$$ −30.0000 −1.80253 −0.901263 0.433273i $$-0.857359\pi$$
−0.901263 + 0.433273i $$0.857359\pi$$
$$278$$ 0 0
$$279$$ −8.00000 −0.478947
$$280$$ 0 0
$$281$$ 15.0000 0.894825 0.447412 0.894328i $$-0.352346\pi$$
0.447412 + 0.894328i $$0.352346\pi$$
$$282$$ 0 0
$$283$$ −15.0000 −0.891657 −0.445829 0.895118i $$-0.647091\pi$$
−0.445829 + 0.895118i $$0.647091\pi$$
$$284$$ 0 0
$$285$$ 0 0
$$286$$ 0 0
$$287$$ 0 0
$$288$$ 0 0
$$289$$ 8.00000 0.470588
$$290$$ 0 0
$$291$$ −13.0000 −0.762073
$$292$$ 0 0
$$293$$ 10.0000 0.584206 0.292103 0.956387i $$-0.405645\pi$$
0.292103 + 0.956387i $$0.405645\pi$$
$$294$$ 0 0
$$295$$ 39.0000 2.27067
$$296$$ 0 0
$$297$$ 20.0000 1.16052
$$298$$ 0 0
$$299$$ 5.00000 0.289157
$$300$$ 0 0
$$301$$ 0 0
$$302$$ 0 0
$$303$$ 19.0000 1.09152
$$304$$ 0 0
$$305$$ −3.00000 −0.171780
$$306$$ 0 0
$$307$$ −3.00000 −0.171219 −0.0856095 0.996329i $$-0.527284\pi$$
−0.0856095 + 0.996329i $$0.527284\pi$$
$$308$$ 0 0
$$309$$ 0 0
$$310$$ 0 0
$$311$$ 12.0000 0.680458 0.340229 0.940343i $$-0.389495\pi$$
0.340229 + 0.940343i $$0.389495\pi$$
$$312$$ 0 0
$$313$$ −5.00000 −0.282617 −0.141308 0.989966i $$-0.545131\pi$$
−0.141308 + 0.989966i $$0.545131\pi$$
$$314$$ 0 0
$$315$$ 0 0
$$316$$ 0 0
$$317$$ 27.0000 1.51647 0.758236 0.651981i $$-0.226062\pi$$
0.758236 + 0.651981i $$0.226062\pi$$
$$318$$ 0 0
$$319$$ −12.0000 −0.671871
$$320$$ 0 0
$$321$$ −12.0000 −0.669775
$$322$$ 0 0
$$323$$ 0 0
$$324$$ 0 0
$$325$$ −20.0000 −1.10940
$$326$$ 0 0
$$327$$ 7.00000 0.387101
$$328$$ 0 0
$$329$$ 0 0
$$330$$ 0 0
$$331$$ 20.0000 1.09930 0.549650 0.835395i $$-0.314761\pi$$
0.549650 + 0.835395i $$0.314761\pi$$
$$332$$ 0 0
$$333$$ −4.00000 −0.219199
$$334$$ 0 0
$$335$$ −15.0000 −0.819538
$$336$$ 0 0
$$337$$ 3.00000 0.163420 0.0817102 0.996656i $$-0.473962\pi$$
0.0817102 + 0.996656i $$0.473962\pi$$
$$338$$ 0 0
$$339$$ 6.00000 0.325875
$$340$$ 0 0
$$341$$ −16.0000 −0.866449
$$342$$ 0 0
$$343$$ 0 0
$$344$$ 0 0
$$345$$ −3.00000 −0.161515
$$346$$ 0 0
$$347$$ 5.00000 0.268414 0.134207 0.990953i $$-0.457151\pi$$
0.134207 + 0.990953i $$0.457151\pi$$
$$348$$ 0 0
$$349$$ −14.0000 −0.749403 −0.374701 0.927146i $$-0.622255\pi$$
−0.374701 + 0.927146i $$0.622255\pi$$
$$350$$ 0 0
$$351$$ 25.0000 1.33440
$$352$$ 0 0
$$353$$ −10.0000 −0.532246 −0.266123 0.963939i $$-0.585743\pi$$
−0.266123 + 0.963939i $$0.585743\pi$$
$$354$$ 0 0
$$355$$ 3.00000 0.159223
$$356$$ 0 0
$$357$$ 0 0
$$358$$ 0 0
$$359$$ 5.00000 0.263890 0.131945 0.991257i $$-0.457878\pi$$
0.131945 + 0.991257i $$0.457878\pi$$
$$360$$ 0 0
$$361$$ 0 0
$$362$$ 0 0
$$363$$ 5.00000 0.262432
$$364$$ 0 0
$$365$$ −27.0000 −1.41324
$$366$$ 0 0
$$367$$ −5.00000 −0.260998 −0.130499 0.991448i $$-0.541658\pi$$
−0.130499 + 0.991448i $$0.541658\pi$$
$$368$$ 0 0
$$369$$ 10.0000 0.520579
$$370$$ 0 0
$$371$$ 0 0
$$372$$ 0 0
$$373$$ −10.0000 −0.517780 −0.258890 0.965907i $$-0.583357\pi$$
−0.258890 + 0.965907i $$0.583357\pi$$
$$374$$ 0 0
$$375$$ −3.00000 −0.154919
$$376$$ 0 0
$$377$$ −15.0000 −0.772539
$$378$$ 0 0
$$379$$ −12.0000 −0.616399 −0.308199 0.951322i $$-0.599726\pi$$
−0.308199 + 0.951322i $$0.599726\pi$$
$$380$$ 0 0
$$381$$ 15.0000 0.768473
$$382$$ 0 0
$$383$$ −15.0000 −0.766464 −0.383232 0.923652i $$-0.625189\pi$$
−0.383232 + 0.923652i $$0.625189\pi$$
$$384$$ 0 0
$$385$$ 0 0
$$386$$ 0 0
$$387$$ 22.0000 1.11832
$$388$$ 0 0
$$389$$ −17.0000 −0.861934 −0.430967 0.902368i $$-0.641828\pi$$
−0.430967 + 0.902368i $$0.641828\pi$$
$$390$$ 0 0
$$391$$ 5.00000 0.252861
$$392$$ 0 0
$$393$$ −15.0000 −0.756650
$$394$$ 0 0
$$395$$ 51.0000 2.56609
$$396$$ 0 0
$$397$$ 35.0000 1.75660 0.878300 0.478110i $$-0.158678\pi$$
0.878300 + 0.478110i $$0.158678\pi$$
$$398$$ 0 0
$$399$$ 0 0
$$400$$ 0 0
$$401$$ 11.0000 0.549314 0.274657 0.961542i $$-0.411436\pi$$
0.274657 + 0.961542i $$0.411436\pi$$
$$402$$ 0 0
$$403$$ −20.0000 −0.996271
$$404$$ 0 0
$$405$$ 3.00000 0.149071
$$406$$ 0 0
$$407$$ −8.00000 −0.396545
$$408$$ 0 0
$$409$$ 7.00000 0.346128 0.173064 0.984911i $$-0.444633\pi$$
0.173064 + 0.984911i $$0.444633\pi$$
$$410$$ 0 0
$$411$$ −5.00000 −0.246632
$$412$$ 0 0
$$413$$ 0 0
$$414$$ 0 0
$$415$$ 48.0000 2.35623
$$416$$ 0 0
$$417$$ 15.0000 0.734553
$$418$$ 0 0
$$419$$ 12.0000 0.586238 0.293119 0.956076i $$-0.405307\pi$$
0.293119 + 0.956076i $$0.405307\pi$$
$$420$$ 0 0
$$421$$ −29.0000 −1.41337 −0.706687 0.707527i $$-0.749811\pi$$
−0.706687 + 0.707527i $$0.749811\pi$$
$$422$$ 0 0
$$423$$ 10.0000 0.486217
$$424$$ 0 0
$$425$$ −20.0000 −0.970143
$$426$$ 0 0
$$427$$ 0 0
$$428$$ 0 0
$$429$$ 20.0000 0.965609
$$430$$ 0 0
$$431$$ −9.00000 −0.433515 −0.216757 0.976226i $$-0.569548\pi$$
−0.216757 + 0.976226i $$0.569548\pi$$
$$432$$ 0 0
$$433$$ −25.0000 −1.20142 −0.600712 0.799466i $$-0.705116\pi$$
−0.600712 + 0.799466i $$0.705116\pi$$
$$434$$ 0 0
$$435$$ 9.00000 0.431517
$$436$$ 0 0
$$437$$ 0 0
$$438$$ 0 0
$$439$$ −7.00000 −0.334092 −0.167046 0.985949i $$-0.553423\pi$$
−0.167046 + 0.985949i $$0.553423\pi$$
$$440$$ 0 0
$$441$$ 14.0000 0.666667
$$442$$ 0 0
$$443$$ 15.0000 0.712672 0.356336 0.934358i $$-0.384026\pi$$
0.356336 + 0.934358i $$0.384026\pi$$
$$444$$ 0 0
$$445$$ 9.00000 0.426641
$$446$$ 0 0
$$447$$ −17.0000 −0.804072
$$448$$ 0 0
$$449$$ −2.00000 −0.0943858 −0.0471929 0.998886i $$-0.515028\pi$$
−0.0471929 + 0.998886i $$0.515028\pi$$
$$450$$ 0 0
$$451$$ 20.0000 0.941763
$$452$$ 0 0
$$453$$ 16.0000 0.751746
$$454$$ 0 0
$$455$$ 0 0
$$456$$ 0 0
$$457$$ 10.0000 0.467780 0.233890 0.972263i $$-0.424854\pi$$
0.233890 + 0.972263i $$0.424854\pi$$
$$458$$ 0 0
$$459$$ 25.0000 1.16690
$$460$$ 0 0
$$461$$ −1.00000 −0.0465746 −0.0232873 0.999729i $$-0.507413\pi$$
−0.0232873 + 0.999729i $$0.507413\pi$$
$$462$$ 0 0
$$463$$ −20.0000 −0.929479 −0.464739 0.885448i $$-0.653852\pi$$
−0.464739 + 0.885448i $$0.653852\pi$$
$$464$$ 0 0
$$465$$ 12.0000 0.556487
$$466$$ 0 0
$$467$$ 12.0000 0.555294 0.277647 0.960683i $$-0.410445\pi$$
0.277647 + 0.960683i $$0.410445\pi$$
$$468$$ 0 0
$$469$$ 0 0
$$470$$ 0 0
$$471$$ −13.0000 −0.599008
$$472$$ 0 0
$$473$$ 44.0000 2.02312
$$474$$ 0 0
$$475$$ 0 0
$$476$$ 0 0
$$477$$ 18.0000 0.824163
$$478$$ 0 0
$$479$$ 27.0000 1.23366 0.616831 0.787096i $$-0.288416\pi$$
0.616831 + 0.787096i $$0.288416\pi$$
$$480$$ 0 0
$$481$$ −10.0000 −0.455961
$$482$$ 0 0
$$483$$ 0 0
$$484$$ 0 0
$$485$$ −39.0000 −1.77090
$$486$$ 0 0
$$487$$ −40.0000 −1.81257 −0.906287 0.422664i $$-0.861095\pi$$
−0.906287 + 0.422664i $$0.861095\pi$$
$$488$$ 0 0
$$489$$ −4.00000 −0.180886
$$490$$ 0 0
$$491$$ 9.00000 0.406164 0.203082 0.979162i $$-0.434904\pi$$
0.203082 + 0.979162i $$0.434904\pi$$
$$492$$ 0 0
$$493$$ −15.0000 −0.675566
$$494$$ 0 0
$$495$$ 24.0000 1.07872
$$496$$ 0 0
$$497$$ 0 0
$$498$$ 0 0
$$499$$ −35.0000 −1.56682 −0.783408 0.621508i $$-0.786520\pi$$
−0.783408 + 0.621508i $$0.786520\pi$$
$$500$$ 0 0
$$501$$ −5.00000 −0.223384
$$502$$ 0 0
$$503$$ 15.0000 0.668817 0.334408 0.942428i $$-0.391463\pi$$
0.334408 + 0.942428i $$0.391463\pi$$
$$504$$ 0 0
$$505$$ 57.0000 2.53647
$$506$$ 0 0
$$507$$ 12.0000 0.532939
$$508$$ 0 0
$$509$$ −5.00000 −0.221621 −0.110811 0.993842i $$-0.535345\pi$$
−0.110811 + 0.993842i $$0.535345\pi$$
$$510$$ 0 0
$$511$$ 0 0
$$512$$ 0 0
$$513$$ 0 0
$$514$$ 0 0
$$515$$ 0 0
$$516$$ 0 0
$$517$$ 20.0000 0.879599
$$518$$ 0 0
$$519$$ −5.00000 −0.219476
$$520$$ 0 0
$$521$$ −6.00000 −0.262865 −0.131432 0.991325i $$-0.541958\pi$$
−0.131432 + 0.991325i $$0.541958\pi$$
$$522$$ 0 0
$$523$$ −29.0000 −1.26808 −0.634041 0.773300i $$-0.718605\pi$$
−0.634041 + 0.773300i $$0.718605\pi$$
$$524$$ 0 0
$$525$$ 0 0
$$526$$ 0 0
$$527$$ −20.0000 −0.871214
$$528$$ 0 0
$$529$$ −22.0000 −0.956522
$$530$$ 0 0
$$531$$ −26.0000 −1.12830
$$532$$ 0 0
$$533$$ 25.0000 1.08287
$$534$$ 0 0
$$535$$ −36.0000 −1.55642
$$536$$ 0 0
$$537$$ −12.0000 −0.517838
$$538$$ 0 0
$$539$$ 28.0000 1.20605
$$540$$ 0 0
$$541$$ 11.0000 0.472927 0.236463 0.971640i $$-0.424012\pi$$
0.236463 + 0.971640i $$0.424012\pi$$
$$542$$ 0 0
$$543$$ 7.00000 0.300399
$$544$$ 0 0
$$545$$ 21.0000 0.899541
$$546$$ 0 0
$$547$$ 7.00000 0.299298 0.149649 0.988739i $$-0.452186\pi$$
0.149649 + 0.988739i $$0.452186\pi$$
$$548$$ 0 0
$$549$$ 2.00000 0.0853579
$$550$$ 0 0
$$551$$ 0 0
$$552$$ 0 0
$$553$$ 0 0
$$554$$ 0 0
$$555$$ 6.00000 0.254686
$$556$$ 0 0
$$557$$ −33.0000 −1.39825 −0.699127 0.714997i $$-0.746428\pi$$
−0.699127 + 0.714997i $$0.746428\pi$$
$$558$$ 0 0
$$559$$ 55.0000 2.32625
$$560$$ 0 0
$$561$$ 20.0000 0.844401
$$562$$ 0 0
$$563$$ 4.00000 0.168580 0.0842900 0.996441i $$-0.473138\pi$$
0.0842900 + 0.996441i $$0.473138\pi$$
$$564$$ 0 0
$$565$$ 18.0000 0.757266
$$566$$ 0 0
$$567$$ 0 0
$$568$$ 0 0
$$569$$ −18.0000 −0.754599 −0.377300 0.926091i $$-0.623147\pi$$
−0.377300 + 0.926091i $$0.623147\pi$$
$$570$$ 0 0
$$571$$ −4.00000 −0.167395 −0.0836974 0.996491i $$-0.526673\pi$$
−0.0836974 + 0.996491i $$0.526673\pi$$
$$572$$ 0 0
$$573$$ −16.0000 −0.668410
$$574$$ 0 0
$$575$$ −4.00000 −0.166812
$$576$$ 0 0
$$577$$ −2.00000 −0.0832611 −0.0416305 0.999133i $$-0.513255\pi$$
−0.0416305 + 0.999133i $$0.513255\pi$$
$$578$$ 0 0
$$579$$ 15.0000 0.623379
$$580$$ 0 0
$$581$$ 0 0
$$582$$ 0 0
$$583$$ 36.0000 1.49097
$$584$$ 0 0
$$585$$ 30.0000 1.24035
$$586$$ 0 0
$$587$$ 25.0000 1.03186 0.515930 0.856631i $$-0.327446\pi$$
0.515930 + 0.856631i $$0.327446\pi$$
$$588$$ 0 0
$$589$$ 0 0
$$590$$ 0 0
$$591$$ 10.0000 0.411345
$$592$$ 0 0
$$593$$ 35.0000 1.43728 0.718639 0.695383i $$-0.244765\pi$$
0.718639 + 0.695383i $$0.244765\pi$$
$$594$$ 0 0
$$595$$ 0 0
$$596$$ 0 0
$$597$$ 3.00000 0.122782
$$598$$ 0 0
$$599$$ −25.0000 −1.02147 −0.510736 0.859738i $$-0.670627\pi$$
−0.510736 + 0.859738i $$0.670627\pi$$
$$600$$ 0 0
$$601$$ −34.0000 −1.38689 −0.693444 0.720510i $$-0.743908\pi$$
−0.693444 + 0.720510i $$0.743908\pi$$
$$602$$ 0 0
$$603$$ 10.0000 0.407231
$$604$$ 0 0
$$605$$ 15.0000 0.609837
$$606$$ 0 0
$$607$$ 0 0 1.00000i $$-0.5\pi$$
1.00000i $$0.5\pi$$
$$608$$ 0 0
$$609$$ 0 0
$$610$$ 0 0
$$611$$ 25.0000 1.01139
$$612$$ 0 0
$$613$$ −1.00000 −0.0403896 −0.0201948 0.999796i $$-0.506429\pi$$
−0.0201948 + 0.999796i $$0.506429\pi$$
$$614$$ 0 0
$$615$$ −15.0000 −0.604858
$$616$$ 0 0
$$617$$ 3.00000 0.120775 0.0603877 0.998175i $$-0.480766\pi$$
0.0603877 + 0.998175i $$0.480766\pi$$
$$618$$ 0 0
$$619$$ −20.0000 −0.803868 −0.401934 0.915669i $$-0.631662\pi$$
−0.401934 + 0.915669i $$0.631662\pi$$
$$620$$ 0 0
$$621$$ 5.00000 0.200643
$$622$$ 0 0
$$623$$ 0 0
$$624$$ 0 0
$$625$$ −29.0000 −1.16000
$$626$$ 0 0
$$627$$ 0 0
$$628$$ 0 0
$$629$$ −10.0000 −0.398726
$$630$$ 0 0
$$631$$ −29.0000 −1.15447 −0.577236 0.816577i $$-0.695869\pi$$
−0.577236 + 0.816577i $$0.695869\pi$$
$$632$$ 0 0
$$633$$ 9.00000 0.357718
$$634$$ 0 0
$$635$$ 45.0000 1.78577
$$636$$ 0 0
$$637$$ 35.0000 1.38675
$$638$$ 0 0
$$639$$ −2.00000 −0.0791188
$$640$$ 0 0
$$641$$ −9.00000 −0.355479 −0.177739 0.984078i $$-0.556878\pi$$
−0.177739 + 0.984078i $$0.556878\pi$$
$$642$$ 0 0
$$643$$ 5.00000 0.197181 0.0985904 0.995128i $$-0.468567\pi$$
0.0985904 + 0.995128i $$0.468567\pi$$
$$644$$ 0 0
$$645$$ −33.0000 −1.29937
$$646$$ 0 0
$$647$$ 0 0 1.00000i $$-0.5\pi$$
1.00000i $$0.5\pi$$
$$648$$ 0 0
$$649$$ −52.0000 −2.04118
$$650$$ 0 0
$$651$$ 0 0
$$652$$ 0 0
$$653$$ −6.00000 −0.234798 −0.117399 0.993085i $$-0.537456\pi$$
−0.117399 + 0.993085i $$0.537456\pi$$
$$654$$ 0 0
$$655$$ −45.0000 −1.75830
$$656$$ 0 0
$$657$$ 18.0000 0.702247
$$658$$ 0 0
$$659$$ 19.0000 0.740135 0.370067 0.929005i $$-0.379335\pi$$
0.370067 + 0.929005i $$0.379335\pi$$
$$660$$ 0 0
$$661$$ −29.0000 −1.12797 −0.563985 0.825785i $$-0.690732\pi$$
−0.563985 + 0.825785i $$0.690732\pi$$
$$662$$ 0 0
$$663$$ 25.0000 0.970920
$$664$$ 0 0
$$665$$ 0 0
$$666$$ 0 0
$$667$$ −3.00000 −0.116160
$$668$$ 0 0
$$669$$ −11.0000 −0.425285
$$670$$ 0 0
$$671$$ 4.00000 0.154418
$$672$$ 0 0
$$673$$ 30.0000 1.15642 0.578208 0.815890i $$-0.303752\pi$$
0.578208 + 0.815890i $$0.303752\pi$$
$$674$$ 0 0
$$675$$ −20.0000 −0.769800
$$676$$ 0 0
$$677$$ −30.0000 −1.15299 −0.576497 0.817099i $$-0.695581\pi$$
−0.576497 + 0.817099i $$0.695581\pi$$
$$678$$ 0 0
$$679$$ 0 0
$$680$$ 0 0
$$681$$ 12.0000 0.459841
$$682$$ 0 0
$$683$$ 0 0 1.00000i $$-0.5\pi$$
1.00000i $$0.5\pi$$
$$684$$ 0 0
$$685$$ −15.0000 −0.573121
$$686$$ 0 0
$$687$$ −22.0000 −0.839352
$$688$$ 0 0
$$689$$ 45.0000 1.71436
$$690$$ 0 0
$$691$$ 36.0000 1.36950 0.684752 0.728776i $$-0.259910\pi$$
0.684752 + 0.728776i $$0.259910\pi$$
$$692$$ 0 0
$$693$$ 0 0
$$694$$ 0 0
$$695$$ 45.0000 1.70695
$$696$$ 0 0
$$697$$ 25.0000 0.946943
$$698$$ 0 0
$$699$$ 11.0000 0.416058
$$700$$ 0 0
$$701$$ 23.0000 0.868698 0.434349 0.900745i $$-0.356978\pi$$
0.434349 + 0.900745i $$0.356978\pi$$
$$702$$ 0 0
$$703$$ 0 0
$$704$$ 0 0
$$705$$ −15.0000 −0.564933
$$706$$ 0 0
$$707$$ 0 0
$$708$$ 0 0
$$709$$ −17.0000 −0.638448 −0.319224 0.947679i $$-0.603422\pi$$
−0.319224 + 0.947679i $$0.603422\pi$$
$$710$$ 0 0
$$711$$ −34.0000 −1.27510
$$712$$ 0 0
$$713$$ −4.00000 −0.149801
$$714$$ 0 0
$$715$$ 60.0000 2.24387
$$716$$ 0 0
$$717$$ 12.0000 0.448148
$$718$$ 0 0
$$719$$ −27.0000 −1.00693 −0.503465 0.864016i $$-0.667942\pi$$
−0.503465 + 0.864016i $$0.667942\pi$$
$$720$$ 0 0
$$721$$ 0 0
$$722$$ 0 0
$$723$$ −21.0000 −0.780998
$$724$$ 0 0
$$725$$ 12.0000 0.445669
$$726$$ 0 0
$$727$$ 13.0000 0.482143 0.241072 0.970507i $$-0.422501\pi$$
0.241072 + 0.970507i $$0.422501\pi$$
$$728$$ 0 0
$$729$$ 13.0000 0.481481
$$730$$ 0 0
$$731$$ 55.0000 2.03425
$$732$$ 0 0
$$733$$ −6.00000 −0.221615 −0.110808 0.993842i $$-0.535344\pi$$
−0.110808 + 0.993842i $$0.535344\pi$$
$$734$$ 0 0
$$735$$ −21.0000 −0.774597
$$736$$ 0 0
$$737$$ 20.0000 0.736709
$$738$$ 0 0
$$739$$ 37.0000 1.36107 0.680534 0.732717i $$-0.261748\pi$$
0.680534 + 0.732717i $$0.261748\pi$$
$$740$$ 0 0
$$741$$ 0 0
$$742$$ 0 0
$$743$$ 29.0000 1.06391 0.531953 0.846774i $$-0.321458\pi$$
0.531953 + 0.846774i $$0.321458\pi$$
$$744$$ 0 0
$$745$$ −51.0000 −1.86850
$$746$$ 0 0
$$747$$ −32.0000 −1.17082
$$748$$ 0 0
$$749$$ 0 0
$$750$$ 0 0
$$751$$ 31.0000 1.13121 0.565603 0.824678i $$-0.308643\pi$$
0.565603 + 0.824678i $$0.308643\pi$$
$$752$$ 0 0
$$753$$ −9.00000 −0.327978
$$754$$ 0 0
$$755$$ 48.0000 1.74690
$$756$$ 0 0
$$757$$ 35.0000 1.27210 0.636048 0.771649i $$-0.280568\pi$$
0.636048 + 0.771649i $$0.280568\pi$$
$$758$$ 0 0
$$759$$ 4.00000 0.145191
$$760$$ 0 0
$$761$$ 38.0000 1.37750 0.688749 0.724999i $$-0.258160\pi$$
0.688749 + 0.724999i $$0.258160\pi$$
$$762$$ 0 0
$$763$$ 0 0
$$764$$ 0 0
$$765$$ 30.0000 1.08465
$$766$$ 0 0
$$767$$ −65.0000 −2.34701
$$768$$ 0 0
$$769$$ 23.0000 0.829401 0.414701 0.909958i $$-0.363886\pi$$
0.414701 + 0.909958i $$0.363886\pi$$
$$770$$ 0 0
$$771$$ −25.0000 −0.900353
$$772$$ 0 0
$$773$$ −25.0000 −0.899188 −0.449594 0.893233i $$-0.648431\pi$$
−0.449594 + 0.893233i $$0.648431\pi$$
$$774$$ 0 0
$$775$$ 16.0000 0.574737
$$776$$ 0 0
$$777$$ 0 0
$$778$$ 0 0
$$779$$ 0 0
$$780$$ 0 0
$$781$$ −4.00000 −0.143131
$$782$$ 0 0
$$783$$ −15.0000 −0.536056
$$784$$ 0 0
$$785$$ −39.0000 −1.39197
$$786$$ 0 0
$$787$$ 20.0000 0.712923 0.356462 0.934310i $$-0.383983\pi$$
0.356462 + 0.934310i $$0.383983\pi$$
$$788$$ 0 0
$$789$$ −19.0000 −0.676418
$$790$$ 0 0
$$791$$ 0 0
$$792$$ 0 0
$$793$$ 5.00000 0.177555
$$794$$ 0 0
$$795$$ −27.0000 −0.957591
$$796$$ 0 0
$$797$$ 42.0000 1.48772 0.743858 0.668338i $$-0.232994\pi$$
0.743858 + 0.668338i $$0.232994\pi$$
$$798$$ 0 0
$$799$$ 25.0000 0.884436
$$800$$ 0 0
$$801$$ −6.00000 −0.212000
$$802$$ 0 0
$$803$$ 36.0000 1.27041
$$804$$ 0 0
$$805$$ 0 0
$$806$$ 0 0
$$807$$ −17.0000 −0.598428
$$808$$ 0 0
$$809$$ −42.0000 −1.47664 −0.738321 0.674450i $$-0.764381\pi$$
−0.738321 + 0.674450i $$0.764381\pi$$
$$810$$ 0 0
$$811$$ −21.0000 −0.737410 −0.368705 0.929547i $$-0.620199\pi$$
−0.368705 + 0.929547i $$0.620199\pi$$
$$812$$ 0 0
$$813$$ 29.0000 1.01707
$$814$$ 0 0
$$815$$ −12.0000 −0.420342
$$816$$ 0 0
$$817$$ 0 0
$$818$$ 0 0
$$819$$ 0 0
$$820$$ 0 0
$$821$$ 11.0000 0.383903 0.191951 0.981404i $$-0.438518\pi$$
0.191951 + 0.981404i $$0.438518\pi$$
$$822$$ 0 0
$$823$$ −5.00000 −0.174289 −0.0871445 0.996196i $$-0.527774\pi$$
−0.0871445 + 0.996196i $$0.527774\pi$$
$$824$$ 0 0
$$825$$ −16.0000 −0.557048
$$826$$ 0 0
$$827$$ 15.0000 0.521601 0.260801 0.965393i $$-0.416014\pi$$
0.260801 + 0.965393i $$0.416014\pi$$
$$828$$ 0 0
$$829$$ 30.0000 1.04194 0.520972 0.853574i $$-0.325570\pi$$
0.520972 + 0.853574i $$0.325570\pi$$
$$830$$ 0 0
$$831$$ −30.0000 −1.04069
$$832$$ 0 0
$$833$$ 35.0000 1.21268
$$834$$ 0 0
$$835$$ −15.0000 −0.519096
$$836$$ 0 0
$$837$$ −20.0000 −0.691301
$$838$$ 0 0
$$839$$ −35.0000 −1.20833 −0.604167 0.796858i $$-0.706494\pi$$
−0.604167 + 0.796858i $$0.706494\pi$$
$$840$$ 0 0
$$841$$ −20.0000 −0.689655
$$842$$ 0 0
$$843$$ 15.0000 0.516627
$$844$$ 0 0
$$845$$ 36.0000 1.23844
$$846$$ 0 0
$$847$$ 0 0
$$848$$ 0 0
$$849$$ −15.0000 −0.514799
$$850$$ 0 0
$$851$$ −2.00000 −0.0685591
$$852$$ 0 0
$$853$$ −21.0000 −0.719026 −0.359513 0.933140i $$-0.617057\pi$$
−0.359513 + 0.933140i $$0.617057\pi$$
$$854$$ 0 0
$$855$$ 0 0
$$856$$ 0 0
$$857$$ 7.00000 0.239115 0.119558 0.992827i $$-0.461852\pi$$
0.119558 + 0.992827i $$0.461852\pi$$
$$858$$ 0 0
$$859$$ −13.0000 −0.443554 −0.221777 0.975097i $$-0.571186\pi$$
−0.221777 + 0.975097i $$0.571186\pi$$
$$860$$ 0 0
$$861$$ 0 0
$$862$$ 0 0
$$863$$ −36.0000 −1.22545 −0.612727 0.790295i $$-0.709928\pi$$
−0.612727 + 0.790295i $$0.709928\pi$$
$$864$$ 0 0
$$865$$ −15.0000 −0.510015
$$866$$ 0 0
$$867$$ 8.00000 0.271694
$$868$$ 0 0
$$869$$ −68.0000 −2.30674
$$870$$ 0 0
$$871$$ 25.0000 0.847093
$$872$$ 0 0
$$873$$ 26.0000 0.879967
$$874$$ 0 0
$$875$$ 0 0
$$876$$ 0 0
$$877$$ 47.0000 1.58708 0.793539 0.608520i $$-0.208236\pi$$
0.793539 + 0.608520i $$0.208236\pi$$
$$878$$ 0 0
$$879$$ 10.0000 0.337292
$$880$$ 0 0
$$881$$ 14.0000 0.471672 0.235836 0.971793i $$-0.424217\pi$$
0.235836 + 0.971793i $$0.424217\pi$$
$$882$$ 0 0
$$883$$ 55.0000 1.85090 0.925449 0.378873i $$-0.123688\pi$$
0.925449 + 0.378873i $$0.123688\pi$$
$$884$$ 0 0
$$885$$ 39.0000 1.31097
$$886$$ 0 0
$$887$$ −45.0000 −1.51095 −0.755476 0.655176i $$-0.772594\pi$$
−0.755476 + 0.655176i $$0.772594\pi$$
$$888$$ 0 0
$$889$$ 0 0
$$890$$ 0 0
$$891$$ −4.00000 −0.134005
$$892$$ 0 0
$$893$$ 0 0
$$894$$ 0 0
$$895$$ −36.0000 −1.20335
$$896$$ 0 0
$$897$$ 5.00000 0.166945
$$898$$ 0 0
$$899$$ 12.0000 0.400222
$$900$$ 0 0
$$901$$ 45.0000 1.49917
$$902$$ 0 0
$$903$$ 0 0
$$904$$ 0 0
$$905$$ 21.0000 0.698064
$$906$$ 0 0
$$907$$ −35.0000 −1.16216 −0.581078 0.813848i $$-0.697369\pi$$
−0.581078 + 0.813848i $$0.697369\pi$$
$$908$$ 0 0
$$909$$ −38.0000 −1.26038
$$910$$ 0 0
$$911$$ −56.0000 −1.85536 −0.927681 0.373373i $$-0.878201\pi$$
−0.927681 + 0.373373i $$0.878201\pi$$
$$912$$ 0 0
$$913$$ −64.0000 −2.11809
$$914$$ 0 0
$$915$$ −3.00000 −0.0991769
$$916$$ 0 0
$$917$$ 0 0
$$918$$ 0 0
$$919$$ −40.0000 −1.31948 −0.659739 0.751495i $$-0.729333\pi$$
−0.659739 + 0.751495i $$0.729333\pi$$
$$920$$ 0 0
$$921$$ −3.00000 −0.0988534
$$922$$ 0 0
$$923$$ −5.00000 −0.164577
$$924$$ 0 0
$$925$$ 8.00000 0.263038
$$926$$ 0 0
$$927$$ 0 0
$$928$$ 0 0
$$929$$ 7.00000 0.229663 0.114831 0.993385i $$-0.463367\pi$$
0.114831 + 0.993385i $$0.463367\pi$$
$$930$$ 0 0
$$931$$ 0 0
$$932$$ 0 0
$$933$$ 12.0000 0.392862
$$934$$ 0 0
$$935$$ 60.0000 1.96221
$$936$$ 0 0
$$937$$ −17.0000 −0.555366 −0.277683 0.960673i $$-0.589566\pi$$
−0.277683 + 0.960673i $$0.589566\pi$$
$$938$$ 0 0
$$939$$ −5.00000 −0.163169
$$940$$ 0 0
$$941$$ 39.0000 1.27136 0.635682 0.771951i $$-0.280719\pi$$
0.635682 + 0.771951i $$0.280719\pi$$
$$942$$ 0 0
$$943$$ 5.00000 0.162822
$$944$$ 0 0
$$945$$ 0 0
$$946$$ 0 0
$$947$$ −35.0000 −1.13735 −0.568674 0.822563i $$-0.692543\pi$$
−0.568674 + 0.822563i $$0.692543\pi$$
$$948$$ 0 0
$$949$$ 45.0000 1.46076
$$950$$ 0 0
$$951$$ 27.0000 0.875535
$$952$$ 0 0
$$953$$ 15.0000 0.485898 0.242949 0.970039i $$-0.421885\pi$$
0.242949 + 0.970039i $$0.421885\pi$$
$$954$$ 0 0
$$955$$ −48.0000 −1.55324
$$956$$ 0 0
$$957$$ −12.0000 −0.387905
$$958$$ 0 0
$$959$$ 0 0
$$960$$ 0 0
$$961$$ −15.0000 −0.483871
$$962$$ 0 0
$$963$$ 24.0000 0.773389
$$964$$ 0 0
$$965$$ 45.0000 1.44860
$$966$$ 0 0
$$967$$ −23.0000 −0.739630 −0.369815 0.929105i $$-0.620579\pi$$
−0.369815 + 0.929105i $$0.620579\pi$$
$$968$$ 0 0
$$969$$ 0 0
$$970$$ 0 0
$$971$$ −3.00000 −0.0962746 −0.0481373 0.998841i $$-0.515328\pi$$
−0.0481373 + 0.998841i $$0.515328\pi$$
$$972$$ 0 0
$$973$$ 0 0
$$974$$ 0 0
$$975$$ −20.0000 −0.640513
$$976$$ 0 0
$$977$$ 30.0000 0.959785 0.479893 0.877327i $$-0.340676\pi$$
0.479893 + 0.877327i $$0.340676\pi$$
$$978$$ 0 0
$$979$$ −12.0000 −0.383522
$$980$$ 0 0
$$981$$ −14.0000 −0.446986
$$982$$ 0 0
$$983$$ 25.0000 0.797376 0.398688 0.917087i $$-0.369466\pi$$
0.398688 + 0.917087i $$0.369466\pi$$
$$984$$ 0 0
$$985$$ 30.0000 0.955879
$$986$$ 0 0
$$987$$ 0 0
$$988$$ 0 0
$$989$$ 11.0000 0.349780
$$990$$ 0 0
$$991$$ −55.0000 −1.74713 −0.873566 0.486705i $$-0.838199\pi$$
−0.873566 + 0.486705i $$0.838199\pi$$
$$992$$ 0 0
$$993$$ 20.0000 0.634681
$$994$$ 0 0
$$995$$ 9.00000 0.285319
$$996$$ 0 0
$$997$$ −25.0000 −0.791758 −0.395879 0.918303i $$-0.629560\pi$$
−0.395879 + 0.918303i $$0.629560\pi$$
$$998$$ 0 0
$$999$$ −10.0000 −0.316386
Display $$a_p$$ with $$p$$ up to: 50 250 1000 Display $$a_n$$ with $$n$$ up to: 50 250 1000
## Twists
By twisting character
Char Parity Ord Type Twist Min Dim
1.1 even 1 trivial 2888.2.a.e.1.1 1
4.3 odd 2 5776.2.a.h.1.1 1
19.7 even 3 152.2.i.a.49.1 2
19.11 even 3 152.2.i.a.121.1 yes 2
19.18 odd 2 2888.2.a.c.1.1 1
57.11 odd 6 1368.2.s.g.577.1 2
57.26 odd 6 1368.2.s.g.505.1 2
76.7 odd 6 304.2.i.b.49.1 2
76.11 odd 6 304.2.i.b.273.1 2
76.75 even 2 5776.2.a.o.1.1 1
152.11 odd 6 1216.2.i.e.577.1 2
152.45 even 6 1216.2.i.i.961.1 2
152.83 odd 6 1216.2.i.e.961.1 2
152.125 even 6 1216.2.i.i.577.1 2
228.11 even 6 2736.2.s.q.577.1 2
228.83 even 6 2736.2.s.q.1873.1 2
By twisted newform
Twist Min Dim Char Parity Ord Type
152.2.i.a.49.1 2 19.7 even 3
152.2.i.a.121.1 yes 2 19.11 even 3
304.2.i.b.49.1 2 76.7 odd 6
304.2.i.b.273.1 2 76.11 odd 6
1216.2.i.e.577.1 2 152.11 odd 6
1216.2.i.e.961.1 2 152.83 odd 6
1216.2.i.i.577.1 2 152.125 even 6
1216.2.i.i.961.1 2 152.45 even 6
1368.2.s.g.505.1 2 57.26 odd 6
1368.2.s.g.577.1 2 57.11 odd 6
2736.2.s.q.577.1 2 228.11 even 6
2736.2.s.q.1873.1 2 228.83 even 6
2888.2.a.c.1.1 1 19.18 odd 2
2888.2.a.e.1.1 1 1.1 even 1 trivial
5776.2.a.h.1.1 1 4.3 odd 2
5776.2.a.o.1.1 1 76.75 even 2 | 18,942 | 33,518 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.546875 | 3 | CC-MAIN-2024-26 | latest | en | 0.391947 |
https://forums.adobe.com/thread/2287417 | 1,537,440,551,000,000,000 | text/html | crawl-data/CC-MAIN-2018-39/segments/1537267156460.64/warc/CC-MAIN-20180920101233-20180920121633-00207.warc.gz | 517,482,407 | 30,583 | 7 Replies Latest reply on Mar 9, 2017 7:38 AM by siegfriedp63497853
# radial transition from haze to solid white
Hello, Community!
I am a novice with PS and I'm trying to prepare the following image:
1. the center of the rectangular picture with a portrait should be abolute perfect sharp (original image)
2. on an imaginary border of an ellipse around this portrait should start a white haze, transitioning to the four edges into a solid white or dense fog
Is someone in the community, who can support me with a detailed procedure?
I apologize for my poor english, sorry!
Best regards, Siegfried
• ###### 1. Re: radial transition from haze to solid white
Add a white Color Fill layer, and in the layer mask, put a radial gradient.
• ###### 2. Re: radial transition from haze to solid white
High, Semaphoric!
Thanks, for your very quick and interesting proposal. But, unfortunately, it doesn't meet my aim not completely.
Getting your image as a template:
I would like to have an inner ellipse without any haze surrounding the complete head, shirt collar and tie. From this imaginary border the haze should start smooth and end with the same intensity (solid/dense fog) on all four edges. In this case, the haze has a non-continuous distribution/density caused by the progress on the different distances from the ellipse to the image borders.
Do you have any idea to realize this?
Best regards, Siegfried
• ###### 3. Re: radial transition from haze to solid white
Not quite sure, but is this more the look you're attempting to achieve, then?
• ###### 4. Re: radial transition from haze to solid white
Hi, S_Gans!
Thank's for your proposal. Unfortunately this is not the result I'm looking for.
Currently I have found three possibilities of frames in PS with more or less haze surrounding the picture:
- the proposal from Semaphoric
- a very similar impression to your proposal, S_Gans, but with an additional smooth radius on the inner edge to soften it
I think, my idea about the design of this type of frame isn't quite easy to perform.
Thank you all, for your effort in supporting me to find a solution for my request!
I think, I have to combine a lot of layers, masks and filters to achieve the desired result.
Best regards, Siegfried
• ###### 5. Re: radial transition from haze to solid white
it may be that the question was a little hard to understand, but as i saw it you you want the gradient to start a little bit mor outside of the middle so that the face is perfectly clear maybe? like this(?):
to be honest i dont know if this is what you want (and it may be a little similar to the others), but the way you do this is as easy as this. you make a new layer and the put a gradient on it. you then make a layer mask and make your brush into an elipse. then you switch the brush colour to black and mark the part you want visible.
• ###### 6. Re: radial transition from haze to solid white
Someway like this one?
Edit your gradient for more or less density and/or the start and end point when applying.
Fenja
• ###### 7. Re: radial transition from haze to solid white
Hello again!
Thank you all, very very much!
I think, the proposal from Hans Martin Kristensen meets my expactations best. I will try out several justifications and I'm sure to get a result very close to my basic idea.
Fenja's proposal I will also try out, just with the definition of the gradient from transparent to white I am a bit unsure - but I am just starting with PS and I'm optimistic to find the correct settings.
Best regards, Siegfried | 817 | 3,571 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.65625 | 3 | CC-MAIN-2018-39 | latest | en | 0.913881 |
http://spotidoc.com/doc/192373/how-to-read-gas-meter-index | 1,539,713,950,000,000,000 | text/html | crawl-data/CC-MAIN-2018-43/segments/1539583510866.52/warc/CC-MAIN-20181016180631-20181016202131-00175.warc.gz | 352,382,048 | 7,022 | # How To Read Gas Meter Index
```How To Read Gas Meter Index
Fig 1 Examples
Gas meter has 4 basic dials.
(1) When reading meter, note location of pointer and
select lowest number between pointer.
If pointer is between 3 and 4, then reading is 3
(2) each dial rotates clockwise or counterclockwise
First dial is clockwise.
Second dial counter-clockwise
Arrow shows above each dial in Fig 1
(3) When dial reaches 0, the count starts over
and dial to left advances by one digit
(4) Each dial is marked with units that it measures
Starting far right:
▪ First dial: 100 cubic feet to 1000 cubic feet
Pointer between 4 and 5 = 400
▪ Second dial: 1000 cubic feet to 10,000 cubic feet
Pointer between 4 and 5 = 4,000
▪ Third dial: 10,000 to 100,000 cubic feet
Pointer between 4 and 5 = 40,000
▪ Fourth dial: 100,000 to 1,000,000 cubic feet
Pointer between 4 and 5 = 400,000
(5) How much gas did you use this month?
Take reading first day of month
Take reading last day of month
Subtract two readings to get total cubic feet used.
Fig 2 shows photo of gas meter
http://waterheatertimer.org/How-to-install-gas-water-heater.html
http://waterheatertimer.org/9-ways-to-save-with-water-heater.html
``` | 344 | 1,188 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.234375 | 3 | CC-MAIN-2018-43 | latest | en | 0.856369 |
https://www.letsplaymaths.com/Class-4-LCM-HCF.html | 1,675,617,282,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764500273.30/warc/CC-MAIN-20230205161658-20230205191658-00455.warc.gz | 895,553,325 | 3,781 | ## LetsPlayMaths.Com
WELCOME TO THE WORLD OF MATHEMATICS
# Class 4 LCM and HCF
Lowest Common Multiples (LCM)
Highest Common Factor (HCF)
LCM / HCF Test
LCM/HCF Worksheet
Answer Sheet
## Lowest Common Multiples (LCM)
The lowest common multiple of two or more numbers is the smallest number that is a multiple of both the numbers. There are different methods by which we can find out the LCM of different numbers.
Listing Method
First list out the multiples of given numbers and then find out the smallest common multiple.
Please have a look at the below given video for better understanding.
Let's have a look at some examples given below.
Example 1. Find out the LCM of 2 and 3?
Solution.
Multiples of 2 = 2, 4, 6, 8, 10, 12, 14, 16, 18, 20
Multiples of 3 = 3, 6, 9, 12, 15, 18, 21, 24, 27, 30
Common multiples of 2 & 3 are 6, 12, 18
So, the lowest multiple of 2 & 3 is 6.
Example 2. Find out the LCM of 3 & 8?
Solution.
Multiples of 3 = 3, 6, 9, 12, 15, 18, 21, 24
Multiples of 8 = 8, 16, 24, 32, 40
Common multiples of 3 and 8 is 24.
So, the lowest multiple of 3 and 8 is 24.
When two numbers have no common factors other than 1, then their Lowest Common Multiple is the product of these two numbers.
Example 1. Find out the LCM of 3 and 7.
Solution. LCM of 3 and 7 = 3 × 7 = 21.
Example 2. Find out the LCM of 5 and 9.
Solution. LCM of 5 and 9 = 5 × 9 = 45
Prime Factor Method
In this method, we take the numbers and divide them with the smallest possible prime factor. If any number gets divided fully then we divide it, else we keep it as it is. In the next step, we divide it by the next possible prime factor. We keep on doing it unless we get all the quotients become 1. In the end, we multiply all the prime factors to get the LCM.
Please have a look at the below given video for better understanding.
Let's have a look at some example given below.
Example 1. Find out the LCM of 4 and 8.
Solution.
Step 1. Divide 4 and 8 by 2 and write the quotient below
Step 2. Divide 2 and 4 by 2 and write the quotient below
Step 3. From 1 and 2, 1 can not be divided by 2, so leave it. Divide 2 by 2, quotient is 1, write both 1s below
Step 4. Now multiply all the prime factors. 2 × 2 × 2 = 8
So, the LCM of 4 and 8 is 8.
Example 2. Find LCM of 4, 6, and 8.
Solution.
LCM = 2 × 2 × 2 × 3 = 24
Example 3. Find the LCM of 10, 15 and 25.
Solution.
LCM = 2 × 5 × 3 × 5 = 150
## Highest Common Factor (HCF)
HCF of two numbers is the highest factor which is common to the given numbers. That means the greatest number which divides the given numbers fully. We have various method to determine HCF of given numbers.
Listing Method
In this method we have to list down all the factors of the given numbers, and then we have to determine the highest common factor out of them.
Please have a look at the below given video for better understanding.
Let's have a look at some examples.
Example 1. Find out the HCF of 8 and 12.
Solution.
Factors of 8 = 1, 2, 4, 8
Factors of 12 = 1, 2, 4, 6, 12
Common factors are 1, 2, and 4, among these common factors 4 is the highest factor.
So, 4 is the HCF of 8 and 12.
Example 2. Find out the HCF of 24 and 36.
Solution.
Factors of 24 = 1, 2, 3, 4, 6, 8, 12, 24
Factors of 36 = 1, 2, 3, 4, 6, 9, 12, 18, 36
Common factors are 1, 2, 3, 4, 6, and 12.
So, 12 is the HCF of 24 and 36.
Prime Factor Method
We need to factorize each number as shown below and multiply all the common prime factors to get the HCF.
Please have a look at the below given video for better understanding.
Let's have a look at some examples.
Example 1. Find out the HCF of 8 and 12.
Solution.
Factorize both the numbers until they have common Factors. So, we have two common prime factors.
2 X 2 = 4
So, the HCF of 8 and 12 is 4.
Example 2. Find out the HCF of 8, 24 and 32?
Solution.
1. Here 2 is the common prime factor of 8, 24 and 32.
2. Here 2 is the common prime factor of 4, 12 and 16.
3. Here 2 is the common prime factor of 2, 6 and 8.
So, we have three common prime factors.
2 X 2 X 2 = 8
HCF of 8, 24 and 32 is 8.
## LCM / HCF Test
LCM & HCF Test - 1
## Class-4 LCM and HCF Worksheet
LCM & HCF Worksheet - 1
LCM & HCF Worksheet - 2
LCM & HCF Worksheet - 3
## Answer Sheet
LCM-HCF-AnswerDownload the pdf
Copyright © 2023 LetsPlayMaths.com. All Rights Reserved.
Email: feedback@letsplaymaths.com | 1,432 | 4,391 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.96875 | 5 | CC-MAIN-2023-06 | latest | en | 0.912785 |
http://stackoverflow.com/questions/7623322/why-do-i-keep-getting-two-of-same-random-values-in-this-code/7623343 | 1,386,304,286,000,000,000 | text/html | crawl-data/CC-MAIN-2013-48/segments/1386163049486/warc/CC-MAIN-20131204131729-00075-ip-10-33-133-15.ec2.internal.warc.gz | 168,924,335 | 12,545 | Why do I keep getting two of same random values in this code? [duplicate]
Possible Duplicate:
Why does it appear that my random number generator isn't random in C#?
I have the following code:
`````` int a;
int aa;
Random aRand = new Random();
Random aaRand = new Random();
a = aRand.Next(20);
aa = aaRand.Next(20);
//if (a == aa)
{
Console.WriteLine(a + " " + aa);
}
``````
I'm assuming that aRand and aaRand would be two different values, but that's not the case. What am I doing wrong? I'm assuming that aRand and aaRand will not always be the same, but they keep coming out the same all the time.
Thanks
-
Can we clarify what language this is, it looks like C#. – Caimen Oct 1 '11 at 22:19
c# correct is the language, sry. – Liger86 Oct 1 '11 at 22:20
marked as duplicate by Michael Petrotta, Adam Maras, svick, Jason, ChrisF♦Oct 1 '11 at 22:25
This is explicitly covered in the docs for Random():
The default seed value is derived from the system clock and has finite resolution. As a result, different Random objects that are created in close succession by a call to the default constructor will have identical default seed values and, therefore, will produce identical sets of random numbers.
-
Why are you creating two different Random variables? You could use just one:
``````int a;
int aa;
Random aRand = new Random();
a = aRand.Next(20);
aa = aRand.Next(20);
//if (a == aa)
{
Console.WriteLine(a + " " + aa);
}
``````
Edit: "The random number generation starts from a seed value. If the same seed is used repeatedly, the same series of numbers is generated. One way to produce different sequences is to make the seed value time-dependent, thereby producing a different series with each new instance of Random. By default, the parameterless constructor of the Random class uses the system clock to generate its seed value, while its parameterized constructor can take an Int32 value based on the number of ticks in the current time. " from http://msdn.microsoft.com/en-us/library/system.random.aspx
-
Yep, I done just one and it worked. Thanks – Liger86 Oct 1 '11 at 22:23
You only need one instance of `Random()` - just call `.Next()` twice.
``````int a;
int aa;
Random aRand = new Random();
a = aRand.Next(20);
aa = aRand.Next(20);
``````
-
You should never have more than One Random variable in your entire application. Get rid of the second
``````Random aaRand = new Random();
``````
- | 613 | 2,434 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.609375 | 3 | CC-MAIN-2013-48 | latest | en | 0.876382 |
https://books.google.gr/books?id=zgwAAAAAYAAJ&vq=%2212+inches+(in.)+%3D+1+foot+(ft.)+3+feet+%3D+1+yard+(yd.)+5J+yards%22&dq=editions:HARVARD32044096994090&lr=&hl=el&output=html_text&source=gbs_navlinks_s | 1,652,764,841,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662515501.4/warc/CC-MAIN-20220517031843-20220517061843-00032.warc.gz | 199,464,075 | 11,094 | # New York State Arithmetic: Years Five, Six, Seven, and Eight
Ginn, 1912 - 337 σελίδες
### Τι λένε οι χρήστες -Σύνταξη κριτικής
Δεν εντοπίσαμε κριτικές στις συνήθεις τοποθεσίες.
### Περιεχόμενα
CHAPTER PAGE I WRITING AND READING NUMBERS 1 ADDITION OF INTEGERS 8 SUBTRACTION OF INTEGERS 17 MULTIPLICATION OF INTEGERS 23 DIVISION OF INTEGERS 35 DECIMAL FRACTIONS 57 DENOMINATE NUMBERS 99 BILLS AND ACCOUNTS 171
PERCENTAGE 231 PROFIT AND Loss 243 THE EQUATION 249 TRADE DISCOUNT 255 COMMISSION 269 INTEREST 277 TABLES OF DENOMINATE NUMBERS 293 INDEX 297
### Δημοφιλή αποσπάσματα
Σελίδα 287 - SQUARE MEASURE 144 square inches (sq. in.) = 1 square foot (sq. ft.) 9 square feet — 1 square yard (sq. yd.) 30^ square yards = 1 square rod (sq. rd.) 160 square rods = 1 acre (A.) 640 acres = 1 square mile (sq.
Σελίδα 288 - CUBIC MEASURE 1728 cubic inches (cu. in.) = 1 cubic foot (cu. ft.) 27 cubic feet = 1 cubic yard (cu. yd.) 128 cubic feet = 1 cord (cd...
Σελίδα 288 - LIQUID MEASURE 4 gills (gi.) = 1 pint (pt.) 2 pints = 1 quart (qt.) 4 quarts = 1 gallon (gal.) 31| gallons = 1 barrel (bbl...
Σελίδα 287 - Square Measure 144 square inches (sq. in.) = 1 square foot (sq. ft.) 9 square feet = 1 square yard (sq. yd.) 30| square yards = 1 square rod (sq. rd.) 160 square rods = 1 acre (A.) 640 acres = 1 square mile (sq. mi.) Cubic Measure 1728 cubic inches (cu.
Σελίδα 133 - Thirty days hath September, April. June, and November; All the rest have thirty.one, Save February, which alone Hath twenty.eight; and one day more We add to it one year in four.
Σελίδα 133 - Time 60 seconds (sec.) = 1 minute (min.) 60 minutes = 1 hour (hr.) 24 hours = 1 day (da.) 7 days = 1 week (wk.) 365 days = 1 common year (yr.) 366 days = 1 leap year...
Σελίδα 47 - Multiplying or dividing both Dividend and Divisor by the same number, does not change the Quotient.
Σελίδα 176 - Divisibility by 3. A number is divisible by 3 if the sum of its digits is so divisible.
Σελίδα 35 - To divide an integer by any power of 10, cut off as many figures from the right of the dividend as there are zeros at the right of the divisor.
Σελίδα 133 - Thirty days hath September, April, June, and November; All the rest have thirty-one, Except the second month alone, Which has but twenty-eight, in fine, Till leap year gives it twenty-nine. | 753 | 2,312 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.8125 | 3 | CC-MAIN-2022-21 | latest | en | 0.467303 |
https://www.statisticshowto.datasciencecentral.com/one-to-one-definition/ | 1,582,794,137,000,000,000 | text/html | crawl-data/CC-MAIN-2020-10/segments/1581875146665.7/warc/CC-MAIN-20200227063824-20200227093824-00430.warc.gz | 849,058,003 | 10,100 | # One to One: Definition
A one to one function is a relation that preserves “distinctness”; Every unique member of the function’s domain is mapped to a unique member of its range. Sometimes this type of mapping is called injective.
Both images below show injective functions. Notice that each element in X is mapped to a distinct element of Y.
The image below is not this type of mapping. Two distinct elements of X, 3 and 4, are mapped to the same element of Y (C).
## The Horizontal Line Test for a One to One Function
• One easy way of determining whether or not a mapping is injective is the horizontal line test.
Graph the function.
• Draw a horizontal line over that graph.
• If any horizontal line intersects the graph of the function more than once, the function is not one to one.
But if there isn’t any straight horizontal line that can be drawn to cross the function more than once, the function is one to one.
Below are images of two different mappings. On the first graph, you could draw a horizontal line that cuts through the hump and crosses the mapping twice. On the second graph, though, there is no way we can draw a horizontal straight line that crosses twice. The function in this second graph is therefore one to one.
## Properties of One to One Functions
Here are some useful properties of injective functions:
• If f and g are both one to one, so is f · g.
• If g · f is one to one, f is as well.
• If f is injective, it has an inverse; a function that undoes what it does.
## References
He, Jiwen. Inverses and More. Lecture 1, Section 7.1 Notes. Retrieved from https://www.math.uh.edu/~jiwenhe/Math1432/lectures/lecture01_handout.pdf
University of Toronto Math. Preparing for Calculus: Functions and Their Inverses. Retrieved from https://www.math.toronto.edu/preparing-for-calculus/4_functions/we_3_one_to_one.html on June 15, 2019.
------------------------------------------------------------------------------
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Statistical concepts explained visually - Includes many concepts such as sample size, hypothesis tests, or logistic regression, explained by Stephanie Glen, founder of StatisticsHowTo. | 519 | 2,332 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.28125 | 4 | CC-MAIN-2020-10 | latest | en | 0.882332 |
http://forums.wolfram.com/mathgroup/archive/1997/May/msg00331.html | 1,618,706,600,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618038464065.57/warc/CC-MAIN-20210417222733-20210418012733-00502.warc.gz | 37,183,168 | 7,479 | How is the performance of Position[] ..
• To: mathgroup at smc.vnet.net
• Subject: [mg7041] How is the performance of Position[] ..
• From: cmchung at se.cuhk.edu.hk (Chung Cheong Ming)
• Date: Sat, 3 May 1997 22:04:37 -0400 (EDT)
• Organization: Engineering Faculty CUHK
• Sender: owner-wri-mathgroup at wolfram.com
```I've got a problem, see anyone can shed some light.
I have an array (a list of sublists) with the following structure,
ARRAY = { { {a11a,a11b}, a12, ...},
{ {a21a,a21b}, a22, ...},
{ {a31a,a31b}, a32, ...},
:
:
{ {am1a,am2b}, am2, ...}}
|<----- n elements ----->|
Thus I have a m by n matrix with the first element in each sublist as
a 2-element list. My question is if I know a particular pair of this
first element eg. {1000,2000}, how will be the computation of Position[]
if I want to use this as a key to extract the row number where
{1000,2000} situates in ARRAY.
If anyone know of any formal literature (maybe Mathematica technical
report), I would be very appreciated.
Thanks
Joseph
```
• Prev by Date: Re: coefficient of constant term (Q:)
• Next by Date: Re: Fit
• Previous by thread: Re: coefficient of constant term (Q:)
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http://mathforum.org/library/drmath/view/61866.html | 1,498,307,695,000,000,000 | text/html | crawl-data/CC-MAIN-2017-26/segments/1498128320261.6/warc/CC-MAIN-20170624115542-20170624135542-00274.warc.gz | 269,591,782 | 4,224 | Associated Topics || Dr. Math Home || Search Dr. Math
### Painting Cubes
```Date: 01/06/2003 at 11:43:04
From: Nia Lynn
Subject: Making Choices
Katie, a student at the world's first orbiting school, noticed that
her unit cubes floated in zero gravity. If she was careful, she could
even build floating geometric shapes with them.
One day Katie used the unit cubes to build a larger floating cube that
was three units long on each edge. She then painted all the exposed
faces of the cube red. Unfortunately, just after she finished, she
sneezed, scattering the unit cubes. No cubes remained attached.
When Katie gathers the cubes, how many cubes will she find that have
exactly 6 faces painted red? 5? 4? 3? 2? 1? 0 faces?
I know that each unit cube does not have the same number of faces
painted red because some are on the corners and they should only have
3 faces painted (the top and the two sides) and there are 6 faces on
the cube.
```
```
Date: 01/06/2003 at 12:35:11
From: Doctor Ian
Subject: Re: Making Choices
Hi Nia,
To find the number of unit cubes, imagine slicing the larger cube into
three layers:
How many cubes are in each layer? How many layers are there? The
total number of cubes has to be the product of those two numbers.
Now, what about the number of painted faces on the unit cubes after
the larger cube is broken up?
Note that for a unit cube to have all 6 of its faces painted, it would
have to have all 6 of those faces exposed while part of the larger
cube. Is there _any_ position in the larger cube where that would be
true?
One way to approach this problem is to figure out how many different
_kinds_ of locations there are. There are 8 corner locations (only 7
are visible in this view), 12 edge locations (only 9 are visible in
this view), and 6 center locations (only 3 are visible in this view).
We can use a table to keep track of this information:
Location unit cubes
-------- ----------
Corner 8
Edge 12
Center 6
The main reason for grouping locations like this is that each location
is equivalent to every other location of the same kind. (If you don't
see why, imagine that while you're looking away, someone picks up the
larger cube and rotates it so that a different corner is facing you.
Would you be able to tell?) And that means that once we figure
something out regarding _any_ unit cube, we've figured it out for
_every_ unit cube in the same kind of location.
For example, look at one of the unit cubes located at a corner of the
larger cube. As you've noticed, it will have three of its faces
painted. But since any corner is just like any other corner, _every_
unit cube located in a corner will have three of its faces painted:
Location unit cubes faces painted
-------- ---------- -------------
Corner 8 3
Edge 12 ?
Center 6 ?
I'll leave the rest of the table for you to fill in.
Finally, note that so far, we've accounted for 26 unit cubes. But
from analyzing the layers, we know we have to have 27 unit cubes. So
where is the final one? And how many of its faces would be painted?
Does this help?
- Doctor Ian, The Math Forum
http://mathforum.org/dr.math/
For a related, interesting problem, see How Many Cube Faces Were Painted?
```
Associated Topics:
High School Euclidean/Plane Geometry
High School Polyhedra
Middle School Polyhedra
Middle School Two-Dimensional Geometry
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Math Forum Home || Math Library || Quick Reference || Math Forum Search | 928 | 3,834 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.21875 | 4 | CC-MAIN-2017-26 | longest | en | 0.953686 |
https://robotics.stackexchange.com/questions/8329/filtering-imu-angle-discontinuities | 1,701,650,007,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100518.73/warc/CC-MAIN-20231203225036-20231204015036-00600.warc.gz | 562,481,565 | 43,033 | # Filtering IMU angle discontinuities
I try to measure Euler angles from an IMU, but some discontinuities happens during measurement, even in vibrationless environment, as shown in the images below.
Can someone explain which type of filter will be the best choice to filter this type discontinuities?
• Check out this question, it may be similar to your problem. Oct 25, 2015 at 10:50
• I couldn't remove spikes with median filter or I dont know how to correctly implement median filter.
– lsn
Oct 25, 2015 at 11:28
• A quick and dirty solution is to choose some maximum change in your reading and reject samples that are greater than that limit. But this could be indicative of a problem with your data acquisition. Oct 25, 2015 at 11:33
• but I cant define tresholds coz it measure angles dynamically between -180º and +180º.
– lsn
Oct 25, 2015 at 11:59
• Sure you can, apply it to the absolute value of the change in your reading. Oct 25, 2015 at 12:00
Here are my two suggestions for dealing with this problem:
Use a median filter, which replaces each value of your signal with the median of the values in a small window around each one. Here is some pseudo-code, where x is your original signal, y is the filtered signal, N is the number of points in your signal, and W is the number of points in the median filter window.
for (k = 1 to N) { y[k] = median of samples x[k-W+1] to x[k] }
If you are using MATLAB then you can use the function medfilt1 to do this, or the function median to make your own filter (see this), whereas if you are using a language like C++ then you may need to write your own functions (see this).
The other option is to simply check the magnitude of the change in the signal and reject any sample whose change is beyond some threshold. Something like this:
if (abs(x[k] - y[k-1]) > threshold) { y[k] = x[k-1] } else { y[k] = x[k] }
EDIT:
Taking a look at your data, it looked suspiciously like an angle wrapping issue, but around 180 deg instead of 360 deg. The spikes disappear if you double the signal then apply an angle wrap (using MATLAB's wrapToPi for example). The plot below shows the doubled signal in blue and the doubled signal after wrapping in red.
Here is the code I used:
sensorData = dlmread('sensor.txt');
t = sensorData(:,1);
x = sensorData(:,2);
x2 = 2*x;
y = wrapToPi(x2*(pi/180))*(180/pi);
figure
hold on;
plot(t,x,'k','linewidth',2);
plot(t,x2,'b','linewidth',2);
plot(t,y,'r','linewidth',2);
• Thnx @Brain but l couldnt achieve yet
– lsn
Oct 25, 2015 at 15:52
• You mean you were able to get those two approaches working but they didn't help? Or you can't get those approaches working at all? Oct 25, 2015 at 22:21
• I think both of them. I also used simulink median filter block with different sample numbers but the result is same.
– lsn
Oct 26, 2015 at 6:24
• When you say "the same", do you mean the result after filtering is exactly the same as your input? Or that it is different but still has spikes? Oct 26, 2015 at 6:45
• Sorrt bro, yes its same as the input. median filter block or matlab function doesn't change the result. I think there is a missing point but l couldnt find it out to correctly implement it. Nonetheless simulink block could do the task u describe but it didnt work also.
– lsn
Oct 26, 2015 at 6:58 | 874 | 3,297 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.25 | 3 | CC-MAIN-2023-50 | latest | en | 0.904262 |
https://www.techiedelight.com/Category/trees/binary-tree/page/3/ | 1,553,369,032,000,000,000 | text/html | crawl-data/CC-MAIN-2019-13/segments/1552912202924.93/warc/CC-MAIN-20190323181713-20190323203713-00492.warc.gz | 917,901,657 | 12,880 | ## Invert Binary Tree | Recursive and Iterative solution
Given a binary tree, write an efficient algorithm to invert binary tree.
## Convert binary tree to Left-child right-sibling binary tree
Given a normal binary tree, convert it to Left-child right-sibling (LC-RS) binary tree.
## Print two-dimensional view of a binary tree
Given a binary tree, write an efficient algorithm to print two-dimensional view of a binary tree.
## Construct a Binary Tree from Ancestor Matrix
Given an ancestor matrix, whose cell (i, j) has value true if i is ancestor of j in a binary tree, construct a binary tree from it where binary tree nodes are labelled from 0 to n-1 where n is the size of the ancestor matrix.
## Calculate height of a binary tree with leaf nodes forming a circular doubly linked list
Write an algorithm to compute the height of a binary tree with leaf nodes forming a circular doubly linked list where the left and right pointers of the leaf node will act as a previous and next pointer of circular doubly linked list respectively. For example, consider below binary tree. The leaf nodes are …
## Huffman Coding Compression Algorithm
Huffman Coding (also known as Huffman Encoding) is a algorithm for doing data compression and it forms the basic idea behind file compression. This post talks about fixed length and variable length encoding, uniquely decodable codes, prefix rules and construction of Huffman Tree.
## Check if given Binary Tree is Height Balanced or not
Given a binary tree, write an efficient algorithm to check if tree is height balanced or not. In a height balanced tree, the absolute difference between height of left subtree and right subtree for every node is 0 or 1.
## Treap Data Structure
A Treap Data Structure is basically a combination of a binary search tree and a heap. Binary Search Trees – Deletions and additions of nodes can make the tree unbalanced (heavier on sides, therefore, the property we value about BSTs, the ability to distribute data by equal divisions, goes out of whack). Therefore we …
## Print Binary Tree Structure with its contents in C++
Given a binary tree, write an efficient algorithm to print binary tree structure in standard output.
## Find the Diagonal Sum of given Binary Tree
Given a binary tree, calculate sum of all nodes for each diagonal having negative slope (\). Assume that the left and right child of a node makes 45 degree angle with the parent.
## Determine if given Binary Tree is a BST or not
Given a Binary Tree, determine if it is a BST or not. This problem has a simple recursive solution. The BST property “every node on the right subtree has to be larger than the current node and every node on the left subtree has to be smaller than the current node” is the key …
## Find maximum sum root to leaf path in a binary tree
Given a binary tree, write an efficient algorithm to find maximum sum root to leaf path i.e. maximum sum path from root node to any leaf node in it. | 620 | 2,984 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.703125 | 4 | CC-MAIN-2019-13 | latest | en | 0.864616 |
http://www.potters.org/subject90999.htm | 1,550,657,268,000,000,000 | text/html | crawl-data/CC-MAIN-2019-09/segments/1550247494694.1/warc/CC-MAIN-20190220085318-20190220111318-00227.warc.gz | 413,000,387 | 4,028 | search current discussion categories events - nceca
## clayart mug exchange at nceca 2006!
### Jennifer Boyer on tue 24 jan 06
So get out your racers pots. It's time to plan for the mug exchange =20
at NCECA.
The 8th Annual Clayart Mug Exchange will take place here in the =20
Clayart Room at
6:45 pm on Friday, March 10.
Here are the particulars:
The Exchange is open to all members of the Clayart Internet =20
Discussion Group.
1. Pick a mug to exchange. Actually, it does not have to be a mug, =20=
but a ceramic object that you
made and think another Clayarter would like. Since all the =20
pieces will be put in bags for the
exchange, it must fit in a paper bag that measures =20
approximately 5=94 X 6=94 X 9=94.
name and contact information
inside the mug. It would be wonderful if you would include =20
mug, such as method of construction, firing temperature, and =20=
glaze on the card.
3. Drop off your mug here at the Clayart room any time on Thursday =20
or Friday. Print your
name neatly in the Mug Exchange Register, which will be in the =20=
Clayart room. We all love
looking at all the mugs during the conference, so bring your =20
piece for this Clayart Members
=93exhibition=94 as early as possible.
4. Come to the Clayart room between 5:30 and 6:30 Friday and =20
Hand it to a Muggette on Duty and they will put it in a =20
paper bag, seal it, and place
it on the table. You will be given a ticket for the Mug =20
Exchange. This time between
bagging your mug and the start of the exchange is a great =20
festive time to visit with
The doors will open at 6:45 for the exchange. You will need =20
enter the room. You will be sent to the table, where you will =20
choose a mug in a bag.
So others may choose, please don=92t hang around the table after =20=
5. After everyone present has picked a mug, try to get together =20
with both the person
picked. Part of
the point, and fun, of the Mug Exchange is to meet new Clayart =20=
friends.
Since there is a very well organized stay-at-home exchange, the one =20
at NCECA is limited to Clayart Members attending NCECA, who are able =20
to meet at 6:30 for the exchange. We find it=92s much more fun when =20
you get to meet your partner face to face. Also, only one item per =20
person may be entered into the exchange.
NOTE!!
So the Muggettes have had an email meeting and it looks like the crew =20=
so far is Rebecca Knight, Jennifer Boyer, Kathleen Gordon, Cheryl =20
Litman, Miki Caldwell and John Jensen.
We are putting out a call for a few new people, to keep a regular =20
rotation of volunteers going.
speak up!
Jennifer
________________________________________________________________________=20=
__________________________________
************************
Jennifer Boyer
Thistle Hill Pottery
Montpelier, VT
http://thistlehillpottery.com
### Victoria E. Hamilton on tue 24 jan 06
Jennifer, et al.
I would love to participate as a new Muggette! Count me in.
Vicki Hamilton
Millennia Antica Pottery
Seattle, WA - 5:09 pm, and still light out!
-----Original Message-----
From: Clayart [mailto:CLAYART@LSV.CERAMICS.ORG] On Behalf Of Jennifer Boyer
Sent: Tuesday, January 24, 2006 16:24
To: CLAYART@LSV.CERAMICS.ORG
Subject: Clayart Mug Exchange at NCECA 2006!
So get out your racers pots. It's time to plan for the mug exchange
at NCECA.
The 8th Annual Clayart Mug Exchange will take place here in the
Clayart Room at
6:45 pm on Friday, March 10.
Here are the particulars:
The Exchange is open to all members of the Clayart Internet
Discussion Group.
1. Pick a mug to exchange. Actually, it does not have to be a mug,
but a ceramic object that you
made and think another Clayarter would like. Since all the
pieces will be put in bags for the
exchange, it must fit in a paper bag that measures
approximately 5" X 6" X 9".
name and contact information
inside the mug. It would be wonderful if you would include
mug, such as method of construction, firing temperature, and
glaze on the card.
3. Drop off your mug here at the Clayart room any time on Thursday
or Friday. Print your
name neatly in the Mug Exchange Register, which will be in the
Clayart room. We all love
looking at all the mugs during the conference, so bring your
piece for this Clayart Members
"exhibition" as early as possible.
4. Come to the Clayart room between 5:30 and 6:30 Friday and
Hand it to a Muggette on Duty and they will put it in a
paper bag, seal it, and place
it on the table. You will be given a ticket for the Mug
Exchange. This time between
bagging your mug and the start of the exchange is a great
festive time to visit with
The doors will open at 6:45 for the exchange. You will need
enter the room. You will be sent to the table, where you will
choose a mug in a bag.
So others may choose, please don't hang around the table after
5. After everyone present has picked a mug, try to get together
with both the person
picked. Part of
the point, and fun, of the Mug Exchange is to meet new Clayart
friends.
Since there is a very well organized stay-at-home exchange, the one
at NCECA is limited to Clayart Members attending NCECA, who are able
to meet at 6:30 for the exchange. We find it's much more fun when
you get to meet your partner face to face. Also, only one item per
person may be entered into the exchange.
NOTE!!
So the Muggettes have had an email meeting and it looks like the crew
so far is Rebecca Knight, Jennifer Boyer, Kathleen Gordon, Cheryl
Litman, Miki Caldwell and John Jensen.
We are putting out a call for a few new people, to keep a regular
rotation of volunteers going.
speak up!
Jennifer
________________________________________________________________________
__________________________________
************************
Jennifer Boyer
Thistle Hill Pottery
Montpelier, VT
http://thistlehillpottery.com
____________________________________________________________________________
__
Send postings to clayart@lsv.ceramics.org
You may look at the archives for the list or change your subscription
settings from http://www.ceramics.org/clayart/
Moderator of the list is Mel Jacobson who may be reached at
### Jennifer Boyer on sat 28 jan 06
So it looks like we have a full list of Muggettes for this year.
There are a few folks who I'll put on a back up list, since 12 seems
like plenty for this year.
Thanks everyone!
Jennifer
On Jan 24, 2006, at 7:23 PM, Jennifer Boyer wrote:
> So get out your racers pots. It's time to plan for the mug exchange
> at NCECA.
>
> The 8th Annual Clayart Mug Exchange will take place here in the
> Clayart Room at
> 6:45 pm on Friday, March 10.
>
### Jennifer Boyer on mon 20 feb 06
In case anyone missed it:
Begin forwarded message:
> From: Jennifer Boyer
> Date: January 24, 2006 7:23:54 PM EST
> To: Ceramic Arts Discussion List Discussion List =20
>
> Subject: Clayart Mug Exchange at NCECA 2006!
>
> So get out your racers pots. It's time to plan for the mug exchange =20=
> at NCECA.
>
> The 8th Annual Clayart Mug Exchange will take place here in the =20
> Clayart Room at
> 6:45 pm on Friday, March 10.
>
> Here are the particulars:
>
> The Exchange is open to all members of the Clayart Internet =20
> Discussion Group.
>
>
> 1. Pick a mug to exchange. Actually, it does not have to be a =20
> mug, but a ceramic object that you
> made and think another Clayarter would like. Since all the =20
> pieces will be put in bags for the
> exchange, it must fit in a paper bag that measures =20
> approximately 5=94 X 6=94 X 9=94.
>
> name and contact information
> inside the mug. It would be wonderful if you would include =20
> mug, such as method of construction, firing temperature, =20
> and glaze on the card.
>
> 3. Drop off your mug here at the Clayart room any time on =20
> Thursday or Friday. Print your
> name neatly in the Mug Exchange Register, which will be in =20
> the Clayart room. We all love
> looking at all the mugs during the conference, so bring your =20=
> piece for this Clayart Members
> =93exhibition=94 as early as possible.
>
> 4. Come to the Clayart room between 5:30 and 6:30 Friday and =20
> Hand it to a Muggette on Duty and they will put it in a =20
> paper bag, seal it, and place
> it on the table. You will be given a ticket for the Mug =20
> Exchange. This time between
> bagging your mug and the start of the exchange is a great =20
> festive time to visit with
>
> The doors will open at 6:45 for the exchange. You will need =20
> enter the room. You will be sent to the table, where you will =20=
> choose a mug in a bag.
> So others may choose, please don=92t hang around the table =20
> after you have your bag.
>
> 5. After everyone present has picked a mug, try to get together =20
> with both the person
> you picked. Part of
> the point, and fun, of the Mug Exchange is to meet new =20
> Clayart friends.
>
> Since there is a very well organized stay-at-home exchange, the one =20=
> at NCECA is limited to Clayart Members attending NCECA, who are =20
> able to meet at 6:30 for the exchange. We find it=92s much more fun =20=
> when you get to meet your partner face to face. Also, only one =20
> item per person may be entered into the exchange.
>
>
>
> If anyone has any comments about the procedure of the exchange =20
> Jennifer
> ______________________________________________________________________=20=
> ____________________________________
>
>
>
>
> ************************
> Jennifer Boyer
> Thistle Hill Pottery
> Montpelier, VT
>
> http://thistlehillpottery.com
>
*****************************
Jennifer Boyer
Thistle Hill Pottery
Montpelier, VT
http://thistlehillpottery.com
*****************************
### Jennifer Boyer on thu 2 mar 06
ONE LAST TIME:
Begin forwarded message:
> From: Jennifer Boyer
> Date: January 24, 2006 7:23:54 PM EST
> To: Ceramic Arts Discussion List Discussion List =20
>
> Subject: Clayart Mug Exchange at NCECA 2006!
>
> So get out your racers pots. It's time to plan for the mug exchange =20=
> at NCECA.
>
> The 8th Annual Clayart Mug Exchange will take place here in the =20
> Clayart Room at
> 6:45 pm on Friday, March 10.
>
> Here are the particulars:
>
> The Exchange is open to all members of the Clayart Internet =20
> Discussion Group.
>
>
> 1. Pick a mug to exchange. Actually, it does not have to be a =20
> mug, but a ceramic object that you
> made and think another Clayarter would like. Since all the =20
> pieces will be put in bags for the
> exchange, it must fit in a paper bag that measures =20
> approximately 5=94 X 6=94 X 9=94.
>
> name and contact information
> inside the mug. It would be wonderful if you would include =20
> mug, such as method of construction, firing temperature, =20
> and glaze on the card.
>
> 3. Drop off your mug here at the Clayart room any time on =20
> Thursday or Friday. Print your
> name neatly in the Mug Exchange Register, which will be in =20
> the Clayart room. We all love
> looking at all the mugs during the conference, so bring your =20=
> piece for this Clayart Members
> =93exhibition=94 as early as possible.
>
> 4. Come to the Clayart room between 5:30 and 6:30 Friday and =20
> Hand it to a Muggette on Duty and they will put it in a =20
> paper bag, seal it, and place
> it on the table. You will be given a ticket for the Mug =20
> Exchange. This time between
> bagging your mug and the start of the exchange is a great =20
> festive time to visit with
>
> The doors will open at 6:45 for the exchange. You will need =20
> enter the room. You will be sent to the table, where you will =20=
> choose a mug in a bag.
> So others may choose, please don=92t hang around the table =20
> after you have your bag.
>
> 5. After everyone present has picked a mug, try to get together =20
> with both the person
> you picked. Part of
> the point, and fun, of the Mug Exchange is to meet new =20
> Clayart friends.
>
> Since there is a very well organized stay-at-home exchange, the one =20=
> at NCECA is limited to Clayart Members attending NCECA, who are =20
> able to meet at 6:30 for the exchange. We find it=92s much more fun =20=
> when you get to meet your partner face to face. Also, only one =20
> item per person may be entered into the exchange.
>
> NOTE!!
> So the Muggettes have had an email meeting and it looks like the =20
> crew so far is Rebecca Knight, Jennifer Boyer, Kathleen Gordon, =20
> Cheryl Litman, Miki Caldwell and John Jensen.
> We are putting out a call for a few new people, to keep a regular =20=
> rotation of volunteers going.
>
> If anyone has any comments about the procedure of the exchange =20
> Jennifer
> ______________________________________________________________________=20=
> ____________________________________
>
>
>
>
> ************************
> Jennifer Boyer
> Thistle Hill Pottery
> Montpelier, VT
>
> http://thistlehillpottery.com
>
*****************************
Jennifer Boyer
Thistle Hill Pottery
Montpelier, VT
http://thistlehillpottery.com
***************************** | 3,448 | 12,969 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.671875 | 3 | CC-MAIN-2019-09 | latest | en | 0.953647 |
https://stackoverflow.com/questions/46067817/comparing-numpy-and-matlab-array-summation-speed | 1,723,555,767,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722641076695.81/warc/CC-MAIN-20240813110333-20240813140333-00644.warc.gz | 418,577,159 | 45,941 | # Comparing Numpy and Matlab array summation speed
I recently converted a MATLAB script to Python with Numpy, and found that it ran significantly slower. I expected similar performance, so I'm wondering if I'm doing something wrong.
As stripped-down example, I manually sum a geometric series:
MATLAB version:
``````function s = array_sum(a, array_size, iterations)
s = zeros(array_size);
for m = 1:iterations
s = a + 0.5*s;
end
end
% benchmark code
array_size = 500
iterations = 500
a = randn(array_size)
f = @() array_sum(a, array_size, iterations);
fprintf('run time: %.2f ms\n', timeit(f)*1e3);
``````
Python/Numpy version:
``````import numpy as np
import timeit
def array_sum(a, array_size, iterations):
s = np.zeros((array_size, array_size))
for m in range(iterations):
s = a + 0.5*s
return s
array_size = 500
iterations = 500
a = np.random.randn(array_size, array_size)
timeit_iterations = 10
t1 = timeit.timeit(lambda: array_sum(a, array_size, iterations),
number=timeit_iterations)
print("run time: {:.2f} ms".format(1e3*t1/timeit_iterations))
``````
On my machine, MATLAB completes in 58 ms. The Python version runs in 292 ms, or 5X slower.
I also tried speeding up the Python code by adding the Numba JIT decorator `@jit('f8[:,:](i8, i8)', nopython=True)`, but the time only dropped to 236 ms (4X slower).
This is slower than I expected. Am I using timeit improperly? Is there something wrong with my Python code?
EDIT: edited so that the random matrix is created outside of benchmarked function.
EDIT 2: I ran the benchmark using Torch instead of Numpy (calculating the sum as `s = torch.add(s, 0.5, a)`) and it runs in just 52 ms on my computer!
• You have `nopython=True`, but aren't you using NumPy funcs there? Commented Sep 6, 2017 at 6:00
• @Divakar I think recent numba versions support some of the array allocation functions. @ LorenzForvang your test only performs element-wise operations, which are not implemented in BLAS as far I know (that is not to say they should be slower in numpy).
– MB-F
Commented Sep 6, 2017 at 6:01
• @Divakar yes, I'm using NumPy functions there. This page (numba.pydata.org/numba-doc/dev/reference/numpysupported.html) lists many NumPy functions are supported by Numba in nopython mode. But the run time is the same whether I set `nopython` to True or False. Commented Sep 6, 2017 at 6:06
• @kazemakase good point! I removed the reference to BLAS Commented Sep 6, 2017 at 6:16
• Btw, fun fact: I changed the iteration to perform a matrix dot product `s = r * s` in Matlab and `s = r @ s` in Python. Matlab was still faster but only by a factor of 1.5.
– MB-F
Commented Sep 6, 2017 at 6:20
From my experience, when using numba's jit function it's usually faster to expand array operations into loops. So I tried to rewrite your python function as:
``````@jit(nopython=True, cache=True)
def array_sum_numba(a, array_size, iterations):
s = np.zeros((array_size, array_size))
for m in range(iterations):
for i in range(array_size):
for j in range(array_size):
s[i,j] = a[i,j] + 0.5 * s[i,j]
return s
``````
And out of curiosity, I've also tested @percusse's version with a little modification on the parameter:
``````def array_sum2(r, array_size, iterations):
s = np.zeros((array_size, array_size))
for m in range(iterations):
s /= 2
s += r
return s
``````
The testing results on my machine are:
• original version run time: 143.83 ms
• numba jitted loop version run time: 26.99 ms
• @percusse's version run time: 61.38 ms
This result is within my expectation. It's worthing mentioning that I've increased timeit iterations to 50, which results in some significant time reduction for numba version.
In summary: The Python code can still be significantly accelerated if you use numba's jit and write the function in loops. I don't have Matlab on my machine to test, but my guess is with numba the python version is faster.
Since you are updating the same variable suitable for inplace operations, you can update your function as
``````def array_sum2(array_size, iterations):
s = np.zeros((array_size, array_size))
r = np.random.randn(array_size, array_size)
for m in range(iterations):
s /= 2
s += r
return s
``````
This has given the following speed benefit on my machine compared to `array_sum`
``````run time: 157.32 ms
run time2: 672.43 ms
``````
• thank you, this sped up my code, though not quite as dramatically. Run time on my machine dropped from 292 ms to 259 ms with Numpy (4X MATLAB), and from 236 ms to 185 ms with Numpy+Numba (3X MATLAB). Commented Sep 6, 2017 at 19:56
• @LorenzForvang That sounds strange indeed. Are you using NumPy+MKL ? I usually go for Cython actually. Commented Sep 6, 2017 at 20:06
• yes, numpy.show_config() shows libraries = ['mkl_intel_lp64', 'mkl_intel_thread', 'mkl_core', 'iomp5', 'pthread'] for blas_mkl_info and lapack_mkl_info Commented Sep 8, 2017 at 1:04
Times include the `randn` call as well as the summation:
``````In [68]: timeit array_sum(array_size, 0)
16.6 ms ± 436 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)
In [69]: timeit array_sum(array_size, 1)
18.9 ms ± 293 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)
In [70]: timeit array_sum(array_size, 20)
55.5 ms ± 131 µs per loop (mean ± std. dev. of 7 runs, 10 loops each)
In [71]: (55-16)/20
Out[71]: 1.95
``````
So it's 16ms for the setup, and 2ms per iteration. Same pattern with 500 iterations.
MATLAB does some JIT compilation. I don't know if that's the case here or not. I don't have MATLAB to test. In Octave (no `timeit`)
``````>> t = time(); array_sum(500,0); (time()-t)*1000
ans = 13.704
>> t = time(); array_sum(500,1); (time()-t)*1000
ans = 16.219
>> t = time(); array_sum(500,20); (time()-t)*1000
ans = 82.346
>> t = time(); array_sum(500,500); (time()-t)*1000
ans = 1610.6
``````
Octave's `random` is faster, but the per iteration sum is slower.
• This means 1000ms for 500 iterations and more or less negligible 16ms for the setup. (In Matlab the picture is similar: 5 ms setup time). So most of the difference comes from the iteration which is the question, isn't it?
– MB-F
Commented Sep 6, 2017 at 6:45
• @kazemakase, Is the MATLAB per iteration time consistent regardless whether you run 1, 20 or 500 iterations in a call? Commented Sep 6, 2017 at 6:53
• @hpaulj with 500 iterations, the 16 ms setup becomes negligible. In any case, I made the random matrix a function parameter so that doesn't impact the test. I got 58 ms in MATLAB, 292 ms in Python, 229 ms in Python + Numba. I do not have Octave set up, but I'll trust that it's slower than MATLAB and Python. Commented Sep 6, 2017 at 6:59
• @hpaulj I get the following timings in Matlab: 0, 1, 20, 500 iterations -> 4.77, 5.15, 10.13, 133.6 ms. (I guess we see jit effects for larger numbers of iterations)
– MB-F
Commented Sep 6, 2017 at 7:12 | 2,046 | 6,833 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.734375 | 3 | CC-MAIN-2024-33 | latest | en | 0.822784 |
https://www.bankrate.com/investing/market-capitalization/ | 1,721,681,017,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763517915.15/warc/CC-MAIN-20240722190551-20240722220551-00481.warc.gz | 588,861,697 | 72,278 | Market capitalization is a term used to describe the size of a company based on the total value of the company’s stock. Market capitalization is an important data point for making informed investment decisions, managing return expectations and building a well-balanced portfolio.
Knowing the total value of stocks can help investors distinguish between risky and conservative investments, or help them to diversify based on their particular goals. For example, large companies might be more stable with less room for growth in their returns, but might be the right choice for a portfolio with a short time horizon or an investor with a low risk tolerance. Market capitalization, or market cap, provides part of the information to make these decisions.
## How market capitalization is calculated
A company’s market cap can be found by multiplying the current stock price by the total number of outstanding shares. Outstanding shares are shares that have been issued and sold to shareholders, including those held by insiders and institutional investors. The calculation does not include treasury shares, which are shares of the company that it has repurchased.
For example, if Company A had 20 million shares outstanding and a share price of \$500, its market cap is as follows:
\$500 x 20,000,000 = \$10,000,000,000 market capitalization
Again, that’s the price of one share multiplied by the total number of outstanding shares.
## How a company’s market cap is classified
So, assuming that Company A has a \$10 billion market cap – what does that actually mean? Companies are typically divided into three major categories based on their size:
• Large-cap: Companies with a market capitalization of \$15 billion or more
• Mid-cap: Companies with a market capitalization between \$3 billion and \$15 billion
• Small-cap: Companies with a market capitalization between \$300 million and \$3 billion
In the example above, Company A with a market cap of \$10 billion could be considered a mid-cap.
Sometimes investors classify stocks that are much larger than large-cap as mega-caps, while those smaller than small-cap are sometimes called micro-caps or even nano-caps.
## What market capitalization means
The market cap of a company often says something about the quality of the business underlying the stock as well as how the stock tends to trade. Below are some of the biggest differences between small-cap and large-caps.
Large-caps are typically known for being stable companies with robust balance sheets. These companies typically show less volatility during market downturns than their mid-and small-cap counterparts.
Companies with large market capitalizations are some of the biggest companies in business. Companies such as Apple, Microsoft, Alphabet, Amazon and Berkshire Hathaway occupy the large-cap market. Large-cap companies often have reputations for producing quality goods, showing steady growth and are often dominant players within established industries.
“In economic downturns, large-cap companies have historically outperformed relative to their small- and mid-cap counterparts, primarily because they represent more established companies with stronger balance sheets,” says Mark Andraos, partner at Regency Wealth Management in New York.
Smaller companies on the other hand can be a mixed bag. While smaller companies may have more room for growth than large-cap companies, their less established position within their industry and generally weaker balance sheets mean investing in these companies comes with more risk but also more potential return if they succeed.
That said, there is a place for mid- and small-caps in certain portfolios. “In any given industry, there might be a handful of major players. It’s the smaller players where you can sometimes find value,” says Holmes Osborne, principal at Osborne Global Investors in Missouri.
Market capitalization is a fundamental piece of information needed to make investment decisions, and gives a big-picture view of the value of a company. However, market cap can fluctuate greatly day-to-day, especially in smaller companies, as the stock bounces around.
Looking at a company based solely on its market capitalization will not provide information on how indebted the company is and the potential risks that come along with that.
## How market capitalization influences investment strategy
Market capitalization can impact how you construct an investment portfolio. Experts generally recommend diversification, meaning owning a combination of small-, mid- and large-cap companies.
“To be fully diversified, one generally should be diversified across market capitalization and across value/growth,” says Robert Johnson, CEO and chair at index development and licensing firm Economic Index Associates in New York.
For example, if your goal is large returns, you can focus on small-caps but also invest in some large-cap companies to reduce volatility. So you could combine the best small-cap ETFs with a selection of the best large-cap ETFs.
“It is much more likely that one can hit a home run by investing in a small-cap stock, but it is also more likely that a strikeout will occur,” says Johnson.
If your goal leans more toward stability, you can focus on large-caps, but you can also include smaller companies with growth potential to provide some extra juice to the portfolio.
Or you could even split the difference by buying mid-cap stocks or the best mid-cap ETFs, enjoying higher growth than large-cap stocks along with higher stability than small-caps.
But be careful if you’re adding individual stocks to a portfolio of index funds since you might be adding in extra exposure to companies that you already own.
“Most index funds today are weighted based on market capitalization. Almost 20 percent of the S&P 500 can be found in technology companies like Apple, Microsoft, Google, Amazon and Tesla, and combining a typical market-weighted approach to investing could over concentrate a portfolio in tech,” says Doug Amis, CEO at Cardinal Retirement Planning in the Chapel Hill, North Carolina area.
Then using your time horizon and risk tolerance as benchmarks, you can build out a diversified selection of investments.
## Bottom line
Understanding market capitalization is important when it comes to selecting your investments because it can help evaluate an investment’s total opportunity. However, it does not provide a well-rounded representation of a company’s prospective returns, and so investors need to carefully evaluate companies when building their portfolios. | 1,245 | 6,591 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.84375 | 3 | CC-MAIN-2024-30 | latest | en | 0.956402 |
https://electnorred.com/law/what-are-the-variables-of-boyles-law.html | 1,675,413,795,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764500044.16/warc/CC-MAIN-20230203055519-20230203085519-00666.warc.gz | 244,232,389 | 15,275 | ### What Are The Variables Of Boyle’S Law?
The variables involved in Boyle’s law are pressure, volume, number of moles and temperature. It simply explains the inverse relationship between pressure and volume.
## What are the 2 variable in Boyle’s law?
Boyle’ Law
Boyle’s Law: Verification Boyle’s Equation Robert Boyle
If the temperature of a gas is constant, as the pressure on the gas increases the volume will decrease. The inverse is also true. If the pressure on the gas decreases then the volume will increase. It was Robert Boyle, in 1662, who was first to fully investigated the pressure-volume relationship of gases.
Boyle used a tube containing a gas and mercury. When he changed the amount of mercury in the tube he changed the pressure, which in turn changed the amount of space occupied by the gas. When the amount of mercury was doubled, the volume decreased by half. Robert Boyle’s observations are summed up in Boyle’s law, which states that for a given mass of gas at constant temperature, the volume of a gas varies inversely with pressure.
Because of the inverse relationship, the product of the two quantities, pressure and volume, is constant. When given any two sets of pressure and volume, at a given temperature, the product will be the constant. When you look at the first set of data plotted on the graph you notice the curve of the line indicating the inverse relationship between pressure and volume. As the pressure was decreasing the volume was increasing. When you take any point on the curve and multiply the pressure value by the volume value the product equals the constant.
This graph is consistent with Boyle’s law. The second graph is showing the relationship between 1/pressure and volume. When the data is plotted in this format the slope of the line is linear. This is also consistent with Boyle’s law. If there is an inverse relationship between two variables, plotting the inverse of one variable will generate a straight line.
Plotting Boyle’s data onto a graph allows us to better see the relationship between pressure and volume and what ultimately led to Boyle’s law. : Boyle’ Law
### What variables must be controlled in Boyle’s law?
What are the variables kept constant in Boyle’s law? Answer Verified Hint: In the above mentioned question, we will discuss Boyle’s gas law as well the mathematical form regarding the law and also about the variables that kept constant in gas law. And also we will discuss some more on it.
Complete step by step answer: Note:
Boyle’s law is a gas law which expresses that the pressing factor applied by a gas (of a given mass, kept at a consistent temperature) is contrarily relative to the volume involved by it. As such, the pressing factor and volume of a gas are contrarily corresponding to one another as long as the temperature and the amount of gas are kept steady.
V \propto \dfrac \$ \$ V = K\dfrac \$ \$ V = \dfrac \$ \$ or,PV = K \$ \$ \$ \$ \therefore = \$ So, the constant variable in Boyle’s law will only be temperature. Albeit the temperature is the lone steady factor; volume, temperature, and pressing factor are contrarily relative to one another. As we probably are aware Boyle’s law expresses that the volume and pressing factor of a gas has a reverse relationship while the temperature is steady.Boyle’s law says that absolute pressure exerted by a specific mass of an ideal gas is inversely proportional to the volume it occupies.
But when we blow up a balloon we do not have one specific mass of gas, because with every blow into the balloon we are increasing the mass of gas inside the balloon. : What are the variables kept constant in Boyle’s law?
### What are the four variables present in Boyle’s law?
Ideal Gases – Ideal gas, or perfect gas, is the theoretical substance that helps establish the relationship of four gas variables, p ressure (P), volume(V), the amount of gas(n) and temperature(T), It has characters described as follow:
1. The particles in the gas are extremely small, so the gas does not occupy any spaces.
2. The ideal gas has constant, random and straight-line motion.
3. No forces between the particles of the gas. Particles only collide elastically with each other and with the walls of container.
#### What are the 2 types of variables?
Text begins A variable is a characteristic that can be measured and that can assume different values. Height, age, income, province or country of birth, grades obtained at school and type of housing are all examples of variables. Variables may be classified into two main categories: categorical and numeric.
## What are 2 variable equations called?
Definition – An equation is said to be linear equation in two variables if it is written in the form of ax + by + c=0, where a, b & c are real numbers and the coefficients of x and y, i.e a and b respectively, are not equal to zero.
• For example, 10x+4y = 3 and -x+5y = 2 are linear equations in two variables.
• The solution for such an equation is a pair of values, one for x and one for y which further makes the two sides of an equation equal.
### What are the independent and dependent variables in Boyle’s Law?
Boyle’s Law. Gases are often characterized by their volume, temperature, and pressure. These characteristics, however, are not independent of each other. Gas pressure is dependent on the force exerted by the molecular collisions and the area over which the force is exerted.
### What are the variables of Boyle’s Law volume and number of molecules of gas?
It states that under a constant temperature when the pressure on a gas increases its volume decreases. In other words according to Boyle’s law volume is inversely proportional to pressure when the temperature and the number of molecules are constant.
## What are the variables in Charles Law?
As per Charles’s law, the volume of a gas is directly proportional to its temperature (in the Kelvin scale) provided the amount of the gas and pressure remain constant. Hence, variables remain constant in Charles’s law: (1) amount of gas and (2) pressure.
#### What are the 3 main variables?
There are three main variables: independent variable, dependent variable and controlled variables.
## What are the 4 variable types?
Qualitative vs. Quantitative Variables – Look at ( Figure 1.1 ) again. On the left hand side you see that there are two larger classifications for the kinds of variables you have been studying. There are qualitative variables and there are quantitative variables,
You can see that the four levels of measure (nominal, ordinal, interval and ratio) fall into these two larger supercategories. So, interval and ratio variables are two kinds of quantitative variables and nominal and ordinal variables are two kinds of qualitative variables. Now one kind of variable isn’t necessarily better than another.
You are a little more used to working with quantitative variables. For example, you can do averages and things like that with quantitative variables, you know there are numbers, you can add them up and divide and things like that. With qualitative variables it’s not so clear cut.
## What is dependent and dependent variable?
Frequently asked questions about independent and dependent variables – What’s the definition of an independent variable? An independent variable is the variable you manipulate, control, or vary in an experimental study to explore its effects. It’s called “independent” because it’s not influenced by any other variables in the study. Independent variables are also called:
Explanatory variables (they explain an event or outcome) Predictor variables (they can be used to predict the value of a dependent variable) Right-hand-side variables (they appear on the right-hand side of a regression equation).
What’s the definition of a dependent variable? A dependent variable is what changes as a result of the independent variable manipulation in experiments, It’s what you’re interested in measuring, and it “depends” on your independent variable. In statistics, dependent variables are also called:
Response variables (they respond to a change in another variable) Outcome variables (they represent the outcome you want to measure) Left-hand-side variables (they appear on the left-hand side of a regression equation)
Can I include more than one independent or dependent variable in a study? Yes, but including more than one of either type requires multiple research questions, For example, if you are interested in the effect of a diet on health, you can use multiple measures of health: blood sugar, blood pressure, weight, pulse, and many more.
#### What variable has only 2 values?
Classifying Types of Variables – LO 4.1: Determine the type (categorical or quantitative) of a given variable. LO 4.2: Classify a given variable as nominal, ordinal, discrete, or continuous. Variables can be broadly classified into one of two types :
Quantitative
Categorical
Below we define these two main types of variables and provide further sub-classifications for each type. Categorical variables take category or label values, and place an individual into one of several groups, Categorical variables are often further classified as either:
Nominal, when there is no natural ordering among the categories,
Common examples would be gender, eye color, or ethnicity.
Ordinal, when there is a natural order among the categories, such as, ranking scales or letter grades.
However, ordinal variables are still categorical and do not provide precise measurements. Differences are not precisely meaningful, for example, if one student scores an A and another a B on an assignment, we cannot say precisely the difference in their scores, only that an A is larger than a B.
Discrete, when the variable takes on a countable number of values.
Most often these variables indeed represent some kind of count such as the number of prescriptions an individual takes daily.
Continuous, when the variable can take on any value in some range of values,
Our precision in measuring these variables is often limited by our instruments. Units should be provided. Common examples would be height (inches), weight (pounds), or time to recovery (days). One special variable type occurs when a variable has only two possible values.
• A variable is said to be Binary or Dichotomous, when there are only two possible levels.
• These variables can usually be phrased in a “yes/no” question.
• Whether nor not someone is a smoker is an example of a binary variable.
• Currently we are primarily concerned with classifying variables as either categorical or quantitative.
Sometimes, however, we will need to consider further and sub-classify these variables as defined above. These concepts will be discussed and reviewed as needed but here is a quick practice on sub-classifying categorical and quantitative variables.
## What is a 3 variable equation?
Linear equations in three variables. If a, b, c and r are real numbers (and if a, b, and c are not all equal to 0) then ax + by + cz = r is called a linear equation in three variables. (The ‘three variables’ are the x, the y, and the z.) The numbers a, b, and c are called the coefficients of the equation.
#### How many variables are there in equation?
Polynomial equations – The solutions –1 and 2 of the polynomial equation x 2 – x + 2 = 0 are the points where the graph of the quadratic function y = x 2 – x + 2 cuts the x -axis. In general, an algebraic equation or polynomial equation is an equation of the form, or where P and Q are polynomials with coefficients in some field (e.g., rational numbers, real numbers, complex numbers ). An algebraic equation is univariate if it involves only one variable, On the other hand, a polynomial equation may involve several variables, in which case it is called multivariate (multiple variables, x, y, z, etc.). For example, is a univariate algebraic (polynomial) equation with integer coefficients and is a multivariate polynomial equation over the rational numbers. Some polynomial equations with rational coefficients have a solution that is an algebraic expression, with a finite number of operations involving just those coefficients (i.e., can be solved algebraically ).
### What is the dependent variable 2 points?
Frequently asked questions about independent and dependent variables – What’s the definition of an independent variable? An independent variable is the variable you manipulate, control, or vary in an experimental study to explore its effects. It’s called “independent” because it’s not influenced by any other variables in the study. Independent variables are also called:
Explanatory variables (they explain an event or outcome) Predictor variables (they can be used to predict the value of a dependent variable) Right-hand-side variables (they appear on the right-hand side of a regression equation).
What’s the definition of a dependent variable? A dependent variable is what changes as a result of the independent variable manipulation in experiments, It’s what you’re interested in measuring, and it “depends” on your independent variable. In statistics, dependent variables are also called:
Response variables (they respond to a change in another variable) Outcome variables (they represent the outcome you want to measure) Left-hand-side variables (they appear on the left-hand side of a regression equation)
Can I include more than one independent or dependent variable in a study? Yes, but including more than one of either type requires multiple research questions, For example, if you are interested in the effect of a diet on health, you can use multiple measures of health: blood sugar, blood pressure, weight, pulse, and many more.
### What 2 variables does Charles law deal with?
Charles’s law, a statement that the volume occupied by a fixed amount of gas is directly proportional to its absolute temperature, if the pressure remains constant.
#### What 2 variables are used in a hot air balloon?
HOW DOES A HOT AIR BALLOON WORK?
To fly a hot air balloon takes a lot of teamwork, air, and heat, to get off the ground. A team of individuals are needed to lay the enormous balloon made of reinforced nylon, a light but strong material, flat on the ground. The balloon is also referred to as an “envelope.” They then use a small gas fan to fill up this envelope. The balloon may look full with this air, but it is still lying on the ground. In order to make it stand tall, and eventually fly, a propane burner is ignited to heat up the air already inside the balloon.
table>
You may have noticed that hot air rises already if you have been in a really warm place, like a sauna, then went to a really cold place, like the snow. If you did this, you would notice steam coming off your body and going up into the air. Another way you may have noticed warm air rises is on a hot day if you live, or have been in a house with two levels. In a house like this, without air conditioning, you may notice as you walk from the bottom level to the top, it feels hotter, and if you walk from the top floor to the bottom floor, then it gets cooler. The reason why the hot air balloon rises is because hot air rises. The inside air that has been heated rises due to the difference of density between the hot and cold air particles. The heated air, the air that is inside the balloon, is less dense than the cool air, the air outside of the balloon. And objects that are less dense rise. Volume = How much space something takes up Mass = How much there is of something Density = Mass / Volume
The reason we know that the hot air is less dense than cool air is due to the Ideal Gas Law. The Ideal Gas Law states that the Pressure times the Volume is equal to the number of molecules times the gas constant ( R ) times the Temperature. Sometimes if you think of it in this equation it is a little easier to understand.
• Since we know that the Volume is getting bigger, we can use this knowledge and see how it will affect the density, and since we know the equation for density is
• Density = Mass / Volume
• You can see that the bigger the volume gets, the smaller the density is.
For example. If you heat the air, your temperature increases, then your volume will increase. Let’s say the volume will increase from 600 to 1000. The mass is 10. Therefor when we plug these numbers into the equation to find density we get
1. Original air density = 10 / 600 which = 1/6
2. Heated air’s density = 10 / 1000 which = 1/ 10
3. And we know that one-tenth is less than one-sixth 1/10 < 1/6 Therefor the air that is heated is less dense, and an object that is less dense will rise
In addition to all this fun information about density, you also need to consider the force of the air pushing on the hot air balloon that gives it a lift. This force is called the Boyant Force. You can also think of the Boyant Force as the pressure water has to hold a boat up, like the water helps the boat float, the air helps the hot air balloon float. This is why when the balloon will fly. Are you interested in concerning hot air balloons? For more about hot air balloons. See a of a balloon This page was created November 1999 by Amanda Renchin I am an elementary education major, mathematics minor at If you have an questions or comments send an e-mail my way at : HOW DOES A HOT AIR BALLOON WORK? | 3,625 | 17,381 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.3125 | 4 | CC-MAIN-2023-06 | latest | en | 0.922639 |
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# If n is a positive integer, what is the remainder when 2n is divided
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Re: If n is a positive integer, what is the remainder when 2n is divided [#permalink]
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Bunuel wrote:
If n is a positive integer, what is the remainder when 2n is divided by 8?
(1) n, when divided by 6, leaves remainder 5.
(2) 3n, when divided by 6, leaves remainder 3.
We need to determine the remainder when 2n is divided by 8.
Statement One Alone:
n, when divided by 6, leaves remainder 5.
Thus, we see n can be a number such as 5, 11, 16, 21, 26, 31, 35, etc.
When 2(5) = 10 is divided by 8, the remainder is 2.
When 2(11) = 22 is divided by 8, the remainder is 6.
Statement one alone is not sufficient to answer the question.
Statement Two Alone:
3n, when divided by 6, leaves remainder 3.
Thus, we see that 3n can be a number such as 3, 9, 15, 21, 28, etc.
When 3n is 3, n is 1; when 3n is 9, n is 3; when 3n is 15, n is 5; etc.
In other words n will always be an odd number: 1, 3, 5, 7, ...
When 2(1) = 2 is divided by 8, the remainder is 2.
When 2(3) = 6 is divided by 8, the remainder is 6.
Statement two alone is not sufficient to answer the question.
Statements One and Two Together:
Using our two statements, we see the first value for n that satisfies both statements is 5. We also see that in statement two, n can be any odd number. So, another number that would match is n = 11.
When 2(5) = 10 is divided by 8, the remainder is 2.
When 2(11) = 22 is divided by 8, the remainder is 6.
We see that the statements together are still not sufficient to answer the question.
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Re: If n is a positive integer, what is the remainder when 2n is divided [#permalink]
Bunuel wrote:
If n is a positive integer, what is the remainder when 2n is divided by 8?
(1) n, when divided by 6, leaves remainder 5.
(2) 3n, when divided by 6, leaves remainder 3.
Asked: If n is a positive integer, what is the remainder when 2n is divided by 8?
(1) n, when divided by 6, leaves remainder 5.
n = 6k + 5
2n = 12k +10
The remainder when 2n is divided by 8 = {6, 2}
NOT SUFFICIENT
(2) 3n, when divided by 6, leaves remainder 3.
3n = 6k + 3
n = 2k + 1
2n = 4k + 2
The remainder when 2n is divided by 8 = {6,2}
NOT SUFFICIENT
(1) + (2)
(1) n, when divided by 6, leaves remainder 5.
n = 6k + 5
2n = 12k +10
The remainder when 2n is divided by 8 = {6, 2}
(2) 3n, when divided by 6, leaves remainder 3.
3n = 6k + 3
n = 2k + 1
2n = 4k + 2
The remainder when 2n is divided by 8 = {6,2}
Combining, we get
The remainder when 2n is divided by 8 = {6,2}
NOT SUFFICIENT
IMO E
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If n is a positive integer, what is the remainder when 2n is divided [#permalink]
Bunuel wrote:
If n is a positive integer, what is the remainder when 2n is divided by 8?
(1) n, when divided by 6, leaves remainder 5.
(2) 3n, when divided by 6, leaves remainder 3.
Hmmm this question was a little tricky.
So basically the two statements are same
1) n = 6a + 5
2) 3n = 6b + 3
we can get statement 2 from 1.
3*n = 3*6a +3*5
3n = 18a + 15
3n = 18a + 12 + 3
3n = 6(3a+2) + 3
3n = 6b + 3 where b=3a +2.
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Re: If n is a positive integer, what is the remainder when 2n is divided [#permalink]
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Remainders by 6 aren't really related to remainders by 8, so there's no reason to think the statements should be sufficient here. Using Statement 1, n can be 5 and 11, and those values also work with Statement 2. So using both Statements, n can be 5 and 11, and 2n can thus have a remainder of 2, when n=5, or 6, when n=11, when we divide by 8, and the answer is E.
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If n is a positive integer, what is the remainder when 2n is divided [#permalink]
Bunuel wrote:
If n is a positive integer, what is the remainder when 2n is divided by 8?
(1) n, when divided by 6, leaves remainder 5.
(2) 3n, when divided by 6, leaves remainder 3.
Question: $$Remainder(\frac{2n}{8}) =$$ ?
Statement 1: n, when divided by 6, leaves remainder 5.
i.e. n = 5, 11, 17, 23, 29... etc
i.e. 2n = 10, 22, 35, 46, 58... etc
i.e. $$Remainder(\frac{2n}{8}) =2, 6, 3...$$ etc.
NOT SUFFICIENT
Statement 2: 3n, when divided by 6, leaves remainder 3
i.e. 3n = 3, 9, 15, 21, 27, ... etc
i.e. n = 1, 3, 5, 7, 9, 11, 13, 15... etc
i.e. 2n = 2, 6, 10, 14, 18... etc
i.e. $$Remainder(\frac{2n}{8}) =2, 6, 3...$$ etc.
NOT SUFFICIENT
COmbining teh statements
i.e. n = 5, 11, 17, 23, 29... etc
i.e. 2n = 10, 22, 35, 46, 58... etc
i.e. $$Remainder(\frac{2n}{8}) =2, 6, 3...$$ etc.
NOT SUFFICIENT
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If n is a positive integer, what is the remainder when 2n is divided [#permalink]
GMATinsight wrote:
Bunuel wrote:
If n is a positive integer, what is the remainder when 2n is divided by 8?
(1) n, when divided by 6, leaves remainder 5.
(2) 3n, when divided by 6, leaves remainder 3.
Question: $$Remainder(\frac{2n}{8}) =$$ ?
Statement 1: n, when divided by 6, leaves remainder 5.
i.e. n = 5, 11, 17, 23, 29... etc
i.e. 2n = 10, 22, 35, 46, 58... etc
i.e. $$Remainder(\frac{2n}{8}) =2, 6, 3...$$ etc.
NOT SUFFICIENT
Statement 2: 3n, when divided by 6, leaves remainder 3
i.e. 3n = 3, 9, 15, 21, 27, ... etc
i.e. n = 1, 3, 5, 7, 9, 11, 13, 15... etc
i.e. 2n = 2, 6, 10, 14, 18... etc
i.e. $$Remainder(\frac{2n}{8}) =2, 6, 3...$$ etc. This is incorrect
NOT SUFFICIENT
COmbining teh statements
i.e. n = 5, 11, 17, 23, 29... etc
i.e. 2n = 10, 22, 35, 46, 58... etc
i.e. $$Remainder(\frac{2n}{8}) =2, 6, 3...$$ etc.
NOT SUFFICIENT
There is a typo that affected your calculation as highlighted
when n =17, then 2n=34..............hence $$Remainder(\frac{2n}{8}) =2...$$
There is no 3 at all in any reminder. It is all 2 &6
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Re: If n is a positive integer, what is the remainder when 2n is divided [#permalink]
If n is a positive integer, what is the remainder when 2n is divided by 8?
(1) n, when divided by 6, leaves remainder 5 --> insuff: n = 6p+5, if p =0, n = 5, 2n =10, so 2n divided by 8, reminder = 2, but if p =1, n = 11, 2n =22, so 2n divided by 8, reminder = 6
(2) 3n, when divided by 6, leaves remainder 3--> insuff: 3n = 6q+3 => n = 2q+1 (=2r+5), if q =0, n = 1, 2n =2, so 2n divided by 8, reminder = 2, but if q =1, n = 3, 2n =6, so 2n divided by 8, reminder = 6
Combining (1) & (2) we get,
n=6p+5, similar as (1), so not sufficient
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Re: If n is a positive integer, what is the remainder when 2n is divided [#permalink]
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Re: If n is a positive integer, what is the remainder when 2n is divided [#permalink]
Moderator:
Math Expert
93696 posts | 3,328 | 9,190 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.21875 | 4 | CC-MAIN-2024-26 | latest | en | 0.899871 |
https://www.thambi.ai/simulate-bankers-algorithm-for-deadlock-avoidance-using-c/ | 1,702,114,580,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100873.6/warc/CC-MAIN-20231209071722-20231209101722-00449.warc.gz | 1,109,523,739 | 12,817 | # Simulate Bankers Algorithm for Deadlock Avoidance Using C
void main()
{
int n,r,i,j,k,p,u=0,s=0,m;
int block[10],run[10],active[10],newreq[10];
int max[10][10],resalloc[10][10],resreq[10][10];
int totalloc[10],totext[10],simalloc[10];
//clrscr();
printf(“Enter the no of processes:”);
scanf(“%d”,&n);
printf(“Enter the no ofresource classes:”);
scanf(“%d”,&r);
printf(“Enter the total existed resource in each class:”);
for(k=1; k<=r; k++)
scanf(“%d”,&totext[k]);
printf(“Enter the allocated resources:”);
for(i=1; i<=n; i++)
for(k=1; k<=r; k++)
scanf(“%d”,&resalloc);
printf(“Enter the process making the new request:”);
scanf(“%d”,&p);
printf(“Enter the requested resource:”);
for(k=1; k<=r; k++)
scanf(“%d”,&newreq[k]);
printf(“Enter the process which are n blocked or running:”);
for(i=1; i<=n; i++)
{
if(i!=p)
{
printf(“process %d:\n”,i+1);
scanf(“%d%d”,&block[i],&run[i]);
}
}
block[p]=0;
run[p]=0;
for(k=1; k<=r; k++)
{
`````` j=0;
for(i=1; i<=n; i++)
{
totalloc[k]=j+resalloc[i][k];
j=totalloc[k];
}
}
for(i=1; i<=n; i++)
{
if(block[i]==1||run[i]==1)
active[i]=1;
else
active[i]=0;
}
for(k=1; k<=r; k++)
{
resalloc[p][k]+=newreq[k];
totalloc[k]+=newreq[k];
}
for(k=1; k<=r; k++)
{
if(totext[k]-totalloc[k]<0)
{
u=1;
break;
}
}
if(u==0)
{
for(k=1; k<=r; k++)
simalloc[k]=totalloc[k];
for(s=1; s<=n; s++)
for(i=1; i<=n; i++)
{
if(active[i]==1)
{
j=0;
for(k=1; k<=r; k++)
{
if((totext[k]-simalloc[k])<(max[i][k]-resalloc[i][k]))
{
j=1;
break;
}
}
}
if(j==0)
{
active[i]=0;
for(k=1; k<=r; k++)
simalloc[k]=resalloc[i][k];
}
}
m=0;
for(k=1; k<=r; k++)
resreq[p][k]=newreq[k];
}
else
{
for(k=1; k<=r; k++)
{
resalloc[p][k]=newreq[k];
totalloc[k]=newreq[k];
}
}
getch();``````
}
OUTPUT:
Enter the no of processes:4
Enter the no ofresource classes:3
Enter the total existed resource in each class:3 2 2
Enter the allocated resources:1 0 0 5 1 1 2 1 1 0 0 2
Enter the process making the new request:2
Enter the requested resource:1 1 2
Enter the process which are n blocked or running:process 2:
1 2
process 4:
1 0
process 5:
1 0 | 817 | 2,037 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.734375 | 3 | CC-MAIN-2023-50 | latest | en | 0.261618 |
https://ask.sagemath.org/question/10063/get-variants-of-complex-cube-root/ | 1,503,414,317,000,000,000 | text/html | crawl-data/CC-MAIN-2017-34/segments/1502886110792.29/warc/CC-MAIN-20170822143101-20170822163101-00634.warc.gz | 752,474,071 | 15,871 | # Get variants of complex cube-root
I found-out that complex cube-root can have 3 variants (see http://en.wikipedia.org/wiki/Cube_root)
But if I try in SageMath to do
(-1)^(1/3)
SageMath return (-1)^(1/3). When I try
(-1)^(1/3).n()
SageMath gives me numerical approximation of the one root (not real)...
How I can get all variants of complex cube-root without numerical approximation?
Thanks! P.S. Sorry for poor English...
edit retag close merge delete
Sort by ยป oldest newest most voted
Fist, when you write
sage: a = (-1)^(1/3)
you define an element of the Symbolic Ring, which is not a safe place:
sage: a.parent()
Symbolic Ring
sage: a^2
1
So it is better to work on QQbar, the set of algebraic complex numbers.
sage: a = QQbar(-1)^(1/3)
sage: a.parent()
Algebraic Field
sage: a^2
-0.500000000000000? + 0.866025403784439?*I
Now, if a is a cubic root of some complex number (in your case -1), the other cube roots are a*j and a*j^2, where j=exp(2*I*pi/3)=(-1+I*sqrt(3))/2. Hence you can define
sage: j = QQbar(-1)^(2/3)
sage: j == QQbar(e^(2*I*pi/3))
True
sage: j == QQbar(-1+I*sqrt(3))/2
True
sage: 1+j+j^2 == 0
True
sage: complex_cube_roots = [a, a*j, a*j^2]
and check
sage: for i in complex_cube_roots:
....: print i, 'whose cube is', i^3
0.500000000000000? + 0.866025403784439?*I whose cube is -1
-1 whose cube is -1
0.500000000000000? - 0.866025403784439?*I whose cube is -1
more
Or you can alternatively use embeddings of a number field as follows. I first show the example for the three cube root of 2. The three cube roots are the solution of the polynomial equation $x^3 - 2 = 0$. You then do
sage: R1.<a> = NumberField(x^3 - 2, 'a')
sage: e1,e2,e3 = R.embeddings(QQbar)
sage: a1 = e1(a)
sage: a2 = e2(a)
sage: a3 = e3(a)
sage: a1
-0.6299605249474365? - 1.091123635971722?*I
sage: a1^3
2
sage: a2
-0.6299605249474365? + 1.091123635971722?*I
sage: a2^3
2
sage: a3
1.259921049894873?
sage: a3^3
2
For the roots of -1 it is different because its polynomial equation $x^3 + 1 = (x + 1)(x^2 - x + 1)$ is not irreducible over QQ. But using the irreducible factor of degree 2 you can get the cube root as well:
sage: R2.<b> = NumberField(x^2 - x + 1, 'b')
sage: e1,e2 = R2.embeddings(QQbar)
sage: b1 = e1(b)
sage: b2 = e2(b)
sage: b1
0.50000000000000000? - 0.866025403784439?*I
sage: b1^3
-1
sage: b2
0.50000000000000000? + 0.866025403784439?*I
sage: b2^3
-1
The b1 and b2 above are exactly the numbers j and j^2 mentionned in tmonteil post. The advantage of this method is that it works for roots of polynomial equations and not only cubic roots.
more
If you want to find all numbers $x$ which satisfy $x^3=-1$ resp. $x^3+1=0$ you can do
sage: QQ[x](x^3+1).roots(QQbar)
[(-1, 1),
(0.500000000000000? - 0.866025403784439?*I, 1),
(0.500000000000000? + 0.866025403784439?*I, 1)]
The x^3+1 is a symbolic polynomial using the symbolic variable x which is already defined by default, but which you could also define using var('x'). The QQ[x] defines a polynomial ring with rational coefficients and using x as the name for its indeterminate. Using that like a function is a type cast, which will turn the symbolic expression into an element of the polynomial ring. And for elements of polynomial rings, you can compute roots. In particular, you can compute all roots exactly by using the filed of algebraic numbers $\bar{\mathbb Q}$. The resulting list contains tuple: the root followed by its multiplicity.
If you want to see some more details along the way, look at this:
sage: x
x
sage: x.parent()
Symbolic Ring
sage: QQx = QQ[x]; QQx
Univariate Polynomial Ring in x over Rational Field
sage: p1 = x^3 + 1; p1
x^3 + 1
sage: p1.parent()
Symbolic Ring
sage: p2 = QQx(p1); p2
x^3 + 1
sage: p2.parent()
Univariate Polynomial Ring in x over Rational Field
sage: p2.roots(AA) # real only
[(-1.000000000000000?, 1)]
sage: p2.roots(QQbar, multiplicities=False)
[-1,
0.500000000000000? - 0.866025403784439?*I,
0.500000000000000? + 0.866025403784439?*I]
more | 1,379 | 3,999 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.71875 | 4 | CC-MAIN-2017-34 | longest | en | 0.737258 |
http://math.stackexchange.com/questions/136595/finding-int-e2x-sin4x-dx | 1,469,583,449,000,000,000 | text/html | crawl-data/CC-MAIN-2016-30/segments/1469257825125.30/warc/CC-MAIN-20160723071025-00136-ip-10-185-27-174.ec2.internal.warc.gz | 110,197,908 | 22,610 | # Finding $\int e^{2x} \sin{4x} \, dx$
Finding $$\int e^{2x} \sin 4x \, dx$$
I think I should be doing integration by parts...
If I let $u=e^{2x} \Rightarrow du = 2e^{2x}$,
$dv = \sin{4x} \Rightarrow v = -\frac{1}{4} \cos{4x}$
$\int{ e^{2x} \sin{4x}} dx = e^{2x}(-\frac{1}{4}\cos 4x) - \color{red}{\int (-\frac{1}{4} \cos 4x) 2e^{2x} \, dx}$ Highlighted in red, I seem to integrating an exponential times trig expression again ... doesn't seem like te right way to go
So I let $u=\sin{4x} \Rightarrow du = 4\cos 4x \, dx$
$dv = e^{2x} dx \Rightarrow v = \frac{1}{2} e^{2x}$
$\sin{4x}(\frac{1}{2} e^{2x}) - \int (\frac{1}{2} e^{2x})(4\cos4x) \, dx$
Again, its an exponential times a trig function?
Am I using the wrong substitution?
-
Yes, after one integration by parts you are getting an exponential times trig expression again, but that doesn't mean it is not the right way to go! Just apply by parts again and enjoy. – Shitikanth Apr 25 '12 at 3:18
This is similar to the integral of $e^x \cos x$, which wikipedia has worked-out. I'd suggest taking a look at that and applying the principles (which "countinghaus" mentioned, more or less) – The Chaz 2.0 Apr 25 '12 at 3:30
I wonder how many times this same question has been asked here. In a way, you could say every instructor who sets this particular problem intends to provoke this particular question. (Unless the instructor is very naive, which certainly does happen.) (Anyway, I've posted an answer below.) – Michael Hardy Apr 25 '12 at 4:04
And, most probably, your calculus textbook has a worked-out example or two where you integrate by parts twice and then solve. – GEdgar Apr 26 '12 at 17:29
I wonder how many times this same question has been asked here, and how many times it is asked every semester in every second-semester calculus course?
If you integrate by parts a second time, you get the same integral you started with, and the naive reaction is that that means you're getting nowhere. But in truth it means you're almost done. You've got $$\int \text{something} = \text{something} - \text{something}\cdot \int [\text{same integral}].$$ So you add the same thing to both sides of the equation and get $$\int \text{something} + \text{something}\cdot\int [\text{same integral}] = \text{something} + C$$ Then you write $$(1+\text{something}) \cdot\int\cdots\cdots = \text{something} + C$$ $$\int\cdots\cdots = \frac{\text{something}}{1+\text{something}} + \text{constant}.$$
-
In general, if you want to find $$\int e^{ax}\cdot \sin{bx}\cdot dx$$ you can argue as follows:
Note that for any $\alpha$ or $\beta$, you have
\eqalign{ & \frac{d}{{dx}}\left( {{e^{\alpha x}}\sin \beta x} \right) = \alpha {e^{\alpha x}}\sin \beta x + \beta {e^{\alpha x}}\cos \beta x \cr & \frac{d}{{dx}}\left( {{e^{\alpha x}}\cos \beta x} \right) = \alpha {e^{\alpha x}}\cos \beta x - \beta {e^{\alpha x}}\sin \beta x \cr}
so that any integral of the form
$$\int e^{\alpha x}\cdot \sin{\beta x}\cdot dx$$
is a linear combination of the former functions. Let's then find $c_1$ and $c_2$ such that
$$\frac{d}{{dx}}\left( {{c_1}{e^{\alpha x}}\sin \beta x + {c_2}{e^{\alpha x}}\cos \beta x} \right) = {e^{\alpha x}}\sin \beta x$$
$${c_1}\alpha {e^{\alpha x}}\sin \beta x + {c_1}\beta {e^{\alpha x}}\cos \beta x + {c_2}\alpha {e^{\alpha x}}\cos \beta x - {c_2}\beta {e^{\alpha x}}\sin \beta x = {e^{\alpha x}}\sin \beta x$$
This means we need
\eqalign{ & {c_1}\alpha - {c_2}\beta = 1 \cr & {c_1}\beta + {c_2}\alpha = 0 \cr}
This will yield with little work
\eqalign{ & {c_1} = \frac{\alpha }{{{\alpha ^2} + {\beta ^2}}} \cr & {c_2} = - \frac{\beta }{{{\alpha ^2} + {\beta ^2}}} \cr}
which means that, in general:
$$\int {{e^{\alpha x}}} \cdot\sin \beta x\cdot dx = {e^{\alpha x}}\frac{{\alpha \sin \beta x - \beta \cos \beta x}}{{{\alpha ^2} + {\beta ^2}}} + C$$
Analogously, you will get that
$$\int {{e^{\alpha x}}} \cdot\cos \beta x\cdot dx = {e^{\alpha x}}\frac{{\alpha \cos \beta x + \beta \sin \beta x}}{{{\alpha ^2} + {\beta ^2}}} + C$$
Hope this helps!
-
Integrating with derivatives and linear algebra, I like it. (+1) – user26872 Apr 26 '12 at 23:10
Take your red integral and integrate it by parts again, making sure to let $u$ be the exponential function again. You will get that
The integral you care about = stuff + the integral you care about*(other stuff)
and therefore
the integral you care about = stuff/(1 - other stuff)
-
I would jump into complex values, and noting that $\sin 4x = \Im e^{4 i x}$ (I wrote \Im there, wanting to get the imaginary part, and got the $\Im$),
$$\int e^{2x} \sin 4x \, dx = \Im \int e^{2x+4ix} \, dx = \Im \int e^{x(2+4i)} \, dx = \Im \frac{e^{x(2+4i)}}{2+4i}.$$
Since $\frac{1}{2+4i} = \frac{2-4i}{20} = \frac{1-2i}{10}$ and $e^{x(2+4i)} = e^{2x}(\cos 4x + i \sin 4x)$, I get, disregarding the $\frac{e^{2x}}{10}$ for a moment,
$\Im (1-2i)(\cos 4x + i \sin 4x) = \Im ((\cos 4x + 2 \sin 4x) + i(-2\cos 4x + \sin 4x))$ $= -2\cos 4x + \sin 4x$
so my final result is $\frac{e^{2x}(-2\cos 4x + \sin 4x)}{10}$.
As a check the derivative of $e^{2x}(-2\cos 4x + \sin 4x)$ (disregarding the 1/10 for now) is
\begin{align} & 2e^{2x}(-2\cos 4x + \sin 4x) + e^{2x}(8\sin 4x + 4\cos 4x) \\[4pt] = {} & e^{2x}(-4\cos 4x + 2\sin 4x + 8\sin 4x + 4\cos 4x) = e^{2x} (10 \sin 4x). \end{align}
-
I would not jump into complex values, in case anyone is curious. – The Chaz 2.0 Apr 25 '12 at 5:04
When I do, I wear a parachute and a portable air bag. This minimizes the landing impact. – marty cohen Apr 26 '12 at 5:05
To find $\int e^{2x}\sin4x\,dx,$ we can let $u=e^{2x}$ and $dv=\sin4x\,dx,$ which implies $du=2e^{2x}\,dx$ and $v=-\frac{1}{4}\cos4x.$ Therefore, \begin{align}\int e^{2x}\sin4x\,dx &= uv-\int v\,du\\ &= e^{2x}\left(-\frac{1}{4}\cos4x\right)-\int-\frac{1}{4}\cos4x\left(2e^{2x}\right)\,dx\\ &= -\frac{1}{4}e^{2x}\cos4x+\frac{1}{2}\int e^{2x}\cos4x\,dx.\end{align} Apply another integration for $\int e^{2x}\cos4x\,dx,$ this time letting $u=e^{2x}$ and $dv=\cos4x\,dx,$ so that $du=2e^{2x}\,dx$ and $v=\frac{1}{4}\sin4x.$ Hence, \begin{align}\int e^{2x}\cos4x\,dx &= uv-\int v\,du\\ &= e^{2x}\left(\frac{1}{4}\sin4x\right)-\int\frac{1}{4}\sin4x\left(2e^{2x}\right)\,dx\\ &= \frac{1}{4}e^{2x}\sin4x-\frac{1}{2}\int e^{2x}\sin4x\,dx.\end{align}
Combining both formulas, we get \begin{align}\int e^{2x}\sin4x\,dx &= -\frac{1}{4}e^{2x}\cos4x+\frac{1}{2}\left(\frac{1}{4}e^{2x}\sin4x-\frac{1}{2}\int e^{2x}\sin4x\,dx\right)\\ &= -\frac{1}{4}e^{2x}\cos4x+\frac{1}{8}e^{2x}\sin4x-\frac{1}{4}\int e^{2x}\sin4x\,dx.\end{align} So, $$\frac{5}{4}\int e^{2x}\sin4x\,dx=-\frac{1}{4}e^{2x}\cos4x+\frac{1}{8}e^{2x}\sin4x.$$ Therefore, multiplying by $\frac{4}{5}$ and adding in the constant of integration, we see that $$\int e^{2x}\sin4x\,dx=-\frac{1}{5}e^{2x}\cos4x+\frac{1}{10}e^{2x}\sin4x+C,$$ which we can rewrite as $$\int e^{2x}\sin4x\,dx=\frac{1}{10}e^{2x}\left(-2\cos4x+\sin4x\right)+C.$$
-
Hi, welcome to Math.SE. Our site supports the use of MathJax for mathematical formulas, and the use of MarkDown for formatting text. I would highly recommend your editing your post to make use of both to improve on the clarity of exposition. – Willie Wong Jan 17 '14 at 16:02
Welcome to math.SE: I have tried to improve the readability of your post by introducing Tex. It is possible that I unintentionally changed the meaning of your post. Please proofread the question to ensure this has not happened. It's also worth noting that this question is nearly two years old, and has an accepted answer, so is probably too late to "help" the original poster. – Cameron Buie Jan 17 '14 at 16:28 | 2,853 | 7,606 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 4, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.09375 | 4 | CC-MAIN-2016-30 | latest | en | 0.898461 |
https://blog.csdn.net/nike0good/article/details/7897259 | 1,561,397,882,000,000,000 | text/html | crawl-data/CC-MAIN-2019-26/segments/1560627999620.99/warc/CC-MAIN-20190624171058-20190624193058-00272.warc.gz | 360,220,892 | 30,373 | # POJ 2828（卡时线段树）
Program P2828;
const
maxn=200000;
var
n,i,j:longint;
pos,val:array[1..maxn] of longint;
sum:array[1..maxn*10] of longint;
ans:array[1..maxn] of longint;
procedure pushup(root:longint);
begin
sum[root]:=sum[root*2]+sum[root*2+1];
end;
procedure build(l,r,root:longint);
var
mid:longint;
begin
if l=r then
begin
sum[root]:=1;
exit;
end;
mid:=(l+r) shr 1;
build(l,mid,root*2);
build(mid+1,r,root*2+1);
pushup(root);
end;
procedure decr(l,r,root,node:longint);
var
mid:longint;
begin
if l=r then
begin
dec(sum[root]);
exit;
end;
mid:=(l+r) shr 1;
if (node<=mid) then decr(l,mid,root shl 1,node)
else decr(mid+1,r,root shl 1+1,node);
pushup(root);
end;
function find(l,r,root,value:longint):longint;
var
mid:longint;
begin
mid:=(l+r) shr 1;
if l=r then
begin
decr(1,n,1,l);
exit(l);
end;
if (sum[root shl 1]>=value) then exit(find(l,mid,root shl 1,value))
else exit(find(mid+1,r,root shl 1+1,value-sum[root shl 1]));
end;
begin
while not seekeof do
begin
for i:=1 to n do read(pos[i],val[i]);
build(1,n,1);
for i:=n downto 1 do
begin
ans[find(1,n,1,pos[i]+1)]:=val[i];
end;
for i:=1 to n-1 do write(ans[i],' ');
writeln(ans[n]);
end;
end. | 446 | 1,153 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.796875 | 3 | CC-MAIN-2019-26 | latest | en | 0.16048 |
https://www.jiskha.com/display.cgi?id=1358027462 | 1,503,422,518,000,000,000 | text/html | crawl-data/CC-MAIN-2017-34/segments/1502886112533.84/warc/CC-MAIN-20170822162608-20170822182608-00039.warc.gz | 926,467,720 | 4,353 | # Math
posted by .
Determine the sum of the first 20 terms of the arithmetic sequence in which ...
f) the 7th term is 43 and the 13th term is 109
I know you can make two equations :
43 = a + 6d and 109 = a + 12 d
And subtract them to get d = 11
But I know there's another way to get d, that you can just quickly do on your calculator. But I can't remember how to do it! It had something to do with dividing the number of terms or something...? I could probably figure it out after some trial and error but I just want to make sure I have the proper method down, so if someone could explain a clear and concise method to me, I would really appreciate it! I know it's really simple... I'm just so burned out right now. Too much studying, too much stress!
• Math -
There are 6 terms from term7 to term13
and the difference in their values is 109-43 = 66
so the common difference is 66/6 = 11
but, ... , notice that you are actually doing the same calculations you would do if you solved the two equations, so nothing gained.
• Math -
I think this way is a lot faster and easier to do by just quickly typing it all into the calculator, so thank you!
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More Similar Questions | 905 | 3,631 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.828125 | 4 | CC-MAIN-2017-34 | latest | en | 0.961272 |
https://uk.mathworks.com/matlabcentral/answers/1595824-how-do-plot-rows-of-your-matrix-into-a-histogram | 1,642,373,019,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320300244.42/warc/CC-MAIN-20220116210734-20220117000734-00180.warc.gz | 642,192,101 | 23,939 | # How do plot rows of your matrix into a histogram?
9 views (last 30 days)
Dameon Solestro on 25 Nov 2021
Answered: Image Analyst on 25 Nov 2021
So usually hist allows you to plot the columns of your matrix into a histogram. i wish to plot the rows. Below is my matrix.
my first column is my ordinal information (so its not counted in the histogram). I also wish to plot for every 2 rows. Is that possible??
0 0.0032 0.0032 0.0032
1.0000 0.0010 0.0028 0.0016
2.0000 0.0012 0.0001 0.0013
3.0000 0.0000 0.0009 0.0025
4.0000 0.0006 0.0000 0.0012
5.0000 0.0006 0.0021 0.0002
6.0000 0.0003 0.0000 0.0003
7.0000 0.0019 0.0016 0.0001
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### Answers (2)
VBBV on 25 Nov 2021
A= [0 0.0032 0.0032 0.0032
1.0000 0.0010 0.0028 0.0016
2.0000 0.0012 0.0001 0.0013
3.0000 0.0000 0.0009 0.0025
4.0000 0.0006 0.0000 0.0012
5.0000 0.0006 0.0021 0.0002
6.0000 0.0003 0.0000 0.0003
7.0000 0.0019 0.0016 0.0001]
bar(A(1:2:end,2:end))
Try with bar function
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Image Analyst on 25 Nov 2021
Not sure what you mean by histogram. Is that your data or some kind of histogram of the data?
Anyway, to plot the rows of your matrix, row-by-row, you can do this:
m=[
0 0.0032 0.0032 0.0032
1.0000 0.0010 0.0028 0.0016
2.0000 0.0012 0.0001 0.0013
3.0000 0.0000 0.0009 0.0025
4.0000 0.0006 0.0000 0.0012
5.0000 0.0006 0.0021 0.0002
6.0000 0.0003 0.0000 0.0003
7.0000 0.0019 0.0016 0.0001];
[rows, columns] = size(m)
rows = 8
columns = 4
for row = 1 : rows
plot(m(row, 2:end), '.-', 'LineWidth', 3, 'MarkerSize', 40)
hold on;
legendStrings{row} = num2str(row);
end
legend(legendStrings)
grid on;
fontSize = 15;
title('m', 'FontSize', fontSize)
xlabel('x', 'FontSize',fontSize)
ylabel('y', 'FontSize',fontSize)
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### Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting! | 846 | 1,991 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.359375 | 3 | CC-MAIN-2022-05 | latest | en | 0.447057 |
http://mathhelpforum.com/discrete-math/136116-recursion-help-print.html | 1,495,904,313,000,000,000 | text/html | crawl-data/CC-MAIN-2017-22/segments/1495463608956.34/warc/CC-MAIN-20170527152350-20170527172350-00490.warc.gz | 277,576,174 | 3,143 | # Recursion help.
• Mar 28th 2010, 12:54 PM
xwanderingpoetx
Recursion help.
I have a general question about recursion. When you're either evaluating a recursive definition like:
$f(0) = 1, f(1) = 0, f(2) = 2, f(n) = 2f(n-3)$ for $n\geq3$
and trying to find a general formula for $f(n)$
or trying to give a recursive definition for a formula like:
$a_n = n(n+1)$
Is there an easier way to do it then look at the sequence and try to find some pattern between the 2 formulas, because I'm really bad at figuring about patters in sequences and forming formulas from them.
Any help would be greatly appreciated. Thank you.
• Mar 28th 2010, 02:23 PM
Plato
In answering this, here is something I do not usually do: give a detailed answer.
But it may help you to see how we work on problems.
I used a computer algebra system to produce some 15 terms of the example.
$\begin{array}{*{20}c} n &\vline & 0 &\vline & 1 &\vline & 2 &\vline & 3 &\vline & 4 &\vline & 5 &\vline & 6 &\vline & 7 &\vline & 8 \\
\hline {a_n } &\vline & 1 &\vline & 0 &\vline & 2 &\vline & 2 &\vline & 0 &\vline & 4 &\vline & 4 &\vline & 0 &\vline & 8 \\ \end{array}$
From that brief table, we see blocks of three and powers of two.
It then took some time to find a closed form of the sequence.
Here is what I found using the ceiling function:
$a_n = \left\{ {\begin{array}{cl}
{0} & {,3\text{ divides }\left( {n - 4} \right)} \\
{2^{\left\lceil {\displaystyle\frac{{n - 1}}
{3}} \right\rceil } } & .\text{else} \\ \end{array} } \right.$ | 520 | 1,508 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 6, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.28125 | 4 | CC-MAIN-2017-22 | longest | en | 0.818376 |
http://mathhelpforum.com/advanced-algebra/13459-linear-algebra-proof.html | 1,529,840,513,000,000,000 | text/html | crawl-data/CC-MAIN-2018-26/segments/1529267866932.69/warc/CC-MAIN-20180624102433-20180624122433-00270.warc.gz | 196,074,198 | 9,717 | 1. ## Linear algebra proof
The following proof was confusing me so any help would be appreciated!
Let V be a vector space and let W_1 and W_2 be subspaces of V. Prove that W_1 (insert upside down U symbol here) W_2 is a subspace of V. Remember to show that W_1 (insert upside down U symbol here) W_2 is nonempty.)
2. Okay...so I think I came up with a solution, but I'm not sure if it's right. Here it is:
Let a and b be vectors in the intersection. This means that a is in W_1 and a is in W_2, and b is in W_1 and b is in W_2. So this means that a + b has to be in W_1, as W_1 under the Closure Property of Addition. The same goes for W_2. Therefore, a + b is in W1 and a + b is in W_2. So a + b must be in the intersection.
Then for scalar multiplication...
Let a be some vector in the intersection. This means that a is in W_1 and a is in W_2. Therefore, a times a scalar x will be in W_1 as well as W_2 by the closure property of scalar multiplication. So a*x must be in the intersection.
3. Here. | 292 | 1,008 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.75 | 4 | CC-MAIN-2018-26 | latest | en | 0.960605 |
https://msofficegeek.com/percentage-calculator/ | 1,726,852,212,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725701419169.94/warc/CC-MAIN-20240920154713-20240920184713-00768.warc.gz | 362,821,342 | 53,317 | # Ready-To-Use Percentage Calculator Excel Template
Ready-to-Percentage Calculator in Excel, OpenOffice Calc, and Google Sheet to find, percent, percentage value, percentage change, and percentage difference.
In all these calculators, insert the value and the percentage you want to find. The “Result” consists of predefined formulas and hence will fetch the result automatically for you.
You can find the percentage value of a given number, a given number is what percentage of a number, and fractional percentages.
Moreover, you can also find the difference in percentage between two values and the increase/decrease in percentage between two values.
Additionally, you can find the actual value of a number that includes a certain amount of percentage already added to it. This template also helps you to convert your decimal values to percent and percent values to decimal.
With the help of this template, you can teach your child or student to learn percentage calculations along with their uses.
Moreover, this template also consists of a printable Percentage Practice Sheet to test the knowledge of your child/student.
We have created a simple and easy Percentage Calculator with predefined formulas to easily calculate the percentage of a number, percentage value, change in percentage, and the percentage difference.
Just insert the numbers and percentages in the desired calculator and the template automatically calculates the result for you.
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Feel free to contact us for the customization of this template as per your requirement. We also design new templates based on your needs. You can hire us for our services on Fiverr or directly contact us at info@msofficegeek.com.
## What is the Percentage?
A Percent is the 100th value of a given base. The sign of Percentage is %. In simple terms, it is the relation between two numbers as a fraction of 100.
A percent is a fractional value of the base whereas the value derived is the percentage value. For example, 10% of 100 books is 10 books. Here 10 is the percentage and 10 books are the percentage value. You can express a percent in decimal or fraction. For example, 50% is 0.5 or 1/2 of a given base number.
The work percent originates from Latin work Per Centum which means per 100. Working with percentages instead of fractions is easier because the base is always the 100th value of a given number. Thus, the calculations are relatively straightforward.
## Formula To Find Percentage
To find the percentage of a given number about the base number divide the percentage value by base value and multiply it by 100.
(Percentage value / Base Value) X 100
### Example Of Percentage
Percentage Value: 100
Base Value: 1000
Applying the above formula:
Percentage = (100/1000) X 100 = 10%
## Formula To Find Percentage Value
To find the percentage value of a given number about the base number multiply the base number with the percentage and divide it by 100.
(Base Value X Percentage) / 100
### Example of Percentage Value
Base Value: 150
Percentage: 25%
Applying the above-mentioned formula:
Percentage Value = (150 X 25) /100 = 37.5
## Formula To Find Percentage Difference
To find the percentage difference, we must first find the difference in value. To find the percentage difference divide the difference by the base value and multiply it by 100.
(Value 1 – Value 2) / Base Value X 100
There are 3 ways to find the percentage difference:
Percentage Difference Taking the Average of Both Numbers As The Base Value
(Value 1 – Value 2) / [(Value 1 + Value 2)/2] X 100
Percentage Difference Taking Value 1 As Base Value
(Value 1 – Value 2) / Value 1 X 100
Percentage Difference Taking Value 2 as Base Value
(Value 1 – Value 2) / Value 2) X 100
### Example of Percentage Difference
Value 1: 125
Value 2: 100
Difference: (Value 1 – Value 2) = 125-100 = 25
Average: (Value 1 + Value 2) / 2 = (125+100)/2 = 112.50
Applying the first formula:
Difference 1: 25 / 112.5 X 100 = 22.22%
Applying the second formula:
Percentage Difference 2: 25 / 125 X 100 = 20.00%
Applying the third formula:
Percentage Difference 2: 25 / 100 X 100 = 25.00%
There is a change in percentage difference due to a change in base value.
## Formula To Find Percentage Change
In percentage change, you can calculate both the increase and decrease of two given values.
Use the following formula to calculate the percentage increase:
Base Value + [(Base Value X Percentage ) / 100]
Use the following formula to calculate the percentage decrease:
Base Value – [(Base Value X Percentage ) / 100]
### Example of Percentage Change
Base Value: 1025
Percentage: 25%
% Increase = 1025 + [(1025 X 25)/100] = 1025 + 256.25 = 1284.25
% Decrease = 1025 – [(1025 X 25)/100] = 1025 – 256.25 = 768.75
## Percentage Calculator Template
This template consists of 5 calculators: Percentage Calculator, Percentage Value Calculator, Percentage Change Calculator, Percentage Difference Calculator, and Actual Value Calculator.
## Percentage Calculator
Insert the percent and the base value to find the percentage value. The formula used here is =D6*B6 where D6 is the base value and B6 is the percentage. As you can see in the image above, 5 percent of 325 is 16.25.
To find that a given number is what percent of a value, insert the number and the final value. The template uses the following formula here: =B7/D7, where B7 is the percentage value and D7 is the final value. As you can see in the example above, 100 is 20% of 1000.
Moreover, if you want to find that a given number is what percent of a number then insert the percentage value and the percentage to find the base value. The formula applied here is =B8*100%/D8, where B8 is the percentage value and D8 is the percentage. In the example given above, 20 is 20% of 100.
You will notice a difference in the formulas given above and formulas entered in excel. The main reason behind this is that excel consists of a preformulated percentage format for a cell. Thus, there is no need to multiply the result by 100 to find the percentage. Excel does this process on its own for the given number in the cell.
## Percentage Value Calculator
In this section, you can find the percentage value of a number in two ways. One is using the percentage another is using the fractional value of the percentage.
You can use the converter given beside the calculator to convert your percent to fraction or fraction to percent.
Insert the value and the percent you want to find the percentage value. The template will automatically display the result.
In the given example, the base value is 10000 and we want to find a 20% value of it. The result is 2000 using the formula =(C11*E11). It is the 1/5th value of 2000. This value is derived by using the fraction to percent converter.
Similarly, in the next calculator, you can insert the fractional value of the percentage. The template automatically displays the result.
In the sample sum given here, the value is 25000 and we want to find the 2/5th value of it. The result is 10000 using =(C14*E14). Basically, the 2/5th value given is 40% of the 25000. This has been derived using the percent to fraction converter.
## Percentage Change Calculator
The Percentage Change Calculator helps you to find the increase or decrease of a given percentage from the base value. Insert the base value and insert the percentage.
The calculator displays the percentage increase value using the =C18+(C18*E18) formula, where C18 is the base value and E18 is the percentage. The increased value is the base value in addition to the percentage value.
The calculator displays the percentage decrease value using the =C18-(C18*E18) formula, where C18 is the base value and E18 is the percentage. The increased value is the base value in addition to the percentage value.
In the sample sum given above, 1025 is the base value and the percentage is 20%. Hence, the increased value is (1025 + 256.25) = 1281.25. Similarly, the decreased value is (1025 – 256.25) = 768.75.
## Percentage Difference Calculator
The percentage difference calculator helps you to find the percentage of a given difference between two values. Insert both the value and the calculator finds the difference and average of both the numbers.
In addition to the above, it calculates the difference percentage using 3 different formulas. The first one calculates the difference taking an average of both numbers as the base amount. The second one uses value 1 as the base amount and the last one uses value 2 as the base amount.
The percentage change using all three values is different depending on the base amount. In simple terms, the base amount decides the difference percentage.
As you can see in the above sample sums, the first result is 22.22% where the base amount is 112.5 which is the average of both the values.
In the second value, the result is 20% which is derived using 125 as the based value. lastly, in the third result, it is 25% which is derived using 100 as the base value.
## Actual Value Calculator
The Actual Value Calculator is useful to find the base value when a given number includes a specific percentage.
Generally, this is useful in finding the product cost from the Maximum Retail Price(MRP) of a product that includes a percentage. These additional percentages can be either applicable taxes or profit margins of a company.
Insert the value of the product and the percentage to find what is the actual price of the product.
As you can see in the example, the MRP of the product is 2345 and it consists of 5% in the price. To final price is 105% of the actual value. Hence, we need to find the Actual price. Divide the price with the total percentage to find the actual value. Thus, 2345/105% = 2233.33.
## Printable Percentage Practice Sheet
This template consists of a Printable Percentage Practice Sheet that you can give your child/student to practice percentage sums.
Download the file and insert random numbers and percentages in each type of sum and print the sheet.
## Uses of Percentage
Usually, the percentage is used in every field. To name some here are a few uses based on their industry:
• To find the Commission of Sales for Sales professionals.
• Knowing the percentage increase or decrease in sales of a product or total sales.
• To know the collection percentage of outstanding invoices.
• Define the Area-wise percentage sales.
• Knowing the traffic source percentage.
• To know the percentage of attendance and the grading system in school uses the percentage value to determine the grades.
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In Depth Information
Using EOMONTH to Calculate the End of the Month
Figure 11.46.
Figure 11.46. You can use
You can use EDATE
EDATE to calculate the maturity date for a se-
to calculate the maturity date for a se-
curity.
curity.
Note that EDATE can be used to back into an investment date from a maturity
date. For example, the records in rows 11 through 16 pass a negative number
for the monthsparameter.
Using
Using EOMONTH
EOMONTH to Calculate the End of the Month
to Calculate the End of the Month
Before Excel 2007, about 89 functions were available only in the Analysis
Toolpack. Some companies had rules that you were not allowed to build
spreadsheets using the functions in the Analysis Toolpack. This rule was
probably created by some corporate executive who didn t know how to turn
on the Analysis Toolpack!
One of my favorite puzzles at MrExcel.com came from someone who worked at
such a company. How can you calculate the end of the month without using
EOMONTH? This is a hard question; the end of the month is the 31st if the
month number is 1, 3, 5, 7, 8, 10, or 12. It is the 30th if the month number is 4, 6, 9,
or 11. If the month number is 2, then you have to look at the year to figure out
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Regular Scientific Notation Worksheet With Answers Pdf. Note that a negative exponent indicates that the number is a fraction less than one. Convert this number into regular form. Negative n indicates a number between zero and one. Move the decimal 4 places to the right 4210.
Scientific Notation Place Value Worksheets Scientific Notation Scientific Notation Worksheet 8th Grade Math From pinterest.com
Write this new number down followed by. Exponent is positive 5. Represents the number of decimal places to be moved. Convert this number into Scientific Notation. Hone in on manipulating the exponents while adding subtracting multiplying and dividing numbers in scientific notation with this free pdf worksheet for high school students. Positive n indicates the standard form is a large number.
### Perform the operation simplify the expression and write your answer in.
Once you find your worksheet s you can either click on the pop-out icon or download button to print or download your desired worksheet. Convert 1500000 to scientific notation. Positive n indicates the standard form is a large number. Some of the worksheets below are Scientific Notation Practice Worksheets with Answers Converting from decimal form into scientific notation Adding subtracting dividing and multiplying scientific notation exercises several fun. Exponent is Negative 5. Convert this number into regular form.
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Positive n indicates the standard form is a large number. Regular Notation Scientific Notation How to Change Regular Notation 7510. Scientific Notation To write a number in scientific notation. Exponent is positive 4. The small number to the right of the 10 in scientific notation is called the exponent.
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February 18 2020 Scientific notation worksheet with answers pdf. Move the decimal 5 places to the right 3810. Use the buttons below to print open or download the PDF version of the Scientific Notation A math worksheet. If there are more versions of this worksheet the other versions will be available below the preview images. 1 0000006 2 5400000 3 60 4 0009 5 67 6 00000002 7 2000000 8.
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Exponent is Negative 3. Once you find your worksheet s you can either click on the pop-out icon or download button to print or download your desired worksheet. Convert 1500000 to scientific notation. Writing in Scientific Notation Date_____ Period____ Write each number in scientific notation. Some of the worksheets below are Scientific Notation Practice Worksheets with Answers Converting from decimal form into scientific notation Adding subtracting dividing and multiplying scientific notation exercises several fun problems with solutions.
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Prior to dealing with Scientific Notation Worksheet Answers make sure you recognize that Schooling is our own factor to an improved down the road as well as learning does not only cease right after the institution bell ringsIn which currently being reported we provide variety of simple still beneficial content along with templates created appropriate for almost any academic purpose. FREE Scientific Notation Worksheets With Answers Pdf updated. Some of the worksheets below are Scientific Notation Practice Worksheets with Answers Converting from decimal form into scientific notation Adding subtracting dividing and multiplying scientific notation exercises several fun problems with solutions. Convert this number into Scientific Notation. Exponent is positive 4.
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Move the decimal 5 places to the right 3810. Move the decimal 5 places to the right 3810. Some of the worksheets below are Scientific Notation Practice Worksheets with Answers Converting from decimal form into scientific notation Adding subtracting dividing and multiplying scientific notation exercises several fun. Move the decimal 3 places to the left. 1 0000006 2 5400000 3 60 4 0009 5 67 6 00000002 7 2000000 8.
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Exponent is Negative 5. Convert this number into Scientific Notation. COUNT how many times you had to move the decimal point use this number as your exponent 5. 23092019 admin September 23 2019. Free worksheetspdf and answer keys on scientific notation.
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Scientific Notation Standard Form questions. Converting Scientific Notation - CommonCoreSheets. Represents the number of decimal places to be moved. FREE Scientific Notation Worksheets With Answers Pdf updated. If there are more versions of this worksheet the other versions will be available below the preview images.
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Scientific Notation Worksheetspdf and Answer Keys. Move the decimal 4 places to the right 4210. What is the order of magnitude of these numbers. Convert this number into regular form. Scientific Notation Worksheetspdf and Answer Keys.
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Exponent is positive 4. Positive n indicates the standard form is a large number. COUNT how many times you had to move the decimal point use this number as your exponent 5. Note that a negative exponent indicates that the number is a fraction less than one. Some of the worksheets below are Scientific Notation Practice Worksheets with Answers Converting from decimal form into scientific notation Adding subtracting dividing and multiplying scientific notation exercises several fun problems with solutions.
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Convert this number into regular form. Scientific Notation Worksheetspdf and Answer Keys. Preview images of the first and second if there is one pages are shown. 1 0000006 2 5400000 3 60 4 0009 5 67 6 00000002 7 2000000 8. February 18 2020 Scientific notation worksheet with answers pdf.
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Create your own worksheets like this one with Infinite Algebra 1. Scientific notation takes the form of M x 1 On where 1 I M. Preview images of the first and second if there is one pages are shown. Convert 1500000 to scientific notation. Exponent is positive 5.
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Scientific notation takes the form of M x 1 On where 1 I M. Convert this number into Scientific Notation. Negative n indicates a number between zero and one. Scientific notation takes the form of M x 1 On where 1 I M. COUNT how many times you had to move the decimal point use this number as your exponent 5.
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Convert 1500000 to scientific notation. We move the decimal point so that there is only. _____ So Much More Online. Convert this number into regular form. Convert this number into Scientific Notation.
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Convert this number into Scientific Notation. The line below shows the equivalent values of decimal notation the way we write numbers usually like 1000 dollars and scientific notation 103 dollars. Each sheet is scaffolder and has model problems explained step by step. Free worksheetspdf and answer keys on scientific notation. The size of the PDF file is 11669 bytes.
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Once you find your worksheet s you can either click on the pop-out icon or download button to print or download your desired worksheet. FREE Scientific Notation Worksheets With Answers Pdf updated. Some of the worksheets below are Scientific Notation Practice Worksheets with Answers Converting from decimal form into scientific notation Adding subtracting dividing and multiplying scientific notation exercises several fun. Convert this number into Scientific Notation. COUNT how many times you had to move the decimal point use this number as your exponent 5.
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For this, I am assuming that -25/2 is not a part of the exponent(although in future questions you should make this clear by using parenthesis.
When I do problems like this, I come to CBAD. where the variables are y=A(Bx+C)+D
CBAD helps significantly in domain and range problems where there is a well known parent function. (y=x^2, y=sqrt(x), y=1/x, y=x, etc). Basically, if there is a parent function with slight additions(y=2*x^2, y=sqrt(x+1), y=1/(5x), y=3x, etc)
This works that in the order, a C term shifts the graph horizontally in the opposite direction expected (a negative number will shift right, a positive number will shift left). For example, if C is -2, the graph of the parent function will shift right 2.
Then, a B term dilates(changes the horizontal size of the graph) by the factor 1/B(every x value is multiplied by 1/B). Thus a B >1 will shrink the graph horizontally, and a 0<B<1 with stretch the graph horizontally. If B is negative, it reflects the graph over the y axis and then either shrinks or stretches horizontally. If B is 1, nothing changes.
Then, a A term dilates(changes the horizontal size of the graph) by the factor A(every y value is multiplied by A). Thus a A >1 will stretch the graph vertically, and a 0<A<1 with shrink the graph vertically. If A is negative, it reflects the graph over the x axis and then either shrinks or stretches vertically. If A is 1, nothing changes.
Finally, a D term shifts the graph vertically in the direction expected (a negative number will shift down, a positive number will shift up). For example, if D is -2, the graph of the parent function will shift down 2.
Warning CBAD must be performed in the order C-B-A-D
Now that you know these simple rules and can memorize the few parent functions, you can figure out the domain and range of almost any function.
In your case, the function x^(2)-25/2 is not much different than x^2, an upwards facing parabola centered at (0,0). In fact, there is only one transformation, D, because C=0 (it is not shown), B=1(it is not shown), and A=1(it is not shown). Thus the only difference between your graph and the parent function is that yours is shifted down by 25/2.
For this reason, because the domain of x^2 is not restricted (it includes all real numbers), the domain of your new function is not restricted (it includes all real numbers). | 635 | 2,399 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.3125 | 4 | CC-MAIN-2017-04 | latest | en | 0.825528 |
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```
At a certain stage of a criminal investigation, the inspector in-charge is 60% convince of the guilt of a certain suspect. Suppose now a piece of evidence that shows the criminal has brown hair is uncovered. If the inspector in-charge is convinced that the suspect is not guilty then there is 20% chance that he/she has brown hair. What is the probability that the inspector in-charge is convinced that the suspect is guilty given he/she has brown hair?```
5 years ago
SHAIK AASIF AHAMED
74 Points
``` Hello student,•G: event that the suspect is guilty•C: event that he possesses the characteristic of the criminal•P(G|C) is ?•P(G|C) = P(GC)/P(C) = P(C|G)P(G) / [P(C|G)P(G) + P(C|Gc)P(Gc)] = 1(.6)/[1(.6) + (.15)(.4)] ≈ .91Thanks and RegardsShaik AasifaskIITians faculty
```
3 years ago
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## Other Related Questions on Algebra
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• View Details | 365 | 1,241 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.609375 | 3 | CC-MAIN-2018-05 | latest | en | 0.887541 |
http://nghiaho.com/?p=437 | 1,524,639,198,000,000,000 | text/html | crawl-data/CC-MAIN-2018-17/segments/1524125947705.94/warc/CC-MAIN-20180425061347-20180425081347-00075.warc.gz | 219,982,901 | 9,962 | # KD-tree on GPU for 3D point searching
This post is a follow up on my previous post about using the GPU for brute force nearest neighbour searching. In the previous post I found the GPU to be slower than using the ANN library, which utilises some sort of KD-tree to speed up searches. So I decided to get around to implementing a basic KD-tree on the GPU for a better comparison.
My approach was to build the tree on the host and transfer it to the GPU. Sounds conceptually easy enough, as most things do. The tricky part was transferring the tree structure. I ended up representing the tree as an array node on the GPU, with each node referencing other nodes using an integer index, rather than using pointers, which I suspect would have got rather ugly. The KD-tree that I coded allows the user to set the maximum depth if they want to, which turned out to have some interesting effects as shown shortly. But first lets see some numbers:
```nghia@nghia-laptop:~/Projects/CUDA_KDtree\$ bin/Release/CUDA_KDtree
Points in the tree: 100000
Query points: 100000
Results are the same!
GPU max tree depth: 99
GPU create + search: 321.849 + 177.968 = 499.817 ms
ANN create + search: 162.14 + 456.032 = 618.172 ms
Speed up of GPU over CPU for searches: 2.56x```
Alright, now we’re seeing better results from the GPU. As you can see I broken the timing down for the create/search part of the KD-tree. My KD-tree creation code is slower than ANN, no real surprise there. However the searches using the GPU is faster, the results I was hoping for. But only 2.56x faster, nothing really to rave about. I could probably get the same speed up, or more, by multi-threading the searches on the CPU. I then started to play around with the maximum depth of my tree and got these results:
```GPU max tree depth: 13
GPU create + search: 201.771 + 56.178 = 257.949 ms
ANN create + search: 163.414 + 403.978 = 567.392 ms
Speed up of GPU over CPU for searches: 7.19x```
From 2.56x to 7.19x, a bit better! A max tree depth of 13 produced the fastest results. It would seem that there is a sweet spot between creating a tree with the maximum depth required, so that each leaf of the tree holds only a single point, versus a tree with a limited depth and doing a linear search on the leaves. I suspect most of the bottleneck is due to the heavy use of global memory, especially randomly accessing them from different threads, which is why we’re not seeing spectacular speed ups.
The code for the CUDA KD-tree can be downloaded here CUDA_KDtree.zip. Its written for Linux and requires the ANN library to be installed.
## 4 thoughts on “KD-tree on GPU for 3D point searching”
1. Junpeng says:
Hello Nghia:
I am a phd student. And my major is computer graphis. I am working on KD-Tree those days. your codes is really helpful. And I think there are several things you can do to improve the performance. So, I want to discuss with you about the following three points.
1. For building KD-Tree process, the most time consuming part is sorting to find median value. This part we can use GPU to do sorting. I try to use thrust library to do that. However, my performance didn’t improve too much. I think it’s because of the data size.
2. For traversing KD-Tree process, I think we can updata radius of the circle while the best_dist is updata. In __device__ void Search(..) function:
if(tmp_dist < best_dist) {
best_dist = tmp_dist;
best_idx = tmp_idx;
}
3. I think this equation: CUDA_STACK = 2^(MAX_TREE_LEVEL) should be satisfied in the code. Because it's tatally possible that one thread would back traverse from the bottom to the top level of the tree.
Thanks very much for sharing the codes.
1. nghiaho12 says:
Hi,
If you manage to improve the code let me know, because I’ve been busy with other projects lately so do not have time to go back to this one.
2. Aaliya says:
Hello Nghia, Thanks a lot for this code! I’m trying to run this by changing the number of dimensions of the points, but it doesn’t seem to be working for any number of dimensions except 3. I get the following error:
terminate called after throwing an instance of std::length_error’
what(): vector::_M_fill_insert
Any help would be much appreciated! Thanks once again!
1. nghiaho12 says:
Your vector might be too large. | 1,047 | 4,288 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.625 | 3 | CC-MAIN-2018-17 | latest | en | 0.915461 |
http://aslstem.cs.washington.edu/topics/list/5547 | 1,511,547,500,000,000,000 | text/html | crawl-data/CC-MAIN-2017-47/segments/1510934808742.58/warc/CC-MAIN-20171124180349-20171124200349-00126.warc.gz | 24,361,870 | 8,239 | ## Viewing Mathematical Analysis topics
The following topics are immediate sub-topics of Mathematical Analysis.
• #### Term: Abel's Limit Theorem
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• Definition: Assume that we have f(x) = sum[(a_n)x^n] as n goes from 0 to infinity, and if -r < x < r. If the series also converges at x = r, then the limit as x -> r- of f(x) exists and we have lim f...
• #### Term: Abel's Test
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• Definition: The series sum[(a_n)(b_n)] converges if sum(a_n) converges and if {b_n} is a monotonic convergent sequence.
• #### Term: Accumulation Point
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• Definition: If S is a subset of R^n and x is an element of R^n, then x is called an accumulation point of S if every n-ball B(x) contains at least one point of S distinct from x.
• #### Term: Adherent Point
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• Definition: Let S be a subset of R^n, and x a point in R^n, x not necessarily in S. Then x is said to be adherent to S if every n-ball B(x) contains at least one point of S.
• #### Term: Almost Everywhere
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• Definition: A property is said to hold almost everywhere on a set S if it holds everywhere on S except for a set of measure 0.
• #### Term: Analytic Function
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• Definition: Let f=u + iv be a complex-valued function defined on an open set S in the complex plane C. Then f is said to be analytic on S if the derivative f' exists and is continuous at every point of S.
• #### Term: Annulus
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• Definition: The region lying between two concentric circles.
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• Definition: Let S be a nonempty set of real numbers with a supremum, say b = sup S. Then for ever a < b there is some x in S such that a < x <= b.
• #### Term: Archimedean Property
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• Definition: If x>0 and if y s an arbitrary real number, there is a positive integer n such that nx > y.
• #### Term: Arcwise Connectedness
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• Definition: A set S in R^n is called arcwise connected if for any two points a and b in S there is a continuous function f:[0, 1] -> S such that f(0) = a and f(1) = b. | 789 | 3,044 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.5 | 4 | CC-MAIN-2017-47 | longest | en | 0.809735 |
https://calculus7.org/tag/lines/ | 1,603,440,454,000,000,000 | text/html | crawl-data/CC-MAIN-2020-45/segments/1603107880878.30/warc/CC-MAIN-20201023073305-20201023103305-00529.warc.gz | 241,048,093 | 21,748 | Linear approximation and differentiability
If a function ${f\colon \mathbb R\rightarrow \mathbb R}$ is differentiable at ${a\in \mathbb R}$, then it admits good linear approximation at small scales. Precisely: for every ${\epsilon>0}$ there is ${\delta>0}$ and a linear function ${\ell(x)}$ such that ${|f(x)-\ell(x)|<\epsilon \,\delta}$ for all ${|x|<\delta}$. Having ${\delta}$ multiplied by ${\epsilon}$ means that the deviation from linearity is small compared to the (already small) scale ${\delta}$ on which the function is considered.
For example, this is a linear approximation to ${f(x)=e^x}$ near ${0}$ at scale ${\delta=0.1}$.
As is done on this graph, we can always take ${\ell}$ to be the secant line to the graph of ${f}$ based on the endpoints of the interval of consideration. This is because if ${L}$ is another line for which ${|f(x)-L(x)|<\epsilon \,\delta}$ holds, then ${|\ell-L|\le \epsilon \,\delta}$ at the endpoints, and therefore on all of the interval (the function ${x\mapsto |\ell(x)-L(x)|}$ is convex).
Here is a non-differentiable function that obviously fails the linear approximation property at ${0}$.
(By the way, this post is mostly about me trying out SageMathCloud.) A nice thing about ${f(x)=x\sin \log |x|}$ is self-similarity: ${f(rx)=rf(x)}$ with the similarity factor ${r=e^{2\pi}}$. This implies that no matter how far we zoom in on the graph at ${x=0}$, the graph will not get any closer to linear.
I like ${x\sin \log |x|}$ more than its famous, but not self-similar, cousin ${x\sin(1/x)}$, pictured below.
Interestingly, linear approximation property does not imply differentiability. The function ${f(x)=x\sin \sqrt{-\log|x|}}$ has this property at ${0}$, but it lacks derivative there since ${f(x)/x}$ does not have a limit as ${x\rightarrow 0}$. Here is how it looks.
Let’s look at the scale ${\delta=0.1}$
and compare to the scale ${\delta=0.001}$
Well, that was disappointing. Let’s use math instead. Fix ${\epsilon>0}$ and consider the function ${\phi(\delta)=\sqrt{-\log \delta}-\sqrt{-\log (\epsilon \delta)}}$. Rewriting it as
$\displaystyle \frac{\log \epsilon}{\sqrt{-\log \delta}+\sqrt{-\log (\epsilon \delta)}}$
shows ${\phi(\delta)\rightarrow 0}$ as ${\delta\rightarrow 0}$. Choose ${\delta}$ so that ${|\phi(\delta)|<\epsilon}$ and define ${\ell(x)=x\sqrt{-\log \delta}}$. Then for ${\epsilon \,\delta\le |x|< \delta}$ we have ${|f(x)-\ell(x)|\le \epsilon |x|<\epsilon\,\delta}$, and for ${|x|<\epsilon \delta}$ the trivial bound ${|f(x)-\ell(x)|\le |f(x)|+|\ell(x)|}$ suffices.
Thus, ${f}$ can be well approximated by linear functions near ${0}$; it’s just that the linear function has to depend on the scale on which approximation is made: its slope ${\sqrt{-\log \delta}}$ does not have a limit as ${\delta\to0}$.
The linear approximation property does not become apparent until extremely small scales. Here is ${\delta = 10^{-30}}$.
Words that contain UIO, and best-fitting lines
In Calculus I we spend a fair amount of time talking about how nicely the tangent line fits a smooth curve.
But truth be told, it fits only near the point of tangency. How can we find the best approximating line for a function ${f}$ on a given interval?
A natural measure of quality of approximation is the maximum deviation of the curve from the line, ${E(f;\alpha,\beta) = \max_{[a, b]} |f(x) - \alpha x-\beta|}$ where ${\alpha,\beta}$ are the coefficients in the line equation, to be determined. We need ${\alpha,\beta}$ that minimize ${E(f;\alpha,\beta)}$.
The Chebyshev equioscillation theorem is quite useful here. For one thing, its name contains the letter combination uio, which Scrabble players may appreciate. (Can you think of other words with this combination?) Also, its statement does not involve concepts outside of Calculus I. Specialized to the case of linear fit, it says that ${\alpha,\beta}$ are optimal if and only if there exist three numbers ${ x_1 in ${[a, b]}$ such that the deviations ${\delta_i = f(x_i) - \alpha x_i-\beta}$
• are equal to ${E(f;\alpha,\beta)}$ in absolute value: ${|\delta_i| = E(f;\alpha,\beta)}$ for ${i=1,2,3}$
• have alternating signs: ${\delta_1 = -\delta_2 = \delta_3}$
Let’s consider what this means. First, ${f'(x_i) =\alpha}$ unless ${x_i}$ is an endpoint of ${[a,b]}$. Since ${x_2}$ cannot be an endpoint, we have ${f'(x_2)=\alpha}$.
Furthermore, ${f(x) - \alpha x }$ takes the same value at ${x_1}$ and ${x_3}$. This gives an equation for ${x_2}$
$\displaystyle f(x_1)-f'(x_2)x_1 = f(x_3)-f'(x_2) x_3 \qquad \qquad (1)$
which can be rewritten in the form resembling the Mean Value Theorem:
$\displaystyle f'(x_2) = \frac{f(x_1)-f(x_3)}{x_1-x_3} \qquad \qquad (2)$
If ${f'}$ is strictly monotone, there can be only one ${x_i}$ with ${f'(x_i)=\alpha}$. Hence ${x_1=a}$ and ${x_3=b}$ in this case, and we find ${x_2}$ by solving (2). This gives ${\alpha = f'(x_2)}$, and then ${\beta}$ is not hard to find.
Here is how I did this in Sage:
var('x a b')
f = sin(x) # or another function
df = f.diff(x)
a = # left endpoint
b = # right endpoint
That was the setup. Now the actual computation:
var('x1 x2 x3')
x1 = a
x3 = b
x2 = find_root(f(x=x1)-df(x=x2)*x1 == f(x=x3)-df(x=x2)*x3, a, b)
alpha = df(x=x2)
beta = 1/2*(f(x=x1)-alpha*x1 + f(x=x2)-alpha*x2)
show(plot(f,a,b)+plot(alpha*x+beta,a,b,color='red'))
However, the algorithm fails to properly fit a line to the sine function on ${[0,3\pi/2]}$:
The problem is, ${f'(x)=\cos x}$ is no longer monotone, making it possible for two of ${x_i}$ to be interior points. Recalling the identities for cosine, we see that these points must be symmetric about ${x=\pi}$. One of ${x_i}$ must still be an endpoint, so either ${x_1=a}$ (and ${x_3=2\pi-x_2}$) or ${x_3=b}$ (and ${x_1=2\pi-x_2}$). The first option works:
This same line is also the best fit on the full period ${[0,2\pi]}$. It passes through ${(\pi,0)}$ and has the slope of ${-0.2172336...}$ which is not a number I can recognize.
On the interval ${[0,4\pi]}$, all three of the above approaches fail:
Luckily we don’t need a computer in this case. Whenever ${|f|}$ has at least three points of maximum with alternating signs of ${f}$, the Chebyshev equioscillation theorem implies that the best linear fit is the zero function.
Partition of the plane by lines
It’s an exercise in induction to prove that $n$ lines in general position divide the plane into latex $M_n=n(n+1)/2+1$ regions, and this number of regions is the maximal possible. Here is a partition that realizes $M_3$:
Three lines can also divide the plane into 6 regions instead of 7: this happens if the triangle collapses to a point, or if two of the lines are made parallel. However, 3 lines can never divide the plane into 5 regions.
Define $\mu_n$ to be the smallest integer such that for any integer $m\in [\mu_n, M_n]$ there is a partition of the plane into $m$ regions by $n$ lines. So, $\mu_3=6$, and of course $\mu_2=3$ and $\mu_1=M_1=2$. Here are a few more values of $\mu_n$, with examples in lieu of proofs.
The last one, unattainability of 23 with 8 lines, isn’t easy to prove.
Does $\mu_n$ have a closed form? 2,3,6,8,12,15,18,24 is not in OEIS, unlike the upper bound.
Update: the 1993 paper Classification of arrangements by the number of their cells by Nicola Martinov gives a complete description of the pairs $(n,f)$ for which there is a partition of (projective) plane by $n$ lines into $f$ regions. In the affine version considered above we should simply says that the ideal line is also a part of arrangement; thus, 3-line affine arrangements correspond to 4-line projective arrangements, etc. However, it is not entirely trivial to get $\mu_n$ out of Martinov’s formulas. | 2,274 | 7,702 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 112, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.15625 | 4 | CC-MAIN-2020-45 | latest | en | 0.898732 |
http://www.oxfordmathcenter.com/drupal7/node/286 | 1,563,850,925,000,000,000 | text/html | crawl-data/CC-MAIN-2019-30/segments/1563195528687.63/warc/CC-MAIN-20190723022935-20190723044935-00402.warc.gz | 246,186,692 | 5,071 | Exercises - Descriptive Statistics
1. Given the following data: 100, 95, 95, 90, 85, 75, 65, 60, 55. Find the median, mean, and mode. Is there a most appropriate measure?
2. Make a sketch of the following, indicating the approximate locations for the mean, median and mode:
1. a normal distribution
2. a skewed distribution
3. a rectangular distribution
3. Given the data set: 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, ''x''. Find the smallest positive integer value for ''x'' such that ''x'' is an outlier. Find the value for ''both'' definitions.
4. Given the following set of golf scores: 67, 70, 72, 74, 76, 76, 78, 80, 82, 85. Find the median, mean, mode, and standard deviation. What percentage of scores are in the interval of one standard deviation from the mean?
5. Give at least five uses and at least five misuses of statistics. Use complete sentences and elaborate as needed for clarity.
6. Give the four levels or categories of data and give an example of each.
7. What amount of data does Chebyshev's Theorem guarantee is within three standard deviations from the mean? Compare this result to the empirical rule. Why are there differences?
8. Given the following grades on a test: 86, 92, 100, 93, 89, 95, 79, 98, 68, 62, 71, 75, 88, 86, 93, 81, 100, 86, 96, 52
1. Make a stem-and-leaf plot to represent this data.
2. Find the mode, median, mean, range, standard deviation, and interquartile range
3. What percentage of scores lie within one standard deviation from the mean? two standard deviations?
4. Are there any outliers? Explain clearly.
9. What is an experimental design and why is it important? Describe a completely randomized experimental design and a rigorously controlled design.
10. Given the following sample of freshman GPA scores:
$$2.2, 2.9, 3.5, 4.0, 3.9, 3.5, 2.9, 2.8, 3.1, 3.5, 3.8, 4.0, 3.8, 2.4, 3.9, 3.4, 2.8, 2.4, 1.8, 3.6, 3.1, 2.9, 3.8, 4.0$$
1. Is there an outlier? (Check both tests and explain)
2. Draw a frequency histogram using 5 to 6 categories. Be consistent with the rules for making histograms.
3. Is the distribution significantly skewed?
4. What percentage of scores is within one standard deviation of the mean? two standard deviations? three standard deviations?
5. Are your findings consistent with the minimum amount of data within two standard deviations guaranteed by Chebyshev's Theorem? | 700 | 2,348 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.09375 | 4 | CC-MAIN-2019-30 | longest | en | 0.856495 |
http://www.mission-utopia.com/10-circle-theorems/ | 1,585,528,935,000,000,000 | text/html | crawl-data/CC-MAIN-2020-16/segments/1585370496330.1/warc/CC-MAIN-20200329232328-20200330022328-00430.warc.gz | 238,462,585 | 13,524 | # 10: CIRCLE Theorems
The Circle is a fascinating entity. It has certain very cool properties. The first theorem we take up is that if two points on the circumference is joined to the centre it forms an angle which is twice that of any angle formed by the same two points on the far side of the circumference. The proof is shown in the diagram above. The trick is to join the centre with the third point forming thereby two triangles. And it must be known that any triangle formed out of two points on the circumference and the centre has to be an isosceles triangle since the two sides of the triangle invariably are two radii (plural of radius) of the circle. Thus as shown, one of the triangles have two angles x and x and the other y and y.
Then by another very important theorem of triangle that any external angle of a triangle is equal to the sum of the two opposite angles in the triangle, we come to the angle at the centre as 2x and 2y. And this proves our theorem since 2x +2y is twice x+y.
what if we change the angle from 45 to something else
The first theorem is extended to show that any two angles formed by two points with a third point on the circumference are equal.
This theorem shows that a perpendicular drawn from the centre of a circle onto a tangent makes a right angle. This is a very unique theorem somewhat difficult to put into words and we recommend that the student try to memorize the theorem visually. The proof depends on the first theorem as shown.
length of tangents drawn from an external point to the circle The length of the tangents drawn from an external point to a circle are equal. The proof is shown by proving that the two triangles are right angled and are shown to be congruent. | 372 | 1,732 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.46875 | 4 | CC-MAIN-2020-16 | latest | en | 0.959704 |
https://eduraa.com/tag/core/ | 1,532,332,107,000,000,000 | text/html | crawl-data/CC-MAIN-2018-30/segments/1531676595531.70/warc/CC-MAIN-20180723071245-20180723091245-00427.warc.gz | 640,814,066 | 11,796 | The Learning Edge
## Core 4 – Partial Fractions (1) Introduction and basic concept – A2 A Level Maths
This channel is managed by up and coming UK maths teachers. Videos designed for the site by Steve Blades, retired Youtuber and owner of m4ths.com to assist learning in UK classrooms. Designed for the Edexcel spec but applicable to AQA, OCR,MEI and WJEC. Video Rating: / 5
## Introduction to Economics for Managers / HBX CORe
In Professor Bharat Anand’s Economics for Managers course, you will learn about topics like customer demand, supplier cost, markets and competition on Harvard Business School’s new interactive online learning platform, HBX. You’ll learn how businesses think about pricing, production, and differentiation. To find out more about Economics for Managers and the HBX CORe curriculum, visit […]
## Core 2 – Coordinate Geometry (The Equation of a Circle) (1) – Basic Introduction
This channel is managed by up and coming UK maths teachers. Videos designed for the site by Steve Blades, retired Youtuber and owner of m4ths.com to assist learning in UK classrooms. Designed for the Edexcel spec but applicable to AQA, OCR,MEI and WJEC. Video Rating: / 5
## COMMON CORE: 2ND GRADER SHOWS US HOW ADDITION AND SUBTRACTION ARE BEING TAUGHT IN SCHOOLS.
Keep in mind that this little girl is in a private school. Common Core is standardized. That means this is nation-wide. The only escape is to pull your kids out of school and to home school. Or you can get all the parents together and fight the school/school district. She is not my daughter. She […]
## January 2015 algebra 1 Common Core New York NY Regents Pt II 6 10
1 The owner of a small computer repair business has one employee, computations. who is paid an hourly rate of . The owner estimates his weekly profit using the function P(x) 8600 22x. In this function, x represents the number of (1) computers repaired per week (2) hours worked per week (3) customers served per […]
## Solving elapsed time word problemsLesson 4 of 5 (Common Core Standard 3.MD.1)
Download resources for this lesson at http://learnzillion.com/lessons/578, try our Common Core tool: http://learnzillion.com/common_core/math/k-8 , or search more than 2,000 other lessons for free at http://LearnZillion.com . In this lesson, you will learn how to solve elapsed time word problems in one-minute intervals by using a number line. Standard 3.MD.1: Tell and write time to the […]
## 5.14 Angle Types and Basic Rules – Basic Maths GCSE Core Skills Level 5 & Grade E
Types of angles and basic Angle rules GCSE Maths and Key Stage 3 revision and help Check out this playlist for all the videos covering level 5 skills This video is part of a series that is linked to a specific system for improving Basic Maths skills. This one is a Level 5 Basic Maths […] | 659 | 2,816 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.171875 | 3 | CC-MAIN-2018-30 | longest | en | 0.917106 |
https://de.mathworks.com/matlabcentral/cody/problems/2020-area-of-an-isoceles-triangle/solutions/534305 | 1,607,123,220,000,000,000 | text/html | crawl-data/CC-MAIN-2020-50/segments/1606141745780.85/warc/CC-MAIN-20201204223450-20201205013450-00640.warc.gz | 255,062,982 | 16,961 | Cody
# Problem 2020. Area of an Isoceles Triangle
Solution 534305
Submitted on 24 Nov 2014 by Sreeram Mohan
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 = 5; y = 8; A_correct = 12; tolerance = 1e-12; assert(abs(isocelesArea(x,y)-A_correct)<tolerance)
2 Pass
%% x = 2; y = 2; A_correct = sqrt(3); tolerance = 1e-12; assert(abs(isocelesArea(x,y)-A_correct)<tolerance)
3 Pass
%% x = 10; y = 2; A_correct = sqrt(99); tolerance = 1e-12; assert(abs(isocelesArea(x,y)-A_correct)<tolerance)
### Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting! | 233 | 747 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.890625 | 3 | CC-MAIN-2020-50 | latest | en | 0.70469 |
https://proofwiki.org/wiki/Jordan_Curve_and_Jordan_Arc_form_Two_Jordan_Curves | 1,597,106,481,000,000,000 | text/html | crawl-data/CC-MAIN-2020-34/segments/1596439738723.55/warc/CC-MAIN-20200810235513-20200811025513-00544.warc.gz | 446,945,896 | 10,903 | # Jordan Curve and Jordan Arc form Two Jordan Curves
Jump to navigation Jump to search
## Theorem
Let $\closedint a b$ denote the closed real interval between $a \in \R, b \in \R: a \le b$.
Let $\gamma: \closedint a b \to \R^2$ be a Jordan curve.
Let the interior of $\gamma$ be denoted $\Int \gamma$.
Let the image of $\gamma$ be denoted $\Img \gamma$.
Let $\sigma: \closedint c d \to \R^2$ be a Jordan arc such that:
$\map \sigma c \ne \map \sigma d$
$\map \sigma c, \map \sigma d \in \Img \gamma$
and:
$\forall t \in \openint c d: \map \sigma t \in \Int \gamma$
Let $t_1 = \map {\gamma^{-1} } {\map \sigma c}$.
Let $t_2 = \map {\gamma^{-1} } {\map \sigma d}$.
Let $t_1 < t_2$.
Define:
$-\sigma: \closedint c d \to \Img \sigma$ by $-\map \sigma t = \map \sigma {c + d - t}$
Let $*$ denote concatenation of paths.
Let $\gamma \restriction_{\closedint a {t_1} }$ denote the restriction of $\gamma$ to $\closedint a {t_1}$.
Define:
$\gamma_1 = \gamma {\closedint a {t_1} } * \sigma * \gamma {\restriction_{\closedint {t_2} b} }$
Define:
$\gamma_2 = \gamma {\closedint a {t_1} } * \paren {-\sigma}$
Then $\gamma_1$ and $\gamma_2$ are Jordan curves such that:
$\Int {\gamma_1} \subseteq \Int \gamma$
and:
$\Int {\gamma_2} \subseteq \Int \gamma$
### Corollary
Define:
$\gamma_1 = \gamma {\restriction_{\closedint a {t_2} } } * \paren {-\sigma} * \gamma {\restriction_{\closedint {t_1} b} }$
Define:
$\gamma_2 = \gamma {\restriction_{\closedint {t_2} {t_1} } } * \sigma$
Then $\gamma_1$ and $\gamma_2$ are Jordan curves such that:
$\map {\operatorname {Int} } {\gamma_1} \subseteq \map {\operatorname {Int} } \gamma$
and:
$\map {\operatorname {Int} } {\gamma_2} \subseteq \map {\operatorname {Int} } \gamma$
## Proof
As:
$\Int \gamma$ and $\Img \gamma$ are disjoint by the Jordan Curve Theorem
and:
$\map \sigma {\openint c d} \subseteq \Int \gamma$
it follows that:
$\Img \gamma \cap \Img \sigma = \set {\map \sigma c, \map \sigma d}$.
As $\gamma$ is a Jordan curve, it follows that $\gamma {\restriction_{\closedint a {t_1} } }$ and $\gamma {\restriction_{\closedint {t_2} b} }$ intersect only in $\map \gamma a$.
It follows that $\gamma_1$ is a Jordan arc.
As the initial point of $\gamma_1$ is $\map \gamma a$, and the final point of $\gamma_1$ is $\map \gamma b = \map \gamma a$, it follows that $\gamma_1$ is a Jordan curve.
As $\Img {-\sigma} = \Img \sigma$, it follows that $\gamma_2$ is a Jordan arc.
As $\map \gamma {t_1} = \map \sigma c = -\map \sigma d$, it follows that $\gamma_2$ is a Jordan curve.
$\Box$
Denote the exterior of $\gamma$ as $\Ext \gamma$.
Let $q_0 \in \Ext \gamma$ be determined.
Let $q \in \Ext \gamma$.
By the Jordan Curve Theorem, $\Ext \gamma$ is unbounded.
Hence for all $N \in \N$ we can choose $q \in \Ext \gamma$ such that:
$\map d {\mathbf 0, q} > N$
where $d$ denotes the Euclidean metric on $\R^2$.
The Jordan Curve Theorem also shows that $\Ext \gamma$ is open and connected.
From Connected Open Subset of Euclidean Space is Path-Connected, there exists a path:
$\rho: \closedint 0 1 \to \Ext \gamma$
joining $q$ and $q_0$.
We have:
$\Img {\gamma_1} \subseteq \Int \gamma \cup \Img \gamma$
This is disjoint with $\Ext \gamma$
Thus it follows that $\rho$ is a path in either $\Ext {\gamma_1}$ or $\Int {\gamma_1}$.
We have that $\map d {\mathbf 0, q}$ can be arbitrary large.
Also, $\Int {\gamma_1}$ is bounded.
So follows that $\rho$ is a path in $\Ext {\gamma_1}$.
In particularly:
$q \in \Ext {\gamma_1}$
Therefore:
$\Ext \gamma \subseteq \Ext {\gamma_1}$
Let $q_1 \in \Int {\gamma_1}$.
Then:
$q_1 \notin \Ext \gamma$
as $\Int {\gamma_1}$ and $\Ext {\gamma_1}$ are disjoint.
It follows that:
$q_1 \in \Int {\gamma_1}$
Therefore:
$\Int {\gamma_1} \subseteq \Int \gamma$
Similarly, it follows that:
$\Int {\gamma_2} \subseteq \Int \gamma$
$\blacksquare$ | 1,281 | 3,871 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.453125 | 3 | CC-MAIN-2020-34 | latest | en | 0.548326 |
http://tutor-usa.com/free/pre-algebra/worksheet/metric-system-metric-conversions | 1,544,706,642,000,000,000 | text/html | crawl-data/CC-MAIN-2018-51/segments/1544376824822.41/warc/CC-MAIN-20181213123823-20181213145323-00597.warc.gz | 304,334,381 | 8,762 | # The Metric System - Metric Conversions
Worksheet (Pre-Algebra)
## The Metric System - Metric Conversions Worksheet
This Free Pre-Algebra Worksheet contains problems on the metric system. Students are asked to choose the appropriate metric unit to use to measure different objects. Problems also require finding equivalent measurements and converting from one metric unit to another.
## Lesson:
To use and apply the metric system to measure objects and make conversions.
22
3
Yes
## Samples:
Free Pre-Algebra Worksheet
The Metric System
Converting Metric Measures
1) A drinking glass holds about ________ of liquid
a) 10 L b) 2 L c) 200 ml d) 1000 ml
2) A car's mass is about ________.
a) 1,500 mg b) 1,500 g c) 1,500 kg d) 100kg
3) A bucket of water can hold about ________ of liquid.
a) 8 L b) 450 ml c) 1 L d) 50 ml
6) Use _________ to measure the mass of a truck.
a) liters b) kiloliters c) grams d) kilograms
8) Use ________ to measure the mass of a feather.
a) liters b) milliliters c) grams d) milligrams
Converting Metric Units.
Complete each statement.
9) 1 g = _____ mg
10) 1 mL = _____ L
11) 1kL = _____ mL
12) 1 km = _____ m
13) 465 cm = _____ m
14) 99 m = _____ mm
15) 92kg = _____ g
16) 8 kL = _____ L
17) 39 km = _____ cm
18) 283 mL = _____ kL
19) 2.4 km = _____ cm
20) 333 mL = _____ kL
21) Laceys books have a mass of 11.2 kg. What is the mass of Laceys books in grams?
22) Cagneys swimming pool can hold up to 3.08 kL of water. What is the capacity of Cagneys pool in millimeters? | 457 | 1,534 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.953125 | 4 | CC-MAIN-2018-51 | latest | en | 0.704309 |
https://www.numberempire.com/91565743 | 1,621,009,767,000,000,000 | text/html | crawl-data/CC-MAIN-2021-21/segments/1620243991428.43/warc/CC-MAIN-20210514152803-20210514182803-00114.warc.gz | 978,005,305 | 7,170 | Home | Menu | Get Involved | Contact webmaster
0 / 12
# Number 91565743
ninety one million five hundred sixty five thousand seven hundred forty three
### Properties of the number 91565743
Factorization 91565743 Divisors 1, 91565743 Count of divisors 2 Sum of divisors 91565744 Previous integer 91565742 Next integer 91565744 Is prime? YES (5302550th prime) Previous prime 91565711 Next prime 91565779 Is a Fibonacci number? NO Is a Bell number? NO Is a Catalan number? NO Is a factorial? NO Is a regular number? NO Is a perfect number? NO Polygonal number (s < 11)? NO Binary 101011101010010111010101111 Octal 535227257 Duodecimal 267b9527 Hexadecimal 5752eaf Square 8384285291142049 Square root 9568.9990594628 Natural logarithm 18.332567774978 Decimal logarithm 7.9617330237905 Sine -0.66756699543633 Cosine -0.74454973413743 Tangent 0.89660497456186
Number 91565743 is pronounced ninety one million five hundred sixty five thousand seven hundred forty three. Number 91565743 is a prime number. The prime number before 91565743 is 91565711. The prime number after 91565743 is 91565779. Number 91565743 has 2 divisors: 1, 91565743. Sum of the divisors is 91565744. Number 91565743 is not a Fibonacci number. It is not a Bell number. Number 91565743 is not a Catalan number. Number 91565743 is not a regular number (Hamming number). It is a not factorial of any number. Number 91565743 is a deficient number and therefore is not a perfect number. Binary numeral for number 91565743 is 101011101010010111010101111. Octal numeral is 535227257. Duodecimal value is 267b9527. Hexadecimal representation is 5752eaf. Square of the number 91565743 is 8384285291142049. Square root of the number 91565743 is 9568.9990594628. Natural logarithm of 91565743 is 18.332567774978 Decimal logarithm of the number 91565743 is 7.9617330237905 Sine of 91565743 is -0.66756699543633. Cosine of the number 91565743 is -0.74454973413743. Tangent of the number 91565743 is 0.89660497456186
### Number properties
0 / 12
Examples: 3628800, 9876543211, 12586269025 | 622 | 2,047 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.953125 | 3 | CC-MAIN-2021-21 | latest | en | 0.645454 |
http://www.infocodify.com/cprog/multidimensional_arrays | 1,561,397,204,000,000,000 | text/html | crawl-data/CC-MAIN-2019-26/segments/1560627999620.99/warc/CC-MAIN-20190624171058-20190624193058-00333.warc.gz | 252,179,408 | 6,418 | # Multidimensional Arrays in C programming
#### Multidimensional arrays definition
Arrays in C can have multiple subscripts. A common use of multiple-subscripted arrays, which the C standard refers to as multidimensional arrays, is to represent tables of values consisting of information arranged in rows and columns.
To identify a particular table element, we must specify two subscripts:
• The first (by convention) identifies the element’s row.
• The second (by convention) identifies the element’s column.
Tables or arrays that require two subscripts to identify a particular element are called double-subscripted arrays. Multidimensional arrays can have more than two subscripts.
Array with three rows and four columns, so it’s said to be a 3-by-4 array.
In general, an array with m rows and n columns is called an m-by-n array.
#### Defining multidimensional array
A multidimensional array can be initialized when it’s defined, much like a single-subscripted array.
Defining a double-subscripted array int b[2][2]:
``int b[ 2 ][ 2 ] = { { 1, 2 }, { 3, 4 } };``
#### Initializing multidimensional arrays.
The program defines three arrays of two rows and three columns (six elements each).
• The definition of array1 provides six initializers in two sublists.
• The definition of array2 provides five initializers.
• The definition of array3 provides three initializers in two sublists.
Note: Uninitialized elements are assigned the value 0.
``````
// Initializing multidimensional arrays.
#include <stdio.h>
void printArray( int a[][ 3 ] ); // function prototype
// function main begins program execution
int main( void ) {
// initialize array1, array2, array3
int array1[ 2 ][ 3 ] = { { 1, 2, 3 }, { 4, 5, 6 } };
int array2[ 2 ][ 3 ] = { 1, 2, 3, 4, 5 };
int array3[ 2 ][ 3 ] = { { 1, 2 }, { 4 } };
puts( "Values in array1 by row are:" );
printArray( array1 );
puts( "Values in array2 by row are:" );
printArray( array2 );
puts( "Values in array3 by row are:" );
printArray( array3 );
} // end main
// function to output array with two rows and three columns
void printArray( int a[][ 3 ] ) {
size_t i; // row counter
size_t j; // column counter
// loop through rows
for ( i = 0; i <= 1; ++i ) {
// output column values
for ( j = 0; j <= 2; ++j ) {
printf( "%d ", a[ i ][ j ] );
} // end inner for
printf( "\n" ); // start new line of output
} // end outer for
} // end function printArray
``````
``````
Output:
Values in array1 by row are:
1 2 3
4 5 6
Values in array2 by row are:
1 2 3
4 5 0
Values in array3 by row are:
1 2 0
4 0 0
``````
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Infocodify is a professional services company providing IT consulting and web development services focused on the Microsoft platform, and tutorials on programming languages such as C, C++, C# and Azure Web Services, covering most aspects of Object Oriented programming and Web Apps development. | 806 | 2,949 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.578125 | 3 | CC-MAIN-2019-26 | latest | en | 0.656811 |
http://nov9blogg9.blogspot.com/2014/04/answering-alec-macandrew.html | 1,531,964,549,000,000,000 | text/html | crawl-data/CC-MAIN-2018-30/segments/1531676590443.0/warc/CC-MAIN-20180719012155-20180719032155-00136.warc.gz | 274,818,429 | 25,836 | ## Friday, 25 April 2014
I used to be good at grasping fairly simple physical explanations when "small" (say, five to ten). You know descriptions of how razor machines and such work. Perhaps they are approximations. I must admit that much of the physics and maths are beyond my level of being able to calculate graphs or analyses to double check. But this does not mean that concept is beyond me. I will partly defend Sungenis based on concept. Then I will pass on to my own explanation, which is not my own, but which I have from St Thomas Aquinas.
Sungenis* speculates about the physics at his hypothetical centre of mass, relying on purely classical mechanics. He claims that there is no reason that the Earth should not coincide with the centre of mass, and in his ball-universe model, indeed there isn’t, temporarily. But he fails to recognize that there is nothing to keep the Earth there.
IF it is centre of mass, earth would be stuck as firmly in there as a ball stuck in the centre of earth if it is hollow. Unless other bodies fell in towards the centre and ousted earth.
One other aspect is also part of it: other bodies cannot oust earth from the centre, because they are involved in a daily orbit around earth. Their velocities sidewards would - on his view at least - be too great for any one of them to fall inwards and oust the earth from the centre.
That is Sungenis' point.
The centre of mass of a system is a point in space and there is no reason to identify it with a particular body – no physical body has to “act as the centre of mass” or “be it” or “occupy it”, all phrases that Sungenis has used in various places in the past, and which show that he fails to understand the concept.
I think you have failed to understand his point. Earth is heavier than air, which is heavier than diffused molecules of hydrogen. If Earth is in the centre of gravity, it will not be ousted from it by its atmosphere, nor by interstellar matter, since none of these are heavy enough. It could only be ousted from there by an equally heavy or heavier body. Since centre of Mass is also centre of gravitation. But the available options would not be getting inwards since already in orbit all of them together around the earth.
I take the point about "no gravitation" as meaning "no gravitation any one direction more than any other direction". Not as meaning "no gravitation inward."
the gravitational field is not zero at the centre of mass (it can be zero in certain symmetrical systems, such as a uniform spherical shell or a spherical ball of perfectly uniform density, or a two body system of exactly equal masses - but it is not generally so)
The point of Sungenis would, as I understand it, be that by daily rotation around earth all other bodies form virtually a symmetrical system around it. So, if the question of "certain symmetrical systems" seems marginal to Alec MacAndrew - he puts it in parentheses - it is rather more like a central theme to "the Sungenis version of Newtonian physics."
Furthermore, Sungenis’s claim that the stars have “gravity [that] will affect how the Sun and Earth react to one another, especially if the Earth is put in the center of that gravity” [my bolding] is wrong, not just because the gravitational field at the Earth of all these stars is vanishingly small compared with that of the Sun, as we have seen, but because gravitational fields of individual bodies are vector-additive—that is, they can cancel each other out if they act from opposite directions—so that if the Earth were to be at the centre, these already minuscule gravitational fields from the stars would tend to sum to zero.
That particular claim is probably behind Sungenis' words about Earth having "zero" gravitation.
Along with definition of "gravitation" as a force tending to acceleration. Question is whether the "zero gravitation" would mean a force - possibly gigantic - preventing accelaration in any one direction, that is how I think Sungenis takes it, or zero force at all. That is how I think Alec MacAndrew takes it.
For my own part, I do not know if gravitation toward centre of mass or cancellation of gravitational directions would take over. I tend to the first one, supposing that Newtonian physics are at all true and even if they are partly false.
Let us sketch up a few scenarios, shall we.
Scenario 1:
There is Earth and there is the sphere of fixed stars (roughly) symmetrically around it.
Corrollary:
Nothing happens.
Scenario 2:
There is Earth and there is the sphere of fixed stars (roughly) symmetrically around it. There is also the Sun and it is bigger and hotter than Earth.
Corrollary:
Nothing happens.
Scenario 3:
There is Earth and there is the sphere of fixed stars (roughly) symmetrically around it. There is also the Sun and it is bigger and hotter than Earth. There is also the Moon.
Corrollary:
Nothing happens.
Scenario 4:
There is Earth and there is the sphere of fixed stars (roughly) symmetrically around it. There is also the Sun and it is bigger and hotter than Earth. There is also the Moon. There is also some sort of gravitation.
Corrollary:
Supposing the gravitation is strictly Newtonian, the most massive object will plummet down into the centre of gravity and stay there. Earth will in that case simultaneously be fried and pushed to the side. Moon will fall into this as well, but Sun will be in the Middle.
This will increase the gravitational pull toward the middle and the Sphere of Fixed Stars will collapse. This is the reason why Newtonian and Mechanistic explanations tend to avoid a Sphere of Fixed Stars. Newton to Olbers, one tried to pretend universe had - at least as a disc of stars with greater tightness - an infinite extension, however unthinkable and counterintuitive and impossible THAT is, so one could pretend the gravitational pull on each stars by each other star was everywhere (not excepting any outer borders even) balanced out by an equal gravitational pull from the other side. After Olbers and through one (but not only possible) explanation of red shift, one now instead sees the inward force of gravitation balanced by an outward thrust from the Big Bang.
Supposing instead the gravitation is Aristotelian, there are two kinds of mass (correponding to light and heavy elements in Aristotle): leptomass and barymass.
Earth is less massive in total mass than the Sun. BUT the Sun may have a surplus of leptomass over barymass. In that case it will not plummet down to the centre of gravity. Even the Moon might have a slight surplus of leptomass over barymass and will not plummet down into it. Earth may be, not indeed the most massive body, but the most massive body with a surplus of barymass over leptomass. This is how Aristotle, Lucretius and a few others motivated Geocentricity physically.
The Fixed Stars would of course be the ones which had the greatest percentual surplus of leptomass, hence they would be furthest up and out. This poses a bit of a problem whether they stay in place at all or whether they expand, further and further up, through their positive lightness.
In either case their staying in relative places would require a Designer. Either for designing a limit beyond which even lightness cannot expand or for mixing precisely enough barymass into the Stars for them to stay at a fixed distance from Earth, or else for the Stars to stay in their angular places relative to each other while their whole sphere is expanding.
Lucretius who did not think there was a God was not very attentive to his own Geocentric astronomy.
Scenario 5:
There is Earth and there is the sphere of fixed stars (roughly) symmetrically around it. There is also the Sun and it is bigger and hotter than Earth. There is also the Moon. There is also some sort of gravitation. There is also rotation.
Corrollary:
In this case even Newtonian gravity might not be incompatible with Geocentricity.
Now - precisely at scenario 5 - look at Alec's resumé of Gary's answer to Robert:
Sungenis has already been shown by Gary Hoge, that there are no observable motions in the universe that could offset the overwhelming gravitational attraction of the Sun, moon, and planets on the Earth.
Is the daily motion of the Universe around the Earth "observable" in MacAndrew's language? It is any way observed, the colloquial terms for those very common observations being "day" and "night".
Is there a problem with this theory? Well, where does the rotation come from. Not how we observe it, we observe it as day and night, but where it comes from.
Lucretius thought Moon and Sun and stuff were just gliding around because they were lighter.
Aristotle and St Thomas find that God can explain the rotation - as in observed daily rotation of the Universe around Earth - as efficient cause thereof.
Newtonian laws of movement?
Problem is, a circular movement is not a uniform movement. Therefore even without any friction, it would not be generated and then automatically ongoing, since changing direction is a kind of acceleration. And each body would be changing direction.
Conservation of angular momentum?
Can it be derived from Newton's laws of motion? As it is experimentally observed, it is usually concerned with solid circular objects rotating around their centres. Not with objects separate from each other rotating around common centres in the void. Even two stones united with a rope (or three with three ropes, like the Bolas of Argentinian Gauchos) as being united by a rope (or three) are in a sense solids. Every observed instance on earth sooner or later stops.
Back to Sungenis' theory now.
If we make a model of the solar system, and use bodily arms between sun model and earth model, so that earth on model is moved by arm from the sun on model, we can indeed fixate earth and this will make sun on model revolve around earth precisely as earth would otherwise revolve around sun.
My take on Sungenis' is that his take on Mach is that this happens in dynamics of Solar System according to Classic Newtonian Model, if and when Earth is for instance fixed into the Centre of Gravity.
My problem with that (and I suspect if Sungenis read this in Mach it is there in Mach) is that this does not take into account that in Newton's model Earth is not moved around the Sun by a physical arm, a body of metal shaped as an arm. But rather by the evasive (and regularly so) equilibrium between its two movements. The one being the movement sidewards and tangential through inertia, the other being the movement inward and radial through gravitation. And both movements or theoretical such equalling out into a movement which is insofar accelerating as it is changing direction.
Now the fixation of earth on a model would make the mechanic arm in the sun move the sun around the earth instead of other way round. But the fixation of Earth in a place in the Universe would involve its lacking a sideways movement relative to the Sun and its lacking an inwards movement relative to the Sun and therefore no movement making an equilibrium between them would arise either.
A movement of the Sun around Earth as per year - apart from this being an abstraction since the main movement of the Sun around Earth is per day, but if we supposed for a moment Earth was rotating so only yearly movement were to reckon with - would in a Newtonian model require this movement to be the result of Sun's sideways movement, determined by its inertia determined by its mass, and the Sun's inwards movement into the Earth, determined by Earth's (or some centre of gravity's situated around it) superior attraction.
However, the Newtonians do find a consistency between the Sun's Mass as used for it being close around Centre of Gravity for Earth and the same as used for it being close around Centre of Gravity for the orbits of Kuiper Belt or Jupiter, Venus or Mars.
For Sungenis to be correct in the main would require that exactly as much inferior as the Earth is by itself to the Sun, so much superior is the Earth with the Centre of Gravitation to the Sun, when it comes to exerting gravitational pulls. Improbable? Yes, if the explanations are purely mechanistic. Possible? Exactly under one precise condition, if a God with a sense of humour made it so.
But this was, as said, ignoring that according to Geocentrism as usually understood, indeed defended by Sungenis himself, as well as me, the main movement of the Sun around the Earth is its daily movement. Its yearly movement is a movement in relation to the Zodiac. This in its turn remaining true whether Sun and each star are every day going West around us and Sun being slower than the stars or Heaven is going West, carrying the Sun with it, while Sun is going East making a full circle around the Zodiac per year.
This would add the further quirk that it is NOT the movement of the Sun which corresponds to what Earth would have been doing if Sun were centre of its system. And yet the correspondence is somehow there.
Now, one can ask oneself, if Geocentrism in Newtonianism is so improbable, why not ditch Geocentrism?
Well, is Heliocentrism in Newtonian explanations very much more probable? If it is sufficiently probable to function without God, maybe there is no God. Then - without God that is - Geocentrism is pretty certain to be impossible.
Could now the Heliocentric and Newtonian system function without God regulating it? Newton did not think so. Laplace is famus for having said he found this hypothesis superfluous. However his cosmology is famous for being less easy to read and check up on (these days at least) than his very famous cosmogony, in which the solar system is supposed to have began as a whirling disc of gas and matter, which then condensed into the planets. That cosmogony (and its prequel in Big Bang) I knew by age eight, before being a Christian, and having momentarily forgot about God. I learnt all this in MARS** where I was the most junior member.
But is this very probable? Look at this:
[ISS] Don Petit, Science Off The Sphere - Water Droplets Orbiting Charged Knitting Needle
SpaceVids.tv·
How many orbits did you count per droplet before the equilibrium of inward and sideward movements changed from orbitting to frankly inward movement and droplet attached to the knitting needle? Not billions, right?
Well, I think the mechanics for movements of heavenly bodies must include, not just a design going on from creation, but also present movers, at least as regulating the movements and stopping collapse by gravitation. And if there are movers capable of that, they are also capable of effecting the quirks - which are very decorous - of Tychonian Geocentrism.
Hans-Georg Lundahl
Bpi, Georges Pompidou
St Marc / Friday of Easter Octave
25-IV-2014
Also 59th Birthday of
Robert Sungenis
* Alec's attack here cited is on:
“Here Comes the Sun”
How the new geocentrists persist in scientific and logical errors
by Alec MacAndrew
** MARS-bulletinen / Malmö astronomi- & rymdfartssällskap
Malmö astronomi- & rymdfartssällskap (utgivare) Malmö : Malmö astronomi- & rymdfartssällskap (MARS), 1966-1980 Svenska.
http://libris.kb.se/bib/1548270
M A R S == M=Malmö, the location, third Swedish city after Stockholm and Gothenburg, A=astronomi- needs no translation, S=-sällskap = society, but what about R=rymdfarts-? Well, it seems the society was originally also about space voyages. Even though Malmö hardly is Cape Canaveral. Did not quite reckon on that in 1966 when the society was founded. I was an active member 1976, after my grandpa's death. By then I was eight. But 1980 when I came back to Malmö, not only was I a Christian but the society was no longer (or not very much longer) in existance. Its traces now is this bulletin preserved in libraries. | 3,468 | 15,783 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.625 | 3 | CC-MAIN-2018-30 | latest | en | 0.976886 |
http://www.gurufocus.com/term/Gross+Profit/AEP/Gross%2BProfit/American%2BElectric%2BPower%2BCo%2BInc | 1,492,926,305,000,000,000 | text/html | crawl-data/CC-MAIN-2017-17/segments/1492917118477.15/warc/CC-MAIN-20170423031158-00227-ip-10-145-167-34.ec2.internal.warc.gz | 542,172,265 | 28,807 | Switch to:
GuruFocus has detected 10 Warning Signs with American Electric Power Co Inc \$AEP.
More than 500,000 people have already joined GuruFocus to track the stocks they follow and exchange investment ideas.
American Electric Power Co Inc (NYSE:AEP)
Gross Profit
\$10,650 Mil (TTM As of Dec. 2016)
American Electric Power Co Inc's gross profit for the three months ended in Dec. 2016 was \$2,431 Mil. American Electric Power Co Inc's gross profit for the trailing twelve months (TTM) ended in Dec. 2016 was \$10,650 Mil.
Gross Margin is calculated as gross profit divided by its revenue. American Electric Power Co Inc's gross profit for the three months ended in Dec. 2016 was \$2,431 Mil. American Electric Power Co Inc's revenue for the three months ended in Dec. 2016 was \$3,790 Mil. Therefore, American Electric Power Co Inc's Gross Margin for the quarter that ended in Dec. 2016 was 64.13%.
American Electric Power Co Inc had a gross margin of 64.13% for the quarter that ended in Dec. 2016 => Durable competitive advantage
During the past 13 years, the highest Gross Margin of American Electric Power Co Inc was 66.41%. The lowest was 60.15%. And the median was 62.88%.
Definition
Gross Profit is the different between the sale prices and the cost of buying or producing the goods.
American Electric Power Co Inc's Gross Profit for the fiscal year that ended in Dec. 2016 is calculated as
Gross Profit (A: Dec. 2016 ) = Revenue - Cost of Goods Sold = 16380.1 - 5730.3 = 10,650
American Electric Power Co Inc's Gross Profit for the quarter that ended in Dec. 2016 is calculated as
Gross Profit (Q: Dec. 2016 ) = Revenue - Cost of Goods Sold = 3790.1 - 1359.6 = 2,431
American Electric Power Co Inc Gross Profit for the trailing twelve months (TTM) ended in Dec. 2016 was 2637.9 (Mar. 2016 ) + 2583.3 (Jun. 2016 ) + 2998.1 (Sep. 2016 ) + 2430.5 (Dec. 2016 ) = \$10,650 Mil.
Gross Profit is the numerator in the calculation of Gross Margin:
American Electric Power Co Inc's Gross Margin for the quarter that ended in Dec. 2016 is calculated as
Gross Margin (Q: Dec. 2016 ) = Gross Profit (Q: Dec. 2016 ) / Revenue (Q: Dec. 2016 ) = (Revenue - Cost of Goods Sold) / Revenue = 2,431 / 3790.1 = 64.13 %
* All numbers are in millions except for per share data and ratio. All numbers are in their local exchange's currency.
A positive Gross Profit is only the first step for a company to make a net profit. The gross profit needs to be big enough to also cover related labor, equipment, rental, marketing/advertising, research and development and a lot of other costs in selling the products.
Explanation
Warren Buffett believes that firms with excellent long term economics tend to have consistently higher margins.
Durable competitive advantage creates a high Gross Margin because of the freedom to price in excess of cost. Companies can be categorized by their Gross Margin
1. Greater than 40% = Durable competitive advantage
2. Less than 40% = Competition eroding margins
3. Less than 20% = no sustainable competitive advantage
Consistency of Gross Margin is key
American Electric Power Co Inc had a gross margin of 64.13% for the quarter that ended in Dec. 2016 => Durable competitive advantage
Related Terms
Historical Data
* All numbers are in millions except for per share data and ratio. All numbers are in their local exchange's currency.
American Electric Power Co Inc Annual Data
Dec07 Dec08 Dec09 Dec10 Dec11 Dec12 Dec13 Dec14 Dec15 Dec16 Gross_Profit 8,413 8,685 8,958 9,398 9,504 9,665 9,254 10,021 10,345 10,650
American Electric Power Co Inc Quarterly Data
Sep14 Dec14 Mar15 Jun15 Sep15 Dec15 Mar16 Jun16 Sep16 Dec16 Gross_Profit 2,632 2,313 2,791 2,471 2,745 2,339 2,638 2,583 2,998 2,431
Get WordPress Plugins for easy affiliate links on Stock Tickers and Guru Names | Earn affiliate commissions by embedding GuruFocus Charts
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https://community.splunk.com/t5/Splunk-Search/Charting-Graphing-Multiple-data-sources-one-chart/td-p/26424 | 1,721,225,531,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514771.72/warc/CC-MAIN-20240717120911-20240717150911-00428.warc.gz | 162,141,324 | 39,453 | Splunk Search
## Charting / Graphing Multiple data sources one chart
Explorer
has anyone experimented with showing statistics for the same time slot over multiple time periods ?
e.g. imagine a chart that shows the number of transaction over a 24 hour period as a line graph. now imagine a second line (of a different color) that has the 24 hour transaction from yesterday, and another that has a weekly average for 24 hour transaction.
i appreciate any insights.
Tags (3)
1 Solution
Splunk Employee
Comparing week-over-week results used to a pain in Splunk, with complex date calculations. No more. Now there is a better way.
I wrote a convenient search command called "timewrap" that does it all, for arbitrary time periods.
``````... | timechart count span=1d | timewrap w
``````
That's it!
http://apps.splunk.com/app/1645/
Splunk Employee
Comparing week-over-week results used to a pain in Splunk, with complex date calculations. No more. Now there is a better way.
I wrote a convenient search command called "timewrap" that does it all, for arbitrary time periods.
``````... | timechart count span=1d | timewrap w
``````
That's it!
http://apps.splunk.com/app/1645/
Explorer
this is very insightful. there are lots of neat 'splunk moves' here that i didn't know existed. i appreciate you taking the time to write this up.
Splunk Employee
For the simple day over day case, the recipe for this solution is pretty well covered in:
To add a series for the weekly average, for each of the previous seven days at each time bucket, the search is significantly more involved. To do this, the first task is to `| append []` some results that are the computation of the average. The details of the inner search depend on what type of aggregate function that you're using, in this case, you're looking at count, so it's not too bad:
``````... | append [search earliest=-7d@d latest=@d ... | eval _time = ((_time - relative_time(now(), "-7d@d")) % 86400) + relative_time(now(), "@d") | bin _time span=15m | chart count by _time | eval metric = count/7 | eval marker = "weekly average"]
``````
Now you just have to glue this right before the final `| chart` in the day_over_day search in the referenced post, and you should have your answer.
Putting it all together:
``````<data> earliest=-1d@d latest=@h
| timechart span=15m count as metric
| eval marker = if(_time < info_min_time + 86400, "yesterday", "today")
| eval _time = if(_time < info_min_time + 86400, _time + 86400, _time)
| eval _time = ((_time - relative_time(now(), "-7d@d")) % 86400) + relative_time(now(), "@d")
| bin _time span=15m
| chart count by _time
| eval metric = count/7
| eval marker = "weekly average"]
| chart median(metric) by _time marker
``````
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https://geo.libretexts.org/Under_Construction/Book%3A_Controversies_in_the_Earth_Sciences_(Richardson)/05%3A_Why_did_it_take_so_long_to_convince%3F/5.9%3A_Specific_Heat_Capacity_of_Water | 1,600,928,585,000,000,000 | text/html | crawl-data/CC-MAIN-2020-40/segments/1600400213454.52/warc/CC-MAIN-20200924034208-20200924064208-00207.warc.gz | 420,208,113 | 22,903 | # 5.9: Specific Heat Capacity of Water
The world's climate system does not respond fully and immediately to externally applied forcing. This lag time has allowed the debate about the effect of human activities on climate to go on longer than it would have if everything we did were instantly noticeable. For example, when a certain amount of extra CO2 is added to the atmosphere, the warming effects on the air, land, and ocean take some time to be fully realized. Even if there was a way to add heat directly to the Earth, it would still take time for the temperatures we measure to increase. In this activity, we'll conduct a simple experiment to observe the specific heat capacity of water. By doing so, we'll be able to gain some insight about the lag time of the climate system's response to external forcing.
The specific heat capacity (Cp) of liquid water at room temperature and pressure is approximately 4.2 J/g°C. This means it takes 4.2 joules of energy to raise 1 gram (or 1 milliliter if you'd rather think of the equivalent volume of 1 gram of water) of water by 1 degree Celsius. This is actually quite large. The specific heat capacity of water vapor at room temperature is also higher than most other materials. Here is a table of the specific heat capacities of various materials:
Specific heat capacity of an assortment of Earth materials
Material \mathbf{C}_{\mathrm{p}}\left(\mathrm{J} / \mathrm{g}^{\circ} \mathrm{C}\right)
liquid water 4.2
air 1.0
water vapor 1.9
granite 0.8
wood 1.7
iron 0.0005
Note that none of the other materials listed above come close to water's ability to absorb heat. (Nitpicker alert: Water does not have the highest known heat capacity. The heat capacity of pure hydrogen gas at room temperature is 14.3 J/g°C, according to the CRC Handbook of Chemistry and Physics. Pure H2 is not a big player in the Earth's climate system, though.)
The high Cp of water is why "a watched pot never boils!" This is also the main reason the climate is slow to respond to external changes. It is lucky for us that the ocean has the ability to absorb a lot of heat before its temperature rises appreciably. The flip side of this is that once an external source of energy is removed, the ocean is similarly slow to respond. Its temperature will not begin to decrease right away. In the next activity, we will observe this phenomenon.
### Optional Fun Lab!: Specific Heat Capacity of Water Lab Experiment
The world climate does not respond immediately to the external forcings applied by human activity. A simple way to show why this is so is to make some simple observations about the heat capacity of water.
#### Materials
These are the materials you need for this lab: water, a pot, a thermometer, a stove or other heat source, a watch or other timer.
#### Directions
1. Measure a volume of water (you can choose the volume) into your pot.
2. Measure the starting temperature of the water.
3. Put the pot on the stove and turn on the stove (you can choose how high to turn it up, but keep the level constant)
4. Measure the temperature of the water at regular intervals. It is up to you to decide how often you need to make measurements.
5. When the water boils, note the time, and remove the pot from the heat.
6. Continue to make regular measurements of the temperature of the water until it cools back down to the initial temperature from step 2.
7. Now pick an experimental variable, such as the initial volume of water in the pot, the kind of pot, how high to turn on the stove. Change this variable and repeat the experiment. Change this same variable at least one more time and do the experiment again.
1. Plot your data for each experiment.
2. How long did it take each pot of water to boil?
3. How long did it take each one to cool back down?
4. Eyeball a best fit function to the water-heating-up data for each experiment. What do these functions look like (linear? curved? could you write an equation down to describe them?) What is indicated by the shapes of these functions? Were they affected by the changes you made between experiments?
5. Eyeball a best fit function to the water-cooling-down data for each experiment. What do these functions look like? Are they all similar to each other? Do they have the same form as the water-heating-up functions? Discuss why or why not.
6. Describe which experimental variable you changed and how the change you made affected your results.
7. Speculate about what would have happened if you had chosen a different experimental variable to change. Predict how changing that other experimental variable would have changed your results.
8. How could you improve this experimental design?
Note:
This is optional, so there is nothing to turn in for this assignment, but this is a fairly simple and instructive experiment, so give it a try if you get a chance!
See my plot below of two experiments in which I varied the initial volume of water in the pot. Note how much time it takes for the water to return to the starting temperature after boiling!
Figure 5.9: Boiling and Cooling of Water
Click for a text description of Figure 5.9.
Time vs. temperature for 2 cups (blue) and 4 cups (green) of water in an open saucepan on a gas stove. During each experiment, the stove was turned on to the "medium" setting and left there until the water boiled, at which point the stove was turned off for the duration of the experiment. | 1,209 | 5,426 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.84375 | 3 | CC-MAIN-2020-40 | latest | en | 0.938599 |
http://steamcommunity.com/app/204240/discussions/0/648814844061025896/?l=swedish | 1,516,499,584,000,000,000 | text/html | crawl-data/CC-MAIN-2018-05/segments/1516084889798.67/warc/CC-MAIN-20180121001412-20180121021412-00298.warc.gz | 323,389,711 | 13,094 | The Bridge
### The Bridge
Visa statistik:
The Bridge > Allmänna diskussioner > Ämnesdetaljer
Urthman 29 dec, 2013 @ 20:28
What's going on with the veil?
The thing that I find unsatisfying about this game is that it's possible to solve most of the puzzles (so far) without really understanding them, just by brute-force randomly wandering around.
I'm only on level 4 and this is the worst so far. I've solved the first 3 puzzles and I still have no idea what's going on when you enter the veil. Is it just that you can rotate the level without the guy moving anywhere? Some of the objects are shaded to look like the veil but I can't see how entering the veil affects them, if it does.
The artwork gets hard to parse in this level. It seems like the game really needs color or some other effect if it wants to have so many different kinds of shading (grey/white/veiled) to distinguish objects.
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Jeratol 29 dec, 2013 @ 23:01
It seems you understand the most basic function of the veil with your question. You can rotate the map without your character moving. But the gravitional relationship between yourself and the "veiled" objects is maintained in the same state that you left it while in the veil.
So for example, when you first start on a map where you (Y↓) and the "veiled" object (O↓) are oriented in a normal manner and observing gravitional pulls in the same, default direction. If you step into the veil and rotate the map clockwise 90°, as you noticed, the object and map will move while you are unaffected. At this point the gravity states change to Y↓ and O→. The moment you exit the veil, that gravitational relationship will stay constant until you change it in the veil again. So after exiting the veil, if you decide to rotate the map 90° counter-clockwise, from a relatively-neutral map perspective, you would look like this: Y← and O↓.
I myself didn't understand the mechanics in the fourth chapter and just rotated in the veil until I came upon the solution, but toward the end of the chapter, you'll need it, or it will take some major plugging-and-entering before you get it.
Sorry if this explanation seems too convoluted or non-sensical, but it was the only way I could think to explain it.
Well anyway, I've blabbed enough. Good luck with the game.
Urthman 30 dec, 2013 @ 0:08
Thanks. That helps.
Emrys 31 dec, 2013 @ 18:05
When you enter the veil, you will notice your "spirit" leaving your body and travel into another object - this is usually the splotchy-shaded object in the level, like a menace or a sliding floor/wall. When you are in the veil, you can change the direction that gravity is pointing for just that veil-linked object, without affecting your body's gravity.
This means that some levels can have three different gravity points to keep track of: regular (your body), inverse (opposite your body gravity, when you turn white), and the veiled object. Often levels will have a gravity indicator for these three points of gravity, like hanging lamps or chains or rolling balls, and color-coded to match (so the white shaded indicator is for your inverse gravity and so on). When you change the veil-linked object's gravity, you will notice that the indicator changes too, so that is one way to keep track of your directions.
Gargish 4 jan, 2014 @ 2:58
I just considered the veil as a switch to change the direction of gravity for certain objects. The main indicator of the direction of gravity are the moving lines all over the screen and the direction the speckles in the objects are moving - this helps when you later have two objects with different gravity directions.
DaneSoul 8 jan, 2014 @ 12:32
Ursprungligen skrivet av Emrys:
When you enter the veil, you will notice your "spirit" leaving your body and travel into another object - this is usually the splotchy-shaded object in the level, like a menace or a sliding floor/wall. When you are in the veil, you can change the direction that gravity is pointing for just that veil-linked object, without affecting your body's gravity.
Thanks a lot! This was realy helpfull comment, using which I finnally finished IV-V "The Intersection" level!
Alcator 27 feb, 2014 @ 1:55
Hallelujah! Didn't click to me, because the "tutorial" level (IV-I) merely looked like I somehow controlled the key.
rad_killer 19 mar, 2014 @ 6:58
it is just a gravity-control thing :) Nothing more. Behind the veil, you can control the gravity. Just observe the chandelier how the gravity behaves.
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The Bridge > Allmänna diskussioner > Ämnesdetaljer
Datum skrivet: 29 dec, 2013 @ 20:28
Inlägg: 7
Fler diskussioner | 1,149 | 4,662 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.28125 | 3 | CC-MAIN-2018-05 | latest | en | 0.925793 |
https://www.tutordale.com/how-to-do-math-problems/ | 1,723,101,092,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640723918.41/warc/CC-MAIN-20240808062406-20240808092406-00682.warc.gz | 788,379,845 | 59,602 | Sunday, August 4, 2024
# How To Do Math Problems
## The Case For Doing Hard Things
Google chrome tips and tricks How to do math problems and even have a scientific calculator
We ask hard questions because so many of the problems worth solving in life are hard. If they were easy, someone else would have solved them before you got to them. This is why college classes at top-tier universities have tests on which nearly no one clears 70%, much less gets a perfect score. Theyâre training future researchers, and the whole point of research is to find and answer questions that have never been solved. You canât learn how to do that without fighting with problems you canât solve. If you are consistently getting every problem in a class correct, you shouldnât be too happy â it means you arenât learning efficiently enough. You need to find a harder class.
The problem with not being challenged sufficiently goes well beyond not learning math as quickly as you can. I think a lot of what we do at AoPS is preparing students for challenges well outside mathematics. The same sort of strategies that go into solving very difficult math problems can be used to tackle a great many problems. I believe weâre teaching students how to think, how to approach difficult problems, and that math happens to be the best way to do so for many people.
## A Closer Look At Sat Math Sections
When working on SAT, the third and fourth parts are dedicated to mathematics. The math part has 54 mathematics questions: 10 student-produced response questions and 44 multiple-choice questions.
The ten multiple-choice questions require students to workout math problems and select answers from the choices provided. But the student-produced response questions are some of the most hard math questions because students have to calculate the right answers: there are no answer choices.
Notably, the SAT math questions are drawn from four main mathematics areas: algebra and functions number and operations data analysis, probability, and statistics and geometry and measurements.
Therefore, you better be well prepared in all of the areas if you want to pass and join college.
## C What Do I Already Have
You know the question. You know what you need in order to solve it. Now you can simply fill in the equation with what you have already been given.
Dont get lost in unimportant details. Math word problems are notorious for giving you too many details. Thats why this step is the last of the three questions.
Some students try to figure out what all they have first. They read the problem, write out all the details they have been given, and then expect to solve it from there. Instead, they often experience detail-overload.
But you will save yourself an enormous amount of time if you know you are looking to answer first. Knowing the question is more important than knowing what details you have. Only when you know the question you are answering and what you need in order to answer it can you then find the right details to answer it correctly.
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## How To Easily Make Your Own Math Word Problems & Word Problems Worksheets
Armed with 120 examples to spark ideas, making your own math word problems can engage your students and ensure alignment with lessons. Do:
• Link to Student Interests: By framing your word problems with student interests, youll likely grab attention. For example, if most of your class loves American football, a measurement problem could involve the throwing distance of a famous quarterback.
• Make Questions Topical: Writing a word problem that reflects current events or issues can engage students by giving them a clear, tangible way to apply their knowledge.
• Include Student Names: Naming a questions characters after your students is an easy way make subject matter relatable, helping them work through the problem.
• Be Explicit: Repeating keywords distills the question, helping students focus on the core problem.
Dont:
• Test Reading Comprehension: Flowery word choice and long sentences can hide a questions key elements. Instead, use concise phrasing and grade-level vocabulary.
• Focus on Similar Interests: Framing too many questions with related interests — such as football and basketball — can alienate or disengage some students.
• Feature Red Herrings: Including unnecessary information introduces another problem-solving element, overwhelming many elementary students.
A key to differentiated instruction, word problems that students can relate to and contextualize will capture interest more than generic and abstract ones.
## Solving Equations Using The Division Property
Consider the equation
3x = 12
The solution to this equation is 4. Also, note that if we divide each member of the equation by 3, we obtain the equations
whose solution is also 4. In general, we have the following property, which is sometimes called the division property.
If both members of an equation are divided by the same quantity, the resulting equation is equivalent to the original equation.
In symbols,
Example 1 Write an equation equivalent to
-4x = 12
Solution Dividing both members by -4 yields
In solving equations, we use the above property to produce equivalent equations in which the variable has a coefficient of 1.
Example 2 Solve 3y + 2y = 20.
We first combine like terms to get
5y = 20
Then, dividing each member by 5, we obtain
In the next example, we use the addition-subtraction property and the division property to solve an equation.
Example 3 Solve 4x + 7 = x – 2.
Solution First, we add -x and -7 to each member to get
4x + 7 – x – 7 = x – 2 – x – 1
Next, combining like terms yields
3x = -9
Last, we divide each member by 3 to obtain
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## B What Do I Need In Order To Find The Answer
After you know what youre being asked, you can then think about what it will take to get that answered. You should have some idea at this point of the equation that will be needed to find a solution.
Specifically, were talking about equations here and the most important variables.
If you know what you are looking for and you can then name the pieces you need to find, even the most difficult problems become extremely manageable.
For a simple example, lets say youve been given a question and you realize you are being asked how tall a ladder youll need to paint a wall .
After discovering whats being asked of you the length of the ladder you realize that the Pythagorean Theorem is what youll need to solve it. This means our final question needed to solve this math word problem will be super easy.
## How To Solve Squares
You can re-write a square equation into numbers that are easier to deal with using this formula
`n^2 = + d^2 `
where n is the number to be squared, and d is the difference
Heres an example
`57^2 = + 3^2# we add 3 to 57, as 60 is easier to multiply than 57, and subtract 3 from the second 57-> 60 x 54 + 9 = 3000 + 240 + 9 = 3249`
The ultimate example is when you are squaring a number ending in 5, then round one number up to the nearest 10, the other number down to the nearest 10, and add 25.
`65^2 = + 5^2 = 4200 + 25 = 4225`
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## Best Ways On How To Learn Math
#### Practice, Practice & More Practice
Practice makes a man able to solve the problems with perfection. Most of the students think that they can solve the math problem by just paying attention to the class. What do you think? Is it possible to solve math problems by just reading and listening? Of Course not, because math is not similar to other subjects. You find problems in math, and you have to solve it using some math formulas.
To solve any problems in math, you need to do a lot of practice. The more you practice to solve the math problems, the more you get better. There are lots of ways to solve math problems in different ways. Each math problem has its characteristics. Therefore these problems have different ways to solve it. Thus the more you explore the best ways to solve the problems, the better you perform in math exams. There is no shortcut to solve math problems without doing the proper practices to solve them.
## Using 1000 Subtraction Rule:
How to do Word Problems Involving Math and Percentages
Its not just addition, when it comes to maths, its like everything that is related to mathematical is troubling.
For example, while subtracting a large number from a thousand, keep note that a person need to subtract the first number from nine and second numbers from nine but except last number.
Instead of subtracting the last number from nine, make use of ten while subtracting.
#### Trick:
Suppose the number is 665, now while subtracting it from 1000 make sure that a person use 9 each time, except for the last time and that is 6-9=3Therefore, the answer is 1000-665=335
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## Strategies For Difficult Math Problems And Beyond
Here are a few strategies for dealing with hard problems, and the frustration that comes with them:
Do something. Yeah, the problem is hard. Yeah, you have no idea what to do to solve it. At some point you have to stop staring and start trying stuff. Most of it wonât work. Accept that a lot of your effort will appear to have been wasted. But thereâs a chance that one of your stabs will hit something, and even if it doesnât, the effort may prepare your mind for the winning idea when the time comes.
We started developing an elementary school curriculum months and months before we had the idea that became Beast Academy. Our lead curriculum developer wrote 100â200 pages of content, dreaming up lots of different styles and approaches we might use. Not a one of those pages will be in the final work, but they spurred a great many ideas for content we will use. Perhaps more importantly, it prepared us so that when we finally hit upon the Beast Academy idea, we were confident enough to pursue it.
Simplify the problem. Try smaller numbers and special cases. Remove restrictions. Or add restrictions. Set your sights a little lower, then raise them once you tackle the simpler problem.
Focus on what you havenât used yet. Many problems have a lot of moving parts. Look back at the problem, and the discoveries you have made so far and ask yourself: âWhat havenât I used yet in any constructive way?â The answer to that question is often the key to your next step.
## Probability And Data Relationships Word Problems
89. Understanding the Premise of Probability: John wants to know his classs favourite TV show, so he surveys all of the boys. Will the sample be representative or biased?
90. Understanding Tangible Probability: The faces on a fair number die are labelled 1, 2, 3, 4, 5 and 6. You roll the die 12 times. How many times should you expect to roll a 1?
91. Exploring Complementary Events: The numbers 1 to 50 are in a hat. If the probability of drawing an even number is 25/50, what is the probability of NOT drawing an even number? Express this probability as a fraction.
92. Exploring Experimental Probability: A pizza shop has recently sold 15 pizzas. 5 of those pizzas were pepperoni. Answering with a fraction, what is the experimental probability that he next pizza will be pepperoni?
93. Introducing Data Relationships: Maurita and Felice each take 4 tests. Here are the results of Mauritas 4 tests: 4, 4, 4, 4. Here are the results for 3 of Felices 4 tests: 3, 3, 3. If Mauritas mean for the 4 tests is 1 point higher than Felices, whats the score of Felices 4th test?
94. Introducing Proportional Relationships: Store A is selling 7 pounds of bananas for \$7.00. Store B is selling 3 pounds of bananas for \$6.00. Which store has the better deal?
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## A For What Am I Looking
This is the biggest question. It shapes your entire time answering the question. Take the following situation for example:
A plane leaves Toronto, Ontario , heads to Newark, New Jersey, and then heads to Seattle, Washington. Along the way to Seattle, a storm forces the plane to head north in order to get around it. The plane ends up crossing the Canadian border, but then has some engine trouble. When the plane is at 30,000 feet, an engine fails, and the plane has to attempt an emergency landing. Unfortunately, the plane crashes and it does so directly on the Canada-United States border. Where will they bury the survivors?
Well let you think about that question for a few minutes. You can see the answer at the bottom of this post if youre curious. But I recommend rechecking the most important detail before you guess: For what I am looking?
Dont get lost in details. Get the question right before anything else. You must know what your math word problem is asking.
If you dont know what you are looking for, youll end up missing it every time.
## Develop The Plan To Solve It
There are four simple steps which one needs to go through in order to develop a plan to solve it. The steps are as mentioned below:
• Firstly one needs to figure out the formula you will need to solve the problem. Here you need to spend some time reviewing the concepts in your textbooks which will help you solve the problem
• You need to write down your need in order to get the answer to your problem. For this, you need to make a step-by-step list of the things which you need to solve the problem and also help you to stay organised
• In case there is an easier problem which is available then you could probably work on that first to solve it. Sometimes, the formulas are repetitive for solving both the problems. This will give you some more time to solve the difficult problem
• You can make an educated guess about the answer so that you can try and get the estimate the answer before you start to solve it. Here you can identify the number and other factors as well that will contribute to the same. Lastly, review the estimate and then check if you haven’t left out on anything
## Solve A Series Of Simple Math Problems
Keep breaking it down. You work the math. Dont let the math work you. Victor writes:
This seems like a lot of steps to go through, but the point is that you can quickly simplify the problem so you can solve it easily in your head. Often is is much easier to solve a long series of simple math problems than a short series of complicated ones. You may not need to break down this problem into as many simple parts as I did, but the process of doing so is important when it comes to quantitative assessment tests.
## How To Solve Trigonometric Equations
Trigonometric functions can be used for various purposes, like for planning urban issues. To solve the basic trigonometric problems, you need to follow the below-mentioned steps:
• First of all, understand the trigonometric identities and reference angles.
• Grasp all necessary ratios that lie between the 0 degrees to 360 degrees of a graph.
• Proceed by using the substitution formula.
• Solve the math problems to get the answer in degrees or radiations.
For example:
#### Solve sin + 2 = 3 in interval 0° x < 360°.
sin + 2 = 3
sin = 3 2 = 1
We know that sine 1 lies in the first quadrant that is between 0 degrees to 90 degrees. Moreover, we are well versed that sin = 1. Therefore, the answer will be:
x = 90 degrees.
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## Multiplying Even And Odd Numbers By Five:
Whenever there is a multiplication problem by five, there is one simple rule or trick that can get one the answer without wasting much time.
And that would be while multiplying any number from five, remember to follow the trick.
For example, while multiplying five with an even number, take the even number and make half of it and add zero at the end of the number to get the answer.
And on other hand, with an odd number, deduct one from the odd number and make half of it, then just add five at the side of the number to get the answer
#### Trick:
With an even number, suppose the number is 5*6Take even number 6, half of it is 3Add 0 aside, therefore, the answer is 5*6=30Trick: With an odd number, suppose the number is 5*7Take odd number, deduct 1 from the number 7-1=6Make half of 6, half is 3Add 5 aside, therefore the answer is 5*7=35.
## Round Off Numbers While Adding:
Algebra Shortcut Trick – how to solve equations instantly (2)
Most of the time adding two or more than two numbers seems difficult and to make addition simple a person can consider rounding off those two numbers and try to get the answer.
For example, if there are numbers which is 535 and 346. While looking at the number it is not that hard to solve, but there is a simple way to solve the problem and that is first, round off both the numbers like this
#### Trick:
540 and 350, now apply addition and the answer is 540+350=890. Then, 540-535=5 , 350-346=4 i.e, 5+4=9Now deduct 9 from 890= 890-9=881 and the answer is 881.
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## From Group To Independent Problem
After rereading the above-grade-level problem as a class, each student receives a word problem printout differentiated based on their ability. Students work through a five-step problem-solving procedure based on the reading protocol they use as a class. They work independently, or in small groups if they need more support. A checklist of the steps guides them through the problem. | 3,890 | 17,476 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.09375 | 3 | CC-MAIN-2024-33 | latest | en | 0.959112 |
https://stats.stackexchange.com/questions/435452/why-do-we-use-pcs-instead-of-transformed-data-as-new-variables-in-linear-regress | 1,586,381,229,000,000,000 | text/html | crawl-data/CC-MAIN-2020-16/segments/1585371824409.86/warc/CC-MAIN-20200408202012-20200408232512-00326.warc.gz | 700,739,660 | 31,853 | # Why do we use PCs instead of transformed data as new variables in linear regression?
PCA and SVD
Using the SVD on matrix X (column as features)
We have X = U\sigmaV* where V contains the PCs and U\sigma would be the transformed data
PCA with linear regression
The application of PCA with linear regression is called Principal component regression (PCR), in which we use PCs as new predictors. With gene expression data, those PCs would be called eigengenes which could represent major expression patterns of genes with similar expression patterns across samples.(PCR has its own problem that PCs with low variation could have a significant effect on the response)
My question is: Why don't we use U\sigma, the transformed data as predictors? In what situations, could we use transformed data as predictors? What do those predictors mean?
• U instead of V contains the PCs for X'X – purod Nov 11 '19 at 3:25
If X is n by p matrix, then the absolutely standard interpetation is that one has n observations of p variables. Therefore $$\Sigma V^*$$ is also a nxp matrix, that is what one has done is created new variables that are linear combinations of the previous variables. That makes good sense. If I have two variables x,y, then $$x+y$$ and $$x-y$$ also make sense as variables. What you suggest $$U\Sigma$$, is taking linear combinations of the observations. One can do it, but for what purpose. The point of SVD is to find a new basis for the set of variables one is observing that is 1) Orthogonal and 2) listed in 'order of importance' as measured by the diagonal entries of $$\Sigma$$. The usual reason for doing so is to throw away variables that are less important. In many situations it may turn out to be the case that a small number of variables in the new basis capture most of the information in the data as measured by variance. Two examples of this come to mind right away. mp3 encoding of music is exactly that. Variables are frequencies measured and the observations are the values at various points in time. Similarly if one represents a picture (black and white for simplicity), the the picture can be thought of a matrix where each pixel is a matrix entry between 0 and 1 where 0 = white and 1 = black. Then pca gives one a way to capture the picture with less information. Google is your friend for more details on either topic. Oh and a third. If one has time series data of various stocks, the doing a pca will almost always produce a situation in which the first principle component is about 60%-80% of the variability and 3-5 components will capture 95% or more of the variance. Incidentally the first component is easily seen to be market direction. | 590 | 2,686 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 5, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.203125 | 3 | CC-MAIN-2020-16 | latest | en | 0.954197 |
https://www.convertunits.com/from/terabit/second/to/tebibit/second | 1,657,038,628,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656104585887.84/warc/CC-MAIN-20220705144321-20220705174321-00181.warc.gz | 755,083,737 | 16,914 | ## ››Convert terabit/second to tebibit/second
terabit/second tebibit/second
## ››More information from the unit converter
How many terabit/second in 1 tebibit/second? The answer is 1.099511627776.
We assume you are converting between terabit/second and tebibit/second.
You can view more details on each measurement unit:
terabit/second or tebibit/second
The main non-SI unit for computer data rate is the bit/second.
1 bit/second is equal to 1.0E-12 terabit/second, or 9.0949470177293E-13 tebibit/second.
Note that rounding errors may occur, so always check the results.
Use this page to learn how to convert between terabits/second and tebibits/second.
Type in your own numbers in the form to convert the units!
## ››Quick conversion chart of terabit/second to tebibit/second
1 terabit/second to tebibit/second = 0.90949 tebibit/second
5 terabit/second to tebibit/second = 4.54747 tebibit/second
10 terabit/second to tebibit/second = 9.09495 tebibit/second
20 terabit/second to tebibit/second = 18.18989 tebibit/second
30 terabit/second to tebibit/second = 27.28484 tebibit/second
40 terabit/second to tebibit/second = 36.37979 tebibit/second
50 terabit/second to tebibit/second = 45.47474 tebibit/second
75 terabit/second to tebibit/second = 68.2121 tebibit/second
100 terabit/second to tebibit/second = 90.94947 tebibit/second
## ››Want other units?
You can do the reverse unit conversion from tebibit/second to terabit/second, or enter any two units below:
## Enter two units to convert
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## ››Metric conversions and more
ConvertUnits.com provides an online conversion calculator for all types of measurement units. You can find metric conversion tables for SI units, as well as English units, currency, and other data. Type in unit symbols, abbreviations, or full names for units of length, area, mass, pressure, and other types. Examples include mm, inch, 100 kg, US fluid ounce, 6'3", 10 stone 4, cubic cm, metres squared, grams, moles, feet per second, and many more! | 584 | 2,004 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.59375 | 3 | CC-MAIN-2022-27 | latest | en | 0.819408 |
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Two congruent right triangles are formed, since the tangent line is perpendicular to the radius.
So line AB has the same length as line BC.
∠ABC Angle x° = 180° - Angle y°, or
∠ADC Angle y° = 180° - Angle x°
Formula:
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sin (x/2) = DA / DB
cos (x/2) = AB / DB
tan (x/2) = DA / AB
Tangent lines to circles. In Euclidean plane geometry, a tangent line to a circle is a line that touches the circle at exactly one point, never entering the circle's interior. Tangent lines to circles form the subject of several theorems, and play an important role in many geometrical constructions and proofs.
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Thanks for your support and do visit nitrio.com for more apps for your iOS devices. | 283 | 1,142 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.921875 | 3 | CC-MAIN-2021-21 | latest | en | 0.897728 |
http://yakasee.com/C3Av0n550/dA30529cU/ | 1,566,042,226,000,000,000 | text/html | crawl-data/CC-MAIN-2019-35/segments/1566027312128.3/warc/CC-MAIN-20190817102624-20190817124624-00117.warc.gz | 350,814,313 | 10,000 | # Worksheet Fraction Multiplication Worksheets Fun Multiplying And Dividing Fractions Grade Improper Pin Equivalent More Fourth
Published at Saturday, 11 May 2019. Color by Number. By .
Color by number printable build fine motor skills, For younger students, coloring “between the lines” helps to build fine motor skills in the hands and fingers. It helps with children’s dexterity, hand-eye coordination, and skills with manipulating tools. This is not as much of a benefit for older kids, but it is definitely something the younger kids should practice. This will eventually lead to future skills with writing letters and writing in cursive. Color by number printable introduce students to colors, Color by number pages are a good way to teach color theory to children. Initial color by number pages can introduce colors like red, yellow, blue, green – the primary colors and secondary colors. Later color by number pages can introduce more complicated colors like magenta, cerulean, and amber. Color by number pages can help introduce kids to fine arts.
If you don’t want to work with the brushes paint-by-number kits come with, you may want to buy your own at a craft store. In particular, you may want to have a larger brush on hand, not just the very small ones offered in the kits. Some canvases come wrinkled. Spray a light mist of water on the canvas and iron on a low setting on the back side to press out any wrinkles before you begin painting. Paint with just one color at a time, beginning with the largest areas meant for the color. “You’ll notice some shapes have two numbers in them, not just one,” stated ThoughtCo. “This indicates that you need to mix two colors together. Equal proportions should give you a suitable color, but don’t dip your brush from one paint container into the next as you’ll contaminate the colors.”
Color by number printable can be differentiated for different students, Color by number pages do not have to be “integers only” pages. We like to use color by number pages to teach mathematics. Instead of giving a number for a space to color, we might give an addition problem: 2 + 3 =. Then, at the bottom of the page, we might say, “5 = purple“. Kids would solve the problem, then color that space in depending on the color given for the answer. This gets kids practicing their skills with mathematics – addition, subtraction, multiplication, division, working with shapes – while completing a fun art project in the classroom. It’s a best of both worlds scenario.
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#### Multiplying And Dividing Fractions Worksheets Grade 8 Gallery
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http://mathforum.org/library/view/19832.html | 1,498,178,667,000,000,000 | text/html | crawl-data/CC-MAIN-2017-26/segments/1498128319933.33/warc/CC-MAIN-20170622234435-20170623014435-00223.warc.gz | 246,923,094 | 2,524 | Constructions with ruler and compass
Visit this site: http://www.zef-damen.myweb.nl/Constructions/Constructions_en.htm Author: Zef Damen Description: A page that answers the question, What are ruler-and-compass constructions? With links to step-by-step drawings for constructions: bisecting a given angle; constructing: a line perpendicular to a given line that divides it into two equal parts; a line perpendicular to a given line passing through a given point on the line (or not on the line); the horizontal and vertical centerlines of a given circle; an equilateral triangle, given one side; an equilateral triangle inscribed in a given circle; a pentagon inscribed in a given circle; a hexagon, given one side, or inscribed in a given circle; a heptagon. Levels: Middle School (6-8), High School (9-12) Languages: English Math Topics: Constructions
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# MTH304 Statics Assignment No 02 Fall 2019 Solution & Discussion Due Date: 10-12-2019
MTH304 Statics Assignment No 02 Fall 2019 Solution & Discussion Due Date: 10-12-2019
MTH304 Statics Assignment 2 Solution & Discussion Fall 2019
Assignment # 2 MTH304 (Fall 2019)
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Question No. 1:
Three coplanar forces of magnitudes FN, 12N and 10N act at a fixed point O and are in equilibrium. The directions of the forces are shown in the diagram. Find the values of and F. MARKS 10
Question No. 2:
A particle of mass 5kg is suspended by two strings. If these strings make angles of 30o and 45o with the horizontal, find the tensions in each string. MARKS 10
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# Problem 753. Solitaire Cipher
Solution 1700117
Submitted on 28 Dec 2018 by Roche de Guzman
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
deck = 1:28; n = 10; out = [3 22 9 4 23 7 25 16 14 14]; assert(isequal(solitaire(deck, n),out))
2 Pass
deck = 1:28; n = 9; out = [3 22 9 4 23 7 25 16 14]; assert(isequal(solitaire(deck, n),out)) | 168 | 466 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.53125 | 3 | CC-MAIN-2020-24 | latest | en | 0.668307 |
https://stats.stackexchange.com/questions/101221/why-use-students-t-distribution-rather-than-students-z-distribution | 1,713,961,871,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296819273.90/warc/CC-MAIN-20240424112049-20240424142049-00789.warc.gz | 483,207,905 | 44,347 | # Why use Student's t distribution rather than Student's z distribution
So suppose I'm trying to decide whether to reject the null hypothesis that the mean of some random variable $X$ is zero based on the sample mean $\hat{\mu}$ computed from $n$ IID draws of this random variable. Then normally I would do a $t$ test $$t=\frac{\hat{\mu}}{\hat{\sigma}\;/ \sqrt{n}}$$
and check to see what the probability is that one would draw a more extreme value than $t$ from the $t$ distribution.
But why even bother scaling by $\sqrt{n}$? Why not just use Student's z-distribution? After all isn't that the true distribution for the sample mean $\hat{\mu}=\frac 1n\sum_{i=1}^nX_i$? The only reason I can see to bother is so that for large $n$ Student's t distribution approximates the standard normal distribution, which might simplify computation, is this the only reason it's done?
• good question. you might want to check out "on the transition from student's z to student's t" eisenhart 1979
– jsk
Jun 5, 2014 at 4:16
• (1) What's the variance of $X_i$? (2) "Why not just use the student's z-distribution? After all isn't that the true distribution for the sample mean?" -- what is this belief based on? Jun 5, 2014 at 10:41
• As a small matter of form, Student was the pseudonym used by W.S. Gosset and so should be capitalised like any surname or family name. (Edited accordingly.) Aug 14, 2014 at 7:53
Generally, when you do a statistical test, you want the null distribution of the statistic to have a stable form, so as to allow easy computation of P-values.
This is partly a historical consideration. In principle, there's no reason you couldn't use the unscaled statistic $\hat{\mu}/\hat{\sigma}$, as opposed to $\hat{\mu}/(\frac{\hat{\sigma}}{\sqrt{n}})$, and compare to the percentage points of a distribution that became successively narrower with sample size. But this would be very cumbersome if you didn't have modern computers to calculate those percentage points for you. Even today, when we do have computers that could do this calculation, having a relatively stable null distribution makes it easier to compare results from different studies involving different sample sizes. A t-statistic of 2.5 is easier to comprehend because it means roughly the same thing whether you have a sample of 100 or 100,000. You can't say the same about the unscaled z-statistic.
• Nice answer. You have not mentionned that using either $t$ or $z$ is strictly equivalent because it is obvious. I would add it because (oddly) it is not clear for everyone, as shown by the upvotes to the first answer. Aug 15, 2014 at 8:02
• This holds when $\mu$=0. Otherwise to get a central t you need to subtract $\mu$. Jun 1, 2017 at 5:10
• You don't get the t distribution with n-1 df if you don't divide by the square root of n. Jun 1, 2017 at 5:23
• In the test the value of $\mu$ used is the value under the null hypothesis. Jun 1, 2017 at 5:24
UPDATE: Added material to respond to some issues that arose in the comments.
As stated by the reference the OP pointed to, $\hat \mu$ does not follow Student's z-distribution - $(\hat \mu -\mu)/s$ does. But when these distributional results hold in finite samples? Only when the $X$'s are normally distributed.
So you have to wonder: Under non-normality the usual t-statistic still converges to a standard normal by the Central Limit Theorem, because under the null it is a sum of random variables centered, and scaled by their standard deviation. Technically, it is the term $\sqrt n$ that prevents it for converging to zero.
So where does the statistic you propose, $(\hat \mu -\mu)/s$, converges to? Isn't it zero? Irrespective of the true value of $\mu$?
So what good is it for a statistic in non-normal samples if it gradually "loses" its finite sample distribution and tends to the same constant?
One can search and verify that if $Z_n$ follows a Student's- $z$ distribution with $n$ degrees of freedom, and $T_{n-1}$ follows a Student's- $t$ distribution with $n-1$ degrees of freedom, then the following relation holds between them:
$$T_{n-1} =Z_{n} \sqrt {n-1} \Rightarrow Z_{n}=\frac 1{\sqrt {n-1}}T_{n-1}$$
Assume that we have a size-$n$ sample of i.i.d. random variables $\{X_i\}$ following $N(\mu, \sigma^2)$, both parameters unknown. Setting
$$\bar X_n = \frac 1n\sum_{i=1}^nX_i,\;\;\; s^2 = \frac 1{n-1}\sum_{i=1}^n(X_i-\bar X_n)^2$$
then we know that the random variable
$$\sqrt n\frac {\bar X_n -\mu}{s} \sim_{\text{finite-sample}} T_{n-1}$$
and also that
$$T_{n-1} \sim_{\text{asympt}} N(0,1)$$
The important thing to note here is that $T_{n-1}$, has a) a finite sample distribution and b) an asymptotic distribution. This means that $T_{n-1}$ remains a non-trivial random variable even asymptotically.
Let's now turn to the $Z_n$ random variable. By previous results we have
$$Z_n = \frac 1{\sqrt {n-1}} \left({\sqrt n}\cdot\frac {\bar X_n -\mu}{s}\right) \sim_{\text{finite-sample}}\frac 1{\sqrt {n-1}}T_{n-1}$$
and this quantity follows a Student's-$z$ distribution. What is its asymptotic distribution? It is obvious that $Z_n$ converges to zero, since
$$\lim_{n\rightarrow \infty} \frac 1{\sqrt {n-1}} = 0,\;\; T_{n-1} \xrightarrow{d}N(0,1)\;\; \Rightarrow Z_n \rightarrow 0$$
What does that tell us? That while $Z_n$ has a finite-sample distribution, asymptotically it collapses to the constant $0$, i.e. it does not remain a non-trivial random variable asymptotically.
So, to respond to a comment, these two statistics are not asymptotically equivalent -the one remains a random variable with a distribution, the other collapses to a constant. Therefore the tests based on $T_{n-1}$ are asymptotically valid while the tests based on $Z_n$ are not.
Should we care? After all, our samples are always of finite size, why should we care what happens asymptotically?
Well, from the above it is obvious that while the distribution of $T_{n-1}$ "stabilizes" as the sample size increases (due to the existence of an asymptotic distribution), the distribution of $Z_n$ does not, as it gradually collapses to a constant. And we do not know how does this affects the shape of the collapsing distribution, and the related probabilities.
This means that, in order for our finite-sample test using $Z_n$ to be valid, for each $n$ we have to calculate the probabilities-critical values that interests us... Amusingly, the OP asked
"why even bother scaling by $\sqrt n$?"
"if you don't bother scaling by $\sqrt n$,then you will have to very much bother computing the CDF of $Z_n$ for the specific $n$"
Which approach is more time-efficient?
Now, when we move from normal to non-normal samples, the centered and scaled sample mean does not follow Student's-$t$ in the first place, and we are left only with the asymptotically valid normality result which now comes from the Central Limit Theorem (and not through Student's -$t$). In this case, using $Z_n$ even for small samples sizes introduces an unknown approximation error, and renders the statistical test results questionable.
PS: Historically, W.S.Gosset derived first Student's -$z$ distribution (since he was interested in very small sample sizes), and laboriously calculated critical values for $n\leq 10$. It was through his collaboration with R.A. Fisher that they turned from it to the Student's -$t$.
• I wonder why the downvote? Could the person who downvoted provide some suggestions for improvement at least? Or point to possible mistakes in this answer? Aug 14, 2014 at 6:36
• @COOLSerdash Thank you for asking the question I would also ask. Since I cannot find any mistake in the answer, perhaps the downvoter considered as negative that I have not provided mathematical formulas and proofs, but only verbal conversation... Apparently (s)he has no interest in communicating the reasons for the downvote, so I guess we will never know. Aug 14, 2014 at 14:31
• @AlecosPapadopoulos The tests are equivalent : one rejects the test based on $t$ if and only if one rejects the test based on $z$. Hence if one is asymptotically valid, the other one is too. Aug 14, 2014 at 20:33
• @AlecosPapadopoulos Sorry I was not the one who downvoted but this time I do, because you're claiming there are some differences between two equivalent tests. Aug 15, 2014 at 6:32
• @AlecosPapadopoulos one has the normal approximation $t_n \approx {\cal N}(0,1)$ equivalent to the normal approximation $z_n \approx {\cal N}(0,1/\sqrt{n-1})$. Try to illustrate using simulations what you are claiming. You can't, because the tests are the same. Aug 15, 2014 at 6:38
If I correctly understand, you ask why one uses $$t=\frac{\hat{\mu}}{\hat{\sigma}\;/ \sqrt{n}}$$ and not $$z= t\;/ \sqrt{n-1}.$$ Using the test based on $t$, one rejects the null hypothesis when $|t|>t^*$ where $t^*$ is a quantile chosen in order that the test achieves the nominal level.
Using the test based on $z$, one would similarly reject the null hypothesis when $|z|>z^*$ where the quantile $z^*$ is simply given by $z^*= t^*\;/ \sqrt{n-1}$.
Thus, using either $t$ or $z$ provides exactly the same test: the conclusion drawn from the test based on $t$ is the same as the one drawn from the test based on $z$. A hypothesis test is nothing but a rejection rule, the rejection rules are the same, hence the tests are exactly the same. This is obvious, but I emphasize this point because the upvoted answer to the OP claims there are some differences !
One advantage of using $t$ is that the values of $t^*$ are more convenient than those of $z^*$. The approximation $t \sim {\cal N}(0,1)$ as $n \to \infty$ is well-known, and it is well known how $t$ "looks like" ${\cal N}(0,1)$ for $n$ growing from $1$ to $\infty$ and in particular we know that $t^*$ decreases as $n$ decreases. For high values of $n$ the values of $t^*$ are close to the quantile of ${\cal N}(0,1)$, and it is not hard to learn by heart the (rounded) values of $t^*$ for small values of $n$.
In other words, the table of the $z$-distribution, giving the $z^*$, would be less digestible than the table of $t$-distribution.
• Can the downvoter explains its downvote ? Aug 15, 2014 at 12:41 | 2,734 | 10,147 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.265625 | 3 | CC-MAIN-2024-18 | latest | en | 0.944359 |
http://www.osti.gov/eprints/topicpages/documents/record/290/1548420.html | 1,406,618,842,000,000,000 | text/html | crawl-data/CC-MAIN-2014-23/segments/1406510266597.23/warc/CC-MAIN-20140728011746-00453-ip-10-146-231-18.ec2.internal.warc.gz | 741,768,552 | 3,079 | Math 501. 6th Homework. Due Friday, November 30, 2007. Homework on "Common discrete and continuous distributions". Summary: Math 501. 6th Homework. Due Friday, November 30, 2007. Homework on "Common discrete and continuous distributions". Name: ............................................................................................................................................... 1. Let {an} n=1 be a sequence of real numbers. Suppose that there exists R > 0 such that n=0 |an|Rn < . Then, (i) For each |x| R, f(x) = n=0 anxn exist. (ii) For each |x| < R, g(x) = n=1 annxn-1 exist. (iii) For each |x| < R, f (x) = g(x). (iv) For each n 0, an = f(n)(0) n! 2. Let 0 < p 1. Let Y be a r.v. Suppose that: (i) P[Y N] = 1. (ii) For each k, n N, Collections: Mathematics | 236 | 773 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.34375 | 3 | CC-MAIN-2014-23 | longest | en | 0.731592 |
https://mathleadershipcorps.com/question/square-1-square-0-how-many-zeroes-will-be-20430425-47/ | 1,638,058,234,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964358323.91/warc/CC-MAIN-20211127223710-20211128013710-00338.warc.gz | 458,269,691 | 14,057 | ## x square + 1 square =0 how many zeroes will be
Question
x square + 1 square =0 how many zeroes will be
in progress 0
6 hours 2021-11-25T04:59:53+00:00 2 Answers 0 views 0
1. Step-by-step explanation:
x2=-1²
x=±1 OK friend this is urs answer
Step-by-step explanation:
x²+1²=0
x²=0-1
x²=-1
x= 1 | 119 | 308 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.71875 | 4 | CC-MAIN-2021-49 | latest | en | 0.674498 |
http://learningfsharp.blogspot.com/2012/01/ | 1,508,738,859,000,000,000 | text/html | crawl-data/CC-MAIN-2017-43/segments/1508187825700.38/warc/CC-MAIN-20171023054654-20171023074654-00227.warc.gz | 193,550,378 | 19,042 | ## Posts
Showing posts from January, 2012
### Project Euler - Problem 30 Redux (using Parallel)
The original post Project Euler - Problem 30 was done some time ago, and I recently read about using parallel methods to improve performance, and in this case, Parallel reduces execution time by 33%.
Description
Surprisingly there are only three numbers that can be written as the sum of fourth powers of their digits. Find the sum of all the numbers that can be written as the sum of fifth powers of their digits.
Solution (using Parallel)
open System
let isNarcissistic num power =
let numAsString = num.ToString()
let arr = [for i=0 to (numAsString.Length - 1) do
yield (float(numAsString.Chars(i).ToString()))**power]
if num = int(List.sum arr) then
num
else
0
let GetAllNarcissistic2_Start = System.DateTime.UtcNow
let GetAllNarcissistic2 =
[|2..999999|]
|> Array.Parallel.map (fun x -> isNarcissistic x 5.0)
|> Array.sum
printfn "%f" ((System.DateTime.UtcNow).Subtract(GetAllNarcissi…
### Project Euler - Problem 26 Redux (using Parallel)
The original post Project Euler - Problem 26 was done some time ago, and I recently read about using parallel methods to improve performance, and in this case, Parallel reduces execution time by 25%.
Description
A unit fraction contains 1 in the numerator. The decimal representation of the unit fractions with denominators 2 to 10 are given:
1/2 = 0.5
1/3 = 0.(3)
1/4 = 0.25
1/5 = 0.2
1/6 = 0.1(6)
1/7 = 0.(142857)
1/8 = 0.125
1/9 = 0.(1)
1/10 = 0.1
Where 0.1(6) means 0.166666..., and has a 1-digit recurring cycle. It can be seen that 1/7 has a 6-digit recurring cycle.
Find the value of d < 1000 for which 1/d contains the longest recurring cycle in its decimal fraction part.
Solution (using Parallel)
open System | 498 | 1,780 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.5625 | 4 | CC-MAIN-2017-43 | longest | en | 0.823003 |
https://www.jiskha.com/display.cgi?id=1315322483 | 1,501,167,028,000,000,000 | text/html | crawl-data/CC-MAIN-2017-30/segments/1500549428300.23/warc/CC-MAIN-20170727142514-20170727162514-00329.warc.gz | 801,533,475 | 3,844 | ipc
posted by .
Suppose all of the apples have a mass of 0.10 kg. If apple (A) is sitting on the ground, what is its potential energy?
• ipc -
relative to ground, zero.
• ipc -
suppose all of the apples have a mass of 0.10 kg. If apple (A) is sitting on the ground, what is its potential energy?
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https://www.hindawi.com/journals/ijmms/2021/5569981/ | 1,702,194,542,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679101282.74/warc/CC-MAIN-20231210060949-20231210090949-00089.warc.gz | 867,739,469 | 115,292 | #### Abstract
This paper mainly focuses on building the fuzzy prime ideal theorem of residuated lattices. Firstly, we introduce the notion of fuzzy ideal generated by a fuzzy subset of a residuated lattice and we give a characterization. Also, we introduce different types of fuzzy prime ideals and establish existing relationships between them. We prove that any fuzzy maximal ideal is a fuzzy prime ideal in residuated lattice. Finally, we give and prove the fuzzy prime ideal theorem in residuated lattice.
#### 1. Introduction
Nonclassical logic is closely related to logic algebraic systems. Many researches have motivated to develop nonclassical logic and also enrich the content of algebra [1, 2]. In modern fuzzy logic theory, residuated lattices and some related algebraic systems play an extremely important role because they provide algebraic frameworks to fuzzy logic and fuzzy reasoning.
Ward and Dilworth [3] initiated the notion of residuated lattice and it interested other authors [48]. The notion of ideal has been introduced in several algebraic structures such as BL-algebras [9] and residuated lattice [5, 8]. Piciu [10] gives and proves the prime ideal theorem in residuated lattices. Dealing with certain and uncertain information is an important task of the artificial intelligence, in order to make computer simulate human being. To handle such information, Zadeh [11] introduced the notion of fuzzy subset of a nonempty set as a function , where is the unit interval of real numbers. Since then, a lot of works have been done on fuzzy mathematical structures and most authors used the above original definition of a fuzzy set. The notion of fuzzy ideal has been studied in several structures such as rings [12], lattices [13, 14], MV-algebras [15], BL-algebras [16], and residuated lattices [6, 8, 17]. However, recent work of Piciu [10] gives the prime ideal theorem in residuated lattices. But that theorem is not yet investigated in fuzzy logic. In this work, we give and demonstrate the fuzzy prime ideal theorem in residuated lattice.
The remainder of this paper is organized as follows: Section 2 is a review on residuated lattices and ideals, whereas Section 3 contains the characterization of a fuzzy ideal generated by a fuzzy subset in residuated lattice. In Section 4, we study the different types of fuzzy prime ideals in residuated lattice, and we give some relations between them. We also define the notion of fuzzy maximal ideal and we give the fuzzy prime ideal theorem of residuated lattice.
#### 2. Review on Residuated Lattices and Ideals
Definition 1. (see [3, 17]). A residuated lattice is an algebraic structure of type satisfying the following axioms:(R1) is a bounded lattice(R2) is a commutative monoid(R3) For all , if and only if Let us give the following notations in a residuated lattice :(i), and (ii),
Definition 2. (see [1, 18]). A residuated lattice is an MTL-algebra if it satisfies the prelinearity condition, that is, , for all .
Definition 3. (see [19]). A De Morgan residuated lattice is a residuated lattice such that, for all , .
Example 1. Let be a lattice defined by the Hass diagram of Figure 1.
Define and as follows:
is an MTL-algebra.
Example 2. Let be a lattice defined by the Hass diagram of Figure 2.
Define and as follows:
is not a De Morgan residuated lattice. We have and ; then .
Any MTL-algebra is a De Morgan residuated lattice but the converse is not always true.
Example 3. Let be a lattice defined by the Hass diagram of Figure 3.
Define and as follows:
is a De Morgan residuated lattice. Since , it follows that is not an MTL-algebra.
Theorem 1 (see [8]). For any residuated lattice , the following properties hold for every :(P1) (P2) If , then (P3) (P4) (P5) , (P6) ;(P7) (P8) (P9) (P10)
Definition 4 (see [6]). Let be a lattice and let be a nonempty subset of . We say that is a filter of , if it satisfies the following conditions:(F1) For every , (F2) For every , if and ,
Definition 5. (see [5, 8, 20]). Let be a residuated lattice and let be a nonempty subset of . We say that is an ideal of if it satisfies the following conditions:(I1) For every , (I2) For every , if and , then We denote by the set of all ideals of . An ideal is called proper if . If , then and if .
There are two types of prime ideal in any residuated lattice.
Definition 6. (see [10, 20]). Let be a proper ideal of a residuated lattice . is said to be a prime ideal, if, for all , if , then and .
Proposition 1 (see [10, 20]). Let be a subset of residuated lattice . is a prime ideal if and only if, for all , if , then or .
Remark 1. (see [19]). If is a De Morgan residuated lattice, then is a prime ideal if, for all , implies or .
Definition 7. (see [20]). Let be a proper ideal of a residuated lattice . is said to be a prime ideal of second kind, if and only if, for any , or .
The next theorem gives the relation between these two types of prime ideal.
Theorem 2 (see [8]). Let be a residuated lattice. Every prime ideal of second kind of is also a prime ideal. If is an MTL-algebra, then prime ideal of and prime ideal of second kind of are equivalent.
Let us recall the notion of maximal ideal.
Definition 8. (see [20]). Let be a proper ideal of a residuated lattice . is said to be a maximal ideal, if, for any ideal of , implies that or .
Proposition 2 (see [10]). Let be an ideal of . If is a maximal ideal of , then it is a prime ideal of .
Theorem 3 (see [10]). Let be a residuated lattice. If is an ideal of and is a filter of the lattice such that , then there is a prime ideal of such that and .
#### 3. Fuzzy Ideal Generated by a Fuzzy Subset of Residuated Lattice
Let be a residuated lattice.
We first recall some definitions and properties of fuzzy subset.
Definition 9. (see [11]). Let be a nonempty set.(i)A map is called a fuzzy subset of (ii)A fuzzy subset of is called proper if it is not a constant map(iii)If is a fuzzy subset of and , then is called the -cut set of
Definition 10. (see [8]). Let be a fuzzy subset of . is a fuzzy ideal of , if it satisfies the following conditions:(FI1) for any , if , then (FI2) for any ,
Example 4. Let be a lattice defined by the Hass diagram of Figure 4.
Define and as follows (Table 1):
is a residuated lattice. The fuzzy subset of , defined byis a fuzzy ideal of .
Definition 11. (see [6, 13]). Let be a fuzzy subset of .(1) is a fuzzy filter of lattice, if it satisfies the following conditions:(i)for any , if , then (ii)for any , (2) is a fuzzy filter of residuated lattice, if it satisfies the following conditions:(i)for any , if , then (ii)for any , Note that any fuzzy filter of residuated lattice is a fuzzy filter of lattice ; but the converse is not always true.
Example 5. Let us consider the residuated lattice defined in Example 3. The fuzzy subset of , defined byis a fuzzy filter of lattice but it is not a filter of a residuated lattice because .
Let be two fuzzy subsets of . We define the order relation by , for all and, for any family of fuzzy ideals (or fuzzy filters) of A, and .
Let and denote, respectively, the set of fuzzy ideals of residuated lattice and the set of fuzzy filters of lattice .
Proposition 3. Let be a family of fuzzy ideals of a residuated lattice . Then is a fuzzy ideal of .
Proof. Let such that . We have ; then .
Moreover, for all , we have ; then . Thus, is a fuzzy ideal of .
In general, of fuzzy ideals of is not always a fuzzy ideal of . Indeed, this following example shows it.
Example 6. Let us consider the fuzzy ideals,of residuated lattice defined in Example 4. For , the fuzzy subset of , we havewhich is not a fuzzy ideal of because .
Let ; we define by , for all .
Proposition 4 (see [7]). Let be a fuzzy ideal of and such that . Then the fuzzy subset of is a fuzzy ideal of .
Theorem 4 (see [8]). Let be a fuzzy subset of . is a fuzzy ideal of if and only if, for each , implies that is an ideal of .
Notation 1. In the remainder of this paper, the following map will be very useful. Let be a nonempty subset of and such that . Define the map as follows: .
Proposition 5 (see [8]). is a proper fuzzy ideal of if and only if is a proper ideal of .
Definition 12. Let be a fuzzy subset of . A fuzzy ideal of is said to be generated by if and, for any fuzzy ideal of , if , then . The fuzzy ideal generated by will be denoted by .
Let be a fuzzy subset of . For all and ; then .
Theorem 5. Let be a fuzzy subset of . Then the fuzzy subset of defined by , for all , is the fuzzy ideal generated by .
Proof. Let . Suppose that ; take , for all and . We have ; then there exists such that and , for all . That is, and , for all . Then , for all , which implies that . Thus, . If , then, for all , ; therefore, for all , , which implies that ; that is, . Thus, . By Proposition 3, is an ideal of . If , then , where or . Therefore, by Theorem 4, is a fuzzy ideal of .
Let ; we have , where ; thus, ; that is, . Therefore, ; that is, .
Let be a fuzzy ideal of such that and . If , then . Suppose that . Then ; that is, for all , . Since , . Hence, .
Example 7. Let us consider the residuated lattice defined in Example 4. Let us consider the fuzzy subset of defined byLet ; if , then . If , then and . If , then . If , then . If , then . Therefore, we haveThus,
#### 4. Fuzzy Prime Ideal Theorem
Definition 13. A proper fuzzy ideal of a residuated lattice is said to be fuzzy prime if or , for any .
Proposition 6. Let be a proper fuzzy subset of a residuated lattice . is a fuzzy prime ideal of if and only if, for any , if , then is a prime ideal of .
Proof. Let be a proper fuzzy subset of . Suppose that is a fuzzy prime ideal. Let such that . By Theorem 4, is an ideal of . Let such that ; that is, ; then or ; that is, or . Therefore, is a prime ideal of .
Conversely, suppose that, for all , if is nontrivial, then is a prime ideal. By Theorem 4, is a fuzzy ideal of . Let . Take ; we have or because if it is not, then , which is absurd ( is proper); that is, or ; that is, or . Thus, . Moreover, because ; then by hypothesis is a prime ideal and or ; that is, or . In addition, and because is a fuzzy ideal. Therefore, or .
If is a De Morgan residuated lattice, then is a fuzzy prime ideal of if or , for any .
Definition 14. A fuzzy ideal of a residuated lattice is said to be fuzzy prime of the second kind if it is nonconstant and or for any .
Example 8. Let be the residuated lattice defined in Example 4. Then, , defined byis a fuzzy prime ideal of the second kind of .
Lemma 1. Let be a residuated lattice. For all , we have and .
Proof. Let . We have by property (P4) of Theorem 1. Since , then . Thus, . In the same way, we show that .
Theorem 6. Any fuzzy prime ideal of the second kind of residuated lattice is a fuzzy prime ideal of . If is an MTL-algebra, then the notions of fuzzy prime ideal of and fuzzy prime ideal of the second kind of are equivalent.
Proof. Let be a fuzzy prime ideal of the second kind of . Let ; we have or . By Lemma 1, and ; then and . If , then . From the fact that is a fuzzy ideal, we have ; then . Identically, if , then . In conclusion, is a fuzzy prime ideal of .
Assume that is an MTL-algebra and let be a fuzzy prime ideal of . Then, satisfies the prelinearity condition; that is, for any , we have ; that is, ; then . Therefore, or , by hypothesis. Thus, is a fuzzy prime ideal of the second kind of .
In general, a fuzzy prime ideal of a residuated lattice is not always a fuzzy prime ideal of the second kind, unless the residuated lattice is an MTL-algebra. The proof of this statement is given by the following counterexample.
Example 9. Let us consider the residuated lattice defined in Example 3 and let be the fuzzy ideal of defined by is a fuzzy prime ideal of which is not a fuzzy prime ideal of the second kind of , because and .
Definition 15. A proper fuzzy ideal of a residuated lattice is prime fuzzy if, for fuzzy ideals and of , implies that or .
Example 10. Let us consider the residuated lattice defined in Example 3.
Consider that the fuzzy subset , defined byis a prime fuzzy ideal of .
Remark 2. A fuzzy prime ideal of is not necessarily a prime fuzzy ideal of as the following example shows.
Example 11. Let us consider the residuated lattice defined in Example 3. Let be a fuzzy subset of defined by is a fuzzy prime ideal of . Let and be two fuzzy ideals of defined byWe haveThen , , and ; therefore, is not a prime fuzzy ideal of .
Definition 16. Let be a proper fuzzy ideal of . is called fuzzy maximal ideal if, for all , implies that is a maximal ideal of .
Example 12. Let us consider the residuated lattice defined in Example 3 and let be a fuzzy subset of defined by is a fuzzy maximal ideal of .
Proposition 7. Let be a fuzzy ideal of . If is a fuzzy maximal ideal of , then is a fuzzy prime ideal of .
Proof. Let be a fuzzy maximal ideal of . Then, for all such that , is a maximal ideal of . By Proposition 2, is a prime ideal of for all such that . Thus, according to Proposition 6, is a fuzzy prime ideal of .
The converse of the above proposition is not true. Let us consider the residuated lattice defined in Example 3 and let be a fuzzy subset of defined by is a fuzzy prime ideal of , which is not a fuzzy maximal ideal of because is not maximal.
Lemma 2. Let , let be a fuzzy ideal of a residuated lattice , and let be a fuzzy filter of the lattice . Then and are, respectively, the ideal of the residuated lattice and the filter of the lattice .
Proof. Let and such that . Then ; that is, . Let . We have . Since is a fuzzy ideal of , . Therefore, because . Thus, . Hence, is an ideal of .
Let and such that . Then ; that is, . Let ; we have . Therefore, . Thus, is a filter of the lattice .
Theorem 7. Let , let be a fuzzy ideal of a residuated lattice , and let be a fuzzy filter of the lattice such that . Then there exists a fuzzy prime ideal of such that and .
Proof. Let and . By Lemma 2, is an ideal of the residuated lattice and is a filter of the lattice .
Let us show that . Suppose that there exists . Then and . Therefore, ; this is a contradiction because, by hypothesis, . Thus, .
Using the fact that is an ideal of the residuated lattice , is a filter of lattice and and, by Theorem 4, there exists a prime ideal of the residuated lattice such that and . LetWe have ; then, by Proposition 5, is a fuzzy ideal of . If , then or ; that is, or ; therefore, or . If , then because . In addition, is a fuzzy ideal of and, for all , ; then, . Therefore, or , since and . Thus, is a fuzzy prime ideal of .
Let ; if , then . Else, ; then and , because ; in this case, . Thus, .
Let ; if , then and (because ). Therefore, . If , then . Thus, .
#### 5. Conclusion
In this paper, we have investigated the notion of fuzzy ideal generated by a fuzzy subset of residuated lattice and established the fuzzy prime ideal theorem of this structure. Besides, we have studied different types of fuzzy prime ideals (fuzzy prime ideal, fuzzy prime ideal of second kind, and prime fuzzy ideal) and fuzzy maximal ideal of residuated lattice. We have proved that any fuzzy maximal ideal is a fuzzy prime ideal, but the converse is not always true. Moreover, any prime fuzzy ideal and any fuzzy prime ideal of the second kind are fuzzy prime ideals and the converse is not also true. We have also given a characterization of fuzzy ideal generated by a fuzzy subset of residuated lattice. The last result of this paper is the fuzzy prime ideal theorem.
#### Data Availability
No data were used to support this study.
#### Conflicts of Interest
The authors declare that they have no conflicts of interest. | 3,864 | 15,650 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.703125 | 3 | CC-MAIN-2023-50 | latest | en | 0.95026 |
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## EU OSS (One Stop Shop) - End pricing issues, wrong price calculated
Excursionist
30 0 21
Since the implementation of OSS in the EU today, we are having issues with how our end prices are being calculated in our online shop. Our tax settings are:
The issue right now is that the end price changes depending on the charged VAT rate. We are registered in Germany with 19% VAT. If we sell to another EU country now, the end price changes instead of adjusting the VAT amount to the sale price (a EU sale of eg. €99 should always be €99, regardless of 19% VAT or 20% VAT or 25% VAT, only the VAT part of the total price would increase if the rate is higher than our 19%). So if we have a french customer right now (France has 20% VAT rate), he/she pays slightly more than our German customers. Has anyone else experienced this issue, and if so, did you manage to properly solve this? We have contacted Shopify support, and they couldn't help us and said our tax settings are correct.
Hopeful for a resolution as having a different end price than listing price is illegal in the EU.
Replies 11 (11)
Excursionist
30 0 21
So, I have managed to figure out what how the price is calculated. Please forward this information to the developers asap.
Here is an example:
Regular price €99. Sale price €69,30. This it how it should be, and how it’s displayed in the shop. But when a customer from another EU country with another VAT rate wants to buy this, Shopify calculates the price like this:
99 - 19% German VAT = 83,19
83,19 is then used to calculate the sale price without VAT, in this case 83,19 * 0,7 = 58,24.
58,24 + 20% French VAT = 69,89
So on the sale (excluding shipping), VAT of 11,65 has been charged to the customer, but the correct VAT charged to the customer should be 20% of 69,30, which equals 11,55.
So the issue with the incorrect prices are due to how Shopify calculates the prices. I tried with a regular priced item as well, and same applies:
Price in the shop = 189 This is what German customers pays, and what everyone in the EU should pay (shipping excluded). However same thing happens, again the French example:
189 - 19% = 158,82
158,82 + 20% = 190,59
Visitor
1 0 1
We have the same issue...
If you untick the second tax settings, it will be correct the pricing and taxes in EU... BUT in that case tax free shopping is not working outside EU. ( USA , CAnada etc) it will still charge them full price and not your Price-VAT.
It seems like its either or currently... I tried to talk to the Shopify support and they seem to not understand the issue still.
I am looking into third party apps now, like Exemptify, to might resolve the issue.
Excursionist
30 0 21
Yes, we have done this as well as we need to comply with EU laws, but our international sales are currently taking a big hit. I agree that it has been really hard explaining to Shopify to even explain the issue, I’ve gotten replies from 3 different support employees and they seem to (at least at first) think that it’s working as it should. If you have any luck with an app, please let us know.
Also @Shopify, how can you mess this up like this and then be so slow reacting? your merchants are loosing money by the minute and you don’t react. I raised this issue on the by 1st, and no solution seems to be in sight. We have a lot of non eu customers and are very upset with your ineptitude. It took you years finding a proper solution to even remove VAT for non EU customers, and now you are at it again… it’s very obvious that your tax team doesn’t have the required knowledge about EU taxation (I guess you are all NA based) which is borderline criminal for a shop provider as big as yourselves.
Tourist
5 0 7
Hi @Robin_K and @HansenGarments (we're one of your retailers )- I'm very relieved to read that we are not alone in identifying this issue.
Like you guys, I've spent many hours chatting with Shopify support people about this, and each time they completely misunderstand the issue we are reporting and tell us the system is operating correctly in respect of prices.
For us there are THREE CRITICAL problems here.
Firstly as @Robin_K correctly pointed out, charging customers a different end price to the one we advertise / display is NOT LEGAL.
Secondly, Shopify SHOULD NOT BE CHANGING THE END PRICE when calculating different EU tax rates. The end price MUST REMAIN THE SAME - it's the proportion of tax inside that amount that should change.
Thirdly, as @HansenGarments correctly pointed out - turning off the option to Include / Exclude tax on the Tax settings isn't a workable solution for EU merchants, because this results in our non EU sales to places such as the US, incorrectly being charged VAT.
Actually I'll add a fourth issue here - the decision on an end price for a product online is a critically important decision and tool - it's not only confusing, illegal and wrong - it's also interfering with one of the most important strategic tools for an online merchant - pricing.
If it's an issue for us, it will be an issue for other EU merchants - so let's hope they see this thread and we can get a solution soon.
J
Excursionist
30 0 21
Excursionist
30 0 21
Does anyone has some new info about this? The only thing we hear from Shopify is that this is taken under consideration, but they cannot say neither if they intend to fix this or when it could actually be fixed. The recommended us to hire an expert to customise the coding, ignoring to mention that to change the checkout coding, we need to have a Shopify Plus membership which starts at several thousand € / month. We are looking at some multiple pricing apps and for us thus far, the most promising for us seems to be "Pricing by Country" from Webrex Studio which we will try today. Anyone else has a better tip or more info, feel free to share.
Shopify Partner
41 1 8
Hi @Robin_K , @Jonathan_Maher ,
Dan here from Vevol Media 🖐 (Shopify Experts), I am not 100% sure this would help you but I'll explain what we did for one of our clients based in the UK.
They had similar issues and they used the Exentify app (which, without upsetting anyone, slowed down their website) so we removed the app and we've added a dropdown on their Home Page that allows the client to switch between seeing the prices with VAT and without VAT. See attached file.
Feel free to get in touch with us if you think this could help you too.
Dan from Vevol Media - Shopify Partners Since 2018
--
Tourist
5 0 7
Either I'm getting something wrong, or Shopify are refusing to acknowledge a serious flaw in how their platform handles prices and vat between EU states.
If I'm not getting this wrong - and according to @Robin_K and @HansenGarments there IS a problem - then there should be thousands of other EU online retailers with the same problem.
If somebody from Shopify is reading this - can you at least explain the logic behind why your system is adjusting the prices of our products when charging different EU vat rates INSTEAD of adjusting the vat paid.
Aside from the LEGALITY of having a product page advertising a product at one price - then the end price being different - it's incredibly frustrating, and unprofessional.
Can we get this discussion heated up a little please, so that perhaps somebody from Shopify can comment.
Shopify Partner
7 0 2
I have the same issue !!! Was working for 1 year without issue!!!
Sudenly stop working and doing the same in our shop ! But funny part on .eu domain ALL WRONG
on .si domain same template & same setup all good!!!
Can somenone from SHOPIFY solve this as we DISABLE now on EU shop calculating taxes which is wrong (as in our acouinting program getting invoices made without each eu country VAT ) ALSO customers dont SEE that price is included in price!
.eu domain have for 3 years-- getting this issue probably from cca december/januar 2022
.si domain runing for 3 months same SETUP / same template/ ALL GOOD if we try for example shipping to germany and checkout ALL GOOD!!
SO please someone solve this as i am being on support tickets for months and NO SOLUTION!!
Shopify Partner
7 0 2
SHOPIFY SUPPORT????? Are you really ignoring this???? We have issue only one one store. on other domain same setup same template ! NO ISSUE???
Visitor
1 0 0
I would like this to be resolved asap @Shopify
We have EU wide RRP (retail price) that is the same price across inclusive of VAT.
This is a common practice in EU, as EU consumers can enjoy the same end product price across single market.
We sell at around retailers across EU but they all list the same price and some retailers in a country with lower VAT make slightly more margin than others with higher VAT. | 2,057 | 8,785 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.59375 | 3 | CC-MAIN-2024-30 | latest | en | 0.931756 |
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## Explore printable Mixed Multiplication and Division worksheets for 1st Class
Mixed Multiplication and Division worksheets for Class 1 are an essential resource for teachers looking to enhance their students' math skills. These worksheets provide a variety of mixed operations problems that challenge young learners to apply their knowledge of multiplication and division in a fun and engaging way. By incorporating these worksheets into their lesson plans, teachers can ensure that their students are developing a strong foundation in math, which is crucial for success in higher grades. Additionally, these worksheets can be easily customized to meet the needs of individual students, making them an ideal tool for differentiated instruction. With Mixed Multiplication and Division worksheets for Class 1, teachers can effectively support their students' growth in math and instill a love for learning.
Quizizz is an innovative platform that offers a wide range of educational resources, including Mixed Multiplication and Division worksheets for Class 1. This platform allows teachers to create interactive quizzes and games that can be used to supplement their existing lesson plans or as a stand-alone activity. With Quizizz, teachers can easily track their students' progress and identify areas where additional support may be needed. In addition to worksheets, Quizizz also offers a variety of other resources, such as flashcards, videos, and presentations, making it a one-stop-shop for all of your educational needs. By incorporating Quizizz into their teaching strategies, teachers can provide their Class 1 students with a dynamic and engaging learning experience that will help them excel in math and other subjects. | 666 | 2,860 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.640625 | 3 | CC-MAIN-2024-30 | latest | en | 0.938507 |
https://www.omnicalculator.com/sports/running-pace | 1,723,329,471,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640826253.62/warc/CC-MAIN-20240810221853-20240811011853-00831.warc.gz | 722,442,864 | 42,111 | # Running Pace Calculator
Created by Wei Bin Loo
Reviewed by Anna Szczepanek, PhD and Adena Benn
Based on research by
Heesch MW, Slivka DR. Running performance, pace strategy, and thermoregulation differ between a treadmill and indoor track.; The Journal of Strength and Conditioning Research; February 2015
Last updated: Jan 18, 2024
We have prepared this cool running pace calculator to help you measure how fast you can run. This incredible running pace calculator can also help you to understand your average speed when completing a run, such as a marathon.
After reading this article, you will understand what a running pace is, how to calculate it, and how to increase your running pace. This cool running pace calculator will also demonstrate some examples to help you understand the concept.
## What is running pace?
Running pace is defined as the time you need to complete a particular running distance, which indirectly tells you how fast you are running. For example, if you need 10 minutes to complete a 1 km run, your running pace will be 10 minutes per km.
It is beneficial to understand metrics, especially if you are preparing for a race, such as completing a marathon. Understanding your running pace will allow you to know how long it will take you to complete the run. Furthermore, when it comes to long-distance running, it is important for you to maintain a constant running pace throughout the run as well.
## How to calculate the running pace?
To understand the running pace calculation, let's take Candy, who is trying to complete a marathon, as an example:
• Event: Marathon
• Distance: 42,195 m
• Time spent: 5 hours
You can calculate running pace in three steps:
1. Determine the distance of the run
The first step of this calculation is to determine the total distance that you will run. In this example, Candy is going to run a marathon, which has a distance of 42,195 m.
2. Calculate the time spent to complete the run
The next step is to determine the amount of time you spent on completing the run. You can use our time duration calculator and race time improvement calculator to help with that.
3. Calculate the running pace
The last step is to calculate the running pace using the following formula:
running pace = time spent / distance run.
Hence, Candy's running pace is 7 minutes and 6 seconds per kilometer.
Our cool running pace calculator also helps you to estimate your race time and the adjusted running pace. You can choose the race that you wish to take and compare it to other runners. We have also provided you with a table for a more precise comparison at different levels.
If you have your speed already calculated but need it in another unit of measurement, you can use our speed conversion calculator. This tool enables you to convert between 5 different units of speed.
## What's a good running pace?
Now, let's talk about what a good running pace is. A good running pace is subjective and depends on individual fitness levels and goals. As a general guideline, a moderate pace for a recreational runner is around 10-12 minutes per mile.
That's why our cool running pace calculator also shows you the running pace of different levels, including world record, world-class, elite, fast amateur, average, and beginner.
## FAQ
### How long does it take to prepare for a marathon?
To complete a marathon, you need to run 26 miles 385 yards, which is 42.195 km. It usually takes about 6 months of constant training to complete this impressive feat.
### Can the running pace be negative?
No, the running pace can never be negative. This is because the distance you run and the time you spend running cannot be negative.
### How do I calculate my running pace?
You can calculate your running pace in three steps:
1. Determine the distance of your run.
2. Calculate the time spent to complete the run.
3. Apply the running pace formula: running pace = time spent / distance run.
### What is my running pace if I finish a marathon in 3 hours?
Your running pace is 0.25595 seconds per meter. The world record for a marathon is 2 hours 1 minute and 9 seconds, and Eliud Kipchoge of Kenya holds it.
### How to increase running pace?
One of the most common questions among runners is how to increase their running pace. To increase the running pace, gradually increase training intensity and volume, incorporate interval training, and prioritize proper nutrition and rest.
Wei Bin Loo
Previous race
Distance
5km (5k)
Time
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Running pace
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Speed
mph
Estimate the finishing time for your next race
Race distance
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Predicted finishing time
0
hrs
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7
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8
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Compare to other runners
Fast Amateur
Compare to...
Female
Show comparison using...
Table (pace)
Units
Auto
LevelPace
World Record04:42 / mile
World Class05:17 / mile
Elite07:03 / mile
YOU08:23 / mile
Fast Amateur08:25 / mile
Average09:31 / mile
Beginner10:56 / mile
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https://www.chegg.com/homework-help/algebra-1-grades-9-12-0th-edition-chapter-6.7-problem-18pa-solution-9780618558827 | 1,550,572,315,000,000,000 | text/html | crawl-data/CC-MAIN-2019-09/segments/1550247489933.47/warc/CC-MAIN-20190219101953-20190219123953-00029.warc.gz | 783,987,996 | 23,880 | Solutions
# Algebra 1, Grades 9-12 (0th Edition) View more editions Solutions for Chapter 6.7 Problem 18PA
• 12073 step-by-step solutions
• Solved by professors & experts
• iOS, Android, & web
Chapter: Problem:
Step-by-Step Solution:
Chapter: Problem:
• Step 1 of 4
Test the zeroby synthetic division.
Write the division in synthetic form.
Here, the divisor is, and the dividend is.
Write the coefficients of the dividend and the divisor in the following form.
Complete the synthetic division as follows:
Bring down the first term in the dividend. Multiply the numbers below the line by the synthetic divisor and carry that product above the line to the next column.
Add the numbers above the horizontal line which completes the division process.
• Chapter , Problem is solved.
Corresponding Textbook
Algebra 1, Grades 9-12 | 0th Edition
9780618558827ISBN-13: 0618558829ISBN: Authors:
This is an alternate ISBN. View the primary ISBN for: Algebra 2, Grades 9-12 0th Edition Textbook Solutions | 246 | 1,003 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.109375 | 3 | CC-MAIN-2019-09 | latest | en | 0.749304 |
https://pybullet.org/Bullet/phpBB3/viewtopic.php?t=12684&p=42112 | 1,582,234,428,000,000,000 | text/html | crawl-data/CC-MAIN-2020-10/segments/1581875145282.57/warc/CC-MAIN-20200220193228-20200220223228-00361.warc.gz | 457,732,258 | 7,868 | ## SoftBody: modeling a rope from trimesh
Dox
Posts: 24
Joined: Thu Oct 18, 2018 5:16 pm
### SoftBody: modeling a rope from trimesh
Hello,
I'm in trouble with SoftBody
I'm trying to modeling a rope starting from a mesh (please, see the attachment RopeMesh.h). This is a cylinder of 10m, with 32 resolution faces and 8 subdivisions on the length. Then, I'm using this code in order to build the SoftBody:
Code: Select all
``````static void Init_RopeMatch(SoftDemo* pdemo)
{
btSoftBody* psb = btSoftBodyHelpers::CreateFromTriMesh(pdemo->m_softBodyWorldInfo, gVerticesRope,
&gIndicesRope[0][0],
ROPE_NUM_TRIANGLES);
psb->m_materials[0]->m_kLST = 0.0; //Linear stiffness coefficient [0,1]
psb->m_materials[0]->m_kAST = 0; //Area/Angular stiffness coefficient [0,1]
psb->m_materials[0]->m_kVST = 0; //Volume stiffness coefficient [0,1]
psb->m_cfg.kMT = 0.5;
psb->m_cfg.piterations = 5;
psb->randomizeConstraints();
psb->scale(btVector3(6, 6, 6));
psb->setTotalMass(1, true);
psb->setMass(0, 0);
psb->setMass(psb->m_nodes.size()-1, 0);
psb->setPose(false, true);
pdemo->getSoftDynamicsWorld()->addSoftBody(psb);
}
``````
Basically, I can see the cylinder that is anchored on the first and last node, but the cylinder is pretty rigid, do not bend freely on the 8 subdivisions. It's like a "high jump pole".
Using the btSoftBodyHelpers::CreateRope the rope behavior is very good but I need to start from my mesh (the btSoftBodyHelpers::CreateRope is just a line without volume and in any case I need an integration with meshes).
Setting material stiffness coefficients to 0 doesen't help. How I can setup the SoftBody in order to make the cylinder mesh like a falling rope under gravity? Maybe I need to build a different mesh?
Basically, in a rope I don't need to make soft the "circular section" (the cylinder volume do not change) but I need to bend the subdivisions....I don't know if it's possible...
RopeMesh.h
Rope Cylinder
(335.83 KiB) Downloaded 37 times
Many thanks,
bye
steven
Posts: 73
Joined: Mon Nov 05, 2018 8:16 am
Location: China
### Re: SoftBody: modeling a rope from trimesh
after removing the "psb->m_cfg.kMT = 0.5;" it will become the one like below which seems that your mesh is constructed of some separate parts.
Collide1.png (60.61 KiB) Viewed 1367 times
Dox
Posts: 24
Joined: Thu Oct 18, 2018 5:16 pm
### Re: SoftBody: modeling a rope from trimesh
Hello steven,
hummm...maybe the triangles indices array has some wrong triangle? Maybe the caps...I can try to remove them but I think that the main problem can be on other point...maybe this behavior is correct considering the first and last node are fixed in space...
Dox
Posts: 24
Joined: Thu Oct 18, 2018 5:16 pm
### Re: SoftBody: modeling a rope from trimesh
... No other idea?
Many thanks, bye
steven
Posts: 73
Joined: Mon Nov 05, 2018 8:16 am
Location: China
### Re: SoftBody: modeling a rope from trimesh
Could you first change your mesh? i would like to work with you to find the root cause.
i also noted that the first and last note are not located at the end of this rope.
Dox
Posts: 24
Joined: Thu Oct 18, 2018 5:16 pm
### Re: SoftBody: modeling a rope from trimesh
Hello steven,
thanks for your interest.
Ok, I've removed the caps from the mesh which I attach here.
I attach also the original fbx (as .txt, you can rename to .fbx, cable_block_cyl_fix_10m8sez_NoCaps_Tri.fbx) , maybe is useful. I attach the .obj converted also (by OpenSceneGraph processing).
I process the fbx mesh via OpenSceneGraph, maybe someone wrong happens here even if the behavior seems the same to me. The mesh generated seems correct...I've noted me too the last node problem...I'm a little bit confused right now...
Thanks again,
bye
Attachments
cable_block_cyl_fix_10m8sez_NoCaps_Tri.obj.txt
Rename to .obj
(211.95 KiB) Downloaded 30 times
cable_block_cyl_fix_10m8sez_NoCaps_Tri.txt
Rename to .fbx
(162.26 KiB) Downloaded 35 times
RopeMesh.h
(327.93 KiB) Downloaded 33 times
steven
Posts: 73
Joined: Mon Nov 05, 2018 8:16 am
Location: China
### Re: SoftBody: modeling a rope from trimesh
i find that there are duplicated vertex in the array of gVerticesRope, the engine will create a node for each vertex when this array is passed to the btSoftBody(), so there will be more than one node for a same point that is not what we expect.
i think you need to re-design your gVerticesRope/gIndicesRope. good luck！ | 1,291 | 4,414 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.5625 | 3 | CC-MAIN-2020-10 | latest | en | 0.769659 |
https://interviewmania.com/data-interpretation/introduction-to-data-interpretation/1/183 | 1,721,805,394,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763518157.20/warc/CC-MAIN-20240724045402-20240724075402-00472.warc.gz | 291,254,182 | 12,275 | ## Introduction to Data Interpretation
#### Data Interpretation
Direction: The following questions are based on the bar graph. Read the graph and answer the questions.
1. What is the percentage increase in the gross traffic receipts in 1995-96 as compared to 1993-94?
1. On the basis of given graph in question ,
The gross traffic receipts in 1995-96 = \$ 7500 million
The gross traffic receipts in 1993-94 = \$ 5300 million
Increase = 7500 - 5300 = \$ 2200 million
Required percent increase = Increase × 100% The gross traffic receipts in 1993-94
##### Correct Option: B
On the basis of given graph in question ,
The gross traffic receipts in 1995-96 = \$ 7500 million
The gross traffic receipts in 1993-94 = \$ 5300 million
Increase = 7500 - 5300 = \$ 2200 million
Required percent increase = Increase × 100% The gross traffic receipts in 1993-94
Required percent increase = 7500 - 5300 × 100 = 41.5% 5300
Direction: Study the bar diagram carefully and answer the following questions.
1. The export to Hongkong is approximately how many times the exports to Germany?
1. According to given bar graph , we have
The export to Hongkong = 10.6
The export to Germany = 1.2
∴ Required answer = The export to Hongkong The export to Germany
##### Correct Option: C
According to given bar graph , we have
The export to Hongkong = 10.6
The export to Germany = 1.2
∴ Required answer = The export to Hongkong The export to Germany
∴ Required answer = 10.6 ≈ 10 1.2
1. The ratio of the total exports to Japan, Belgium and Hongkong to the export to rest of the countries in 1991–92 is nearly :
1. From the given bar diagram , we see
Total exports to Japan, Belgium and Hongkong = 12.5 + 12.1 + 10.6 = 35.2
Total exports = \$ (22.6 + 12.5 + 12.1 + 10.6 + 3.3 + 2.5 + 1.6 × 3 + 1.2) billion = \$ 69.6 billion
Total exports of remaining countries = 69.6 - 35.2 = 34.4
Required ratio = Total exports to Japan, Belgium and Hongkong : Total exports of remaining countries
##### Correct Option: A
From the given bar diagram , we see
Total exports to Japan, Belgium and Hongkong = 12.5 + 12.1 + 10.6 = 35.2
Total exports = \$ (22.6 + 12.5 + 12.1 + 10.6 + 3.3 + 2.5 + 1.6 × 3 + 1.2) billion = \$ 69.6 billion
Total exports of remaining countries = 69.6 - 35.2 = 34.4
Required ratio = Total exports to Japan, Belgium and Hongkong : Total exports of remaining countries
∴ Required ratio = 35.2 : 34.4 ≈ 35 : 34
1. The country to which twice the export is nearly equal to the average exports in 1991–92 is
1. According to given bar graph , we have
Total exports = \$ (22.6 + 12.5 + 12.1 + 10.6 + 3.3 + 2.5 + 1.6 × 3 + 1.2) billion = \$ 69.6 billion
Number of countries = 10
∴ Average exports = Total exports Number of countries
table>∴ Average exports =
69.6
= \$ 6.96 billion 10
##### Correct Option: D
According to given bar graph , we have
Total exports = \$ (22.6 + 12.5 + 12.1 + 10.6 + 3.3 + 2.5 + 1.6 × 3 + 1.2) billion = \$ 69.6 billion
Number of countries = 10
∴ Average exports = Total exports Number of countries
table>∴ Average exports =
69.6
= \$ 6.96 billion 10
Exports to UAE = \$ 3.3 billion
Hence , required answer is UAE .
1. The ratio of the sum of the exports to the bottom six countries to the total exports to all the given countries in 1991 – 1992 is approximately :
1. On the basis of given graph in question ,
Total exports = \$ (22.6 + 12.5 + 12.1 + 10.6 + 3.3 + 2.5 + 1.6 × 3 + 1.2) billion = \$ 69.6 billion
Total exports to bottom six countries = \$ (3.3 + 2.5 + 1.6 × 3 + 1.2) billion = \$ 11.8 billion
Required Ratio = Total exports to bottom six countries Total exports
##### Correct Option: A
On the basis of given graph in question ,
Total exports = \$ (22.6 + 12.5 + 12.1 + 10.6 + 3.3 + 2.5 + 1.6 × 3 + 1.2) billion = \$ 69.6 billion
Total exports to bottom six countries = \$ (3.3 + 2.5 + 1.6 × 3 + 1.2) billion = \$ 11.8 billion
Required Ratio = Total exports to bottom six countries Total exports
∴ Required Ratio = 11.8 ≈ 1 69.6 6 | 1,290 | 3,990 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.90625 | 4 | CC-MAIN-2024-30 | latest | en | 0.872936 |
http://oeis.org/A200503 | 1,571,070,199,000,000,000 | text/html | crawl-data/CC-MAIN-2019-43/segments/1570986653876.31/warc/CC-MAIN-20191014150930-20191014174430-00415.warc.gz | 146,003,382 | 5,234 | This site is supported by donations to The OEIS Foundation.
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A200503 Record (maximal) gaps between prime sextuplets (p, p+4, p+6, p+10, p+12, p+16). 11
90, 15960, 24360, 1047480, 2605680, 2856000, 3605070, 4438560, 5268900, 17958150, 21955290, 23910600, 37284660, 40198200, 62438460, 64094520, 66134250, 70590030, 77649390, 83360970, 90070470, 93143820, 98228130, 117164040, 131312160, 151078830, 154904820 (list; graph; refs; listen; history; text; internal format)
OFFSET 1,1 COMMENTS Prime sextuplets (p, p+4, p+6, p+10, p+12, p+16) are densest permissible constellations of 6 primes. Average gaps between sextuplets (and, more generally, between prime k-tuples) can be deduced from the Hardy-Littlewood k-tuple conjecture and are O(log^6(p)). Maximal gaps may be significantly larger than average gaps; this sequence suggests that maximal gaps are O(log^7(p)). A200504 lists initial primes in sextuplets preceding the maximal gaps. A233426 lists the corresponding primes at the end of the maximal gaps. LINKS Alexei Kourbatov, Table of n, a(n) for n = 1..56 G. H. Hardy and J. E. Littlewood, Some problems of 'Partitio numerorum'; III: on the expression of a number as a sum of primes, Acta Mathematica, Vol. 44, pp. 1-70, 1923. Alexei Kourbatov, Maximal gaps between prime k-tuples A. Kourbatov, Maximal gaps between prime k-tuples: a statistical approach, arXiv preprint arXiv:1301.2242 [math.NT], 2013 and J. Int. Seq. 16 (2013) #13.5.2 Alexei Kourbatov, Tables of record gaps between prime constellations, arXiv preprint arXiv:1309.4053 [math.NT], 2013. Alexei Kourbatov, The distribution of maximal prime gaps in Cramer's probabilistic model of primes, arXiv preprint arXiv:1401.6959 [math.NT], 2014. Alexei Kourbatov and Marek Wolf, Predicting maximal gaps in sets of primes, arXiv preprint arXiv:1901.03785 [math.NT], 2019. Eric W. Weisstein, k-Tuple Conjecture FORMULA (1) Conjectured upper bound: gaps between prime sextuplets (p, p+4, p+6, p+10, p+12, p+16) are smaller than 0.058*(log p)^7, where p is the prime at the end of the gap. (2) Estimate for the actual size of the maximal gap that ends at p: maximal gap = a(log(p/a)-1/3), where a = 0.058*(log p)^6 is the average gap between sextuplets near p, as predicted by the Hardy-Littlewood k-tuple conjecture. Formulas (1) and (2) are asymptotically equal as p tends to infinity. However, (1) yields values greater than all known gaps, while (2) yields "good guesses" that may be either above or below the actual size of maximal gaps. Both formulas (1) and (2) are derived from the Hardy-Littlewood k-tuple conjecture via probability-based heuristics relating the expected maximal gap size to the average gap. Neither of the formulas has a rigorous proof (the k-tuple conjecture itself has no formal proof either). In both formulas, the constant ~0.058 is reciprocal to the Hardy-Littlewood 6-tuple constant 17.2986... EXAMPLE The gap of 15960 between sextuplets with initial primes 97 and 16057 is a maximal gap - larger than any preceding gap; therefore a(2)=15960. CROSSREFS Cf. A022008 (prime sextuplets), A113274, A113404, A200504, A201596, A201598, A201062, A201073, A201051, A201251, A202281, A202361, A008407, A002386, A233426. Sequence in context: A276352 A203779 A295594 * A199229 A278411 A278631 Adjacent sequences: A200500 A200501 A200502 * A200504 A200505 A200506 KEYWORD nonn,hard AUTHOR Alexei Kourbatov, Nov 18 2011 STATUS approved
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Last modified October 14 12:21 EDT 2019. Contains 328006 sequences. (Running on oeis4.) | 1,197 | 3,823 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.515625 | 3 | CC-MAIN-2019-43 | latest | en | 0.686438 |
https://www.convertunits.com/from/decaliter/to/centilitre | 1,656,284,180,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656103322581.16/warc/CC-MAIN-20220626222503-20220627012503-00679.warc.gz | 811,742,626 | 17,235 | ## ››Convert decalitre to centiliter
decaliter centilitre
How many decaliter in 1 centilitre? The answer is 0.001.
We assume you are converting between decalitre and centiliter.
You can view more details on each measurement unit:
decaliter or centilitre
The SI derived unit for volume is the cubic meter.
1 cubic meter is equal to 100 decaliter, or 100000 centilitre.
Note that rounding errors may occur, so always check the results.
Use this page to learn how to convert between decaliters and centiliters.
Type in your own numbers in the form to convert the units!
## ››Quick conversion chart of decaliter to centilitre
1 decaliter to centilitre = 1000 centilitre
2 decaliter to centilitre = 2000 centilitre
3 decaliter to centilitre = 3000 centilitre
4 decaliter to centilitre = 4000 centilitre
5 decaliter to centilitre = 5000 centilitre
6 decaliter to centilitre = 6000 centilitre
7 decaliter to centilitre = 7000 centilitre
8 decaliter to centilitre = 8000 centilitre
9 decaliter to centilitre = 9000 centilitre
10 decaliter to centilitre = 10000 centilitre
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## ››Definition: Decalitre
The SI prefix "deca" represents a factor of 101, or in exponential notation, 1E1.
So 1 decalitre = 101 liters.
The definition of a litre is as follows:
The litre (spelled liter in American English and German) is a metric unit of volume. The litre is not an SI unit, but (along with units such as hours and days) is listed as one of the "units outside the SI that are accepted for use with the SI." The SI unit of volume is the cubic metre (m³).
## ››Definition: Centilitre
A centilitre (cL or cl) a metric unit of volume that is equal to one hundredth of a litre and is equal to a little more than six tenths (0.6102) of acubic inch, or one third (0.338) of a fluid ounce.
## ››Metric conversions and more
ConvertUnits.com provides an online conversion calculator for all types of measurement units. You can find metric conversion tables for SI units, as well as English units, currency, and other data. Type in unit symbols, abbreviations, or full names for units of length, area, mass, pressure, and other types. Examples include mm, inch, 100 kg, US fluid ounce, 6'3", 10 stone 4, cubic cm, metres squared, grams, moles, feet per second, and many more! | 671 | 2,436 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.703125 | 4 | CC-MAIN-2022-27 | latest | en | 0.75862 |
https://www.halfbakery.com/idea/Perfect_20Length_20Tie_20Calculator | 1,638,949,101,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964363445.41/warc/CC-MAIN-20211208053135-20211208083135-00480.warc.gz | 840,025,812 | 6,924 | h a l f b a k e r y
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# Perfect Length Tie Calculator
Queery for the straight tie
(+26) [vote for, against]
Any properly attired gentleman knows that the length of your tie should be such that the point rests in the middle of your belt buckle (while standing).
The resting point of the tie is dependant on the height of the wearer, the distance the tie has to travel over a belly, the thickness of the neck, the type of knot used, the taper of the tie and thickness of material used in its construction.
I propose a handy calculator that accepts all these variables and gives you a measurement indicating where the first loop of the knot should start. This could then be marked on the tie so that it could be tied to the perfect length first time - every time.
— MikeyTheBikey, May 12 2005
The low down on ties. http://www.straight...lassics/a2_155.html
From the 'Straight Dope' archive. [DrBob, May 12 2005]
[theircompetitor, May 12 2005]
Finally my "tie, retie, retie, retie, adjust" procedure is a thing of the past! [+]
— DocBrown, May 12 2005
a mechanical solution would be nice
— po, May 12 2005
For people of non corpulent build, the belt buckle rule is indeed correct. For the fat, though, things are more complicated. If the tie point is to meet the belt buckle exactly and not hang plumb from the tangent point, to overcome the overhang it requires to be either attached to the belt buckle itself (perhaps with magnets) or worn too long and tucked into the trousers. This latter option has the effect of leaving the wearer looking like he has been giftwrapped.
The popular alternative, to wear the tie short, looking like a silken expeditionary force run out of courage or provisions at the edge of the Great Belly Tundra, is also unacceptable.
So, the answer is: magnets.
— calum, May 12 2005
Serious question: Why does anyone wear a tie?
As a draught excluder for your shirt buttons?
As a reminder and pointer to your belt buckle? (even then, it only works reliably when you're standing up straight)
C'mon, it is the single most pointless and ridiculous item of "clothing" ever inflicted. OK, second most ridiculous - bow-ties come top - all they do is hide a collar button, and face it, collars are only there to hide ties. </rant>
— coprocephalous, May 12 2005
i remember reading that the neck tie derived from the bib.
i've been told that a good way to measure is to make the front of the tie, when draped around the neck in the beginning, reach down to the bottom of your scrotum. i'm six foot one and overweight, this procedure will take into account any gut, but i still have to be sure to buy long enough ties.
works for me.
— -wess, May 12 2005
But what if you just got out of the pool?
— RayfordSteele, May 12 2005
Where can I subscribe to the newsletter?! [+]
— contracts, May 12 2005
I've always assumed the tie is there to distract from the fact that you've worn the same shirt three days in a row.
— moomintroll, May 12 2005
I've always been taught that the tie should reach the bottom of the belt buckle, and that they should be to-the-millimeter even. The line of the overlayer of the button area should also extend straight down to the belt, be even with the right edge of the belt, and be even with the right edge of the flap of the fly.
EDIT: Crap, I never realized what a total tool I am.
— shapu, May 12 2005
I always though some of you people had to go to the SD too.
Ties? they're ok, but figety little things.
— finrod, May 12 2005
// reach down to the bottom of your scrotum // - Is that why not many women wear ties?
I find that the right length is acheived when the front of the tie hangs about knee height. Either I tie a different knot to you - or I could gain employment in a circus as an anatomical feak. I'll leave it to your imagination which...
— MikeyTheBikey, May 13 2005
Nobody's got it right yet. The tie is the right length when the widest part of the tie is even with the top of the belt buckle.
I recommend not tying in front of a mirror. Works every time for me. If I use a mirror I almost always have to retie.
This idea would not work, since there are varying places that the first loop can be started and still work out to be the same length. The loop starting point does not determine the length of tie, it's how far up you draw the knot.
— waugsqueke, May 13 2005
I can't understand why no one's suggested selling ties with a matching set of little velcro-covered tie-extenders to enable men to always get their ties to the right length without any tedious retying.
— hippo, May 13 2005
oh wonderful tc. thank you.
— po, May 13 2005
Where can I buy one? (+)
— PauloSargaco, Jun 02 2006
[+] for the sub heading. I could use this for the once a year that I do put on a tie.
— energy guy, Jun 02 2006
YES YES YES +
— elhigh, Jun 07 2006
+ ah, the lighthearted years of the 'bakery...
— xandram, Jan 17 2013
[annotate]
back: main index | 1,311 | 5,142 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.6875 | 3 | CC-MAIN-2021-49 | latest | en | 0.920472 |
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