content stringlengths 86 994k | meta stringlengths 288 619 |
|---|---|
Pascal Getreuer
I am an applied mathematician specialized in image processing. I am a software engineer at Google, a part time collaborator to CMLA, ENS Cachan, and an editorial board and tech team member of the
journal Image Processing On Line.
My interests: image and signal processing, inverse problems, scientific computing, software development
Disclaimer: The opinions stated here are my own, not necessarily those of my employer.
→All publications | {"url":"http://www.getreuer.info/home","timestamp":"2014-04-21T07:03:33Z","content_type":null,"content_length":"22990","record_id":"<urn:uuid:a4770a51-f0bb-4073-b279-420ba5098be6>","cc-path":"CC-MAIN-2014-15/segments/1397609539665.16/warc/CC-MAIN-20140416005219-00035-ip-10-147-4-33.ec2.internal.warc.gz"} |
Post a reply
ive figured out a good method that renders the objects perfectly, at arbitrary depth, and works for any implicit function
Basicly: i split the viewing area into equally sized squares.
for each square:
i evaluate the function for each corner, and also whether that point is inside (if the value is less than 0)
if all are inside, or all are outside, i finish here:
then i go through the following 6 cases:
top left is inside ONLY or outside ONLY
top right is inside ONLY or outside ONLY
bottom left is inside ONLY or outside ONLY
bottom right is inside ONLY or outside ONLY
(top is inside AND bottom is outside) OR (top is outside AND bottom is inside)
(left is inside AND right is outside) OR (left is outside AND right is inside)
then for each case, i use the function values to interpolate linearly for where it is assumed f(x,y) would be 0 from the two function values (not perfect, but a good approximation) then same for the
other side for whatever case it is, and draw a line.
the red graph is for the square size of 50pixels (pretty terrible)
the green graph is for the square size of 25pixels (not too bad)
the blue graph is for the square size of 10pixels (almost perfect)
the black graph is for teh square size of 5pixels (perfect for all itensive purposes)
0.02x³ - 2x² + y² + 2xy - 100 = 0
20cos(0.08y) - 50sin(0.04x) - 0.004xy= 0
i absolutely adore the second graph, and third graph is just immense
thinking about it, this could be directly extended into 3D aswell, only you would have a few more cases to check for, but otherwise it is directly extendable | {"url":"http://www.mathisfunforum.com/post.php?tid=7829&qid=77284","timestamp":"2014-04-21T02:09:27Z","content_type":null,"content_length":"24403","record_id":"<urn:uuid:223705f8-4cb8-4488-a412-317160e03aaa>","cc-path":"CC-MAIN-2014-15/segments/1397609539447.23/warc/CC-MAIN-20140416005219-00026-ip-10-147-4-33.ec2.internal.warc.gz"} |
Domestic biogas plants Sizes and dimensions Chapter 1 to 3
Domestic biogas plants Sizes and dimensions ( Chapter 1)
1 Plant size range.
As each situation differs in terms of e.g. gas requirement or available feeding material, a unique plant size could be calculated for each household. For larger dissemination programmes, however,
this would be impractical. Therefore, most such programmes work with a limited number of plant sizes that are expected to cover the demand of (most) households. The following example shows how to
develop a plants size range with the lowest number of sizes covering a certain demand scope.
1.1 Plant size parameters.
Although size calculations can become very complicated, for domestic application the following parameters suffice to arrive at a practical plant size range
│Parameter │Explanation │Values used │
│Dung/water│ Theoretically, the dung / water ratio depends on the total solids (TS) concentration of the dung, whereby │The TS values suggest a dung / water ratio of a little under 1 : 1 for cattle │
│ratio │optimum fermentation results are claimed at 6 to 7% TS. The TS of dung varies considerably, for livestock │dung and 1 : 2 for pig dung. For practical reasons.A 1 : 1 ratio has the │
│ │in development countries TS values in the 10 to 15% (cattle) and 15 to 20% (pigs) range are reported. │advantage that households can easily measure the amount of required process │
│ │ │water. │
│Specific │The specific gas production of dung depends on the type and quality of dung. │For cattle, typically 1 kg of dung fed to a digester produces about 40 litres │
│gas │ │of biogas per day. Values for other substrates will differ; pigs, poultry and │
│production│ │human excreta typically have higher yields. │
│Minimum │Depending on construction costs and gas demand pattern, below a certain nominal gas production the │One cubic meter of biogas daily will render 2.5 to 3.5 stove hours. This │
│gas │investment becomes uninteresting for the household. │could, depending on family size, suffice for e.g. breakfast and lunch │
│production│ │preparation, and would then provide a meaningful contribution. │
│Hydraulic │The hydraulic retention time (HRT) is the period the dung/water mix fed to the installation remains in the │Typical HRTs for domestic (simple) biogas plants are 40 to 60 days for warm │
│Retention │plant. As the fermentation process works better at higher ambient temperatures, installations in warmer │climates and 50 to 75 days for temperate climates. │
│Time │climates can work with a shorter HRT and vice versa. As a longer HRT requires a larger digester volume, │ │
│ │plants become more expensive to construct. │ │
│Gas │Biogas is generated more or less continuously, but consumption in households typically takes place during 3│For the gas storage volume, a fixed share of the maximum amount of daily │
│storage │or 4 periods during the day. The generated gas needs to be stored in the installation. │generated gas, 60% is taken │
│volume │ │ │
The table summarizes the values of the main design parameters. As the aim is to develop a plant size range, the hydraulic retention time is applied as a minimum and maximum value.
Calculation examples hereunder use parameter values for a warm climate
Domestic biogas plants Sizes and dimensions ( Chapter 1
Plant size range calculations (Chapter 2)
Biogas Plant Design (Chapter 3)
Plant size range calculations (Chapter 2)
1.2 Plant size range calculations
1.2.1 For the smallest plant size (“size 1”) in the range:
i. The minimum daily substrate feeding (min sub fee[1]) is equal to the minimum required gas production for the smallest installation divided by the specific gas production:min sub fee[1] =
min gas prod[1] / spec gas prod.
Or: min sub fee[1] = 1.00 / 0.040 = 25 [kg dung day]
ii. A feeding of dung of 25 [kg dung / day] requires with a 1:1 dung to water ratio an equal amount of water. The minimum feeding(min fee[1]) to the plant thus arrives at 50 [ltr / day].
iii. For the situation in which the daily feeding corresponds with the minimum feeding amount for which the plant will be designed, the hydraulic retention time is maximal (HRT max). The
required digester volume (dig vol[1]) is equal to the hydraulic retention time multiplied by the daily feeding: dig vol[1] = HRTmaxxmin fee[1].
Or: dig vol[1] = 60 x 50 = 3,000 [ltr]
iv. For the situation in which the daily feeding corresponds with the maximum feeding amount for which the plant will be designed, the hydraulic retention time is minimal (HRTmin). The
maximum feeding (max fee[2]), then, equals the digester volume divided by the minimum hydraulic retention time: max fee[1] = dig vol / HRTmin
Or: max fee[1] = 3,000 / 40 = 75 [ltr/day]
v. A maximum feeding of 75 [ltr/day], with a dung / water ratio of 1:1, then requires a maximum substrate feeding (max sub fee[1]) of 37.5 [kg/day].
vi The maximum gas production of this installation equals the maximum substrate feeding (max sub fee[1]) multiplied by the specific gas production: max gas prod[1] = max sub fee[1] x spec
gas prod.
Or: max gas prod[1] = 37.5 x 0.040 = 1.5 [m^3 biogas/day]
vii The required gas storage volume for this plant then is 60% of the maximum daily gas production.
Or: gas stor vol[1] = 0.6 x 1.5 = 0.9 [m^3]
viii The resulting main dimensions of plant size 1 then are:
Digester volume: 3,00 m^3
Gas storage volume: 0.90 m^3
Total plant volume: 3.90 m^3
1.2.2 For the second smallest plant size (“size 2”) in the range:
i. For a range of plant sizes, the minimum daily feeding for the next size (min sub fee[2]) is equal to the maximum feeding for the next smallest installation (max sub fee[1]). The minimum
feeding for plant size 2 then equals 75 [ltr/day] or, for the maximum substrate feeding (max fee[2]), 37.5 [kg dung/day].
ii. The minimum gas production for size 2, consequently, is equal to the maximum gas production of size 1, and equal to the minimum feeding multiplied by the specific gas production: min gas
prod[2] = min sub fee[2] x spec gas prod.
Or: min gas prod[2] = 37.5 x 0.040 = 1.5 [m^3 biogas/day]
ii. Similar to 1.21. ii, for the situation in which the daily feeding corresponds with the minimum feeding amount for which the plant will be designed, the hydraulic retention time is
maximal (HRT max). The required digester volume (dig vol[2]) is equal to the maximum hydraulic retention time multiplied by the daily feeding: dig vol[2] = HRTmax x min fee[2].
Or: dig vol[2] = 60 x 75 = 4,500 [ltr]
iii. For the situation in which the daily feeding corresponds with the maximum feeding amount for which the plant will be designed, the hydraulic retention time is minimal (HRTmin). The
maximum feeding (max fee[2]), then, equals the digester volume divided by the minimum hydraulic retention time: max fee[2] = dig vol[2] / HRTmin
Or: max fee[1] = 4,500 / 40 = 112,5 [ltr/day]
This corresponds with a maximum substrate feeding (max sub fee[2]) for this size of 56.25 [kg dung/day]
vi The maximum gas production of this installation equals the maximum substrate feeding (max sub fee[2]) multiplied by the specific gas production: max gas prod[2] = max sub fee[2] x spec
gas prod.
Or: max gas prod[2] = 56.25 x 0.040 = 2.25 [m^3 biogas/day]
vii The required gas storage volume for this plant then is 60% of the maximum daily gas production.
Or: gas stor vol[2] = 0.6 x 2.25 = 1.35 [m^3]
viii The resulting main dimensions of plant size 2 then are:
Digester volume: 4,50 m^3
Gas storage volume: 1.35 m^3
Total plant volume: 5.85 m^3
[1]This is an approximation; in reality a similar amount of feeding will generate slightly more biogas in a larger installation as the retention time is longer.
Domestic biogas plants Sizes and dimensions ( Chapter 1
Plant size range calculations (Chapter 2)
Biogas Plant Design (Chapter 3)
Biogas Plant Design
Biogas Plant Design (Chapter 3)
2 Plant design
A simple plant design similar to the designs in Vietnam (KT), Cambodia (modified Dheenbandu) or Tanzania (modified Camartec) is used for this example.
In this hemi-spherical design, the digester volume is the volume under the lower slurry level(LSL), and the gas storage volume is the volume between the lower and higher slurry level (HSL).
For all plants with internal gas storage, the gas storage volume in the plant is equal to the volume of the compensation volume
2.1 Total plant volume
digester volume dimensions
As pic 1 shows, part of the dome volume, over the higher slurry level, is not used by a well functioning installation. The volume, often referred to as “dead volume” is required, however, to
accommodate the floating layer on top of the slurry. In addition, when gas production is less then nominal (cold seasons) or when gas is slowly leaking, the higher slurry level can rise (up to
overflow level). For that reason, the total plant volume used for dimensioning should be higher than the plant size range volume results. For this example, 20% addition is allowed for this dead
volume. Hence, taking plant size 1 for the example, the total plant volume (digester + gas storage + dead volume) arrives at 3.90 x 1.2 = 4.68 m^3[.]
2.2 Dome radius
The dome radius, R dome, follows from: R dome = (V tot / 2/3 π)^1/3. For plant size 1, then:
R dome[1] = (4.68 / ^2/[3] π)^1/3 = 1.3 [m]
2.2 Calculating digester volume dimensions
This calculation serves to find the upper level of the digester volume, or the height of the lower slurry level (LSL) in the dome. For this, the digester volume (calculated at 3.00 m^3 in 1.2.1 iii)
can be seen as the total dome volume minus the volume of the “dome cap”. In this way V dig cap[1] = V dome[1] – V dig[1].
Or: V dig cap[1] = 4.68 – 3.00 = 1.68 [m^3]
To apply the formula V dome cap equals ^π/[6] x h x (3a^2 + h^2), the dome (R dome = 1.3 [m]) has to be draw precisely on scale. Through “trial and error”, then, you should find when h = 0.7 [m] a
will be 1.66 [m],
and V dome cap[1] = ^π/[6] x 0.7 x (3 x 1.66^2 + 0.7^2) = 1.66 [m^3] .
The digester volume, then, results in V dig[1] = Vdome[1] – Vdome cap[1]
= 4.68 – 1.66 = 3,02 [m
] which is close to the design volume of 3.00 [m
]. The LSL equals Rdome
minus a
; LSL
= 1.30 – 0.70 = 0.60 [m]
gas storage volume dimensions
2.3 Calculating gas storage volume dimensions
The gas storage volume should be at least 0.90 [m^3] (see 1.2.1 vii).
The volume of a segment of a sphere is:
V segment = ^π/[6 ]x (3R1^2 + 3R2^2 + h^2) x h
R1 is equal to a(from 2.3); R1 thus equals to 1.66 [m].
To find R2 you’ll have to use the drawing again, and measure h and R2 . You’ll find that when h = 0.25 [m] then R2 = 0,97 [m] and
Vseg1 = ^π/[6 ]x (3 x 1.16^2 + 3 x 0.97^2 x 0.25^2) x 0.25 = 0.91 [m^3].
2.4 Plant dimensions
From the above follows for plant size 1 that:
R dome = 0.130 [m]
LSL = 0.130 – 0.70 = 0.60 [m]
HSL = 0.60 + 0.25 = 0.85 cm
The LSL is also the height of the manhole entry in the plant (beam height or, for Vietnam, outlet pipe height)
The HSL is also to floor level height of the compensation chamber.
2.4 Overflow height.
The height of the overflow determines:
- the maximum pressure in the plant;
- the extent to which slurry can reach into the gas dome pipe, and;
- of course, the height determines the dimensions of the compensation chamber.
For the positioning of the overflow, there are two schools of thought:
1 The overflow should be positioned under the bottom of the dome pipe. This will avoid slurry reaching the bottom of the gas dome pipe. Slurry can reach the bottom of the dome
pipe when plants are leaking gas or when, for temperature reasons or other, the gas production is significantly lower than the gas consumption over a prolonged period of time.
2 The overflow should be positioned higher than the bottom of the dome pipe. This allows a higher maximum pressure in the plant and makes the compensation chamber dimensions more
economic. Slurry entering the dome pipe, then, is an indication of a mistake in the construction of the operation of the installation, and should be remedied.
In this example the overflow is placed 5 cm under the top of the dome.
overflow and pressure height
2.5 Pressure height check
The pressure height is the maximum pressure that the installation can produce. This maximum pressure is limited by the LSL; when pressure increases to the point whereby the LSL is pushed down further
below the beam / outletpipe level, biogas will escape through the compensation chamber.
As shown in the picture, the pressure height (ph) is the difference between overflow height (oh) and LSL.
Compensation chamber dimensions
ph[1] = 1.25 - 0.60 = 0.65 [m].
2.6 Compensation chamber dimensions
The volume of the compensation chamber (V cc) shall be equal to the plant’s gas storage volume. In case of “size 1”, then, V cc shall be 0.9 [m^3]. Following the earlier position that the overflow
level should be lower that the top of the dome, the compensation chamber height (cch) is the difference between the overflow height (oh) and the higher slurry level (HSL) (= compensation chamber
floor level).
For the example size 1, the compensation chamber height then is 1.25 – 0.85 = 0.40 [m].
Assuming a cylindrical compensation chamber, the radius of the compensation chamber (R cc) follows from:
R cc = (V cc / (π x cch))^1/2.
Or: R cc[1] = (0.90 / (π x 0.40))^1/2 = 0.84 [m]
2.7 Inlet floor and inlet pipe
To avoid reflux, the inlet floor height (ifh) should be higher than the overflow height (oh). For the example for plant size 1, ifh is 0.15 [m] higher than oh.
To avoid biogas escaping through the inlet pipe (toilet connection!), the top of the inlet entering the dome should be below the LSL. At the same time the inlet pipe height (iph) should not be too
close to digester floor to prevent obstruction by debri. Typically, the iph should be about 0.30 [m] above the digester floor.
Finally, the inlet / pipe layout should allow entering a long stick in case of inlet pipe blockage.
Inlet floor and inlet pipe
Domestic biogas plants Sizes and dimensions ( Chapter 1
Plant size range calculations (Chapter 2)
Biogas Plant Design (Chapter 3)
Thanks for such information Felix ter Heegde (http://www.ppre.uni-oldenburg.de/download/Biogas/Biogas2011/03_20110427_Biogas_plant_sizes_and_dimensions.docx)
• http://Enteryourwebsite...
Dear sir,
I am currently building a anaerobic digester and I wish to generate Electricity from it.
Please, I need your professional advice.
• http://bio-gas-plant.blogspot.com
Respected Sir,
I have agriculture land in Dist Okara Tehsil Diplapur. I intend constructing a Bio Gas plant and electricity generation unit at my village. Please guide me to a vendor or organization that could
provide me with the technical advice and support. Thanx | {"url":"http://blog.paksc.org/2011/08/23/biogas-plant-design/","timestamp":"2014-04-16T10:09:38Z","content_type":null,"content_length":"82348","record_id":"<urn:uuid:6348c6b5-15f6-4ecf-8375-3cb0790f39c0>","cc-path":"CC-MAIN-2014-15/segments/1397609523265.25/warc/CC-MAIN-20140416005203-00105-ip-10-147-4-33.ec2.internal.warc.gz"} |
ranking compounds in terms of basicity in Organic Help Forum
Could someone please help me with the following question: Rank each compound in terms of increasing basicity: aniline, pyrrole, pyridine. The answer is: pyrrole < aniline < pyridine. What is the
logic behind this ranking? Why is aniline a weaker base than pyridine? I thought that aniline's lone pair of electrons is in a sp^3 orbital while pyridine's is in a sp^2 orbital... so wouldn't
pyridine's electrons be more tightly held due to the higher s character and therefore be a weaker base than aniline? Why is this reasoning wrong? If anyone could help me with this, it would be much
appreciated! Thank you!
Let's look at this problem differently. Because the electrons of pyrrole are needed to make the ring aromatic, it is reasonable that it should be the least acidic.
Comparing pyridine and aniline, they are not vastly different in the acidities (of their conjugate acids). Your reasoning to this point is reasonable. We need to add that aniline is connected to an
sp2 carbon. The effect of the sp2-carbon is to withdraw electrons from the aniline nitrogen. The nitrogen itself is sp2-hybridized. The pi-electrons participate in the aromatic character of the ring.
The basic electrons are orthogonal to the aromatic electrons. Who would predict that pyridine will be slightly more basic than aniline?
While it appears to be true that pyridine is slightly more basic than aniline, the difference is small. Replacing hydrogens of aniline with methyl groups results in a substituted aniline that is more
basic than pyridine. Were we predicting that also?
I would simply rationalize the results to match the data, namely the sp2-carbons are better electron withdrawers than a pyridine ring. | {"url":"http://organicchemistrymessageboard.yuku.com/topic/1535/ranking-compounds-in-terms-of-basicity","timestamp":"2014-04-21T02:03:09Z","content_type":null,"content_length":"35200","record_id":"<urn:uuid:e8e01ad9-9516-4fdc-90a9-cf601d0fbd16>","cc-path":"CC-MAIN-2014-15/segments/1397609539447.23/warc/CC-MAIN-20140416005219-00388-ip-10-147-4-33.ec2.internal.warc.gz"} |
Existence of a chain map lifting the identity; Alexander-Whitney/Eilenberg-Zilber maps
up vote 0 down vote favorite
Some preliminary definitions:
Let $\Pi = \langle \alpha | \alpha^2 = 1\rangle$ be the cyclic group of order $2$ and let $\mathbb{Z}\Pi$ denote the group ring of $\Pi$ over $\mathbb{Z}$. Embed $\Pi$ in $\mathbb{Z}\Pi$ by letting
$1:\Pi \rightarrow \mathbb{Z}$ be the map defined by $1(1) = 1$ and $1(\alpha) = 0$ and letting $\alpha:\Pi \rightarrow \mathbb{Z}$ be the map defined by $\alpha(1) = 0$ and $\alpha(\alpha) = 1$.
Now let $(W,\partial)$ be the chain complex defined as follows. Set $W_n$ as the free $\mathbb{Z}\Pi$-module on the single generator $e_n$ if $n \geq 0$ and set $W_n = 0$ otherwise. Let $\partial_n$
be the zero homomorphism for all $n \leq 0$ and let $\partial_n$ be the homomorphism defined by $\partial_n(r) = (\alpha+(-1)^n)(r)$ for all $n > 0$. Lastly, let $\varepsilon:W_0 \rightarrow \mathbb
{Z}$ be the augmentation map defined by setting $\varepsilon(1) = \varepsilon(\alpha) = 1$.
The problem:
I have managed to show that the augmented complex $(W,\partial,\varepsilon)$ is acyclic and am currently trying to find a chain map $\nabla:W \rightarrow W \otimes_{\mathbb{Z}} W$ which lifts the
identity $\mathbb{Z} \rightarrow \mathbb{Z} \otimes_{\mathbb{Z}} \mathbb{Z}$. I have been told that it is possible to take $\nabla(e_0) = e_0 \otimes e_0$ and $\nabla(e_1) = e_1 \otimes \alpha e_0 +
e_0 \otimes e_1$ and so on, but I can't seem to figure out how this works. Alexander-Whitney maps and Eilenberg-Zilber maps seem like they might be relevant, but I am uncertain how to use them here.
Would someone help orient me in the right direction here?
at.algebraic-topology homological-algebra
so you need to define a chain map by specifying its values on each $e_i$. Use the fact that is a chain map so you know... this should help. Also, this question is more appropriate for
math.stackexchange.com – Sean Tilson Jun 3 '12 at 3:35
Incidentally, it seems you are confusing the group ring $\mathbb{Z}\Pi$ (the group of formal linear combinations of elements of $\Pi$) with its dual (the group of functions from $\Pi$ to $\mathbb
{Z}$). For a finite group like $\Pi=\mathbb{Z}/2\mathbb{Z}$ there is not a huge difference, but for infinite groups the distinction is very important. For example, if $\Pi$ is countably infinite,
the group ring $\mathbb{Z}\Pi$ will be countably infinite, while its dual is uncountable! – Tom Church Jun 3 '12 at 5:02
If you consider only the mappings $\Pi \rightarrow \mathbb{Z}$ with finite support, then the two definitions of group ring ought to coincide. – James Miller Jun 3 '12 at 5:44
add comment
1 Answer
active oldest votes
Chapter XI "Products" of Cartan and Eilenberg's 1956 classic book "Homological Algebra" includes an explicit formula on page 219.
up vote 2 down vote accepted
add comment
Not the answer you're looking for? Browse other questions tagged at.algebraic-topology homological-algebra or ask your own question. | {"url":"http://mathoverflow.net/questions/98693/existence-of-a-chain-map-lifting-the-identity-alexander-whitney-eilenberg-zilbe?sort=newest","timestamp":"2014-04-18T18:44:20Z","content_type":null,"content_length":"56148","record_id":"<urn:uuid:212b4f81-80fd-4fa7-ba01-17dbf378887d>","cc-path":"CC-MAIN-2014-15/segments/1398223206118.10/warc/CC-MAIN-20140423032006-00158-ip-10-147-4-33.ec2.internal.warc.gz"} |
Resilience Science
A special report on the future of finance in The Economist Fallible mathematical models: In Plato’s cave:
… although the normal distribution closely matches the real world in the middle of the curve, where most of the gains or losses lie, it does not work well at the extreme edges, or “tails”. In
markets extreme events are surprisingly common—their tails are “fat”. Benoît Mandelbrot, the mathematician who invented fractal theory, calculated that if the Dow Jones Industrial Average
followed a normal distribution, it should have moved by more than 3.4% on 58 days between 1916 and 2003; in fact it did so 1,001 times. It should have moved by more than 4.5% on six days; it did
so on 366. It should have moved by more than 7% only once in every 300,000 years; in the 20th century it did so 48 times.
In Mr Mandelbrot’s terms the market should have been “mildly” unstable. Instead it was “wildly” unstable. Financial markets are plagued not by “black swans”—seemingly inconceivable events that
come up very occasionally—but by vicious snow-white swans that come along a lot more often than expected.
This puts VAR in a quandary. On the one hand, you cannot observe the tails of the VAR curve by studying extreme events, because extreme events are rare by definition. On the other you cannot
deduce very much about the frequency of rare extreme events from the shape of the curve in the middle. Mathematically, the two are almost decoupled.
The drawback of failing to measure the tail beyond 99% is that it could leave out some reasonably common but devastating losses. VAR, in other words, is good at predicting small day-to-day losses
in the heart of the distribution, but hopeless at predicting severe losses that are much rarer—arguably those that should worry you most.
When David Viniar, chief financial officer of Goldman Sachs, told the Financial Times in 2007 that the bank had seen “25-standard-deviation moves several days in a row”, he was saying that the
markets were at the extreme tail of their distribution. The centre of their models did not begin to predict that the tails would move so violently. He meant to show how unstable the markets were.
But he also showed how wrong the models were.
Modern finance may well be making the tails fatter, says Daron Acemoglu, an economist at MIT. When you trade away all sorts of specific risk, in foreign exchange, interest rates and so forth, you
make your portfolio seem safer. But you are in fact swapping everyday risk for the exceptional risk that the worst will happen and your insurer will fail—as AIG did. Even as the predictable
centre of the distribution appears less risky, the unobserved tail risk has grown. Your traders and managers will look as if they are earning good returns on lower risk when part of the true risk
is hidden. They will want to be paid for their skill when in fact their risk-weighted returns may have fallen. | {"url":"http://rs.resalliance.org/tag/var/","timestamp":"2014-04-23T20:26:13Z","content_type":null,"content_length":"51737","record_id":"<urn:uuid:c46c1d56-33cb-41b5-b99a-9813c7f7feda>","cc-path":"CC-MAIN-2014-15/segments/1398223203422.8/warc/CC-MAIN-20140423032003-00235-ip-10-147-4-33.ec2.internal.warc.gz"} |
Got Homework?
Connect with other students for help. It's a free community.
• across
MIT Grad Student
Online now
• laura*
Helped 1,000 students
Online now
• Hero
College Math Guru
Online now
Here's the question you clicked on:
help please(: 1 over 3 (2x − 8) = 4
• one year ago
• one year ago
Your question is ready. Sign up for free to start getting answers.
is replying to Can someone tell me what button the professor is hitting...
• Teamwork 19 Teammate
• Problem Solving 19 Hero
• Engagement 19 Mad Hatter
• You have blocked this person.
• ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.
This is the testimonial you wrote.
You haven't written a testimonial for Owlfred. | {"url":"http://openstudy.com/updates/50e348cbe4b0e36e35142b74","timestamp":"2014-04-20T03:14:03Z","content_type":null,"content_length":"98107","record_id":"<urn:uuid:e0c482bc-7d27-4883-9211-1a211dad605b>","cc-path":"CC-MAIN-2014-15/segments/1397609537864.21/warc/CC-MAIN-20140416005217-00003-ip-10-147-4-33.ec2.internal.warc.gz"} |
trig help
April 11th 2009, 06:04 PM #1
Apr 2009
trig help
the terminal arm of an angle in standard position passes through (3,11). Find the radian value of the angle in the interval [0,2pi], to the nearest hundredth.
i dont understand how to solve this problem. help is appreciated
answer is:1.30
From the information we know that there is a point 3 away from start and in the y direction there is a point 11 away.
This will form a triangle with opposite 11 and adjacent 3. From there should be able to use a basic trig function to solve
why do i use tan and not the other trig ratios?
Because the (3,11) refers to an x and a y value. As it must be in the first quadrant that means the angle is measured from the horizontal (as is convention). If you sketch the diagram it is
You could use pythagoras theorem to find the hypoteneuse and use another ration but that could introduce rounding errors
ooh, okay thank you for your help ! (:
April 11th 2009, 06:14 PM #2
April 11th 2009, 06:16 PM #3
Apr 2009
April 11th 2009, 06:32 PM #4
April 11th 2009, 06:36 PM #5
Apr 2009 | {"url":"http://mathhelpforum.com/trigonometry/83293-trig-help.html","timestamp":"2014-04-17T22:17:57Z","content_type":null,"content_length":"42495","record_id":"<urn:uuid:faf7e498-2d5e-4f0c-8856-8c8e5e0c731f>","cc-path":"CC-MAIN-2014-15/segments/1397609532128.44/warc/CC-MAIN-20140416005212-00366-ip-10-147-4-33.ec2.internal.warc.gz"} |
Pauline's Ideas for Mixing Math and Food
"A good education for every child does not mean the same education for every child. "
Please use what works for you and your child and ignore the rest. Every child is different. If this information does not seem to be a good fit for your particular child, then keep looking and
experimenting until you find what works.
This Web Page by Pauline Harding for Art Nurk.
Contents may be copied for personal use if credit is given. Please ask for permission before any other use. Do not copy this information onto your own web site without permission.
● Geometry – shapes – spherical oranges, circles of salami, squares of cheese, cubes of cheese
● Arranging patterns on the plate- alternating tomatoes and onions on a platter for a party
● Using cookie cutters to make shapes – which shapes tessellate?
● Counting everything –jellybeans, carrots, forks, peas
● Sorting and counting a bag of M&M’s – how many browns? How many oranges? Make a graph. Now subtract!
● Grouping by twos or five or tens – each person gets two cookies – how many do we have? Do we have enough?
● Putting out a fork for each person, a spoon for each plate.
● Buying snacks at the pool. How much money do we have? How much do things cost? How can we get something for each person? What about the change?
● How long until dinner? How will we know? What does the clock say?
● Measuring cups and spoons. How many half cups in a cup? How many quarter cups?
● Weight. A pound of pasta, a pound of butter, five pounds of flour, an ounce of cheese.
● Adding an extra person for dinner. How many plates? How many kids? How many adults? How many in all?
● Subtraction of cookies.
● Making patterns on Christmas cookies.
● Making a food pyramid chart, a chore chart, a what’s for dinner chart.
● Surveying the family – rice or potatoes? Tallying the vote.
● What does a hundred look like? A hundred cheerios, jellybeans, chocolate chips. How many more make a thousand?
● How many chocolate chips in each cookie? How many in the whole batch?
● Measuring the temperature of the candy or fudge.
● Estimating which container has the capacity for the leftovers.
● Dropping food coloring into glasses of water. Ratio of red to yellow when making orange.
● Does this cup hold more or less water? What if we use a shorter, fatter cup?
● If four people eat one slice of a pizza cut into eighths, how much pizza is left?
● How can we make twice as many cookies in one batch?
● Who gets the first pancake? Who gets the second? Third?
● Estimating how many beans in the jar. Were you right? Count and see!
● Reading stories about food. Reading recipes. Following the directions on the cake mix box.
● How many pieces do we need from this rectangular cake? What are the different ways we can cut it?
● If the muffin pan has four cups in one direction and three in the other, how many muffins can we make?
● How many pieces of pizza do you think the people at our party will eat?
● Decorating round cakes using radial symmetry.
● Using toothpicks in marshmallows to make geometric models.
● Sculpting bread dough.
● Dividing the cookies among the number of kids present. Deciding what to do with the remainder.
● Half of the jellybeans is how many? How many is one third?
● Cooking with metric units.
● Throwing eggs out the window wrapped in straws and Popsicle sticks and tape. Do they break?
● Saving seeds from apples and melons and tomatoes. Sorting them. Planting them – how high do they grow? Make a chart.
● Cut a stalk of celery in half at the bottom. Put each end in water with food coloring. What happens to the leaves?
● Shake the can of mixed nuts. Now open it. Where are the big nuts? Where are the small ones? Why?
● How much sugar is in this cereal? How much fiber? What about that one? Which costs more? Why?
● Sorting cans and boxes in the cabinets. How do they fit together? Which one is taller? Cylinders and rectangular solids.
● Properties of matter: solids, liquids, gasses. Ice, water, steam.
● How hot is the oven? How cold is the freezer? How do we know? How does it work? | {"url":"http://home.comcast.net/~askpauline/math/mathandfood.html","timestamp":"2014-04-19T22:18:56Z","content_type":null,"content_length":"19170","record_id":"<urn:uuid:1927db53-f6bd-4235-965f-ddcd6484ccef>","cc-path":"CC-MAIN-2014-15/segments/1397609537754.12/warc/CC-MAIN-20140416005217-00483-ip-10-147-4-33.ec2.internal.warc.gz"} |
Hubble's Dark Constant
published Feb 8, 2013
Numb: The Hubble Constant is currently estimated in 2009 to be 74.2 ± 3.6 (km/s)/Mpc.
Number: The Hubble Constant is currently estimated in 2009 to be 74.2 ± 3.6 (km/s)/Mpc. This constant of proportionality represents the ratio of the velocity of a galaxy receding from us to its
distance from Earth.
Num-best: Based on astronomical observations, Edwin Hubble first formulated his law in 1929 that the velocity (kilometers per second) of a galaxy receding from Earth is proportional to its distance
(Megaparsecs) from us. Hubble's original estimate, 500 (km/s)/Mpc, indicates the difficulty in measuring this constant of proportionality which is currently calculated to be 74.2 ± 3.6 (km/s)/Mpc,
using the latest data from the Hubble Space telescope.
The basic idea seems simple enough, astronomers Vesto Slipher (1917) and Edwin Hubble (1929) analyze light spectrum from distant celestial objects and notice the tell-tale shift of the signature
spectral lines towards the red end of the spectrum:
See my column on Planck's constant for more information on the light spectrum. The reason Hubble has a constant, and not poor Vesto, is that Hubble was specifically looking for such redshift to
confirm that the universe was expanding as boldly predicted by Lemaitre. This prediction was bold because it contradicted the great Einstein, who started the modern era in cosmology with his theory
of general relativity, which describes gravitation as a consequence of the curvature of space-time. Einstein believed the universe was static (the prevailing view at the time) and thus was concerned
that gravity would make the universe collapse according to his field equations; so he introduced the infamous "cosmological constant"into his equations to prevent such collapse in 1917:
The capital lambda represents his "anti-gravity" term. Can you find the capital lambda? This reminds me of the old math joke, where a question on a math test asks the students to find x ( an all too
common question!) and a bright student simply circles the letter x. This innocuous looking equation actually contains tensors, and if there was one thing that drove me to fits in grad school it was
tensor notation. So all we are going to do with this equation is admire its simplicity, be amazed that it can describe the fate of the entire universe, and circle lambda!
Of interest to us, is that following Einstein's published solutions to his equation, the Russian Alexander Alexandrovich Friedman showed that solutions exist which would indicate an expanding or
contracting universe depending on how matter was distributed. This was just so much mathematical theoretical hogwash to the physicists until the Belgian priest and astronomer Georges Henri Joseph
Edouard Lemaitre independently published the Friedman solutions in 1927 with the additional important caveat: an expanding universe could be empirically verified! Lemaitre noted that every galaxy in
an expanding universe would be receding from every other galaxy at a rate that increases with distance separation. This is what Hubble directly verified in his 1929 paper and partly why he is so
famous; also note that this expanding universe scenario is where the "big bang theory" comes from. The following is a graphic, illustrating the proportionality of distance and velocity:
There are several gedankenexperiments ("thought experiments") that need to be more fully explored before moving on to the darker side of this column. The first is trying to imagine how every galaxy
can recede from every other galaxy. A standard visual aid is a balloon being blown up with dots on it, as the balloon expands all the dots get farther away from each other. Martin Gardner, the math
puzzle guy, offers a more substantial image of a ball of dough with raisins in it expanding as it cooks. I prefer a 4 dimensional glob of dark matter with isolated singularities dispersing into a 12
dimensional space time manifold due to the mysterious quintessence of a dark energy field... but that's just me. Next gedanken, the redshift of the light from the galaxies moving apart is due to the
actual stretching of the universe itself (think ball of dough stretching as it expands), not from the traditional Doppler shift of moving objects. Finally try to imagine why the velocity of raisins
in the doughball is proportional to the distance between them, i.e. why is a raisin moving twice as fast from another raisin if it is twice as far? Mathematically this is a consequence of the theorem
that states any two points which are moving away from the origin, each along straight lines and with speed proportional to distance from the origin, will be moving away from each other with a speed
proportional to their distance apart (like that actually helps).
The velocities of the receding galaxies are easier to calculate, using the observed redshift in spectral lines, than the distances of the galaxies, which require knowing how bright the celestial
object is supposed to be and comparing that with the observed brightness. Astronomers have refined their distance estimations using improved data collection, as well as finding more easily measurable
bright objects in the sky. Type Ia supernovae, exploding stars that briefly shine as brightly as 10 billion suns, are rare examples. In 1998 two teams of astronomers studying just such supernovae
determined that they were dimmer than expected, leading them to conclude that they were farther than estimated using Hubble's constant... meaning that the expansion of the universe must be speeding
up! This completely unexpected result has been further supported with more observations, leading to cosmologists looking for some kind of anti-gravity term that could account for this acceleration...
Amazingly they have turned back to Einstein's cosmological constant, which he had called "his biggest blunder" after seeing Hubble's results.
Physicists now refer to the needed "stuff" to cause the accelerating expansion: "dark energy." To be precise (if possible when speculating), there are two competing forms of this hypothetical energy:
one a constant energy density field versus a variable energy density field called "quintessence." For those of you who still remember the hypothesized "aether" from the turn of the last century
(actually the idea dates back to Aristotle, with aether being the fifth element with air, earth, fire and water, hence the "quint" in quintessence), the following statement should be eerily familiar:
"The exact nature of dark energy is a matter of speculation. It is known to be very homogeneous, uniformly filling otherwise empty space, not very dense and it is not known to interact through any of
the fundamental forces other than gravity." This is incredibly exciting news!!! For those of you who thought you missed the boat when Einstein replaced the Newtonian paradigm with relativity theory,
and quantum theory blew the socks off of just about everyone, this is a chance to make a real contribution to our knowledge of the universe. Just as the "aether" and problems related to it led to a
deeper understanding of light and quantum phenomena, dark energy will certainly push young scientists to come up with creative explanations. A word of caution is warranted here, note that I say young
"scientists" in the previous sentence. Lots of people like to "speculate" but to really contribute to the field of physics requires years of hard work, persistence and study in said field. Getting
back to dark energy, scientists have calculated how much of this stuff is out there, even though they have no idea what it actually might be:
Wow, three-quarters of the pie is made up of something that we didn't know about 10 years ago. Some of you may have noticed that almost all of the rest of the pie is "dark matter" which has at least
been around since 1934. Fritz Zwicky coined this term to help explain observed galaxy rotation and galaxy cluster formation. So only 4% of the universe consists of stuff we can actually "explain",
which is in quotes because quantum theory still has many conundrums that leave our understanding of ordinary matter with much to be desired. Jermey Bernstein's nice little book, Quantum Leaps,
provides an excellent insider look at the development of the quantum theory with particular focus on John Bell and his theorem related to the EPR paradox and the measurement problem. On page 172 he
asks: "But what is the explanation for the correlations of the widely separated entangled photons or electrons?", and then answers: "If you accept the usual interpretation of the quantum theory,
there isn't one." In order to avoid taking one of his "quantum leaps" I will say no more on this subject here, except that someone now needs to write a book called, Cosmic Leaps.
Finishing up our story on Hubble and his erstwhile "constant", which now is accepted as varying according to the cosmological model being used, we find ourselves in a universe comprised largely of
energy and matter that we cannot explain. Given such uncertainties in the science of cosmology it might come as no surprise that a Scientific American article was published in March 2008 titled, The
End of Cosmology. The authors do not propose that all cosmologists should pack up their bags since we cannot account for most of what is being observed, but rather posit that this is the golden age
of cosmology. The accelerating universe actually implies that all galaxies will eventually get so far away from us, that in roughly 100 billion years their light will not be able to overcome the
ever-increasing distance and reach us! This will leave us with only light from the Milky Way and nearby galaxies which will have since formed a ball-like super- galaxy, turning our night sky and
constellations into one big white smudge. Our sun is expected to burn out in 5 billion years so humans will probably be someplace else by then. What is interesting is that there will be no traces of
any other galaxies, thus no evidence of the expanding universe, leading future astronomers to believe in the static universe that Einstein had originally tried to derive with his cosmological
Sapere Aude!
Comment? Start the discussion about Hubble's Dark Constant | {"url":"http://serc.carleton.edu/nnn/blog_posts/hubble.html","timestamp":"2014-04-18T08:11:47Z","content_type":null,"content_length":"31873","record_id":"<urn:uuid:407e3907-6832-4897-b78b-75f07a4224e1>","cc-path":"CC-MAIN-2014-15/segments/1398223205137.4/warc/CC-MAIN-20140423032005-00138-ip-10-147-4-33.ec2.internal.warc.gz"} |
Got Homework?
Connect with other students for help. It's a free community.
• across
MIT Grad Student
Online now
• laura*
Helped 1,000 students
Online now
• Hero
College Math Guru
Online now
Here's the question you clicked on:
Using substitution, could someone please help me with this system? –y + 3x = 6 y = –6x + 12 So far I've gotten: Substituting the second equation into the first you get -(-6x+12) + 3x = 6 Which can be
solved for x after finding x you can use the second equation above directly to find y
• one year ago
• one year ago
Your question is ready. Sign up for free to start getting answers.
is replying to Can someone tell me what button the professor is hitting...
• Teamwork 19 Teammate
• Problem Solving 19 Hero
• Engagement 19 Mad Hatter
• You have blocked this person.
• ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.
This is the testimonial you wrote.
You haven't written a testimonial for Owlfred. | {"url":"http://openstudy.com/updates/50d27cbce4b052fefd1d8a37","timestamp":"2014-04-24T12:36:49Z","content_type":null,"content_length":"111296","record_id":"<urn:uuid:592e9aae-7d5b-4900-9b87-7f19e4588078>","cc-path":"CC-MAIN-2014-15/segments/1398223206120.9/warc/CC-MAIN-20140423032006-00376-ip-10-147-4-33.ec2.internal.warc.gz"} |
Braingle: 'How Old, Again?' Brain Teaser
How Old, Again?
Math brain teasers require computations to solve.
Puzzle ID: #49426
Category: Math
Submitted By: fishmed
Suppose I am a grandfather. My grandson is about as old in days as my son is in weeks. He is also as many months old as I am in years. Together, the three of us are 140 years old. How old am I?
Show Hint
Show Answer
What Next?
RRAMMOHAN Very cleverly worded. A good one. Took a few minutes to solve.
Oct 07, 2012
SaraGrace Well done and tricky
Aug 01, 2013
Streuther I took "My grandson is about as old in days as my son is in weeks. He is also as many months old as I am in years."
Dec 02, 2013 to mean
MY SON is as many months old as I am in years...but when that gave my age as being over 140 years old, I reread the question and got the correct anwer
Proph perfect :-P
Jan 06, 2014
Jimmygu3 My way to solve: Say grandson is 1. Son would be 7 and grandfather 12. Adds up to 20. So multiply everything by 7 to make the sum 140: 7, 49, 84.
Feb 23, 2014 | {"url":"http://www.braingle.com/brainteasers/teaser.php?id=49426&comm=1","timestamp":"2014-04-16T16:16:32Z","content_type":null,"content_length":"27781","record_id":"<urn:uuid:6eb3585a-464e-4723-a193-cd6b3df1fada>","cc-path":"CC-MAIN-2014-15/segments/1398223201753.19/warc/CC-MAIN-20140423032001-00464-ip-10-147-4-33.ec2.internal.warc.gz"} |
Inferential Statistics Help for Types of Statistics - Transtutors
To clarify it bit further suppose we have the data of a good and its consumption in various months. Even though we don’t get direct implications from it but what we can infer is which good has more
demand in which month and why as it has not been mentioned in the raw data. So we just came with the conclusion from the raw data what common people will think. So to make assessment in any data
which has been obtained in groups as a chance factor can be judged with this statistics. So to conclude to make some general conclusions we use this statistics.
As inference means taking out conclusion and statistics means compilation of data collected from various sources and validating any hypothesis through it. So the definition only tells that the
statistics where we infer the conclusion even though from the raw data what we have is known as inferential statistics.
Now coming on to its application part it has found great application in the experimental and quasi experimental. To just clarify on the topic suppose we have two groups in a single class and we have
to see who performed well then we have to take average of their marks and see the difference and it can also be done by the t test which is a inferential statistics. This category of statistics is
mostly derive from the family of statistics known as the general linear model which includes many of the statistical analytical methods such as t test, ANACOVA, ANOVA regression analysis and many
other factors too such as discriminant function assessment, multidimensional scaling and so. Coming on to the main areas where inferential statistics plays a part is:
Experimental analysis- where there is a comparison between the two groups where sampling is done in randomized manner in the groups and the data’s are processed through the statistics methods of t
test and one way ANOVA. The experiments which were designed on the basis of factorial are done with the help of ANOVA by which we infer the conclusion. Randomized Block Analysis usually uses this
ANOVA which uses dummy variables. These all are experimental designs which are validated with this method.
Quasi experimental- Here there is no random assignments of any units the groups as mostly done in experimental approach like t test and f test. Since there is no randomized assignments so here the
analysis done is bit complicated. And if we do Non equivalent group design then we have to adjust first the pretest scores to measure the error and it is known as the Reliability corrected analysis
of covariance model.
So to conclude it’s a broad area of statistics where we can have many measures and methods to carry out and validate many experiments.
We at transtutors.com are aware of the complexities of this topic, and therefore our online Statistics tutors are available round the clock to help you. We offer a range of services
• Online homework help - to help students with the homework related to statistics
• Online assignment help – to help students with their assignments.
So if you are you in need of Homework Help/ Humanities Assignment Help? Our competent pools of statistics professionals are available 24*7 to give you a solution to your problem!
Related Questions
more assignments » | {"url":"http://www.transtutors.com/homework-help/statistics/inferential-statistics/","timestamp":"2014-04-19T04:43:35Z","content_type":null,"content_length":"77651","record_id":"<urn:uuid:149a2689-f323-4ce3-ae72-265d71fb2a9c>","cc-path":"CC-MAIN-2014-15/segments/1397609535775.35/warc/CC-MAIN-20140416005215-00018-ip-10-147-4-33.ec2.internal.warc.gz"} |
Transactions of the American Mathematical SocietyBibliography
ISSN 1088-6850(online) ISSN 0002-9947(print)
Hierarchical zonotopal spaces
Authors: Olga Holtz, Amos Ron and Zhiqiang Xu
Journal: Trans. Amer. Math. Soc. 364 (2012), 745-766
MSC (2010): Primary 13F20, 13A02, 16W50, 16W60, 47F05, 47L20, 05B20, 05B35, 05B45, 05C50, 52B05, 52B12, 52B20, 52C07, 52C35, 41A15, 41A63
Published electronically: September 8, 2011
MathSciNet review: 2846351
Full-text PDF
Abstract | References | Similar Articles | Additional Information
Abstract: Zonotopal algebra interweaves algebraic, geometric and combinatorial properties of a given linear map
In this paper we show that the fundamental principles of zonotopal algebra as described in the previous paragraph extend far beyond the setup of external, central and internal ideals by building a
whole hierarchy of new combinatorially defined zonotopal spaces.
Similar Articles
Retrieve articles in Transactions of the American Mathematical Society with MSC (2010): 13F20, 13A02, 16W50, 16W60, 47F05, 47L20, 05B20, 05B35, 05B45, 05C50, 52B05, 52B12, 52B20, 52C07, 52C35, 41A15,
Retrieve articles in all journals with MSC (2010): 13F20, 13A02, 16W50, 16W60, 47F05, 47L20, 05B20, 05B35, 05B45, 05C50, 52B05, 52B12, 52B20, 52C07, 52C35, 41A15, 41A63
Additional Information
Olga Holtz
Affiliation: Department of Mathematics, University of California-Berkeley, Berkeley, California 94720 – and – Institut für Mathematik, Technische Universität Berlin, Berlin, Germany
Email: holtz@math.berkeley.edu
Amos Ron
Affiliation: Department of Mathematics and Computer Sciences, University of Wisconsin-Madison, Madison, Wisconsin 53706
Email: amos@cs.wisc.edu
Zhiqiang Xu
Affiliation: LSEC, Academy of Mathematics and Systems Sciences, Chinese Academy of Sciences, Beijing, 100190, People’s Republic of China
Email: xuzq@lsec.cc.ac.cn
DOI: http://dx.doi.org/10.1090/S0002-9947-2011-05329-8
PII: S 0002-9947(2011)05329-8
Keywords: Zonotopal algebra, multivariate polynomials, polynomial ideals, duality, grading, Hilbert series, kernels of differential operators, polynomial interpolation, box splines, zonotopes,
hyperplane arrangements, matroids
Received by editor(s): October 28, 2009
Received by editor(s) in revised form: February 11, 2010
Published electronically: September 8, 2011
Additional Notes: The work of the first author was supported by the Sofja Kovalevskaja Research Prize of Alexander von Humboldt Foundation and by the National Science Foundation under agreement
DMS-0635607 and was performed in part at the Institute for Advanced Study, Princeton
The work of the second author was supported by the National Science Foundation under grants DMS-0602837 and DMS-0914986, and by the National Institute of General Medical Sciences under Grant
The work of the third author was supported in part by NSFC grant 10871196 and was performed in part at Technische Universität Berlin
Article copyright: © Copyright 2011 American Mathematical Society | {"url":"http://www.ams.org/journals/tran/2012-364-02/S0002-9947-2011-05329-8/home.html","timestamp":"2014-04-23T14:01:51Z","content_type":null,"content_length":"43843","record_id":"<urn:uuid:896b09d3-b0e0-4745-96b6-216f17589137>","cc-path":"CC-MAIN-2014-15/segments/1398223202548.14/warc/CC-MAIN-20140423032002-00096-ip-10-147-4-33.ec2.internal.warc.gz"} |
CGTalk - Subsurface Scattering Using Depth Maps
09-21-2005, 09:35 AM
How do I go about to complete this program? The part I'm confused with is how to compute the 2D lightDepthTex from the float4 dist? Does it go like... the vertex position (x,y,z) is projected to
texture space (s,t) then the depth (dist) is stored there? Thanks so much!!!
If you have a better way to do subsurface scattering please tell me. Just not the texture diffusion method (e.g. blur shaders).
//VERTEX PROG
struct a2v {
float4 pos : POSITION;
float3 normal : NORMAL;
struct v2f {
float4 hpos : POSITION;
float dist : TEXCOORD0;
v2f main(a2v IN,
uniform float4x4 modelViewProj,
uniform float4x4 modelView,
uniform float grow)
v2f OUT;
float4 P = IN.pos;
P.xyz += IN.normal * grow;
OUT.hpos = mul(modelViewProj, P);
OUT.dist = length(mul(modelView, IN.pos));
return OUT;
//FRAGMENT PROG (SEPARATE FILE)
float4 main(float dist : TEX0) : COLOR
return dist;
float trace(float3 P,
uniform float4x4 lightTexMatrix,
uniform float4x4 lightMatrix,
uniform sampler2D lightDepthTex)
float4 texCoord = mul(lightTexMatrix, float4(P, 1.0));
float d_i = tex2Dproj(lightDepthTex, texCoord.xyw);
float4 Plight = mul(lightMatrix, float4(P, 1.0));
float d_o = length(Plight);
float s = d_o - d_i;
return s; | {"url":"http://forums.cgsociety.org/archive/index.php/t-279140.html","timestamp":"2014-04-19T23:04:18Z","content_type":null,"content_length":"10280","record_id":"<urn:uuid:01355291-7fe3-4cf5-b140-2193150d1ae7>","cc-path":"CC-MAIN-2014-15/segments/1397609537754.12/warc/CC-MAIN-20140416005217-00343-ip-10-147-4-33.ec2.internal.warc.gz"} |
Re: "exponent (expo) overflow"
Karim Belabas on Thu, 27 Dec 2007 00:15:10 +0100
[Date Prev] [Date Next] [Thread Prev] [Thread Next] [Date Index] [Thread Index]
Re: "exponent (expo) overflow"
* Alessandro Languasco [2007-12-26 11:52]:
> I'm writing a GP script in which I have to compute a quite large
> power of a matrix. This is fine until I use the exponent L<=2^(23)
> but for larger L>=2^(24) I have a
> *** for: exponent (expo) overflow
> (which in fact is not too surprising; the exponent is quite large....).
> But I really need to use larger exponents.
> Do you know if is it possible to solve this problem ?
> Maybe some default() instruction could be used in this situation,
> but I don't know if it is possible and how to do this.
The allowed range of exponents is hard-coded into the t_REAL representation:
BITS_IN_LONG - 2 bits are used to code the exponent, and one of them is a sign.
So the simplest solution to your problem is to use a 64bit machine for
your computation. (Assuming you won't need exponents larger than 2^61.)
Another possibility is to split your computation and renormalize your matrix
(say divide it by 2^N for some suitable N) after each part of the
Another is to do the computation in a completely different way, but I'll
need more details ( is the matrix diagonalizable ? If not, perhaps a
suitable power of A^t A might suit you just as well ? )
Karim Belabas Tel: (+33) (0)5 40 00 26 17
IMB, Universite Bordeaux 1 Fax: (+33) (0)5 40 00 69 50
351, cours de la Liberation http://www.math.u-bordeaux.fr/~belabas/
F-33405 Talence (France) http://pari.math.u-bordeaux.fr/ [PARI/GP] | {"url":"http://pari.math.u-bordeaux.fr/archives/pari-users-0712/msg00010.html","timestamp":"2014-04-19T06:57:06Z","content_type":null,"content_length":"5842","record_id":"<urn:uuid:8cbf722d-8738-4938-806a-2bc3cbf8aabb>","cc-path":"CC-MAIN-2014-15/segments/1397609536300.49/warc/CC-MAIN-20140416005216-00044-ip-10-147-4-33.ec2.internal.warc.gz"} |
Geometry Tutors
Powell, OH 43065
Well-Known Chinese And Math Teacher With 20 Years' Teaching Experience
...I was awarded several teaching awards during my teaching career as well in China. Using my knowledge and my teaching experiences, I tutored my older son
when he was in high school in the USA. He had high rank in his school, and still enjoys the benefit...
Offering 7 subjects including geometry | {"url":"http://www.wyzant.com/Columbus_OH_geometry_tutors.aspx","timestamp":"2014-04-16T07:54:01Z","content_type":null,"content_length":"64712","record_id":"<urn:uuid:654297c9-0cac-4746-a7a8-6b50ce18a88a>","cc-path":"CC-MAIN-2014-15/segments/1397609521558.37/warc/CC-MAIN-20140416005201-00330-ip-10-147-4-33.ec2.internal.warc.gz"} |
International Workshop on QCD Green's Functions, Confinement and Phenomenology
Table of contents
Gluon mass through ghost synergy
PoS(QCD-TNT-II)001 pdf
Ghost-gluon and ghost-quark bound states and their role in BRST quartets
PoS(QCD-TNT-II)002 pdf
On the infrared behaviour of QCD Green functions in the Maximally Abelian gauge
PoS(QCD-TNT-II)003 pdf
Regge-like quark-antiquark excitations in the effective-action formalism
PoS(QCD-TNT-II)004 pdf
The dynamical equation of the effective gluon mass
PoS(QCD-TNT-II)006 pdf
AdS/QCD and Light-Front Holography: A Novel Approach to Non-Perturbative QCD
PoS(QCD-TNT-II)008 pdf
Tetraquark and the flux tube recombination
PoS(QCD-TNT-II)009 pdf
Renormalization group invariance in the Pinch Technique
PoS(QCD-TNT-II)010 pdf
The role of the running coupling constant in the unveiling of the hadronic structure
PoS(QCD-TNT-II)011 pdf
Quark masses in two-flavor QCD
PoS(QCD-TNT-II)012 pdf
Ghost dissection
PoS(QCD-TNT-II)015 pdf
Mass spectra and Regge Trajectories of Heavy Mesons and Baryons
PoS(QCD-TNT-II)016 pdf
Baryon form factors from Dyson-Schwinger equations
PoS(QCD-TNT-II)017 pdf
Applications of Dyson-Schwinger equations to heavy flavours
PoS(QCD-TNT-II)018 pdf
On the zero-mass limit for nonabelian gauge theories
PoS(QCD-TNT-II)054 pdf
Yang-Mills Theory at Non-Vanishing Temperature
PoS(QCD-TNT-II)021 pdf
Exploring Gauge-Invariant Vacuum Wave Functionals for Yang-Mills Theory
PoS(QCD-TNT-II)022 pdf
Aspects of Gribov-Zwanziger theory and QCD
PoS(QCD-TNT-II)023 pdf
Testing proposals for the Yang-Mills vacuum wavefunctional
PoS(QCD-TNT-II)024 pdf
The pole part of the three gluon vertex
PoS(QCD-TNT-II)025 pdf
Quark and gluon confinement from an effective model of Yang-Mills theory
PoS(QCD-TNT-II)026 pdf
Dense but confined matter as a new state in QCD
PoS(QCD-TNT-II)027 pdf
The structure of the residual gauge orbit
PoS(QCD-TNT-II)028 pdf
Gauge invariant parton subamplitudes
PoS(QCD-TNT-II)029 pdf
Massive gluon propagator at zero and finite temperature
PoS(QCD-TNT-II)030 pdf
Gauge invariant mass terms and wave functions
PoS(QCD-TNT-II)031 pdf
Chiral symmetry breaking with a confining propagator and dynamically massive gluons
PoS(QCD-TNT-II)032 pdf
Soft Gluon resummation and total cross-sections phenomenology
PoS(QCD-TNT-II)033 pdf
Schwinger mechanism in QCD
PoS(QCD-TNT-II)034 pdf
Effective theory for QCD at finite temperature and density from strong coupling expansion
PoS(QCD-TNT-II)035 pdf
Nonperturbative study of the four-point heavy quark Green's functions in Coulomb gauge
PoS(QCD-TNT-II)036 pdf
Background field dependence from the Slavnov-Taylor identities in (non-perturbative) Yang-Mills theory
PoS(QCD-TNT-II)037 pdf
Hamiltonian Approach to QCD in Coulomb Gauge
PoS(QCD-TNT-II)038 pdf
Hadron physics and dynamical chiral symmetry breaking
PoS(QCD-TNT-II)039 pdf
The dimension-two gluon condensate, the ghost-gluon vertex and the Taylor theorem
PoS(QCD-TNT-II)040 pdf
Properties of Delta and Omega baryons in a covariant Faddeev approach
PoS(QCD-TNT-II)041 pdf
Beyond QCD: A Composite Universe
PoS(QCD-TNT-II)042 pdf
Excited charmonium states from Bethe-Salpeter equation
PoS(QCD-TNT-II)043 pdf
Gauge-Invariant Determination of Relevant Gluonic Component for Nonperturbative QCD
PoS(QCD-TNT-II)044 pdf
Schwinger-Dyson equations in models with disorder
PoS(QCD-TNT-II)045 pdf
Fermion RG blocking transformations and IR structure
PoS(QCD-TNT-II)046 pdf
Properties of the electroweak versus the QCD-vacuum in strong magnetic fields
PoS(QCD-TNT-II)047 pdf
Heavy meson interquark potential
PoS(QCD-TNT-II)048 pdf
Schwinger-Dyson equations for manifestly gauge invariant correlators
PoS(QCD-TNT-II)049 pdf
Leading order QCD in Coulomb gauge
PoS(QCD-TNT-II)051 pdf
Vertex Sensitivity in the Schwinger-Dyson Equations of QCD
PoS(QCD-TNT-II)052 pdf
Exact bounds on the free energy in QCD
PoS(QCD-TNT-II)053 pdf | {"url":"http://pos.sissa.it/cgi-bin/reader/conf.cgi?confid=136","timestamp":"2014-04-16T04:17:22Z","content_type":null,"content_length":"21052","record_id":"<urn:uuid:ed3e5c89-53a0-488e-bfde-dfc47e6bf231>","cc-path":"CC-MAIN-2014-15/segments/1398223206672.15/warc/CC-MAIN-20140423032006-00332-ip-10-147-4-33.ec2.internal.warc.gz"} |
Patent application title: METHOD FOR DETERMINING THE PERMITTED WORKING RANGE OF A NEURAL NETWORK
Inventors: Georg Mogk (Kurten-Bechen, DE) Thomas Mrziglod (Bergisch Gladbach, DE) Peter Hubl (Koln, DE)
Assignees: BAYER AKTIENGESELLSCHAFT
IPC8 Class: AG06F1518FI
USPC Class: 706 21
Class name: Neural network learning task prediction
Publication date: 2009-08-20
Patent application number: 20090210370
Sign up to receive free email alerts when patent applications with chosen keywords are published SIGN UP
A method for checking whether an input data record is in the permitted working range of a neural network in which a definition of the complex envelope which is formed by the training input records of
the neural network, and of its surroundings as the permitted working range of a neural network and checking whether the input data record is in the convex envelope.
A method for making a neural network prediction for an input data record being manufacturing process data selected from the group comprising data related to the materials used, composition data,
parameters of the production system, pressure data and/or temperature data; the prediction being based on whether the input data record is in a working range of a neural network, wherein the working
range is defined by a convex envelope formed by training input data records of the neural network, comprising the following steps:(a) storing training input data records, forming the convex envelope
using the training input data records,(b) checking whether the input data record is in the convex envelope, in(i) selecting a number (d+1) of non-collinear points from the set of training input
records,(ii) forming a first simplex (S
) from the selected points,(iii) selecting a point (x
) from the interior of the first simplex (S
),(iv) defining a path between the input data record and the selected point,(v) checking whether there is an intersection point (x
+1) between the path and a facet of the first simplex, and(vi) checking whether a second simplex (S
+1) which contains the intersection point and a section of the path can be formed from the number of points from the training input data records,(c) delivering a result that input data record is
inside or outside the working range of the used neural network through confirming that the input data is respectively inside or outside the convex envelope; and(d) making the neural network
prediction by disregarding the input data record if it is outside the working range of the used neural network and processing by the used neural network as output data records the input data record
if it is inside the working range, wherein the output data records represent a subset of the input data records.
The method according to claim 1, further comprising the steps for checking whether a second simplex may be formed:(vii) determining vertices of a facet of the first simplex on which the intersection
point is located,(viii) selecting a further, non-collinear point from the training input data records,(ix) forming a simplex (S') from the vertices and the further point,(x) checking whether the
simplex contains a section of straight line, and outputting the simplex as a second simplex, if this is the case,(xi) exchanging the further point for another, non-collinear point from the set of
training input data records and renewed checking.
A computer digital storage medium program product for carrying out a method according to claim
1. 4.
A method for making a neural network prediction for an input data record being manufacturing process data selected from the group comprising data related to the materials used, composition data,
parameters of the production system, pressure data and/or temperature data; the prediction being based on whether the input data record is in a working range of a neural network, and wherein the
working range is defined by a convex envelope formed by training input data records of the neural network, comprising the following steps:(a) storing training input data records, forming the convex
envelope using the training input data records,(b) checking whether the input data record is in the convex envelope, in(i) selecting an initial vector λ.sup.(0)=(λ
, . . . , λ
) with λ
+ . . . +λ
=1 and λ
≧0 (j=1, . . . , n), where preferably λ
=1/n is selected,(ii) selecting a matrix M in such a way that the lines matrix {circumflex over (P)}.sup.(i):=MP.sup.(i) are orthonomal,(iii) calculating λ=λ.sup.(i)+{circumflex over (P)}.sup.(i)T
({circumflex over (x)}-{circumflex over (x)}.sup.(i)), where {circumflex over (x)}.sup.(i):={circumflex over (P)}.sup.(i)λ.sup.(i),(iv) checking whether all λ
≧0 (for j=1, . . . , n),(v) deleting all components from the matrix P.sup.(i) and from the vector λ.sup.(i), which infringe the secondary condition λ
≧0 (for j=1, . . . , n),(vi) renewing calculating of λ, and(c) delivering result that input data record is within or without the working range of used neural network through confirming that the input
data is respectively inside or outside the convex envelope; and(d) making the neural network prediction by disregarding the input data record if it is outside the working range of the used neural
network and processing by the used neural network as output data records the input data record if it is inside the working range, wherein the output data records represent a subset of the input data
This application is a continuation-in-part application of U.S. patent application Ser. No. 10/758,322, filed Jan. 15, 2004, which is herein incorporated by reference in its entirety.
BACKGROUND OF THE INVENTION [0002]
1. Field of the Invention
The invention relates to a method for checking whether an input data record is in the permitted working range of a neural network, and to a corresponding computer program product and system.
2. Description of the Related Art
A plurality of application possibilities for neural networks are known from the prior art. Neural networks are used for data-driven model formation, for example for physical, biological, chemical and
technical processes and systems, cf. Babel W.: "Einsatzmoglichkeiten neuronaler Netze in der Industrie: Mustererkennung anhand uberwachter Lernverfahren--mit Beispielen aus der Verkehrs- und
Medizintechnik"[translation of title: "Possibilities of use of neural networks in industry: pattern recognition using monitored learning methods--with examples from transportation and medicine
technology"], Expert Verlag, Renningen-Malmsheim, 1997. In particular, the fields of use of neural networks include process optimization, image processing, pattern recognition, robot control and
medicine technology.
Before a neural network can be used for predictive or optimization purposes, it must be trained. This usually involves adapting the weights of the neurons by means of an iterative method using
training data, cf. Barmann F.: "Prozessmodellierung: Modellierung von Kontianlagen mit neuronalen Netzen" [translation of title: "Process modelling: modelling of continuous systems with neural
networks"], Internet page NN-Tool, www.baermann.de and Barmann F.: "Neuronale Netze". Skriptum zur Vorlesung, FH-Gelsenkirchen, Fachbereich Physikalische Technik, Fachgebiet Neuroinformatik
[translation of title: "Neural networks". Lecture paper. Technical University of Gelsenkirchen, Department of Physico-Technology, Subject area of neurocomputing], 1998.
The so-called back propagation method is particularly suitable for training a neural network. A further approach is implemented in the program "NN-Tool 2000". This program is commercially available
from Professor Frank Barmann, Technical University of Gelsenkirchen, Department of Physical Technology. The corresponding training method is also described in the publication "Neural Network", Volume
5, pages 139 to 144, 1992, "On a class of efficient learning algorithms for neural networks", Frank Barmann, Friedrich Biegler-Konig.
DE 195 31 967 discloses a method for training a neural network with the non-deterministic behaviour of a technical system. The neural network is integrated here into a control loop in such a way that
the neural network outputs, as output variable, a manipulated variable to the technical system, and the technical system generates a controlled variable from the manipulated variable supplied by the
neural network and said controlled variable is fed to the neural network as an input variable. Noise with a known noise distribution is superimposed on the manipulated variable before it is fed to
the technical system. Further methods for training neural networks are known from DE 692 28 412 T2 and DE 198 38 654 C1.
In addition, a method for estimating the confidence level of the prediction which is output by a neural network is known from the prior art: Protzel P., Kindermann L., Tagscherer M., Lewandowski A.
"Abschatzung der Vertrauenswurdigkeit von Neuronalen Netzprognosen bei der Prozessoptimierung [translated title: "Estimating the confidence level of neural network predictions when optimizing
processes"], VDI Berichte [Reports] No. 1526, 2000. EP 0 762 245 B1 also discloses a method for recognizing faulty predictions in a neural-model-supported or neural process control system.
A common disadvantage of these methods known from the prior art is that they can only permit conclusions to be drawn about the sensitivity, in terms of variations of the training data, of the model
which is made available by the neural network. However, it is thus not possible to draw conclusions about the confidence level of a prediction which is made by the neural network.
F. Barmann; Handbuch zu NN-Tool 98 [translated title: "Neural Network Tool Manual"], 1998, discloses an approach in which it is attempted to estimate the prediction error at a specific point using
the known prediction error at adjacent data points.
All these methods have in common the fact that it is not possible to draw a conclusion as to whether an input data record is at all in the permitted working range of the neural network. However, an
incorrect estimation is possible only in this case.
The invention is therefore based on the object of providing a method which makes it possible to check whether an input data record is in the permitted working range of a neural network. In addition,
the invention is based on the object of providing a corresponding computer program product.
The object on which the invention is based is achieved in each case with the features of the independent patent claims. Preferred embodiments of the invention are given in the dependent patent
SUMMARY OF THE INVENTION [0015]
The present invention makes it possible to check an input data record for a neural network to determine whether it is in the permitted working range of the neural network. The invention is based on
the realization that structural information is not input into neural networks but instead only training input data records are used which have been, for example, obtained by measuring means. Due to
this recognition, such models can provide trustworthy predictions only in the areas in which the models have been trained.
Between the given training data points it is possible to interpolate very efficiently using such models. However, in contrast to corresponding rigorous models, data-driven models cannot extrapolate,
or can extrapolate only to a very restricted degree. Therefore, in particular for monitoring and/or controlling critical applications, it is advantageous that it is possible to check whether the
model used is utilized in the permitted working range.
This applies to a greater degree also to hybrid models in which a plurality of neural networks is connected to rigorous models. Although hybrid models are capable of extrapolation as an overall
model, the interpolation area must be checked for each individual data-driven subcomponent, that is to say, for the neural networks which are contained in the hybrid model.
According to the invention, the working range of a neural network is defined by the convex envelope formed by the training input data records of the neural network. For example, a neural network has
a number of a inputs and a number of b outputs. To form models, data records for the a input parameters and the b output parameters are acquired by measuring means.
If, for example, a model is to be formed for a manufacturing process, the input parameters can be data relating to the materials used, their composition and/or parameters of the production system,
for example pressures, temperatures and the like. The resulting product properties, for example, are then measured for the output parameters. In this way, training data records which each contain a
set of input parameters and associated output parameters are obtained. The neural network is trained using these training data records, that is to say the weightings of the neurons are adapted
According to one preferred embodiment of the invention, the following definition of the convex envelope is applied as a way of defining the permitted working range:
P is assumed to be a given, finite set of n points p
, . . . , p
. The points p
(for i=1, . . . , n) of the set P are formed by means of the training input data records with which the neural network has been trained. A point x, that is to say a specific input data record, is
associated with the convex envelope which is formed by P and is referred to as conv(P) if it yields real numbers λ
, . . . , λ
≧0 where λ
+ . . . +λ
=1 so that λ
+ . . . +λ
=x for p
εP (for i=1, . . . , n). (On the theory of convex envelopes, see also: Dieter Jungnickel: "Optimierungsmethoden" [translated title: "Optimization methods"]; Springer, Heidelberg; 1999; ISBN:
According to one preferred embodiment of the invention, the direct surroundings of the convex envelope are also considered as a permitted working range as neural networks can also supply appropriate
results in the direct vicinity of the convex envelope. However, the working range is alternatively restricted directly to the convex envelope as it is not possible to draw a precise conclusion as to
where the "direct vicinity" ends. In particular for critical applications which relate, for example, to continuous production, the working range is therefore restricted to the interior of the convex
envelope, the external surroundings in the direct vicinity of the convex envelope being excluded from the working range.
In the practical application, in particular in time-critical applications, it is of particular significance to use efficient procedures in order to determine whether an input data record is in the
permitted working range of the associated neural network.
The algorithms Quickhull (see C. B. Barber, D. P. Dobkin and H. T. Huhdanpaa: "The Quickhull Algorithm for Convex Hulls"; ACM Transaction Mathematical Software; Vol. 22, No. 4; 1996; p 469-483,
Simplex Algorithm) as well as the simplex algorithm (see, for example, Dieter Jungnickel: "Optimierungsmethoden" [translated title: "Optimization methods"]; Springer, Heidelberg; 1999; ISBN:
3540660577) are known per se from the literature. These methods are inefficient in highly dimensional spaces (i.e. input dimensions greater than 9) because they require extremely long computing times
and fail on commercially available computers due to the memory requirement. On the other hand, in preferred embodiments of the invention it is possible to have recourse to three basically different,
very efficient methods.
According to one preferred embodiment of the invention, firstly a simplex composed of a number of d+1 non-collinear points from the set P is formed in order to check whether an input data record is
in the convex envelope, d being the dimension of the space formed by P. A point is then selected from the interior of this simplex. To do this, it is possible to use, for example, the center of
gravity of the simplex which is calculated from the vertices of the simplex. This point is referred to below by x
In the next step, the path [x, x
] between the point x defined by the input data record and the point x
selected from the simplex is considered. It is then checked whether there is an intersection point of the path [x, x
] with a facet of the simplex. The facets are the "side faces" of the simplex.
If there is no such intersection point, this means that the point x is in the interior of the convex envelope.
If the opposite is the case, this results in the point x being outside the simplex. However, this does not yet answer the question as to whether the point x is inside or outside the convex envelope.
It is therefore checked whether it is possible to form a further simplex from d+1 non-collinear points from the set P in such a way that the further simplex contains the intersection point with the
facet and a section of the path [x, x
If this is not possible, the result of this is that the point x is outside the convex envelope (Caratheodory's set). If such a simplex can be formed, the check is carried out again to determine
whether an intersection point of the path [x, x
] exists with a facet of the further simplex. As there is only a finite number of points in P after a finite number of iterations, this method indicates whether or not x is in the convex envelope as
all the simplices can be checked successively.
According to one preferred embodiment of the invention, the check to determine whether it is possible to form a further simplex which contains a section of the path [x, x
] is carried out as follows: firstly, the vertices of the facet which is intersected by the path [x, x
] are determined. Then a further point is selected from the set P. This can be any desired point which is not associated with the vertices of the facet.
A further simplex is then formed on a trial basis from the further point and the vertices. If this further simplex which is formed on a trial basis contains a section of the path [x, x
], this further simplex which is formed on a trial basis is used as the simplex for a further iteration of the method.
If such a section of the path [x, x
] is not contained in the simplex formed on a trial basis, the further point which is selected from P is replaced by another point in order to form a further simplex on a trial basis, and in order to
carry out again the subsequent check to determine whether a section of path [x, x
] is in the simplex formed on a trial basis.
This method is carried out until either a further simplex has been found or all the points which are possible from the set P have been selected without a simplex which fulfils the secondary condition
of containing a section of the path [x, x
] having been formed. In this case, the method ends with the conclusion that it is not possible to form a further simplex which contains a section of the path [x, x
], that is to say x is outside the convex envelope.
According to a further preferred embodiment of the invention, a different geometric property of the convex envelope is used. This property is as follows:
If there is a hyper-plane through the point x to be investigated so that all the p
εP are located on one side of the plane, the point x is outside the convex envelope formed by P (set of Hahn-Banach). If there is no such plane, the point lies in the interior.
According to a further preferred embodiment of the invention, in order to answer the question as to whether or not a point x is in the convex envelope, it is checked whether the equation system given
by the analytical definition of the convex envelope can be solved. For this purpose, an iterative method is used.
According to a further preferred embodiment of the invention, a model for checking whether an input data record is in a permitted working range of the neural network is positioned before a neural
network. If the respective system is a system with a plurality of neural networks and/or a system with rigorous model components, that is to say a so-called hybrid model, such a module is preferably
positioned before each neural network of the system. If a plurality of neural networks is used, these modules can be logically linked to a logic "AND" in order to ensure that an input data record is
in the permitted working range of all these neural networks. This is significant in particular in hybrid models.
BRIEF DESCRIPTION OF THE DRAWINGS [0038]
In what follows, preferred embodiments of the invention are explained in more detail with reference to the drawings, in which:
FIG. 1 shows a flowchart of a first embodiment of a method for checking whether an input data record is in the convex envelope,
[0040]FIG. 2
shows a development of the method from FIG. 1 for determining a further simplex,
[0041]FIG. 3
shows a further embodiment of a method according to the invention for checking whether an input data record is in the convex envelope,
[0042]FIG. 4
shows a graphic illustration of the method in
FIG. 3
[0043]FIG. 5
shows a further embodiment of the method for checking whether an input data record is in the convex envelope, based on a check as to whether there is a solution for the equation system provided by
the analytical definition of the convex envelope,
[0044]FIG. 6
shows a block diagram of an embodiment of a system according to the invention.
FIG. 1 illustrates a first embodiment of the method for checking whether an input data record is in the convex envelope. This method starts from a point x
in the interior of the convex envelope and checks whether the path [x, x
] is in the interior of the convex envelope.
Here, x is the point which is determined by the input record and it is desired to determine whether this point also lies within the interior of the convex envelope.
For this purpose it is tested whether the path [x, x
] intersects one of the facets of the convex envelope. If this is the case, the point x lies outside. Here, use is made of the geometric property of the convex envelope that any linear connection
between any two points on the convex envelope lies completely in the convex envelope.
This method is based on the procedure described below:
A d
-dimensional space R
is assumed, d being the number of non-collinear training input data records of the neural network. The set P of points includes all the training input data records with which the neural network has
been trained. This set P of points is therefore contained completely in the space R
. In addition, a point x
will be assumed from the interior of the convex envelope which is formed by P, with a known representation as a convex linear combination of the points from P, i.e. there are λ
.sup.(0), . . . , λ
.sup.(0)≧0 where λ
.sup.(0)+ . . . +λ
.sup.(0)=1 and λ
+ . . . +λ
. According to the set of Caratheodory, the coefficients λ
(i=1, . . . , n) can be selected such that all are equal to 0 with the exception of d+1. In addition, it is assumed xεR
is a point for which it is to be investigated whether or not it lies in the interior of conv(P).
It is assumed that [x, x
] is the path between the points x and x
. The known coefficients λ
.sup.(0) (i=1, . . . , n) are then modified in such a way that a linear combination with the new coefficients yields a point x
which is located on the path [x
x]. This procedure is repeated until finally the point x is found, or one of the lateral boundaries of the convex envelope is reached.
In order to modify the coefficients λ
.sup.(0), a suitable solution of the following, underdetermined, linear equation system is searched for:
= 1 n i p i = x - x 0 i = 1 n i = 0 Equation 1 ##EQU00001##
Then, a factor c>0 is determined such that λ
≧0 for i=1, . . . , n applies. It will then be assumed that λ
. Then, x
+ . . . +λ
, is a convex linear combination for a point x
εconv(P) which is nearer to x than x
. If the above-described linear combination in which at maximum d+1 coefficients are unequal to 0 has been assumed for x
, and if the largest possible c is selected, in this way the intersection point of the path [x
,x] with a facet of the simplex which is formed by the points from P which are associated with the coefficients is obtained.
However, the equation system of the equation 1 cannot be solved in a uniquely defined way and a c>0 that fulfils the abovementioned requirements in order to determine new coefficients λ
.sup.(1) cannot be found for each solution.
Furthermore, an iterative method is specified which makes it possible to answer the question as to whether a solution of the equation system exists, and thus whether or not the point x is in the
convex envelope:
Initialization step: it is assumed that d is the dimension of the space in which the convex envelope is located. In order to determine a starting value x
, d random, linearly independent points q
.sup.(0)εP for j=1, . . . , d are selected. Then, a
.sup.(0)+ . . . +λ
.sup.(0), . . . , q
will be selected as the starting value
. Furthermore, we assume i=0.
Iteration step: a (d-1)-dimensional hyper-area is uniquely defined in the R
through the points q
.sup.(i), . . . , q
.sup.(i). This hyper-area will be expanded by adding a further point from the set P to form a d-dimensional simplex. It will now be assumed that q
+1.sup.(i)εP is the point with the property that the longest possible part of the path x
is in the interior of the simplex q
.sup.(i), . . . , q
In order to discover this point, it is necessary to solve virtually the same equation system repeatedly, which can be carried out efficiently. If it is not possible to find a further vertex of the
simplex, the point x is outside the convex envelope and the method is aborted. Otherwise, the equation system
= 1 d + 1 j q j ( i ) = x - x i j = 1 d + 1 i = 0 ##EQU00002##
has a uniquely defined solution
, . . . , ε
+1) and it is possible to select a c>0 with the properties described above. It is then assumed that:
λ j ( i + 1 ) := λ j ( i ) + c j ##EQU00003## x i + 1 := j = 1 d + 1 λ j ( i + 1 ) q j ( i ) ##EQU00003.2##
for j
=1, . . . , d+1. It is possible to select c in such a way that either one of the λ
.sup.(i+1) (for j=1, . . . , d+1) is equal to 0 or c=1 applies. If the case c=1 occurs, the point x is in the interior of the convex envelope and the method can be terminated.
Otherwise, a further iteration step has to be carried out. As points q
.sup.(i+1), . . . , q
.sup.(i+1) the d points will be selected from the set {q
.sup.(i), . . . , q
+1.sup.(i)} in which the respectively associated (j=1, . . . , d+1) is unequal to 0. Then, i is increased by 1.
If the point x is in the interior of the convex envelope of P, the algorithm supplies a convex linear combination to represent the point. If the point lies outside, d points through which a
hyper-plane E is determined, which separates the point set P from the point x, are obtained. This means that all the points of the R
which lie on the same side of E as the point x cannot be associated with the convex envelope. This can be utilized for a multiple evaluation in order to speed up the entire evaluation considerably.
One form of implementation of this method is illustrated in FIG. 1:
In step 100, an input data record for which a prediction is to be created is input. This input data record for the neural network determines a point x. The input data records are stored as training
input data records that form the convex envelope.
In step 101, a number of d+1 non-collinear points is selected from the set P.
In step 102, the index l is set to zero. In step 103, a simplex S
is formed from the points selected in step 101.
In step 104, a point x
is selected from the interior of the simplex S
. The center of gravity is calculated, for example, from the vertices of the simplex S
in order to obtain the point x
In step 105, a path [x
] is defined between x and x
In step 106 it is checked whether an intersection point x
+1 of the path [x
] with a facet of the simplex S
is located between x and x
. It is therefore checked whether, starting from x
on the straight line in the direction of x, firstly x or a facet of the simplex S
is reached.
If there is such an intersection point x
+1 of the path [x
] with a facet of S
, this means that the point x is not within the simplex S
. If the opposite is the case, in step 107 there is an output indicating that x is in the convex envelope, as of course it has been determined that x is in the simplex S
, and this is in turn completely inside the convex envelope.
If, on the other hand, the point x is outside the simplex S
, in step 108 it is checked whether it is possible to find a further simplex S
+1 in P which includes both the intersection point x
+1 and a section of the straight line g. If this is not possible, in step 109 there is an output indicating that x is outside the convex envelope. The neural network predication is made by
disregarding the input data record if it is outside the working range of the used neural network (outside the convex envelope) and processing by the used neural network as output data records the
input data record if it is inside the working range (inside the convex envelope), wherein the output data records represent a subset of the input data records.
In the opposite case, the index l is increased by one in step 110, and step 106 is carried out again with respect to the further simplex.
[0069]FIG. 2
shows a development of the method in FIG. 1 for carrying out the check in step 108. In order to carry out this check, the vertices of the facet of S
, on which the intersection point x
+1 is located, are firstly determined in step 200.
In step 201, a further point is selected from P which is not already a vertex of the facet of S
, and which is not collinear with respect to the vertices of the facet.
In step 202, a simplex S' is formed from the vertices and the further point from P.
In step 203 it is checked whether the simplex S' includes a section of the path [x
]. If this is the case, the further simplex S
+1 which is being searched for is made equal to the simplex S' in step 204. This then also answers the question that it is actually possible to form such a simplex S
If the check in step 203 reveals that the simplex S' does not contain a section of the straight line g, in step 205 it is checked whether all the possible points from P have already previously been
selected in step 201. If this is not the case, in step 201a further point from P which has not yet been previously selected is selected in order to carry out a further iteration of the method.
If it was not possible to find a simplex S
+1 after "trying out" all the points which are possible from P, a corresponding item of information is output in step 206. This means at the same time, that the point x is outside the convex
In this embodiment it is of particular advantage that in all cases, after a finite number of steps, the method indicates whether or not the input data record is in the convex envelope, and thus in
the working range.
[0076]FIG. 3
shows a further embodiment of a method for checking whether an input data record is in the convex envelope. This method is not obtained directly from the definition of the convex envelope as a linear
combination of the support points. Instead, a different geometric property of the convex envelope is used here, and is also illustrated graphically in
FIG. 4
If there is a hyper-plane through the point x to be investigated so that all the p
εP are on one side of the plane, the point x then lies outside the convex envelope formed by P. If there is no such plane, the point is in the interior.
If the plane is represented by means of the normal vector k, the condition "all points p
εP lie on one side of the plane" can be expressed as follows:
>0, i=1 . . . n
-x being the position vectors of the data points in a coordinate system which has the data point to be investigated at the origin.
Without restricting the generality, inequality can be interrogated with respect to "greater" as the normal vector -k represents the same hyper-plane as k. Points on the facets of the convex envelope
lead to a scalar product equal to 0 and are thus a component of the convex envelope.
An optimization method is preferably used for searching for a hyper-plane.
Here, the following target function is minimized when the normal vector k varies:
= - min ( k r i k ) ##EQU00004##
If the optimum of F is smaller than 0, the point to be investigated lies outside the convex envelope. For points within the convex envelope it is not possible to find a hyper-plane for which F<0
For the use as an optimization method, various methods are possible, for example the MATLAB routine fminsearch as well as gradient methods, Levenberg-Marquard algorithm or an evolution strategy which
can also be used in combination with local methods.
An advantage which is significant for the running time behaviour of the algorithm is that, if a corresponding hyper-plane has been found for a data point, said hyper-plane also constitutes a solution
for all the points on the side of the plane lying opposite the convex envelope. If the investigation for membership of the convex envelope is to be carried out simultaneously for a plurality of data
points, the method can thus be considerably speeded up.
[0085]FIG. 3
illustrates this method by reference to a flowchart. In step 300, the input data record, that is to say the point x, is input.
In step 301, it is checked by means of one or more of the aforesaid methods, whether there is a hyper-plane which contains x and for which kr
>0, i=1, . . . , n applies, where k is the normal vector of the searched-for hyper-plane, and r
is the difference vector between a point p
and x provided by a training input data record.
If there is such a hyper-plane, it follows in step 302 that x is in the convex envelope. In the opposite case, in step 303 information is output according to which x is outside the convex envelope.
The check in step 301 to determine whether there is a suitable hyper-plane is illustrated in
FIG. 4
. The points p
located in the grey-hatched area of
FIG. 4
form a convex envelope 400. The point x is located outside the convex envelope 400. Between the point x and the points p
there are the difference vectors r
A hyper-plane 401, which is described by the normal vector k, runs through x. As all the points p
of the convex envelope 400 are located on the same side of the hyper-plane 400, it follows from this that x is actually outside the convex envelope 400.
[0090]FIG. 5
illustrates a further method for checking whether an input data record x is in the convex envelope.
In this method it is checked whether there is a solution for the equation system which is obtained from the analytical definition of the convex envelope.
+ . . . +λ
={tilde over (x)}
+ . . . +λ
=1 Equation 2
Here, a solution is searched for so that the secondary conditions λ
≧0 are fulfilled. In the following method, successive attempts are made to achieve this.
As in the method in FIGS. 1 and 2, in this case also an initial solution for λ.sup.(0):=(λ
.sup.(0), . . . , λ
.sup.(0)) is assumed for which in general the inequality secondary conditions are not fulfilled.
Equation 2 is written below in matrix form. Then,
.sup.(0)λ=x Equation 3
is obtained where a line of ones has been added to the vector x and to the
dot matrix P.sup.(0) respectively.
Initialization Step
We assume i
=0 and select a random n-dimensional vectorλ.sup.(0)=(λ
, . . . , λ
) where λ
+ . . . +λ
=1and λ
Iteration Step
Firstly, we transform the equation 3 by multiplying it on both sides by a matrix M. The matrix M is to be selected here in such a way that the lines of the matrix {circumflex over (P)}.sup.(i):=
MP.sup.(i) are ortho-normed (if such a matrix M does not exist, dependent lines in the matrix P.sup.(i) can be omitted). In addition, it is assumed that {circumflex over (x)}:=Mx. It is not attempted
now to solve the equation system {circumflex over (P)}.sup.(i)λ={circumflex over (x)} directly but instead we start from the known coefficient vector λ.sup.(i) and assume {circumflex over (x)}.sup.
(i):={circumflex over (P)}.sup.(i)λ.sup.(i). We then search for a solution of the equivalent equation system
{circumflex over (P)}.sup.(i)(λ-λ.sup.(i))={circumflex over (x)}-{circumflex over (x)}.sup.(i).
As in most cases this equation system is underdetermined, we search for the solution λ so that ∥λ-λ.sup.(i)∥ is minimal (where ∥•∥ designates the Euclidean norm). Here, we can make use of the fact
that the matrix {circumflex over (P)}.sup.(i) is ortho-normed. The following applies
λ=λ.sup.(i)+{circumflex over (P)}.sup.(i)T({circumflex over (x)}-{circumflex over (x)}.sup.(i)),
{circumflex over (P)}.sup.(i)T being the transponent of the matrix {circumflex over (P)}.sup.(i). If all the components of the coefficient vector λ which is found in this way fulfil the secondary
conditions λ
≧0, a convex linear combination for the point x has been found, and the point x is therefore in the interior of the convex envelope. Otherwise, we set all the coefficients which infringe the
secondary condition for the rest of the method to zero, and attempt to correct the components which do not infringe the secondary condition in such a way that this step is compensated for. In
practical terms, this is brought about by the fact that all the components which infringe the secondary condition, from the vector λ, and all the associated columns are eliminated from the matrix
{circumflex over (P)}.
The vector which is obtained in this way with a relatively small dimension and the matrix which is obtained in this way with fewer columns are designated by λ.sup.(i+1) and P.sup.(i+1).
For the correction, the (smaller) equation system
.sup.(i+1)λ={circumflex over (x)}
must be solved
. To do this, a further iteration step is then carried out, i being increased by one, this time with λ.sup.(i+1) as the starting value. If the equation system cannot be solved, there is no convex
linear combination for the point x and the point is located outside the convex envelope.
As at least one column is always eliminated at each iteration step, the method comes to a result after a maximum of n steps.
One embodiment of this method is illustrated in
FIG. 5
In step 500, the index i is set to be equal to zero. In step 501, a starting value for the n-dimensional vector λ.sup.(0) which fulfils the secondary conditions is selected. For this purpose, it is
possible to select, for example, λ
In step 502, the matrix M is calculated. On the basis of this, in step 503 the matrix {circumflex over (P)}.sup.(i) and the vectors {circumflex over (x)} and {circumflex over (x)}.sup.(i) are
On the basis of this, in step 504, λ=λ.sup.(i)+{circumflex over (P)}.sup.(i)
({circumflex over (x)}-{circumflex over (x)}.sup.(i)) is calculated.
In step 505, it is checked whether all λ
(j=1, . . . , n) of the vector calculated in step 504 are λ≧0. If this is the case, in step 506, it follows that the point provided by the input data record is within the convex envelope.
If the opposite is the case, in step 507 all the components of the vector λ and the corresponding columns matrix P.sup.(i) which infringe the secondary condition are deleted. This results in the
smaller equation system P.sup.(i+1)λ={circumflex over (x)}.
In step 508, the index i is incremented in order to carry out a further iteration of the method.
[0110]FIG. 6
shows a block diagram of an embodiment of a system 600 according to the invention. The system 600 has an input module 601 for inputting an input data record which is composed of a=3 parameters in the
example considered here.
The input module 601 is logically linked to a module 602 (e.g., a computer) which is used for checking whether an input data record lies within the convex envelope of the neural network 603. This
checking is carried out, for example, according to a method which is described with respect to FIGS. 1 to 5, or according to another method. Module 602 has a memory 606 and a processor 607 for
processing of a computer digital storage medium program product stored in the memory 606.
The module 602 is logically linked to the neural network 603. If the module 602 determines that an input data record is in the permitted working range of the neural network which is provided by the
convex envelope, this input data record is input into the neural network 603, which then outputs at least one predicted value at its output 604. On the other hand, if the module 602 determines that
the input data record is not in the permitted working range, a corresponding signal is emitted at the output 605, after which no reliable prediction is possible for the current input data record.
In addition to the neural network 603, the system 600 can also contain further neural networks (hybrid model), each of which having in turn arranged upstream of it a module which corresponds to the
module 602. The results of the individual modules 602 must then be logically linked to a logic "AND". This ensures that all the neural networks of the hybrid model 600 are operated in a permitted
working range for a specific input data record of the input module 601. In addition, the system 600 can also contain rigorous model components.
Patent applications by Thomas Mrziglod, Bergisch Gladbach DE
Patent applications by BAYER AKTIENGESELLSCHAFT
Patent applications in class Prediction
Patent applications in all subclasses Prediction
User Contributions:
Comment about this patent or add new information about this topic: | {"url":"http://www.faqs.org/patents/app/20090210370","timestamp":"2014-04-16T21:28:19Z","content_type":null,"content_length":"73586","record_id":"<urn:uuid:576223c5-bebc-4552-8662-2108ac46831e>","cc-path":"CC-MAIN-2014-15/segments/1398223202548.14/warc/CC-MAIN-20140423032002-00502-ip-10-147-4-33.ec2.internal.warc.gz"} |
Microsoft Office Access 2007 VBA: Using Built-In Functions
< Back Page 3 of 10 Next >
Working with Date Functions
VBA has many functions that help you deal with dates. As long as you understand how Access stores Date/Time values, you should have no problem in working with date functions and values.
→ For a description of the Date/Time datatype see "VBA DataTypes," p. 28.
In this section we go over most of the functions you use when dealing with dates.
Returning the Current Date
To return the current date (as stored on your system) use the following function, which gives you a number counting the days from 12/30/1899:
How this value is displayed depends on your regional settings. You can use the Date$() function to return a 10-character string representing the date. This string uses the format mm-dd-yyyy. The Date
() function returns only the system date; if you need to include the time use the Now() function. As noted earlier, a date/time value is a number where the integer portion represents the date and the
decimal portion represents the time. So the Now() function will return an integer and decimal that represents the current date and time. The Now() function defaults to displaying its value according
to the regional settings on your PC. On my PC it displays 7/25/2007 5:06:34 PM.
Performing Date Arithmetic
Because dates are stored as numbers, you can do date arithmetic simply by adding or subtracting date values. However, VBA gives you a better way, the DateAdd function. Using this function, you can
add 14 days, 14 weeks, 14 months, or 14 years to any date. Or you can find a time 60 hours earlier than the specified date and time.
The following is the syntax for DateAdd, where interval is a string that indicates the type of time period that you want to calculate:
DateAdd(interval, value, date)
Table 4.1 shows the various strings that can be entered as intervals. The number argument is a value or expression that specifies the number of intervals you want to calculate. The number used is an
integer. If a decimal value is included, it's rounded to the nearest whole number, before performing the calculation. The date argument is a Date/Time value that is the base value to use in the
Table 4.1. Interval Settings
String Setting Description
yyyy Years
q Quarters
m Months
y Day of year
d Days
w Weekdays
ww Weeks
h Hours
n Minutes
s Seconds
The y, d, and w intervals work interchangeably in the DateAdd function but have more meaning in other Date/Time functions. If the interval evaluates to a negative number, it returns an earlier date/
time; a positive number returns a future date/time.
Determining the Difference Between Two Dates
The DateDiff function is used to determine the number of intervals between two date/time values. The following is the syntax for the DateDiff function, where interval is a string that indicates the
type of time period used to calculate the difference between the first and second dates represented by date1 and date2 (refer to Table 4.1):
DateDiff(interval, date1, date2[,firstdayofweek[, firstweekofyear]])
Also included in the DateDiff function are two optional arguments: firstdayofweek and firstdayofyear. These are numerical constants that can be used to adjust the first day of a week or year when
using the DateDiff function. Tables 4.2 and 4.3 show a list of the values for each constant. The default values are Sunday and January 1, respectively.
Table 4.2. First Day of Week Constants
Constant Description Integer Value
vbSunday Sunday (the default) 1
vbMonday Monday 2
vbTuesday Tuesday 3
vbWednesday Wednesday 4
vbThursday Thursday 5
vbFriday Friday 6
vbSaturday Saturday 7
Table 4.3. First Week of Year Constants
Constant Description Integer Value
vbFirstJan1 Use the week in which January 1 occurs (the default). 1
vbFirstFourDays Use the first week that has at least four days in the new year. 2
vbFirstFullWeek Use the first full week of the new year. 3
The results from this function might not always be as expected:
• If date2 falls before date1 , the function yields a negative value.
• The DateDiff function calculates a year has passed when a new year falls between the two dates, even if there are fewer than 365 days. So when using 12/31 and 1/1 as date1 and date2,
respectively, the function returns a 1.
Figure 4.1 shows how these guidelines affect the function in the Immediate window.
Extracting Parts of Dates
The DatePart function is used to extract a portion of a date from a date value. A Date/Time data type contains several components that correspond to the intervals listed in Table 4.1. For example,
the following expressions return the values 4, 1, and 2007, respectively:
The DatePart function uses the following syntax, where interval is a String value that defines the part of the date you want to extract and date is a valid Date/Time value (refer to Table 4.1 for a
list of interval values):
DatePart(interval, date[,firstdayofweek[, firstweekofyear]])
Also included in the DatePart function are two optional arguments: firstdayofweek and firstdayofyear. These are numerical constants that can be used to adjust the first day of a week or year when
using the DatePart function. Tables 4.2 and 4.3 show a list of the values for each constant. The default values are Sunday and January 1, respectively.
Creating Dates from the Individual Parts
With DatePart you extract part of a date; conversely, with the DateSerial function you combine the parts of a date to return a date value. The DateSerial function uses the following syntax, where
Year, Month, and Day can be any expression that evaluates to an integer value that represents the respective date part:
DateSerial(Year, Month, Day)
There are some rules for each of the arguments:
• Year is required and must be equal to an integer from 100 to 9999.
• Month is required, and integers from 1 to 12 (positive or negative) are considered.
• Day is required, and integers from 0 to 31 (positive or negative) are considered.
The DateSerial function can take integer values outside those ranges and calculate the difference to return a date value. This makes it very powerful if you use expressions for the arguments. For
example, the following expression returns June 5, 2008 because the 18th month from the start of 2007 is June:
Similarly, the following returns May 15, 2007, by using the 30 days in April and adding the difference of 15 days to the next month:
Although this shouldn't be used as a substitute for DateAdd or DateDiff, it can make it easy to create dates from calculated values.
Creating Dates from String Values
The DateValue function can be used to return a date value from a string value; it uses the following syntax, where stringexpression must conform to the formats used by the system's Regional settings:
The following three expressions return the date June 1, 2007:
DateValue("June 1, 2007")
DateValue("1 Jun 07")
Extracting a Specific Date or Time Portion
Table 4.4 lists several functions that return a specific portion of a date or time value. The syntax for these functions is simple:
Table 4.4. Date Component Functions
Function Result
Day(date) Returns the day of the month as an integer between 1 and 31
Hour(time) Returns the hour as an integer between 0 and 23
Minute(time) Returns the minute as an integer between 0 and 59
Second(time) Returns the second as an integer between 0 and 59
Month(date) Returns the month as an integer between 1 and 12
Year(date) Returns the year as an integer between 100 and 9999
A Conversion and Date Example
Sometimes you might need to round a time value to the nearest quarter hour or hour. This example uses some of the conversion and date/time functions previously discussed to accomplish that task.
1. Create a blank form and put two text boxes on it. Label the boxes txtTime and txtResult.
2. Add an option group to the form with the options Hour and Quarter Hour. Name the group optType.
3. Add a button to the form (turn off the wizard first). Name the button cmdRound.
4. Set the Record Selectors and Navigation buttons to No. Set Scroll Bars to neither.
5. In the On Click event of the button use the following code:
Private Sub cmdRound_Click()
Dim intHrs As Integer, intMin As Integer
Dim dteTime As Date
' convert entered time to Time value
dteTime = CDate(Me.txtTime)
'extract parts of time
intHrs = DatePart("h", dteTime)
intMin = DatePart("n", dteTime)
If Me.optType = 1 Then 'test for nearest type
'Round to nearest hour
If intMin >= 30 Then
dteTime = DateAdd("h", 1, dteTime)
dteTime = DateAdd("n", -intMin, dteTime)
dteTime = DateAdd("n", -intMin, dteTime)
End If
'Round to quarter hour
Select Case intMin
Case Is < 8
intMin = 0
Case 8 To 23
intMin = 15
Case 24 To 38
intMin = 30
Case 39 To 53
intMin = 45
Case Else
intHrs = intHrs + 1
intMin = 0
End Select
dteTime = TimeSerial(intHrs, intMin, 0)
End If
'Populate Result control
Me.txtResult = dteTime
End Sub
6. Save form as frmRound (see Figure 4.2).
Figure 4.2 The completed frmRound showing an example of input and result.
< Back Page 3 of 10 Next > | {"url":"http://www.quepublishing.com/articles/article.aspx?p=1143872&seqNum=3","timestamp":"2014-04-20T18:25:36Z","content_type":null,"content_length":"48207","record_id":"<urn:uuid:e4c09082-92a8-44ca-879e-83f1ec7d1ce3>","cc-path":"CC-MAIN-2014-15/segments/1398223211700.16/warc/CC-MAIN-20140423032011-00095-ip-10-147-4-33.ec2.internal.warc.gz"} |
parabola question
May 26th 2009, 07:56 PM #1
Junior Member
Apr 2009
parabola question
the graph of y=x^2 is reflected in the x axis,then stretched vertically by a factor of 2,and then translated 3 units to the left and one unit down. What is the equation of this parabola,standard
and general form
I will give you a hint to start you off:
y= a(x-h)2+k
From the info., you can determine that the centre (your h,k) will be (3,-1). There is a stretch, so your A would be 2 and add your reflection and you are done.
Your h, is your horizontal shifts, and your k value is your vertical shifts. In your case, we had shift to the left of 3 which translates to "3". The shift up 1, is your k. The stretch is your A
value and for any reflections in the X axis, it should be negative.
Hope this helps.
May 26th 2009, 08:44 PM #2
Junior Member
May 2009 | {"url":"http://mathhelpforum.com/pre-calculus/90653-parabola-question.html","timestamp":"2014-04-17T03:53:28Z","content_type":null,"content_length":"31037","record_id":"<urn:uuid:1a8b8403-afc1-464f-8ed1-c02302c1769d>","cc-path":"CC-MAIN-2014-15/segments/1397609526252.40/warc/CC-MAIN-20140416005206-00062-ip-10-147-4-33.ec2.internal.warc.gz"} |
: Path Optimization for Debugging
BTrace: Path Optimization for Debugging
This research was conducted by Akash Lal, Junghee Lim, Marina Polishchuk, and Ben Liblit.
We present and solve a path optimization problem on programs. Given a set of program nodes, called critical nodes, we find a shortest path through the program’s control flow graph that touches the
maximum number of these nodes. Control flow graphs over-approximate real program behavior; by adding dataflow analysis to the control flow graph, we narrow down on the program’s actual behavior and
discard paths deemed infeasible by the dataflow analysis. We derive an efficient algorithm for path optimization based on weighted pushdown systems. We present an application for path optimization by
integrating it with the Cooperative Bug Isolation Project (CBI), a dynamic debugging system. CBI mines instrumentation feedback data to find suspect program behaviors, called bug predictors, that are
strongly associated with program failure. Instantiating critical nodes as the nodes containing bug predictors, we solve for a shortest program path that touches these predictors. This path can be
used by a programmer to debug his software. We present some early experience on using this hybrid static/dynamic system for debugging.
Full Paper
The full paper is available as a single PDF document. A suggested BibTeX citation record is also available. | {"url":"http://pages.cs.wisc.edu/~liblit/tr-1535/","timestamp":"2014-04-18T02:58:23Z","content_type":null,"content_length":"3803","record_id":"<urn:uuid:6f5f1b40-5229-44ef-9d35-4ca929560f9e>","cc-path":"CC-MAIN-2014-15/segments/1397609532480.36/warc/CC-MAIN-20140416005212-00363-ip-10-147-4-33.ec2.internal.warc.gz"} |
Second Evaluation
No. 1 In a workshop two machines are driven by an electric motor
Power consumption: motor A = 1 KW
motor B = 2 KW
For a new machine of which its motor C receives the power of 4 KW, the maximum main power of Pmax = 4 KW is sufficient only if the two other machines (A and B) are not switched on. Design a circuit
which supplies a warning signal Q if a power higher than 4 KW is taken from the main power system by the three machines.
- Truth table
- Karnaugh map
- Logic circuit
No. 2 For a display of the figures 1-6, in the well known spot configuration of a game die, design the code conversion circuit. At the input the figures are binary coded.
T, U, V, W, X, Y, Z are output variables (LED's)
- Truth table
- Karnaugh map
- Logic circuit
No. 3 Design a JK Flip Flop with NAND and AND gates.
- Logic circuit
- Truth table
No. 4 Design a D Flip Flop. Use only NAND gates.
- Logic circuit
- Truth table | {"url":"http://www.nzdl.org/gsdlmod?e=d-00000-00---off-0gtz--00-0----0-10-0---0---0direct-10---4-------0-0l--11-en-50---20-about---00-0-1-00-0--4----0-0-11-10-0utfZz-8-00&a=d&c=gtz&cl=CL2.1&d=HASH01bfa821af920541897c1078.12","timestamp":"2014-04-19T00:57:51Z","content_type":null,"content_length":"20662","record_id":"<urn:uuid:cd2ba7e6-9a29-4936-b8a7-8431f33ca9b6>","cc-path":"CC-MAIN-2014-15/segments/1397609535535.6/warc/CC-MAIN-20140416005215-00136-ip-10-147-4-33.ec2.internal.warc.gz"} |
Hello everybody!
Re: Hello everybody!
Welcome to the forum.
I would love to hear why you are studying maths!
I think we do not choose math it chooses us.
In mathematics, you don't understand things. You just get used to them.
I have the result, but I do not yet know how to get it.
All physicists, and a good many quite respectable mathematicians are contemptuous about proof. | {"url":"http://www.mathisfunforum.com/viewtopic.php?pid=254619","timestamp":"2014-04-18T19:06:00Z","content_type":null,"content_length":"12068","record_id":"<urn:uuid:d9213725-57b9-4a7e-8fe9-01a2f6bf031c>","cc-path":"CC-MAIN-2014-15/segments/1398223206120.9/warc/CC-MAIN-20140423032006-00514-ip-10-147-4-33.ec2.internal.warc.gz"} |
Pinole Prealgebra Tutor
...I enjoy taking on new students and working with their specific needs and interests. I score in the 99th percentile in English in the US, and have a strong writing and speaking/rhetoric
background. I have also edited several texts and assisted many fellow students in polishing their papers.
32 Subjects: including prealgebra, reading, chemistry, English
...I also teach calculus to individuals outside of any class. I supervised small groups with aspiring math teachers learning calculus and I taught for several years my own university courses that
built heavily on calculus. More than one third of my tutored hours covers calculus. "Great Tutor" - Alan K.
41 Subjects: including prealgebra, calculus, geometry, statistics
...As an undergraduate, I began by taking advantage of the free physics tutoring center at CSUS. Soon enough, I was tutoring in the same tutoring center! I fell in love with tutoring very quickly
after that and began tutoring high school and college students privately.
14 Subjects: including prealgebra, chemistry, calculus, geometry
...I have helped students succeeded from the most difficult to the most elementary Math courses. I have an unmatched Math (with technical) background and know many issues of learning Math in
general and any particular class. I was admitted to UCB Math PhD program (but chose Purdue PhD for some rea...
15 Subjects: including prealgebra, calculus, GRE, algebra 1
I graduated from Cornell University and Yale Law. I have several years of experience tutoring in a wide variety of subjects and all ages, from small kids to junior high to high school, and kids
with learning disabilities. I am also available to tutor adults who are preparing for the GRE, LSAT, or wish to learn a second language.
48 Subjects: including prealgebra, Spanish, reading, English | {"url":"http://www.purplemath.com/Pinole_prealgebra_tutors.php","timestamp":"2014-04-20T16:38:26Z","content_type":null,"content_length":"23899","record_id":"<urn:uuid:91756191-a76a-4fbd-b878-b026343e4e32>","cc-path":"CC-MAIN-2014-15/segments/1398223202457.0/warc/CC-MAIN-20140423032002-00157-ip-10-147-4-33.ec2.internal.warc.gz"} |
Working with angles
October 17th 2011, 01:40 PM #1
Oct 2011
Working with angles
1) It is 4.7km from Lighthouse A to Port B. The bearing of the port from the lighthouse is N73 degrees E. A ship has sailed due west from the port and its bearing from the lighthouse is N31
degrees E. How far has the ship sailed from the port? Round to the nearest tenth.
2) Points A and B are on opposite sides of a lake. Point C is 97.4 meters from A. The measure of angle BAC is 76 degrees 40', and the measure of angle ACB is determined to be 34 degrees 30'. Find
the distance between points A and B (to the nearest meter)
Need help in explanation... be as detailed as possible. Thanks
Last edited by Ackbeet; October 17th 2011 at 05:00 PM.
Re: Working with angles
1) It is 4.7km from Lighthouse A to Port B. The bearing of the port from the lighthouse is N73 degrees E. A ship has sailed due west from the port and its bearing from the lighthouse is N31
degrees E. How far has the ship sailed from the port? Round to the nearest tenth.
2) Points A and B are on opposite sides of a lake. Point C is 97.4 meters from A. The measure of angle BAC is 76 degrees 40', and the measure of angle ACB is determined to be 34 degrees 30'. Find
the distance between points A and B (to the nearest meter)
Need help in explanation... be as detailed as possible. Thanks
What have you done to complete any one of these problems?
Last edited by mr fantastic; October 18th 2011 at 03:42 AM.
Re: Working with angles
Last edited by Ackbeet; October 17th 2011 at 05:00 PM.
Re: Working with angles
October 17th 2011, 01:49 PM #2
October 17th 2011, 02:25 PM #3
Oct 2011
October 17th 2011, 02:32 PM #4 | {"url":"http://mathhelpforum.com/trigonometry/190633-working-angles.html","timestamp":"2014-04-17T01:19:02Z","content_type":null,"content_length":"41796","record_id":"<urn:uuid:c99aa9aa-9a9e-41b6-864f-11ad53a3ec6b>","cc-path":"CC-MAIN-2014-15/segments/1397609526102.3/warc/CC-MAIN-20140416005206-00201-ip-10-147-4-33.ec2.internal.warc.gz"} |
188 helpers are online right now
75% of questions are answered within 5 minutes.
is replying to Can someone tell me what button the professor is hitting...
• Teamwork 19 Teammate
• Problem Solving 19 Hero
• Engagement 19 Mad Hatter
• You have blocked this person.
• ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.
This is the testimonial you wrote.
You haven't written a testimonial for Owlfred. | {"url":"http://openstudy.com/users/kinggeorge/medals/1","timestamp":"2014-04-18T23:22:26Z","content_type":null,"content_length":"141614","record_id":"<urn:uuid:cb389874-54ad-4ca0-b73c-694fcc52a181>","cc-path":"CC-MAIN-2014-15/segments/1398223205137.4/warc/CC-MAIN-20140423032005-00530-ip-10-147-4-33.ec2.internal.warc.gz"} |
The Minimum block identifies the value and/or position of the smallest element in each row or column of the input, along vectors of a specified dimension of the input, or of the entire input. The
Minimum block can also track the minimum values in a sequence of inputs over a period of time. The Mode parameter specifies the block's mode of operation, and can be set to Value, Index, Value and
Index, or Running.
The Minimum block supports real and complex floating-point, fixed-point, and Boolean inputs. Real fixed-point inputs can be either signed or unsigned, while complex fixed-point inputs must be signed.
The data type of the minimum values output by the block match the data type of the input. The index values output by the block are double when the input is double, and uint32 otherwise.
For the Value, Index, and Value and Index modes, the Minimum block produces identical results as the MATLAB^® min function when it is called as [y I] = min(u,[],D), where u and y are the input and
output, respectively, D is the dimension, and I is the index.
Value Mode
When the Mode parameter is set to Value, the block computes the minimum value in each row or column of the input, along vectors of a specified dimension of the input, or of the entire input at each
sample time, and outputs the array y. Each element in y is the minimum value in the corresponding column, row, vector, or entire input. The output y depends on the setting of the Find the minimum
value over parameter. For example, consider a 3-dimensional input signal of size M-by-N-by-P:
● Each row — The output at each sample time consists of an M-by-1-by-P array, where each element contains the minimum value of each vector over the second dimension of the input. For an input that
is an M-by-N matrix, the output at each sample time is an M-by-1 column vector.
● Each column — The output at each sample time consists of a 1-by-N-by-P array, where each element contains the minimum value of each vector over the first dimension of the input. For an input that
is an M-by-N matrix, the output at each sample time is a 1-by-N row vector.
In this mode, the block treats length-M unoriented vector inputs as M-by-1 column vectors.
● Entire input — The output at each sample time is a scalar that contains the minimum value in the M-by-N-by-P input matrix.
● Specified dimension — The output at each sample time depends on Dimension. If Dimension is set to 1, the output is the same as when you select Each column. If Dimension is set to 2, the output is
the same as when you select Each row. If Dimension is set to 3, the output at each sample time is an M-by-N matrix containing the minimum value of each vector over the third dimension of the
For complex inputs, the block selects the value in each row or column of the input, along vectors of a specified dimension of the input, or of the entire input that has the minimum magnitude squared
as shown below. For complex value , the magnitude squared is .
Index Mode
When Mode is set to Index, the block computes the minimum value in each row or column of the input, along vectors of a specified dimension of the input, or of the entire input, and outputs the index
array I. Each element in I is an integer indexing the minimum value in the corresponding column, row, vector, or entire input. The output I depends on the setting of the Find the minimum value over
parameter. For example, consider a 3-dimensional input signal of size M-by-N-by-P:
● Each row — The output at each sample time consists of an M-by-1-by-P array, where each element contains the index of the minimum value of each vector over the second dimension of the input. For
an input that is an M-by-N matrix, the output at each sample time is an M-by-1 column vector.
● Each column — The output at each sample time consists of a 1-by-N-by-P array, where each element contains the index of the minimum value of each vector over the first dimension of the input. For
an input that is an M-by-N matrix, the output at each sample time is a 1-by-N row vector.
In this mode, the block treats length-M unoriented vector inputs as M-by-1 column vectors.
● Entire input — The output at each sample time is a 1-by-3 vector that contains the location of the minimum value in the M-by-N-by-P input matrix. For an input that is an M-by-N matrix, the output
will be a 1-by-2 vector.
● Specified dimension — The output at each sample time depends on Dimension. If Dimension is set to 1, the output is the same as when you select Each column. If Dimension is set to 2, the output is
the same as when you select Each row. If Dimension is set to 3, the output at each sample time is an M-by-N matrix containing the indices of the minimum values of each vector over the third
dimension of the input.
When a minimum value occurs more than once, the computed index corresponds to the first occurrence. For example, when the input is the column vector [-1 2 3 2 -1]', the computed one-based index of
the minimum value is 1 rather than 5 when Each column is selected.
Value and Index Mode
When Mode is set to Value and Index, the block outputs both the minima and the indices.
Running Mode
When Mode is set to Running, the block tracks the minimum value of each channel in a time sequence of M-by-N inputs. In this mode, you must also specify a value for the Input processing parameter:
● When you select Elements as channels (sample based), the block outputs an M-by-N array. Each element y[ij] of the output contains the minimum value observed in element u[ij] for all inputs since
the last reset.
● When you select Columns as channels (frame based), the block outputs an M-by-N matrix. Each element y[ij] of the output contains the minimum value observed in the jth column of all inputs since
the last reset, up to and including element u[ij] of the current input.
Running Mode for Variable-Size Inputs
When your inputs are of variable size, and you set the Mode to Running , there are two options:
● If you set the Input processing parameter to Elements as channels (sample based), the state is reset.
● If you set the Input processing parameter to Columns as channels (frame based), then there are two cases:
○ When the input size difference is in the number of channels (i.e., number of columns), the state is reset.
○ When the input size difference is in the length of channels (i.e., number of rows), there is no reset and the running operation is carried out as usual.
Resetting the Running Minimum
The block resets the running minimum whenever a reset event is detected at the optional Rst port. The reset sample time must be a positive integer multiple of the input sample time.
When a reset event occurs while the Input processing parameter is set to Elements as channels (sample based), the running minimum for each channel is initialized to the value in the corresponding
channel of the current input. Similarly, when the Input processing parameter is set to Columns as channels (frame based), the running minimum for each channel is initialized to the earliest value in
each channel of the current input.
You specify the reset event by the Reset port parameter:
● None — Disables the Rst port
● Rising edge — Triggers a reset operation when the Rst input does one of the following:
○ Rises from a negative value to a positive value or zero
○ Rises from zero to a positive value, where the rise is not a continuation of a rise from a negative value to zero (see the following figure)
● Falling edge — Triggers a reset operation when the Rst input does one of the following:
○ Falls from a positive value to a negative value or zero
○ Falls from zero to a negative value, where the fall is not a continuation of a fall from a positive value to zero (see the following figure)
● Either edge — Triggers a reset operation when the Rst input is a Rising edge or Falling edge (as described above)
● Non-zero sample — Triggers a reset operation at each sample time that the Rst input is not zero
│ Note: When running simulations in the Simulink^® MultiTasking mode, reset signals have a one-sample latency. Therefore, when the block detects a reset event, there is a one-sample delay at │
│ the reset port rate before the block applies the reset. For more information on latency and the Simulink tasking modes, see Excess Algorithmic Delay (Tasking Latency) and Scheduling in the │
│ Simulink Coder™ User's Guide. │
ROI Processing
To calculate the statistical value within a particular region of interest (ROI) of the input, select the Enable ROI processing check box. This applies to any mode other than the running mode and when
you set the Find the minimum value over parameter to Entire input and you select the Enable ROI processing check box. ROI processing applies only for 2-D inputs.
│ Note Full ROI processing is only available to users who have a Computer Vision System Toolbox™ license. If you only have a DSP System Toolbox™ license, you can still use ROI processing, but │
│ are limited to the ROI type Rectangles. │
Use the ROI type parameter to specify whether the ROI is a rectangle, line, label matrix, or binary mask. A binary mask is a binary image that enables you to specify which pixels to highlight, or
select. In a label matrix, pixels equal to 0 represent the background, pixels equal to 1 represent the first object, pixels equal to 2 represent the second object, and so on. When the ROI type
parameter is set to Label matrix, the Label and Label Numbers ports appear on the block. Use the Label Numbers port to specify the objects in the label matrix for which the block calculates
statistics. The input to this port must be a vector of scalar values that correspond to the labeled regions in the label matrix. For more information about the format of the input to the ROI port
when the ROI is a rectangle or a line, see the Draw Shapes block reference page.
For rectangular ROIs, use the ROI portion to process parameter to specify whether to calculate the statistical value for the entire ROI or just the ROI perimeter.
Use the Output parameter to specify the block output. The block can output separate statistical values for each ROI or the statistical value for all specified ROIs. This parameter is not available
if, for the ROI type parameter, you select Binary mask.
If, for the ROI type parameter, you select Rectangles or Lines, the Output flag indicating if ROI is within image bounds check box appears in the dialog box. If you select this check box, the Flag
port appears on the block. The following tables describe the Flag port output based on the block parameters.
Output = Individual statistics for each ROI
┃ Flag Port Output │ Description ┃
┃ 0 │ ROI is completely outside the input image. ┃
┃ 1 │ ROI is completely or partially inside the input image. ┃
Output = Single statistic for all ROIs
┃ Flag Port Output │ Description ┃
┃ 0 │ All ROIs are completely outside the input image. ┃
┃ 1 │ At least one ROI is completely or partially inside the input image. ┃
If the ROI is partially outside the image, the block only computes the statistical values for the portion of the ROI that is within the image.
If, for the ROI type parameter, you select Label matrix, the Output flag indicating if input label numbers are valid check box appears in the dialog box. If you select this check box, the Flag port
appears on the block. The following tables describe the Flag port output based on the block parameters.
Output = Individual statistics for each ROI
┃ Flag Port Output │ Description ┃
┃ 0 │ Label number is not in the label matrix. ┃
┃ 1 │ Label number is in the label matrix. ┃
Output = Single statistic for all ROIs
┃ Flag Port Output │ Description ┃
┃ 0 │ None of the label numbers are in the label matrix. ┃
┃ 1 │ At least one of the label numbers is in the label matrix. ┃
Fixed-Point Data Types
The parameters on the Data Types pane of the block dialog are only used for complex fixed-point inputs. The sum of the squares of the real and imaginary parts of such an input are formed before a
comparison is made, as described in Value Mode. The results of the squares of the real and imaginary parts are placed into the product output data type. The result of the sum of the squares is placed
into the accumulator data type. These parameters are ignored for other types of inputs. | {"url":"http://www.mathworks.se/help/dsp/ref/minimum.html?nocookie=true","timestamp":"2014-04-23T12:08:31Z","content_type":null,"content_length":"68043","record_id":"<urn:uuid:eaaad24e-93ad-44f1-bbbf-0b8d4f037674>","cc-path":"CC-MAIN-2014-15/segments/1398223202548.14/warc/CC-MAIN-20140423032002-00025-ip-10-147-4-33.ec2.internal.warc.gz"} |
Help me!
June 28th 2006, 09:15 AM
Help me!
student has a car that gets, on average, 25.0 mi/gal of gasoline. She plans to spend the a year in Europe and take the car with her.
What should she expect the car's average gas mileage to be in kilometers per liter?
During the year there, she drove 6000 km. Assuming that gas costs $5.00/gal in Europe, how much did she spend on fuel? (Compute to the nearest dollar)
June 28th 2006, 12:54 PM
Km to Miles
Originally Posted by babygirl
student has a car that gets, on average, 25.0 mi/gal of gasoline. She plans to spend the a year in Europe and take the car with her.
What should she expect the car's average gas mileage to be in kilometers per liter?
During the year there, she drove 6000 km. Assuming that gas costs $5.00/gal in Europe, how much did she spend on fuel? (Compute to the nearest dollar)
Okay, I'm new here but let's see:
It seems the key to this puzzle is finding out how many miles in a km.
Hope the following helps:
Let us know how u get on
July 21st 2006, 02:08 AM
25.0mi/gal of gasoline= 1.609km/gal of gasoline
.: 6000km of travel= 6000/1.609= no. of gal of gasoline
= ~3729.02 gallons
~3729.02 gallons= $18,645 (to nearest dollar) (3729.02 x 5)
July 21st 2006, 06:57 PM
Originally Posted by babygirl
student has a car that gets, on average, 25.0 mi/gal of gasoline. She plans to spend the a year in Europe and take the car with her.
What should she expect the car's average gas mileage to be in kilometers per liter?
During the year there, she drove 6000 km. Assuming that gas costs $5.00/gal in Europe, how much did she spend on fuel? (Compute to the nearest dollar)
I would do this problem with ratios: When I convert measurements I write things out in ratios so that they equal 1, such as: $\frac{1y}{3ft}=\frac{1(3ft)}{3ft}=\frac{3ft}{3ft}= 1$
now let's try solving for how many feet are in 5 yards...
now let's do your problem:
Student's car gets: $\frac{25m_i}{1g}$
Add conversions: $\frac{25m_i}{1g}\times1\times1=\frac{25m_i}{1g}<br /> \times<br /> \frac{1g}{3.785l}\times\frac{1km}{0.62m_i}$
$\frac{25m_i}{1g}=\frac{25m_i\times1g\times1km}{1g<br /> \times<br /> 3.785l\times0.62m_i}$
$\frac{25m_i}{1g}=\frac{25ot{m_i}<br /> \times<br /> ot{g}<br /> \times<br /> 1km}{ot{g}\times3.785l\times0.62ot{m_i}}$
$\frac{25m_i}{1g}=\frac{25\times1km}{3.785l\times0. 62}$
~ $Q\!u\!i\!c\!k$
July 23rd 2006, 12:58 AM
Originally Posted by babygirl
student has a car that gets, on average, 25.0 mi/gal of gasoline. She plans to spend the a year in Europe and take the car with her.
What should she expect the car's average gas mileage to be in kilometers per liter?
During the year there, she drove 6000 km. Assuming that gas costs $5.00/gal in Europe, how much did she spend on fuel? (Compute to the nearest dollar)
Hello, babygirl,
you only have to change the imperial units into metric units, using the conversion factors:
$\frac{25 mi}{1 gal}=\frac{25 mi\cdot 1.609\frac{km}{mi}}{1 gal\cdot 3.785\frac{l}{gal}}\approx 10.6\frac{km}{l}$. That's the result Quick has already calculated.
She'll travel 6000 km. So she had to buy 6000 km / 10.6 (km/l) = 566 l.
566 l is the same as 566 / 3.785 (l/gal) = 149.54 gal.
So she has to pay: 149.54 * 5 $ = 747.70 $
By the way: Today in Germany 1 gal of gas costs 6.86 $. So she will go a long distance by foot.
August 19th 2006, 01:14 AM
Originally Posted by earboth
Hello, babygirl,
you only have to change the imperial units into metric units, using the conversion factors:
$\frac{25 mi}{1 gal}=\frac{25 mi\cdot 1.609\frac{km}{mi}}{1 gal\cdot 3.785\frac{l}{gal}}\approx 10.6\frac{km}{l}$. That's the result Quick has already calculated.
Not imperial units, the imperial gallon is 20 fl-oz while the US gallon is 16 fl-oz.
You have used the correct conversion factor for US gallons to litres. | {"url":"http://mathhelpforum.com/algebra/3880-help-me-print.html","timestamp":"2014-04-20T21:52:29Z","content_type":null,"content_length":"12590","record_id":"<urn:uuid:870a1761-0cab-4207-ad58-321988c4a089>","cc-path":"CC-MAIN-2014-15/segments/1397609539230.18/warc/CC-MAIN-20140416005219-00345-ip-10-147-4-33.ec2.internal.warc.gz"} |
Optimal Computer Crash Performance Precaution
Efraim Laksman, Håkan Lennerstad, Lars Lundberg
For a parallel computer system with m identical computers, we study optimal performance precaution for one possible computer crash. We want to calculate the cost of crash precaution in the case of no
crash. We thus define a tolerance level r meaning that we only tolerate that the completion time of a parallel program after a crash is at most a factor r + 1 larger than if we use optimal allocation
on m - 1 computers. This is an r-dependent restriction of the set of allocations of a program. Then, what is the worst-case ratio of the optimal r-dependent completion time in the case of no crash
and the unrestricted optimal completion time of the same parallel program? We denote the maximal ratio of completion times f(r,m) - i.e., the ratio for worst-case programs. In the paper we establish
upper and lower bounds of the worst-case cost function f(r,m) and characterize worst-case programs.
Full Text:
PDF PostScript | {"url":"http://www.dmtcs.org/dmtcs-ojs/index.php/dmtcs/article/viewArticle/1840","timestamp":"2014-04-21T04:47:33Z","content_type":null,"content_length":"12032","record_id":"<urn:uuid:d3b1c0d8-fd59-4b46-8a6b-518681146d8f>","cc-path":"CC-MAIN-2014-15/segments/1398223203235.2/warc/CC-MAIN-20140423032003-00296-ip-10-147-4-33.ec2.internal.warc.gz"} |
Math Forum: Teacher2Teacher - Q&A #6122
View entire discussion
From: Jeanne (for Teacher2Teacher Service)
Date: Apr 13, 2001 at 13:00:03
Subject: Re: Using money
Hi Shannon,
I have a couple of ideas which might help you meet his IEP.
Every algebra and pre-algebra text I have seen has problems I call, "money"
problems. For example, "Shannon has 12 coins in her piggybank. Some of
them are nickels and some are dimes. If the total value is $0.70, how many
of each coin does she have?" This problem can be solved using "guess and
check". In the process of using guess and check, I sometimes pull out my
play money (or poker chips labeled with the needed values) for my students
to use and/or for me to use while demonstrating. This might be a good place
to help your student learn/reinforce identification/counting money.
When I was asked to so something similar, I was in the middle of teaching
problem solving strategies. I chose to help my student with this part of
his IEP while we worked on "making a systematic list." Some of the
questions we worked on: "How many different ways are there to make 52 cents
using pennies, nickels, dimes, and quarters?" "I have a certain combination
of coins written down on a piece of paper which has a value of 25 cents. If
you had to make a list of all of the different combination of coins which
have a total value of 25 cents, what would you do?" Again, while the class
worked on these questions, I had play money available.
Hope this helps.
-Jeanne, for the T2T service
Post a public discussion message
Ask Teacher2Teacher a new question | {"url":"http://mathforum.org/t2t/message.taco?thread=6122&message=3","timestamp":"2014-04-16T22:51:05Z","content_type":null,"content_length":"5687","record_id":"<urn:uuid:3ad62252-cfd8-4a40-b32e-89c306bb38d3>","cc-path":"CC-MAIN-2014-15/segments/1397609525991.2/warc/CC-MAIN-20140416005205-00392-ip-10-147-4-33.ec2.internal.warc.gz"} |
= Preview Document = Member Document = Pin to Pinterest
Circle the item that does not belong in each group (square, orange, soap, cards).
• [member-created using abctools] From "circle" to "rectangle". These vocabulary building word strips are great for word walls.
• [member-created document] This sample was made with the abcteach word search puzzle tool.
Explains flip, turn, and slide, and then has a worksheet for students to identify which transformation has occurred.
Explains the concept of rotation in geometry, and then reviews it with a worksheet.
• Three pages of complex math problems draw on critical thinking skills and logic in order to find the correct solution.
Describes lines (line segments, rays, points, and more) and then provides worksheets that can be used as quizzes or review pages.
Identify the outside shape of each shield by circling the correct name. Write the number of sides.
• Help students review geometry terms with this fun bingo game. Includes game boards and calling cards.
Design a quilt following examples, and then use geometry to answer questions about the quilt.
Cut out the two dimensional shape and fold it to make a three dimensional triangular prism.
Spin the wheel and advance to the corresponding shape. This castle-themed game is a fun way to master basic shapes.
Five pages of colorful shapes -trapezoid, oval, pentagon, hexagon- each in five different colors, unlabeled.
Five pages of colorful shapes -square, rectangle, circle, and triangle- each in five different colors, unlabeled.
Use this sheet to check which three-dimensional shapes from our upper elementary set (from CONE to PENTAGONAL PRISM) have been called.
Brief introductions to basic shapes: quadrilateral shapes, triangles, and curved shapes; one set per page, each with a picture and a definition.
Four three-dimensional shapes to a page. These colorful flashcards match the upper elementary set of our three-dimensional shape bingo cards.
Use this sheet to check which three-dimensional shapes from our elementary set (from CONE to SPHERE) have been called.
Cut out the two dimensional shape and fold it to make a three dimensional pyramid with a square base.
Three math posters that help teach the geometric concepts of flip, slide, and turn.
• Describes (with pictures) two dimensional figures, and then provides worksheets that can be used as quizzes or review pages.
Cut out the two dimensional shape and fold it to make a three dimensional cylinder.
Cut out the two dimensional shape and fold it to make a three dimensional rectangular prism.
Use this sheet to check which three-dimensional shapes from our elementary set (from CONE to SPHERE) have been called.
Spin the wheel and advance to the corresponding shape. This colorful castle-themed game is a fun way to master basic shapes.
Use Euler's formula to determine the number of edges of a simply connected polyhedron such as a soccer ball. Common Core_Math_6.EE.A.1, 6.EE.A.2, 6.EE.A.3,
Describes rays and angles, and then provides worksheets that can be used as quizzes or review pages.
Trace the line of symmetry. Then copy the pattern to make the heart symmetrical.
"Place a TRIANGLE at 2,4." Place the colorful pictures (included) in this simple (5x5) grid by following the the correct co-ordinates. This grid game is great for developing early map skills and
shape recognition, and for practicing following directions.
Set of twenty colorful bingo cards for three-dimensional shapes, from CONE to SPHERE.
Rules and practice for understanding rotation, reflection, and translation in geometry.
Circle the item that does not belong in each group (circle, ball, orange, soap). | {"url":"http://www.abcteach.com/directory/subjects-math-geometry-655-7-2","timestamp":"2014-04-16T11:02:12Z","content_type":null,"content_length":"147705","record_id":"<urn:uuid:043f99ee-d9dd-4e53-9c01-c9f0cb6c9038>","cc-path":"CC-MAIN-2014-15/segments/1397609523265.25/warc/CC-MAIN-20140416005203-00132-ip-10-147-4-33.ec2.internal.warc.gz"} |
A Circular Ring Of Radius
Image text transcribed for accessibility: A circular ring of radius "a" carries a uniform line charge charge rho L C/m and is placed on the xy-plane with axis the same as the z-axis. Show that (a) E
(0,0,h) = rho Lah/2epsilon 0(h2 + a2)2 a2 (b) What value of h gives the maximum value of E ?
Electrical Engineering | {"url":"http://www.chegg.com/homework-help/questions-and-answers/circular-ring-radius-carries-uniform-line-charge-charge-rho-l-c-m-placed-xy-plane-axis-z-a-q2201881","timestamp":"2014-04-20T13:09:47Z","content_type":null,"content_length":"20632","record_id":"<urn:uuid:a38126ac-1698-4877-b105-dadeef28c817>","cc-path":"CC-MAIN-2014-15/segments/1397609538423.10/warc/CC-MAIN-20140416005218-00297-ip-10-147-4-33.ec2.internal.warc.gz"} |
Is Simple Extension Minimal?
May 21st 2006, 09:29 AM #1
Global Moderator
Nov 2005
New York City
Is Simple Extension Minimal?
If you have fields $F\leq E$ with $\alpha \in E$ algebraic over $F$, show that,
$F(\alpha)$ is the minimal field containing $F$ and $\alpha$.
(Note: $F(\alpha)=\phi_{\alpha}[F[x]]$-a simple extension).
The map $\phi_\alpha : F[X] \rightarrow E$ given by evaluation at $\alpha$ is a ring homomorphism and hence its image, which you denote $F(\alpha)$, is a subring of E. All you need to show is
that it is closed under taking multiplicative inverses. Note for use in a moment that a subring of a field is an integral domain. The fact that $\alpha$ is algebraic is equivalent to saying that
$F(\alpha)$ is a finite-dimensional F-vector space. Let $\beta$ be a non-zero element of $F(\alpha)$. Multiplication by $\beta$ is an F-linear endomorphism and has trivial kernel (integral
domain), hence is surjective by finiteness of the dimension and the rank--nullity formula. So 1 is in the image, hence a multiple of $\beta$, which is thus invertible.
I was thinking about this today, maybe you can do this?
You have $F\leq E$ with $\alpha$ algebraic over $F$. Let there exist a field $K$ with $\alpha \in K$ such as,
$F\leq K\leq F(\alpha)\leq E$,
Any element $\beta \in F(\alpha)$ can be expressed as,
$\beta =a_0+a_1\alpha +...+ a_{n-1}\alpha ^{n-1}$ with $a_i \in K$ because every finite dimensional vector basis has a finite basis. Therefore $\beta \in K$ thus, $K\leq F$ and $F\leq K$ thus, $K
Am I missing something?
I realise that I didn't actually address the issue of being the minimal field, just that it actually was a field. But any field containing $\alpha$ clearly contains any polynomial in $\alpha$.
Hence the minimality.
$\alpha$ clearly contains any polynomial in $\alpha$. Hence the minimality.
What do you mean by that?
Any field that contains F and contains $\alpha$ must contain any polynomial in $\alpha$ with coefficients in F. The previous part of my argument showed that the ring comprised of polynomials in $
\alpha$ with coefficients in F is actually a field. Hence it is the minimal subfield of E containing both F and $\alpha$.
Any field that contains F and contains $\alpha$ must contain any polynomial in $\alpha$ with coefficients in F. The previous part of my argument showed that the ring comprised of polynomials in $
\alpha$ with coefficients in F is actually a field. Hence it is the minimal subfield of E containing both F and $\alpha$.
You mean a polynomial F[x] evaluated at $\alpha$.
Isn't this all circular? I thought $F(\alpha)$ was defined to be the smallest the field containing both F and $\alpha$.
Isn't this all circular? I thought $F(\alpha)$ was defined to be the smallest the field containing both F and $\alpha$.
Maybe in some texts. The version I seen is that F(a) is defined as above.
May 21st 2006, 10:59 AM #2
May 21st 2006, 01:29 PM #3
Global Moderator
Nov 2005
New York City
May 21st 2006, 01:34 PM #4
May 21st 2006, 01:36 PM #5
Global Moderator
Nov 2005
New York City
May 21st 2006, 09:56 PM #6
May 22nd 2006, 02:12 PM #7
Global Moderator
Nov 2005
New York City
May 23rd 2006, 02:47 AM #8
Junior Member
Mar 2006
May 23rd 2006, 01:17 PM #9
Global Moderator
Nov 2005
New York City | {"url":"http://mathhelpforum.com/advanced-algebra/3047-simple-extension-minimal.html","timestamp":"2014-04-18T01:22:59Z","content_type":null,"content_length":"63797","record_id":"<urn:uuid:ebe8a77b-eaa5-4b6a-a0f6-f5700790f43c>","cc-path":"CC-MAIN-2014-15/segments/1397609532374.24/warc/CC-MAIN-20140416005212-00065-ip-10-147-4-33.ec2.internal.warc.gz"} |
Jason Rosenhouse and Laura Taalman - Inspired by Math #3
Jason Rosenhouse is Associate Professor of Mathematics at James Madison University and author of The Monty Hall Problem: The Remarkable Story of Math's Most Contentious Brain Teaser.
Laura Taalman is Professor of Mathematics at James Madison University and co-founder of Brainfreeze Puzzles. She is the the author of Integrated Calculus and co-author of three books of original
Sudoku puzzles.
More information about Taking Sudoku Seriously is available at the Oxford University Press web-site.
Comments (1) Trackbacks (0) ( subscribe to comments on this post )
1. another great job Sol! (when I first saw 49 mins. I thought it sounded long, but it went by quickly). Funny that I almost mentioned Jason in that list of potential interviewees I sent along,
because I love his volume on the Monty Hall Problem (his blog is also good), but left him off in the end since I was already tossing forth so many names. Glad to see you already had him covered,
even though I’m not familiar with his newer book. | {"url":"http://wildaboutmath.com/2012/02/24/jason-rosenhouse-and-laura-taalman-inspired-by-math-3/comment-page-1/","timestamp":"2014-04-19T09:32:04Z","content_type":null,"content_length":"34733","record_id":"<urn:uuid:d4192a63-65d0-442d-b4ac-6e56e31b822f>","cc-path":"CC-MAIN-2014-15/segments/1397609537097.26/warc/CC-MAIN-20140416005217-00491-ip-10-147-4-33.ec2.internal.warc.gz"} |
The Initial Voltage On A .2μF Capacitor Is -60.6V. ... | Chegg.com
The initial voltage on a .2μF capacitor is -60.6V. The capacitor current has the wave form
i(t)=100e^-1000t mA, for t>0 and i(t)=0for t<0.
a)find the voltage across the capacitor for t>0.
b) is there an instantaneous change in the voltage across thecapacitor at t=0?
c)Is there an instantaneous change in the current across thecapacitor at t=0
d) how much energy is storedc in the capacitor at t=250μs
Electrical Engineering | {"url":"http://www.chegg.com/homework-help/questions-and-answers/initial-voltage-2-f-capacitor-606v-capacitor-current-wave-form-t-100e-1000t-ma-t-0-t-0for--q539342","timestamp":"2014-04-20T01:18:58Z","content_type":null,"content_length":"18736","record_id":"<urn:uuid:2fd19c13-7b34-48aa-9863-b4f9ded3e5eb>","cc-path":"CC-MAIN-2014-15/segments/1397609537804.4/warc/CC-MAIN-20140416005217-00201-ip-10-147-4-33.ec2.internal.warc.gz"} |
Chapter 4. Recursive Enumerability
Recursive Enumerability 93 4.2 Parameters Next we want to turn our attention to the subject of calculating programs. That is, suppose we want a program that will meet some particular need. Possibly
we have a good reason to know that a program exists that meets the need. But we might want more than that; we might want actually to find such a program. For example, we know that every total
constant function is computable. (As noted in item 2 in Chapter 2, every such function is primitive recursive.) But even more is true. Given a constant k, we can actually compute an index of the
one-place function that is constantly equal to k. See Exercise 4 in Chapter 3. For another example, we have recently seen that the range of any computable partial one-place function [[e]] is an r.e.
set, and therefore is W y for some y. That is, there exists some index y of a function whose domain is the set we want. But a stronger fact is true: Given e, we can actually compute such a number y.
(We will see a proof of this later.) That is, there is a computable function f such that ran [[e]] = W f (e) for every e. For a third example, suppose we have a two-place computable partial function
f . Then, clearly the one-place function g obtained by holding the second variable fixed | {"url":"http://my.safaribooksonline.com/book/-/9780123849588/chapter-4dot-recursive-enumerability/42_parameters","timestamp":"2014-04-17T13:12:54Z","content_type":null,"content_length":"45926","record_id":"<urn:uuid:2fc35bd9-37e4-465f-aa82-b3ae9a6b8934>","cc-path":"CC-MAIN-2014-15/segments/1397609530131.27/warc/CC-MAIN-20140416005210-00088-ip-10-147-4-33.ec2.internal.warc.gz"} |
and Yongzhi Xu, Decoupling of the quasistatic system of thermoelasticity on the unit disk Jour. of Elasticity 31
- Jour. Math. Anal. Appl , 1998
"... . Existence, uniqueness and regularity theory is developed for a general initial-boundary-value problem for a system of partial differential equations which describes the Biot consolidation
model in poroelasticity as well as a coupled quasistatic problem in thermoelasticity. Additional effects of se ..."
Cited by 10 (7 self)
Add to MetaCart
. Existence, uniqueness and regularity theory is developed for a general initial-boundary-value problem for a system of partial differential equations which describes the Biot consolidation model in
poroelasticity as well as a coupled quasistatic problem in thermoelasticity. Additional effects of secondary consolidation and pore fluid exposure on the boundary are included. This quasi-static
system is resolved as an application of the theory of linear degenerate evolution equations in Hilbert space, and this leads to a precise description of the dynamics of the system. 1. Introduction We
shall consider a system modeling diffusion in an elastic medium in the case for which the inertia effects are negligible. This quasi-static assumption arises naturally in the classical Biot model of
consolidation for a linearly elastic and porous solid which is saturated by a slightly compressible viscous fluid. The fluid pressure is denoted by p(x; t) and the displacement of the structure by u
(x; t). ...
"... We report on some recent progress in the mathematical theory of nonlinear fluid transport and poro-mechanics, specifically, the design, analysis and application of mathematical models for the
flow of fluids driven by the coupled pressure and stress distributions within a deforming heterogeneous p ..."
Cited by 2 (0 self)
Add to MetaCart
We report on some recent progress in the mathematical theory of nonlinear fluid transport and poro-mechanics, specifically, the design, analysis and application of mathematical models for the flow of
fluids driven by the coupled pressure and stress distributions within a deforming heterogeneous porous structure. The goal of this work is to develop a set of mathematical models of coupled flow and
deformation processes as a basis for fundamental research on the theoretical and numerical modeling and simulation of flow in deforming heterogeneous porous media. | {"url":"http://citeseerx.ist.psu.edu/showciting?cid=2675537","timestamp":"2014-04-16T15:01:49Z","content_type":null,"content_length":"15027","record_id":"<urn:uuid:ea283895-48b9-4ece-8718-4f0ff3a39beb>","cc-path":"CC-MAIN-2014-15/segments/1397609523429.20/warc/CC-MAIN-20140416005203-00232-ip-10-147-4-33.ec2.internal.warc.gz"} |
Symbolic Programming: Computationally Active Language
June 15, 2007 — Kovas Boguta, Special Projects Group
In this blog and elsewhere, you’ll often see the statement that some advanced Mathematica feature is just another application of symbolic programming. It’s the kind of idea that seems too powerful to
explain in a single blog post, yet simple enough that I am tempted to try. So, here goes.
Symbolic programming is based on the concept of recasting core features of human language into a computationally active form.
What does it mean to have a human-language-oriented programming language?
Our cognitive model of computation is typically a three-stage process: 1) describing the computation, 2) executing that description, and 3) outputting the results.
The “language” part of most programming languages begins and ends with stage one. Linguistic structures are erected to describe the program. But the execution of the program is typically oriented
around an entirely different system of types and objects; and likewise, the program’s output structure tends to resemble nothing particularly language-like.
Symbolic programming uses linguistic structures as the foundation of all aspects of computation. From a computation’s description, to how the computation executes, to how humans interface with the
results, the exact same basic tree structure is used throughout.
This is a powerful unification, making possible many useful computations that in other systems range from cumbersome to practically impossible. We’ll see examples along the way, but let me first
describe what these linguistic structures actually are.
Symbols: The Atoms of Language
Symbols are the basic atoms of language and also of symbolic programming. The point of symbols is to have distinct pieces of notation–be they words or mathematical functions or musical notes–which
are then assigned an interpretation within the overall system.
In programming, the defining feature of symbols is that they can stand for themselves. This is in contrast to variables, which must stand for some other value or be considered an error. For instance,
JavaScript cannot perform the operation x + x unless x has a value:
while Mathematica has no trouble:
By existing simply as distinct entities, symbols serve as lightweight carriers of meaning. They are inert, but can acquire meaning through combination with other symbols. For example the symbol for
Pi is generically just another symbol, but when it appears “within” the symbol N, it takes on its special numerical meaning:
Because Mathematica manipulates things abstractly, it is able to do more-precise computation than some other systems. As the above example makes clear, symbols are only the building blocks: the real
action is with the arrangements of symbols.
Symbolic Expressions: Combining Symbols into Meaning
An English phrase is built of subphrases. A mathematical formula is built of subformulas. A musical score is built of bars, which contain notes. A web page layout is built from sublayouts. When
humans want to communicate nontrivial structures, they use hierarchical nesting.
Across different languages and domains, these structures, or trees, take many syntactic forms. However, they can always be reduced to very simple data structures known as symbolic expressions.
Symbolic expressions simply capture the concept of arranging symbols into trees. Here are a few trees built from the symbol “e”:
Like the symbols themselves, trees of symbols may simply exist, without a need to correspond to something else:
Symbolic expressions are the fabric from which we construct meaning. Different tree structures correspond to different meanings.
A tree with symbols like Plus and Power may represent an arithmetic expression, while the symbol SetterBar can be combined with a subtree indicating the current selection and a list of possible
Everything in Mathematica is ultimately represented using this uniform simple structure. At the formal level, it ensures a degree of structural compatibility across all parts of the far-flung system.
It also provides a system for introducing your own lightweight yet sophisticated structures.
Pure functions, for example, are the original killer apps for lightweight structures of this vein. In Mathematica, they can be represented with Function, but let’s roll our own using other generic
MyPureFunction[var_, expr_][arg_] := expr /. var -> arg
Let’s describe a function that adds 1 to its argument, and attach it to the symbol f. Notice the structure stays inert, and therefore can be passed around as data:
When brought into contact with an argument, the definition fires.
In symbolic programming, it is commonplace for a programmer to create abstractions that in another language would have required a modification of the core specification. This is the beauty of
language: that creating new concepts does not require inventing new atomic “types” of things–only new arrangements of structure.
Computation: The Transformation of Meaning
The most important and unique aspect of symbolic programming is that these tree structures can be made computationally active. They don’t just describe things–they actually execute.
The basic idea is that because meaning is encoded by structure, computation–the transformation of one meaning into another–corresponds to the transformation of one structure into another.
For instance, position 1 in the SetterBar expression represents the current selection. By modifying the tree at that position, we modify the meaning of the expression as a whole:
A computation begins and ends with unique, domain-specific structures. The process of computation, however, is executed using extremely general methods that are exactly the same across domains:
So, computation proceeds by very generic tree transformations. Rather conveniently, trees are also how we specify the computations themselves.
This reduces the process of computation into a linguistic puzzle: using trees to describe how other trees should transform and intersect. Not only does this allow computational primitives to have an
immediate and clear meaning, it also allows the process of language design to proceed organically as regularities in these tree structures are discovered and abstracted.
Representing and Interacting with Meaning
Symbolic programming’s comprehensive emphasis on meaning allows a unique perspective on input, output, and the whole interaction cycle between human and machine computation.
The basic idea is that symbolic expressions are the canonical, abstract containers of meaning. Given a symbolic expression, its meaning can then be embodied in a number of useful ways.
For example, the expression Plus[2,2] is rendered as 2+2. And Graphics[Rectangle[]] renders as:
And it works similarly for all sorts of domain-specific mathematics notation, documents, and even sound and music primitives.
On their own, these are useful representations. But the fact that they are still really the same underlying expression leads to a powerful consequence: human computation and programmatic computation
become structurally equivalent.
If I interactively create a Disk using a drawing palette:
its underlying data structure is the same as it would be if I had typed in the code, or if it had been generated by a program:
Graphics[Disk[{0.49, 0.49}, {0.35, 0.35}],
ImageSize -> {36.335, Automatic},
PlotRange -> {{0, 1}, {0, 1}}]
Completely different representations–in this case text and graphics–are immediately and transparently compatible, because in the end they are all just expressions. Here we have some “zero-overhead”
code to color just the Disks of a graphic:
The equivalence between these representations, and between human and programmatically generated structures, opens the door to exceptionally powerful ways of interacting with computation. It is the
basis of Mathematica‘s dynamic interactivity framework.
The implications, however, go much further than that. Symbolic programming is the first abstraction that can uniformly span all three stages of computation, from specification to execution to output,
and as a result it can leverage each aspect against the others.
Programs can be created by traditional programming, by the programs themselves, or by interaction with what is considered output. The process of computation can be equivalently executed by the user
or by the machine. Output is simply exposing a window into a piece of a (possibly running) program.
It’s pretty impressive that so many useful consequences can flow from the simple idea of symbolic programming: fitting computation to language, rather than the other way around. | {"url":"http://blog.wolfram.com/2007/06/15/symbolic-programming-computationally-active-language/?year=2007&monthnum=06","timestamp":"2014-04-21T09:37:54Z","content_type":null,"content_length":"61110","record_id":"<urn:uuid:349f0525-d4e0-485f-aa4b-d658bdb093d6>","cc-path":"CC-MAIN-2014-15/segments/1397609539705.42/warc/CC-MAIN-20140416005219-00337-ip-10-147-4-33.ec2.internal.warc.gz"} |
How Does Your Tree Measure Up?
Brief Description
Complete the chart that shows locations and sizes of some of the biggest trees in the United States.
Students will
• follow directions.
• work with partners or in small groups.
• measure accurately.
• accurately calculate data about a tree.
Calculate, data, tree, measure, circumference, transpiration, calculator, height, Arbor Day, spring
Materials Needed
• ruler
• stick or dowel, about 4 feet in length
• 8 small sticks (ice pop sticks will work well) per team/group
• roll of string
• square centimeter sheet
• plastic bag (optional)
Lesson Plan
In this lesson, students work in pairs or small groups to gather data about a tree. Each group might gather data about a different tree; all groups might collect data about the same tree; or two
teams might gather data about each tree and compare their results.
Measuring the Height of a Tree
Explain to students that they can measure the height of a tree in the following ways:
• Stand at the base of the tree, and hold a ruler straight out in front of you in a vertical position. Close one eye and back away from the tree until you reach the point at which the ruler and the
tree appear to be the same size. Stop and have a partner measure the distance between the tree and the ruler. That is the approximate height of the tree.
• One of the simplest ways to measure a tree's height requires a sunny day. Pound a stick or dowel into the ground. Measure the length of the stick above the ground, then measure its shadow. (For
example, the stick might be 3 feet tall; its shadow might be 2 feet long.) Now you know that a 3-foot stick casts a 2-foot shadow. Measure the shadow cast by the tree. If the tree's shadow is 10
feet long, you can use a little algebra to figure out the height of the tree: If 2/3 = 10/x, then 2x =10 X 3 or 2x = 30, so x = 15; the tree is approximately 15 feet high.
• Students in the upper elementary grades and above can figure out the height of the tree by making an inclinometer and using it to calculate the tree's height.
• For additional tree-measuring methods, see Following Fall: Measuring Tree Height, How to Measure a Big Tree, or Measuring Guide.
Measuring the Tree's Canopy (Leaf Cover)
Provide students with 8 small sticks (ice pop sticks or small dowels work well), and explain that they should visualize the tree's leaf cover as a clock. Have students place sticks in the ground at
the edge of the leaf cover at 12:00, 6:00, 3:00, and 9:00. Then have them place sticks halfway between each of those four sticks. When all eight sticks are inserted into the ground, students use
string to create a large circle around all the sticks, and then measure the length of the string. The length of the string is the circumference of the tree canopy.
Estimating the Number of Leaves on the Tree
Have students count as accurately as possible the number of leaves on a small twig from the tree. Then have them count the number of twigs on an average branch, and calculate the approximate number
of leaves on a branch. Then have them count the number of branches on the tree, and multiply the number of leaves on a branch by the number of branches on the tree. (Students might use calculators to
complete this activity.) Have each student in the group calculate the number of leaves on the tree; the group then uses the average of its members' counts as its official "leaf census" number. Older
students should produce a more accurate count than younger students.
Estimating the Age of the Tree
The approximate age of some trees can be estimated by measuring the distance around the trunk of the tree at a point five feet off the ground. If the tree's girth at that point measures 24 inches,
the tree is approximately 24 years old. This method is not accurate for all trees. Contact a local tree expert or your university's extension service for additional information.
Calculating the Average Size of a Leaf
Students select four leaves of varying sizes from a tree. They use the square centimeter sheet to approximate the size of each leaf. That can be done by tracing the leaf on the centimeter sheet and
counting how many squares (square centimeters) the leaf covers. Alternatively, you might provide students with the square centimeter sheets photocopied onto transparencies. Have students place a
transparency over the leaf, color in the blocks that cover the leaf, then count the blocks.
Extension Activity: Determining a Tree's Transpiration
Older students can figure the amount of transpiration (water released) by a tree. (A mature tree can transpire more than 200 gallons per day!) In one method of determining the amount of
transpiration, students tie a plastic bag over a group of leaves and measure the amount of water. For another more detailed method, see Determining the amount of transpiration from a schoolyard tree.
Students share how they arrived at each of their calculations. You might have students write in their journals explanations of the math processes they used.
Lesson Plan Source
Education World
Submitted By
Gary Hopkins
National Standards
LANGUAGE ARTS: English
GRADES K - 12
NL-ENG.K-12.12 Applying Language Skills
MATHEMATICS: Number and Operations
GRADES 3 - 5
NM-NUM.3-5.3 Compute Fluently and Make Reasonable Estimates
GRADES 6 - 8
NM-NUM.6-8.3 Compute Fluently and Make Reasonable Estimates
GRADES 9 - 12
NM-NUM.9-12.3 Compute Fluently and Make Reasonable Estimates
MATHEMATICS: Algebra
GRADES 6 - 8
NM-ALG.6-8.3 Use Mathematical Models to Represent and Understand Quantitative Relationships
GRADES 9 - 12
NM-ALG.9-12.3 Use Mathematical Models to Represent and Understand Quantitative Relationships
MATHEMATICS: Measurement
GRADES 3 - 5
NM-MEA.3-5.1 Understand Measurable Attributes of Objects and the Units, Systems, and Processes of Measurement
NM-MEA.3-5.2 Apply Appropriate Techniques, Tools, and Formulas to Determine Measurements
GRADES 6 - 8
NM-MEA.6-8.1 Understand Measurable Attributes of Objects and the Units, Systems, and Processes of Measurement
NM-MEA.6-8.2 Apply Appropriate Techniques, Tools, and Formulas to Determine Measurements
GRADES 9 - 12
NM-MEA.9-12.1 Understand Measurable Attributes of Objects and the Units, Systems, and Processes of Measurement
NM-MEA.9-12.2 Apply Appropriate Techniques, Tools, and Formulas to Determine Measurements
MATHEMATICS: Problem Solving
GRADES Pre-K - 12
NM-PROB.PK-12.2 Solve Problems That Arise in Mathematics and in Other Contexts
NM-PROB.PK-12.3 Apply and Adapt a Variety of Appropriate Strategies to Solve Problems
NM-PROB.PK-12.4 Monitor and Reflect on the Process of Mathematical Problem Solving
MATHEMATICS: Communications
GRADES Pre-K - 12
NM-COMM.PK-12.2 Communicate Their Mathematical Thinking Coherently and Clearly to Peers, Teachers, and Others
MATHEMATICS: Connections
GRADES Pre-K - 12
NM-CONN.PK-12.3 Recognize and Apply Mathematics in Contexts Outside of Mathematics
GRADES K - 4
NS.K-4.3 Life Science
NS.K-4.4 Earth and Space Science
GRADES 5 - 8
NS.5-8.3 Life Science
NS.5-8.4 Earth and Space Science
GRADES 9 - 12
NS.9-12.3 Life Science
NS.9-12.4 Earth and Space Science
Find more great springtime lessons on Education World's Spring Lesson Plans page.
Click to return to this week's Lesson Planning article, Trees Sprout Classroom Lessons Throughout the Year.
Originally published 04/18/2003
Last updated 03/20/2010 | {"url":"http://www.educationworld.com/a_lesson/03/lp309-01.shtml","timestamp":"2014-04-19T07:41:45Z","content_type":null,"content_length":"94075","record_id":"<urn:uuid:cdfc2fd2-099b-43bd-b2d8-c45375d0b8c6>","cc-path":"CC-MAIN-2014-15/segments/1398223203841.5/warc/CC-MAIN-20140423032003-00459-ip-10-147-4-33.ec2.internal.warc.gz"} |
Integral Equations and Applications
Author: , Date: 10 Apr 2010, Views: ,
C. Corduneanu, quot;Integral Equations and Applicationsquot;
Cambridge University Press | 2008 | ISBN: 052109190X | 380 pages | File type: PDF | 2,6 mb
The purpose of this book is threefold: to be used for graduate courses on integral equations; to be a reference for researchers; and to describe methods of application of the theory. The author
emphasizes the role of Volterra equations as a unifying tool in the study of functional equations, and investigates the relation between abstract Volterra equations and other types of
functional-differential equations.
[Fast Download] Integral Equations and Applications
Copyright Disclaimer:
This site does not store any files on its server. We only index and link to content provided by other sites. Please contact the content providers to delete copyright contents if any and email us,
we'll remove relevant links or contents immediately.
Astronomy and Cosmology Physics
Philosophy Medicine
Mathematics DSP
Cryptography Chemistry
Biology and Genetics Psychology and Behavior | {"url":"http://www.ebook3000.com/Integral-Equations-and-Applications_52755.html","timestamp":"2014-04-19T19:34:50Z","content_type":null,"content_length":"13885","record_id":"<urn:uuid:ddec3932-c398-4845-a9ac-175052dc8c37>","cc-path":"CC-MAIN-2014-15/segments/1397609537376.43/warc/CC-MAIN-20140416005217-00391-ip-10-147-4-33.ec2.internal.warc.gz"} |
Differential eq problem and area problem
May 24th 2009, 04:14 AM
Differential eq problem and area problem
State a differential equation that has the solution:
decide the number k that the area beneath the curve y=ke^-x on the first square will be 2e areaunits
Would be great if you showed how to solve these problems, step by step. thanks
June 5th 2009, 04:46 PM
A. $y=3\sin(4x+v) \rightarrow y'=12\cos(4x+v) \rightarrow y''=-48\sin(4x+v)=-16y$, so the differential equation $y''=-16y$ has the solution you seek, where v is a constant of integration.
B. Area= $\int_a^b ke^{-x}dx=-ke^{-x}|_a^b=-k(e^{-b}-e^{-a})$ . I am not sure what is meant by "on the first square" here. You are seeking a k for which the Area equals $2e$, but I do not know
what this phrase is implying for choosing a and b. | {"url":"http://mathhelpforum.com/calculus/90269-differential-eq-problem-area-problem-print.html","timestamp":"2014-04-19T23:33:40Z","content_type":null,"content_length":"4821","record_id":"<urn:uuid:16ec5935-9098-40cb-997c-7a761e2226b9>","cc-path":"CC-MAIN-2014-15/segments/1397609537754.12/warc/CC-MAIN-20140416005217-00390-ip-10-147-4-33.ec2.internal.warc.gz"} |
How to Write an Equation without LateX in Wordpress
How to Write an Equation without LateX in WordPress
(written by: Willy Yanto Wijaya)
For those who are not familiar with LateX, writing mathematical equations in WordPress might be something troublesome and confusing. Furthermore, many of you may not have much time to learn about the
syntax of LateX.
So is there other way to write mathematical equations without LateX?
Yes, there is.
By converting the mathematical equation into a graphic (picture) file (such as .jpg, .png, etc)
How to do that?
Here are simple steps to do it:
1. Write the equation (such as using Microsoft Equation) in a Ms. Powerpoint slide.
2. Right-click the equation, and then choose “Save as Picture…”
3. Choose save as type .png (I recommend to save as .png since generally it has better quality than .jpg. Actually so far as I know .tiff is best but unfortunately this file type is not supported
by WordPress)
Can we just directly uploaded the .png file? Yes, we can. But the result will be like this:
For those you who are perfectionist, I know that you will not accept an image with such a quality. Well, so let’s do another step:
4. Using a simple image software (such as Ms. Office Picture Manager), choose “edit picture”, then compress picture to document size. After that, just save the modified picture as a new file (.jpg or
.png). (You will notice that the new file will have much bigger size, in my case it became about 29 kB from previously only 3 kB — but 29 kB is still reasonably a small size for webpage purpose).
When uploaded to WordPress, it looks approximately like this (.jpg file):
I suppose that this is already good enough. But if you still feel not satisfied, you can do this alternative step:
5. Using the picture (.png) file of step 3, increase the dpi of the .png file (using for example Adobe Photoshop or other software). When the dpi is increased, the file size will also increase. But
we can reduce the size (using such as simple Ms. Office Picture Manager – just choose the “edit pictures” and then “resize” and then adjust the “percentage of original width and height” to fit our
desired file size).
The above equation (.png) has the size just 26 kB (smaller than the previous 29 kB .jpg file and also better quality). I recommend you to do step 1, 2, 3 then 5 for best quality mathematical equation
visual for WordPress.
If you have any ideas, comments, suggestions or other insights, please feel free to write them down in the comments section of this page.
Hope that this little tip/ trick can be useful for you.. =)
Btw, the above equation is the derivation form of van’t Hoff equation to determine the equilibrium constant of chemical reactions with respect to temperature change.
7 Responses to How to Write an Equation without LateX in WordPress
1. I think this method is far more troublesome than latex, you need more than one (and not free) softwares to do this, and also it will be a nightmare if we need to edit the equations one day,
especially in a case we have more than 20 equations in our post, and also how about fully-automatic equation numbering and referencing? I don’t think we can do that by this method. Furthermore,
actually wordpress is not suitable to write scientific post with many equations, wiki-powered site would be the best.
Just learn wiki+latex, it’s fun! :mrgreen:
□ That’s true, if you are going to write a lot of equations in your wordpress blog =)
But for those who just occasionally want to write math equation and no idea about LateX at all, this is one good option. Furthermore, things like Ms. Powerpoint, Ms. Equation, Ms. Picture
Manager are just standard stuffs basically included in the Windows OS package nowadays.
2. Well, doing this is very great however, I think uploading images in wordpress is not so convenient.
if only they had the same uploading style just like the facebook.
□ Yeah.. uploading images in wordpress is not so convenient since wordpress is designed as a blog-style website, not an images hosting website.
Facebook part for keeping images in albums is quite convenient; but if you want to add images in Facebook notes, it is also not convenient.. much more troublesome than in wordpress (since
sometimes the position of the image in a Facebook note can shift to other location).
3. Well…. I’d better spend my time to concentrate on latex expression I write rather than waiting hundreds of small images finish uploading…..
But is there any wordpress plugin to help? If so, please do share…..:)
□ yes, that’s true, if you are going to write hundreds of equations, better learn latex expressions. :)
this is just an alternative for those who rarely write equations on their blog..
4. Thanks for the good tips. I prefer this method since I’m already familiar with image editing software. | {"url":"http://willyyanto.wordpress.com/2010/02/21/how-to-write-an-equation-without-latex-in-wordpress/","timestamp":"2014-04-16T10:09:33Z","content_type":null,"content_length":"92352","record_id":"<urn:uuid:54a5ec43-f445-4b06-add0-be5f866fbe10>","cc-path":"CC-MAIN-2014-15/segments/1397609523265.25/warc/CC-MAIN-20140416005203-00537-ip-10-147-4-33.ec2.internal.warc.gz"} |
A Ladder Of Weight 200 N And Length 10 Meters Leans ... | Chegg.com
A ladder of weight 200 N and length 10 meters leans against a smooth wall (no friction on wall). The ladder makes an angle 50.0° with the ground. A firefighter of weight 600 N climbs a distance x up
the ladder. The coefficient of friction between the ladder and the ground is 0.5. What is the maximum value of x if the ladder is not to slip. | {"url":"http://www.chegg.com/homework-help/questions-and-answers/ladder-weight-200-n-length-10-meters-leans-smooth-wall-friction-wall--ladder-makes-angle-5-q1079039","timestamp":"2014-04-16T17:28:46Z","content_type":null,"content_length":"20828","record_id":"<urn:uuid:61afbc78-e8cc-4053-b186-abc2bb44dede>","cc-path":"CC-MAIN-2014-15/segments/1397609524259.30/warc/CC-MAIN-20140416005204-00387-ip-10-147-4-33.ec2.internal.warc.gz"} |
Posts by
Total # Posts: 117
3.The Lo Company earned $2.60 per share and paid a dividend of $1.30 per share in the year that just ended. Earning and dividend per share are expected to grow at a rate of 5% per year in the future.
Determine the value of the stock:
contract the expression,use the property of logarithms to write each expression as a single logarithm with a coefficient of 1
which of the following determines who holds the position of speaker of the house? a the president b the house c the senate d the joint houses of congress
An ethanol-water solution is prepared by dissolving 10.00 ml ethanol C2H5OH (density=0.789g/ml) in sufficient water to produce 100.00 ml of solution with a density of 0.982g/ml. What is the
concentration of ethanol in this solution? Expressed as ml. a.)%volume b.)%mass c.)%m/v...
how to write solving problems scene of three characters which started with prologue
whats the answer?
I also need answers.
I also need answers.
physical mod
What mass (in grams) of steam at 100°C must be mixed with 462 g of ice at its melting point, in a thermally insulated container, to produce liquid water at 40.0°C? The specific heat of water is 4186
J/kg·K. The latent heat of fusion is 333 kJ/kg, and the latent ...
how to factorise?
physical pharmacy
Hoy does KH2PO4 induce controlled flocculation..?
no one knows?
A solution contains 25.5 g of NaCl dissolved in sufficient water to give a total mass of 325.0 g. What is the molality of the solution
A solution of ethanol, C2H5OH, is prepared by dissolving 25.0 mL of ethanol in enough water to give a total volume of 250.0 mL. What is the percent-by-volume concentration of ethanol?
wheels on a skateboard are 3.63 inches. If the skateboarder is travelling down a hill at 23 miles per hour what is the angular velocity.
Two identical point charges are separated by a distance of 3.0 cm and they repel each other with a force of 4 X 10^-5 N. What is the new force if the distance between the point charges is doubled?
Language art
I am new to this also.
The height of a rocket is launched from a platform 12 meters above the ground is modeled by the function h(t)=-4.9t^2+25t+12 where h(t) represents the height of the rocket in meters and t represents
the number of seconds after launching. How many seconds after launching will t...
The height of a rocket is launched from a platform 12 meters above the ground is modeled by the function h(t)=-4.9t^2+25t+12 where h(t) represents the height of the rocket in meters and t represents
the number of seconds after launching. How many seconds after launching will t...
The height of a rocket is launched from a platform 12 meters above the ground is modeled by the function h(t)=-4.9t^2+25t+12 where h(t) represents the height of the rocket in meters and t represents
the number of seconds after launching. How many seconds after launching will t...
The height of a rocket is launched from a platform 12 meters above the ground is modeled by the function h(t)=-4.9t^2+25t+12 where h(t) represents the height of the rocket in meters and t represents
the number of seconds after launching. How many seconds after launching will t...
btw thats for you ms.sue
i told some of my classmates about this site and they said it really help but they also added that you all shall show examples when you are finished and an explanation and then give the person a sum
similar to that.
On July 1, 2011, Jackson Company exercises a $5,000 call option (plus par value) on its outstanding bonds that have a carrying value of $208,000 and par value of $200,000. The company exercises the
call option after the semiannual interest is paid on June 30, 2011. Record the ...
HOW YOU DO THAT?
0.3 grams NaOH dissolved in 2.5litres. How many moles?
help me in direct and indirect
how do you do this table? (gx2)-3 ?
A mass of m1 = 6.00 kg is sitting on an inclined plane, which meets the horizantal at an angle of 22.0, which has a coefficient of sliding friction of 0.290, which is 3.50 m = L. a string is attached
to m1, it is strung over a pulley, and attached to mass m2 = 7.0kg which is i...
Identify the appositive or complete appositive phrase. Sam, our cat, is too lazy to catch mice.
Ms.Susu's math
someone tell me what website to go on and find x+z-h=___
umm i will help you just go to google look up the question
world history
idk look it up on google com then find it is youu a girl (chicka) or a boy {chicko}
In science wwhy is that when they say what is the meaning of friction?
Perform the indicated operations and simplify: i^-31
divide the left and right side of the equation by ð r^2 V / (ð r ^2) = h ?
im not sure how to solve this problem at all
C B D?
A rectangular prism has a length of 100 cm, a height of 10 cm and a width of 5 cm. What is the volume of the rectangular prism? A. 3100 square centimeters B. 1550 square centimeters C. 5000 square
centimeters D. 1200 square centimeters 2. A cylinder has a diameter of 5 meters ...
1. Which of these chemical formulasCl2, KCl, Ar, or NO2represents both an element and a molecule? 2. A miner finds a small mineral fragment with a volume of 5.74 cm3 and a mass of 14.4 g. What is the
density of that mineral fragment?
thank you !
Pentane gas (C5H12) combusts with oxygen gas (O2) to form water (H2O) and carbon dioxide (CO2). Write a balanced chemical equation for this reaction and explain the scientific principle (statement)
that requires the balancing of an equation to make it conform to reality.
Explain in one or two sentences how new evidence affects scientific models
Use the balanced equation below to answer the following questions. 2Al(NO3)3 + 3Na2CO3 Al2(CO3)3(s) + 6NaNO3 - What is the ratio of moles of Al(NO3)3 to moles Na2CO3? (1 point) - If 852.04 g Al(NO3)3
reacted with the 741.93 g Na2CO3, which would be the limiting reagent? Show y...
Use the following steps to balance the redox reaction below: Mg + Au+ Mg2+ + Au -Write the oxidation and reduction half-reactions. Make sure each half-reaction is balanced for number of atoms and
charge. -Multiply each half-reaction by the correct number, in order to balance c...
A stock solution is prepared by adding 20mL of 0.2M MgF2 to enough water to make 80mL. What is the F concentration of 20 mL of the stock solution?
if offspring exhibit a 3:1 phenotype ratio, what are the genotypes of the parents?
what latin word is the phrase strong distaste derived rom
Please help me errors in my paper so I can correct them. The Lottery takes place on a beautiful summer day in a small town. Although the location of the town, is never given, it appears to be in a
rural community. The story begin as a regural summer day, but has a surising en...
The Lottery takes place on a beautiful summer day in a small town. Although the location of the town, is never given, it appears to be in a rural community. The story begin as a regural summer day,
but has a surising ending. In the short story "The Lottery" by Shirl...
Please check for any errors in my paper. And crrect the errors for me. Let me know if there is anything I need to change. The Lottery takes place on a beautiful summer day in a small day in a small
town.However, the the location of the town, is never given, it appears to be in...
physics 12
An 884 kg satellite in orbit around a planet has a gravitational potential energy of -5.44*10^10 J. The orbital radius of the satellite is 8.52*10^6 m and its speed is 7.84*10^3 m/s. A) What is the
mass of the planet? B) What is the kinetic energy of the satellite? C) What is ...
3 times 3
write an expression to represent each total and then simplify it cost of 2 pens and 3 notebooks the note books costs 179 each nd the pens 289 each
classify each statement as sometimes, always, or never true. give examples or properties to support your answer a+ (bc) =( a+b) .(a+c)
Find the perimeter of the triangle with vertices at A (0, 0), B (-4, -3) and C (-5, 0). If it is not possible, write not possible and explain why it is not possible.
our cat
Part 1. The area of a circular pool is approximately 7,850 square feet. The owner wants to replace the tiling at the edge of the pool. The edging is 6 inches wide, so he plans to use 6-inch square
tiles to form a continuous inner edge. How many tiles will he need to purchase? ...
chemistry 32
Calculate the concentration, in molarity, of a solutiion prepared by adding 9 mL of water to 1mL of 0.1 M of HCl
algebra 1
four less than half a number is 17.find the number
A bond price of $987.50 has a face value of $1000, pays 5% semiannually, and will repay the face value in 15 years. 5% tables Present Values Pv Annuity year 13 .53032 9.39357 year 14 .50507 9.89864
year 15 .48102 10.37966 What is the yield to maturity of the loan a) 4.9% b) 5....
A bond price of $987.50 has a face value of $1000, pays 5% semiannually, and will repay the face value in 15 years. 5% tables Present Values Pv Annuity year 13 .53032 9.39357 year 14 .50507 9.89864
year 15 .48102 10.37966 What is the yeild to maturity of the loan a) 4.9% b) 5....
Calculate the concentration, in molarity, of a solutiion prepared by adding 9 mL of water to 1mL of 0.1 M of HCl
4th grade math input output tables
how do you do this on the table ??? expression (gx2)-3
calculate the frequency of light that has a wavelength of 103 nm
what is the mathematical expression of 9+4x5 divided by 2 ?
4th grade
what is a question with the word there force
How did dalton discover the atom?
math 116
How do you know if a value is a solution for an inequality? How is this different from determining if a value is a solution to an equation? If you replace the equal sign of an equation and put an
inequality sign in its place, is there ever a time when the same value will be a ...
math 116
How do you know if a value is a solution for an inequality? How is this different from determining if a value is a solution to an equation? If you replace the equal sign of an equation and put an
inequality sign in its place, is there ever a time when the same value will be a ...
look at the answer for number 13 in 7.2 in the book or the solutions manual if you have it
college algebra
(5,1) is a solution to the equation 2x-3y=7. what does this means
u.s historyy
brittish officals issued warrents called _____ when they wanted to search for smuggled goods
What is the representative particle for a salt, nacl2?
Good luck
Factor completely x³-8x²+11x+20
Post your response to the following: Explain whether or not you believe CHAMPVA provides an adequate level of health insurance for participants.
**********Correct easy answer************* See this last answer would've been right up to one point I got a different answer than all of you guys :P HINT: To actually visualize the problem, draw it
out! :) Obviously the dimensions of the painting are: w and 3+w We know tha...
**********Correct easy answer************* See this last answer would've been right up to one point I got a different answer than all of you guys :P HINT: To actually visualize the problem, draw it
out! :) Obviously the dimensions of the painting are: w and 3+w We know tha...
College Algebra
Knowing the given conditions, write the equation of the circle. Center at (5,3) and tangent to the y-axis.
Kevin hit a golf ball 123 yards. How far did he hit the golf ball in feet?
what is the difference between to convert and conversion ex. convert 623.5 F to K ex. conversion 203.77miles/sec to cm/nsec
5th grade math
whats 5.0 pluse o.90
Principles Of Business
explain potential for growth internally and externally.
what happen to cereal when heated?
AP Chem!
oh ok, thank you. I was pretty much asking for some type of situation. Such as what happens to a balloon when. blah. but i see what you mean. My teacher often relates back to old AP tests for his
tests. I have one more quesiton: what are the units of the rates in the law of ef...
AP Chem!
Hey, I have a test tomorrow in ap chemistry on gases. My teacher said that the test would be many open ended questions that are sort of like "explain what happens when a balloon is lowered into
77degrees C of liquid nitrogen and why" Can you guys give me some other e...
reading 4th grade
unscramble draleun
Does a change in price create curve shifts?
I have two poems that i have to analyze for poetic devices / figurative language such as (Alliteration, Metaphor, Hyperbole, Simile, Etc) I also need the theme of these poems, Not just "Love" but the
more specific theme Sonnet 18 Shall I compare thee to a summer'...
i need to have sex in two days what do i need to buy
You have recently found a location for your bakery and have begun implementing the first phases of your business plan. Your budget consists of an $80,000 loan from your family and a $38,250 small
business loan. These loans must be repaid in full within 10 years
Pages: 1 | 2 | Next>> | {"url":"http://www.jiskha.com/members/profile/posts.cgi?name=Adrian","timestamp":"2014-04-19T17:07:30Z","content_type":null,"content_length":"25266","record_id":"<urn:uuid:fc0147af-046d-4ed1-b524-85deea194cf1>","cc-path":"CC-MAIN-2014-15/segments/1397609537308.32/warc/CC-MAIN-20140416005217-00484-ip-10-147-4-33.ec2.internal.warc.gz"} |
Mathematical statements are exactly the same as fashion statements. Except instead of clothes, we have mathematical formulas. Hopefully we won't get chilly walking down the runway.
The simplest kind of mathematical statement is an explanation of how numbers are related. For example, you might say, "x = 5" or, "4 + 7 = 35" or, "58 is the sum of two prime numbers." As you can
see, some statements are true, some are false, and some are as clear as a mud smoothie.
What all of these statements have in common is that they can't be split into simpler statements—they are indivisible (with liberty and justice for all). The Greek word for indivisible is atomos, so
we call these statements atoms. Unlike the atoms in chemistry, mathematical atoms can make only statements, not bombs.
Does any value of s make the statement "s > 5 and s < 2" true?
Does any food satisfy, "This food is a pie or it tastes like a blueberry"?
Is there any value of x that makes "x < 4 and not(x^2 < 20)" true?
Sandra loves skydiving, pepperoni pizza, and the number 17. She doesn't like Tuesdays or the color purple (sorry, Alice Walker). Is the statement "Sandra loves pepperoni pizza and she loves Tuesdays"
Sandra loves skydiving, pepperoni pizza, and the number 17. She doesn't like Tuesdays or the color purple (sorry, Alice Walker). Is the statement "Sandra loves 17 or she does not love 17" true?
Sandra loves skydiving, pepperoni pizza, and the number 17. She doesn't like Tuesdays or the color purple (sorry, Alice Walker). Is the statement "Sandra does not love purple or she loves skydiving"
Is there a value for x that will satisfy the statement "x = 7 or x^2 = 25"?
Is there a value that will satisfy the statement "x^0 = 1 and x + 1 = 4,623"? | {"url":"http://www.shmoop.com/logic-proof/mathematical-statements-help.html","timestamp":"2014-04-18T13:23:17Z","content_type":null,"content_length":"43423","record_id":"<urn:uuid:6aad36c7-7bd6-4a90-b2d5-b51cc8033e84>","cc-path":"CC-MAIN-2014-15/segments/1398223202457.0/warc/CC-MAIN-20140423032002-00646-ip-10-147-4-33.ec2.internal.warc.gz"} |
week beginning
Monday April 26^th
Seminar: Probability Seminar
Location and Time: Meeting Room 12 (CMS) at 2.00pm
Speaker: Dilip Madan (University of Maryland)
Title: Correlation and the pricing of Risk
Seminar: CMS Colloquia
Location and Time: Lecture Room 1 (INI) at 5.00pm
Speaker: Professor D. Frenkel (Amsterdam)
Title: Revisiting the basis of molecular simulations
Tuesday April 27^th
Seminar: Kuwait Lecture
Location and Time: Meeting Room 2 (CMS) at 5.00pm
Speaker: Professor Bernard Malgrange
Title: Non-linear differential Galois Theory
Seminar: Probability Seminar
Location and Time: Meeting Room 12 (CMS) at 2.00pm
Speaker: Serguei Novak (Middlesex University Business School)
Title: A probabilistic version of Taylor's formula and a moment inequality
Wednesday April 28^th
Seminar: Study Group
Location and Time: Meeting Room 13 (CMS) at 1.30pm
Speaker: Professor A.J. Scholl
Title: Study Group on Heegner Points
Seminar: Geometry Seminar
Location and Time: Meeting Room 13 (CMS) at 3.00pm
Speaker: Bernard Malgrange (Grenoble)
Title: Isomonodromic deformations
Seminar: Algebra Seminar
Location and Time: Meeting Room 12 (CMS) at 4.30pm
Speaker: Eric Friedlander (Evanston)
Title: A semi-topological view of K-theory and cohomology
Thursday April 29^th
Seminar: Combinatorics Seminar
Location and Time: Meeting Room 12 (CMS) at 2.00pm
Speaker: Professor Svante Janson (Uppsala)
Title: Displacements for linear probing hashing | {"url":"https://www.dpmms.cam.ac.uk/Seminars/Weekly/2003-2004/seminars-26-Apr-2004.html","timestamp":"2014-04-16T16:47:49Z","content_type":null,"content_length":"5977","record_id":"<urn:uuid:e3e275e1-5d31-42ff-b42a-eebb29b386e1>","cc-path":"CC-MAIN-2014-15/segments/1398223207046.13/warc/CC-MAIN-20140423032007-00043-ip-10-147-4-33.ec2.internal.warc.gz"} |
Department of Mathematics
We strongly encourage applications from strong candidates interested in pursuing a PhD in probability or statistics. Below are some suggested projects that are on offer; we also warmly welcome
suggestions for other projects that you might be interested in, and which have connections with our research interests.
Dr Stephen Connor
Dr Degui Li
Please contact Dr Li directly if you are interested in pursuing a PhD with him.
Dr Samer Kharroubi
Dr Sonia Mazzi
Please contact Dr Mazzi directly if you are interested in pursuing a PhD with her.
Prof. Wenyang Zhang
Prof. Wenyang Zhang is interested in taking on PhD students to work on Nonparametric Statistics, Nonlinear Time Series, Survival Analysis, Functional Data Analysis, Spatial Data Analysis, Multi-level
Modelling, Structural Equation Models. The following is one example of the PhD projects he would like to supervise. Please contact him directly for more PhD projects.
• Coupling of Markov chains
-- Dr Stephen Connor
Let X and Y be two copies of a Markov chain, begun from different states. We are often interested in constructing X and Y so that if they meet one another then they remain together from then onwards
- such a construction is known as a coupling. Couplings are interesting for a number of reasons, not least because they can provide information about the rate at which X converges to its stationary
distribution (if it has one). They also have a number of practical uses, in particular in the construction of perfect simulation algorithms. There are lots of interesting open problems in coupling
theory, many of which would make good PhD projects for students interested in stochastic processes.
• Statistical methods in health economics
-- Dr Samer Kharroubi
The main concern of the field of health economics is to examine the cost-effectiveness of medical technologies. Whenever new drugs, treatments or medical devices (collectively called medical
technologies) are proposed, their sponsors have to demonstrate that they are safe and effective before they can be licensed for public use. Demonstration of safety and efficacy is generally done by
conducting a substantial clinical trial, followed by statistical analysis of the data. This is a well-developed area of medical statistics and so a fruitful area for ongoing research.
• Estimation of Time Varying High Dimensional Covariance Matrix and Its Application in Optimal Portfolio Allocation in Finance
-- Prof. Wenyang Zhang
This project is stimulated by the optimal portfolio allocation in finance. It is well know that the optimal portfolio allocation can be expressed in terms of the covariance matrix of the returns of
the stocks under consideration. The estimation of covariance matrix would not be a big deal when sample size is much larger than the size of the covariance matrix, for example, the sample covariance
matrix would be a good estimator. However, when the sample size and the size of the covariance matrix are comparable, which is the case in portfolio allocation as the number of stocks is typically of
the same order as the sample size, the sample covariance matrix as an estimator of the covariance matrix would run into trouble. This is because the estimated optimal portfolio allocation depends on
the inverse of the estimator of the covariance matrix, when the size of the covariance is large, the random errors of the estimators of the elements in the covariance matrix will accumulate, which
will make the estimator of the optimal portfolio allocation very poor.
Most existing literature about the estimation of the covariance matrix is based on the assumption that the covariance matrix is a constant matrix. This is clearly not a realistic assumption in
portfolio allocation as today's optimal portfolio allocation may not be optimal next month. What is more realistic is to assume the optimal portfolio allocation depends on time and estimate it
through the estimated time-varying covariance matrix of the returns of the stocks.
This project is going to investigate the dynamic structure of the time-varying covariance matrix of large size and construct the estimation procedure for the time-varying covariance matrix and the
time-varying optimal portfolio allocation. In summary, this project is going to develop a new estimation procedure for time-varying covariance matrix of large size, establish asymptotic properties to
justify the estimation method, and apply the method to analyse some financial data sets from London Stock Market and Shanghai Stock Market. | {"url":"http://maths.york.ac.uk/www/StatisticsPhDProjects","timestamp":"2014-04-17T18:27:25Z","content_type":null,"content_length":"21149","record_id":"<urn:uuid:daaac274-8184-4402-aed2-113d8b7b13a2>","cc-path":"CC-MAIN-2014-15/segments/1397609530895.48/warc/CC-MAIN-20140416005210-00627-ip-10-147-4-33.ec2.internal.warc.gz"} |
, 2005
"... We discuss questions related to the non-existence of gaps in the series defining modular forms and other arithmetic functions of various types, and improve results of Serre, Balog & Ono and
Alkan using new results about exponential sums and the distribution of B-free numbers. ..."
Cited by 4 (3 self)
Add to MetaCart
We discuss questions related to the non-existence of gaps in the series defining modular forms and other arithmetic functions of various types, and improve results of Serre, Balog & Ono and Alkan
using new results about exponential sums and the distribution of B-free numbers.
- Proceedings of the Millennial Conference on Number Theory , 2002
"... This paper presents lower bounds and upper bounds on the distribution of smooth integers; builds an algebraic framework for the bounds; shows how the bounds can be computed at extremely high
speed using FFT-based power-series exponentiation; explains how one can choose the parameters to achieve ..."
Cited by 3 (1 self)
Add to MetaCart
This paper presents lower bounds and upper bounds on the distribution of smooth integers; builds an algebraic framework for the bounds; shows how the bounds can be computed at extremely high speed
using FFT-based power-series exponentiation; explains how one can choose the parameters to achieve any desired level of accuracy; and discusses several generalizations.
"... V. Vu has recently shown that when k ≥ 2 and s is sufficiently large in terms of k, then there exists a set X(k), whose number of elements up to t is smaller than a constant times (t log t) 1/s,
for which all large integers n are represented as the sum of s kth powers of elements of X(k) in order lo ..."
Add to MetaCart
V. Vu has recently shown that when k ≥ 2 and s is sufficiently large in terms of k, then there exists a set X(k), whose number of elements up to t is smaller than a constant times (t log t) 1/s, for
which all large integers n are represented as the sum of s kth powers of elements of X(k) in order log n ways. We establish this conclusion with s ∼ k log k, improving on the constraint implicit in
Vu’s work which forces s to be as large as k 4 8 k. Indeed, the methods of this paper show, roughly speaking, that whenever existing methods permit one to show that all large integers are the sum of
H(k) kth powers of natural numbers, then H(k) + 2 variables suffice to obtain a corresponding conclusion for “thin sets, ” in the sense of Vu. 1.
, 2012
"... Abstract. We are interested in classifying those sets of primes P such that when we sieve out the integers up to x by the primes in P c we are left with roughly the expected number of unsieved
integers. In particular, we obtain the first general results for sieving an interval of length x with prime ..."
Add to MetaCart
Abstract. We are interested in classifying those sets of primes P such that when we sieve out the integers up to x by the primes in P c we are left with roughly the expected number of unsieved
integers. In particular, we obtain the first general results for sieving an interval of length x with primes including some in ( √ x, x], using methods motivated by additive combinatorics. 1.
Introduction and | {"url":"http://citeseerx.ist.psu.edu/showciting?cid=1678702","timestamp":"2014-04-19T23:04:54Z","content_type":null,"content_length":"20500","record_id":"<urn:uuid:7a6daeb8-21bd-4dbb-bfc8-6127bffd8f11>","cc-path":"CC-MAIN-2014-15/segments/1398223207985.17/warc/CC-MAIN-20140423032007-00508-ip-10-147-4-33.ec2.internal.warc.gz"} |
Got Homework?
Connect with other students for help. It's a free community.
• across
MIT Grad Student
Online now
• laura*
Helped 1,000 students
Online now
• Hero
College Math Guru
Online now
Here's the question you clicked on:
the product of some negative number and 4 less than twice that number is 240. find the number.
Best Response
You've already chosen the best response.
solve \[x(2x-4)=240\] for x and take the negative solution
Best Response
You've already chosen the best response.
x = 10 x = -12 so the negative number is -12
Best Response
You've already chosen the best response.
ooo okay thanks :)
Your question is ready. Sign up for free to start getting answers.
is replying to Can someone tell me what button the professor is hitting...
• Teamwork 19 Teammate
• Problem Solving 19 Hero
• Engagement 19 Mad Hatter
• You have blocked this person.
• ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.
This is the testimonial you wrote.
You haven't written a testimonial for Owlfred. | {"url":"http://openstudy.com/updates/4e7a9fae0b8b2cb239c53bb2","timestamp":"2014-04-21T04:45:25Z","content_type":null,"content_length":"32369","record_id":"<urn:uuid:d8aac30b-8b6b-4217-8ac0-3f0cf20550e8>","cc-path":"CC-MAIN-2014-15/segments/1397609539493.17/warc/CC-MAIN-20140416005219-00435-ip-10-147-4-33.ec2.internal.warc.gz"} |
Bonita, CA Trigonometry Tutor
Find a Bonita, CA Trigonometry Tutor
...I lived in France and Mexico and I know both cultures very well. Having a strong knowledge of these foreign languages, for twelve years I have taught elementary school through university as
well as to private students and corporate. Give yourself or your child the opportunity that learning Fren...
8 Subjects: including trigonometry, Spanish, prealgebra, algebra 2
...I can give examples of why the subject is useful as well as explain the best way to apply the different concepts in a way that is easy to understand. I am also familiar with both AP Calculus
exams as I was able to receive a top mark on both of them when I took them. I can help to prepare any student looking to do well on the AP Calculus AB or BC exam.
19 Subjects: including trigonometry, chemistry, calculus, physics
...I passed AP World History with a 5. I then went to UCSD's Eleanor Roosevelt College where I used the same textbook as in AP World History for two years as part of the Making of the Modern
World sequence. I passed each of my courses with flying colors, and have become more familiar with that textbook than almost any other book.
42 Subjects: including trigonometry, Spanish, reading, writing
...My major is currently in math, and I'm working to become a math teacher My strength is breaking down what seems like big concepts and relating it to stuff students have seen before. I'm
patient and personable, and I work really well with students who may have anxiety toward this subject. I hav...
13 Subjects: including trigonometry, chemistry, statistics, calculus
...I graduated with a Bachelor of Science in theoretical mathematics, which uses logic as its backbone. As a result, I took several courses in logic as part of my degree and had to be extremely
fluent in its content (both theory and syntax) to complete my degree. Finally, I have tutored in the subject many times.
26 Subjects: including trigonometry, chemistry, physics, geometry
Related Bonita, CA Tutors
Bonita, CA Accounting Tutors
Bonita, CA ACT Tutors
Bonita, CA Algebra Tutors
Bonita, CA Algebra 2 Tutors
Bonita, CA Calculus Tutors
Bonita, CA Geometry Tutors
Bonita, CA Math Tutors
Bonita, CA Prealgebra Tutors
Bonita, CA Precalculus Tutors
Bonita, CA SAT Tutors
Bonita, CA SAT Math Tutors
Bonita, CA Science Tutors
Bonita, CA Statistics Tutors
Bonita, CA Trigonometry Tutors | {"url":"http://www.purplemath.com/Bonita_CA_trigonometry_tutors.php","timestamp":"2014-04-16T22:31:23Z","content_type":null,"content_length":"24315","record_id":"<urn:uuid:9a43584f-f63e-4366-93a7-d5ac7d943e64>","cc-path":"CC-MAIN-2014-15/segments/1397609525991.2/warc/CC-MAIN-20140416005205-00288-ip-10-147-4-33.ec2.internal.warc.gz"} |
Quantum blog
Goodbye Matlab, hello Python !
This blog will not be updated any more (except for the post comments).
New location is
Please update your bookmarks and feeds.
A couple of observant readers already figured out that in spite of the url of this blog, past posts have been based on Python code. The gradual transfer of all of my research code from Matlab to
Python is now complete. After working with Python for more than two years, I can state that it an excellent tool for research and data crunching. When it comes to collecting data from the web,
cleaning and aligning datasets with missing data or building GUIs, it is the best tool I have ever come across. And don't forget that Python is open source and free.
The only downside that I found is that it can be challenging for a new user to set up the needed tools and libraries and the information is scattered around the web. This is the reason that I decided
to create a 'Trading With Python' course which is focused on trading strategy research and automation of boring daily tasks. The course will take you all the way from installation & data acquisition
to strategy design and backtesting. After just four weeks you'll be able to do most of the things you have seen on this blog yourself.
In my previous post I came to a conclusion that close-to-close pairs trading is not as profitable today as it used to be before 2010. A reader pointed out that it could be that mean-reverting nature
of spreads just shifted towards shorter timescales. I happen to share the same idea, so I decided to test this hypothesis.
This time only one pair is tested: 100$ SPY vs -80$ IWM. Backtest is performed on 30-second bar data from 11.2011 to 12.2012.
The rules are simple and similar to strategy I tested in the last post:
if bar return of the pair exceeds 1 on z-score, trade the next bar.
The result looks very pretty:
I would consider this to be enough proof that there is still plenty of mean-reversion on 30-second scale.
If you think that this chart is too good to be true, that is unfortunately indeed the case. No transaction costs or bid-ask spread were taken into account. In fact, I would doubt that there would be
any profit left after subtracting all trading costs.
Still, this kind of charts is the carrot dangling in front of my nose, keeping me going...
Bad news everybody, according to my calculations, ( which I sincerely hope are incorrect) the classical pairs trading is dead. Some people would strongly disagree, but here is what I found:
Let's take a hypothetical strategy that works on a basket of etfs:
From these etfs 90 unique pairs can be made. Each pair is constructed as a market-neutral spread.
Strategy rules:
On each day, for each pair, calculate z-score based on 25-day standard deviation.
If z-score > threshold, go short, close next day
If z-score < -threshold go long, close next day
To keep it all simple, the calculation is done without any capital management (one can have up to 90 pairs in portfolio on each day) . Transaction costs are not taken into account either.
To put it simply, this strategy tracks one-day mean reverting nature of market neutral spreads.
Here are the results simulated for several thresholds:
No matter what threshold is used, the strategy is highly profitable in 2008, pretty good throuh 2009 and completely worthless from early 2010.
This is not the first time I came across this change in mean-reverting behavior in etfs. No matter what I've tried, I had no luck in finding a pairs trading strategy that would work on ETFs past
2010. My conclusion is that these types of simple stat-arb models just don't cut it any more.
I suppose that my previous post did not provide insights on how PCA really works. Here is another try at the subject, using a simple pair as an example.
Let's take SPY and IWM, which are highly correlated. If daily returns of IWM are plotted against daily returns of SPY, the relationship is highly linear (see left chart).
Applying PCA on this data gives two principal component vectors, plotted in red (first) and green (second). These two vectors are orhogonal, with the first one pointing in the direction of highest
variance. Transformed data is nothing more than the original data projected on the new coordinate axis formed by these two vectors. The transformed data is shown in the right chart. As you can
clearly see, all points are still there, but the dataset is rotated.
The second vector is in this case -0.78 SPY + 0.62 IWM which produces a market-neutral spread. Of course the same result would be achieved by using the beta of IWM.
The fun thing about PCA is that it is useful in building three- and more legged spreads. The procedure is exactly the same as above, but the transformation is done in a higer dimensional space.
Classical pairs trading usually involves building a pair consisting of two legs, which ideally should be market-neutral or in other words, pair returns should have zero correlation with market
returns. The process of building a 'good' pair is pretty standard. A typical way of building a pair (spread) involve choosing two correlated securities and forming a market-neutral pair using stock
Multi-legged spreads are more advanced and very difficult to build using the traditional method.
However, there is a mathematical method called Principal Component Analysis that can be easily used to create stable (=tradeable?) spreads. All the linear algebra is luckily hidden inside the
princomp function, but if you'd like to understand how PCA really works, take a look at this tutorial. The transformed data can be described as : 1-st component: 'max volatility portfolio', which is
usually very highly correlated with the market. 2-nd component: 'market-neutral' portfolio, having maximum variance. 3-d and further components have decreasing degrees of variance. Note that by
design, PCA produces orthogonal components, meaning that all portfolios are not correlated to each other. So 2nd and further portfolios are market-neutral.
Here is an example of applying PCA on some correlated etfs in the energy sector:
The upper chart shows raw prices, the lower char are the cumulative returns of principal components. To compute the principal components I only used first 250 days of data. It seems that the
principal components, which are linear combinations of each security returns are quite stable out-of-sample, which is a pleasant surprise. First (blue) component has most of the variance, and it is
clearly correlated to the movement of the prices in the upper chart.
Let's take a closer look at the last two components: these seem to be quite stable and tradeable even far out-of-sample.
The gap fading strategy from previous posts looked all right, but my worry is that Yahoo data does not provide accurate quotes. To check the strategy performance, I've generated a new OHLC dataset
based on the Weighted Average Price (wap) of 30-second intraday data. So the opening quote is the wap of first 30 seconds of trading and close is the last 30-second wap. To make sure that my dataset
is correct, I have compared it to the yahoo quotes. As shown in the chart below, the difference between the two quotes is ~5ct which seems very reasonable.
Now, testing the gap fade strategy on the OHLC data that I generated myself produces much less favorable result:
One look at the pnl chart is enough to say that this strategy would be rubbish.
This brings me to a conclusion that I already was aware of: Yahoo opening quotes are not suitable for strategy backtesting. | {"url":"http://matlab-trading.blogspot.com/","timestamp":"2014-04-16T16:17:29Z","content_type":null,"content_length":"96021","record_id":"<urn:uuid:690e4842-d666-4035-b4dd-17c2b313a2cd>","cc-path":"CC-MAIN-2014-15/segments/1398223202774.3/warc/CC-MAIN-20140423032002-00436-ip-10-147-4-33.ec2.internal.warc.gz"} |
Commutation of differentiation and averaging operations
I've been studying Turbulence, and there's a lot of averaging of differential equations involved. The books I've seen remark offhandedly that differentiation and averaging commute
for eg. < [tex]\frac{df}{dt}[/tex] > = [tex]\frac{d<f>}{dt}[/tex]
Here < > is temporal averaging. If the differentiation is w.r.t. a spatial coordinate it makes sense, but could someone help me with the above equation? | {"url":"http://www.physicsforums.com/showthread.php?t=487048","timestamp":"2014-04-16T16:05:37Z","content_type":null,"content_length":"19560","record_id":"<urn:uuid:b82993e0-13eb-4d27-b596-bae34d6fa8dd>","cc-path":"CC-MAIN-2014-15/segments/1397609524259.30/warc/CC-MAIN-20140416005204-00533-ip-10-147-4-33.ec2.internal.warc.gz"} |
Steam Engine Efficiency
From Open Source Ecology
Main > Energy > Steam Engine
[edit] Introduction
Start with definition of terms. Here is a chart showing relationships of various efficiency standards for a steam engine: (source:[1], Chapter 10. There are some compounds over 26% overall efficiency
and some singles running over 14%. Reportedly, 20% efficiency was obtained (another reference), but it had very low cutoff ratio and couldn't tolerate load changes. That is where the electronic steam
valve comes in to achieve this.)
The same reference has these typical Rankine cycle ratios for engines of different types:
[edit] Steam Engine Efficiency Predictions for the Factor e Farm Solar Power Generator
Updated 2.1.09
The missing link to date on the Factor e Farm solar power generator is the heat engine. Our goal is to develop a steam engine with Milestone 1 thermal efficiency of 17%, and Milestone 2 efficiency of
26%. The latter involves the use of electronic valving for total control of steam injection. Steam injection timing is called the cutoff ratio. As an analogy, this is comparable to electronic fuel
injection in internal combustion engines.
[edit] Efficiency Predictions of Steam Engine
The basic efficiency of a heat engine is governed by the Carnot Cycle. For our case, we will operate at 650 K [2] (377C, or 710F), the practical upper bound due to blackbody radiation losses for our
This is the practical upper limit:
For our case, we take Carnot efficiency = (650K - 363K)/650K = 44%
for the case of going from 377C to 90C. The Rankine cycle ratio of .4 is proven for simple steam engines - so our worst case efficiency scenario is 17.6%. If we subtract 10% from this for mechanical
efficiency losses - then we obtain about 16% efficiency. With steam injection electronic valving, we can approach the triple-expansion engine efficiency - .6 Rankine cycle ratio - or steam engines
with 26.4% thermal efficiency. Cylinder insulation should cover cylinder heat loss, and perfect control of steam injection should allow for near-complete extraction of usable energy from steam.
As the last step, we have a 95% efficienct generator, for a total of about 15% efficiency for the steam - electrical cycle.
[edit] Overall Efficiency of Solar Generator
We have shown in Solar_Collector_Losses_Summary that the solar collector/boiler efficiency is 47%. We multiply this by 16% efficiency of the steam engine cycle - and obtain an overall system
efficiency of 7.5%. This will give us 3.6 kW of power as milestone that reaches the $1/watt materials cost prediction.
[edit] Comparison to Others
• The Solar One power tower showed an overall 14% efficiency.
• The Nevada Solar One parabolic trough system showed an overall efficiency of 6.5% and a cost of $3.9 per watt.
• Others - here
□ The most relevant for our work is the Shenandoah steam power plant, with cycle efficiency of 17%, or 40% of Carnot efficiency.
• Ausra compact linear Fresnel reflectors showed up to at 69 bar and 285C - peak 39% overall efficiency from sun to electricity. See First Results from Compact Linear Fresnel Reflector Installation
, by D.R. Mills et al - downloadable here
□ Paper shows over 300 kW obtained from about 1350 kW of solar intercept
□ Note: carnot efficiency here = (658K-348K)/658K = 47%. Ausra is operating with near perfect reflection and near zero thermal losses.
□ Ausra results are spectacular, and Ausra made a claim that it could power the entire USA
[edit] Relevance to Factor e Farm Prototype
By utilizing the same strategy of compact linear fresnel reflectors and similar operating temperatures, we are aiming to achieve an efficiency about 1/5 that of Ausra, while still making the
economics work out.
Power towers are likely to be more expensive because each heliostat must be supported and controlled individually. Solar troughs don't produce sufficient concentration to gain the advantages of CLFR
Main > Energy > Solar Power > Solar Turbine
Main > Energy > Steam Engine | {"url":"http://opensourceecology.org/wiki/Steam_Engine_Efficiency","timestamp":"2014-04-16T21:54:17Z","content_type":null,"content_length":"26631","record_id":"<urn:uuid:4298304f-33e5-4dbf-9466-a37f7cfdf822>","cc-path":"CC-MAIN-2014-15/segments/1397609525991.2/warc/CC-MAIN-20140416005205-00601-ip-10-147-4-33.ec2.internal.warc.gz"} |
General information - Bending of statically indeterminate (continuous) beams
Note that in the above formulas only those members that correspond to the load components located to the left side of the considered cross section with coordinate x should be taken into account:
values of a[i], b[i], c[i], d[i], l[i] must be less than x. Formulas can be simplified in the case of uniformly distributed load discarding members with factors (q[Ri]- q[Li]), which are equal to 0. | {"url":"http://soft4structures.com/Beam/st_general_info.html","timestamp":"2014-04-20T21:22:47Z","content_type":null,"content_length":"11241","record_id":"<urn:uuid:8136bcf8-e22b-417f-ac5f-cc5d7d2c749f>","cc-path":"CC-MAIN-2014-15/segments/1397609539230.18/warc/CC-MAIN-20140416005219-00339-ip-10-147-4-33.ec2.internal.warc.gz"} |
Re: 4th "Atelier PARI/GP" (january 2014)
My number one request is a forfactored() function which uses a sieve to produce factored values efficiently. I plan to write code for this, actually -- even poorly-optimized code will beat 'factoring
each individual integer' by miles.
Along those lines, support for arithmetic functions on factored numbers would be nice. One can always do
prod(i=1, #f~, (f[i,1]-1)*f[i,1]^(f[i,2]-1))
but it would be much nicer to write eulerphi(f).
Another improvement that would be nice: a combinatorial algorithm for primepi() so that for large inputs the code wouldn't need to count the primes individually. (The analytic methods are too slow to
be practical at this point...)
It seems that the GP syntax would allow ^^ to be added for tetration (repeated exponentiation); if this is interesting I'd be happy to code the implementation for t_INTMODs (the only interesting | {"url":"http://pari.math.u-bordeaux.fr/archives/pari-dev-1308/msg00002.html","timestamp":"2014-04-19T19:37:53Z","content_type":null,"content_length":"8718","record_id":"<urn:uuid:ca4a06eb-aaa8-4777-ba25-70b422ac132f>","cc-path":"CC-MAIN-2014-15/segments/1397609537376.43/warc/CC-MAIN-20140416005217-00270-ip-10-147-4-33.ec2.internal.warc.gz"} |
Want to work at Google? Answer these questions (Wired UK)
Want to work at Google? Answer these questions
11. How much would you charge to wash all the windows in Seattle?
The first step here is to estimate the population of Seattle. The U.S. Census recently put it at 594,000 (city limits) or 3.26 million (metro area). In a job interview, no one would fault you for
saying Seattle has about a million people.
How many windows are there per Seattle resident? In Manhattan, young people count themselves lucky to have one window. Seattle is different; apartments are larger, and more people live in houses with
panoramic windows overlooking evergreen forests. Many houses and townhouses are two story. A decent guess, favoring the convenient round number, is ten residential windows per Seattleite.
There are also windows at places of work, Starbucks, department stores, airports, concert halls, and so forth. This probably doesn't add all that much to the per capita total. The average cubicle has
no windows. A big-box store has little surface area (and few windows) relative to its volume. The windows in public spaces like restaurants and airports are shared among the huge mass of people using
Don't forget windows in cars. (You might ask the interviewer whether to count them.) A car is going to have four windows at bare minimum, often twice that. But big 4x4s are driven by big families and
don't add many windows per capita.
A reasonable guess is that windows outside the home account for another ten per person. This comes to twenty windows per Seattle resident. Assuming a population of a million, there are about twenty
million windows to be cleaned.
How much should you charge to wash a window? With the windows in your home, it takes a few spritzes of Windex, a few paper towels, and a few seconds. Some of the windows in Seattle are huge, like
those in the restaurant atop the Space Needle, and they're high up, requiring special crews, special equipment, high workers' comp rates, and considerable guts.
Someone knowing what he's doing could probably clean one side of a typical window a minute, given that most are small. That means cleaning "a window" (both sides) in two minutes. That comes to thirty
windows an hour.
Say the average window washer makes $10 an hour. Throw in another $5 an hour for supplies and insurance. That's $15 for an hour's work, which cleans thirty windows. Cost per window: 50 cents. Twenty
million windows times 50 cents is $10 million. This question is used at Amazon and Google. Just so you don't miss the joke, such as it is, Windows is another company's registered trademark.
12. How much toilet paper would it take to cover the entire state?
A square of toilet paper is about 4 X 4 inches. Nine squares, in a 3 X 3 grid, would then make a square foot. Let's call that "about 10" squares to a square foot. A roll of toilet paper has maybe 300
squares. Then a roll is about 30 square feet.
A mile is about 5,000 feet. A square mile is therefore 5,000 X 5,000 or 25 million square feet. The number of rolls of toilet paper needed to cover a square mile would be 25 million divided by 30.
For Fermi question purposes, 25 is practically the same as 30. Call it a million rolls of toilet paper to the square mile.
Greater London is more or less the area contained by the M25 motorway, which is almost circular. It takes about two hours to go round it at 50mph. This makes its circumference 100 miles. As a
circle's diameter is its circumference divided by pi (roughly 3), the M25's diameter is 100 ÷ 3, which is about 30 miles. So, its radius is 15 miles. A circle's area is given by πr2, which means the
area contained is approximately 3 X 15² = 675 square miles. To cover it, you'd need 675 X 1 million rolls. That's 675 million rolls.
13. How do you find the closest pair of stars in the sky?
This is the "closest pair" problem, well known to computer scientists. The human eye can often spot the closest pair of stars (or points in a plane) at a glance, just as the eye can tell whether a
face is male or female or solve a Captcha. Microprocessors can't readily do any of these things. Algorithms for solving the closest-pair problem were intensively studied in the 1970s and are now a
building block of the digital world. Even your smartphone has apps that use them, for purposes ranging from mapping to games to cameras. It's part of how our clever devices are starting to "see."
This means that the interviewer is really asking a technically trained candidate, "Were you paying attention in algorithm class?" More specifically: given the co-ordinates of a large number of random
points in a plane, how can a computer, with no visual cortex, tell which pair is closest by pure computation?
One inefficient approach is to compute the distance between every pair of stars. That's a lot of calculations when N is a large number. But it's not actually necessary to check the distance between
every pair of stars, only those that are "reasonably close." Unfortunately, computers are stupid. They can't tell which pairs are "reasonably close" unless they do the maths.
The optimal closest-pair algorithm goes like this. Mentally split the sky in two. There's a right half and a left half, each with N/2 stars. Keep partitioning the sky into quarters, eighths,
sixteenths, thirty-seconds, and so forth. The cuts are all to be made vertically (or from celestial north to south).
A suburbanite will see maybe a thousand stars in his hazy, light-polluted sky. It therefore takes about ten halvings to end up with strips of sky so thin that they contain about two naked-eye stars
each (210 is 1,024).
The diagram on page 250 gives you the basic idea. Compute the distance between each strip's star pair. That's vastly less work than computing distances between all the stars.
A complete doofus might think that you're almost done. Find the strip with the closest pair of all, and that's it! Hardly -- look at the diagram. Because the strips are so long and narrow, the
closest pair within a strip may not be so close at all. The closest pair in the whole sky is likely to be two stars straddling two strips. If you look below, I've circled a straddling pair in the
It's necessary to have an algorithm for knitting together two adjacent slices. You give the algorithm the closest pair in the left slice and the closest pair in the right slice, and then it deduces
the closest pair in the combined left- plus-right slice. This algorithm would then be applied, over and over, to create slices twice as wide, four times as wide, eight times, and so forth. After ten
stages of knitting together, we'd be back to a "slice" consisting of the whole sky.
We'd also know the closest pair in the whole sky. The knitting-together algorithm works like this. Given two adjacent slices, inspect a zone centered on the dividing line to see whether it's got a
pair closer than any in either half. If it does, then that will be the closest pair in the combined slice.
Take the closest pair wholly within the right or left slice and call the distance between its stars d. Think of d as the distance to beat. We need to search within distance d of the dividing line for
possible straddling pairs.
The zone is a 2d- wide strip. This can be efficiently done. You might be thinking, "Sure, whatever." Computer scientists have a way of worrying about worst-case scenarios (which the user experiences
as bugs). The stars in the zone can be readily sorted by their vertical coordinates. For each such star, it's necessary to check distances to stars within d of it, up or down. A simple diagram
(facing page) proves that there can be only six stars to check at most. That's nicely manageable.
Here we want to make sure that Betelgeuse, just to the left of the dividing line, doesn't form a close line-straddling pair with a star on the right. This defines a d X 2d box to check. Since we
already know that no two stars on the right are within d of each other, there can be only six stars, at the very most, that fit within the box.
If you haven't studied computer science, your head is probably spinning. For coders, this question should be easier than the brainteasers -- it simply asks them to recite what they learned at
There is an offbeat answer that rates mention. Astronomy buffs may correctly point out that there are two kinds of close star pairs. They are binary stars (two stars orbiting around each other, the
way the earth orbits the sun) and optical doubles (unrelated stars that just happen to look close in the sky, as seen from the earth). Given the profound emptiness of space, it's all but a certainty
that the closest pair of stars in the earth's sky will be a binary.
Not only that, but there are eclipsing binaries, where the orbital plane is so close to our line of sight that the two stars actually pass behind and in front of each other. The most famous example
is Algol, a visible star in the constellation Perseus. Algol looks like a single star to the eye and to the most powerful telescopes. Every three days, its brightness varies. Astronomers have
determined that Algol is really two stars whose orbits look like this, to scale:
When one star partly blocks the other, Algol looks dimmer, even to the unaided eye. During an eclipse, there is no distance between the two stars' disks. They actually touch and overlap in the sky.
That's as close as a pair of stars can get.
How do you find the closest pair? The astronomer's answer is that eclipsing binaries are identified by their regularly varying brightness and spectroscopic signatures.
14. Can you swim faster through water or syrup?
Isaac Newton and Christiaan Huygens debated this question in the 1600s, without resolving it. Three centuries later, two University of Minnesota chemists, Brian Gettelfinger and Edward Cussler, did a
syrup-versus-water experiment. Maybe it's not so surprising that it took so long. Cussler said he had to obtain twenty-two approvals, including permission to pour massive quantities of syrup down a
drain. He had to say no thanks to enough free corn-syrup to fill twenty lorries because it was judged hazardous to the Minneapolis sewer system. Instead he used guar gum, an edible thickener used in
ice cream, shampoo, and salad dressing. About six hundred pounds of the stuff transformed a swimming pool into something that "looked like snot."
Gettelfinger, an Olympic- hopeful swimmer, had the unique experience of diving in. As to the outcome, I will leave you in suspense for a bit longer. The findings were published in a 2004 article in
the American Institute of Chemical Engineers Journal. The next year, Gettelfinger and Cussler won the 2005 Ig Nobel Prize in Chemistry. The Ig Nobels are the silly counterpart to the better-known
prizes awarded in Stockholm, and they automatically get news-of- the- weird coverage. The media attention was apparently responsible for the syrup riddle's revival as a uniquely sadistic interview
In the great syrup swim-off, the guargum glop's viscosity was about twice that of water. Its density was virtually the same as that of plain water. That was important because, as swimmers have long
known, people swim faster in denser saltwater. Like a ship, a swimmer's body rides higher in saltwater, encountering less drag. Gettelfinger and other Minnesota students did laps through both water
and "syrup." They tried the standard strokes: backstroke, breaststroke, butterfly, and freestyle. In no case did the speeds vary between fluids by more than a few percentage points. There was no
overall pattern favoring the glop or the water.
This meant that Newton was wrong: he thought syrup's viscosity would slow swimmers. Huygens correctly predicted there would be no difference in speed. Gettelfinger and Cussler's paper supplies the
reason. Think of the way smoke rises from a cigarette. For a few inches, the smoke rises in a sleek vertical column.
Above that, it breaks up into complicated swirls and eddies. The swirls are turbulence. Turbulence is bad for jet planes, speedboats, and anything that wants to cut quickly through a fluid. Because
the human body is not optimized for swimming, we create a ridiculous amount of turbulence, which we then fight in our struggle to pull ourselves through the water. The turbulence produces much more
drag than the viscosity does. Relatively speaking, the viscosity hardly matters. As the turbulence is similar in water and syrup, the swimming speed is about the same, too.
The flow of water is much less turbulent for a fish, and even less so for a bacterium -- which would swim slower in syrup.
Is this a fair interview question? Cussler told me that a background in computer science "is probably of no use" in answering the syrup question. But he added, "Anyone with sophomore physics should
be able to answer it." As a physics graduate, I'm willing to bet that's optimistic. At any rate, most of those confronting this question on a job interview won't know much about the physics. Good
responses invoke simple, intuitive analogies to explain why the answer needs to be determined by experiment. Here are four points:
1. You can't swim in the La Brea Tar Pits. Some fluids are too thick to swim in. Ask the mastodons in the tar pits. Imagine trying to swim in liquid cement or quicksand. Clearly, swimming is slower
in very thick fluids than in water. Even though there's more to push against, you swim slower, or not at all.
2. "Syrup" covers a lot of territory. The question doesn't ask about pitch or quicksand but rather "syrup." There is maple syrup, cough syrup, chocolate syrup, high-fructose corn syrup, and the kind
they sip in nightclubs, with consistencies ranging from watery thin to the crud left in the bottom of an old tin of golden syrup. The question is unanswerable unless you know the exact type of syrup
-- or can prove that swimming is slower in every thicker-than-water fluid.
3. Cite Darwin. Suppose there's an optimum level of viscosity in which swimming speed is at a maximum. Is there any reason to believe that H20 just happens to be at that optimum level for swimming?
You might say yes, were you an insightful fish. Evolution has streamlined fish to "fit" the way water flows over their sleek bodies. Humans are not very fishlike, and the way we swim is not much like
the way a fish swims. Neither people nor our immediate ancestors have spent much gene-pool quality time in the pool -- or in rivers, lakes, and oceans. Swimming is something we do, like hang gliding,
but we're not built for it. A creature built for doing the Australian crawl would look very different from a human. Edward Cussler noted, "The best swimmer should have the body of a snake and the
arms of a gorilla." It might not be surprising to find that people could swim faster in something with a different viscosity than water. Nor would it be surprising to find that the speed was the same
over a wide range of viscosities.
4. Swimming is chaos. The dynamics of liquids and gases is a textbook example of chaos. It is so dependent on the granular details as to defy prediction. That's why aircraft designers need wind
tunnels to test their designs. The unstreamlined human body, with its relatively awkward movements in water, only complicates things further. This is a question that needs to be tested with an
experiment, using the exact type of "syrup" in question.
Cussler's Ig Nobel Prize acceptance speech boiled all this down to six words: "The reasons for this are complicated."
15. It is difficult to remember what you read, especially after many years. How would you address this?
In a few years, practically all reading will be on digital screens of some kind. It's possible to envision a personal reading agent keeping track of everything you've read, on any device -- from
e-mails and tweets to electronic books and magazines. This agent or its data would migrate from hardware device to device and follow you throughout life.
• The agent would index all that text so that you'd be able to do a keyword search.
• The agent would let you annotate things you read anywhere (as you can with e-book readers). The simplest annotation would be a highlight, essentially meaning "I may want to remember this later!"
The highlighted passages should get extra weight in searches. You should be able to search the text of annotations, too.
• This agent could become a built-in function of Google search or its future counterparts. While Google "remembers" that newspaper article you read last October, and (via Google Books) everything in
many books, the profusion of hits in a given search can be overwhelming. If Google (via this agent) kept track of what you read, on any and all devices, you could filter searches by "stuff I've
• You can search for something you remember you've forgotten, but not for something you've completely forgotten or aren't thinking about. Maybe you read the complete short stories of Nikolay Gogol at
school and loved them but hardly remember a character, plot, or phrase now. The agent might address that by periodically reminding you of highlighted passages, those you've identified as worth the
attention of your future self. Perhaps the agent would have a Twitter account and tweet random snippets of material you want to remember but don't otherwise have time to go back to.
• The agent would also "remember" podcasts, movies, and TV shows -- provided transcripts or subtitles exist or could be generated.
16. You're in a car with a helium balloon tied to the floor. The windows are closed. When you step on the gas pedal, what happens to the balloon? Does it move forward, backwards, or stay put?
(generic the company question, including Google)
The near-universal intuition is that the balloon leans backward as you accelerate. Well, the intuition is wrong. Your job is to deduce how the balloon does move and to explain it to the interviewer.
One good response is to draw an analogy to a spirit level. For the not so handy, a spirit level is the little gizmo carpenters use to make sure a surface is horizontal. It contains a narrow glass
tube of colored liquid with a bubble in it. Whenever the spirit level rests on a perfectly horizontal surface, the bubble hovers in the middle of the tube. When the surface isn't so level, the bubble
migrates to the higher end of the tube. The takeaway here is that the bubble is simply a "hole" in the liquid. When the surface isn't level, gravity pulls the liquid toward the lower end. This pushes
the bubble wherever the liquid isn't -- toward the opposite end.
Untie the helium balloon and let it hit the sunroof. It becomes a spirit level. The balloon is a "bubble" of lower-density helium in higher-density air, all sealed in a container (the car). Gravity
pulls the heavy air downward, forcing the light balloon against the sunroof.
When the car accelerates, the air is pushed backward, just as your body is. This sends a lighter-than-air balloon forward. When the car brakes suddenly, the air piles up in front of the windshield.
This sends the balloon backward. Centrifugal force pushes the air away from the turn and sends the balloon toward the center of the turn. Of course, the same applies when the balloon is tied to
something; it's just less free to move. The short answer to this question is that the balloon nods in the direction of any acceleration.
Don't believe it? Put the book down right now. Go to the supermarket, buy a helium balloon, and tie the string to the gear stick or hand brake. Drive back home (no lead-footing necessary). You'll be
astonished. The balloon does exactly the opposite of what you'd expect. When you step on the accelerator, it bobs forward, like it's trying to race the car to the next light. Brake hard enough to
throw the kids' toys out of the backseat, and the balloon pulls backward. In a high-speed turn, as your body leans outward, the crazy balloon veers inward. It's so freaky that there are videos of
this on YouTube.
Why are our intuitions right about spirit levels and wrong about helium balloons? In a spirit level, the heavy liquid is dyed a fluorescent sports-drink hue, while the bubble is a ghostly void. We
associate color with density, and transparency with nothingness. That instinct is completely wrong with balloons. Air is invisible, and we ignore it 99+ per cent of the time. The balloon, on the
other hand, is dressed up in pretty colors or Mylar and screams, "Look at me!" We forget that, masswise, it's a partial vacuum within the surrounding air. A helium balloon does the opposite of what a
mass does because it's a deficit of mass. The real mass -- the air -- is invisible.
Interviewers who ask this question don't expect you to know much physics. But there is an alternate response that makes use of the theory of relativity. Seriously.
It relates to Albert Einstein's famous thought experiment about the lift. Imagine you're in a lift going to your tax accountant's office, and a mischievous extraterrestrial decides it would be fun to
teleport you and the lift into intergalactic space. The lift is sealed, so there's enough air inside to keep you alive long enough to amuse the alien for a few minutes. There are no windows, so you
can't look out and see where you are. The alien puts the lift in a tractor beam and tows it at a constant acceleration exactly matching that of Earth's gravity. Is there anything you could do in the
sealed lift to determine whether you're experiencing real Earth gravity or "fake" gravity, mimicked by acceleration?
Einstein said no. Should you take your keys out of your pocket and drop them, they would accelerate toward the lift floor exactly as they would on Earth. Let go of a helium balloon's string, and it
would float upward, just as on Earth. Things would appear perfectly normal. The Einstein equivalence principle says that there is no (simple) physics experiment that can distinguish between gravity
and acceleration. This assumption is the foundation of Einstein's theory of gravity, known as general relativity. Physicists have been trying to punch holes in the equivalence principle for nearly a
century now. They haven't been able to. It's safe to assume that Einstein's premise is right, at least for any experiments you can do in a car with a fifty-pee balloon.
All right, here's a physics experiment. Tie the string of a plumb bob (carpenter's weight) to your right index finger. Tie a helium balloon to the same finger. Note the angle between the two strings.
In a lift, a parked car, or a cruising jetliner, the outcome will be the same. The plumb bob points straight down. The balloon points straight up. The two strings, joined at your finger, form a
straight line. This is the outcome whenever you are subjected to gravity. Now picture what happens when you start driving. As you accelerate, your body sinks back in the seat. Fallible intuition may
tell you that the plumb bob and the balloon will each lean backward a bit from your finger. During acceleration, there will be an angle between the two strings (if this intuition is right). This
would provide a way of distinguishing between gravity and acceleration. When the car is subject to gravity alone, the two strings form a straight line. But when it's subject to centrifugal force or
other forms of acceleration, the strings form an angle with your finger as the vertex. That's all you need to prove that general relativity is wrong. Forget about getting a job at Google -- this
would be worthy of a Nobel Prize.
But since the equivalence principle has been rigorously tested and shown to be true, this doesn't happen, and you can use it to answer this question. Physics must be the same in an accelerating car
as in a car subject to gravity alone. In both cases, the balloon, your finger, and the plumb bob form a straight line. In answer to the question, then, the helium balloon does the exact opposite of
what you'd expect of an object with mass. It goes forward rather than backward… left rather than right… and, of course, up rather than down.
17. You have a choice of two wagers: One, you're given a basketball and have one chance to sink it for £1,000. Two, you have to make two out of three shots, for the same £1,000. Which do you prefer?
Call the probability of making a basket p. With the first bet, you have a p chance of winning £1,000. Otherwise, you get nothing. On the average, you can expect to win £1,000 X p.
With the second wager, you shoot three times and have to make the basket twice to be in the money. The chance of making the basket on any given attempt is still p. Your chance of missing on any
attempt is 1 - p.
There are 23, or 8, scenarios for the second wager. Let's list them. A check mark means you make the shot; a blank means you miss.
The first scenario is the one where you're really off your game. You miss all three shots. The chance of that is 1 - p, multiplied by itself three times. You don't get the money.
In four of the eight scenarios, you win the money. In three of them you miss once. These scenarios have the probability of p2(1 - p). In the case where you make all three shots, the probability is
p3. Add all of them up. Three times p2(1 - p) comes to 3p2 - 3p3. Add p3 to that, to get 3p2 - 2p3. The expectation is £1,00 x (3p2 - 3p3).
So which wager is best?
First wager's expectation: £1,000 × p
Second wager's expectation: £1,000x (3p2 - 2p3)
You may be a complete klutz (p is roughly 0) or an NBA baller (p approaches 1). For reference, I've done what you can't do in the interview: plugged the formulas into a worksheet and made a chart.
The chart shows how expected winnings vary with p.
The straight diagonal line represents the first bet, and the more S-shaped curve is the second. The first bet is better if your chance of making the shot is less than 50 percent. Otherwise, you're
better off picking the second wager.
This makes sense. A poor player cannot expect to win either wager. He must pin his hopes on a freak lucky shot, which is obviously more likely to happen once than twice ("lightning never strikes
twice"). The bad player is better off with wager 1.
The very good player ought to win either wager, though there is a small chance he'll miss the money shot. Two out of three is a better gauge of his talents, and that's what he wants. It's like the
legal maxim: if you're guilty, you want a jury trial (because anything can happen); if you're innocent, you want a judge.
Assuming you get this far, the interviewer's follow-up question is, "What value of p makes you switch bets?" To answer that, set the probability of winning the two bets equal. This represents the
skill level where it's a toss-up which you pick.
p = 3p2 - 2p3
Divide by p:
1 = 3p - 2p2
and then get
2p2 - 3p + 1 = 0
From there you can use the quadratic formula, warming the heart of your old algebra teacher. The interviewer will be looking for brio as much as book learning. You know p, a probability, has to be
between 0 and 1. It's better style to experimentally try a rea-sonable value. "Okay, I need a number between 0 and 1. Let's try 0.5." Works like a charm.
18. You have N companies and want to merge them into one big company. How many different ways are there to do it?
In the proper sense of "merger," two companies surrender their identities and fuse into a brand-new entity. The pharmaceutical giants Glaxo Wellcome and SmithKline Beecham merged in 2000 to form the
pharma colossus GlaxoSmithKline. (You guessed it -- both parent companies were themselves merger spawn.)
CEO egos being what they are, true mergers are fairly uncommon. Mergers require a near-exact match of bargaining power. More typically, one company's management has the stronger hand, and it's not
about to let the weaker company's leaders forget it. The deal is likely to be an acquisition, in which company A gulps up company B, and B ceases to exist as a separate entity (though it often
survives as a brand). An example is Google's 2006 acquisition of YouTube.
Mergers are symmetrical. There is only one way for two companies to merge as equals. Acquisitions are asymmetrical. There are two ways for two companies to acquire or be acquired -- Google buying
YouTube is not the same as YouTube buying Google. Most people outside investment banking gloss over the distinction between mergers and acquisitions. Any melding of corporations is loosely called a
"merger." The point is, you need to ask the interviewer what is meant by "merge." Fortunately, most of the reasoning applies whatever the interviewer's answer.
Start with acquisitions because they're more common (and a little easier to work with). You can visualize the companies as draughts, and the acquisitions as the moves in the game. Start with N
pieces. A move consists of putting one piece on top of another to signify that the top piece is "acquiring" the bottom piece. After an acquisition, you manipulate the pieces involved as if they were
glued together (like a "kinged" piece in the regular game).
Every move diminishes the number of pieces (or stacks) by one. Eventually, you will be placing stacks of pieces on top of other stacks to create yet- taller stacks. It will take exactly N - 1 moves
to achieve the game's goal, a single tall stack consisting of all N pieces combined into one. How many different scenarios can lead to that outcome?
The simplest case involves two companies. Company A can gulp up B, or B can gulp up A. That's two possible scenarios. With three companies, you have to decide which company first acquires what other
company. There are six possibilities for that first acquisition, corresponding to the six possible ordered pairs of three items (AB, AC, BA, BC, CA, and CB). After the initial acquisition, you're
left with two companies. The situation is then exactly as in the paragraph above. The number of possible acquisition histories for three companies is therefore 6 x 2 = 12.
With four companies, there are twelve possibilities for the first acquisition: AB, AC, AD, BA, BC, BD, CA, CB, CD, DA, DB, and DC. That decided, you have three companies and, as we already know,
twelve histories. There must be 12 x 6 x 2, or 144, acquisition histories for four companies. Let's generalize. With N companies, the number of possible initial acquisitions is
N(N - 1) This just means that any of the N companies can be the first acquirer, and any of the remaining N - 1 companies can be the first acquiree. After the first acquisition, there will remain N -
1 distinct companies, and there will be (N - 1)(N - 2) possibilities for the second acquisition. Then there will be (N - 2) companies and (N - 2)(N - 3) possible acquisitions. We're going to be
multiplying the ever- decreasing numbers of possible acquisitions until we're left with 2 x 1 possibilities for the final acquisition.
It's easy to see that, using factorial notation, the product will come to N! x (N - 1)! acquisition histories. What if you want true mergers instead of acquisitions? The above analysis overcounts the
possibilities by a factor of two -- for each of the N - 1 mergers. That means the number of proper merger histories is N! x (N - 1)! divided by 2N - 1.Finally, if "merger" can mean merger or
acquisition, you simply add the two answers.
19. You've got an analogue watch with a second hand. How many times a day do all three of the watch's hands overlap?
This is an update to a classic Microsoft interview question that asks how many times a day a clock's hour and minute hands overlap. Because that one's become pretty well known, interviewers have
started using this variant.
The Microsoft answer: First, figure when the hour and minute hands overlap. Everyone knows the minute and hour hands overlap at 12:00 midnight and at approximately 1:05, 2:10, 3:15, and so forth.
There's an overlap in every hour except 11:00 to 12:00. At 11:00, the faster minute hand is at 12 and the slower hour hand is on 11. They won't meet until 12:00 noon -- ergo, there is no overlap in
the 11:00 hour.
There are thus eleven overlaps in each 12-hour period. They are evenly spaced in time (since both hands travel at constant speeds). That means that the interval between overlaps of hour and minute
hands is 12/11 of an hour. This comes to 1 hour, 5 minutes, and 27 3/11 seconds. The eleven alignments of minute and hour hands in each 12-hour cycle take place at:
1:05:27 3/11
2:10:54 6/11
3 :16:21 9/11
4:21:49 1/11
5:27:14 4/11
6:32:43 7/11
7: 38:10 10/11
8:43:38 2/11
9:49:05 5/11
10:54:32 8/11
How can we determine if any of these times is a three-way overlap? Though the question is about an analog clock, think of a digital clock that gives time in hours, minutes, and seconds:
There is an overlap between minute and second hands only when the minutes figure (00 here) equals the seconds figure (00). There is a precise three-way overlap at 12:00:00. In general, the minute-
and second-hand overlap will occur at a fractional second. For example, here,
the second hand would be at 37 past the minute, while the minute hand would be between 37 and 38 past the hour. The instant of overlap would come a split- second later. But the hour hand wouldn't be
near the others, so this isn't a three-way overlap.
None of the hour- and minute-hand overlaps in the list above passes this test except for 12:00:00. That means all three hands align just twice every day, at midnight and noon.
The Google answer: The second hand is intended for timing short intervals, not for telling time with split-second accuracy. It's normally not in sync with the other two hands. "In sync" would mean
that all three hands point to 12 at the stroke of midnight and noon. Most analog watches and clocks do not let you set the second hand from the stem. (I've never seen one that does.) A workaround
would be to take the battery out (or let a windup watch run down), set the minute and hour hands in sync with where the second hand stopped, and wait until it becomes the time shown to replace the
battery or wind up the watch. It would take a maniacal analogue watch fetishist to do that. But unless you do this, the second hand will not show the "real" time. It will be offset from the accurate
seconds by a random interval of up to 60 seconds. Given a random offset, the odds are overwhelming that the three hands would never align precisely.
20. You work in a 100-storey building and are given two identical eggs. You have to determine the highest floor from which an egg can be dropped without breaking. You are allowed to break both eggs
in the process. How many drops would it take you to do it?
Lest there be any confusion, the building and eggs are strictly imaginary. This is an algorithm question. It's a test of your ability to craft a smart, practical way of doing something. That's
important in engineering, in management, and in everything else. Every cook knows that a raw egg dropped from counter height onto tile is history. But if Google's 100-story building is surrounded
with something softer than concrete and harder than grass, the answer is not obvious.
To answer in the spirit intended, you have to assume that it is possible that the maximum egg-safe floor could be any floor at all, from 1 to 100. In fact, you should allow for the possibility that
no floor is egg-safe (seconding the wisdom of cooks). You are permitted to ignore the strong element of chance in egg defenestration (demonstrated in the 1970 experiments). Pretend that the outcome
of dropping an egg from a given floor will always be the same. Either it breaks or it doesn't. The interviewer is not expecting you to name the egg-safe floor. There are no eggs and there is no
100-story building: it's all a fictional situation, okay? You are asked only to devise an efficient method for determining the floor, while explaining your thought process.
The one question you are expected to answer with an actual number is, how many drops would you need? Scoring is like golf: the fewer, the better.
Bits and Eggs Dropping an egg is a simple experiment that yields one bit of information. To get the most bang for the bit, you'd start in the middle of the building. That would be floor 50 or 51, as
there's no "middle" floor in a building with an even number of floors.
Try dropping an egg from the 50th floor. Let's say it breaks. This would mean that the desired egg-safe floor must be below the 50th floor. Split the difference again by dropping the next egg from
the 25th floor. Oops! It broke again. Now you're out of eggs. You can now conclude that the highest egg-safe floor is below the 25th. You don't know which floor, and that means that the method has
failed. It's possible to keep reusing an egg so long as it doesn't break.
Start on the bottom floor and drop the first egg. If it survives, go to the 2nd floor and try that. Then the 3rd, the 4th, the 5th, and so on until the egg breaks. That will tell the highest floor
the egg can be dropped from without breaking. You will have determined it with just one egg. Call this the "slow algorithm." It's a miser of eggs and a spendthrift of egg drops. Every single floor
might have to be tested, but it gets the job done.
The challenge is to create a solution that makes good use of both eggs. Suppose that Google's optimal algorithm for the egg-drop experiment was written in a book somewhere. You don't even have to
suppose -- it is written in a book, namely this one. The algorithm reads something like this:
1. Go to floor N and drop the first egg.
How do I know the algorithm begins this way? Well, I'm not going out on much of a limb. An algorithm is a list of idiotproof instructions, starting with instruction 1. It tells you to drop an egg,
naturally, because that's the modus operandi here. There's nothing else to do but drop eggs. The only interesting part (floor N) is currently concealed under an algebraic veil. The real algorithm
gives a specific floor, like 43, in place of the N.
Further deduction: because the experiment in instruction 1 has two outcomes, it has to have follow-up instructions for both eventualities. Call them 2a (what to do if the egg breaks) and 2b (what to
do if it survives). Once you break the first egg, you will have to play it safe with the second. You can't take the risk of skipping any floor, lest you break the second egg and not be able to deduce
the correct floor.
Instruction 2a of Google's algorithm has to say this, in so many words:
2a. (To be used if the egg breaks in 1.) Go down to floor 1.
Adopt the "slow algorithm" with the remaining egg. Test it from every floor, working your way up the building until the egg breaks. The maximum egg-safe floor is the one below that. For instance,
imagine the first drop is from the 50th floor, and the egg breaks. You can't risk trying the second egg from floor 25 because it might break, too. Instead, you have to try floors 1, 2, 3 . . .
potentially all the way up to 49. And since we started with the 50th, that could mean making fifty drops in all. It doesn't take much coder's intuition to see that a search needing fifty tests to
find one thing out of a hundred is not optimal. It's lousy. It's better to count on making the first drop from a lower floor. If we start from floor 10, and the egg breaks, we might need as many as
ten drops. This is the key aha! moment of the puzzle. In general, when the egg breaks on a first drop from floor N, it will take as many as N drops total to identify the right floor. This strongly
suggests that the first drop should be from a floor much lower than 50. This choice has the appealing feature that we use the first egg to deduce the tens digit of the egg-safe floor, and the second
egg to find the ones digit. For example, test the egg from floors 10, 20, 30, 40, and 50. Say it breaks when dropped from floor 60. This tells us the maximum egg-safe floor is 50-something. Move down
to floor 51 and work upward floor-by-floor with the other egg. If the second egg breaks at floor 58, that means the egg-safe floor is 57.How efficient is this procedure? The worst-case scenario would
be to try 10, 20, 30, all the way up to 100, where the egg finally breaks. Then you'd backtrack to 91 and work up. You could end up needing nineteen drops in all to determine that 99 is the correct
floor. This isn't a bad approach. But it's not the best.
Crash Test Remember, the question asks, "How many drops would it take you to do it?" That's a strong hint that you're being graded on how few drops you need. (More exactly, you're to minimize the
number of drops required in a worst- case scenario. Sometimes you'll get lucky and arrive at the answer in just a few drops.)
Because the first egg's role is as crash-test dummy, you want to put it in high-risk situations; that's how you learn as much as possible as quickly as possible. The second egg is a backup. Once it's
the sole remaining egg, you have to play safe with it. It's the crash-test-dummy egg that's crucial to achieving a good solution. It's the egg that can eliminate many floors in a single drop. The
question is, how many? A bit of mental gymnastics is required to answer that. It throws many smart people.
I'll begin with an analogy. You're a professional golfer on the eighteenth hole, vying for a big prize. To win, you've got to make the hole within three strokes. That necessity dictates which clubs
you choose and whether you risk a bunker rather than playing it safe. It will require you to aim for the hole on the third stroke (rather than being satisfied with making the green). The three-stroke
limit constrains your strategy throughout. Google's perfect algorithm also has a limit: a maximum number of drops needed to determine the correct floor. Call this number D. The D-drop limit
constrains your strategy. For the sake of concreteness, imagine the limit is ten drops. Then you might as well drop the first egg from the 10th floor. See why? You want to choose a floor as high as
possible, to rule out as many floors as possible. The 10th floor is the highest option, for this reason: should the first egg break, you could end up needing all ten permitted drops to determine the
correct floor. (The above paragraph is the hardest thing to understand, I promise.)
Everything else follows from this insight. After the initial drop, you've got nine drops left. Assuming the egg survives, you will again want to move up as many floors as possible for the second
drop. You might think you'd move up another 10 floors. Not quite. Since you've only got nine drops left, you can move up 9 floors at most. That's because the egg could break on the second drop,
forc-ing you to use a floor-by-floor search thereafter. You might have to test every floor between 10 and the one you just tried, 19, using up all your allotted drops. Had you gone up even a floor
higher, you could get caught short and not be able to single out the correct floor.Or say the egg survives the first two drops. That leaves eight drops. You would go up 8 floors for the next drop.The
floors you test, assuming a series of unbroken eggs, form a simple series.
10 + 9 =19
10 + 9 + 8 = 27
10+ 9 +8 +7 = 34
Wait a minute. The highest floor you can possibly reach is 10 + 9 + 8 + 7 + 6 + 5 + 4 + 3 + 2 + 1. That comes to 55. This scheme would work perfectly for a 55-story building. But the question says
it's a 100- story building.
That's easily fixed. Remember, I just picked 10 out of the air. Replace 10 with D, the required number of drops in the best algorithm. The highest floor an optimal method can reach is
D + (D - 1) + (D - 2) + (D - 3)+ . . . + 3 + 2 + 1
This must equal or exceed 100.
From here on it's just algebra. The sum above is D plus every whole number smaller than it. This is a triangular number. Picture a rack of billiard balls. It's 5 + 4 + 3 + 2 + 1 balls. You may recall
from school that the total can be computed by multiplying 5 by 5 + 1 and dividing by 2. So you've got 5 x 6/2 = 15, and that's the number of balls in a rack.
In this case, the sum of D and every smaller whole number is equal to D times (D + 1) divided by 2. So:
D x (D + 1)/2 ≥ 100
Multiply both sides by 2, and you get
D2 + D ≥ 200
Focus on D2 and ignore the much smaller D. This equation says that D2 is at least as big as 200 or so. The square root of 200 is just over 14. Try that for D.
142 + 14 = 196 + 14 = 210 ≥ 200
Bingo. It fits. Just to be sure, try 13.
132 + 13 = 169 +13 =182
Nope, that's not greater than or equal to 200. Fourteen it is. You drop the first egg from floor 14, and you're guaranteed to find the answer in fourteen drops or less. The recap: First, drop the egg
from floor 14. If it breaks, go down to floor 1 and work your way upward floor-by-floor. This will get the answer in no more than fourteen drops total.
Should the first drop not break the egg, go up to floor 27 (14 - 1 floors above floor 14) and try again. Should it break this time, you'll have to go down to 15 and work up. This will also yield the
answer in fourteen drops total or less.
Given a string of unbroken eggs, you'd test floors 39, 50, 60, 69, 77, 84, 90, 95, 99, and then 100 (actually, were the building higher, the next floor would be 102). This means it would take twelve
drops, and no broken eggs, to deduce that the egg can survive every floor of the building. Should the egg break at any point in the process, you'd be shunted into the slow algorithm and might need
all fourteen permitted drops.
21. Add any standard arithmetic signs to this equation to make it true: 3 1 3 6 = 8
Starting from the left, you see 3 and 1. The first maths sign you learned was probably a plus sign. 3 = 1 gives 4. Conveniently, 4 is half of 8, the number that's the goal. Moving righr, there's a 3
and a 6. Well, 3 is half of 6. Put a division sign between them and you've got 3/6 or 1 / 2. Dividing by 1/2 is the same as multiplying by 2. This gives the answer: (3 + 1) / (3 / 6) = 8.
That wasn't so hard, was it?
1. This is really interesting stuff, it reminds me of a piece of information given to me on a walking tour of Oxford; All Souls College is an invitation only graduate college, and is credited as
having the most difficult entrance exam/interview of all the Oxford colleges. Candidates will typically be presented with one word exam questions (amongst other questions), e.g. 'Ethics' and
expected to give a suitable, and no doubt erudite, response.
2. I suspect there aren't wrong answers to many of the questions, they are wanting to find out how your mind works, can it make sensible approximations and can you base robust conclusions based on
those. A certain amount of general knowledge always helps get the right order of magnitude for calculations and self checking.
3. Just a note to say that you don't need to let a mechanical watch wind down to stop it, just pull out the crown.
4. Sorry you've got Question 5 wrong there. The logic to calculate the number of girls to boys born is correct but you've forgotten to add the parents into the equation.
1. In reply to NickyB
A pair of child bearing parents will be a man and a woman for sake of convenience. It can also be noted that a man is not a boy, nor is a woman a girl.
2. In reply to NickyB
1Man+1Women= (>=1child)so if we add parents to the equation, it will remain same, balancing on both of the sides!
3. In reply to NickyB
The question asked about the ratio of girls to boys, not male to female. And the ratio will still be 1:1 even if we count parents.
shashank nagamalla
Dec 5th 2012
5. So now they just reprint the questions from the books about getting a job at Microsoft, and change one word in the title.
6. You might want to check your answer to this question:
Using only a four-minute hourglass and seven-minute hourglass, measure exactly nine minutes.
In your answer you talk about using a 9 minute hourglass... if there was a 9 minute one available... just use that one duuurrr!
7. I see now I made a mistake in previous comment while energy in muscles proportional also to mass not the cross section, the nice explanation of many issues and example of data are in the book
"The conclusion is that similar animals of different masses should jump to the same height provided that their muscles contracts with the same force" (p.177-179)
8. I'm an engineer at Google, and this is wildly inaccurate. We ask questions about programming and algorithms, not questions like this.
1. In reply to Kelly
sir,,how to apply for google.i am soft engg. In india.(just passed)
2. In reply to Kelly
Hii, so how can i work at google as well as i'm an Assistant Lecturer in the faculty of computer science!! thxx
9. I interviewed at Google last year, and in fact have another interview coming up. None of these questions were asked before and I don't anticipate them being asked again. Like Kelly said, it's
about algorithms and programming languages for the engineering job. I don't doubt there are some fringe jobs that might have these questions though.
10. "When the nine-minute glass runs out, flip it over. "
11. Your RSA example is horribly wrong.
First off, you get the decryption operation wrong- it isn't Cd mod N, it's C^d mod N.
Secondly, moduli up to 768 bits have been publicly factored for something like a decade, and a 20 bit modulus could be factored on a desktop machine in no time- on mine, your example factorizes
in less than 3 seconds. This means that from the public key I can obtain your private key- a total break.
Thirdly, even if you correctly chose a large random modulus (>2048 bits) the value you're encrypting is >> c = pow(1234, 3, 5053366937341834823) # signature is pow(message, exponent, modulus)
>>> round(c**(.33333)) # pow(x, 3) and pow(x, 1/3) are inverse
(example is in Python, for the curious).
This leads me to the fourth problem in your example: using textbook RSA. RSA is widely known in the security community to be insecure without use of an appropriate padding method (eg, OAEP),
which could almost certainly not be performed by hand.
Problem number five is that you assume Eve can't modify what's written on the card, or exchange it for another card. If this were the case, she would just hand Bob a card with a public key she
generated on it, let Bob run through the rigamarole, and decrypt it (obtaining your phone number) then re-encrypt it under the key you gave so you wouldn't know it was intercepted.
Finally, the set of 7-digit phone numbers can be trivially enumerated on a computer. All Eve would have to do would be to write a script that performed the same operation Bob did for every
possible phone number and then stop when it matched the one he wrote back, and she would still get your phone number- no trickiness required.
Also, +1 to Kelly's comment.
1. In reply to G
Since part of the example got swallowed by a lame attempt to prevent XSS, let me restate it:Thirdly, even if you correctly chose a large random modulus (>2048 bits) the value you're
encrypting is less than 25 bits. This means that m^3 will be very small relative to the modulus (~75 bits). Because of this, all an attacker needs to do is take the cube root of the
ciphertext (over the integers) to decrypt it.(The remainder of the example stands)
12. I interviewed at Google last year, and in fact have another interview coming up. None of these questions were asked before and I don't anticipate them being asked again. Like Kelly said, it's
about algorithms and programming languages for the engineering job. I don't doubt there are some fringe jobs that might have these questions though.
13. That's not true. I had an interview at Google and the only question they asked me was why I was late.
1. In reply to krystian
why were you late to an interview with google ?????????????????????????????
14. When I interviewed at Google, they asked me how many tennis balls would it take to completely fill a room.
I suck at Math, which resulted in my incoherently blabbering some random formula. Needless to say, I didn't get the job.
1. In reply to Iris
Answer is the number of balls is approx 0.0771 times the room volume.Closest possible packing of spheres is hexagonally close packed with packing density p ~74%.To be exact:p= pi()/root(18) =
number of balls x ball volume /volume of roomequation is: p=N(b)*V(b)/V(r). where ball vol=V(b)=pi()*cube(d)/6 , ball diameter=d & room vol=V(r)=h*l*wMax poss Number of balls is: N(b)=p*V
(r)/V(b)=root(2)*wlh/cube(d) [pi()'s cancel] So:N~1.4142*wlh/cube(d)Average tennis ball diameter is about 2.6375" & its cube is 18.347521484375so N(b)~0.0771*wlh ...QED...[eg 12x12x12
inch box has N~133 from formula which roughly agrees with the Face-centered cubic packed approx of 125=5x5x5 i.e 12/2.6375~4.5498~5]
Aug 12th 2012
15. why cloning minutes with sandglasses.
measure one minute as explained let it run and then just flip the four minute hourglass twice giving 4+4+1=9
1. In reply to x
To get sand worth 1 minute in the bowl, you have to take 7 minutes. As per your method it would take 7+(1+4+4)=16 minutes from starting. That means you have to wait 7 minutes to get sand
worth 1 minute in the bowl and then start counting 9 minutes. By the given method , it would take 7 minutes required to get sand worth 1 minute in the bowl + 1 minute that you cloned twice =
9 minutes exactly from start to measure 9 minutes..
Read more on Google
The search and advertising company behind Gmail, Android and Chrome | {"url":"http://www.wired.co.uk/magazine/archive/2012/05/start/want-to-work-at-google/page/2","timestamp":"2014-04-20T21:40:08Z","content_type":null,"content_length":"167555","record_id":"<urn:uuid:4bc767ce-22a0-4efe-8cbe-810ed91c8b56>","cc-path":"CC-MAIN-2014-15/segments/1398223205375.6/warc/CC-MAIN-20140423032005-00116-ip-10-147-4-33.ec2.internal.warc.gz"} |
MP3 of the Riemann Zeta Function on the Critical Line
MP3 of the Riemann Zeta Function on the Critical Line
The Riemann zeta function, for values of its argument located on the critical line (z = 1/2 + i t for real t), looks so much like a sound wave that I decided I ought to listen to it.
Here is a C++ program I use to compute (approximate) values of the function for positive arguments (the negative side is the same: zeta(a+bi) = zeta(a-bi)) and write them to an existing WAV file. I
use an existing file to avoid having to know the WAV header format. The value of the Z function is a complex number; it writes the imaginary component to the left channel and the real component to
the right.
The actual computation of the zeta function isn't accurate enough to locate zero-crossings, but is adequate for creating a sound wave.
It computes a series sum based on the following rather fundamental formula for the Zeta function:
zeta(z) = 1/(1-2^1-z) SIGMA[[n=1..inf]] -1^n-1 n^-z
I split the odd and even terms to get:
zeta(z) = 1/(1-2^1-z) ( SIGMA[[odd n>0]] n^-z - SIGMA[[even n>0]] n^-z )
and combine pairs of terms to get:
zeta(z) = 1/(1-2^(1-z)) SIGMA[[n=1..inf]] ( (2n - 1)^-z - (2n)^-z )
The primary component of the sound (the loudest tone you hear) is the term
2 cos(theta(t))
in the first term of the infinite series for the Riemann-Siegel Z function. It rises in pitch at the rate
freq ≈ K ln(t)
(which is related to the known result concerning the average spacing of the complex zeros). You can hear this tone "drop out" after about 15 seconds because the Zeta function was computed using 1000
terms of the above series for zeta(z), and the higher-frequency components are the ones that take longest to converge. The next-loudest component drops out some time later. I actually kind of like
this because if the loud components did not drop out, it would be hard to hear the patterns in the other components (which is where most of the structure is).
Riemann-Siegel Formulas
The Riemann-Siegel Z and theta functions define zeta(z) in terms of its argument (theta) and absolute value (Z) for a value of z equal to 1/2 + i t, by the relation:
Z(t) = zeta(1/2 + i t) e^i theta(t)
For real t, Z(t) and theta(t) are both real, and we have:
theta(t) = (0 or pi) - arg (zeta(1/2 + i t))
magnitude( Z(t) ) = magnitude( zeta(1/2 + i t) )
Z and theta are defined in terms of each other, the Gamma function, power series coefficients, and a few other things.
Z(t) and theta(t) should be a really good way to calculate the Zeta function on the critical line. However, I have not been able to find sufficiently accurate descriptions of Z() and theta() to
actually use in a computer program. (If anyone knows of one, please email me!)
If you like this sort of thing, you might also be interested in these other pages having to do with recreational math:
My numbers and large numbers pages, the RIES program (which finds algebraic equations given their solution); Hypercalc (the calculator that doesn't overflow); and my table of Classical Sequences. | {"url":"http://mrob.com/pub/ries/zeta.html","timestamp":"2014-04-20T05:41:38Z","content_type":null,"content_length":"7420","record_id":"<urn:uuid:7b45b28a-d27c-4f17-9567-203151614a95>","cc-path":"CC-MAIN-2014-15/segments/1398223206118.10/warc/CC-MAIN-20140423032006-00060-ip-10-147-4-33.ec2.internal.warc.gz"} |
Conditional probability, deviation from the uniform distribution
up vote 1 down vote favorite
Let $N\in\mathbb{N}$ and $G$ the group $\mathbb{Z}/n\mathbb{Z}$.
Let $q< N$ and:
• $a_1, ..., a_q$ pairwise distinct elements of $G$
• $b_1, ..., b_q$ pairwise distinct elements of $G$
• $x_1, ..., x_q$ pairwise distinct elements of $G$
• $y_1, ..., y_q$ pairwise distinct elements of $G$.
Let $P$ be a random permutation of $G$ verifying $P(a_i)=b_i, \forall i\leq q$.
Let $Q$ be a random permutation of $G$.
Let $k_0, k_1$ be random elements of $G$.
Let $E$ denote the function $E(x)=P(x + k_0)+k_1$
I want to prove that $$\Pr[E(x_q)=y_q|E(x_\ell )=y_\ell, \forall \ell < q]$$ is greater than $$\left(1-\frac{2q}{N}\right)\Pr[Q(x_q)=y_q|Q(x_\ell)=y_\ell, \forall \ell < q]$$
The first probability is over the randomness of $k_0, k_1$ and $P$. The second probability is over the randomness of $Q$.
If we hadn't equations $P(a_i)=b_i$ on $P$, then the two probabilites would be equal. So this result would show that, adding conditions to $P$ doesn't change too much the distribution of $(E(x_1),
..., E(x_q))$, it should still be close of the uniform distribution.
Thank you
add comment
Know someone who can answer? Share a link to this question via email, Google+, Twitter, or Facebook.
Browse other questions tagged pr.probability or ask your own question. | {"url":"http://mathoverflow.net/questions/116143/conditional-probability-deviation-from-the-uniform-distribution","timestamp":"2014-04-21T15:53:25Z","content_type":null,"content_length":"45629","record_id":"<urn:uuid:c526907d-e0f5-4ad7-8aee-e4e7e49c168e>","cc-path":"CC-MAIN-2014-15/segments/1397609540626.47/warc/CC-MAIN-20140416005220-00587-ip-10-147-4-33.ec2.internal.warc.gz"} |
Confidence Intervals - Statistics Teaching Tools
What is a confidence interval?
A confidence interval is a range around a measurement that conveys how precise the measurement is. For most chronic disease and injury programs, the measurement in question is a proportion or a rate
(the percent of New Yorkers who exercise regularly or the lung cancer incidence rate). Confidence intervals are often seen on the news when the results of polls are released. This is an example from
the Associate Press in October 1996:
The latest ABC News-Washington Post poll showed 56 percent favored Clinton while 39 percent would vote for Dole. The ABC News-Washington Post telephone poll of 1,014 adults was conducted March
8-10 and had a margin of error of plus or minus 3.5 percentage points. (Emphasis added).
Although it is not stated, the margin of error presented here was probably the 95 percent confidence interval. In the simplest terms, this means that there is a 95 percent chance that between 35.5
percent and 42.5 percent of voters would vote for Bob Dole (39 percent plus or minus 3.5 percent). Conversely, there is a 5 percent chance that fewer than 35.5 percent of voters or more than 42.5
percent of voters would vote for Bob Dole.
The precise statistical definition of the 95 percent confidence interval is that if the telephone poll were conducted 100 times, 95 times the percent of respondents favoring Bob Dole would be within
the calculated confidence intervals and five times the percent favoring Dole would be either higher or lower than the range of the confidence intervals.
Instead of 95 percent confidence intervals, you can also have confidence intervals based on different levels of significance, such as 90 percent or 99 percent. Level of significance is a statistical
term for how willing you are to be wrong. With a 95 percent confidence interval, you have a 5 percent chance of being wrong. With a 90 percent confidence interval, you have a 10 percent chance of
being wrong. A 99 percent confidence interval would be wider than a 95 percent confidence interval (for example, plus or minus 4.5 percent instead of 3.5 percent). A 90 percent confidence interval
would be narrower (plus or minus 2.5 percent, for example).
What does a confidence interval tell you?
he confidence interval tells you more than just the possible range around the estimate. It also tells you about how stable the estimate is. A stable estimate is one that would be close to the same
value if the survey were repeated. An unstable estimate is one that would vary from one sample to another. Wider confidence intervals in relation to the estimate itself indicate instability. For
example, if 5 percent of voters are undecided, but the margin of error of your survey is plus or minus 3.5 percent, then the estimate is relatively unstable. In one sample of voters, you might have 2
percent say they are undecided, and in the next sample, 8 percent are undecided. This is four times more undecided voters, but both values are still within the margin of error of the initial survey
On the other hand, narrow confidence intervals in relation to the point estimate tell you that the estimated value is relatively stable; that repeated polls would give approximately the same results.
How are confidence intervals calculated?
Confidence intervals are calculated based on the standard error of a measurement. For sample surveys, such as the presidential telephone poll, the standard error is a calculation which shows how well
the poll (sample point estimate) can be used to approximate the true value (population parameter), i.e. how many of the people surveyed said they would vote for Dole versus how many people actually
would vote for Dole in the election. Generally, the larger the number of measurements made (people surveyed), the smaller the standard error and narrower the resulting confidence intervals.
Once the standard error is calculated, the confidence interval is determined by multiplying the standard error by a constant that reflects the level of significance desired, based on the normal
distribution. The constant for 95 percent confidence intervals is 1.96. | {"url":"http://www.health.ny.gov/diseases/chronic/confint.htm","timestamp":"2014-04-17T00:59:45Z","content_type":null,"content_length":"13826","record_id":"<urn:uuid:03252766-e327-445d-abc3-93f5dee20586>","cc-path":"CC-MAIN-2014-15/segments/1397609539066.13/warc/CC-MAIN-20140416005219-00503-ip-10-147-4-33.ec2.internal.warc.gz"} |
Longwood University
2007 Contest
A total of 949 teams registered for the 2007 Mathematical Contest in Modeling. Each team had to chose one of three problems (Problem A, B, or C) which they would attempt to solve during the second
weekend of February. Teams from all over the world competed in this contest.
A total of 1222 teams successfully completed the contest earning the following possible designations (from highest to lowest): Outstanding, Meritorious, Honorable Mention, and Successful Participant.
All of the competing teams are to be congratulated for their excellent work and enthusiasm for scientific and mathematical modeling and interdisciplinary problem solving.
Problems A and B constitute the MCM (Mathematical Contest in Modeling) and Problem C constitutes the Interdisciplinary Contest in Modeling (ICM). Below are the statistics for the MCM and ICM.
Team Omega - MCM Problem A Successful Participant
• Andrew McFayden, Freshman - Mathematics w/ Music Minor
• Michael Souza, Sophomore, CS w/ Mathematics Minor
• Ashley Swandby, Junior - Mathematics
Team Theta - MCM Problem A Successful Participant
• Kimberly Bowman, Senior - Mathematics
• Zach Johnson, Junior - Mathematics w/ CS Minor
• Brandon Taylor, Senior - Mathematics
The Problems
PROBLEM A: Gerrymandering
The United States Constitution provides that the House of Representatives shall be composed of some number (currently 435) of individuals who are elected from each state in proportion to the state's
population relative to that of the country as a whole. While this provides a way of determining how many representatives each state will have, it says nothing about how the district represented by a
particular representative shall be determined geographically. This oversight has led to egregious (at least some people think so, usually not the incumbent) district shapes that look "unnatural" by
some standards.
Hence the following question: Suppose you were given the opportunity to draw congressional districts for a state. How would you do so as a purely "baseline" exercise to create the "simplest" shapes
for all the districts in a state? The rules include only that each district in the state must contain the same population. The definition of "simple" is up to you; but you need to make a convincing
argument to voters in the state that your solution is fair. As an application of your method, draw geographically simple congressional districts for the state of New York.
PROBLEM B: Wheel Chair Access at Airports
Airlines are free to seat passengers waiting to board an aircraft in any order whatsoever. It has become customary to seat passengers with special needs first, followed by first-class passengers (who
sit at the front of the plane). Then coach and business-class passengers are seated by groups of rows, beginning with the row at the back of the plane and proceeding forward.
Apart from consideration of the passengers' wait time, from the airline's point of view, time is money, and boarding time is best minimized. The plane makes money for the airline only when it is in
motion, and long boarding times limit the number of trips that a plane can make in a day.
The development of larger planes, such as the Airbus A380 (800 passengers), accentuate the problem of minimizing boarding (and deboarding) time.
Devise and compare procedures for boarding and deboarding planes with varying numbers of passengers: small (85-210), midsize (210-330), and large (450-800).
Prepare an executive summary, not to exceed two single-spaced pages, in which you set out your conclusions to an audience of airline executives, gate agents, and flight crews.
An article appeared in the NY Times Nov 14, 2006 addressing procedures currently being followed and the importance to the airline of finding better solutions. The article can be seen at: http://
PROBLEM C: Organ Transplant: The Kidney Exchange Problem
Organ Transplant: The Kidney Exchange Problem (pdf)
This SlideShowPro photo gallery requires the Flash Player plugin and a web browser with JavaScript enabled.
Overall Results
A total of 949 teams registered for the 2007 Mathematical Contest in Modeling. Each team had to chose one of three problems (Problem A, B, or C) which they would attempt to solve during the second
weekend of February. Teams from all over the world competed in this contest. A total of 1222 teams successfully completed the contest earning the following possible designations (from highest to
lowest): Outstanding, Meritorious, Honorable Mention, and Successful Participant. All of the competing teams are to be congratulated for their excellent work and enthusiasm for scientific and
mathematical modeling and interdisciplinary problem solving.
Problems A and B constitute the MCM (Mathematical Contest in Modeling) and Problem C constitutes the Interdisciplinary Contest in Modeling (ICM). Below are the statistics for the MCM and ICM.
│ Classification │MCM (Problems A and B)│ICM (Problem C)│
│Total Number of Participating Teams │ 949 │ 273 │
│High School Teams │ 13 (1%) │ 2 (1%) │
│United States Teams │ 283 (30%) │ 36 (13%) │
│Foreign Teams form Australia, Canada, China, England, Finland, Indonesia, Ireland, Jamaica, Korea, New Zealand, Singapore, and South Africa│ 666 (70%) │ 233 (87%) │
│Outstanding Winners │ 14 (1%) │ 4 (2%) │
│Meritorious Winners │ 122 (13%) │ 42 (16%) │
│Honorable Mentions │ 255 (27%) │ 169 (61%) │
│Successful Participants │ 558 (59%) │ 58 (21%) │
Virginia Schools
• James Madison University, Harrisonburg, David B Walton, Problem A, Meritorious
• University of Richmond, Richmond, Kathy W Hoke, Problem A, Meritorious
• Radford University, Radford, Laura J Spielman, Problem B, Meritorious
• Godwin High School Science, Mathematics, and Technology Center, Richmond, Ann W Sebrel, Problem B, Honorable Mention
• James Madison University, Harrisonburg, Anthony Tongen, Problem B, Honorable Mention
• Maggie Walker Governor's School, Richmond, John A. Barnes, Problem B, Honorable Mention
• Maggie Walker Governor's School, Richmond, John A. Barnes, Problem B, Honorable Mention
• Godwin High School Science, Mathematics, and Technology Center, Richmond, Ann W Sebrell, Problem C, Honorable Mention
• James Madison University, Harrisonburg, Anthony Tongen, Problem C, Honorable Mention
• James Madison University, Harrisonburg, Hasan N Hamdan, Problem C, Honorable Mention
• Maggie Walker Governor's School,Richmond, Harold Houghton, Problem C, Honorable Mention
• Eastern Mennonite University, Harrisonburg, Leah Shao Boyer, Problem A, Successful Participant
• Eastern Mennonite University, Harrisonburg, Leah Shao Boyer, Problem A, Successful Participant
• Longwood University, Farmville, M. Leigh Lunsford, Problem A, Successful Participant
• Longwood University, Farmville, M. Leigh Lunsford, Problem A, Successful Participant
• Virginia Western Community College, Roanoke, Steve T Hammer, Problem A, Successful Participant
• Maggie Walker Governor's School, Richmond, Harold Houghton, Problem B, Successful Participant
• Radford University, Radford, Laura J Spielman, Problem B, Successful Participant
• Virginia Western Community College, Roanoke, Ruth A Sherman, Problem B, Successful Participant
• Virginia Western Community College, Roanoke, Gerald D. Benson, Problem B, Successful Participant | {"url":"http://longwood.edu/mathematics/COMAP2007.htm","timestamp":"2014-04-16T13:03:58Z","content_type":null,"content_length":"24490","record_id":"<urn:uuid:9caa63cb-0daf-46bd-835b-a9f9955d2220>","cc-path":"CC-MAIN-2014-15/segments/1397609523429.20/warc/CC-MAIN-20140416005203-00466-ip-10-147-4-33.ec2.internal.warc.gz"} |
2.5.2 Cohen-Sutherland Line-Clipping Algorithm
2.5.2 Cohen-Sutherland Line-Clipping Algorithm
Cf. [FvDFH90, 3.12.3].
First we test whether both endpoints are inside (and hence draw the line segment) or whether both are left of , right of , below , or above (then we ignore line segment). Otherwise we split the line
segment into two pieces at a clipping edge (and thus reject one part). Now we proceed iteratively.
A rather simple accept-reject test is the following:
Divide the plane into 9 regions and assign a 4 bit code to each:
...above top edge
...below bottom edge
...right of right edge
...left of left edge
mathcalculate the corresponding bit-codes for both endpoints.
If both codes are zero then the line segment is completely inside the rectangle. If the bitwise-and of these codes is not zero then the line does not hit since both endpoints lie on the wrong side of
at least one boundary line (corresponding to a bit equal to 1). Otherwise take a line which is met by the segment (for this find one non-zero bit), divide the given line at the intersection point in
two parts and reject the one lying in the outside halfplane.
Andreas Kriegl 2003-07-23 | {"url":"http://www.mat.univie.ac.at/~kriegl/Skripten/CG/node31.html","timestamp":"2014-04-16T16:39:06Z","content_type":null,"content_length":"7259","record_id":"<urn:uuid:e353fea3-8fe1-4f1c-a715-f144454600b7>","cc-path":"CC-MAIN-2014-15/segments/1398223206120.9/warc/CC-MAIN-20140423032006-00314-ip-10-147-4-33.ec2.internal.warc.gz"} |
Interval and Related Software
Interval and Verified Software
• ACETAF, a C-XSC library for the verified automatic computation of estimates for Taylor coefficients of analytic functions.
• AERN, distributed arbitrary precision packages for verifiably Approximating Exact Real Numbers
• ALIAS Library, Algorithms Library of Interval Analysis for Systems; see also this link
• AQCS, a free software package for Approximate Quantified Constraint Solving - computes box approximations of the solution set of formulae of the first-order predicate language over the reals.
• Ariadne, a package for verified reachability analysis of hybrid systems that includes a library for verified set-based computation
• AWA, a software package for verified solution of ordinary differential equations.
• BARON, Global Optimization Software.
Purely continuous, purely integer, and mixed-integer nonlinear problems can be solved with the software.
• BOOST interval C++ library implemented within the popular BOOST project
• CGAL, Computational Geometry Algorithms Library, written in C++, uses interval computations to make geometric computations robust and efficient; the related part of CGAL manual can be found at
the following site the CLIP
• CLIP
Among other applications, it allows rigorous modeling of hybrid systems
• CLP(BNR), a Constraint Interval-Arithmetic Package
This package is actually a language for Constraint Logic Programming based on Interval Arithmetic.
• CoStLy, a library of complex interval standard functions.
• COSY. This software package is based on Taylor model and interval methods. It is intended for verified solution of such problems as ordinary differential equations, quadrature, range bounding,
etc. It can either be used as a stand-alone program or via C++ and Fortran 90/95 frontends.
• DAEPACK, MIT-produced symbolic and numeric library for general numerical calculations which uses interval computations.
• FADBAD/TADIFF, an automatic differentiation package, with applications to verified solution of ordinary differential equations.
• filib++, Fast Interval LIRrary, a C++ library on extended interval arithmetic.
• Interval arithmetic library for a special programming language Frink (implemented in Java)
• General Interval Arithmetic package for manipulating finite sets of (possibly unbounded) intervals; there is an applet interface
• HSolver
This package allows analysis of hybrid systems with non-linear ODEs over an unbounded time horizon, uses intervals as its main work-horse
• HySAT, A bounded model checker for hybrid systems
The package allows bounded-time analysis of hybrid systems with complex discrete structure, also uses intervals for reasoning about numerical constraints
• ICE, Interval Calculator for Engineers
Intervals are to be entered like min_max or mean%tol, for example: 1_2, 4_3, 4%2. The help section of this ICE is full of examples and explanations to use properly the syntax which looks like
• ICOS, Interval COnstraints Solver
• InC++ libraries for interval computations and constraint satisfaction.
• INTBLAS: C++ Implementation
An implementation of INTerval Basic Linear Algebra Software.
• Interval arithmetic for ADA
• Interval library is a part of Intel Math Kernel Library (MKL).
It includes interval arithmetic (standard and two complex) and interval linear solvers that compute enclosures (sharp or fast) of the solution set to interval linear systems of equations,
check properties of interval matrices; some part of the solvers works with complex intervals. Questions and suggestions are welcome at MKL forum
• Interval Software in Fortran-90
This Fortran-90 module is actually a language for interval computations
• Interval based Java applets:
□ Rigorous Gradient Calculator
Computes the gradient of a function using rigorous interval arithmetic
□ Rigorous Graphing Applet
Graph functions and finds their roots. All results are verified, no second guess necessary.
□ Cascade Applet
Rigorously solves the initial value problem for discrete, planar maps. A powerful new algorithm produces guaranteed error bounds in form of zonotope enclosures.
• Libaffa; a C++ affine arithmetic library for GNU/Linux
• Maple: Interval Software in Maple and its clones:
• Mathematica based web site supports dynamic interval web computations and visualization.
• Matlab: Interval Software in Matlab:
• MISO, a software package containing a set of solvers based on Modal Interval Analysis. These solvers are:
□ FSTAR solver: Allows doing computations with Modal Intervals.
□ QRCS solver: Allows proving the satisfiability of a class of quantified real constraints.
□ QSI solver: Allows obtaining inner and outer approximations of the solution set of a class of quantified real constrains.
□ MINIMAX solver: Allows solving constrained minimax optimization problems.
□ SQUALTRACK solver: Allows detecting faults in dynamic systems.
• MPFI, Multiple Precision Floating-Point Interval Library
• MuPAD, integrates interval computation in its kernel; both Maple and MuPad are integrated into Scientific Workplace
• ParaGlobSol
Parallel/distributed implementations of the interval global optimization FORTRAN 90 package GlobSol, which solves global optimization problems with the interval branch-and-bound algorithm
together with interval Newton/generalized bisection method.
• ParLinSys, software for solving parametric interval linear systems
• PROFIL/BIAS
A fast C++ library for interval arithmetic. Speed and comparisons can be found in:
O. Knueppel: PROFIL/BIAS - A Fast Interval Library, COMPUTING, Vol. 53, No. 3-4, p. 277-287.
• Pytaylor, a Python package implementing Taylor models and associated guaranteed ODE solvers.
Pytaylor uses mpmath for interval arithmetic and sympy for symbolic manipulation.
• RealPaver
Interval software implementing constraint satisfaction techniques for solving nonlinear systems.
• Range software of Oliver Aberth
A suite of demonstration programs, using variable precision interval arithmetic, that can solve the typical problems of elementary numerical analysis to desired accuracy.
• SMMS
• Rsolver
solves quantified inequality constraints
• smathlib
A library of C routines for interval arithmetic and constraint narrowing
• Interface between the interval and constraint propagation library smathlib and a functional programming language Objective Caml
• Statool
Software for interval-based arithmetic on random variables.
• SvLis
a free C++ geometric modeller that is substantially based on interval arithmetic
• UniCalc
A solver based on interval constraint propagation. It allows to tackle nonlinear algebraic systems with real and/or integer variables.
• VALENCIA-IVP
software for VALidation of state ENClosures using Interval Arithmetic for Initial Value Problems
• VNODE-LP
A C++ based Interval Solver for Initial Value Problems in Ordinary Differential Equations
• XSC software (for more detailed information about XSC languages, see the following page)
XSC languages and fi_lib
Related Software
See also Languages for Interval Analysis.
See also MathTools.net, a free portal covering computing tools for science and engineering. This portal contains over 20,000 useful links, covering software in Matlab, Java, Excel, C/C+, Fortran, | {"url":"http://www.cs.utep.edu/interval-comp/intsoft.html","timestamp":"2014-04-20T16:01:32Z","content_type":null,"content_length":"17030","record_id":"<urn:uuid:b7c59f3a-f939-4639-8da2-0a1896e22371>","cc-path":"CC-MAIN-2014-15/segments/1398223206770.7/warc/CC-MAIN-20140423032006-00495-ip-10-147-4-33.ec2.internal.warc.gz"} |
Two capacitors, C1 = 20 µF and C2 = 5.0 µF, are connected in parallel and charged with a 150 V power supply
Number of results: 93,101
MITx: 6.002x Circuits and Electronics
to add capacitors in parallel Ceq = C1 + C2 because C = Q/V V the same, charges add to add capacitors in series q is the same on both in general V = C Q here V = V1 + V2 = C1 q + C2 q V/q = (C1+C2)
so Ceq = q/V = 1/(C1 + C2)
Saturday, February 8, 2014 at 11:05am by Damon
Three parallel plate capacitors (C1,C2 and C3 ) are in series with a battery of 100 V. C1=2500 micro Farad; C2=2C1 , C3=3C1 . (a) What is the potential difference over each capacitor and how much
charge is on each capacitor? Across C1 :V1 (in Volts) = incorrect Q1 (in C) = ...
Wednesday, March 13, 2013 at 12:19am by P
This is the situation I have been given: You are given three capacitors, one with a capacitance of 0.025 micro-farads, another with a capacitance of 0.05 micro farads, and the third with a
capacitance of 0.015 micro-farads. Using only these capacitors, arranged in series, ...
Thursday, October 11, 2012 at 7:04pm by Rachel
lets say two capacitors C1 and C2 1uF each are connected in series with a voltage source 5v. One of the capacitors C2 has an initial voltage of 1v and the other has no charge on it initially.What is
the final voltages across the capacitor C2?
Monday, September 10, 2012 at 4:37am by Venkatarama
lets say two capacitors C1 and C2 1uF each are connected in series with a voltage source 5v. One of the capacitors C2 has an initial voltage of 1v and the other has no charge on it initially.What is
the final voltages across the capacitor C2?
Monday, September 10, 2012 at 8:36am by Venkatarama
Capacitors in series subract. Capacitors in parallel add. #2. !/c = 1/c1 + 1/c2 You know c and c2, substitute and solve for c2.
Tuesday, October 15, 2013 at 12:44am by DrBob222
Two capacitors are connected in parallel as shown above. A voltage V is applied to the pair. What is the ratio of charge stored on C1 to the charge stored on C2, when C1 = 1.5C2 ? how do you do this?
( The answer is 3/2) Thanks for helping ; ) I appreciate that
Friday, February 4, 2011 at 2:12am by Amanda C.
Two capacitors are connected to a battery. The battery voltage is V = 60 V, and the capacitances are C1 = 2.00 μF and C2 = 4.00 μF. Determine the total energy stored by the two capacitors when they
are wired (a) in parallel and (b) in series.
Saturday, February 2, 2013 at 12:53pm by PLEASE HELP
Put the capacitors in series with the 4.5 V of batteries in series. Although no DC current will flow, the voltage drop across the two capacitors will be in inverse proportion to the capacitances. C1/
C2 = V2/V1 With one known capacitance C1, you can solve for C2. You do not ...
Saturday, February 13, 2010 at 3:47am by drwls
physics - capacitors
Capacitors C1 = 5.0 µF and C2 = 2.0 µF are charged as a parallel combination across a 150 V battery. The capacitors are disconnected from the battery and from each other. They are then connected
positive plate to negative plate and negative plate to positive plate. Calculate ...
Saturday, July 10, 2010 at 11:39pm by lo
A)Two identical resistors are connected in parallel across a 30-V battery, which supplies them with a total power of 7.0 W. While the battery is still connected, one of the resistors is heated so
that its resistance doubles. The resistance of the other resistor remains ...
Saturday, October 22, 2011 at 8:33pm by Kristi
A)Two identical resistors are connected in parallel across a 30-V battery, which supplies them with a total power of 7.0 W. While the battery is still connected, one of the resistors is heated so
that its resistance doubles. The resistance of the other resistor remains ...
Monday, October 24, 2011 at 12:56pm by Kristi
Vm = Xc Im Xc = 1/ωC therefore for fixed V I2/I1 = C2 /C1 C2 = 4.2C1 + C1 = 5.2C1 (they are connected in parallel) I2 = (C2/C1)I1 = 5.2*0.22 = 1.144 A
Tuesday, November 13, 2012 at 1:33pm by Kay
Three capacitors with C1=C2=C3= 10.00 uF are connected to battery with voltage V= 194 V. What is the voltage across capacitor C1
Wednesday, November 23, 2011 at 11:50pm by JL
Two capacitors, C1 = 27.0 µF and C2 = 35.0 µF, are connected in series, and a 21.0 V battery is connected across them. (a) Find the equivalent capacitance, and the energy contained in this equivalent
capacitor. equivalent capacitance= µF total energy stored= (b) Find the ...
Friday, January 27, 2012 at 11:08am by Help
By symmetry this is 1/3! P(X,Y,Z are all different) P(X,Y,Z are all different) = 1- P(X,Y,Z are not all different) X,Y,Z are not all different if and only if one or more of the three conditions are
satisfied: c1: X = Y c2: X = Z c3: Y = Z Terefore: P(X,Y,Z are not all ...
Tuesday, November 20, 2007 at 2:36am by Count Iblis
Two fully charged capacitors (C1 = 2.00uF, q1=6.00uC,C2=8.00uF, q2 = 12.0 uC). The switch is closed, and the charge flowed until equilibrium is re-established, which occurs when both capacitors have
the same voltage across their plates. Find the resulting voltage across either...
Wednesday, February 17, 2010 at 8:09pm by Eric
Two capacitors, C1=25.0x10^-6 and C2=5.0x10^-6 are connected in parallel and charged with a 100-V power supply. a)calculate the total energy stored in the 2 capacitors b) what potential difference
would be required across the same 2 capacitors connected in series in order that...
Monday, February 4, 2008 at 2:33am by nat
two circles c1 and c2 meet at the points A and B. CD is a common tangent to these circles where C and D lie on the circumference of C1 and C2 respectively. CA is the tangent to c2 at A. When produced
DB meets the circumference of C1 at p. Prove that PC is parallel to AD
Thursday, August 18, 2011 at 6:51am by JJ
in general q/C here V = V1 + V2 = q/C1 + q/C2 V/q = (1/C1+1/C2) so Ceq = q/V = 1/(1/C1 + 1/C2) = C1 C2 /(C1+C2)
Saturday, February 8, 2014 at 11:05am by Damon
Calculus - Second Order Differential Equations
Solve the initial-value problem. y'' - 2y' + y = 0 , y(2) = 0 , y'(2) = 1 r^2-2r+1=0, r1=r2=1 y(x)=c1*e^x+c2*x*e^x y(2)=c1*e^2+c2*2*e^2=0 c1=-(2*c2*exp(2))/exp(2) c1=-2*c2 y'(x)=-2*c2*e^x+c2*e^x*
(x-1) y'(2)=-2*c2*e^2+c2*e^2*(2-1)=1 c2(-2e^2+e^2)=1 c2=1/(-2e^2+e^2)=1/(-e^2) c1...
Tuesday, July 10, 2007 at 4:15am by COFFEE
social studies
What is the total capacitance of capacitors connected in series, C1=444F, C2=621F
Monday, March 25, 2013 at 5:43am by Trevor
y=c1 e^x cos(x)+ c2 e^x sin(x) y(0) = 1 , so 1 = c1 e^0 cos 0 + c2 e^0 sin 0 1 = c1 + 0 c1 = 1 y' = c1 e^x (-sin(x)) + c1 e^x cosx + c2 e^x cos(x) + c2 e^x sinx = e^x( - c1 y'(0) = -1 -1 = c1 e^0
(-sin 0) + c2 e^0 cos 0 -1 = 0 + c2 c2 = -1 y'' = c1 e^x (-cos(x)) + c2 e^x (-sin...
Tuesday, July 2, 2013 at 9:08pm by Reiny
y = c1 e^-6x + c2 e^x Plug in your boundary conditions, and you get c1 + c2 = 0 c1/e^12 + c2/e^2 = 1 Now just solve for c1 and c2
Sunday, March 16, 2014 at 6:39pm by Steve
What is the total capacitance of capacitors connected in series, C1=25F, C2=125F ? Round result to whole number.
Tuesday, February 26, 2013 at 6:24am by mano
Calculus - Second Order Differential Equations
Solve the boundary-value problem. y''+5y'-6y=0, y(0)=0, y(2)=1 r^2+5r-6=0, r1=1, r2=-6 y=c1*e^x + c2*e^-6x y(x)=c1*e^x+c2*e^-6x y'(x)=c1*e^x-6*c2*e^-6x y(0)=c1+c2=0, c1=-c2 y(2)=c1*e^2+c2*e^(-12)=1
-c2*e^2-6c2*e^(-12)=1 -c2(e^2-6*e^-12)=1 c2=-1/(e^2-6e^-12) c1=1/(e^2-6e^-12) y...
Monday, July 9, 2007 at 8:19pm by COFFEE
I have a circuit with three capacitors and a battery. The battery supplies 15V. The battery is in series with C3. C3 is in series with ( C23). C3= 3*10^-6 F C2= 2 *10^-6 F C1= 1*10^-6 F (respectively
C2 +C3 are is parallel withe ach other). I know the equivalent capacitance ...
Wednesday, February 10, 2010 at 6:57pm by Sara
y = e^x (c1 + c2*x) y' = e^x (c1 + c2 + c2*x) Now plug in the initial conditions to get e^2 (c1 + 2c2) = 0 e^2 (c1 + c2 + 2c2) = 1 Now just solve for c1 and c2
Sunday, March 16, 2014 at 6:40pm by Steve
Solve 1/C=1/C1 + 1/C2 for C1 This multiple choice and the listed answers are: a.)C1= CC2/(C2-C) b.)C1= C-C2/C c.)C1= C+C2/CC2 d.)C1= C+C2/C2
Sunday, October 12, 2008 at 11:35am by Shawn
physics - capacitors
C1 acquires C1*V = 7.5*10^-4 C and C2 acquires C2*V = 3.0*10^-4 C, initially. When rewired, there is a total of 4.5*10^-4 C on the positive-plate sides and -4.5*10^-4 C on the negative-plate sides.
The charges redistribute to maintain the same potential on each capacitors. (2/...
Saturday, July 10, 2010 at 11:39pm by drwls
Four capacitors are connected in series with a battery, as in the figure below, where C1 = 3.24 µF, C2 = 6.11 µF, C3 = 12.1 µF, C4 = 25.0 µF, V = 18.4 V. (a) Calculate the capacitance of the
equivalent capacitor. µF (b) Compute the charge on C3. µC (c) Find the voltage drop ...
Friday, September 17, 2010 at 12:07am by Max
question 9 a: Ec1=Voltage/plate separation b: because the battery is connected the E field does not change c: when the battery is disconnected then Q1+Q2 must equal Q1'+Q2' Ec1=Q1'/(k*C1*d) and Ec2=
Q2'/(C2*d) where Q1'=(Q1+Q2)/(1+(C2/K*C1) and Q2'=C2/(K*C1)*(Q1+Q2)/(1+(C2/K*C1))
Saturday, June 15, 2013 at 4:36am by boss
Math another problem
Subtract 1/C2 from each side. 1/C1 = 1/C - 1/C2 = (C2-C)/(C*C2) Now turn both fractions upside down. This is called "taking the reciprocal of both sides" C1 = C*C2/(C2 - C) = 1/[(1/C) - (1/C2)]
Sunday, October 12, 2008 at 12:25am by drwls
Discrete Math
"So the characteristic roots are 1?" Correct! For _n+1 = -8a_n – 16a_n - 1, n ≥ 1, given a₀ = 5, a₁ = 17. you have correctly solved for λ=-4, and the solution is an=C1(-4)^n+C2*n*(-4)^n which
simplifies to: an=(C1+nC2)(-4)^n a0=5=(c1+0*C2)(-4)^0=C1, so...
Tuesday, April 5, 2011 at 4:11pm by MathMate
you have 8microC total charge. You have 23microF capacitance. Voltage= Qtotal/C=8/23 volts. on C1, q=C1*8/23=7micro*8/23=56/23 micro F on C2, q=16micro*8/23=128/23 microC adding q1+q2, 184/23 microF=
Tuesday, February 14, 2012 at 9:45pm by bobpursley
I meant C3 is in series with C2 and C1 which means that C2 and C1 are in parallel
Wednesday, February 10, 2010 at 6:57pm by Sara
Two capacitors C1 = 7.00 µF and C2 = 11.0 µF are connected in series to a 9.00 V battery. (a) Find the equivalent capacitance of the combination. µF (b) Find the potential difference across each
capacitor. (c) Find the charge on each capacitor.
Monday, October 14, 2013 at 1:47pm by HELP!!!
Two circles C1 and C2 touch externally at A. The tangent to C1 drawn at a point B on the circumference of C1 meets C2 in P and Q so that P is between B and Q. QA is produced and meets C1 at X. Prove
that angle XAB = angle PAB.
Wednesday, September 19, 2012 at 7:48am by Jun
Science ; Micro-organisms
i have to finished this cross word , and i need help on the last two . So thank you if you can solve it :) A substance that micro organisms can be grown on . its a 4 letter word. A group Of Micro
Organisms , its a 6 letter word.
Thursday, April 16, 2009 at 1:55pm by Saffron.
Ceq(c1,c2):3uF. This combination is in series with the 3uF capacitor and both are connected to the 15v battery. The voltage will get divided equally since they are both of the same value(7.5v). Hence
energy would be. Both c1 and c2 have 7.5v across them since they are in ...
Wednesday, February 10, 2010 at 6:57pm by Venkatarama
What is the force exerted on 40 micro coulombs by 20 micro coulombs? (charges are separated by 80cm)
Wednesday, May 2, 2012 at 6:16pm by paul
Calculate the magnitude and direction of the Coulomb force on each of the three charges connected in series: 1st - 6 micro Coulombs (positive charge) 2nd - 1.5 micro Coulombs (positive charge) 3dr -
2 micro Coulombs (negative charge) Distance between 1st and 2nd - 3cm Distance...
Wednesday, May 16, 2012 at 11:19pm by andrew
Calculate the magnitude and direction of the Coulomb force on each of the three charges connected in series: 1st - 6 micro Coulombs (positive charge) 2nd - 1.5 micro Coulombs (positive charge) 3dr -
2 micro Coulombs (negative charge) Distance between 1st and 2nd - 3cm Distance...
Monday, May 21, 2012 at 7:35am by andrew
A C1= 2.50uF capacitor is charged to 862 V and a C2= 6.80uF capacitor is charged to 660 V . These capacitors are then disconnected from their batteries. Next the positive plates are connected to each
other and the negative plates are connected to each other. What will be the ...
Friday, March 8, 2013 at 6:32pm by Steve
An infinite sheet with a surface charge density of -2.1 micro-coulombs/m2 runs parallel to the x axis. Another infinite sheet a surface charge density of +2.1 micro-coulombs/m2 runs parallel to the y
axis. A charge of -4.1 micro-coulombs moves at a 45 degree angle towards the ...
Friday, February 10, 2012 at 8:33am by Mandaleigh
The drawing shows two fully charged capacitors (C1 = 7 µF, q1 = 4 µC; C2 = 16 µF, q2 = 4 µC). The switch is closed, and charge flows until equilibrium is reestablished (i.e., until both capacitors
have the same voltage across their plates). Find the resulting voltage across ...
Tuesday, February 14, 2012 at 9:45pm by Laura
two charges of +2.6 micro c and -5.4 micro C experience an attractive force of 6.5 mN.what is separation between the charges?
Sunday, February 12, 2012 at 10:43am by sheldan
Suppose that there is a common resource of size y in a two period society. Each of two citizens, one and two, can withdraw a nonnegative amount c1 or c2 for consumption in period one, provided that
c1+c2 <=y . In the event that they attempt to consume in excess of what is ...
Wednesday, September 14, 2011 at 3:11pm by quynh
Capacitors in parallel have the same voltage V applied to each. The stored charge on each is CV. Since V is the same the ratio of charges is the ratio of capacitances, C. Q1/Q2 = C1/C2 = 1.5
Friday, February 4, 2011 at 2:12am by drwls
Two small conducting spheres are identical except that sphere x has a charge of -10 micro coulombs and sphere y has a charge of + 6 micro coulombs . After the spheres are brought in contact and then
separated,what is the charge on each sphere ,in micro coulombs ?
Thursday, August 18, 2011 at 5:21pm by Troy
What is the potential halfway between two point charges spaced 1 mm apart if q1=10 micro culumbs and q2= -5 micro culumbs
Wednesday, February 15, 2012 at 8:28pm by physics
Calculus Antiderivative Problem
v = 1/3 (t+2)^2 + c1 x = 1/6 (t+2) + c1*t + c2 now plug in the given conditions when t=0 to find c1 and c2
Friday, November 8, 2013 at 7:29pm by Steve
physics HELP MEEE SO CONFUSED
What is the potential halfway between two point charges spaced 1 mm apart if q1=10 micro culumbs and q2= -5 micro culumbs
Wednesday, February 15, 2012 at 9:03pm by bel
The capacitor with the dielectric (#2) will have 4.2 times the C-value of the air-gap capacitor (#1). Let's call the capacitances C1 and C2 = 4.2 C1. Since the energies (E) are equal, E1 = (1/2) C1 V
^2 = E2 = (1/2) C2 V2^2 C1 * 18^2 = (4.2 C1) * V2^2 V2^2 = (1/4.2) * 324 Solve...
Monday, September 3, 2007 at 11:02am by drwls
A series circuit contains a resistor with R = 24 ohms, an inductor with L = 2 H, a capacitor with C = 0.005 F, and a generator producing a voltage of E(t) = 12 sin(10t). The initial charge is Q =
0.001 C and the initial current is 0. (a) Find the charge at time t. 2Q"+24Q'+...
Thursday, July 12, 2007 at 12:49am by COFFEE
a. C = C1*C2/(C1+C2) = 30 uF 33.6*C2/(33.6+C2) = 30 33.6C2 = 1008+30C2 33.6C2-30C2 = 1008 3.6C2 = 1008 C2 = 280 uF In series. b. C1 + C2 = 30 uF 29.3 + C2 = 30 C2 = 30 - 29.3 = 0.7 uF in parallel.
Tuesday, October 15, 2013 at 11:33pm by Henry
What media? Micro what? This sounds like a question from a lab experiment you were supposed to perform. We need the details
Tuesday, February 19, 2008 at 11:28am by drwls
m1=2 kg, c1= 0.107 cal/(g*K)= 448 J/kg•K, T1=273+70=343.15 K, m2=1kg, c2=4185 J/kg•K, T2=273+15=288.15 K Q1 =m1•c1(T1-Tₒ) = m1•c1•T1- m1•c1•Tₒ Q2 = m2•c2• ( Tₒ-T2) = m2•c2•Tₒ- m2•c2•T2 Q1=Q2
m1•c1•T1- m1•c1•Tₒ= m2•c2•Tₒ- m2•c2•T2 m1•c1•T1+ ...
Sunday, July 1, 2012 at 10:06pm by Elena
The solution will be c1 e^k1x + c2 e^k2x where k1 and k2 are the roots of 2x^2+x-4=0. x = (-1±√33)/4 So, y = c1 e(-1+√33)/4 x + c2 e(-1-√33)/4 x y(0) = 0, so c1+c2=0 y'(0) = 1 so (-1+√33) c1 +
(-1-√33) c2 = 1 c1 = 1/(2√33) c2 = -1/(2√33)
Saturday, July 20, 2013 at 9:10pm by Steve
What is the potential halfway between two point charges spaced 1 mm apart if q1=10 micro coulumbs and q2= -5 micro coulumbs?
Thursday, February 16, 2012 at 11:38am by anne
Capacitors in parallel: Give as clear as explanation as possible as to why it is physically reasonable to expect that two identical parallel plate capacitors are placed in parallel ought to have
twice the capacitance as one capacitor.
Sunday, March 30, 2014 at 7:04pm by ikine4
Determine the equivalent capacitance between A and B for the group of capacitors in the drawing. Let C1 = 12 µF and C2 = 4.0 µF.
Friday, March 4, 2011 at 10:37pm by YANA
Calculus II
There IS an antiderivative (integral) of 1/x. It is the natural log-base-e (ln) function. In your case f'(x) = -1/x + C1 f"(x) = - ln x + C1 x + C2 where C1 and C2 are arbitrary constants
Saturday, January 24, 2009 at 11:07pm by drwls
m1 = 10.9•10^-3 kg, v =400 m/s, t(o) = 20oC, m2 = 0.4 kg,. specific heat of the iron and wood, respectively, c1 = 444 J/kg, and c2 = 1700 J/kg. (a) m1•v^2/2 = 10.9•10^-3•(400)^2/ 2 =872 J = QQ/ Q =
m1•c1•Δt, Δt = Q/ m1•c1 = 872/10.8•10^-3•444 = 180oC. t(new) = 20oC...
Thursday, April 26, 2012 at 1:12am by Elena
Great! I understand now that C0=f(a) c1=f'(a) c2=f"(a)/2 Now, If I am to find the parabolization of the equation x^2-x at x=2, then c0=x^2-x=2^2-2=2 c1=2x-1=2(2)-1=3 c2=2/2=1 So, the equation (taken
from c0+c1(x-a)+c2(x-a)^2) is >> 2+3(x-2)+1(x-2)^2?? Is this correct? ...
Monday, November 13, 2006 at 8:17pm by Matt
You can count the number of (micro) craters on the surface of the rock. Knowing the age of te rock, you then know the flux of micro meteorites. Using this information you can deduduce how old a
surface is by counting the number of (micro) craters per unit area
Wednesday, March 12, 2008 at 5:50pm by Count Iblis
Physics please help
What is the potential halfway between two point charges spaced 1 mm apart if q1=10 micro coulombs and q2= -5 micro coulombs? Attempt at solution: no idea but need to get this right our homework is
Wednesday, February 15, 2012 at 9:30pm by bel
The potential diffrence across a series combination of 4(micro)F and 12(micro)F capacitor is 240V. What is the potential diffrence across 4(micro)F capacitor?
Wednesday, January 11, 2012 at 11:33am by Nuwaya
Ct = C1*C2/(C1+C2) = 8.3*2.9/(8.3+2.9) = 2.15 uF Qt = Ct*E = 2.15 * 24 = 51.6 uC.=Q1=Q2. Energy=0.5C*E^2 = 0.5*2.15*24^2 =619.2 Joules Parallel Combination: Ct = C1 + C2 = 8.3 + 2.9 = 11.2 uF. Energy
= 0.5*Ct*E^2 = 619.2 Joules 0.5*11.2*E^2 = 619.2 5.6E^2 = 619.2 E^2 = 110.6 E...
Wednesday, October 16, 2013 at 2:01pm by Henry
Discrete Math
Solve the recurrence relation: an+1 = 7an – 10an-1, where n ≥ 2, and given a₁ = 10, a₂ = 29. You have done the first step to give: λ1=2, λ2=5. So the function is defined by the powers of λ1 and λ2,
with the associated constants C1 ...
Tuesday, April 5, 2011 at 4:11pm by MathMate
electronics technician
There is no way I will answer all of these. If you try yourself someone might help. The first one is just common sense. Capacitance is charge per unit voltage. With two capacitors in parallel they
are both at the same voltage, so each has charge corresponding to that voltage, ...
Tuesday, June 18, 2013 at 9:52am by Damon
A point charge of +10 micro-coulombs lies at x = -0.39 m, y = 0 m, a point charge of -5.7 micro-coulombs lies at x = +0.39 m, y = 0 m, and a point charge of +14.9 micro-coulombs lies at x = 0 m, y =
-0.3 m. A point charge of +10 micro-coulombs is moves from x = 0 m, y = +0.11 ...
Friday, February 10, 2012 at 8:32am by Mandaleigh
Electrons circuits
Hi everyone! I'm new in here. Recently I bought a movement monitor which detects micro movement. If the sensor does not detect a movement in 20 seconds, the build-in speaker will beep until the
sensor detects a movement. I'm using it for my project and I have a question about ...
Friday, April 4, 2008 at 11:26pm by Jacky
Solve the initial-value problem. Am I using the wrong value for beta here, 2sqrt(2) or am I making a mistake somewhere else? Thanks. y''+4y'+6y=0, y(0)=2, y'(0)=4 r^2+4r+6=0, r=(-4 +/- sqrt(16-4(1)
(6))/2 r=-2 +/- sqrt(2)*i , alpha = -2, beta = 2(sqrt(2)) y=e^-2x*(c1*cos(sqrt(2...
Thursday, July 12, 2007 at 12:15am by COFFEE
Please look at my work below: Solve the initial-value problem. y'' + 4y' + 6y = 0 , y(0) = 2 , y'(0) = 4 r^2+4r+6=0, r=(16 +/- Sqrt(4^2-4(1)(6)))/2(1) r=(16 +/- Sqrt(-8)) r=8 +/- Sqrt(2)*i, alpha=8,
Beta=Sqrt(2) y(0)=2, e^(8*0)*(c1*cos(0)+c2*sin(0))=c2=2 y'(0)=4, c2=4 y(x)=e^(...
Tuesday, July 10, 2007 at 2:08pm by COFFEE
2 circles C1 and C2 meet at the points A and B. Points C and Dare points on the circumferences of C1 and C2 respectively such that angle CAB = angle DAB. CD cuts C1 and C2 at the points Y and X
respectively. if Z is the foot of the perpendicular from X to BY, prove that angle ...
Wednesday, September 19, 2012 at 7:30am by Jun
1) 2 spheres of equal diameters have 15 micro coulomb charge and -10 micro coulomb charge respectively. if they are placed 5cm apart what will be the force of attraction between them? 2) a charge of
.5 micro coulomb is placed in an electric field whose intensity is 4x10^5n/c...
Sunday, October 10, 2010 at 2:13pm by Deyo
Do you think lotteries have both micro and macro economic effects or only micro? How do lotteries change what and for whom goods and services are produced?
Sunday, September 18, 2011 at 5:20pm by Emily
Discrete Math
For 9:05 pm an=(C1+nC2)1^n=C1+nC2 a0=40 => C1=40 a1=31 => 31=(40+1*C2) => C2=-9 Maybe you should show your work how you got C2=-3
Tuesday, April 5, 2011 at 4:11pm by MathMate
Differential equations, initial value problem. The general equation of motion is: mx"+Bx'+kx=f(t), where the independent variable is t, and the displacement x is the dependent variable. In this case,
external force f(t)=0, so mx"+Bx'+kx=0 substitute m=0.25, k=4, B=1 we have a ...
Wednesday, March 16, 2011 at 1:54pm by Anonymous
Okay I understand it now I have to make them have the same denominators so I multiply 1/C by the numerator and denominator by C2 so it would be C2/(C2*C) and the other part is 1/C2 times C which
would be C/(C2*C) which would make the problem 1/C1= C2-C/(C2*C) what do I do next?
Sunday, October 12, 2008 at 11:35am by Shawn
A charge of -6.5 micro-Coulombs is placed at the origin of a coordinate system. Another charge of -7.6 micro-Coulombs is placed at x = +0.29 m, y = +0.17 m. A third charge of +14 micro-Coulombs is
placed at x = -0.29 m, y = 0 m. What is the magnitude of the total electric ...
Saturday, January 29, 2011 at 12:06pm by kelly
Two charges are arranged on the x axis so that there is a positive charge of 3.6 micro C at the orgin and a negative charge of -2.5 micro C located 2 centimeters to its right. to the nearest
centimeter, where on the x axis should a positive charge be placed so that the net ...
Friday, March 21, 2008 at 12:44pm by Maria Lewis
Two charges are arranged on the x axis so that there is a positive charge of 3.6 micro C at the orgin and a negative charge of .2.5 micro C located 2.5 centimeters to its right. to the nearest
cetimeter, where on the x axis should a positive charge be placed so that the net ...
Friday, March 21, 2008 at 1:04pm by Maria Lewis
Find f'(x) and see if there is a zero (root) on the interval. If not, the extrema are the two limits of the interval. Check the sign of f'(x) on the interval. If there is no zero on the interval, the
sign should not change (+ OR -). If f'(x) is + on the interval, it is ...
Thursday, November 4, 2010 at 7:00am by MathMate
It is understood to be mass/volume for ease of dosage calculations. Therefore whether it is 0.9% saline or simply water, the proportions are the same. You can do the calculations as follows: Suppose
we need v ml of c1% concentrated dextrose diluted with 1000 ml of liquid to ...
Saturday, September 18, 2010 at 4:38pm by MathMate
Replace C2 by 2C1 E = 1.4 log 2C1/C1 = 1.4 log 2 If C1<C2 then C1/C2 < 1 and the log is < 0 I assume negative energy needed means energy is released.
Thursday, March 8, 2012 at 8:35pm by Steve
what is the electric field at the position 20 cm from Q2 and 60 cm from Q1? Q2=40 micro coulombs Q1=20 micro coulombs Q2 and Q1 are separated by 80 cm
Wednesday, May 2, 2012 at 7:20pm by quinton
Physics 2
2 spheres of equal diameters have a positive 15 micro column and negative 10 micro column charge if they are place 5 cm apart what would be the force of attraction between them.
Tuesday, January 28, 2014 at 6:40am by schenncyl
c1•m1 •(310-25.4) =c2•m2•(25.4-20.2) m1= { c2•m2•(25.4-20.2)}/ c1•(310-25.4) = ={4180•1.1•5.2}/385•284.6 = 0.218 kg
Thursday, August 9, 2012 at 10:38pm by Elena
Calculate the coulomb force on q. They are all in a straight line q1, q2,q3. Q1:1.05 micro coulombs Q2:5.51 micro coulombs Q3:-2.15 micro coulombs The distance, r1, between q1 and q2 is 20.1cm. The
distance, r2, between q2 and q3 is 15.3 cm. Also, calculate e coulomb force on q2
Tuesday, July 9, 2013 at 9:21am by Amalia
take integrals across the equation 2ff' - 2gg' = 0 ∫2f df - ∫2g dg = 0 (f^2 + c1) - (g^2 + c2) = 0 f^2 - g^2 = -(c1+c2) = c
Sunday, October 21, 2012 at 2:13pm by Steve
physics 2
A charge of -6.8 micro-Coulombs is placed at x = 0, y = 0. A charge of -18.5 micro-Coulombs is placed at x = 0, y = +37 cm. Another charge of -12.9 micro-Coulombs is placed at x = +17 cm, y = 0 cm.
What is the angle the total electric force on the charge at x = +17 cm, y = 0 ...
Friday, January 18, 2013 at 2:18pm by Jessica
Discrete Math
What method are you expected to use to solve the recurrence relation? Have you covered characteristic polynomials or generating functions? You seem to give me the hint that you have done
characteristic polynomials, so I will solve this by characteristic polynomials. The key is...
Tuesday, April 5, 2011 at 4:11pm by MathMate
Diff eqn- IVP
Differential equations, initial value problem. The general equation of motion is: mx"+Bx'+kx=f(t), where the independent variable is t, and the displacement x is the dependent variable. In this case,
external force f(t)=0, so mx"+Bx'+kx=0 substitute m=0.25, k=4, B=1 we have a ...
Thursday, March 10, 2011 at 9:45pm by MathMate
I assume in ax=bt, x is a subscript and not a variable. 'a' will be used from here on. Since initial conditions are all known, the function X(t) can be found by integrations. b=3.0 m/s³ a(t) = bt V
(t) = ∫a(t) + C1 = (b/2)t² + C1 since V(0)=Vo=0, C1=0, therefore V...
Thursday, October 8, 2009 at 7:34am by MathMate
Here goes... 1. Consider a model in which an individual lives two periods: this period (time one) and next period (time two). This period his budget constraint requires that his consumption, c1; plus
his saving, s; equals his income, y1: c1 + s = y1: Next period his budget ...
Sunday, September 30, 2012 at 9:25pm by Wesley
What is the electric field at a point midway between a -7.34 micro coulombs and a +4.49 micro coulombs charge apart 31.5m apart?
Tuesday, July 9, 2013 at 9:17am by Amalia
what is the electric field at the center of the square? q1 and q2 = -10 micro coulombs q3 and q4= 5 micro coulombs each side is 0.10 m. How do i set this up?
Thursday, July 7, 2011 at 6:18pm by Sarah (C)
- Physic college
The equation for the change of position of a train starting at x = 0 is given by x=1/2at^2 + bt3 . Find the dimensions of a and b. If y = C1 sin (C2 t) where y is a distance and t is the time. What
are the dimensions of C1 and C2 ?.
Friday, October 1, 2010 at 7:03am by meera
Pages: 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 | 11 | 12 | 13 | 14 | 15 | Next>> | {"url":"http://www.jiskha.com/search/index.cgi?query=Two+capacitors%2C+C1+%3D+20+%26micro%3BF+and+C2+%3D+5.0+%26micro%3BF%2C+are+connected+in+parallel+and+charged+with+a+150+V+power+supply","timestamp":"2014-04-18T07:11:03Z","content_type":null,"content_length":"44762","record_id":"<urn:uuid:e2324aa6-ae01-4213-bd04-185cdedf42c5>","cc-path":"CC-MAIN-2014-15/segments/1398223206770.7/warc/CC-MAIN-20140423032006-00023-ip-10-147-4-33.ec2.internal.warc.gz"} |
Uniform Circular Motion.
P: 21 Hi there. I need some help with this question. Can anyone help me......
A centrifuge is a device in which a small container of material is rotated at a high speed on a circular path. Suppose the centripetal acceleration of the
sample is 6.25 X 103 times as large as the acceleration due to gravity. How many revolutions per minute is the sample making, if it is located at a radius
of 5.00cm from the axis of rotation?
Any help would be appreciated.
Emeritus Could you please show some work or thoughts?
Sci Advisor HINT: What is the equation for centripetal acceleration?
PF Gold ~H
P: 9,789
P: 21 The equation is a = v2/r. But what do they mean when they say the centripetal acceleration is 6.25x10 3 times as large as the acceleration due to gravity????
The equation is a = v2/r. But what do they mean when they say the centripetal acceleration is 6.25x10 3 times as large as the acceleration due to gravity????
Emeritus Uniform Circular Motion.
Sci Advisor [tex]a = \left( 6.25\times 10^{3} \right)g[/tex]
PF Gold ~H
P: 9,789
P: 21 So then that probably means that v2/r = (6,25x10 3)g
And the circumference of the circle its rotating in is 0,314m or 31,4cm
Emeritus But you want to find revolutions per minute, so your next step would be calculating the angular velcoity ([itex]\omega[/itex]). You will need to use;
Sci Advisor [tex]v = \omega r[/tex]
PF Gold ~H
P: 9,789
P: 21 But i can get (v) also with v=(2)(pie)(r)\T
So i still need T
Emeritus They are effectively the same thing, but you don't need to work out v;
Sci Advisor [tex]a = \frac{v^2}{r}[/tex]
PF Gold [tex]v = \omega r = \frac{2\pi r}{T}[/tex]
[tex]a = \frac{\omega^2 r^2}{r}[/tex]
P: 9,789 [tex]\omega^{2} = \frac{a}{r}[/tex]
[tex]\frac{2\pi}{T} = \sqrt{\frac{a}{r}}[/tex]
Related Discussions
Uniform Circular Motion, Rotational Motion, Torque, and Inertia General Physics 1
Uniform Circular Motion Help Introductory Physics Homework 2
Uniform Circular Motion Introductory Physics Homework 7
non-uniform circular motion (i think) Introductory Physics Homework 3
help IN UNIFORM CIRCULAR MOTION Introductory Physics Homework 3 | {"url":"http://www.physicsforums.com/showthread.php?t=118010","timestamp":"2014-04-20T05:56:23Z","content_type":null,"content_length":"40559","record_id":"<urn:uuid:082967ff-3de8-48de-b3d0-8d31e47f3a1c>","cc-path":"CC-MAIN-2014-15/segments/1397609538022.19/warc/CC-MAIN-20140416005218-00019-ip-10-147-4-33.ec2.internal.warc.gz"} |
188 helpers are online right now
75% of questions are answered within 5 minutes.
is replying to Can someone tell me what button the professor is hitting...
• Teamwork 19 Teammate
• Problem Solving 19 Hero
• Engagement 19 Mad Hatter
• You have blocked this person.
• ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.
This is the testimonial you wrote.
You haven't written a testimonial for Owlfred. | {"url":"http://openstudy.com/users/blingblong/asked","timestamp":"2014-04-17T16:10:51Z","content_type":null,"content_length":"86337","record_id":"<urn:uuid:24cc1e3c-c73b-4fd0-b5bb-abb39c8df386>","cc-path":"CC-MAIN-2014-15/segments/1398223206770.7/warc/CC-MAIN-20140423032006-00340-ip-10-147-4-33.ec2.internal.warc.gz"} |
About This Tool
The online Outlier Calculator is used to calculate the outliers of a set of numbers.
An outlier in a distribution is a number that is more than 1.5 times the length of the box away from either the lower or upper quartiles. Specifically, if a number is less than Q1 - 1.5×IQR or greater
than Q3 + 1.5×IQR, then it is an outlier.
Interquartile Range (IQR)
In descriptive statistics, the interquartile range (IQR) is a measure of statistical dispersion, being equal to the difference between the third quartile (Q3) and first quartile (Q1), that is, IQR =
Q3 - Q1.
First Quartile And Third Quartile
The first quartile, also called lower quartile, is equal to the data at the 25th percentile of the data. The third quartile, also called upper quartile, is equal to the data at the 75th percentile of
the data.
There are several different methods for calculating quartiles. This calculator uses a method described by Moore and McCabe to find quartile values. The same methord also used by The TI-83 to
calculate quartile values. With this method, the first quartile is the median of the numbers below the median, the third quartile is the median of the numbers above the median. | {"url":"http://www.miniwebtool.com/outlier-calculator/","timestamp":"2014-04-18T13:09:27Z","content_type":null,"content_length":"5876","record_id":"<urn:uuid:cd60df05-42fb-4ae9-8054-0dae3aed9164>","cc-path":"CC-MAIN-2014-15/segments/1398223207985.17/warc/CC-MAIN-20140423032007-00041-ip-10-147-4-33.ec2.internal.warc.gz"} |
Four coins
If you were to toss four coins, what are the odds in favor of at least two landing heads up?
What is the sample space of four coins being tossed: HHHH, HHHT, HHTH, HTHH, THHH. . .etc. Now of those how many have at least two H's in them? Or, if you have studied the binomial formula, you
recognize this as a binomial distribution with four independent trials (the individual coins being tossed).
Let X be the probability of tossing heads Then $P(X\geq 2) = P(X= 2)+P(X= 3)+P(X= 4)$ | {"url":"http://mathhelpforum.com/statistics/120818-four-coins.html","timestamp":"2014-04-17T10:48:10Z","content_type":null,"content_length":"34553","record_id":"<urn:uuid:60e58c5c-479b-4aea-aed2-2ba3cc4f42f8>","cc-path":"CC-MAIN-2014-15/segments/1397609535775.35/warc/CC-MAIN-20140416005215-00136-ip-10-147-4-33.ec2.internal.warc.gz"} |
Two capacitors, C1 = 20 µF and C2 = 5.0 µF, are connected in parallel and charged with a 150 V power supply
Number of results: 93,101
MITx: 6.002x Circuits and Electronics
to add capacitors in parallel Ceq = C1 + C2 because C = Q/V V the same, charges add to add capacitors in series q is the same on both in general V = C Q here V = V1 + V2 = C1 q + C2 q V/q = (C1+C2)
so Ceq = q/V = 1/(C1 + C2)
Saturday, February 8, 2014 at 11:05am by Damon
Three parallel plate capacitors (C1,C2 and C3 ) are in series with a battery of 100 V. C1=2500 micro Farad; C2=2C1 , C3=3C1 . (a) What is the potential difference over each capacitor and how much
charge is on each capacitor? Across C1 :V1 (in Volts) = incorrect Q1 (in C) = ...
Wednesday, March 13, 2013 at 12:19am by P
This is the situation I have been given: You are given three capacitors, one with a capacitance of 0.025 micro-farads, another with a capacitance of 0.05 micro farads, and the third with a
capacitance of 0.015 micro-farads. Using only these capacitors, arranged in series, ...
Thursday, October 11, 2012 at 7:04pm by Rachel
lets say two capacitors C1 and C2 1uF each are connected in series with a voltage source 5v. One of the capacitors C2 has an initial voltage of 1v and the other has no charge on it initially.What is
the final voltages across the capacitor C2?
Monday, September 10, 2012 at 4:37am by Venkatarama
lets say two capacitors C1 and C2 1uF each are connected in series with a voltage source 5v. One of the capacitors C2 has an initial voltage of 1v and the other has no charge on it initially.What is
the final voltages across the capacitor C2?
Monday, September 10, 2012 at 8:36am by Venkatarama
Capacitors in series subract. Capacitors in parallel add. #2. !/c = 1/c1 + 1/c2 You know c and c2, substitute and solve for c2.
Tuesday, October 15, 2013 at 12:44am by DrBob222
Two capacitors are connected in parallel as shown above. A voltage V is applied to the pair. What is the ratio of charge stored on C1 to the charge stored on C2, when C1 = 1.5C2 ? how do you do this?
( The answer is 3/2) Thanks for helping ; ) I appreciate that
Friday, February 4, 2011 at 2:12am by Amanda C.
Two capacitors are connected to a battery. The battery voltage is V = 60 V, and the capacitances are C1 = 2.00 μF and C2 = 4.00 μF. Determine the total energy stored by the two capacitors when they
are wired (a) in parallel and (b) in series.
Saturday, February 2, 2013 at 12:53pm by PLEASE HELP
Put the capacitors in series with the 4.5 V of batteries in series. Although no DC current will flow, the voltage drop across the two capacitors will be in inverse proportion to the capacitances. C1/
C2 = V2/V1 With one known capacitance C1, you can solve for C2. You do not ...
Saturday, February 13, 2010 at 3:47am by drwls
physics - capacitors
Capacitors C1 = 5.0 µF and C2 = 2.0 µF are charged as a parallel combination across a 150 V battery. The capacitors are disconnected from the battery and from each other. They are then connected
positive plate to negative plate and negative plate to positive plate. Calculate ...
Saturday, July 10, 2010 at 11:39pm by lo
A)Two identical resistors are connected in parallel across a 30-V battery, which supplies them with a total power of 7.0 W. While the battery is still connected, one of the resistors is heated so
that its resistance doubles. The resistance of the other resistor remains ...
Saturday, October 22, 2011 at 8:33pm by Kristi
A)Two identical resistors are connected in parallel across a 30-V battery, which supplies them with a total power of 7.0 W. While the battery is still connected, one of the resistors is heated so
that its resistance doubles. The resistance of the other resistor remains ...
Monday, October 24, 2011 at 12:56pm by Kristi
Vm = Xc Im Xc = 1/ωC therefore for fixed V I2/I1 = C2 /C1 C2 = 4.2C1 + C1 = 5.2C1 (they are connected in parallel) I2 = (C2/C1)I1 = 5.2*0.22 = 1.144 A
Tuesday, November 13, 2012 at 1:33pm by Kay
Three capacitors with C1=C2=C3= 10.00 uF are connected to battery with voltage V= 194 V. What is the voltage across capacitor C1
Wednesday, November 23, 2011 at 11:50pm by JL
Two capacitors, C1 = 27.0 µF and C2 = 35.0 µF, are connected in series, and a 21.0 V battery is connected across them. (a) Find the equivalent capacitance, and the energy contained in this equivalent
capacitor. equivalent capacitance= µF total energy stored= (b) Find the ...
Friday, January 27, 2012 at 11:08am by Help
By symmetry this is 1/3! P(X,Y,Z are all different) P(X,Y,Z are all different) = 1- P(X,Y,Z are not all different) X,Y,Z are not all different if and only if one or more of the three conditions are
satisfied: c1: X = Y c2: X = Z c3: Y = Z Terefore: P(X,Y,Z are not all ...
Tuesday, November 20, 2007 at 2:36am by Count Iblis
Two fully charged capacitors (C1 = 2.00uF, q1=6.00uC,C2=8.00uF, q2 = 12.0 uC). The switch is closed, and the charge flowed until equilibrium is re-established, which occurs when both capacitors have
the same voltage across their plates. Find the resulting voltage across either...
Wednesday, February 17, 2010 at 8:09pm by Eric
Two capacitors, C1=25.0x10^-6 and C2=5.0x10^-6 are connected in parallel and charged with a 100-V power supply. a)calculate the total energy stored in the 2 capacitors b) what potential difference
would be required across the same 2 capacitors connected in series in order that...
Monday, February 4, 2008 at 2:33am by nat
two circles c1 and c2 meet at the points A and B. CD is a common tangent to these circles where C and D lie on the circumference of C1 and C2 respectively. CA is the tangent to c2 at A. When produced
DB meets the circumference of C1 at p. Prove that PC is parallel to AD
Thursday, August 18, 2011 at 6:51am by JJ
in general q/C here V = V1 + V2 = q/C1 + q/C2 V/q = (1/C1+1/C2) so Ceq = q/V = 1/(1/C1 + 1/C2) = C1 C2 /(C1+C2)
Saturday, February 8, 2014 at 11:05am by Damon
Calculus - Second Order Differential Equations
Solve the initial-value problem. y'' - 2y' + y = 0 , y(2) = 0 , y'(2) = 1 r^2-2r+1=0, r1=r2=1 y(x)=c1*e^x+c2*x*e^x y(2)=c1*e^2+c2*2*e^2=0 c1=-(2*c2*exp(2))/exp(2) c1=-2*c2 y'(x)=-2*c2*e^x+c2*e^x*
(x-1) y'(2)=-2*c2*e^2+c2*e^2*(2-1)=1 c2(-2e^2+e^2)=1 c2=1/(-2e^2+e^2)=1/(-e^2) c1...
Tuesday, July 10, 2007 at 4:15am by COFFEE
social studies
What is the total capacitance of capacitors connected in series, C1=444F, C2=621F
Monday, March 25, 2013 at 5:43am by Trevor
y=c1 e^x cos(x)+ c2 e^x sin(x) y(0) = 1 , so 1 = c1 e^0 cos 0 + c2 e^0 sin 0 1 = c1 + 0 c1 = 1 y' = c1 e^x (-sin(x)) + c1 e^x cosx + c2 e^x cos(x) + c2 e^x sinx = e^x( - c1 y'(0) = -1 -1 = c1 e^0
(-sin 0) + c2 e^0 cos 0 -1 = 0 + c2 c2 = -1 y'' = c1 e^x (-cos(x)) + c2 e^x (-sin...
Tuesday, July 2, 2013 at 9:08pm by Reiny
y = c1 e^-6x + c2 e^x Plug in your boundary conditions, and you get c1 + c2 = 0 c1/e^12 + c2/e^2 = 1 Now just solve for c1 and c2
Sunday, March 16, 2014 at 6:39pm by Steve
What is the total capacitance of capacitors connected in series, C1=25F, C2=125F ? Round result to whole number.
Tuesday, February 26, 2013 at 6:24am by mano
Calculus - Second Order Differential Equations
Solve the boundary-value problem. y''+5y'-6y=0, y(0)=0, y(2)=1 r^2+5r-6=0, r1=1, r2=-6 y=c1*e^x + c2*e^-6x y(x)=c1*e^x+c2*e^-6x y'(x)=c1*e^x-6*c2*e^-6x y(0)=c1+c2=0, c1=-c2 y(2)=c1*e^2+c2*e^(-12)=1
-c2*e^2-6c2*e^(-12)=1 -c2(e^2-6*e^-12)=1 c2=-1/(e^2-6e^-12) c1=1/(e^2-6e^-12) y...
Monday, July 9, 2007 at 8:19pm by COFFEE
I have a circuit with three capacitors and a battery. The battery supplies 15V. The battery is in series with C3. C3 is in series with ( C23). C3= 3*10^-6 F C2= 2 *10^-6 F C1= 1*10^-6 F (respectively
C2 +C3 are is parallel withe ach other). I know the equivalent capacitance ...
Wednesday, February 10, 2010 at 6:57pm by Sara
y = e^x (c1 + c2*x) y' = e^x (c1 + c2 + c2*x) Now plug in the initial conditions to get e^2 (c1 + 2c2) = 0 e^2 (c1 + c2 + 2c2) = 1 Now just solve for c1 and c2
Sunday, March 16, 2014 at 6:40pm by Steve
Solve 1/C=1/C1 + 1/C2 for C1 This multiple choice and the listed answers are: a.)C1= CC2/(C2-C) b.)C1= C-C2/C c.)C1= C+C2/CC2 d.)C1= C+C2/C2
Sunday, October 12, 2008 at 11:35am by Shawn
physics - capacitors
C1 acquires C1*V = 7.5*10^-4 C and C2 acquires C2*V = 3.0*10^-4 C, initially. When rewired, there is a total of 4.5*10^-4 C on the positive-plate sides and -4.5*10^-4 C on the negative-plate sides.
The charges redistribute to maintain the same potential on each capacitors. (2/...
Saturday, July 10, 2010 at 11:39pm by drwls
Four capacitors are connected in series with a battery, as in the figure below, where C1 = 3.24 µF, C2 = 6.11 µF, C3 = 12.1 µF, C4 = 25.0 µF, V = 18.4 V. (a) Calculate the capacitance of the
equivalent capacitor. µF (b) Compute the charge on C3. µC (c) Find the voltage drop ...
Friday, September 17, 2010 at 12:07am by Max
question 9 a: Ec1=Voltage/plate separation b: because the battery is connected the E field does not change c: when the battery is disconnected then Q1+Q2 must equal Q1'+Q2' Ec1=Q1'/(k*C1*d) and Ec2=
Q2'/(C2*d) where Q1'=(Q1+Q2)/(1+(C2/K*C1) and Q2'=C2/(K*C1)*(Q1+Q2)/(1+(C2/K*C1))
Saturday, June 15, 2013 at 4:36am by boss
Math another problem
Subtract 1/C2 from each side. 1/C1 = 1/C - 1/C2 = (C2-C)/(C*C2) Now turn both fractions upside down. This is called "taking the reciprocal of both sides" C1 = C*C2/(C2 - C) = 1/[(1/C) - (1/C2)]
Sunday, October 12, 2008 at 12:25am by drwls
Discrete Math
"So the characteristic roots are 1?" Correct! For _n+1 = -8a_n – 16a_n - 1, n ≥ 1, given a₀ = 5, a₁ = 17. you have correctly solved for λ=-4, and the solution is an=C1(-4)^n+C2*n*(-4)^n which
simplifies to: an=(C1+nC2)(-4)^n a0=5=(c1+0*C2)(-4)^0=C1, so...
Tuesday, April 5, 2011 at 4:11pm by MathMate
you have 8microC total charge. You have 23microF capacitance. Voltage= Qtotal/C=8/23 volts. on C1, q=C1*8/23=7micro*8/23=56/23 micro F on C2, q=16micro*8/23=128/23 microC adding q1+q2, 184/23 microF=
Tuesday, February 14, 2012 at 9:45pm by bobpursley
I meant C3 is in series with C2 and C1 which means that C2 and C1 are in parallel
Wednesday, February 10, 2010 at 6:57pm by Sara
Two capacitors C1 = 7.00 µF and C2 = 11.0 µF are connected in series to a 9.00 V battery. (a) Find the equivalent capacitance of the combination. µF (b) Find the potential difference across each
capacitor. (c) Find the charge on each capacitor.
Monday, October 14, 2013 at 1:47pm by HELP!!!
Two circles C1 and C2 touch externally at A. The tangent to C1 drawn at a point B on the circumference of C1 meets C2 in P and Q so that P is between B and Q. QA is produced and meets C1 at X. Prove
that angle XAB = angle PAB.
Wednesday, September 19, 2012 at 7:48am by Jun
Science ; Micro-organisms
i have to finished this cross word , and i need help on the last two . So thank you if you can solve it :) A substance that micro organisms can be grown on . its a 4 letter word. A group Of Micro
Organisms , its a 6 letter word.
Thursday, April 16, 2009 at 1:55pm by Saffron.
Ceq(c1,c2):3uF. This combination is in series with the 3uF capacitor and both are connected to the 15v battery. The voltage will get divided equally since they are both of the same value(7.5v). Hence
energy would be. Both c1 and c2 have 7.5v across them since they are in ...
Wednesday, February 10, 2010 at 6:57pm by Venkatarama
What is the force exerted on 40 micro coulombs by 20 micro coulombs? (charges are separated by 80cm)
Wednesday, May 2, 2012 at 6:16pm by paul
Calculate the magnitude and direction of the Coulomb force on each of the three charges connected in series: 1st - 6 micro Coulombs (positive charge) 2nd - 1.5 micro Coulombs (positive charge) 3dr -
2 micro Coulombs (negative charge) Distance between 1st and 2nd - 3cm Distance...
Wednesday, May 16, 2012 at 11:19pm by andrew
Calculate the magnitude and direction of the Coulomb force on each of the three charges connected in series: 1st - 6 micro Coulombs (positive charge) 2nd - 1.5 micro Coulombs (positive charge) 3dr -
2 micro Coulombs (negative charge) Distance between 1st and 2nd - 3cm Distance...
Monday, May 21, 2012 at 7:35am by andrew
A C1= 2.50uF capacitor is charged to 862 V and a C2= 6.80uF capacitor is charged to 660 V . These capacitors are then disconnected from their batteries. Next the positive plates are connected to each
other and the negative plates are connected to each other. What will be the ...
Friday, March 8, 2013 at 6:32pm by Steve
An infinite sheet with a surface charge density of -2.1 micro-coulombs/m2 runs parallel to the x axis. Another infinite sheet a surface charge density of +2.1 micro-coulombs/m2 runs parallel to the y
axis. A charge of -4.1 micro-coulombs moves at a 45 degree angle towards the ...
Friday, February 10, 2012 at 8:33am by Mandaleigh
The drawing shows two fully charged capacitors (C1 = 7 µF, q1 = 4 µC; C2 = 16 µF, q2 = 4 µC). The switch is closed, and charge flows until equilibrium is reestablished (i.e., until both capacitors
have the same voltage across their plates). Find the resulting voltage across ...
Tuesday, February 14, 2012 at 9:45pm by Laura
two charges of +2.6 micro c and -5.4 micro C experience an attractive force of 6.5 mN.what is separation between the charges?
Sunday, February 12, 2012 at 10:43am by sheldan
Suppose that there is a common resource of size y in a two period society. Each of two citizens, one and two, can withdraw a nonnegative amount c1 or c2 for consumption in period one, provided that
c1+c2 <=y . In the event that they attempt to consume in excess of what is ...
Wednesday, September 14, 2011 at 3:11pm by quynh
Capacitors in parallel have the same voltage V applied to each. The stored charge on each is CV. Since V is the same the ratio of charges is the ratio of capacitances, C. Q1/Q2 = C1/C2 = 1.5
Friday, February 4, 2011 at 2:12am by drwls
Two small conducting spheres are identical except that sphere x has a charge of -10 micro coulombs and sphere y has a charge of + 6 micro coulombs . After the spheres are brought in contact and then
separated,what is the charge on each sphere ,in micro coulombs ?
Thursday, August 18, 2011 at 5:21pm by Troy
What is the potential halfway between two point charges spaced 1 mm apart if q1=10 micro culumbs and q2= -5 micro culumbs
Wednesday, February 15, 2012 at 8:28pm by physics
Calculus Antiderivative Problem
v = 1/3 (t+2)^2 + c1 x = 1/6 (t+2) + c1*t + c2 now plug in the given conditions when t=0 to find c1 and c2
Friday, November 8, 2013 at 7:29pm by Steve
physics HELP MEEE SO CONFUSED
What is the potential halfway between two point charges spaced 1 mm apart if q1=10 micro culumbs and q2= -5 micro culumbs
Wednesday, February 15, 2012 at 9:03pm by bel
The capacitor with the dielectric (#2) will have 4.2 times the C-value of the air-gap capacitor (#1). Let's call the capacitances C1 and C2 = 4.2 C1. Since the energies (E) are equal, E1 = (1/2) C1 V
^2 = E2 = (1/2) C2 V2^2 C1 * 18^2 = (4.2 C1) * V2^2 V2^2 = (1/4.2) * 324 Solve...
Monday, September 3, 2007 at 11:02am by drwls
A series circuit contains a resistor with R = 24 ohms, an inductor with L = 2 H, a capacitor with C = 0.005 F, and a generator producing a voltage of E(t) = 12 sin(10t). The initial charge is Q =
0.001 C and the initial current is 0. (a) Find the charge at time t. 2Q"+24Q'+...
Thursday, July 12, 2007 at 12:49am by COFFEE
a. C = C1*C2/(C1+C2) = 30 uF 33.6*C2/(33.6+C2) = 30 33.6C2 = 1008+30C2 33.6C2-30C2 = 1008 3.6C2 = 1008 C2 = 280 uF In series. b. C1 + C2 = 30 uF 29.3 + C2 = 30 C2 = 30 - 29.3 = 0.7 uF in parallel.
Tuesday, October 15, 2013 at 11:33pm by Henry
What media? Micro what? This sounds like a question from a lab experiment you were supposed to perform. We need the details
Tuesday, February 19, 2008 at 11:28am by drwls
m1=2 kg, c1= 0.107 cal/(g*K)= 448 J/kg•K, T1=273+70=343.15 K, m2=1kg, c2=4185 J/kg•K, T2=273+15=288.15 K Q1 =m1•c1(T1-Tₒ) = m1•c1•T1- m1•c1•Tₒ Q2 = m2•c2• ( Tₒ-T2) = m2•c2•Tₒ- m2•c2•T2 Q1=Q2
m1•c1•T1- m1•c1•Tₒ= m2•c2•Tₒ- m2•c2•T2 m1•c1•T1+ ...
Sunday, July 1, 2012 at 10:06pm by Elena
The solution will be c1 e^k1x + c2 e^k2x where k1 and k2 are the roots of 2x^2+x-4=0. x = (-1±√33)/4 So, y = c1 e(-1+√33)/4 x + c2 e(-1-√33)/4 x y(0) = 0, so c1+c2=0 y'(0) = 1 so (-1+√33) c1 +
(-1-√33) c2 = 1 c1 = 1/(2√33) c2 = -1/(2√33)
Saturday, July 20, 2013 at 9:10pm by Steve
What is the potential halfway between two point charges spaced 1 mm apart if q1=10 micro coulumbs and q2= -5 micro coulumbs?
Thursday, February 16, 2012 at 11:38am by anne
Capacitors in parallel: Give as clear as explanation as possible as to why it is physically reasonable to expect that two identical parallel plate capacitors are placed in parallel ought to have
twice the capacitance as one capacitor.
Sunday, March 30, 2014 at 7:04pm by ikine4
Determine the equivalent capacitance between A and B for the group of capacitors in the drawing. Let C1 = 12 µF and C2 = 4.0 µF.
Friday, March 4, 2011 at 10:37pm by YANA
Calculus II
There IS an antiderivative (integral) of 1/x. It is the natural log-base-e (ln) function. In your case f'(x) = -1/x + C1 f"(x) = - ln x + C1 x + C2 where C1 and C2 are arbitrary constants
Saturday, January 24, 2009 at 11:07pm by drwls
m1 = 10.9•10^-3 kg, v =400 m/s, t(o) = 20oC, m2 = 0.4 kg,. specific heat of the iron and wood, respectively, c1 = 444 J/kg, and c2 = 1700 J/kg. (a) m1•v^2/2 = 10.9•10^-3•(400)^2/ 2 =872 J = QQ/ Q =
m1•c1•Δt, Δt = Q/ m1•c1 = 872/10.8•10^-3•444 = 180oC. t(new) = 20oC...
Thursday, April 26, 2012 at 1:12am by Elena
Great! I understand now that C0=f(a) c1=f'(a) c2=f"(a)/2 Now, If I am to find the parabolization of the equation x^2-x at x=2, then c0=x^2-x=2^2-2=2 c1=2x-1=2(2)-1=3 c2=2/2=1 So, the equation (taken
from c0+c1(x-a)+c2(x-a)^2) is >> 2+3(x-2)+1(x-2)^2?? Is this correct? ...
Monday, November 13, 2006 at 8:17pm by Matt
You can count the number of (micro) craters on the surface of the rock. Knowing the age of te rock, you then know the flux of micro meteorites. Using this information you can deduduce how old a
surface is by counting the number of (micro) craters per unit area
Wednesday, March 12, 2008 at 5:50pm by Count Iblis
Physics please help
What is the potential halfway between two point charges spaced 1 mm apart if q1=10 micro coulombs and q2= -5 micro coulombs? Attempt at solution: no idea but need to get this right our homework is
Wednesday, February 15, 2012 at 9:30pm by bel
The potential diffrence across a series combination of 4(micro)F and 12(micro)F capacitor is 240V. What is the potential diffrence across 4(micro)F capacitor?
Wednesday, January 11, 2012 at 11:33am by Nuwaya
Ct = C1*C2/(C1+C2) = 8.3*2.9/(8.3+2.9) = 2.15 uF Qt = Ct*E = 2.15 * 24 = 51.6 uC.=Q1=Q2. Energy=0.5C*E^2 = 0.5*2.15*24^2 =619.2 Joules Parallel Combination: Ct = C1 + C2 = 8.3 + 2.9 = 11.2 uF. Energy
= 0.5*Ct*E^2 = 619.2 Joules 0.5*11.2*E^2 = 619.2 5.6E^2 = 619.2 E^2 = 110.6 E...
Wednesday, October 16, 2013 at 2:01pm by Henry
Discrete Math
Solve the recurrence relation: an+1 = 7an – 10an-1, where n ≥ 2, and given a₁ = 10, a₂ = 29. You have done the first step to give: λ1=2, λ2=5. So the function is defined by the powers of λ1 and λ2,
with the associated constants C1 ...
Tuesday, April 5, 2011 at 4:11pm by MathMate
electronics technician
There is no way I will answer all of these. If you try yourself someone might help. The first one is just common sense. Capacitance is charge per unit voltage. With two capacitors in parallel they
are both at the same voltage, so each has charge corresponding to that voltage, ...
Tuesday, June 18, 2013 at 9:52am by Damon
A point charge of +10 micro-coulombs lies at x = -0.39 m, y = 0 m, a point charge of -5.7 micro-coulombs lies at x = +0.39 m, y = 0 m, and a point charge of +14.9 micro-coulombs lies at x = 0 m, y =
-0.3 m. A point charge of +10 micro-coulombs is moves from x = 0 m, y = +0.11 ...
Friday, February 10, 2012 at 8:32am by Mandaleigh
Electrons circuits
Hi everyone! I'm new in here. Recently I bought a movement monitor which detects micro movement. If the sensor does not detect a movement in 20 seconds, the build-in speaker will beep until the
sensor detects a movement. I'm using it for my project and I have a question about ...
Friday, April 4, 2008 at 11:26pm by Jacky
Solve the initial-value problem. Am I using the wrong value for beta here, 2sqrt(2) or am I making a mistake somewhere else? Thanks. y''+4y'+6y=0, y(0)=2, y'(0)=4 r^2+4r+6=0, r=(-4 +/- sqrt(16-4(1)
(6))/2 r=-2 +/- sqrt(2)*i , alpha = -2, beta = 2(sqrt(2)) y=e^-2x*(c1*cos(sqrt(2...
Thursday, July 12, 2007 at 12:15am by COFFEE
Please look at my work below: Solve the initial-value problem. y'' + 4y' + 6y = 0 , y(0) = 2 , y'(0) = 4 r^2+4r+6=0, r=(16 +/- Sqrt(4^2-4(1)(6)))/2(1) r=(16 +/- Sqrt(-8)) r=8 +/- Sqrt(2)*i, alpha=8,
Beta=Sqrt(2) y(0)=2, e^(8*0)*(c1*cos(0)+c2*sin(0))=c2=2 y'(0)=4, c2=4 y(x)=e^(...
Tuesday, July 10, 2007 at 2:08pm by COFFEE
2 circles C1 and C2 meet at the points A and B. Points C and Dare points on the circumferences of C1 and C2 respectively such that angle CAB = angle DAB. CD cuts C1 and C2 at the points Y and X
respectively. if Z is the foot of the perpendicular from X to BY, prove that angle ...
Wednesday, September 19, 2012 at 7:30am by Jun
1) 2 spheres of equal diameters have 15 micro coulomb charge and -10 micro coulomb charge respectively. if they are placed 5cm apart what will be the force of attraction between them? 2) a charge of
.5 micro coulomb is placed in an electric field whose intensity is 4x10^5n/c...
Sunday, October 10, 2010 at 2:13pm by Deyo
Do you think lotteries have both micro and macro economic effects or only micro? How do lotteries change what and for whom goods and services are produced?
Sunday, September 18, 2011 at 5:20pm by Emily
Discrete Math
For 9:05 pm an=(C1+nC2)1^n=C1+nC2 a0=40 => C1=40 a1=31 => 31=(40+1*C2) => C2=-9 Maybe you should show your work how you got C2=-3
Tuesday, April 5, 2011 at 4:11pm by MathMate
Differential equations, initial value problem. The general equation of motion is: mx"+Bx'+kx=f(t), where the independent variable is t, and the displacement x is the dependent variable. In this case,
external force f(t)=0, so mx"+Bx'+kx=0 substitute m=0.25, k=4, B=1 we have a ...
Wednesday, March 16, 2011 at 1:54pm by Anonymous
Okay I understand it now I have to make them have the same denominators so I multiply 1/C by the numerator and denominator by C2 so it would be C2/(C2*C) and the other part is 1/C2 times C which
would be C/(C2*C) which would make the problem 1/C1= C2-C/(C2*C) what do I do next?
Sunday, October 12, 2008 at 11:35am by Shawn
A charge of -6.5 micro-Coulombs is placed at the origin of a coordinate system. Another charge of -7.6 micro-Coulombs is placed at x = +0.29 m, y = +0.17 m. A third charge of +14 micro-Coulombs is
placed at x = -0.29 m, y = 0 m. What is the magnitude of the total electric ...
Saturday, January 29, 2011 at 12:06pm by kelly
Two charges are arranged on the x axis so that there is a positive charge of 3.6 micro C at the orgin and a negative charge of -2.5 micro C located 2 centimeters to its right. to the nearest
centimeter, where on the x axis should a positive charge be placed so that the net ...
Friday, March 21, 2008 at 12:44pm by Maria Lewis
Two charges are arranged on the x axis so that there is a positive charge of 3.6 micro C at the orgin and a negative charge of .2.5 micro C located 2.5 centimeters to its right. to the nearest
cetimeter, where on the x axis should a positive charge be placed so that the net ...
Friday, March 21, 2008 at 1:04pm by Maria Lewis
Find f'(x) and see if there is a zero (root) on the interval. If not, the extrema are the two limits of the interval. Check the sign of f'(x) on the interval. If there is no zero on the interval, the
sign should not change (+ OR -). If f'(x) is + on the interval, it is ...
Thursday, November 4, 2010 at 7:00am by MathMate
It is understood to be mass/volume for ease of dosage calculations. Therefore whether it is 0.9% saline or simply water, the proportions are the same. You can do the calculations as follows: Suppose
we need v ml of c1% concentrated dextrose diluted with 1000 ml of liquid to ...
Saturday, September 18, 2010 at 4:38pm by MathMate
Replace C2 by 2C1 E = 1.4 log 2C1/C1 = 1.4 log 2 If C1<C2 then C1/C2 < 1 and the log is < 0 I assume negative energy needed means energy is released.
Thursday, March 8, 2012 at 8:35pm by Steve
what is the electric field at the position 20 cm from Q2 and 60 cm from Q1? Q2=40 micro coulombs Q1=20 micro coulombs Q2 and Q1 are separated by 80 cm
Wednesday, May 2, 2012 at 7:20pm by quinton
Physics 2
2 spheres of equal diameters have a positive 15 micro column and negative 10 micro column charge if they are place 5 cm apart what would be the force of attraction between them.
Tuesday, January 28, 2014 at 6:40am by schenncyl
c1•m1 •(310-25.4) =c2•m2•(25.4-20.2) m1= { c2•m2•(25.4-20.2)}/ c1•(310-25.4) = ={4180•1.1•5.2}/385•284.6 = 0.218 kg
Thursday, August 9, 2012 at 10:38pm by Elena
Calculate the coulomb force on q. They are all in a straight line q1, q2,q3. Q1:1.05 micro coulombs Q2:5.51 micro coulombs Q3:-2.15 micro coulombs The distance, r1, between q1 and q2 is 20.1cm. The
distance, r2, between q2 and q3 is 15.3 cm. Also, calculate e coulomb force on q2
Tuesday, July 9, 2013 at 9:21am by Amalia
take integrals across the equation 2ff' - 2gg' = 0 ∫2f df - ∫2g dg = 0 (f^2 + c1) - (g^2 + c2) = 0 f^2 - g^2 = -(c1+c2) = c
Sunday, October 21, 2012 at 2:13pm by Steve
physics 2
A charge of -6.8 micro-Coulombs is placed at x = 0, y = 0. A charge of -18.5 micro-Coulombs is placed at x = 0, y = +37 cm. Another charge of -12.9 micro-Coulombs is placed at x = +17 cm, y = 0 cm.
What is the angle the total electric force on the charge at x = +17 cm, y = 0 ...
Friday, January 18, 2013 at 2:18pm by Jessica
Discrete Math
What method are you expected to use to solve the recurrence relation? Have you covered characteristic polynomials or generating functions? You seem to give me the hint that you have done
characteristic polynomials, so I will solve this by characteristic polynomials. The key is...
Tuesday, April 5, 2011 at 4:11pm by MathMate
Diff eqn- IVP
Differential equations, initial value problem. The general equation of motion is: mx"+Bx'+kx=f(t), where the independent variable is t, and the displacement x is the dependent variable. In this case,
external force f(t)=0, so mx"+Bx'+kx=0 substitute m=0.25, k=4, B=1 we have a ...
Thursday, March 10, 2011 at 9:45pm by MathMate
I assume in ax=bt, x is a subscript and not a variable. 'a' will be used from here on. Since initial conditions are all known, the function X(t) can be found by integrations. b=3.0 m/s³ a(t) = bt V
(t) = ∫a(t) + C1 = (b/2)t² + C1 since V(0)=Vo=0, C1=0, therefore V...
Thursday, October 8, 2009 at 7:34am by MathMate
Here goes... 1. Consider a model in which an individual lives two periods: this period (time one) and next period (time two). This period his budget constraint requires that his consumption, c1; plus
his saving, s; equals his income, y1: c1 + s = y1: Next period his budget ...
Sunday, September 30, 2012 at 9:25pm by Wesley
What is the electric field at a point midway between a -7.34 micro coulombs and a +4.49 micro coulombs charge apart 31.5m apart?
Tuesday, July 9, 2013 at 9:17am by Amalia
what is the electric field at the center of the square? q1 and q2 = -10 micro coulombs q3 and q4= 5 micro coulombs each side is 0.10 m. How do i set this up?
Thursday, July 7, 2011 at 6:18pm by Sarah (C)
- Physic college
The equation for the change of position of a train starting at x = 0 is given by x=1/2at^2 + bt3 . Find the dimensions of a and b. If y = C1 sin (C2 t) where y is a distance and t is the time. What
are the dimensions of C1 and C2 ?.
Friday, October 1, 2010 at 7:03am by meera
Pages: 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 | 11 | 12 | 13 | 14 | 15 | Next>> | {"url":"http://www.jiskha.com/search/index.cgi?query=Two+capacitors%2C+C1+%3D+20+%26micro%3BF+and+C2+%3D+5.0+%26micro%3BF%2C+are+connected+in+parallel+and+charged+with+a+150+V+power+supply","timestamp":"2014-04-18T07:11:03Z","content_type":null,"content_length":"44762","record_id":"<urn:uuid:e2324aa6-ae01-4213-bd04-185cdedf42c5>","cc-path":"CC-MAIN-2014-15/segments/1397609532573.41/warc/CC-MAIN-20140416005212-00023-ip-10-147-4-33.ec2.internal.warc.gz"} |
R Programming/Factor Analysis
From Wikibooks, open books for an open world
Factor analysis is a set of techniques to reduce the dimensionality of the data. The goal is to describe the dataset with a smaller number of variables (ie underlying factors). Factor Analysis was
developed in the early part of the 20th century by L.L. Thurstone and others. Correspondence analysis was originally developed by Jean-Paul Benzécri in the 60's and the 70's. Factor analysis is
mainly used in marketing, sociology and psychology. It is also known as data mining, multivariate data analysis or exploratory data analysis.
There are three main methods. Principal Component Analysis deals with continuous variables. Correspondence Analysis deals with a contingency table (two qualitative variables) and Multiple
correspondence analysis is a generalization of the correspondence analysis with more than two qualitative variables. The major difference between Factor Analysis and Principal Components Analysis is
that in FA, only the variance which is common to multiple variables is analysed, while in PCA, all of the variance is analysed. Factor Analysis is a difficult procedure to use properly, and is often
misapplied in the psychological literature. One of the major issues in FA (and PCA) is the number of factors to extract from the data. Incorrect numbers of factors can cause difficulties with the
interpretation and analysis of the data.
There are a number of techniques which can be applied to assess how many factors to extract. The two most useful are parallel analysis and the minimum average partial criterion. Parallel analysis
works by simulating a matrix of the same rank as the data and extracting eigenvalues from the simulated data set. The point at which the simulated eigenvalues are greater than those of the data is
the point at which the "correct" number of factors have been extracted. The Minimum Average Partial criterion uses a different approach but can often be more accurate. Simulation studies have
established these two methods as the most accurate. Both of these methods are available in the psych package under the fa.parallel and the VSS commands.
Another issue in factor analysis is which rotation (if any) to choose. Essentially, the rotations transform the scores such that they are more easily interpretable. There are two major classes of
rotations, orthogonal and oblique. Orthogonal rotations assume that the factors are uncorrelated, while oblique rotations allow the factors to correlate (but do not force this). Oblique rotations are
recommended by some (e.g. MacCallum et al 1999) as an orthogonal solution can be obtained from an oblique rotation, but not vice versa.
One of the issues surrounding factor analysis is that there are an infinite number of rotations which explain the same amount of variance, so it can be difficult to assess which model is correct. In
response to such concerns, Structural Equation Modelling (SEM), which is also known as Confirmatory Factor Analysis (CFA) was developed by Joreskeg in the 1970's. The essential principle of SEM is
that given a model, it attempts to reproduce the observed covariance matrix seen in the data. The ability of a model to reproduce the data can be used as a test of that model's truth. SEM is
implemented in R in the sem and lavaan packages, as well as the OpenMx package (which is not available on CRAN).
See the following packages : FactoMineR (website), amap, ade4, anacor, vegan, '"psych"'
Principal Component Analysis (PCA)[edit]
PCA deals with continuous variables
• prcomp() in the stats package.
• princomp() in the stats package.
• PCA() (FactoMineR)
• See also factanal()
• See also fa and prcomp in the psych package
N <- 1000
factor1 <- rnorm(N)
factor2 <- rnorm(N)
x1 <- rnorm(N) + factor1
x2 <- rnorm(N) + factor1
x3 <- rnorm(N) + factor2
x4 <- rnorm(N) + factor2
mydat <- data.frame(x1,x2,x3,x4)
pca <- prcomp(mydat)
plot(pca) # plot the eigenvalues
biplot(pca) # A two dimensional plot
pca2 <- princomp(mydat)
pca2 <- princomp(~ x1 + x2 + x3 + x4, data = mydat) # princomp with a formula syntax
Correspondence Analysis (CA)[edit]
Correspondence analysis is a tool for analyzing contingency tables.
• corresp() MASS
• Michael Greenacre's ca package (JSS article)
• Correspondence Analysis and Related Network (link)
• Quick-R's page (link)
• Simple and Canonical Correspondence Analysis Using the R Package anacor (pdf, JSS article)
• multiv
Multiple Correspondence Analysis (MCA)[edit]
This section is a stub.
You can help Wikibooks by expanding it. | {"url":"http://en.wikibooks.org/wiki/R_Programming/Factor_Analysis","timestamp":"2014-04-20T00:53:23Z","content_type":null,"content_length":"66446","record_id":"<urn:uuid:719b9de2-cd75-4398-b18a-fecccddd0af3>","cc-path":"CC-MAIN-2014-15/segments/1397609537804.4/warc/CC-MAIN-20140416005217-00317-ip-10-147-4-33.ec2.internal.warc.gz"} |
Take This Math Job and Love It.
Why Math?
Last week, a friend of a student was visiting on campus. She told me that although she really liked math, she had never really considered being a math major. Her reason astounded me. She had been
told by her father, a mathematics professor at a state university, that math was worthwhile preparation only for a teaching career! In the lingo of our times, that is SO NOT TRUE!
Unfortunately, though, it is an impression shared by all too many students. They just don't realize that in addition to being an absolutely fascinating and captivating subject, math is also excellent
preparation for a large number of career paths, for a wide range of job titles other than "mathematician."
The MAA has done a good deal over the years to publicize the broad range of career opportunities awaiting those with mathematical training, including publishing some very popular books (101 Careers
in Mathematics, and She Does Math). But in this electronic age, we can all benefit from the efforts of thoughtful colleagues around the country to explain the attractions and benefits of studying our
subject. Here is a sampling of web sites that I turned up recently: | {"url":"http://www.stolaf.edu/people/humke/jobs-info.html","timestamp":"2014-04-18T00:34:11Z","content_type":null,"content_length":"3155","record_id":"<urn:uuid:ca9331da-f351-446f-89be-702bc5118be1>","cc-path":"CC-MAIN-2014-15/segments/1397609532374.24/warc/CC-MAIN-20140416005212-00205-ip-10-147-4-33.ec2.internal.warc.gz"} |
Algebra of sets
From Encyclopedia of Mathematics
2010 Mathematics Subject Classification: Primary: 03E15 Secondary: 28A05 [MSN][ZBL]
Algebra of sets
Also called Boolean algebra or field of sets by some authors. A collection $\mathcal{A}$ of subsets of some set $X$ which contains the empty set and is closed under the set-theoretic operations of
finite union, finite intersection and taking complements, i.e. such that
• $A\in\mathcal{A}\Rightarrow X\setminus A\in \mathcal{A}$;
• $A,B\in \mathcal{A}\Rightarrow A\cup B\in\mathcal{A}$;
• $A,B\in \mathcal{A}\Rightarrow A\cap B\in\mathcal{A}$
(see Section 4 of [Ha]). Indeed it is sufficient to assume that $\mathcal{A}$ satisfies the first two properties to conclude that also the third holds.
It follows easily that an algebra is also closed under the operation of taking differences. Algebras are special classes of rings of sets (also called Boolean rings). A ring of sets is a nonempty
collection $\mathcal{R}$ of subsets of some set $X$ which is closed under the set-theoretic operations of finite union and difference. An algebra can be characterized as a ring containing the set
The algebra generated by a family $\mathcal{B}$ of subsets of $X$ is defined as the smallest algebra $\mathcal{A}$ of subsets of $X$ containing $\mathcal{B}$. A simple procedure to construct $\
mathcal{A}$ is the following. Define $\mathcal{A}_0$ as the set of all elements of $\mathcal{B}$ and their complements. Define $\mathcal{A}_1$ as the elements which are intersections of finitely many
elements of $\mathcal{A}_0$. $\mathcal{A}$ consists then of finite unions of arbitrary elements of $\mathcal{A}_1$.
An algebra of sets that is also closed under countable unions, cp. with Section 40 of [Ha] (also called Boolean $\sigma$-algebra or $\sigma$-field). Analogously one defines $\sigma$-rings (also
called Boolean $\sigma$-rings) as rings of sets which are closed under countable unions (see Section 40 of [Ha]). $\sigma$-algebras of $X$ can be characterized as $\sigma$-rings which contain $X$.
As a corollary a $\sigma$-algebra is also closed under countable intersections. As above, given a collection $\mathcal{B}$ of subsets of $X$, the $\sigma$-algebra generated by $\mathcal{B}$ is
defined as the smallest $\sigma$-algebra of subsets of $X$ containing $\mathcal{B}$ (also called Borel field generated by $\mathcal{B}$). A construction can be given using transfinite numbers. As
above, $\mathcal{A}_0$consists of all elements of $\mathcal{B}$ and their complements. Given a countable ordinal $\alpha$, $\mathcal{A}_\alpha$ consists of those sets which are countable unions or
countable intersections of elements belonging to \[ \bigcup_{\beta<\alpha} \mathcal{A}_\beta\, . \] $\mathcal{A}$ is the union of the classes $\mathcal{A}_\alpha$ where the index $\alpha$ runs over
all countable ordinals (cp. with Exercise 9 of Section 5 in [Ha] where the same construction is outlined for $\sigma$-rings).
Relations to measure theory
Algebras (respectively $\sigma$-algebras) are the natural domain of definition of finitely-additive ($\sigma$-additive) measures. Therefore $\sigma$-algebras play a central role in measure theory,
see for instance Measure space.
According to the theorem of extension of measures, any $\sigma$-finite, $\sigma$-additive measure, defined on an algebra $A$, can be uniquely extended to a $\sigma$-additive measure defined on the $\
sigma$-algebra generated by $A$.
1) Let $X$ be an arbitrary set. The collection of finite subsets of $X$ and their complements is an algebra of sets (so-called finite-cofinite algebra). The collection of subsets of $X$ which are at
most countable and of their complements is a $\sigma$-algebra (so-called countable-cocountable σ-algebra).
2) The collection of finite unions of intervals of the type \[ \{x\in\mathbb R : a\leq x <b\} \qquad \mbox{where } -\infty \leq a <b\leq \infty \] is an algebra.
3) If $X$ is a topological space, the elements of the $\sigma$-algebra generated by the open sets are called Borel sets.
4) The Lebesgue measurable sets of $\mathbb R^k$ form a $\sigma$-algebra (so-called Lebesgue σ-algebra, see Lebesgue measure).
5) Let $T$ be an arbitrary set and consider $X = \mathbb R^T$ (i.e. the set of all real-valued functions on $\mathbb R$). Let $A$ be the class of sets of the type \[ \{\omega\in \mathbb R^T: (\omega
(t_1), \ldots,\omega (t_k))\in E\} \] where $k$ is an arbitrary natural number, $E$ an arbitrary Borel subset of $\mathbb R^k$ and $t_1,\ldots, t_k$ an arbitrary collection of distinct elements of
$T$. $A$ is an algebra of subsets of $\mathbb R^T$ (so-called cylindrical algebra). In the theory of random processes a probability measure is often originally defined only on an algebra of this
type, and then subsequently extended to the $\sigma$-algebra generated by $A$.
[Bo] N. Bourbaki, "Elements of mathematics. Integration", Addison-Wesley (1975) pp. Chapt.6;7;8 (Translated from French) MR0583191 Zbl 1116.28002 Zbl 1106.46005 Zbl 1106.46006 Zbl 1182.28002 Zbl
1182.28001 Zbl 1095.28002 Zbl 1095.28001 Zbl 0156.06001
[DS] N. Dunford, J.T. Schwartz, "Linear operators. General theory", 1, Interscience (1958) MR0117523 Zbl 0635.47001
[Ha] P.R. Halmos, "Measure theory", v. Nostrand (1950) MR0033869 Zbl 0040.16802
[Ne] J. Neveu, "Mathematical foundations of the calculus of probability", Holden-Day, Inc., San Francisco, Calif.-London-Amsterdam 1965 MR0198505 Zbl 0137.1130
How to Cite This Entry:
Algebra of sets. Encyclopedia of Mathematics. URL: http://www.encyclopediaofmath.org/index.php?title=Algebra_of_sets&oldid=30098
This article was adapted from an original article by V.V. Sazonov (originator), which appeared in Encyclopedia of Mathematics - ISBN 1402006098.
See original article | {"url":"http://www.encyclopediaofmath.org/index.php/Algebra_of_sets","timestamp":"2014-04-20T10:49:09Z","content_type":null,"content_length":"29102","record_id":"<urn:uuid:7bb75243-9037-4b9c-a10f-75418d34abf8>","cc-path":"CC-MAIN-2014-15/segments/1397609539705.42/warc/CC-MAIN-20140416005219-00389-ip-10-147-4-33.ec2.internal.warc.gz"} |
Attleboro SAT Tutor
Find an Attleboro SAT Tutor
...I taught Fortran for four semesters at URI/Providence while I was a graduate student in Computer Science at URI. As an undergraduate I wrote a compilation of student programming errors. My
major professor reported that my project speeded up his teaching of the course by 20%. I have a Masters Degree in Computer Science from URI.
14 Subjects: including SAT math, calculus, geometry, statistics
...The Cooperative Admissions Exam (COOP) and the High School Placement Test (HSPT) are entrance exams for admission to Catholic high schools. The COOP consists of seven sections which test a
student's verbal reasoning, reading, and language arts skills, as well as math skills. The HSPT is divided into five sections, and it also tests verbal and math skills.
30 Subjects: including SAT reading, SAT math, SAT writing, English
...Helping students do this at Salve Regina University's Academic Development/Writing Center was the initial step I would take with them when assisting them with extra-curricular help in
Literature courses. Far and away, my best skill and talent is in writing. I am excellent at communicating my thoughts through the written word.
11 Subjects: including SAT reading, SAT writing, English, reading
...I also have lots of hands-on experience with circuits; I have designed circuits for school projects and for my professional work, built my own loudspeaker crossovers, and upgraded the wiring
in my vintage car. I occasionally tutor introductory electrical engineering courses; when I do, students ...
8 Subjects: including SAT math, physics, calculus, differential equations
...I also taught a small group of 7th graders who struggled with math concepts and helped them to become very successful by giving them real-world applications of math: checking accounts,
budgeting, figuring out taxes on their "paychecks", etc. My strengths in working with math students are helpin...
23 Subjects: including SAT writing, SAT reading, Spanish, reading
Nearby Cities With SAT Tutor
Attleboro Falls SAT Tutors
Central Falls SAT Tutors
Cumberland, RI SAT Tutors
East Providence SAT Tutors
Easton, MA SAT Tutors
Franklin, MA SAT Tutors
Mansfield, MA SAT Tutors
North Attleboro SAT Tutors
Norton, MA SAT Tutors
Pawtucket SAT Tutors
Plainville, MA SAT Tutors
Providence, RI SAT Tutors
Rehoboth, MA SAT Tutors
Taunton, MA SAT Tutors
Woonsocket, RI SAT Tutors | {"url":"http://www.purplemath.com/attleboro_sat_tutors.php","timestamp":"2014-04-16T19:03:47Z","content_type":null,"content_length":"23880","record_id":"<urn:uuid:058645ea-b869-4235-aa52-0410b5caae35>","cc-path":"CC-MAIN-2014-15/segments/1397609524644.38/warc/CC-MAIN-20140416005204-00571-ip-10-147-4-33.ec2.internal.warc.gz"} |
Please check here regularly for announcements (especially if you missed class).
Click an item to expand.
In Reverse Chronological Order:
Sat Dec 15, Open book final exam, but no laptops or phones
Unlike the midterm, we will not allow any communication devices for the final exam. So please prepare accordingly.
Fri Dec 14, New Room for Final Exam on Dec 18
Note that the room has been moved to WWH 317 from our regular classroom.
The time is from 3:30 to 7:30.
Tue Oct 2, Fall Break Oct 15-16
There is no class on Oct 16 because of NYU's Fall Break (Oct 15 and 16).
Tue Oct 2, Midterm Reminder
Recall that we had announced a midterm for Oct 11 in the INFO page. To prepare for this, be sure to study our solutions for homeworks. The exam is open book.
Thu Sep 20, Policy for credit on reporting errors in lecture notes
I had announced awarding one point for first report of any error of ANY kind in the lecture notes (more points for significant errors). Now how should I incorporate that into your course grade?
You noticed (see Course Info) that 5\% of the course grade is for class participation, etc. Well, at the end of semester I will convert these points into this 5\%.
Fri Sep 7, Improved DIC
I mentioned this in class yesterday. Define the Default Initial Condition (DIC) this way: there exists C>=0, n_1 >=0 such that the recurrence holds for T(n) holds for n>n_1 and for all n<=n_1, T
(n) is assigned arbitrary values <= C. This is a slight variation of the lecture notes, meant to prevent unbounded values of T(n) when n<=n_1.
Wed Sep 5, Wiki's Toggle Feature fixed
This announcement page is a main user of this toggling feature! Basically, I had to upgrade my pmWiki.
Tue Sep 4, First Class
Please note that webpage has been slightly re-organized (simplified). Note that the classroom is WWH 412, not WWH 312.
Wed Jul 18, Work in Progress
This class page is work in progress. Feel free to send me an email if you have questions about this course. | {"url":"http://cs.nyu.edu/~yap/wiki/class/index.php/Algo/2012f-ann","timestamp":"2014-04-20T00:38:26Z","content_type":null,"content_length":"12904","record_id":"<urn:uuid:6fba6579-ec94-4ca4-a9e5-3da38c1e4047>","cc-path":"CC-MAIN-2014-15/segments/1398223206118.10/warc/CC-MAIN-20140423032006-00589-ip-10-147-4-33.ec2.internal.warc.gz"} |
Got Homework?
Connect with other students for help. It's a free community.
• across
MIT Grad Student
Online now
• laura*
Helped 1,000 students
Online now
• Hero
College Math Guru
Online now
Here's the question you clicked on:
Time (hours) Number of Cells 0 200 .25 400 .5 800 .75 1600 1 3200 Write the rule of the function that gives the number of C cells at time t hours. C(t) =
Your question is ready. Sign up for free to start getting answers.
is replying to Can someone tell me what button the professor is hitting...
• Teamwork 19 Teammate
• Problem Solving 19 Hero
• Engagement 19 Mad Hatter
• You have blocked this person.
• ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.
This is the testimonial you wrote.
You haven't written a testimonial for Owlfred. | {"url":"http://openstudy.com/updates/4eb877c9e4b07fdfbfff029b","timestamp":"2014-04-19T04:47:37Z","content_type":null,"content_length":"77992","record_id":"<urn:uuid:41b2fb78-d9d9-4a97-9d6b-8b69f1c7463d>","cc-path":"CC-MAIN-2014-15/segments/1397609535775.35/warc/CC-MAIN-20140416005215-00621-ip-10-147-4-33.ec2.internal.warc.gz"} |
Physics Forums - View Single Post - Self-learning Trig? calculus?
I'm a sophmore in high school, right now im in honors chemistry and algebra 2, (if you are familiar with science bowl, my school, mira loma, won last year)
I want to learn trig and eventually calculus on my own, with the hopes of understanding highschool and college level physics. I have an old textbook from my dad's study, its by Mcgraw hill and it
says "Plane Trigonometry with Tables"(1974) is this a good textbook to learn from? If not what is a better method and what would be better for learning calculus and physics on my own?
Also, what are the REAL pre-req.s to learn calculus and trig because theres a lot of stuff I hear like Pre Calculus is not really related or used in Calculus, so what do I need to learn then? | {"url":"http://www.physicsforums.com/showpost.php?p=2535234&postcount=1","timestamp":"2014-04-21T14:57:41Z","content_type":null,"content_length":"9028","record_id":"<urn:uuid:3c7ce431-0341-4151-a2ae-c53d5ebb2a2e>","cc-path":"CC-MAIN-2014-15/segments/1397609540626.47/warc/CC-MAIN-20140416005220-00329-ip-10-147-4-33.ec2.internal.warc.gz"} |
A thought on Linear Models on Stocks
April 16, 2012
By enguyen
Timely Portfolio has a nice post about linear models sytems for stock.
The idea follows from the steps below:
• Get the weekly closing values of the S&P 500.
• Choose a time window (i.e. 25 weeks) and for each window, linearly regress the subset of closing values
• Choose an investment strategy based on the residuals, the running average of slope coefficients, or the running average of correlation with data points
The idea is quite simple, and so far, from Timely Portfolio’s post, it looks like the drawdown is behaving nicely.
It seems like the idea could be extended to a non-linear method. The residuals are getting larger and larger, and this indicates that linear methods are less reliable as time goes by.
# code from Timely Portfolio
# http://timelyportfolio.blogspot.ca/2011/08/unrequited-lm-love.html
GSPC <- to.weekly(GSPC)[,4]
width = 25
for (i in (width+1):NROW(GSPC)) {
linmod <- lm(GSPC[((i-width):i),1]~index(GSPC[((i-width):i)]))
ifelse(i==width+1,signal <- coredata(linmod$residuals[length(linmod$residuals)]),
signal <- rbind(signal,coredata(linmod$residuals[length(linmod$residuals)])))
signal <- as.xts(signal,order.by=index(GSPC[(width+1):NROW(GSPC)]))
plot(signal, main="Residuals through time")
plot(log(signal), main="Log of Residuals through time")
for the author, please follow the link and comment on his blog:
DataPunks.com » R
daily e-mail updates
news and
on topics such as: visualization (
), programming (
Web Scraping
) statistics (
time series
) and more...
If you got this far, why not
subscribe for updates
from the site? Choose your flavor:
, or | {"url":"http://www.r-bloggers.com/a-thought-on-linear-models-on-stocks/","timestamp":"2014-04-18T10:42:01Z","content_type":null,"content_length":"36878","record_id":"<urn:uuid:49d26ba3-a266-4c60-91e9-47d9c94f269e>","cc-path":"CC-MAIN-2014-15/segments/1398223205375.6/warc/CC-MAIN-20140423032005-00541-ip-10-147-4-33.ec2.internal.warc.gz"} |
Predicting and controlling infectious disease epidemics using temporal networks
• We are sorry, but NCBI web applications do not support your browser and may not function properly.
More information
F1000Prime Rep. 2013; 5: 6.
Predicting and controlling infectious disease epidemics using temporal networks
Infectious diseases can be considered to spread over social networks of people or animals. Mainly owing to the development of data recording and analysis techniques, an increasing amount of social
contact data with time stamps has been collected in the last decade. Such temporal data capture the dynamics of social networks on a timescale relevant to epidemic spreading and can potentially lead
to better ways to analyze, forecast, and prevent epidemics. However, they also call for extended analysis tools for network epidemiology, which has, to date, mostly viewed networks as static
entities. We review recent results of network epidemiology for such temporal network data and discuss future developments.
Infectious diseases are a major threat to public health—estimated to account for 43% of the global burden of disease (where the burden of health is measured in the number of years of healthy life
lost http://www.who.int/trade/glossary/story036/en/). To mitigate their spreading, we need to understand pathogenesis, environmental factors, and the social structure of contagion. Network theory has
become a valuable framework to study the last one—the role of the contact patterns in epidemics. Social networks among individuals (i.e. humans and animals), as the one shown in Fig. 1 affect the
possibility, extent, and speed of epidemic spreading. Lately, contact pattern data with precise temporal information of the contacts (i.e., timing and duration of contacts) between individuals have
been collected in various situations. The aim in this report is to discuss the relevance of such data, collectively called the temporal networks [1,2], to theoretical epidemiology.
Network epidemiology has proved an indispensable approach for understanding epidemics of infectious disease, often acknowledged in medical epidemiology, mathematical biology, and more recently
network science [3-15]. A recent practical example of network epidemiology is the GLEAMviz platform that succeeded in forecasting the 2009 H1N1 pandemic [16] (also see [17,18]). An implicit
assumption behind most network epidemiology studies is that a link (i.e., contact) between two individuals is an incessant pathway for contagion. This is clearly a strong simplification of reality.
In fact, data possessing precise temporal information of the events of a (potentially contagious) contact between individuals, collected from humans and other animals, are available nowadays. In a
temporal network approach to such datasets, we view a link as dynamic entity and assume that contagion can occur only inside the time window in which the two individuals interact. With this view in
mind, the static approach of network epidemiology may miss a great deal of what is happening in reality.
An early example of such temporal network epidemiology is found in the analysis of sexually transmitted infections (STIs). First of all, STIs lend themselves naturally to network epidemiological
approaches because a contact is well-defined and in many cases stable on the timescale of the epidemics. In a conventional survey, one asks respondents about the sexual partners over some certain
time frame, often spanning several years, to generate a static sexual contact network [19-22]. However, if we consider the temporal aspect of contacts of a given individual, a disease can be
transmitted from an old partner to a new partner, via the focal individual, but not vice versa [21,23-25]. An early attempt to account for the difference between sexual contact patterns—like people
being serial monogamists or having concurrent partnerships—was the concurrency measure proposed by Kretzschmar and Morris [26,27]. This measure is based on the network of contacts that have happened
by a given time. Because the concurrency measure is based on a network of accumulated contacts, it does not incorporate all the temporal information that a temporal network could. Nevertheless, the
philosophy behind it rests on a temporal network thinking and thus Kretzschmar and Morris’s work is a precursor of temporal network epidemiology.
As mentioned above, these days we have access to an increasing amount of temporal network data. The most common instance is data of online social contacts [28-31], although it is perhaps not directly
related to epidemic spreading (other than computer and email viruses). Another temporal network data source of human communication that people have investigated is mobile phone calls [25,32-37]. Yet
more relevant, as it logs actual physical proximity (and thus potential disease-spreading events), is data recorded from humans wearing Bluetooth sensors of mobile phones [38-40], Radio-Frequency
Identification Devices [41,42], and infrared modules [43-46]. In particular, the SocioPatterns project (http://www.sociopatterns.org/) [41,42] collected data from a community hospital [47] and a
primary school [48], where epidemic spreading is an important issue. On a slower timescale, airport networks, which mediate global epidemic outbreaks, are also dynamic [49]. Pertaining to infections
of animals, data obtained from, for example, sheep [50], cattle [51-53], ants [54], and zebra [55] have been analysed in different contexts. As mentioned, this is probably only the beginning—we will
have access to an increasing number of such temporal network datasets in the near future. Then, we will also need a temporal network approach to understand the role of contact patterns in epidemics.
By exploiting the information in temporal networks, we can better describe, predict, and control epidemics than by only relying on static-network epidemiology. Essentially, ignoring temporal
correlations, as a static network approach does, can lead to an over- or underestimation of the speed and extent of an outbreak. Notwithstanding, temporal-network epidemiology, both analytical and
computational, is still in its infancy, and more powerful methods are urgently needed. We will expand on this topic and discuss emergent issues and unsolved problems. We focus on the following two
specific problems in this report.
Effects of temporal networks on epidemic spreading and the need of developing analysis methods
Recent analysis, mainly numerical, showed that statistics of outbreaks (such as their final size and speed of spreading) differ considerably between empirical temporal networks and corresponding
static networks. However, how and when they are different is not sufficiently understood. We describe the results of some key studies, compare them, and identify the problems inherent in the current
temporal network approaches to epidemics.
Disease prevention methods
A basic issue in epidemiology, both in theory and practice, is to develop methods to mitigate disease spreading by immunizing individuals or restrict their interaction with others. By exploiting
temporal information of the contacts, it may be possible to contrive disease prevention strategies that outperform existing ones. We describe very recent approaches in this direction (there are only
a few papers though) and suggest where this kind of study should be directed.
Our main thesis is that the use of temporal contact networks makes it possible to better describe, predict, and control epidemic spreading in many practical situations. In the first main section, we
describe temporal networks and illustrate differences between them and the conventional static networks. In the second main section, we briefly introduce mathematical epidemic models that are often
used in theoretical and numerical studies of epidemic spreading on networks. The third main section describes recent results on epidemic spreading on temporal networks. The fourth main section is
devoted to (preliminary studies of) prevention measures on temporal networks. Finally, we will discuss unsolved issues and future directions.
Temporal networks
From a static social-network viewpoint, a link between a pair of individuals usually indicates that the two individuals interact at least once within a sampling period. The collection of links and
the participating individuals form a network (Fig. 1 shows an example). Epidemic spreading on networks has been an intensively investigated topic involving different research fields [3-15]. In this
section, we will discuss some interesting examples.
The temporal network is an emerging modelling framework for understanding epidemic spreading on social networks [1,2]. The “atoms” making up a temporal network are the contacts (sometimes called
events). A contact is the interaction between two individuals such that a disease can spread from one to another. The contact is either treated as lasting for a certain amount of time or as
instantaneous. A link may contain multiple contacts between the same pairs of individuals occurring at different times. In fact, different contacts on the same link may have different impacts on
dynamics on networks and hence on epidemic spreading [46]. In addition to the time, the duration of contact is an important component of a temporal network that affects epidemic spreading.
Peer-to-peer infection would occur if the contact has a long duration.
An example of a temporal network is depicted in Fig. 2(A) and even this small example clearly shows the importance of considering how networks vary with time. In Fig. 2(A), an infectious disease may
spread from individual A to D via B. However, the converse cannot occur. This is because there is at least one “temporal path” from A to D but not vice versa [1,56-58].
Temporal network and its aggregated counterpart
If we disregard the timing information in the temporal network shown in Fig. 2(A), we obtain the static, aggregated network shown in Fig. 2(B). If we simulate an outbreak on the aggregated network, a
chain of infection from A to D may occur, and vice versa. This is because the static network shown in Fig. 2(B) is undirected (i.e., only bidirectional links are assumed), which corresponds to the
fact that the contacts in Fig. 2(A) are implicitly assumed to be bidirectional. There is no way of encoding all the information contained in Fig. 2(A) into Fig. 2(B). The lesson here is that we may
be misguided if we only use aggregated static networks. One may argue that a real temporal network lasts much longer than that shown in Fig. 2(A), so that temporal paths could exist between any
ordered pair of individuals to provide pathways for pathogens. In fact, this is generally true. However, even in such a case, the likelihood with which a chain of infection from A to D occurs and
that of the inverse chain (i.e. from D to A) are usually different. Furthermore, the fact that pathogens may mutate or other factors affecting the disease spreading may change on a fairly short
timescale is a reason to restrict the sampling time.
Another well-documented phenomenon that occurs in real temporal networks and is relevant to epidemiology is the “bursty” nature of events (i.e. a mixture of short and long intervals between
contacts). In many types of empirical data, the intervals between the contacts involving a certain pair of individuals follow a long-tailed distribution [30,59,60], as shown in Fig. 3(A). Most such
inter-contact times are relatively short, but a few are very long. In addition, successive inter-contact times are often positively correlated. In contrast, an implicit but crucial assumption
underlying most theoretical and numerical analysis of epidemic dynamics in well-mixed populations and static networks is that inter-contact times obey an exponential distribution without correlation
between different contacts. An example contact sequence generated under this assumption is shown in Fig. 3(B). The two sequences shown in Fig. 3 have the same number of contacts. The sequence shown
in Fig. 3(B) lacks bursts, which contrasts with the sequence shown in Fig. 3(A), which shows the bursty nature of real temporal network data. The bursty nature of data may change the current
understanding of epidemic dynamics on networks [61].
Bursty nature of inter-contact times
Epidemiological models
The SIR model
The mathematical epidemiological model that is probably the most widely used for theorizing about and emulating epidemics is the so-called Susceptible-Infected-Recovered (or, succinctly, the SIR)
model [9,62-65]. In the individual-based version of the SIR model, with which we are concerned in the present report, each individual belongs to either a susceptible (S), infected (I), or recovered
(R) state at any given time. The simplest version of the SIR model without demographic factors is defined as follows. When a susceptible individual and an infected individual interact, the former may
be infected at an infection rate, denoted by β (Fig. 4(A)). Precisely speaking, the susceptible individual contracts infection with probability β Δt within a short time interval Δt. With probability
1 – β Δt, the susceptible individuals stays in the susceptible state throughout the interval Δt. Δt can be identified with a discrete simulation time step. If a susceptible individual interacts with
k infected individuals at the same time, which is usually assumed in the case of epidemic simulations on static networks (see Fig. 4(B) for the case of k =3), the susceptible individual is infected
with probability 1 – (1 – β Δt)^k ≈ k β Δt within small time Δt. The interval Δt is chosen to be small enough to make this probability sufficiently small, to avoid the effect of time discretization.
An infected individual is assumed to recover with rate μ, irrespectively of the state (i.e., susceptible, infected, or recovered) of other individuals in the neighborhood. In other words, each
infected individual transits to the recovered state with probability μΔt within time Δt.
In the SIR model, “R” corresponds to the recovered state with immunity. Therefore, once an individual has entered the recovered state, the individual will never be infected again. When the initial
population is a mixture of susceptible and infected individuals (usually the number of infected individuals is initially set to one for modelling purposes), the SIR dynamics stops with a mixture of
susceptible and recovered individuals. Then, the final number of the recovered individuals is called the final (epidemic) size and represents the total damage to the population caused by an epidemic
wave. Mathematically, the “R” state can also signify death if the rate of encounters that an individual makes with other individuals is independent of the population size; the dead individual does
not infect anybody and is not infected by anybody, which is functionally equivalent to a recovered and immune individual. For this reason, the SIR model is widely used to represent infectious
diseases in which one-shot epidemic waves can occur and the mortality is high or the recovery implies immunity. Examples include influenza, SARS, measles, and mumps.
An important contribution of mathematical epidemiology to medicine is the concept of basic reproduction number and epidemic threshold [5,9,64]. The basic reproduction number, conventionally denoted
by R[0], is the expected number of secondary infections of an infected individual in a population of all susceptible individuals. If R[0] exceeds unity, the epidemics can spread to a finite fraction
of the population, meaning that epidemic spreading is a threshold phenomenon. The epidemic threshold is equivalent to R[0] = 1, where R[0] = βk/μ. Many studies have measured R[0] for different
diseases and societies even though it is only directly related to an epidemic threshold in the context of the well-mixed SIR model. In temporal networks both the temporal structure and the network
topology can move the threshold away from R[0] = 1.
The SEIR model
To explore more realistic models, one often uses extensions and variants of the SIR model, generally termed compartmental models [5,9,64]. One such example is the
susceptible-exposed-infected-recovered (SEIR) model, whereby a new state “E” (for exposed) is added between susceptible and infected. The interaction between susceptible and infected individuals
causes the former to transit to the latent exposed state. An exposed individual is infected but not infectious. After staying in the exposed state for some time, the exposed individual transits to
the infected state to be able to infect others.
The SI model
For theoretical purposes, the susceptible-infected (SI) model is also used. The SI model is unrealistic in that an individual stays in the I state forever, but it captures the incipient stage of an
outbreak. The SI model is devoid of the R state. Therefore, if there is at least one infected individual in the population, the SI dynamics necessarily ends up with the entire infected population.
The SI model is used to investigate transient dynamics, in particular, the speed of infection in the initial and middle stages of epidemic spreading. For example, the analysis of the SI model on
static networks revealed that the initial epidemic spreading is faster when the individuals have different numbers of contacts; (see Fig. 1 for an example) than when they all have the same number of
contacts [66,67]. When infected individuals are rare in an early stage, the SI and SIR models behave almost the same because there are few recovered individuals.
Epidemic spreading on temporal networks
In this section, we describe the main findings regarding epidemic spreading on temporal networks. Explanation of this important topic has only recently started using a combination of real data,
numerical simulations, and theory. A major research strategy for now is to compare the results (e.g. final size and spreading speed) obtained for temporal networks with those for randomized networks.
Randomized networks are temporal or static networks generated by randomizing the original temporal networks, with some of the properties of the original networks conserved. A network obtained by
randomizing the times of each contact is such an example. This randomization preserves the number of contacts on each link and the structure of the aggregated network, whereas it destroys the
temporal structure of the contact sequence on each link. Randomized networks serve as null models. If epidemic spreading is different between the original and randomized networks, we can infer the
properties of temporal networks that are responsible and those that are not responsible for the observed difference [2].
In the following, we will introduce the results obtained in the last few years. We note that the fact that temporal network effects influence epidemic spreading has been recognized longer than this,
see e.g. [26,27,50,52,68-71]. However, systematically exploring different effects of temporal networks, or the corresponding aggregated networks, on epidemic spreading is a more recent theme. We
survey some key studies that explored effects of temporal networks on the final size and time evolution of epidemic spreading. In order not to digress from the main points, we do not mention other
aspects (e.g., temporal networks yield longer persistence of infection than the aggregated network in late stages of spreading [61,72-74]). We also briefly mention mathematical modelling (see [75]
and references therein).
Karsai et al. numerically simulated the deterministic SI dynamics on the temporal network data of phone calls between people [76]. By the deterministic SI dynamics, we mean that in this model a
contact between a pair of a susceptible individual and an infected individual always causes infection; this assumption reduces the number of parameters to be investigated. The mobile phone data
contain about 4.5 million individuals, 9 million links, and 31 million events (i.e. phone calls). They compared transient spreading dynamics in temporal networks and that in randomized networks,
whereby their different randomization methods destroy temporal structure of the data to different extents. The authors showed that epidemic spreading slows down on temporal networks compared with
different randomized networks. They also corroborated their results with two other datasets, but without going into so much detail as the phone-call data. Similar conclusions are supported by their
follow-up study [77], an approach based on the so-called average temporal path length [58], and different numerical simulations of the susceptible-infected-susceptible (SIS) model (whereby the
recovered individual turns back to the susceptible state and is possibly reinfected, representing endemic diseases) on artificial temporal network data [78]. In contrast, Rocha et al. [79] observed
that randomizing the time stamps sped up disease spreading in a network of sexual contacts between prostitutes and their sex buyers. The cause of this difference is still an open question. One
hypothesis to be tested is that while the links of the dataset analysed in [76,77] are typically active throughout the sampling time, the links in Rocha et al.’s data last only for a short period
(still, within this period, the contacts can be bursty as seen in Fig. 3(A)). An interesting general question is to identify temporal network structure that speeds up or slows down disease spreading.
The results in [58], with a dataset different from the one analysed in [76,77], also suggest that the temporal nature of networks enhances rather than suppresses epidemic spreading (figure 6b in [58
]). Also in a variant of the SI model, in which multiple infection attempts by infected individuals within a short time is necessary for a susceptible individual to be infected, epidemic spreading is
more facilitated by empirical temporal structure compared to randomized reference data [80].
Miritello et al. numerically investigated stochastic SIR dynamics with a fixed recovery time on a mobile phone dataset containing 20 million individuals [37]. They observe that, when the per-contact
infection probability (corresponding to infection rate β shown in Fig. 4(A)) is large, the final size of the infected population for the temporal network data is smaller than that for the randomized
temporal network, consistent with the results in [76]. Because of the bursty nature of the data (Fig. 3(A)), newly infected individuals have to wait longer for the next encounter with a susceptible
individual, which opens a pathway for infection. Therefore, the authors conclude, a global outbreak is suppressed on this type of temporal network structure [37]. This mechanism should be common to
the system studied in [76]. In contrast, at small per-contact infection probabilities, where the final size of the infected population is presumably small, temporal networks yield a larger final size
than the randomized networks do. Different numerical simulations with the SI model support similar conclusions [81]. Miritello et al. [37] conclude that group conversation contacts (i.e.,
correlations between the contact sequences of adjacent links–see Fig. 5–which are also treated in [77]) inherent in the data are responsible for the promotion of local contagion at a low per-contact
infection probability. Because of the correlated occurrence of the contacts on adjacent links, a newly infected individual tends to meet a susceptible “prey” sooner in the original temporal network
than in the case of the randomized networks.
Correlated contact sequences on adjacent links
Stehlé et al. investigated stochastic SEIR epidemics on the temporal network recorded in a two-day medical conference [82]. They numerically showed that the final size of the infected population does
not differ between the original temporal network and the aggregated network. The origin of the discrepancy in the results in different studies is unclear. Even mathematical analysis of epidemic
spreading models with dynamic contacts (without using real data) yields opposite results. In a model based on the SIR dynamics, temporal dynamics of links speeds up epidemic spreading [83], whereas
another SIS-based model with dynamic contact changes concludes the converse [84].
Prevention of epidemic spreading using temporal networks
One potentially important application of temporal network studies is to construct efficient immunization protocols. Faced with an evolving outbreak, society might only have resources to produce a
limited amount of vaccine. This might not be such a big problem in practice because only a fraction of individuals in a population needs to be immunized for stopping epidemics (in other words, to
achieve herd immunity). Mathematically we can phrase the problem as “how can we immunize a given fraction of the population to mitigate the disease as effectively as possible?” Two things are
important to understand in coping with the immunization problem.
Firstly, in practice, we can only use local information. In other words, we can obtain information about specified individuals and their experience, but hardly map out all contacts of the entire
population. In [85], Cohen et al. proposed an elegant method for static networks where one picks individuals at random, asks them to name an acquaintance that they meet regularly (such that the
disease can be transmitted from one to the other), and immunize the acquaintance. With this procedure, so-called the neighbor immunization (sometimes, this procedure is known as ring vaccination, but
this terminology is not always consistent), the probability of immunizing individuals with many network neighbors—obviously important for preventing disease spreading—increases.
Secondly, the immunization problem would benefit from accurately predicting future contacts and is therefore a temporal problem. Therefore, temporal information can be exploited in containing an
outbreak. To this end, Lee et al. [86] extended the neighbor immunization protocol for static networks [85] to temporal networks. One of their protocols works by asking a randomly chosen individual
to name the acquaintance that he or she met most recently. Lee et al. showed that this protocol outperforms the standard neighbor immunization on several empirical temporal networks [86].
In the context of mobile phone malware epidemics, Tang et al. examined a containment strategy based on temporal networks [87]. Their intervention method was to propagate a patch message, the
reception of which immunized a mobile phone against the malware. In the context of infectious disease spreading of humans, this protocol would be analogous to word-of-mouth type information spreading
combined with voluntary means of lowering the probability of acquiring the infection, such as vaccination, pre-exposure prophylaxis [88], use of condoms, and hand washing. They showed that the
so-called temporal closeness centrality, which is a score calculated for each node and intuitively quantifies the abundance of epidemic pathways from/to the node, is an efficient measure for
prioritizing the mobile phones into which the patch message is injected.
We are not aware of other immunization methods that exploit both the network topology and the temporal structure of contacts. As the studies mentioned above suggest, exploiting temporal information
likely leads to improvement of network-based immunization protocols. This idea is currently an understudied question. Various link-prediction techniques, in which occurrence of contacts in the future
is predicted based on the temporal network data at hand [89-91], may offer a promising approach to this problem.
What does the future hold?
The results of studying disease spreading and its prevention on temporal networks seem to be model dependent. In addition, they are likely to also depend on datasets. So far, most studies have
employed different models and datasets and this complicates the comparison between different studies and any comprehensive understanding of them. To better understand and intervene in epidemic
spreading using temporal networks, we need a more systematic comparison across different epidemic models and datasets. To this end, platforms that make network data available for academic researchers
are very valuable. Some projects generously make their data, including temporal network data, open for academic use (e.g. SocioPatterns project http://www.sociopatterns.org/datasets/ and Stanford
Large Network Data Collection http://snap.stanford.edu/data/).
Other future challenges include exploring the intersection between temporal networks and metapopulation networks. The metapopulation network is a modeling framework in which a node (called the
metapopulation) in a network is a place (like home, school, or workplace) rather than an individual. Individuals move from one metapopulation to another, and at the same time, disease transmission
can occur between individuals in the same metapopulation. Such a framework has been investigated in mathematical epidemiology for a couple of decades [64, 92, 93], and more recent network
epidemiology has deepened the results and also established strong linkage between data and models [15]. This framework was also used by the GLEAMviz project to predict the H1N1 pandemic [16]. Human
travel certainly has temporal components [33, 94], and links of metapopulation networks may also be dynamic, but on a slower timescale [49]. A recent study reports that introduction of dynamics to
links changes the final size of the SIR model on metapopulation networks [95]. Applying temporal network analysis of epidemic spreading to metapopulation networks may thus be a fruitful direction to
Related to the intersection between the metapopulation framework and temporal networks is the analysis of the effects of travel restriction. A majority of such studies, many of which are based on
metapopulation networks, suggest that travel restriction is a poor strategy to contain epidemic outbreaks [96-103]. There are, however, dissonant results that find travel restriction effective [103,
104]. Traveling is a highly temporal phenomenon, such that travel restriction implemented at right moments of time may be effective in containing epidemic spreading.
It may also be an important question to determine the equivalent of the basic reproduction number R[0] and epidemic threshold for temporal networks. It is recently debated that R[0] is misused in
situations to which its original definition does not apply or when its reliable estimation is difficult in practice [65,105-107]. Another interesting question may be what the traditional basic
reproduction number tells us about outbreaks in temporal networks. Even more fundamental perhaps, it would be interesting to examine how much improvement temporal-network methods can make on
static-network models in fitting to empirical outbreak data.
Another challenge for applications is to identify specific pathogens and situations for which temporal network approaches are useful. Although social contacts are dynamic in general, analytical and
computational methods for static network epidemiology are more powerful than those for temporal networks. Therefore, we can be content with results derived from static network epidemiology in the
limit of slow network dynamics relative to the epidemic dynamics. In contrast, typical childhood diseases such as mumps, measles, and pertussis can transmit infection upon short and non-intense
contacts between individuals, such that descriptions under the temporal network framework may be useful. Another possible example is foot-and-mouth disease, which is highly contagious among
livestocks [108]. Contact patterns of livestocks such as cattle and sheep form temporal networks because they move among premises because of trading [50-53]. It could furthermore be interesting to
study improvements of medical practice that today operate on the contact network, such as contact tracing (where individuals testing positive for a pathogen have to name their recent contacts so that
they can be called for testing). Perhaps temporal-network methods can give improved estimates of how far back in time it is necessary to pursue the tracing.
We acknowledge Taro Takaguchi for critical reading of the manuscript. We acknowledge financial supports provided through Grants-in-Aid for Scientific Research (No. 23681033) from MEXT, Japan (NM) the
Swedish Research Council and the WCU program through NRF Korea funded by MEST R31–2008–10029–0 (PH).
The authors declare that they have no disclosures.
1. Kempe D, Kleinberg J, Kumar A. Connectivity and inference problems for temporal networks. Proc STOC. 2000:504–13.
Holme P, Saramäki J. Temporal networks. Phys Rep. 2012;519:97–125. doi: 10.1016/j.physrep.2012.03.001. [Cross Ref]
Albert R, Barabási A-L. Statistical mechanics of complex networks. Rev Mod Phys. 2002;74:47–97. doi: 10.1103/RevModPhys.74.47. [Cross Ref]
4. Barabási A-L. Linked – The New Science of Networks. Cambridge: Perseus Publishing; 2002.
5. Giesecke J. Modern Infectious Disease Epidemiology. 2. London: Arnold; 2002.
Newman MEJ. The structure and function of complex networks. Siam Rev. 2003;45:167–256. doi: 10.1137/S003614450342480. [Cross Ref]
9. Keeling MJ, Rohani P. Modeling Infectious Diseases in Humans and Animals. Princeton: Princeton University Press; 2007.
Barrat A, Barthélemy M, Vespignani A. Dynamical Processes on Complex Networks. Cambridge: Cambridge University Press; 2008. [Cross Ref]
12. Cohen R, Havlin S. Complex Networks – Structure, Robustness and Function. Cambridge: Cambridge University Press; 2010.
Easley D, Kleinberg J. Networks, Crowds, and Markets – Reasoning about a Highly Connected World. Cambridge: Cambridge University Press; 2010. [Cross Ref]
14. Newman MEJ. Networks – An Introduction. Oxford: Oxford University Press; 2010.
Vespignani A. Modelling dynamical processes in complex socio-technical systems. Nat Phys. 2012;8:32–9. doi: 10.1038/nphys2160. [Cross Ref]
Balcan D, Hu H, Goncalves B, Bajardi P, Poletto C, Ramasco JJ, Paolotti D, Perra N, Tizzoni M, Van den Broeck W, Colizza V, Vespignani A. Seasonal transmission potential and activity peaks of the new
influenza A(H1N1): a Monte Carlo likelihood analysis based on human mobility. BMC Medicine. 2009;7:45. doi: 10.1186/1741-7015-7-45. [PMC free article] [PubMed] [Cross Ref] http://f1000.com/prime/
Van den Broeck W, Gioannini C, Gonçalves B, Quaggiotto M, Colizza V, Vespignani A. The GLEaMviz computational tool, a publicly available software to explore realistic epidemic spreading scenarios at
the global scale. BMC Infect. Dis. 2011;11:37. doi: 10.1186/1471-2334-11-37. [PMC free article] [PubMed] [Cross Ref] http://f1000.com/prime/717978483
Bajardi P, Poletto C, Balcan D, Hu H, Goncalves B, Ramasco JJ, Paolotti D, Perra N, Tizzoni M, Van den Broeck W, Colizza V, Vespignani A. Modeling vaccination campaigns and the Fall/Winter 2009
activity of the new A(H1N1) influenza in the Northern Hemisphere. Emerg Health Threats J. 2009;2:e11. [PMC free article] [PubMed]
Haraldsdottir S, Gupta S, Anderson RM. Preliminary studies of sexual networks in a male homosexual community in Iceland. J Acquir Immune Defic Syndr. 1992;5:374–81. [PubMed]
Liljeros F, Edling CR, Amaral LAN. Sexual networks: implications for the transmission of sexually transmitted infections. Microbes Infect. 2003;5:189–96. doi: 10.1016/S1286-4579(02)00058-8. [PubMed]
[Cross Ref]
Bearman PS, Moody J, Stovel K. Chains of affection: the structure of adolescent romantic and sexual networks. Am J Sociol. 2004;110:44–91. doi: 10.1086/386272. [Cross Ref] http://f1000.com/prime/
Dietz K, Hadeler KP. Epidemiological models for sexually transmitted diseases. J Math Biol. 1988;26:1–25. doi: 10.1007/BF00280169. [PubMed] [Cross Ref]
Watts CH, May RM. The influence of concurrent partnerships on the dynamics of HIV/AIDS. Math Biosci. 1992;108:89–104. doi: 10.1016/0025-5564(92)90006-I. [PubMed] [Cross Ref]
Rocha LEC, Liljeros F, Holme P. Information dynamics shape the sexual networks of Internet-mediated prostitution. P Natl Acad Sci USA. 2010;107:5706–11. doi: 10.1073/pnas.0914080107. [PMC free
article] [PubMed] [Cross Ref]
Danowski JA, Edison-Swift P. Crisis effects on intraorganizational computer-based communication. Comm Res. 1985;12:251–70. doi: 10.1177/009365085012002005. [Cross Ref]
Balthrop J, Forrest S, Newman MEJ, Williamson MM. Technological networks and the spread of computer viruses. Science. 2004;304:527–9. doi: 10.1126/science.1095845. [PubMed] [Cross Ref]
Eckmann J-P, Moses E, Sergi D. Entropy of dialogues creates coherent structures in e-mail traffic. P Natl Acad Sci USA. 2004;101:14333–7. doi: 10.1073/pnas.0405728101. [PMC free article] [PubMed] [
Cross Ref]
Villani A, Frigessi A, Liljeros F, Nordvik MK, de Blasio BF. A characterization of Internet dating network structures among Nordic men who have sex with men. PloS one. 2012;7:e39717. doi: 10.1371/
journal.pone.0039717. [PMC free article] [PubMed] [Cross Ref]
Onnela J-P, Saramäki J, Hyvönen J, Szabó G, Lazer D, Kaski K, Kertész J, Barabási A-L. Structure and tie strengths in mobile communication networks. P Natl Acad Sci USA. 2007;104:7332–6. doi: 10.1073
/pnas.0610245104. [PMC free article] [PubMed] [Cross Ref]
González MC, Hidalgo CA, Barabási A-L. Understanding individual human mobility patterns. Nature. 2008;453:779–82. doi: 10.1038/nature06958. [PubMed] [Cross Ref]
Wang P, González MC, Hidalgo CA, Barabási A-L. Understanding the spreading patterns of mobile phone viruses. Science. 2009;324:1071–6. doi: 10.1126/science.1167053. [PubMed] [Cross Ref]
Song C, Qu Z, Blumm N, Barabási A-L. Limits of predictability in human mobility. Science. 2010;327:1018–21. doi: 10.1126/science.1177170. [PubMed] [Cross Ref]
Jo H-H, Pan RK, Kaski K. Emergence of bursts and communities in evolving weighted networks. PloS one. 2011;6:e22687. doi: 10.1371/journal.pone.0022687. [PMC free article] [PubMed] [Cross Ref]
Eagle N, Pentland AS. Reality mining: sensing complex social systems. Pers Ubiquit Comput. 2006;10:255–68. doi: 10.1007/s00779-005-0046-3. [Cross Ref]
Kim T, Chang A, Holland L, Pentland AS. Meeting mediator: enhancing group collaboration using sociometric feedback. Proc 2008 ACM Conference on Computer Supported Cooperative Work. 2008:457–66. doi:
10.1145/1460563.1460636. [Cross Ref]
40. Wu L, Waber B, Aral S, Brynjolfsson E, Pentland AS. Mining face-to-face interaction networks using sociometric badges: predicting productivity in an it configuration task. Proc International
Conference on Information Systems. 2008:14–7.
Cattuto C, Van den Broeck W, Barrat A, Colizza V, Pinton J-F, Vespignani A. Dynamics of person-to-person interactions from distributed RFID sensor networks. PloS one. 2010;5:e11596. doi: 10.1371/
journal.pone.0011596. [PMC free article] [PubMed] [Cross Ref] http://f1000.com/prime/717978490
Isella L, Stehlé J, Barrat A, Cattuto C, Pinton J-F, Van den Broeck W. What's in a crowd? Analysis of face-to-face behavioral networks. J Theor Biol. 2011;271:166–80. doi: 10.1016/j.jtbi.2010.11.033.
[PubMed] [Cross Ref] http://f1000.com/prime/8311960
43. Yano K, Ara K, Moriwaki N, Kuriyama H. Measurement of human behavior: creating a society for discovering opportunities. Hitachi Review. 2009;58:139–44.
44. Wakisaka Y, Ara K, Hayakawa M, Horry Y, Moriwaki N, Ohkubo N, Sato N, Tsuji S, Yano K. Beam-scan sensor node: reliable sensing of human interactions in organization. Proc 6^th International
Conference on Networked Sensing Systems. 2009:58–61.
Takaguchi T, Nakamura M, Sato N, Yano K, Masuda N. Predictability of conversation partners. Phys. Rev. X. 2011;1:011008. doi: 10.1103/PhysRevX.1.011008. [Cross Ref]
Takaguchi T, Sato N, Yano K, Masuda N. Importance of individual events in temporal networks. New J Phys. 2012;14:093003. doi: 10.1088/1367-2630/14/9/093003. [Cross Ref]
Isella L, Romano M, Barrat A, Cattuto C, Colizza V, Van den Broeck W, Gesualdo F, Pandolfi E, Ravà L, Rizzo C, Tozzi AE. Close encounters in a pediatric ward: measuring face-to-face proximity and
mixing patterns with wearable sensors. PloS one. 2011;6 doi: 10.1371/journal.pone.0017144. article No. e17144. [PMC free article] [PubMed] [Cross Ref]
Stehlé J, Voirin N, Barrat A, Cattuto C, Isella L, Pinton J-F, Quaggiotto M, Van den Broeck W, Régis C, Lina B, Vanhems P. High-resolution measurements of face-to-face contact patterns in a primary
school. PloS one. 2011;6:e23176. doi: 10.1371/journal.pone.0023176. [PMC free article] [PubMed] [Cross Ref]
Kao RR, Green DM, Johnson J, Kiss IZ. Disease dynamics over very different time-scales: foot-and-mouth disease and scrapie on the network of livestock movements in the UK. J R Soc Interface. 2007;4
:907–16. doi: 10.1098/rsif.2007.1129. [PMC free article] [PubMed] [Cross Ref]
Heath MF, Vernon MC, Webb CR. Construction of networks with intrinsic temporal structure from UK cattle movement data. BMC Vet Res. 2008;4 doi: 10.1186/1746-6148-4-11. article No. 11. [PMC free
article] [PubMed] [Cross Ref]
Vernon MC, Keeling MJ. Representing the UK's cattle herd as static and dynamic networks. P R Soc B. 2009;276:469–76. doi: 10.1098/rspb.2008.1009. [PMC free article] [PubMed] [Cross Ref]
Bajardi P, Barrat A, Natale F, Savini L, Colizza V. Dynamical patterns of cattle trade movements. PloS one. 2011;6 doi: 10.1371/journal.pone.0019869. article No. e19869. [PMC free article] [PubMed] [
Cross Ref]
Blonder B, Dornhaus A. Time-ordered networks reveal limitations to information flow in ant colonies. PloS one. 2011;6 doi: 10.1371/journal.pone.0020298. article No. e20298. [PMC free article] [PubMed
] [Cross Ref]
Tantipathananandh C, Berger-Wolf T, Kempe D. A framework for community identification in dynamic social networks. Proc. KDD'07. 2007:717–26. doi: 10.1145/1281192.1281269. [Cross Ref]
Holme P. Network reachability of real-world contact sequences. Phys Rev E. 2005;71 doi: 10.1103/PhysRevE.71.046119. article No. 046119. [PubMed] [Cross Ref]
Tang J, Musolesi M, Mascolo C, Latora V. Temporal distance metrics for social network analysis. Proc. WOSN'09. 2009:31–6. doi: 10.1145/1592665.1592674. [Cross Ref]
Pan RK, Saramäki J. Path lengths, correlations, and centrality in temporal networks. Phys Rev E. 2011;84:016105. doi: 10.1103/PhysRevE.84.016105. [PubMed] [Cross Ref]
Barabási A-L. The origin of bursts and heavy tails in human dynamics. Nature. 2005;435:207–11. doi: 10.1038/nature03459. [PubMed] [Cross Ref]
Vázquez A, Oliveira JG, Dezsö Z, Goh K-I, Kondor I, Barabási A-L. Modeling bursts and heavy tails in human dynamics. Phys Rev E. 2006;73 doi: 10.1103/PhysRevE.73.036127. article No. 036127. [PubMed]
[Cross Ref]
Vazquez A, Rácz B, Lukács A, Barabási A-L. Impact of non-Poissonian activity patterns on spreading processes. Phys Rev Lett. 2007;98:158702. doi: 10.1103/PhysRevLett.98.158702. [PubMed] [Cross Ref]
Kermack WO, McKendrick AG. A contribution to the mathematical theory of epidemics. P R Soc Lond A. 1927;115:700–21. doi: 10.1098/rspa.1927.0118. [Cross Ref]
63. Bailey NTJ. The simulation of stochastic epidemics in two dimensions. Proc 5th Berkeley Symposium on Mathematical Statistics and Probability. 1967;4:237–7.
64. Anderson RM, May RM. Infectious Diseases of Humans – Dynamics and Control. Oxford: Oxford University Press; 1991.
Barthélemy M, Barrat A, Pastor-Satorras R, Vespignani A. Velocity and hierarchical spread of epidemic outbreaks in scale-free networks. Phys Rev Lett. 2004;92:178701. doi: 10.1103/
PhysRevLett.92.178701. [PubMed] [Cross Ref] http://f1000.com/prime/717978493
Barthélemy M, Barrat A, Pastor-Satorras R, Vespignani A. Dynamical patterns of epidemic outbreaks in complex heterogeneous networks. J Theor Biol. 2005;235:275–88. doi: 10.1016/j.jtbi.2005.01.011. [
PubMed] [Cross Ref]
Eames KTD, Keeling MJ. Monogamous networks and the spread of sexually transmitted diseases. Math Biosci. 2004;189:115–30. doi: 10.1016/j.mbs.2004.02.003. [PubMed] [Cross Ref]
Fefferman NH, Ng KL. How disease models in static networks can fail to approximate disease in dynamic networks. Phys Rev E. 2007;76 doi: 10.1103/PhysRevE.76.031919. article No. 031919. [PubMed] [
Cross Ref]
Smieszek T, Fiebig L, Scholz RW. Models of epidemics: when contact repetition and clustering should be included. Theor Biol Med Model. 2009;6 doi: 10.1186/1742-4682-6-11. article No. 11. [PMC free
article] [PubMed] [Cross Ref]
Iribarren JL, Moro E. Impact of human activity patterns on the dynamics of information diffusion. Phys Rev Lett. 2009;103 doi: 10.1103/PhysRevLett.103.038702. article No. 038702. [PubMed] [Cross Ref]
Karrer B, Newman MEJ. Message passing approach for general epidemic models. Phys Rev E. 2010;82:016101. doi: 10.1103/PhysRevE.82.016101. [PubMed] [Cross Ref]
Min B, Goh K-I, Vazquez A. Spreading dynamics following bursty human activity patterns. Phys Rev E. 2011;83:036102. doi: 10.1103/PhysRevE.83.036102. [PubMed] [Cross Ref]
Bansal S, Read J, Pourbohloul B, Meyers LA. The dynamic nature of contact networks in infectious disease epidemiology. J Biol Dyn. 2010;4:478–89. doi: 10.1080/17513758.2010.503376. [PubMed] [Cross
Ref] http://f1000.com/prime/717978495
Karsai M, Kivelä M, Pan RK, Kaski K, Kertész J, Barabási A-L, Saramäki J. Small but slow world: how network topology and burstiness slow down spreading. Phys Rev E. 2011;83 doi: 10.1103/
PhysRevE.83.025102. article No. 025102(R) [PubMed] [Cross Ref] http://f1000.com/prime/717978496
Kivelä M, Pan RK, Kaski K, Kertész J, Saramäki J, Karsai M. Multiscale analysis of spreading in a large communication network. J Stat Mech. 2012;3:P03005. doi: 10.1088/1742-5468/2012/03/P03005. [
Cross Ref]
Perra N, Gonçalves B, Pastor-Satorras R, Vespignani A. Activity driven modeling of time varying networks. Sci. Rep. 2012;2 doi: 10.1038/srep00469. article No. 469. [PMC free article] [PubMed] [Cross
Rocha LEC, Liljeros F, Holme P. Simulated epidemics in an empirical spatiotemporal network of 50,185 sexual contacts. Plos Comput Biol. 2011;7 doi: 10.1371/journal.pcbi.1001109. article No. e1001109.
[PMC free article] [PubMed] [Cross Ref]
Rocha LEC, Decuyper A, Blondel VD. Epidemics on a stochastic model of temporal network. Preprint. 2012 Mar; http://arxiv.org/abs/1204.5421 arXiv:1204.5421.v1.
Stehlé J, Voirin N, Barrat A, Cattuto C, Colizza V, Isella L, Régis C, Pinton J-F, Khanafer N, Van den Broeck W, Vanhems P. Simulation of an SEIR infectious disease model on the dynamic contact
network of conference attendees. BMC Medicine. 2011;9 doi: 10.1186/1741-7015-9-87. article No. 87. [PMC free article] [PubMed] [Cross Ref]
Volz E, Meyers LA. Epidemic thresholds in dynamic contact networks. J R Soc Interface. 2009;6:233–41. doi: 10.1098/rsif.2008.0218. [PMC free article] [PubMed] [Cross Ref]
Risau-Gusman S. Influence of network dynamics on the spread of sexually transmitted diseases. J R Soc Interface. 2012;9:1363–72. doi: 10.1098/rsif.2011.0445. [PMC free article] [PubMed] [Cross Ref]
Cohen R, Havlin S, ben-Avraham D. Efficient immunization strategies for computer networks and populations. Phys Rev Lett. 2003;91 doi: 10.1103/PhysRevLett.91.247901. article No. 247901. [PubMed] [
Cross Ref] http://f1000.com/prime/717978497
Lee S, Rocha LEC, Liljeros F, Holme P. Exploiting temporal network structures of human interaction to effectively immunize populations. PloS one. 2012;7 article No. e36439. [PMC free article] [PubMed
87. Tang J, Mascolo C, Musolesi M, Latora V. Exploiting temporal complex network metrics in mobile malware containment. Presented at Proc IEEE International Symposium on a World of Wireless, Mobile
and Multimedia Networks (WoWMoM) 2011:1–9 June 2011; Lucca, Italy.
Grant RM, Lama JR, Anderson PL, McMahan V, Liu AY, Vargas L, Goicochea P, Casapía M, Guanira-Carranza JV, Ramirez-Cardich ME, Montoya-Herrera O, Fernández T, Veloso VG, Buchbinder SP, Chariyalertsak
S, Schechter M, Bekker L-G, Mayer KH, Kallás EG, Amico KR, Mulligan K, Bushman LR, Hance RJ, Ganoza C, Defechereux P, Postle B, Wang F, McConnell JJ, Zheng J-H, Lee J, et al. Preexposure
chemoprophylaxis for HIV prevention in men who have sex with men. N Engl J Med. 2010;363:2587–99. doi: 10.1056/NEJMoa1011205. [PMC free article] [PubMed] [Cross Ref] http://f1000.com/prime/6581956
Liben-Nowell D, Kleinberg J. The link-prediction problem for social networks. J Am Soc Inf Sci Tec. 2007;58:1019–31. doi: 10.1002/asi.20591. [Cross Ref]
Hasan MA, Zaki MJ. A survey of link prediction in social networks. Soc Netw Data Anal. 2011:243–75. doi: 10.1007/978-1-4419-8462-3_9. [Cross Ref]
Lü L, Zhou T. Link prediction in complex networks: a survey. Physica A. 2011;390:1150–70. doi: 10.1016/j.physa.2010.11.027. [Cross Ref]
Rvachev LA, Longini IM., Jr A mathematical model for the global spread of influenza. Math Biosci. 1985;75:3–22. doi: 10.1016/0025-5564(85)90064-1. [Cross Ref]
Diekmann O, Heesterbeek JAP, Metz JAJ. On the definition and the computation of the basic reproduction ratio R[0] in models for infectious diseases in heterogeneous populations. J Math Biol. 1990;28
:365–82. doi: 10.1007/BF00178324. [PubMed] [Cross Ref]
Liu S, Baronchelli A, Perra N. Contagion dynamics in time-varying metapopulation networks. Preprint. 2012 Oct; http://arxiv.org/abs/1210.2776 arXiv:1210.2776.v1.
Cooper BS, Pitman RJ, Edmunds WJ, Gay NJ. Delaying the international spread of pandemic influenza. PLoS Med. 2006;3:e212. doi: 10.1371/journal.pmed.0030212. [PMC free article] [PubMed] [Cross Ref]
Ferguson NM, Cummings DAT, Fraser C, Cajka JC, Cooley PC, Burke DS. Strategies for mitigating an influenza pandemic. Nature. 2006;442:448–52. doi: 10.1038/nature04795. [PubMed] [Cross Ref]
Hollingsworth TD, Ferguson NM, Anderson RM. Will travel restrictions control the international spread of pandemic influenza? Nat Med. 2006;12:497–9. doi: 10.1038/nm0506-497. [PubMed] [Cross Ref]
Colizza V, Barrat A, Barthelemy M, Valleron A-J, Vespignani A. Modeling the worldwide spread of pandemic influenza: Baseline case and containment interventions. PLoS Med. 2007;4:e13. doi: 10.1371/
journal.pmed.0040013. [PMC free article] [PubMed] [Cross Ref]
Epstein JM, Goedecke DM, Yu F, Morris RJ, Wagener DK, Bobashev GV. Controlling pandemic flu: the value of international air travel restrictions. PloS one. 2007;2:e401. doi: 10.1371/
journal.pone.0000401. [PMC free article] [PubMed] [Cross Ref]
Colizza V, Vespignani A. Epidemic modeling in metapopulation systems with heterogeneous coupling pattern: theory and simulations. J Theor Biol. 2008;251:450–67. doi: 10.1016/j.jtbi.2007.11.028. [
PubMed] [Cross Ref]
Bajardi P, Poletto C, Ramasco JJ, Tizzoni M, Colizza V, Vespignani A. Human mobility networks, travel restrictions, and the global spread of 2009 H1N1 pandemic. PloS one. 2011;6:e16591. doi: 10.1371/
journal.pone.0016591. [PMC free article] [PubMed] [Cross Ref]
Wang B, Cao L, Suzuki H, Aihara K. Safety-information-driven human mobility patterns with metapopulation epidemic dynamics. Sci. Rep. 2012;2:887. doi: 10.1038/srep00887. [PMC free article] [PubMed] [
Cross Ref]
Brownstein JS, Wolfe CJ, Mandl KD. Empirical evidence for the effect of airline travel on inter-regional influenza spread in the United States. PLoS Med. 2006;3:e401. doi: 10.1371/
journal.pmed.0030401. [PMC free article] [PubMed] [Cross Ref] http://f1000.com/prime/717978498
Heffernan JM, Smith RJ, Wahl LM. Perspectives on the basic reproductive ratio. J R Soc Interface. 2005;2:281–93. doi: 10.1098/rsif.2005.0042. [PMC free article] [PubMed] [Cross Ref]
Wallinga J, Lipsitch M. How generation intervals shape the relationship between growth rates and reproductive numbers. P R Soc B. 2007;274:599–604. doi: 10.1098/rspb.2006.3754. [PMC free article] [
PubMed] [Cross Ref]
Li J, Blakeley D, Smith RJ. The failure of R[0]. Comput Math Methods Med. 2011;2011:527610. doi: 10.1155/2011/527610. [PMC free article] [PubMed] [Cross Ref]
Haydon DT, Kao RR, Kitching RP. The UK foot-and-mouth disease outbreak – the aftermath. Nat Rev Micro. 2004;2:675–81. doi: 10.1038/nrmicro960. [PubMed] [Cross Ref]
Articles from F1000Prime Reports are provided here courtesy of Faculty of 1000 Ltd
Your browsing activity is empty.
Activity recording is turned off.
See more... | {"url":"http://www.ncbi.nlm.nih.gov/pmc/articles/PMC3590785/?tool=pubmed","timestamp":"2014-04-17T04:50:17Z","content_type":null,"content_length":"188463","record_id":"<urn:uuid:7bc3e779-3b38-43e8-9aec-c40815fe509c>","cc-path":"CC-MAIN-2014-15/segments/1397609526252.40/warc/CC-MAIN-20140416005206-00305-ip-10-147-4-33.ec2.internal.warc.gz"} |
Gill, CO Algebra 2 Tutor
Find a Gill, CO Algebra 2 Tutor
...Additionally I have tutored many high school and middle schoolers through the year including my own daughter currently in Algebra. Additionally I have refreshed myself on Standardized test
(ACT, SAT and GRE) in preparation for job application with Princeton Review. I have been a Quickbooks user since 2000 doing the books for my wife's photography business.
14 Subjects: including algebra 2, calculus, physics, GRE
...I'd like to generate some extra income to help support their school and other activities by applying my knowledge in math and/or science in helping others interested in and/or struggling in
either of these subject areas.I was horribly afraid of giving speeches growing up until I attended Toastmas...
14 Subjects: including algebra 2, reading, algebra 1, grammar
...I have learned a lot through my own experience and lectures about good study habits. I know a lot about learning styles and strategies for teaching students based on their specific needs. I
believe that my success as a bachelor of arts student, tutor, and future teacher, qualifies me to tutor in the area of study skills.
20 Subjects: including algebra 2, English, writing, algebra 1
...I feel very comfortable with the material and effective ways to explain the algorithms and practical applications of all precalculus topics. Those topics include: logarithmic and exponential
functions, polynomial and rational functions, vectors, conic sections, complex numbers, trigonometry and ...
14 Subjects: including algebra 2, reading, geometry, algebra 1
...I write test harness software to facilitate job distribution and management of tasks under the Linux/Unix environment as well as coordination with a Windows (Cygwin) environment. I also use
LSF (Load Sharing Facility) to handle job scheduling and do so under a Linux/Unix environment. Management...
47 Subjects: including algebra 2, chemistry, calculus, physics
Related Gill, CO Tutors
Gill, CO Accounting Tutors
Gill, CO ACT Tutors
Gill, CO Algebra Tutors
Gill, CO Algebra 2 Tutors
Gill, CO Calculus Tutors
Gill, CO Geometry Tutors
Gill, CO Math Tutors
Gill, CO Prealgebra Tutors
Gill, CO Precalculus Tutors
Gill, CO SAT Tutors
Gill, CO SAT Math Tutors
Gill, CO Science Tutors
Gill, CO Statistics Tutors
Gill, CO Trigonometry Tutors | {"url":"http://www.purplemath.com/gill_co_algebra_2_tutors.php","timestamp":"2014-04-18T16:02:41Z","content_type":null,"content_length":"23912","record_id":"<urn:uuid:58018214-d0ff-47ed-9716-8b325acba0c8>","cc-path":"CC-MAIN-2014-15/segments/1398223206147.1/warc/CC-MAIN-20140423032006-00641-ip-10-147-4-33.ec2.internal.warc.gz"} |
Bargaineering » Calculating Discounted Retirement Asset Value - Bargaineering » Print
A lot of bloggers, and those keeping score at home, put their retirement assets in their net worth considerations but one of my friends, Dimitri, asked if anyone actually discounts it because of time
and future taxes. Personally, I do not because it complicates the calculation and takes more work than it does right now (Login, copy and paste number) but the idea does have merit. First, a little
about discounting cash flows and then a proposal as to how personal finance bloggers (anyone really) can be a tad more accurate on their current net worth with respect to retirement assets.
With respect to discounting, the idea is that you’re assessing the current value of a future sum of money. A dollar today is worth more than a dollar tomorrow because of the time value of money.
Simplistically, it’s worth more is because you can invest the money today (earning interest or profit of one day) and because inflation hasn’t taken a tiny piece away yet. That being said, your
retirement assets aren’t available to you (unless you take the 10% penalty) until you are [DEL:64.5:DEL] 59.5 (thanks qw!), and even then only a portion is available to you, so it’s current value is
lower than the dollar amount.
Reduce Asset Value by Marginal Tax Rate
This was Dimitri’s original suggestion and is by far the easiest way to be a little more accurate. He suggests that you ignore the fact that you can’t touch the money until retirement and simply
decrease its value by your marginal tax rate. At the very least, if you do nothing then you must decrease your pre-tax retirement accounts (IRAs, 401ks) by your marginal tax rate when you compare
them with your post-tax retirement accounts (Roth IRAs) so you’re comparing apples to apples.
Why Discounting Really Doesn’t Matter
When you discount your assets, you’re assuming a certain value in the future, hopefully a value greater than your 401(k) balance right now. You reach that value by assuming a rate of return (a
popular but overly optimistic one is 10-12%) and then calculating what your asset would be worth when you could take disbursements. Normally, you’d then discount that final asset value by a rate of
return, which would still be 10-12%, and so you would arrive at the value you have started with. In reality, it’s not that simple because you would still contribute every year and there’s inflation,
but for our purposes your current value is good enough with respect to discounting.
Verdict: Reduce Pre-Tax Retirement Asset Worth by Tax Rate
For a better picture of your pre-tax retirement asset value, I think you should reduce it by your marginal tax rate because at the very least your 401(k) can be compared on even ground with your Roth
IRA (for example). While it’s not exactly perfect, it’s good enough.
What do you all think? Good idea? Bad idea? Do you have an alternative that you use? I look forward to reading all and any criticisms, thoughts, rants, anythings on the matter. Thanks! | {"url":"http://www.bargaineering.com/articles/calculating-discounted-retirement-asset-value.html/print/","timestamp":"2014-04-17T12:50:41Z","content_type":null,"content_length":"6947","record_id":"<urn:uuid:094b7bc3-c806-4b6d-be62-d036f17cae7f>","cc-path":"CC-MAIN-2014-15/segments/1397609530131.27/warc/CC-MAIN-20140416005210-00339-ip-10-147-4-33.ec2.internal.warc.gz"} |
Calculating the angle between a plane and the x-y plane.
May 17th 2009, 08:07 AM #1
May 2009
Calculating the angle between a plane and the x-y plane.
I've been asked:
What does the plane 2x + 3y -z = 1 make with the x-y plane.
I'm completely stuck, never been any good with vectors, I'm sure the solution is simple using cos theta = n1 x n2/ modulus n1 x modulus n2 but can't quite work it out.
Thanks plato,
if I could just ask you to explain.
1. why is x - y plane, 0, 0, 1 (this makes sense to me in a way but I can't picture it)
2. what is this significance of vector 'A' = 2x + 3y -z =1 (equaling one)
That makes no sense to me at all! I think you meant to ask "why is <0, 0, 1> the normal vector to the x-y plane?" The xy-plane is given by the equation z= 0. Any point in it is of the form
(x,y,0) and its "position vector"
is given by <x, y, 0>. In order that <a, b, c> be normal to every such vector we must have the dot product <a, b, c>.<x, y, 0>= ax+ by+ 0= 0. Taking x= 1, y= 0, a(1)+ b(0)+ 0= 0 so we must have a
= 0. Taking x= 0, y= 1, a(0)+ b(1)+ 0= 0 so we must have b= 0. A vector is normal to the xy-plane if and only if it is of the form <0, 0, z> and a unit vector normal to the xy-plane is <0, 0, 1>.
2. what is this significance of vector 'A' = 2x + 3y -z =1 (equaling one)
"'A'= 2x+ 3y- z= 1" is not a vector at all! If you are asking about the difference between the planes 2x+ 2y- z= 1 and, say, 2x+ 3y- z= 2, they are parallel planes. The first crosses the z axis
at (0, 0, -1) (x=y= 0 so the equation becomes 2(0)+ 2(0)- z= 1 or z= -1) and the second at (0, 0, -2).
May 17th 2009, 08:43 AM #2
May 17th 2009, 09:13 AM #3
May 2009
May 17th 2009, 09:46 AM #4
May 17th 2009, 10:34 AM #5
May 17th 2009, 11:51 AM #6
MHF Contributor
Apr 2005 | {"url":"http://mathhelpforum.com/geometry/89352-calculating-angle-between-plane-x-y-plane.html","timestamp":"2014-04-21T13:02:35Z","content_type":null,"content_length":"50782","record_id":"<urn:uuid:d45ad4ea-7572-496d-b220-32317d4b34bc>","cc-path":"CC-MAIN-2014-15/segments/1397609539776.45/warc/CC-MAIN-20140416005219-00350-ip-10-147-4-33.ec2.internal.warc.gz"} |
Trying to figure out recursive algorithm to find greatest common divisor of two integers
Author Trying to figure out recursive algorithm to find greatest common divisor of two integers
Ranch Hand
I have not searched the forums on this because I want to try to work out as much of this on my own as I can.
Joined: Jul 23, 2009 I know how to solve this with with a loop. If the the larger integer can be divided by the smaller integer with a remainder
Posts: 41 of 0, then the smaller integer is the greatest common divisor. Otherwise, iterate from (smaller - 1) down to 1 and check
for divisibility for both larger and smaller integer.
I know I should be thinking in terms of a base case and a simplification of the original problem, but I don't see it.
Any small hint?
Ranch Hand
Joined: Jul 23, 2009 I remember now that the solution is the Euclidean algorithm. Never mind. I will try some other recursive problem.
Posts: 41
Here's the general non-recursive algorithm:
Joined: Oct 27, 2005 Now try and convert that into a recursive call. You already have the finished condition: max % mid == 0. Another possibility is a == b.
Posts: 19543
To make it easier, the first step can be to make sure that a is larger than b:
16 Hint: the actual Euclidian algorithm described here already is a recursive call, with the finished condition being b == 0.
SCJP 1.4 - SCJP 6 - SCWCD 5 - OCEEJBD 6
I like... How To Ask Questions How To Answer Questions
subject: Trying to figure out recursive algorithm to find greatest common divisor of two integers | {"url":"http://www.coderanch.com/t/484673/java/java/figure-recursive-algorithm-find-greatest","timestamp":"2014-04-19T17:35:49Z","content_type":null,"content_length":"24551","record_id":"<urn:uuid:57c7a8d8-02c2-47a7-ba6c-07e87d1c548c>","cc-path":"CC-MAIN-2014-15/segments/1397609537308.32/warc/CC-MAIN-20140416005217-00272-ip-10-147-4-33.ec2.internal.warc.gz"} |
Updated: June 2002
Table Of Contents
pgmabel - create cross section using Abel Integration for Deconvolution
pgmabel [-help] [-axis axis] [-factor factor] [-pixsize pixsize] [-left | -right] [-verbose] [filespec]
This program is part of Netpbm.
pgmabel reads as input a PGM image, which it assumes to be an image of a rotational symmetric transparent object. The image must have a vertical symmetry axis. pgmabel produces as output an image of
a cross-section of the image. pgmabel does the calculation by performing the Abel Integration for Deconvolution of an axial-symmetrical image by solving the system of linear equations. After
integration, pgmabel weights all gray-values of one side by the surface area of the calculated ring in square pixels divided by 4*factor multiplied by the size of one pixel (pixsize). With the
-verbose option, pgmabel prints the weighting factors.
Where the calculation generates a negative result, the output is black.
The computation is unstable against periodic structures with size 2 in the vertical direction.
You can abbreviate any option to its shortest unique prefix.
Prints a help message.
-axis axis
Position of the axis of symmetry in the image in pixels from the left edge of the image. Default is the center of the image.
-factor factor
User defined factor for enhancement of the output. Use a factor less than 1 for decreasing gary values. Default is 1.0.
-pixsize pixsize
The size of a pixel for getting scale invariant. Default is 0.1.
-left Calculate only the left side of the image. You cannot specify both left and right.
-right Analogous to -left.
print information about the calculation.
Rotate a PGM image to get an image with a vertical axis of symmetry, then calculate the cross section:
pnmrotate 90 file.pgm | pgmabel -axis 140 >cross_section.pgm
pnmrotate, pgm,
This program was added to Netpbm in Release 10.3 (June 2002).
Volker Schmidt (lefti@voyager.boerde.de)
Copyright (C) 1997-2002 German Aerospace research establishment
Table Of Contents | {"url":"http://netpbm.sourceforge.net/doc/pgmabel.html","timestamp":"2014-04-20T15:52:41Z","content_type":null,"content_length":"3727","record_id":"<urn:uuid:b1481c04-589f-440f-aca2-fc6d0d91ed11>","cc-path":"CC-MAIN-2014-15/segments/1397609538824.34/warc/CC-MAIN-20140416005218-00589-ip-10-147-4-33.ec2.internal.warc.gz"} |
proving a beta identity
July 5th 2008, 04:31 AM #1
proving a beta identity
Hey All:
I was looking at a Beta function identity in a math physics book I have and
it said to show ${\beta}(n,1/2)=2^{2n-1}{\beta}(n,n)$.
I done that OK. But....how does one prove
I know the various gamma and beta identities, but got a little stuck on
where this comes from.
I know ${\Gamma}(n+1)=n!, \;\ {\Gamma}(2n+1)=(2n)!$ and so forth.
I tried various ways, but failed to get it to come together.
Where is the world do those factorials come from in that beta identity. I am missing something. Probably obvious. Always is.
What about using the identity $\beta(m,n)=2 \int_0^{\pi/2} (\sin \theta)^{2m-1} (\cos \theta)^{2n-1} \ d\theta$ (I haven't seen a proof of it yet, but perhaps it is possible by substituting $t=\
cos^2 \theta$ in $\beta(m,n)=\int_0^1 t^{m-1}(1-t)^{n-1} \ dt$)
--> $\beta(n,n)=2 \int_0^{\pi/2} \left(\frac{\sin 2 \theta}{2}\right)^{2n-1} \ d\theta$
$\beta(n,1/2)=2^{2n} \int_0^{\pi/2} \left(\frac{\sin 2 \theta}{2}\right)^{2n-1} \ d\theta$
I think you can prove what you want from it (by induction e.g.)
Edit : héhé, I stress the "I think"
Last edited by Moo; July 5th 2008 at 05:02 AM.
Ok, here is another (and, in my opinion, simpler !) way.
We want to show this :
$=\frac{2^{2n} \Gamma(n+1)\Gamma(n)}{\Gamma(2n+1)}$
But we know that $\beta(x,y)=\frac{\Gamma(x) \Gamma(y)}{\Gamma(x+y)}$ (there is a proof of it in wikipedia).
Therefore, we want to prove that $\beta(n,1/2)=2^{2n} \beta(n,n+1)$
We know that $\beta(n,1/2)=2^{2n-1} \beta(n,n)$
So if we can prove that $\beta(n,n)=2 \beta(n,n+1)$, we're done.
$\beta(n,n+1)=\int_0^1 t^n(1-t)^{n-1} \ dt=\int_0^1 t^{n-1}(1-t)^n \ dt$ (symmetry of beta function).
So \begin{aligned} 2 \beta(n,n+1)&=\int_0^1 t^n(1-t)^{n-1} + t^{n-1}(1-t)^n \ dt \\ \\<br /> &=\int_0^1 t^{n-1}(1-t)^{n-1} [t+1-t] \ dt \\ \\<br /> &=\beta(n,n) \quad \blacksquare \end{aligned}
Thanks Moo, I actually got that far and didn't think of induction. Thanks.
I was looking at the gamma identities.
I knew that $\frac{(n!)^{2}}{(2n)!}=\frac{n{\Gamma}(n){\Gamma}( n+1)}{{\Gamma}(2n+1)}$
I tried plugging in ${\beta}(n,1/2)=\frac{{\Gamma}(n){\Gamma}(1/2)}{{\Gamma}(n+1/2)}$
And knew that ${\Gamma}(n+1/2)=\frac{{\Gamma}(n){\Gamma}(1/2)}{{\beta}(n,1/2)}={\Gamma}(n+1/2)$
I had trouble tying them together with that factorial-laden identity.
Simplest one
$\beta(n,1/2)=2^{2n-1} \beta(n,n)$
$\beta(n,n)=\frac{\Gamma(n) \Gamma(n)}{\Gamma(2n)}=\frac{(n-1)!(n-1)!}{(2n-1)!}$
$=\frac{(n!)^2}{n^2} \cdot \frac{2n}{(2n)!}=\frac{2(n!)^2}{n(2n)!}$
Thanks Moo, I actually just got the same result that way. My rpoblem was, for some reason, I failed to see that
$(2n-1)!=\frac{(2n)!}{2n}$ right off the bat. That made all the difference. I knew it was something in front of my nose. Always is.
I am glad to see we jive. Appears to be the best way to go about it. I like all your methods. Good girl.
Regarding your second post, I hadn't thought of it that way. That was a good
Last edited by galactus; July 6th 2008 at 09:04 AM.
July 5th 2008, 04:41 AM #2
July 5th 2008, 05:41 AM #3
July 5th 2008, 05:52 AM #4
July 6th 2008, 06:57 AM #5
July 6th 2008, 08:54 AM #6 | {"url":"http://mathhelpforum.com/advanced-math-topics/43053-proving-beta-identity.html","timestamp":"2014-04-19T20:39:24Z","content_type":null,"content_length":"56845","record_id":"<urn:uuid:66089fd8-0d28-4538-b661-b2163520b6d3>","cc-path":"CC-MAIN-2014-15/segments/1397609537376.43/warc/CC-MAIN-20140416005217-00628-ip-10-147-4-33.ec2.internal.warc.gz"} |
Automatic asymptotics
This thesis is part of a project to automate asymptotic analysis. We study systems of functional equations, which, besides the usual algebraic operations, involve the asymptotic relations of
equivalence and inequality. We develop an algebraic and effective theory for solving such systems.
In part A, we develop the algebraic formalism, using the language of transseries. This enables us to describe very violent asymptotic behaviours, which are for instance encountered in the study of
non-linear differential equations.
In part B, we give algorithms for the asymptotic resolution of algebraic differential equations and systems of transseries in several variables. In a first stage, our results only apply to functions
with strongly monotonic asymptotic behaviours, i.e. functions which do not present any direct or indirect oscillatory phenomena at infinity: from an analytical point of view, this corresponds to
considering only Hardy field functions. But we have already extended some of our results to the weakly oscillatory case.
Keywords: asymptotic analysis, computer algebra, transseries, differential equation, functional equation, exp-log function, Hardy field
View: Gzipped Postscript, Pdf, BibTeX | {"url":"http://www.texmacs.org/joris/phd/phd-abs.html","timestamp":"2014-04-19T17:01:21Z","content_type":null,"content_length":"4383","record_id":"<urn:uuid:60ca9c8c-6ed1-40ed-8f88-34601ddd5018>","cc-path":"CC-MAIN-2014-15/segments/1398223205375.6/warc/CC-MAIN-20140423032005-00649-ip-10-147-4-33.ec2.internal.warc.gz"} |
Math Forum Discussions - Re: Why Dehaene is Wrong
Date: Nov 2, 2012 1:08 AM
Author: Robert Hansen
Subject: Re: Why Dehaene is Wrong
On Nov 2, 2012, at 12:55 AM, Louis Talman <talmanl@gmail.com> wrote:
> That's even more unbelievable. If it were so, the mathematically successful population would be demographically identical with the mathematically unsuccessful population. I think it can be easily demonstrated that this isn't so.
No, I proved it. With millions of exams.
If you look at two exams, both scoring, say, 70% correct, you cannot tell ANYTHING about the students who took the exam. When a girl is as successful as a boy on the exam their successes are indistinguishable from each other when you examine the problems they got right and the problems they got wrong. It doesn't even matter if the girl is black, poor, rich, or lives in Siberia.
Read this again very carefully Lou, I really don't think you understood it.
Bob Hansen | {"url":"http://mathforum.org/kb/plaintext.jspa?messageID=7916474","timestamp":"2014-04-21T07:47:48Z","content_type":null,"content_length":"1830","record_id":"<urn:uuid:d46a649e-8cb1-4a63-84e2-27af1b8125b0>","cc-path":"CC-MAIN-2014-15/segments/1397609539665.16/warc/CC-MAIN-20140416005219-00224-ip-10-147-4-33.ec2.internal.warc.gz"} |
This visualization shows how fast you're spinning around Earth's axis
Ever wonder how fast you're spinning around the imaginary rod connecting the North and South Poles? Wonder no more. By finding your location on this map and searching for where your line of latitude
intersects with the bold, black curve, you can determine the speed at which you're currently zipping around Earth's axis.
"I was initially going to make the graphic for Mars," planetary geologist Seth Kadish tells io9, " but people always enjoy a data visualization more when they can relate it to themselves. As no one
lives on Mars (yet), I opted to use the Earth."
The result was the visualization you see above – a graph that depicts the speed of a point on Earth's surface for a given latitude due to the rotation of the Earth about its axis. Note that this
visualization does not take our rotation around the Sun into account (those interested can factor in our roughly 67,000-mph orbital speed, if they're so inclined).
The speed at which you circle the Earth's axis, of course, depends upon your location on the planet's surface. An object on the Earth's equator travels once around the Earth's circumference (about
40,075 kilometers, or 24,901 miles) each day. Divide that distance by 24 hours (or 23 hours 56 minutes, if you're going by sidereal days – which we'll get to shortly) and you get a speed of about
1,670 km/h, or roughly 1,040 mph. After that, all you need to do to calculated speed due to rotation at any other point on the Earth is multiply the speed at the equator by the cosine (remember
trigonometry?) of the latitude at the point. [Trigonometric diagram via]
But hang on a second. If we're going to bring simple trig into this calculation, it helps to assume that the Earth is a perfect sphere – but it's not. You'd never know it from images like this one,
but our planet is actually an oblate spheroid (a sphere that's been smooshed a little bit). It's also not perfectly smooth; the radius of the Earth measured to the base of Mount Everest is smaller
than it is measured to the mountain's peak. And what about the fact that a full rotation of the Earth doesn't actually take 24 hours, but roughly 23 hours and 56 minutes – wouldn't that affect our
calculations, as well? Well... yes and no. As Kadish explains:
Though this is an approximation, in an effort to be as accurate as possible, I used the length of a sidereal day (23 hrs, 56 min, 4 sec), which is a full 360° rotation of Earth. Because Earth is
an oblate spheroid rather than a sphere, I varied the radius as a function of latitude when calculating the tangential speed. The polar radius is 3950 miles and the equatorial radius is 3963
miles; I approximated the radius at other latitudes via a linear interpolation. This has no visible effect on the curve, though. Using the average radius of the earth (3959 miles) as a constant
changes the global tangential speeds by <1 mph. Topography of the Earth is equally unimportant for this level of accuracy because the difference between a mountain peak and the bottom of the
ocean is trivial compared to the radius of the Earth. If, hypothetically, Mt. Everest's peak (5.5 miles above datum) and the deepest part of the Mariana Trench (6.8 miles below datum) were both
located along the equator, the difference in tangential speed caused by the 12.3 mile elevation difference would only be about 3 mph, or less than a third of a percent of the equator's 1040 mph
tangential speed.
As I told Kadish in an e-mail, I've seen this visualization done a couple of times in the past, but I've never seen anyone take the time to explain the considerations that one might take into account
when making such a chart (i.e. elevation differences, the fact that Earth is an oblate spheroid, solar vs. sidereal time etc.), let alone explain why those considerations give rise to such negligible
differences in one's calculations. Really great work all around.
Check out more of Kadish's visualizations on his utter time-suck of a tumblr, Vizual Statistix.
11 66Reply | {"url":"http://io9.com/how-fast-are-you-spinning-around-earths-axis-1508810529","timestamp":"2014-04-20T00:49:14Z","content_type":null,"content_length":"95327","record_id":"<urn:uuid:76af7f1c-152b-4f70-bd1f-872e32e96274>","cc-path":"CC-MAIN-2014-15/segments/1397609537804.4/warc/CC-MAIN-20140416005217-00659-ip-10-147-4-33.ec2.internal.warc.gz"} |
Re: st: -tabout- excess space between caption and first line
Notice: On March 31, it was announced that Statalist is moving from an email list to a forum. The old list will shut down at the end of May, and its replacement, statalist.org is already up and
[Date Prev][Date Next][Thread Prev][Thread Next][Date Index][Thread Index]
Re: st: -tabout- excess space between caption and first line
From Eric Booth <ebooth@ppri.tamu.edu>
To "<statalist@hsphsun2.harvard.edu>" <statalist@hsphsun2.harvard.edu>
Subject Re: st: -tabout- excess space between caption and first line
Date Thu, 14 Apr 2011 00:41:49 +0000
I cannot replicate the extra space you describe with the code you have provided. Can you please provide the -tabout- syntax you used as well as the entire LaTeX syntax?
Using the -tabout- code below, I got the LaTeX code below that (using top.tex and bot.tex from Ian Watson's documentation), and compiled it using pdflatex with texshop for Mac OSX 10.6 -- the resulting table did not have a large gap between the caption and the \toprule line. I've put a copy of the .tex, .do, and resulting .pdf in my dropbox (links provided below).
The gap might be due to (1) something in your -tabout- syntax or (2) something in your LaTeX document that is dictating the space between the caption and the label (e.g., something in your preamble or a \usepackage{} that you included (?)), but it's hard to tell.
**************************BEGIN Stata -tabout- code
sysuse nlsw88.dta, clear
tabout coll smsa south race using table1.tex, replace ///
c(col) f(1p) clab(Col%) stats(chi2) npos(lab) ptotal(both) ///
style(tex) bt font(bold) cl1(2-5) ///
topf(top.tex) botf(bot.tex) topstr(12cm) botstr(10cm)
type table1.tex
**************************END Stata -tabout- code
*********************************BEGIN LATEX CODE
\usepackage{geometry, booktabs, tabularx, graphicx, epstopdf}
\caption{test caption table title here }
\begin{tabularx} {12cm} {@{} l X Y Y Y Y Y Y Y Y Y Y Y Y Y Y Y@{}} \\
& \multicolumn{4}{c}{\textbf{race}} \\
&\textbf{white}&\textbf{black}&\textbf{other}&\textbf{Total} \\
&Col\%&Col\%&Col\%&Col\% \\
\textbf{college graduate}&&&& \\
not college grad (n=1,714)&74.3\%&82.3\%&65.4\%&76.3\% \\
college grad (n=532)&25.7\%&17.7\%&34.6\%&23.7\% \\
\textbf{Total} (n=2,246)&100.0\%&100.0\%&100.0\%&100.0\% \\
Pearson chi2(2) =& 16.9189 &Pr =& 0.000& \\
\textbf{lives in SMSA}&&&& \\
nonSMSA (n=665)&31.3\%&24.9\%&26.9\%&29.6\% \\
SMSA (n=1,581)&68.7\%&75.1\%&73.1\%&70.4\% \\
\textbf{Total} (n=2,246)&100.0\%&100.0\%&100.0\%&100.0\% \\
Pearson chi2(2) =& 8.7161 &Pr =& 0.013& \\
\textbf{lives in south}&&&& \\
0 (n=1,304)&65.4\%&36.0\%&88.5\%&58.1\% \\
1 (n=942)&34.6\%&64.0\%&11.5\%&41.9\% \\
\textbf{Total} (n=2,246)&100.0\%&100.0\%&100.0\%&100.0\% \\
Pearson chi2(2) =& 162.6246 &Pr =& 0.000& \\
******************************END LATEX CODE
Download these files (in a zip archive from my DropBox) with this link:
- Eric
Eric A. Booth
Public Policy Research Institute
Texas A&M University
Office: +979.845.6754
On Apr 13, 2011, at 6:15 PM, Michael Crain wrote:
> I am using Ian Watson's -tabout- to create a table in Tex format. After compiling the Tex file in LaTex, the table looks fine except the spacing under the caption to the first horizontal line is too great. The gap is not trivial. It is about 0.3 inches and obviously too much. I've got other tables in my Tex document created using -esttab_ where the caption spacing appears normal. Even if I had to live with the excess spacing in this table, it makes my other tables appear inconsistent.
> I can't figure out how to control the caption spacing in the table created by -tabout-. The first part of the relevant Tex code generated by -tabout- is below. Has anyone else encountered this issue and been able to overcome it?
> begin{table}[htbp]
> \centering
> \footnotesize
> \caption{My caption is here}
> \begin{center}
> \footnotesize
> \newcolumntype{Y}{>{\raggedleft\arraybackslash}X}
> \begin{tabularx} {12cm} {@{} l Y Y Y Y Y Y Y Y Y Y Y Y Y Y Y Y@{}} \\
> \toprule
> *
> * For searches and help try:
> * http://www.stata.com/help.cgi?search
> * http://www.stata.com/support/statalist/faq
> * http://www.ats.ucla.edu/stat/stata/
* For searches and help try:
* http://www.stata.com/help.cgi?search
* http://www.stata.com/support/statalist/faq
* http://www.ats.ucla.edu/stat/stata/ | {"url":"http://www.stata.com/statalist/archive/2011-04/msg00697.html","timestamp":"2014-04-16T04:27:18Z","content_type":null,"content_length":"12534","record_id":"<urn:uuid:e5e09e67-c9e0-4b44-8dc2-de4ed5821826>","cc-path":"CC-MAIN-2014-15/segments/1398223206120.9/warc/CC-MAIN-20140423032006-00226-ip-10-147-4-33.ec2.internal.warc.gz"} |
number theory
Prove that if gcd(a,b)=1 and if c divides b, then gcd(a,c)=1.
Suppose gcd(a,b) = 1. That is, b=1k and a = 1m for some integers k,m. Suppose b|c that is, c = bx for some integers x. Then c = (1k)x where 1k is an integer. I think now you need to show that k = 1
because a = 1m (and so 1 is the largest divisor of a). That's all I've got...it might not be complete or correct. Can someone double check it?
Might be too wordy, I'm not really good at these, but I think it includes everything it needs to: *if gcd(a,b)=1 then a and b have no common elements in their prime factorizations *if c|b then c must
be entirely composed of elements from b's prime factorization *therefore there are no common elements in a and c's prime factorization *therefore the greatest common denominator of a and c is 1. ====
============================================== Attempt at a formal proof: Let A be the set of prime integers which occur in a's prime factorization. Let B be the set of prime integers which occur in
b's prime factorization. Let $A_k \in A$ and $B_j \in B$ where k and j are integers used to denote which element is being taken from the set. Let n be the cardinality of A (n = # of elements in a's
prime factorization) Let m be the cardinality of B (m = # of lements in b's prime factorization) Then $a = A_1*A_2*A_3*...*A_n$ and $b = B_1*B_2*B_3*...*B_m$ If there were any elements such that $B_j
= A_k$ then said element would divide both A and B, because it is common to both of their prime factorizations. As such, the sets A and B would intersect at that element. Therefore the $gcd(a,b) =
B_j = A_k ot{=} 1$ So the sets A and B cannot intersect. --------- Let c be an integer, and C be the set of prime integers which occur in c's prime factorization. If c|b then C must be a subset of B
(every element in C is an element in B) Because C is a subset of B, and B does not intersect A, C must not intersect A (there can be no common elements in c's prime factorization and a's prime
factorization). Therefore the gcd(a,c) = 1 | {"url":"http://mathhelpforum.com/discrete-math/33909-number-theory-print.html","timestamp":"2014-04-18T14:37:57Z","content_type":null,"content_length":"10943","record_id":"<urn:uuid:af23bf25-86ed-4f37-afb1-7328d85f21ac>","cc-path":"CC-MAIN-2014-15/segments/1397609533689.29/warc/CC-MAIN-20140416005213-00371-ip-10-147-4-33.ec2.internal.warc.gz"} |
l'Hopital's Rule
01-16-2003 #1
l'Hopital's Rule
I'm having a discussion (read: argument) with my calculus teacher about this.
lim cos(x) - .5
h->PI/6 PI/6 - x
I say that if you evaluate this limit it turns out undefined. She says that since the expression results in division by zero, l'Hopital's rule can be used.
lim sin(x)
h->PI/6 1
and this results in 0.5
Who is right? I think I am, but I want a general concensus here so I don't keep fighting a losing battle here.
your teacher is corect, however you are confused.
The limit has the potential of existing not because the denominator is 0, but because BOTH the numerator AND the denominator are 0.
In other circumsatnces (IE without trigonometric functions) you can do things like factoring and cancelation (logic being that 0/0 means both the numerator and the denominator have a common
factor) or use other methods.
I think you're both wrong.
Shouldn't it be x->pi/6 instead of h->pi/6
Anyways, shouldn't the answer be infinity since the numerator isn't zero at x=pi/6?
Try not.
Do or do not.
There is no try.
- Master Yoda
Cshot, I'm not sure because you would end up with .366/0, and that isn't possible. also, yes, it should be x-> pi/6, probably just a typo.
Yes, your teacher is right though, hopefully not with this problem. L'Hopitals Rule can be used with these following inderterminate forms:
infiniti - infinity
i think there is more, i just can't remember them off the top of my head right now, and my calc notes are at school.
>> Cshot, I'm not sure because you would end up with .366/0, and that isn't possible.
You're right that .366/0 is undefined. But this is a limit problem. As x->pi/6, the limit goes to infinity.
And yea, I was just joking about the h->pi/6 part
Try not.
Do or do not.
There is no try.
- Master Yoda
this particular limit does not classify as being able to use L'Hopital's Rule, this limit is undefined. I ran this through my calculator, and it gave me an answer of undefined (wonderful uses of
the TI-89). If cos(x) - .5 as x -> pi/6 evaluated to zero, then i can see this working. as i hope this particular problem is not the one being discussed and does have typos.
I agree with you alpha that this isn't a l'hopital problem. Just solve it like any other limit problem. The numerator goes to a constant while the denominator approaches 0, therefore the entire
limit approaches infinity.
If the problem were to say what is (cos(x)-0.5)/(pi/6 - x) at x=pi/6, then the answer would be undefined.
Try not.
Do or do not.
There is no try.
- Master Yoda
ha, you're right, i didn't notice the numerator wasn't 0. You'd think that after I specifically said the numerator had to be 0 that would have checked if it was in this example
cos( pi/3 ) is 1/2 NOT cos( pi/6 )
Cshot, it still doesn't make sense. The calculator does do limits and it says the answer is undefined. if the denominator approached 1, then i could see how it approaches a constant; but it
approaches 0, and once pi/6 is plugged in just like any limit problem, it is evaluated with the values, and .366/0 is undefined.
I'm still in high school Calc BC, so maybe I'm missing something here...but I don't think I am.
edit: Actually, come to think of it, if the limit was as x approaches infinity, then the limit would evaluate to infinity...
Last edited by alpha; 01-16-2003 at 06:31 PM.
My Mathmatical Theory (which you people are free to shoot down and you probably will):
I think that inf/inf should equal 1. x/x=1 for non-zero integers. Even the lim as h->inf of h/h=1. Therefore, since infinity is just a very very big number, shouldn't the same be true? This would
make the integration of divergent functions very easy. For instance, with this idea, int from -2 to 2 of 1/x dx would equal 0.
Originally posted by golfinguy4
My Mathmatical Theory (which you people are free to shoot down and you probably will):
I think that inf/inf should equal 1. x/x=1 for non-zero integers. Even the lim as h->inf of h/h=1. Therefore, since infinity is just a very very big number, shouldn't the same be true? This would
make the integration of divergent functions very easy. For instance, with this idea, int from -2 to 2 of 1/x dx would equal 0.
you can't take the ln of negative numbers, so the integral from -2 to 2 of 1/x dx would not equal zero. and inf/inf is indeterminate, because infinity is a concept. inf/inf could be 1, could be
5, etc.
Why do you mention ln? Yes, ln is the anti-derivative of 1/x. However, if you read the FToC carefully, you will see that the antidifferentiation rule is only guaranteed to work if f is continuous
along the interval.
>> I think that inf/inf should equal 1. x/x=1 for non-zero integers.
Nah, you can't apply normal math operations with infinity. I think they're called indeterminants and normal math rules don't apply.
Hmm, alpha you may be right.
I looked it up here:
It says:
If lim f(x) != 0 and lim g(x) = 0 then one of the following is true:
a) lim f(x)/g(x) = inf or lim f(x)/g(x) = -inf
b) lim f(x)/g(x) doesn't exist
Seems like if you approach pi/6 from the right, you'll approach negative infinity. If you approach pi/6 from the left, you'll approach positive infinity. Therefore, the limit doesn't exist for
this case.
I think you are right. This is because you can only use lhr
when 0/0, inf/inf or -inf/inf.
You might want to show this to
your teacher
1/h as h->0 is certainly undefined. What follows gives
a similar expression
(cos(x) - .5) / (pi/6 -x) as x->pi/6 is equivalent to
(cos(x + h) - .5)/h as h->0 where x is fixed to be pi/6.
This is the same as
(cos(x + h) - cos(x) + (sqrt(3) - 1)/2) / h as h->0 which comes to
-sin(x) + (sqrt(3) - 1)/2h as h->0
I think that inf/inf should equal 1. x/x=1 for non-zero integers. Even the lim as h->inf of h/h=1. Therefore, since infinity is just a very very big number, shouldn't the same be true? This would
make the integration of divergent functions very easy. For instance, with this idea, int from -2 to 2 of 1/x dx would equal 0.
If inf/inf equals 1 then x / (2x) as x->inf would be 1.
Wouldn't you just integrate 1/x and get ln |x| from -2 to 2 and
so then you have 0. integral(1/x) is not ln x.
01-16-2003 #2
01-16-2003 #3
01-16-2003 #4
Registered User
Join Date
Nov 2002
01-16-2003 #5
01-16-2003 #6
Registered User
Join Date
Nov 2002
01-16-2003 #7
01-16-2003 #8
01-16-2003 #9
Registered User
Join Date
Nov 2002
01-16-2003 #10
¡Amo fútbol!
Join Date
Dec 2001
01-16-2003 #11
Registered User
Join Date
Nov 2002
01-16-2003 #12
¡Amo fútbol!
Join Date
Dec 2001
01-16-2003 #13
01-16-2003 #14
Join Date
Aug 2001
01-16-2003 #15
Join Date
Aug 2001 | {"url":"http://cboard.cprogramming.com/brief-history-cprogramming-com/32537-l%27hopital%27s-rule.html","timestamp":"2014-04-17T13:03:19Z","content_type":null,"content_length":"97942","record_id":"<urn:uuid:3ccb3036-78db-4be3-89eb-bdefd7fecf31>","cc-path":"CC-MAIN-2014-15/segments/1397609530131.27/warc/CC-MAIN-20140416005210-00594-ip-10-147-4-33.ec2.internal.warc.gz"} |
Middle School Level Questions on Rate and Time
from the Dr. Math Archives
Middle School Dr. Math Questions on Measurement
Below are some questions and answers from our archive of questions sent via e-mail by K12 students and teachers to Ask Dr. Math. You may also browse the entire middle school level archive by
Given the distance (4.260 km) and time (1.15.989 min/sec), how do you use the d/t=average speed formula?
If I was born on 01/02/84, when will I have been alive for a billion seconds?
Nancy ran a distance of 3.8km in 55 min. Calculate her speed in kilometers per hour and metres per second.
I have a clock that is "measured by 100", meaning 25=15 minutes, 50=30 minutes, etc. How do I convert this into "real time"?
On Voyager, the new Star Trek show, they are 75 years from home. Their top speed is Warp 9.75. Can you define "Warp" in lightyears?
What is wrong with D' = sqrt(X^2 - X'2)?
Bernice is cycling around a track at 15 mph. Betty starts at the same time, but only goes 12 mph. How many minutes after they start will Bernice pass Betty if the track is 1/2 mile long?
Ian travelled by train at 80km/h and then by car at 90 km/h. It took him 3 hours to travel the total distance of 265km. How far did he travel during each phase of his journey, and how long
did each phase take?
Liza and Tamar leave the same camp and run in opposite directions. Liza runs 2m/s and Tamar runs 3m/s. How far apart are they after 1 hour?
Should the earth's rotational speed be referred to in revolutions per... or in miles per hour?
Ships travel in knots (which is a speed). How fast is a knot?
Inside a rectangular room a spider is at a point on the middle of one of the end walls and a fly is on the opposite wall. What is the shortest distance the spider must crawl to reach the fly?
How many nanoseconds are there in one year?
If you drive one mile in 45 seconds, how fast are you driving?
What's the difference between statute miles and nautical miles, and why do they use statute miles for space shuttle launches?
A train passes completely through a tunnel in 5 minutes. A second train, twice as long, passes through the tunnel in 6 minutes...
An electric train is climbing a hill at 45 kph...
If a train left Baltimore at 6:00 pm and travels at 19 m.p.h. and another train left Philadelphia travelling at 85 mph in the opposite direction, where would they meet?
You are traveling 55 mph over a bridge that is 4260 ft. long. How long does it take to cross the bridge?
A storm is 50 miles offshore and its path is perpendicular to a straight shoreline. It is approaching the shore at a rate of 4mph....
If a car moves at 44 mph for 50 minutes, how many kilometers does it travel?
Back to Middle School Dr. Math Questions
Back to Contents
Mara Landers
[Privacy Policy] [Terms of Use]
Home || The Math Library || Quick Reference || Search || Help
© 1994-2014 Drexel University. All rights reserved.
The Math Forum is a research and educational enterprise of the Drexel University School of Education. | {"url":"http://mathforum.org/paths/measurement/m.drmath.drt.html","timestamp":"2014-04-19T00:22:38Z","content_type":null,"content_length":"7877","record_id":"<urn:uuid:3fc96ef8-bdd9-4612-b707-9c3ea6ea3672>","cc-path":"CC-MAIN-2014-15/segments/1398223206118.10/warc/CC-MAIN-20140423032006-00285-ip-10-147-4-33.ec2.internal.warc.gz"} |
scipy.interpolate.barycentric_interpolate(xi, yi, x, axis=0)[source]¶
Convenience function for polynomial interpolation.
Constructs a polynomial that passes through a given set of points, then evaluates the polynomial. For reasons of numerical stability, this function does not compute the coefficients of the
This function uses a “barycentric interpolation” method that treats the problem as a special case of rational function interpolation. This algorithm is quite stable, numerically, but even in a
world of exact computation, unless the x coordinates are chosen very carefully - Chebyshev zeros (e.g. cos(i*pi/n)) are a good choice - polynomial interpolation itself is a very ill-conditioned
process due to the Runge phenomenon.
xi : array_like
1-d array of x coordinates of the points the polynomial should pass through
yi : array_like
The y coordinates of the points the polynomial should pass through.
Parameters :
x : scalar or array_like
Points to evaluate the interpolator at.
axis : int, optional
Axis in the yi array corresponding to the x-coordinate values.
y : scalar or array_like
Returns :
Interpolated values. Shape is determined by replacing the interpolation axis in the original array with the shape of x.
Construction of the interpolation weights is a relatively slow process. If you want to call this many times with the same xi (but possibly varying yi or x) you should use the class
BarycentricInterpolator. This is what this function uses internally. | {"url":"http://docs.scipy.org/doc/scipy-dev/reference/generated/scipy.interpolate.barycentric_interpolate.html","timestamp":"2014-04-21T02:04:17Z","content_type":null,"content_length":"10522","record_id":"<urn:uuid:01e23b9f-e3f1-4b05-9853-d3a266b4b826>","cc-path":"CC-MAIN-2014-15/segments/1398223203422.8/warc/CC-MAIN-20140423032003-00187-ip-10-147-4-33.ec2.internal.warc.gz"} |
Got Homework?
Connect with other students for help. It's a free community.
• across
MIT Grad Student
Online now
• laura*
Helped 1,000 students
Online now
• Hero
College Math Guru
Online now
Here's the question you clicked on:
2. Washburn is seeking a Sales and Marketing Coordinator with a bachelor's degree or equivalent experience, knowledgeable in Microsoft Office. This position pays $25,000 to $35,000, depending on
experience. Assume a person is paid weekly and earns $32,500. Using the percentage method, what would be the taxes withheld for a married person who claims 3 exemptions?
Best Response
You've already chosen the best response.
Can someone help me understand this problem?
Best Response
You've already chosen the best response.
There must be a table that goes with this question that tells us what the tax rates are? We need to know the table to answer the question.
Your question is ready. Sign up for free to start getting answers.
is replying to Can someone tell me what button the professor is hitting...
• Teamwork 19 Teammate
• Problem Solving 19 Hero
• Engagement 19 Mad Hatter
• You have blocked this person.
• ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.
This is the testimonial you wrote.
You haven't written a testimonial for Owlfred. | {"url":"http://openstudy.com/updates/4ee112ade4b0a50f5c55a449","timestamp":"2014-04-20T21:13:40Z","content_type":null,"content_length":"30373","record_id":"<urn:uuid:973db97d-2a5a-4aa8-8aa3-1b1741c0df00>","cc-path":"CC-MAIN-2014-15/segments/1397609539230.18/warc/CC-MAIN-20140416005219-00145-ip-10-147-4-33.ec2.internal.warc.gz"} |
Got Homework?
Connect with other students for help. It's a free community.
• across
MIT Grad Student
Online now
• laura*
Helped 1,000 students
Online now
• Hero
College Math Guru
Online now
Here's the question you clicked on:
I need help... Evaluate 4g^2+g for g=6 When g=6, 4g^2+g=
• one year ago
• one year ago
Best Response
You've already chosen the best response.
suppose i ask you this evaluate x+2 when x= 4 what'll be the ans?
Best Response
You've already chosen the best response.
4g^2 + g Put 6 instead of g, find the answer.
Your question is ready. Sign up for free to start getting answers.
is replying to Can someone tell me what button the professor is hitting...
• Teamwork 19 Teammate
• Problem Solving 19 Hero
• Engagement 19 Mad Hatter
• You have blocked this person.
• ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.
This is the testimonial you wrote.
You haven't written a testimonial for Owlfred. | {"url":"http://openstudy.com/updates/507cf354e4b040c161a2b356","timestamp":"2014-04-18T10:40:45Z","content_type":null,"content_length":"30000","record_id":"<urn:uuid:714ecd6a-653d-41cd-8c06-4c6bafe3047a>","cc-path":"CC-MAIN-2014-15/segments/1397609533308.11/warc/CC-MAIN-20140416005213-00139-ip-10-147-4-33.ec2.internal.warc.gz"} |
Summary: SYMMETRIC EXTENSION FOR TWO-CHANNEL QUINCUNX FILTER BANKS
Yi Chen, Michael D. Adams, and Wu-Sheng Lu
Dept. of Elec. and Comp. Eng., University of Victoria, Victoria, BC, CANADA
In the case of one-dimensional filter banks, symmetric extension is
a commonly used technique for constructing nonexpansive trans-
forms of finite-length sequences. In this paper, we show how sym-
metric extension can be extended to the case of two-dimensional
filter banks based on quincunx sampling. In particular, we show
how, for filter banks of this type, one can construct nonexpansive
transforms for input sequences defined on arbitrary rectangular re-
The two-dimensional (2-D) two-channel filter bank shown in Fig. 1
can be used to compute a class of transforms that has proven ex-
tremely useful in many image processing applications. Often, such
a filter bank is defined so as to operate on sequences of infinite
extent. In practice, however, we almost invariably deal with se-
quences of finite extent. Therefore, we usually require some means
for adapting filter banks to such sequences. This leads to the well | {"url":"http://www.osti.gov/eprints/topicpages/documents/record/953/1490844.html","timestamp":"2014-04-16T04:59:51Z","content_type":null,"content_length":"8423","record_id":"<urn:uuid:b12c0b87-6fe7-4440-b174-6efc0a11e399>","cc-path":"CC-MAIN-2014-15/segments/1397609521512.15/warc/CC-MAIN-20140416005201-00362-ip-10-147-4-33.ec2.internal.warc.gz"} |
Estabilishing a Statistically Based Causal Relationship
Judea Pearl has done quite a bit of work on causality, especially through Bayesian networks. Googling his name, you will find quite a few general-audience articles that might be interesting.
More mathematically, we may consider Bayesian Networks through graphical models and consider "interventions" in the model. In particular, see "Causal inference in statistics:
An overview" by Pearl at
Everyone always like to say "Correlation does not imply Causation", so it is nice to be able to think about the other direction!
Math and Pi: this is so close to what I was looking for it's not even funny. Thank you
Stephen tashi: yes I suppose you are right. I may need to revise my range of outcome to 0 to 1.
Mfb: I am talking about a metric in which you conclude some analog truth value to "x causes y" using both time series for all k.
Edit* ImaLooser: Based on MathandPi's Post (after actually starting to read the material from Pearl), Causation does not imply correlation since it's actually possible that the causation is nonlinear
(from my understanding since correlation would imply, if anything at all, a linear causation between [itex] X [/itex] and [itex] Y [/itex]). There is no reason for causation to be an inherently
linear operation in general. | {"url":"http://www.physicsforums.com/showthread.php?p=4185788","timestamp":"2014-04-20T16:01:37Z","content_type":null,"content_length":"38597","record_id":"<urn:uuid:41c3aa56-0410-4f73-a94e-d1ba94b2b390>","cc-path":"CC-MAIN-2014-15/segments/1398223205375.6/warc/CC-MAIN-20140423032005-00501-ip-10-147-4-33.ec2.internal.warc.gz"} |
Summary: The Annals of Probability
2009, Vol. 37, No. 4, 14591482
DOI: 10.1214/08-AOP439
© Institute of Mathematical Statistics, 2009
BY JONATHAN E. TAYLOR1,2 AND ROBERT J. ADLER1
Stanford University and Technion
We consider vector valued, unit variance Gaussian processes defined over
stratified manifolds and the geometry of their excursion sets. In particular, we
develop an explicit formula for the expectation of all the LipschitzKilling
curvatures of these sets. Whereas our motivation is primarily probabilistic,
with statistical applications in the background, this formula has also an inter-
pretation as a version of the classic kinematic fundamental formula of integral
geometry. All of these aspects are developed in the paper.
Particularly novel is the method of proof, which is based on a an approxi-
mation to the canonical Gaussian process on the n-sphere. The n limit,
which gives the final result, is handled via recent extensions of the classic
Poincaré limit theorem.
1. Introduction. The central aim of this paper is to describe a new result in | {"url":"http://www.osti.gov/eprints/topicpages/documents/record/677/2437786.html","timestamp":"2014-04-18T09:39:09Z","content_type":null,"content_length":"8344","record_id":"<urn:uuid:dbe0cd35-a371-4a44-a480-acb0b8907d56>","cc-path":"CC-MAIN-2014-15/segments/1397609533121.28/warc/CC-MAIN-20140416005213-00138-ip-10-147-4-33.ec2.internal.warc.gz"} |
Problems with retrieving forst three bit
Problems with retrieving forst three bits of a number
Hey,I am trying to retrieve the first three bits of a number. The code that I am using should work but it isn't giving me the correct result when trying certain numbers. Below is the code I am using:
1 unsigned short num1, num2 = 0;
2 unsigned short num = 65535// binary 111111111111111
3 num1 = num && 0x07;// gives me 1 but should give 7(111)
4 num2 = num >>3;//gives me 8191, which is correct
Can anyone tell me why I am not getting the first three correct bits(111)?
Thanks In Advance
num1 = num & 0x07 ;
&& is the logical and.
Topic archived. No new replies allowed. | {"url":"http://www.cplusplus.com/forum/general/100257/","timestamp":"2014-04-19T14:33:05Z","content_type":null,"content_length":"7856","record_id":"<urn:uuid:94355e3e-5a2c-4af4-9206-3d190a9254b5>","cc-path":"CC-MAIN-2014-15/segments/1397609537271.8/warc/CC-MAIN-20140416005217-00266-ip-10-147-4-33.ec2.internal.warc.gz"} |
Physics Forums - View Single Post - kinetic energy of matter
The p=mv equation should not be used. The v in your mv is not simply related to the v which is the velocity of the wave in f lambda=v. The wave function is
exp[p.x-(p^2/2m)t/hbar]. You can find the wave and group velocity from that. | {"url":"http://www.physicsforums.com/showpost.php?p=2237155&postcount=3","timestamp":"2014-04-16T16:06:12Z","content_type":null,"content_length":"7292","record_id":"<urn:uuid:f085e7b1-29d0-4181-8bcd-87e4daee304b>","cc-path":"CC-MAIN-2014-15/segments/1398223206147.1/warc/CC-MAIN-20140423032006-00498-ip-10-147-4-33.ec2.internal.warc.gz"} |
the encyclopedic entry of Ajutage
Gr. ubpops avuca
) is the science of the mechanics of water and fluids in general, including
or the mathematical theory of fluids in equilibrium, and hydrodynamics, the theory of fluids in motion. The practical application of hydromechanics forms the province of
The fundamental principles of hydrostatics were first given by Archimedes in his work IIepi r%uv oxouuiewv, or De its quae vehuntur in humido, about 250 BC, and were afterwards applied to experiments
by Marin Getaldić (1566-1627) in his Promotus Archimedes (1603). Archimedes maintained that each particle of a fluid mass, when in equilibrium, is equally pressed in every direction ; and he inquired
into the conditions according to which a solid body floating in a fluid should assume and preserve a position of equilibrium.
In the Greek school at Alexandria, which flourished under the auspices of the Ptolemies, the first attempts were made at the construction of hydraulic machinery, and about 120 BC the fountain of
compression, the siphon, and the forcing-pump were invented by Ctesibius and Hero. The siphon is a simple instrument; but the forcing-pump is a complicated invention, which could scarcely have been
expected in the infancy of hydraulics. It was probably suggested to Ctesibius by the Egyptian Wheel or Noria, which was common at that time, and which was a kind of chain pump, consisting of a number
of earthen pots carried round by a wheel. In some of these machines the pots have a valve in the bottom which enables them to descend without much resistance, and diminishes greatly the load upon the
wheel; and, if we suppose that this valve was introduced so early as the time of Ctesibius, it is not difficult to perceive how such a machine might have led to the invention of the forcing-pump.
Notwithstanding these inventions of the Alexandrian school, its attention does not seem to have been directed to the motion of fluids; and the first attempt to investigate this subject was made by
Sextus Julius Frontinus, inspector of the public fountains at Rome in the reigns of Nerva and Trajan. In his work De aquaeductibus urbis Romae commentaries, he considers the methods which were at
that time employed for ascertaining the quantity of water discharged from ajutages, and the mode of distributing the waters of an aqueduct or a fountain. He remarked that the flow of water from an
orifice depends not only on the magnitude of the orifice itself, but also on the height of the water in the reservoir; and that a pipe employed to carry off a portion of water from an aqueduct
should, as circumstances required, have a position more or less inclined to the original direction of the current. But as he was unacquainted with the law of the velocities of running water as
depending upon the depth of the orifice, the want of precision which appears in his results is not surprising.
Benedetto Castelli, and Evangelista Torricelli, two of the disciples of Galileo, applied the discoveries of their master to the science of hydrodynamics. In 1628 Castelli published a small work,
Della misura dell' acque correnti, in which he satisfactorily explained several phenomena in the motion of fluids in rivers and canals; but he committed a great paralogism in supposing the velocity
of the water proportional to the depth of the orifice below the surface of the vessel. Torricelli, observing that in a jet where the water rushed through a small ajutage it rose to nearly the same
height with the reservoir from which it was supplied, imagined that it ought to move with the same velocity as if it had fallen through that height by the force of gravity, and hence he deduced the
proposition that the velocities of liquids are as the square root of the head, apart from the resistance of the air and the friction of the orifice. This theorem was published in 1643, at the end of
his treatise De motu gravium projectorum, and it was con-firmed by the experiments of Raffaello Magiotti on the quantities of water discharged from different ajutages under different pressures
In the hands of Blaise Pascal hydrostatics assumed the dignity of a science, and in a treatise on the equilibrium of liquids (Sur l'equilibre des liqueurs), found among his manuscripts after his
death and published in 1663, the laws of the equilibrium of liquids were demonstrated in the most simple manner, and amply confirmed by experiments.
The theorem of Torricelli was employed by many succeeding writers, but particularly by Edme Mariotte (1620-1684), whose Traité du mouvement des eaux, published after his death in the year 1686, is
founded on a great variety of well-conducted experiments on the motion of fluids, performed at Versailles and Chantilly. In the discussion of some points he committed considerable mistakes. Others he
treated very superficially, and in none of his experiments apparently did he attend to the diminution of efflux arising from the contraction of the liquid vein, when the orifice is merely a
perforation in a thin plate; but he appears to have been the first who attempted to ascribe the discrepancy between theory and experiment to the retardation of the water's velocity through friction.
His contemporary Domenico Guglielmini (1655-1710), who was inspector of the rivers and canals at Bologna, had ascribed this diminution of velocity in rivers to transverse motions arising front
inequalities in their bottom. But as Mariotte observed similar obstructions even in glass pipes where no transverse currents could exist, the cause assigned by Guglielmini seemed destitute of
foundation. The French philosopher, therefore, regarded these obstructions as the effects of friction. He supposed that the filaments of water which graze along the sides of the pipe lose a portion
of their velocity; that the contiguous filaments, having on this account a greater velocity, rub upon the former, and suffer a diminution of their celerity; and that the other filaments are affected
with similar retardations proportional to their distance from the axis of the pipe. In this way the medium velocity of the current may be diminished, and consequently the quantity of water discharged
in a given time must, from the effects of friction, be considerably less than that which is computed from theory.
The effects of friction and viscosity in diminishing the velocity of running water were noticed in the Principia of Sir Isaac Newton, who threw much light upon several branches of hydromechanics. At
a time when the Cartesian system of vortices universally prevailed, he found it necessary to investigate that hypothesis, and in the course of his investigations he showed that the velocity of any
stratum of the vortex is an arithmetical mean between the velocities of the strata which enclose it; and from this it evidently follows that the velocity of a filament of water moving in a pipe is an
arithmetical mean between the velocities of the filaments which surround it. Taking advantage of these results, Italian-born French engineer Henri Pitot afterwards showed that the retardations
arising from friction are inversely as the diameters of the pipes in which the fluid moves.
The attention of Newton was also directed to the discharge of water from orifices in the bottom of vessels. He supposed a cylindrical vessel full of water to be perforated in its bottom with a small
hole by which the water escaped, and the vessel to be supplied with water in such a manner that it always remained full at the same height. He then supposed this cylindrical column of water to be
divided into two parts,the first, which he called the "cataract," being an hyperboloid generated by the revolution of an hyperbola of the fifth degree around the axis of the cylinder which should
pass through the orifice, and the second the remainder of the water in the cylindrical vessel. He considered the horizontal strata of this hyperboloid as always in motion, while the remainder of the
water was in a state of rest, and imagined that there was a kind of cataract in the middle of the fluid.
When the results of this theory were compared with the quantity of water actually discharged, Newton concluded that the velocity with which the water issued from the orifice was equal to that which a
falling body would receive by descending through half the height of water in the reservoir. This conclusion, however, is absolutely irreconcilable with the known fact that jets of water rise nearly
to the same height as their reservoirs, and Newton seems to have been aware of this objection. Accordingly, in the second edition of his Principia, which appeared in 1713, he reconsidered his theory.
He had discovered a contraction in the vein of fluid (vena contracta) which issued from the orifice, and found that, at the distance of about a diameter of the aperture, the section of the vein was
contracted in the subduplicate ratio of two to one. He regarded, therefore, the section of the contracted vein as the true orifice from which the discharge of water ought to be deduced, and the
velocity of the effluent water as due to the whole height of water in the reservoir; and by this means his theory became more conformable to the results of experience, though still open to serious
objections. Newton was also the first to investigate the difficult subject of the motion of waves.
In 1738 Daniel Bernoulli published his Hydrodynamica seu de viribus et motibus fluidorum commentarii. His theory of the motion of fluids, the germ of which was first published in his memoir entitled
Theoria nova de motu aquarum per canales quocunque fluentes, communicated to the Academy of St Petersburg as early as 1726, was founded on two suppositions, which appeared to him conformable to
experience. He supposed that the surface of the fluid, contained in a vessel which is emptying itself by an orifice, remains always horizontal; and, if the fluid mass is conceived to be divided into
an infinite number of horizontal strata of the same bulk, that these strata remain contiguous to each other, and that all their points descend vertically, with velocities inversely proportional to
their breadth, or to the horizontal sections of the reservoir. In order to determine the motion of each stratum, he employed the principle of the conservatio virium vivarum, and obtained very elegant
solutions. But in the absence of a general demonstration of that principle, his results did not command the .confidence which they would otherwise have deserved, and it became desirable to have a
theory more certain, and depending-solely on the fundamental laws of mechanics. Colin Maclaurin and John Bernoulli, who were of this opinion, resolved the problem by more direct methods, the one in
his Fluxions, published in 1742, and the other in his Hydraulica nunc primum detecta, et demonstrata directe ex furulamentis pure mechanicis, which forms the fourth volume of his works. The method
employed by Maclaurin has been thought not sufficiently rigorous; and that of John Bernoulli is, in the opinion of Lagrange, defective in clearness and precision.
The theory of Daniel Bernoulli was opposed also by Jean le Rond d'Alembert. When generalizing the theory of pendulums of Jacob Bernoulli he discovered a principle of dynamics so simple and general
that it reduced the laws of the motions of bodies to that of their equilibrium. He applied this principle to the motion of fluids, and gave a specimen of its application at the end of his Dynamics in
1743. It was more fully developed in his Traité des fluides, published in 1744, in which he gave simple and elegant solutions of problems relating to the equilibrium and motion of fluids. He made use
of the same suppositions as Daniel Bernoulli, though his calculus was established in a very different manner. He considered, at every instant, the actual motion of a stratum as composed of a motion
which it had in the preceding instant and of a motion which it had lost; and the laws of equilibrium between the motions lost furnished him with equations re-presenting the motion of the fluid. It
remained a desideratum to express by equations the motion of a particle of the fluid in any assigned direction. These equations were found by d'Alembert from two principles--that a rectangular canal,
taken in a mass of fluid in equilibrium, is itself in equilibrium, and that a portion of the fluid, in passing from one place to another, preserves the same volume when the fluid is incompressible,
or dilates itself according to a given law when the fluid is elastic. His ingenious method, published in 1752, in his Essai sur la resistance des fluides, was brought to perfection in his Opuscules
mathematiques, and was adopted by Leonhard Euler.
The resolution of the questions concerning the motion of fluids was effected by means of Euler's partial differential coefficients. This calculus was first applied to the motion of water by
d'Alembert, and enabled both him and Euler to represent the theory of fluids in formulae restricted by no particular hypothesis.
One of the most successful labourers in the science of hydrodynamics at this period was Pierre Louis Georges Dubuat (1734-1809). Following in the steps of the Abbé Charles Bossut (Nouvelles
Experiences sur la resistance des fluides, 1777), he published, in 1786, a revised edition of his Principes d'hydraulique, which contains a satisfactory theory of the motion of fluids, founded solely
upon experiments. Dubuat considered that if water were a perfect fluid, and the channels in which it flowed infinitely smooth, its motion would be continually accelerated, like that of bodies
descending in an inclined plane. But as the motion of rivers is not continually accelerated,and soon arrives at a state of uniformity, it is evident that the viscosity of the water, and the friction
of the channel in which it descends, must equal tke accelerating force. Dubuat, therefore, assumed it as a proposition of fundamental importance that, when water flows in any channel or bed, the
accelerating force which obliges it to move is equal to the sum of all the resistances which it meets with, whether they arise from its own viscosity or from the friction of its bed. This principle
was employed by him in the first edition of his work, which appeared in 1779. The theory contained in that edition was founded on the experiments of others, but he soon saw that a theory so new, and
leading to results so different from the ordinary theory, should be founded on new experiments more direct than the former, and he was employed in the performance of these from 1780 to 1783. The
experiments of Bossut were made only on pipes of a moderate declivity, but Dubuat used declivities of every kind, and made his experiments upon channels of various sizes.
The theory of running water was greatly advanced by the researches of Gaspard Riche de Prony (1755-1839). From a collection of the best experiments by previous workers he selected eighty-two
(fifty-one on the velocity of water in conduit pipes, and thirty-one on its velocity in open canals); and, discussing these on physical and mechanical principles, he succeeded in drawing up general
formulae, which afforded a simple expression for the velocity of running water.
JA Eytelwein of Berlin, who published in 1801 a valuable compendium of hydraulics entitled Handbuch der Mechanik und der Hydraulik, investigated the subject of the discharge of water by compound
pipes, the motions of jets and their impulses against plane and oblique surfaces; and he showed theoretically that a water-wheel will have its maximum effect when its circumference moves with half
the velocity of the stream.
JNP Hachette in 1816-1817 published memoirs containing the results of experiments on the spouting of fluids and the discharge of vessels. His object was to measure the contracted part of a fluid
vein, to examine the phenomena attendant on additional tubes, and to investigate the form of the fluid vein and the results obtained when different forms of orifices are employed. Extensive
experiments on the discharge of water from orifices (Experiences hydrauliques, Paris, 1832) were conducted under the direction of the French government by JV Poncelet (1788-1867) and JA Lesbros
(1790-1860). PP Boileau (1811-1891) discussed their results and added experiments of his own (Traité de la mesure des eaux courantes, Paris, 1854). KR Bornemann re-examined all these results with
great care, and gave formulae expressing the variation of the coefficients of discharge in different conditions (Civil Ingenieur, 1880). Julius Weisbach (1806-1871) also made many experimental
investigations on the discharge of fluids. The experiments of JB Francis (Lowell Hydraulic Experiments, Boston, Mass., 1855) led him to propose variations in the accepted formulae for the discharge
over weirs, and a generation later a very complete investigation of this subject was carried out by H Bazin. An elaborate inquiry on the flow of water in pipes and channels was conducted by HGP Darcy
(1803-1858) and continued by H Bazin, at the expense of the French government (Recherches hydrauliques, Paris, 1866).
German engineers have also devoted special attention to the measurement of the flow in rivers; the Beiträge zur Hydrographie des Königreiches Bohmen (Prague, 1872-1875) of AR Harlacher contained
valuable measurements of this kind, together with a comparison of the experimental results with the formulae of flow that had been proposed up to the date of its publication, and important data were
yielded by the gaugings of the Mississippi made for the United States government by AA Humphreys and HL Abbot, by Robert Gordon's gaugings of the Ayeyarwady River, and by Allen JC Cunningham's
experiments on the Ganges canal. The friction of water, investigated for slow speeds by Coulomb, was measured for higher speeds by William Froude (1810-1879), whose work is of great value in the
theory of ship resistance (Brit. Assoc. Report., 1869), and stream line motion was studied by Professor Osborne Reynolds and by Professor HS Hele-Shaw. | {"url":"http://www.reference.com/browse/Ajutage","timestamp":"2014-04-16T13:39:34Z","content_type":null,"content_length":"98288","record_id":"<urn:uuid:7c2ba06f-0b24-44cb-ac89-433bf781ae47>","cc-path":"CC-MAIN-2014-15/segments/1398223203841.5/warc/CC-MAIN-20140423032003-00491-ip-10-147-4-33.ec2.internal.warc.gz"} |